IGNOU - B.Sc. - PHE15 : Astronomy and Astrophysics

UNIT 1 ASTRONOMICAL SCALES

Astronomical Scales

Structure 1.1

Introduction Objectives

1.2 1.3 1.4

Astronomical Distance, Mass and Time Scales Brightness, Radiant Flux and Luminosity Measurement of Astronomical Quantities Astronomical Distances Stellar Radii Masses of Stars Stellar Temperature

1.5 1.6 1.7

Summary Terminal Questions Solutions and Answers

1.1

INTRODUCTION

You have studied in Units 9 to 11 of the Foundation Course in Science and Technology (FST-1) that the universe is vast. You know that the Sun is one amongst billions of stars situated in as many galaxies. You have also learnt that the distances between planets and stars are huge, and so are their masses. For example, the distance between the Sun and the Earth is of the order of 1.5×1011m. The radius of the Sun itself is about 7×108m, which is almost 100 times the Earth’s radius. The mass of the Earth is of the order of 1024kg and the Sun’s mass is a million times larger. The time scales involved are also huge. For example, the estimated age of the Sun is about 5 billion years, compared to the lifetime of a human being, which is less than 100 years in most cases. All these numbers are very large compared to the lengths, masses and time scales we encounter everyday. Obviously, we need special methods to measure them and represent them. The distances and masses of celestial objects are of fundamental interest to astronomy. Does a star in the night sky seem bright to us because it is closer, or is it so because it is intrinsically bright? The answer can be obtained if we know the distance to a star. You have also learnt in Unit 10 of FST-1 that the mass of a star determines how it will evolve. In this unit we introduce you to some important physical quantities of interest in astronomy, such as distance, size, mass, time, brightness, radiant flux, luminosity, temperature and their scales. You will also learn about some simple methods of measuring these quantities. In the next unit, you will learn about the various coordinate systems used to locate the positions of celestial objects. Objectives After studying this unit, you should be able to: •

describe the distance, mass, time and temperature scales used in astronomy and astrophysics;

•

compare the brightness and luminosity of astronomical objects; and

•

determine the distance, size and mass of astronomical objects from given data. 5

Basics of Astronomy

1.2

ASTRONOMICAL DISTANCE, MASS AND TIME SCALES

In astronomy, we are interested in measuring various physical quantities, such as mass, distance, radius, brightness and luminosity of celestial objects. You have just learnt that the scales at which these quantities occur in astronomy are very different from the ones we encounter in our day-to-day lives. Therefore, we first need to understand these scales and define the units of measurement for important astrophysical quantities. We begin with astronomical distances. Astronomical Distances You have studied in your school textbooks that the Sun is at a distance of about 1.5 × 1011m from the Earth. The mean distance between the Sun and the Earth is called one astronomical unit. Distances in the solar system are measured in this unit. Another unit is the light year, used for measuring distances to stars and galaxies. The parsec is a third unit of length measurement in astronomy. We now define them.

Units of measurement of distances •

1 Astronomical Unit (AU) is the mean distance between the Sun and the Earth. 1 AU = 1.496 × 1011m

•

1 Light Year (ly) is the distance travelled by light in one year. 1 ly = 9.460 × 1015m = 6.323 × 104 AU

•

1 Parsec (pc) is defined as the distance at which the radius of Earth’s orbit subtends an angle of 1″ (see Fig.1.1). 1 pc = 3.262 ly = 2.062 × 105AU = 3.085 × 1016m

EARTH

angle = 1 arc second

1 AU

SUN STAR

distance = 1 parsec

6

Fig.1.1: Schematic diagram showing the definition of 1 parsec. Note that 1°° ≡ 60′′ and 1′′ = 60″″. Thus, 1″″ = 1/3600 degree

Astronomical Scales

Dimensions of Astronomical Objects The sizes of stars or stellar dimensions are usually measured in units of solar radius RΘ. For example, Sirius ( ), the brightest star in the sky, has radius 2RΘ. The radius of the star Aldebaran ( ) in Taurus is 40RΘ and that of Antares ( ) in Scorpius is 700 RΘ.

Unit of measurement of size 8

1 solar radius, RΘ = 7 × 10 m

Mass Stellar masses are usually measured in units of solar mass MΘ. We know that 11 30 MΘ = 2 × 10 kg. For example, the mass of our galaxy is ~ 10 MΘ. The mass of a globular cluster is of the order of 105 − 106 MΘ. S. Chandrasekhar showed (Unit 11) that the mass of a white dwarf star cannot exceed 1.4 MΘ. This is called the Chandrasekhar limit.

Unit of measurement of mass 1 solar mass MΘ = 2 × 1030 kg

Time Scales The present age of the Sun is about 5 billion years. It has been estimated that it would live for another 5 billion years in its present form. The age of our galaxy may be around 10 billion years. Various estimates of the age of the universe itself give a figure between 12 and 16 billion years. On the other hand, if the pressure inside a star is insufficient to support it against gravity, then it may collapse in a time, which may be measured in seconds, rather than in millions of years. In Table 1.1, we list the distances, sizes and masses of some astronomical objects. Table 1.1: Distance, radii and masses of astronomical objects

Sun Earth Jupiter

Distance

Radius

Mass

Remarks

1 AU

1 RΘ

1 MΘ

−

−

0.01 RΘ

−6

−

−3

10 MΘ

4 AU (5 AU from the Sun)

0.1 RΘ

10 MΘ

Largest planet

Proxima Centauri

1.3 pc

0.15 RΘ

0.12 MΘ

Nearest star

Sirius A

2.6 pc

2 RΘ

3 MΘ

Brightest star

Sirius B

2.6 pc

0.02 RΘ

1 MΘ

First star identified as white dwarf

Antares

150 pc

700 RΘ

15 MΘ

Super giant star 7

Basics of Astronomy

You may like to express the distances and sizes of some astronomical objects in various units introduced here. Spend 10 min.

SAQ 1 a) Express the distance between Jupiter and Sun in parsecs, and the distance between the Earth and the Sun in light years. b) Express the radius of the Earth in units of RΘ. Next time when you look at the familiar stars in the night sky, you will have some idea of how far these are from us, and also how massive they are. An important problem in astronomy is to find out how much energy is emitted by celestial objects. It is expressed in terms of the luminosity and is related to the radiant flux and brightness of the object. You may have noticed that some stars in the night sky appear bright to us, some less bright and others appear quite faint. How do we estimate their real brightness? Let us find out.

1.3

BRIGHTNESS, RADIANT FLUX AND LUMINOSITY

It is a common experience that if we view a street lamp from nearby, it may seem quite bright. But if we see it from afar, it appears faint. Similarly, a star might look bright because it is closer to us. And a really brighter star might appear faint because it is too far. We can estimate the apparent brightness of astronomical objects easily, but, if we want to measure their real or intrinsic brightness, we must take their distance into account. The apparent brightness of a star is defined in terms of what is called the apparent magnitude of a star. Apparent Magnitude In the second century B.C., the Greek astronomer Hipparchus was the first astronomer to catalogue stars visible to the naked eye. He divided stars into six classes, or apparent magnitudes, by their relative brightness as seen from Earth. He numbered the apparent magnitude (m) of a star on a scale of 1 (the brightest) to 6 (the least bright). This is the scale on which the apparent brightness of stars, planets and other objects is expressed as they appear from the Earth. The brightest stars are assigned the first magnitude (m = 1) and the faintest stars visible to the naked eye are assigned the sixth magnitude (m = 6).

Apparent Magnitude Apparent magnitude of an astronomical object is a measure of how bright it appears. According to the magnitude scale, a smaller magnitude means a brighter star.

The magnitude scale is actually a non-linear scale. What this means is that a star, two magnitudes fainter than another, is not twice as faint. Actually it is about 6.3 times fainter. Let us explain this further. The response of the eye to increasing brightness is nearly logarithmic. We, therefore, need to define a logarithmic scale for magnitudes in which a difference of 5 magnitudes is equal to a factor of 100 in brightness. On this scale, the brightness ratio corresponding to 1 magnitude difference is 1001/5 or 2.512. Therefore, a star of magnitude 1 is 2.512 times brighter than a star of magnitude 2. 8

It is (2.512)2 = 6.3 times brighter than a star of magnitude 3.

Astronomical Scales

How bright is it compared to stars of magnitude 4 and 5? It is (2.512)3 = 16 times brighter than a star of magnitude 4. And (2.512)4 = 40 times brighter than a star of magnitude 5. As expected, it is 2.5125 = 100 times brighter than a star of magnitude 6. For example, the pole star (Polaris, Dhruva) has an apparent magnitude +2.3 and the star Altair has apparent magnitude 0.8. Altair is about 4 times brighter than Polaris. Mathematically, the brightness b1 and b2 of two stars with corresponding magnitudes m1 and m2 are given by the following relations. Relationship between brightness and apparent magnitude

m1 − m2 = 2.5 log10

b2 b1

b2 = 100 ( m1 − m2 ) / 5 b1

(1.1) b1 = 100 −(m1 −m2 ) / 5 b2

(1.2)

In Table 1.2, we give the brightness ratio for some magnitude differences. Table 1.2: Brightness ratio corresponding to given magnitude difference Magnitude Difference

Brightness Ratio

0.0

1.0

0.2

1.2

1.0

2.5

1.5

4.0

2.0

6.3

2.5

10.0

3.0

16.0

4.0

40.0

5.0

100.0

7.5

1000.0

10.0

10000.0

Modern astronomers use a similar scale for apparent magnitude. With the help of telescopes, a larger number of stars could be seen in the sky. Many stars fainter than the 6th magnitude were also observed. Moreover, stars brighter than the first magnitude have also been observed. Thus a magnitude of zero or even negative magnitudes have been assigned to extend the scale. A star of −1 magnitude is 2.512 times brighter than the star of zero magnitude. The brightest star in the sky other than the Sun, Sirius A, has an apparent magnitude of − 1.47.

9

Basics of Astronomy

The larger magnitude on negative scale indicates higher brightness while the larger positive magnitudes indicate the faintness of an object. The faintest object detectable with a large modern telescope in the sky currently is of magnitude m = 29. Therefore, the Sun having the apparent magnitude m = − 26.81, is 1022 times brighter than the faintest object detectable in the sky. In the following table we list the apparent magnitudes of some objects in the night sky. Table 1.3: Apparent magnitudes of some celestial objects Object

Indian Name

Apparent Magnitude

Sun

Surya

−26.81

Full Moon

Chandra

−12.73

Venus

Shukra

−4.22

Jupiter

Guru

−2.60

Sirius A

Vyadha

−1.47

Canopus

Agastya

−0.73

α-Centauri

−0.10

Betelgeuse

Ardra

+0.80

Spica

Chitra

+0.96

Polaris

Dhruva

+2.3

Uranus

Varuna

+5.5

Sirius B

+8.68

Pluto

+14.9

Faintest Star (detected by a modern telescope)

+29

Let us now apply these ideas to a concrete example. Example 1: Comparison of Brightness Compare the brightness of the Sun and α-Centauri using the apparent magnitudes listed in the Table 1.3. Solution From Table 1.3, mSun − mαC = − 26.81 − (− 0.10) = −26.71. Therefore, using Eq. (1.2), we obtain bαC bSun

= 100 −( 26.71) / 5 = 10 −10.7

or bSun bαC 10

= 1010.7 , i.e. the Sun is about 10 times brighter than α-Centauri. 11

You may now like to solve a problem based on what you have studied so far.

Astronomical Scales

SAQ 2 a) The apparent magnitude of the Sun is − 26.81 and that of the star Sirius is − 1.47. Which one of them is brighter and by how much?

Spend 10 min.

b) The apparent magnitudes of the stars Arcturus and Aldebaran are 0.06 and 0.86, respectively. Calculate the ratio of their brightness. The apparent magnitude and brightness of a star do not give us any idea of the total energy emitted per second by the star. This is obtained from radiant flux and the luminosity of a star.

Luminosity and Radiant Flux The luminosity of a body is defined as the total energy radiated by it per unit time. Radiant flux at a given point is the total amount of energy flowing through per unit time per unit area of a surface oriented normal to the direction of propagation of radiation.

The unit of radiant flux is erg s−1 cm−2 and that of luminosity is erg s−1. In astronomy, it is common to use the cgs system of units. However, if you wish to convert to SI units, you can use appropriate conversion factors. Note that here the radiated energy refers to not just visible light, but includes all wavelengths. The radiant flux of a source depends on two factors: (i) the radiant energy emitted by it, and (ii) the distance of the source from the point of observation. Suppose a star is at a distance r from us. Let us draw an imaginary sphere of radius r round the star. The surface area of this sphere is 4π r2. Then the radiant flux F of the star, is related to its luminosity L as follows:

F=

L 4πr 2

(1.3)

The luminosity of a stellar object is a measure of the intrinsic brightness of a star. It is expressed generally in the units of the solar luminosity, LΘ, where LΘ = 4 × 10 26 W = 4 × 10 33 erg s −1

For example, the luminosity of our galaxy is about 1011 LΘ. Now, the energy from a source received at any place, determines the brightness of the source. This implies that F is related to the brightness b of the source: the brighter the source, the larger would be the radiant flux at a place. Therefore, the ratio of brightness in Eq. (1.2) can be replaced by the ratio of radiant flux from two objects at the same place and we have F2 = 100 ( m1 − m2 ) / 5 F1

(1.4) 11

Basics of Astronomy

You know from Eq. (1.3) that the flux received at a place also depends on its distance from the source. Therefore, two stars of the same apparent magnitude may not be equally luminous, as they may be located at different distances from the observer: A star’s apparent brightness does not tell us anything about the luminosity of the star. We need a measure of the true or intrinsic brightness of a star. Now, we could easily compare the true brightness of stars if we could line them all up at the same distance from us (see Fig. 1.2). With this idea, we define the absolute magnitude of a star as follows:

Absolute Magnitude The absolute magnitude, M, of an astronomical object is defined as its apparent magnitude if it were at a distance of 10 pc from us.

Fig.1.2: Absolute magnitude of astronomical objects

Let us now relate the absolute magnitude of a star to its apparent magnitude. Let us consider a star at a distance r pc with apparent magnitude m, intrinsic brightness or luminosity L and radiant flux F1. Now when the same star is placed at a distance of 10 pc from the place of observation, then its magnitude would be M and the corresponding radiant flux would be F2. From Eq. (1.4), we have F2 = 100 ( m − M ) / 5 F1

(1.5)

Since the luminosity is constant for the star, we use Eq. (1.3) to write F2 F1

=

r pc 10 pc

2

(1.6)

Using Eq. (1.6) in Eq. (1.5), we get the difference between the apparent magnitude (m) and absolute magnitude (M ). 12

Astronomical Scales

It is a measure of distance and is called the distance modulus (see Fig. 1.3).

Distance modulus m − M = 5 log10

r pc 10 pc

= 5 log10 r − 5

(1.7)

Farther away

Closer

Fig.1.3: Star cluster showing distance modulus as a measure of distance. For the star farther away, m = 12.3, M = 2.6, r = 871 pc. For the closer star, m = 8.0, M = 5.8, r = 28 pc

We can also relate the absolute magnitudes of stars to their luminosities. From Eq. (1.3), we know that the ratio of radiant flux of two stars at the same distance from the point of observation is equal to the ratio of their luminosities. Thus, if M1 and M2 are the absolute magnitudes of two stars, using Eq. (1.5), we can relate their luminosities to M1 and M2. Relationship between Luminosity and Absolute Magnitude L2 = 100 ( M1 − M 2 ) / 5 L1

(1.8)

or M 1 − M 2 = 2.5 log10

L2 L1

(1.9)

Thus, the absolute magnitude of a star is a measure of its luminosity, or intrinsic brightness. Often if we know what kind of star it is, we can estimate its absolute magnitude. We can measure its apparent magnitude (m) directly and solve for distance using Eq. (1.7). For example, the apparent magnitude of Polaris (pole star) is +2.3. Its absolute magnitude is −4.6 and it is 240 pc away. The apparent magnitude of Sirius A is −1.47, its absolute magnitude is +1.4 and it is at a distance of 2.7 pc. You may now like to stop for a while and solve a problem to fix these ideas. SAQ 3

Spend 5 min.

a) The distance modulus of the star Vega is −0.5. At what distance is it from us? b) If a star at 40 pc is brought closer to 10 pc, i.e., 4 times closer, how bright will it appear in terms of the magnitude? 13

Basics of Astronomy

We now discuss some simple methods of measuring astronomical distances, sizes, masses and temperatures.

1.4

MEASUREMENT OF ASTRONOMICAL QUANTITIES

Since the brightness of heavenly objects depends on their distances from us, the measurement of distance is very important in astronomy. You must have measured the lengths of several objects in your school and college laboratories. But how do we measure astronomical distances? Obviously, traditional devices like the metre stick or measuring tapes are inadequate for such measurements. Other less direct ways need to be used. We now discuss some common methods of measuring astronomical distances. Since stars have been studied most extensively, we will focus largely on them in our discussion.

1.4.1

Astronomical Distances

You may be familiar with the method of trigonometric parallax. To get an idea of what it is, perform the following activity. Activity 1: Trigonometric Parallax

Extend your arm and hold your thumb at about one foot or so in front of your eyes. Close your right eye and look at your thumb with your left eye. Note its position against a distant background. Now close your left eye and look at your thumb with your right eye. Do you notice that the position of the thumb has shifted with respect to the background? Your thumb has not moved. However, since you have looked at it from different point (left and right eyes), it seems to have shifted. The shift in the apparent position of the thumb can be represented by an angle θ (Fig. 1.4). Parallax is the apparent change in the position of an object due to a change in the location of the observer.

θ d

b Fig.1.4: Parallax angle and baseline

We call θ/2, the parallax angle. The distance b between the points of observation (in this case your eyes), is called the baseline. From simple geometry, for small θ b angles, = , where d is the distance from the eyes to the thumb. 2 d

14

The parallax method can be used to measure the distances of stars and other objects in the sky. The principle of the method is similar to the one used in finding the height of mountain peaks, tall buildings, etc.

Astronomical Scales

Let us now find out how this method can be used to measure astronomical distances. Stellar Parallax

For measuring the distance of a star, we must use a very long baseline. Even for measuring the distance to the nearest star, we require a baseline length greater than the Earth’s diameter. This is because the distance of the star is so large that the angle measured from two diametrically opposite points on the Earth will differ by an amount which cannot be measured. Therefore, we take the diameter of the Earth’s orbit as the baseline, and make two observations at an interval of six months (see Fig. 1.5). One half of the maximum change in angular position (Fig. 1.5) of the star is defined as its annual parallax. From Fig. 1.5, the distance r of the star is given by d SE = tan θ r

(1.10a)

where dSE is the average distance between the Sun and the Earth. Since the angle θ is very small, tan θ ≅ θ, and we can write r=

d SE θ

(1.10b)

Remember that this relation holds only when the parallax angle θ is expressed in radians.

r

dSE

Fig.1.5: Stellar parallax

Since, dSE = 1 AU, we have 1AU r= θ

(1.10c) 15

Basics of Astronomy

If we measure θ in arc seconds, then the distance is said to be in parsecs.

One parsec is the distance of an object that has a parallax of one second of an arc (1″).

The nearest star Proxima Centauri has a parallax angle 0.77 ″. Thus its distance is 1.3 pc. Since the distance is proportional to 1/θ, the more distant a star is, the smaller is its parallax. In Table 1.4 we give the parallax angles and distances of some stars. Table 1.4: Parallaxes and distances of some bright stars Star

θ (in arc-seconds)

distance (r pc)

α-CMa

0.375

2.67

αCMi

0.287

3.48

αAquila

0.198

5.05

αTauri

0.048

20.8

αVirginis

0.014

71.4

αScorpii

0.008

125

Note that the angle θ cannot be measured precisely when the stellar object is at a large distance. Therefore, alternative methods are used to determine distances of stellar objects. You could now try an exercise to make sure you have grasped the concept of parallax. Spend 10 min.

SAQ 4

a) The parallax angles of the Sun’s neighbouring stars (in arc-seconds) are given below. Calculate their distances. Star

Parallax

Alpha Centauri

0.745

Barnard’s star

0.552

Altair

0.197

Alpha Draco

0.176

b) A satellite measures the parallax angle of a star as 0.002 arc-second. What is the distance of the star? You have just learnt that the parallax method helps us in finding the distances to nearby stars. But how can we find out which stars are nearby? We can do this by observing the motion of stars in the sky over a period of time. Proper Motion

All celestial objects, the Sun, the Moon, stars, galaxies and other bodies are in relative motion with respect to one another. Part of their relative motion is also due to the Earth’s own motion. However, the rate of change in the position of a star is very slow. It is not appreciable in one year or even in a decade. For example, if we photograph a small area of the sky at an interval of 10 years, we will find that some of the stars in the photograph have moved very slightly against the background objects (Fig.1.6). 16

Astronomical Scales

Fig.1.6: Motion of a star with respect to distant background objects

The motion of a star can be resolved along two directions: i) Motion along the line of sight of the observer, (either towards or away from the observer) is called the radial motion. ii) Motion perpendicular to the line of sight of the observer is called proper motion (see Fig. 1.7). Radial motion

Space motion

Fig.1.7: Radial and proper motion of a star

Radial motion causes the spectral lines of a star to shift towards red (if the motion is away from the observer) or towards blue (if the motion is towards the observer). This shift is the well-known Doppler shift. The proper motion is very slow. It is measured over an interval of 20 to 30 years. It is expressed in arc seconds per year. The average proper motion for all naked eye stars is less than 0.1 arc second/yr. The proper motion is denoted by µ. For a star at a distance r from the Earth it is related to its transverse velocity as follows: proper motion =

or

transverse velocity distance of the star

v µ= θ , r

(1.11a)

where vθ is the transverse velocity. Hence, vθ = µr

(1.11b) 17

Basics of Astronomy

If µ is measured in units of arc-seconds per year and r in pc, the transverse velocity is given by vθ ( km s −1 ) = 4.74µr

(1.11c)

If we add the radial velocity vector and the proper motion vector, we obtain the space velocity of a star (Fig.1.7). We can locate stars that are probably nearby by looking for stars with large proper motions (see Fig. 1.8a). Proper motion of a star gives us statistical clues to its distance. If we see a star with a small proper motion, it is most likely to be a distant star. However, we cannot be absolutely certain since it could also be a nearby star moving directly away from us or toward us (see Fig. 1.8b).

(a)

(b)

Fig.1.8: a) If two stars have the same space velocity and move perpendicular to the line of sight, the one with the larger proper motion will be nearer; b) Two stars at the same distance with the same velocity may have different proper motions, if one moves perpendicular to the line of sight and the other is nearly parallel to the line of sight

We know that the Sun itself is not stationary. The space velocity vector of a star must be corrected by subtracting from it the velocity vector of the Sun. The space velocity of a star corrected for the motion of the Sun is termed as the peculiar velocity of the star. The peculiar velocities of stars are essentially random and their typical magnitude is such that in a time of about 106 years the shape of the present constellations will change completely and they would not be recognisable (Fig. 1.9). 100,000 years ago

Present

100,000 years from now

Fig.1.9: Change in the shape of the Big Dipper due to peculiar velocities

You may now like to solve a problem based on these concepts. 18

SAQ 5

The star η CMa is at a distance of 800 pc. If the proper motion of the star is 0.008″/yr, calculate its transverse velocity in km s−1.

Spend 2 min.

Astronomical Scales

So far you have learnt how we can find distances of stars. In astronomy, it is equally important to know the sizes of stars. Are they all the same size, or are some of them smaller or larger than the others? Let us now find out how stellar radii may be measured.

1.4.2

Stellar Radii

There are several ways of measuring the radii of stars. Here we describe two methods: •

the direct method, and

•

the indirect method.

Direct Method

We use this method to measure the radius of an object that is in the form of a disc. In this method, we measure the angular diameter and the distance of the object from the place of observation (see Fig. 1.10).

D θ

r

Fig.1.10: Direct method

If θ (rad) is the angular diameter and r is the distance of the object from the observer then the diameter of the stellar object will be D=θ×r

(1.12)

This method is useful for determining the radii of the Sun, the planets and their satellites. Since stars are so far that they cannot be seen as discs even with the largest telescopes, this method cannot be used to find their radii. For this we use other methods. In Table 1.5 we give the radius of some stars. Table 1.5: Radius of some stars Star

θ (in arc seconds)

Radius (in RΘ)

α Tau

0.020

48

α Ori

0.034

214

α Sco

0.028

187

The luminosity of a star can also reveal its size since it depends on the surface area and temperature of star. This provides a basis for the indirect method of determining stellar radii. 19

Basics of Astronomy

Indirect Method

To obtain stellar radii, we can also use Stefan-Boltzmann law of radiation F = σ T4

(1.13)

where F is the radiant flux from the surface of the object, σ, Stefan’s constant and T, the surface temperature of the star. You have learnt in Sec. 1.3 that the luminosity L of a star is defined as the total energy radiated by the star per second. Since 4π R2 is the surface area, we can write L = 4π R 2 F

where R is the radius of the star. If the star’s surface temperature is T, using Eq. (1.13), we obtain L = 4πR 2 σT 4

(1.14)

The knowledge of L and T gives R. Now let us consider two stars of radii R1 and R2 and surface temperatures T1 and T2, respectively. The ratio of luminosities of these two stars will be L1 R = 1 L2 R2

2

T1 T2

4

(1.15)

But from Eq. (1.8), we have L2 = 100 ( M1 − M 2 ) / 5 L1

(1.16)

where M1 and M2 are the absolute magnitudes. Therefore, from Eqs. (1.15) and (1.16), we get R22T24 R12T14

= 100.4(M 1 − M 2 )

(1.17)

Using Eq. (1.17) let us now determine the ratio of radii of Sirius A and Sirius B.

Example 2: Determining stellar radii

The surface temperatures of Sirius A and Sirius B are found to be equal. The absolute magnitude of Sirius B is larger than that of Sirius A by 10. Thus, M1 − M2 = −10 and we have R2 = 0.01 × R1 Thus the radius of Sirius A is 100 times that of Sirius B. You may like to attempt an exercise now. Spend 5 min.

SAQ 6

The luminosity of a star is 40 times that of the Sun and its temperature is twice as much. Determine the radius of the star. Mass is also a fundamental property of a star, like its luminosity and its radius. Unfortunately, mass of a single star cannot be found directly. If, however, two stars revolve round each other, it is possible to estimate their masses by the application of Kepler’s laws.

20

1.4.3

Masses of Stars

Astronomical Scales

Two stars revolving around each other form a binary system. Fortunately, a large fraction of stars are in binary systems and therefore their masses can be determined. Binary stars can be of three kinds: 1. Visual binary stars: These stars can be seen moving around each other with the help of a telescope. If both the stars have comparable masses, then both revolve around their common centre of mass in elliptic orbits. If, however, one is much more massive than the other, then the less massive star executes an elliptic orbit around the more massive star (Fig.1.11).

Fig.1.11: Orbit of visual binary stars

2. Spectroscopic binary stars: The nature of these stars being binary is revealed by the oscillating lines in their spectra. Consider the situation in Fig.1.12a. Here star 1 is moving towards the observer and star 2 is moving away from the observer. The spectral lines of star 1 are, therefore, shifted towards blue region from their original position due to Doppler Effect. The lines of star 2 are shifted towards red. Half a period later, star 1 is moving away from the observer and star 2 is moving towards the observer (Fig.1.12b). Now the spectral lines of the two stars are shifted in the directions opposite to the earlier case. In this way the spectral lines oscillate.

1

2

To Earth

(a)

2

1

To Earth

(b)

Fig.1.12: Spectroscopic binary stars

21

Basics of Astronomy

Observations of oscillating lines indicate that the stars are binary stars. If only one of the stars is bright, then only one set of oscillating lines is observed. If both the stars are bright, then two sets of oscillating lines are seen. 3. Eclipsing binary stars: If the orbits of two stars are such that the stars pass in front of each other as seen by an observer (Fig.1.13), then the light from the group dips periodically. The periodic dips reveal not only the binary nature of the stars, but also give information about their luminosities and sizes.

d a

c

Total light output

b

c

a

a

b

b

Due to limb darkening

d Time

Fig.1.13: Eclipsing binary stars

Now suppose M1 and M2 are the masses of the two stars and a is the distance between them, then we can write Kepler’s third law as GP 2 4π

2

(M 1 + M 2 ) = a 3

(1.18)

where P is the period of the binary system and G is the constant of gravitation. This relation gives us the combined mass of the two stars. However, if the motion of both the stars around the common centre of mass can be observed, then we have M 1a1 = M 2 a 2

(1.19)

where a1 and a 2 are distances from the centre of mass. Then both these equations allow us to estimate the masses of both the stars. Masses of stars are expressed in units of the solar mass, MΘ = 2 × 1030 kg. Most stars have masses between 0.1 MΘ and 10 MΘ. A small fraction of stars may have masses of 50 MΘ or 100 MΘ. So far we have discussed the ways of measuring stellar parameters such as distance, luminosity, radii and mass. Stellar temperature is another important property of a star.

1.4.4

Stellar Temperature

The temperature of a star can be determined by looking at its spectrum or colour. The radiant flux (Fλ) at various wavelengths (λ) is shown in Fig.1.14. This figure is quite similar to the one obtained for a black body at a certain temperature. Assuming the star to be radiating as a black body, it is possible to fit in a Planck’s curve to the observed data at temperature T. This temperature determines the colour of the star. 22

Astronomical Scales

Eλ

T

λmax

λ

Fig.1.14: Total energy flux at different wavelengths.

The temperature of a star (corresponding to a black body) may be estimated using Wien’s law: λmaxT = 0.29 cm K

(1.20)

Such a temperature is termed as surface temperature, Ts. In general it is difficult to define the temperature of a star. For instance the temperature obtained from line emission is indicative of temperature from a region of a star where these lines are formed. Similarly the effective temperature of a star corresponds to the one obtained using Stefan-Boltzman law, i.e., F = σTe4 . In Table 1.6 we give the range of values of stellar parameters of interest in astronomy such as mass, radius, luminosity and stellar temperature. Table 1.6: Range of Stellar Parameters Stellar Parameters

Range

0.1 − 100 MΘ

Mass

0.01* − 1000 RΘ

Radius

10−5 − 105 LΘ

Luminosity Surface Temperature

3000 − 50,000 K

*It is difficult to put any lower limit on the radii of stars. As you will learn later, a neutron star has a radius of only 10 km. The radius of a black hole cannot be defined in the usual sense.

We can find various empirical relationships among different stellar parameters, e.g., mass, radius, luminosity, effective temperatures, etc. Observations show that the luminosity of stars depends on their mass. We find that the larger the mass of a star, the more luminous it is. For most stars, the mass and luminosity are related as L LΘ

=

M

3.5

MΘ

(1.21)

In this unit we have introduced you to a number of astronomical quantities and described some simple ways of measuring them. We now summarise the contents of this unit.

1.5 •

SUMMARY The astronomical units of distance, size, mass and luminosity are defined as follows:

− 1 astronomical unit (AU) is the mean distance between the Sun and the Earth. 1 AU = 1.496 × 1011m.

− 1 light year (ly) is the distance travelled by light in one year. 1 ly = 9.4605 × 1015m = 6.32 × 104 AU.

23

− 1 parsec (pc) is defined as the distance at which the radius of Earth’s orbit

Basics of Astronomy

subtends an angle of 1″. 1 pc = 3.262 ly = 2.06 × 105AU = 3.084 × 1016m.

− 1 solar radius RΘ = 7 × 108 m. − 1 solar mass MΘ = 2 × 1030 kg. − 1 solar luminosity LΘ = 4 × 1026 W. •

Apparent magnitude of an astronomical object is a measure of how bright it appears. Its absolute magnitude is defined as its apparent magnitude if it were at a distance of 10 pc from us.

•

The difference in apparent and absolute magnitude is called the distance modulus and is a measure of the distance of an astronomical object: m − M = 5 log10

•

rpc 10pc

Radiant flux is the total amount of energy flowing per unit time per unit area oriented normal to the direction of its propagation. The luminosity of a body is defined as the total energy radiated per unit time by it.

•

Brightness and radiant flux of an object are related to its apparent magnitude as follows: b2 = 100 ( m1 − m2 ) / 5 b1 F2 = 100 ( m1 − m2 ) / 5 F1

•

The absolute magnitude and luminosity are related as follows: M 1 − M 2 = 2.5 log10

•

If θ is the parallax of an object in arc seconds, then its distance in parsecs is given by r=

•

L2 L1

1 AU θ

The motion of an object can be resolved into two components: radial motion and proper motion. The proper motion µ of a star is related to its transverse velocity vθ as follows: v µ= θ r

where r is its distance. •

Stellar radii are related to the absolute magnitudes and temperatures of stars: R22T24

24

R12T14

= antilog{+ 0.4(M 1 − M 2 )}

•

The masses M1 and M2 of stars in a binary system can be estimated from the following relations: GP 2 4π

2

Astronomical Scales

(M 1 + M 2 ) = a 3

M 1a1 = M 2 a 2

where P is the time period of the binary system, G the constant of gravitation, a the distance between them and a1, a2, their distances from the centre of mass, respectively. •

The temperature of a star can be estimated by fitting observed data to Planck’s black body radiation curve or using Wein’s law: λ max T = 0.29 cm The temperature T of a star can also be estimated from Stefan-Boltzmann law: F = σT 4 .

1.6

TERMINAL QUESTIONS

Spend 30 min.

1. The apparent magnitude of full moon is − 12.5 and that of Venus at its brightest is − 4.0. Which is brighter and by how much? 2. The apparent magnitude of the Sun is − 26.8. Find its absolute magnitude. Remember that the distance between the Sun & the Earth is 1.5 × 1013 cm. 3. After about 5 billion years the Sun is expected to swell to 200 times its present size. If its temperature becomes half of what it is today, find the change in its absolute magnitude. 4. The mass of star Sirius is thrice that of the Sun. Find the ratio of their luminosities and the difference in their absolute magnitudes. Taking the absolute magnitude of the Sun as 5, find the absolute magnitude of Sirius.

1.7

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Jupiter is 5 AU from the Sun. 1 pc = 2.06 × 105 AU 5 AU = 2.43 × 10−5 pc Distance between Earth and the Sun = 1 AU. 1 ly = 6240 AU 1 AU = 1.6 × 10−4 ly b) The radius of the Earth is 0.01 RΘ . 2. a) The Sun is brighter mSun − mSi = − 26.81 − (− 1.47) = − 25.34 25

Basics of Astronomy

b1 b2

= 100 25.34 / 5 = 100 5.07 = 1.38 × 1010

The Sun is about 1010 times brighter than Sirius. b1

b)

b2

= 100 −(0.06−0.86) / 5 = 100

+.80/5

m − M = − 0.5 = 5 log10

3. a)

r

log10

= −

10 pc

r

0 .5

= (10)0.32 = 2.09

r 10 pc

= −0.1

5

= (10 )−0.1

10 pc

r = 7.9 pc

F2

b)

F1

=

40

2

10

= 16

It will appear 16 times brighter, which corresponds to m = 3 from Table 1.2.

4. a)

r=

1 AU θ

pc ;

Alpha Centauri 1.34 pc; Barnard’s star 1.81 pc; Altair

5.07 pc;

Alpha Draco

5.68 pc

b) Distance = 500 pc 5.

vθ = 4.74 µ r = 30.34 km s−1

6.

M1 − M2 = 40 R22

T4 L 1 = 1 . 2 = 2 4 L 2 R1 T2 1

( )

R22 = R12 × 2.5 R2 = 1.58RΘ .

26

4

× 40

Astronomical Scales

Terminal Questions

1. Apparent magnitude of moon is lower (larger negative number), than that of Venus. Therefore, moon is brighter than Venus. Moreover, bmoon

= 10 −0.4( mmoon − mvenus )

bvenus

= 10 −0.4( −12.5+ 4.0) = 10 −0.4−8.5 = 10 3.4

= 2.5 × 10 3.

2. The relation between apparent magnitude m and absolute magnitude M is M = m − 5 log r + 5 where the distance r is in parsec. Distance of the Sun in parsec is 1.5 × 1013/3 × 1018 = 5 × 10−6. So, M = − 26.8 − 5 (log 5 − 6) + 5 = − 26.8 − 5 log 5 + 30 + 5 = 8.4 − 3.5 = 4.9 3. According to Eq. (1.17) (200) 2 ( 2)

4

= 10 0.4( M1 − M 2 )

where M1 is the present absolute magnitude of the Sun. Therefore,

M 1 − M 2 = 2.5 log

200 × 200 16

= 2.5 log (2500) = 2.5 × 3.4 = 8.5 So, the absolute magnitude of the Sun will decrease by 8.5 and it will, therefore, become much more luminous. 4. Using Eq. (1.21) LSirius LΘ

= (3) 3.5

27

Basics of Astronomy

Now using Eq. (1.16), (3) 3.5 = 10 0.4( M Θ −M Sirius )

where MΘ and MSirius are absolute magnitudes of the Sun and Sirius. (MΘ − MSirius) = 2.5 log (46.8)

So,

= 2.5 × 1.7 = 4.25. ∴

28

MSirius = 5 − 4.25 = 0.75.

UNIT 2 BASIC CONCEPTS OF POSITIONAL ASTRONOMY

Basic Concepts of Positional Astronomy

Structure 2.1

Introduction

2.2

Celestial Sphere

Objectives Geometry of a Sphere Spherical Triangle

2.3

Astronomical Coordinate Systems Geographical Coordinates Horizon System Equatorial System Diurnal Motion of the Stars Conversion of Coordinates

2.4

Measurement of Time Sidereal Time Apparent Solar Time Mean Solar Time Equation of Time Calendar

2.5 2.6 2.7

Summary Terminal Questions Solutions and Answers

2.1

INTRODUCTION

In Unit 1, you have studied about the physical parameters and measurements that are relevant in astronomy and astrophysics. You have learnt that astronomical scales are very different from the ones that we encounter in our day-to-day lives. In astronomy, we are also interested in the motion and structure of planets, stars, galaxies and other celestial objects. For this purpose, it is essential that the position of these objects is precisely defined. In this unit, we describe some coordinate systems (horizon, local equatorial and universal equatorial) used to define the positions of these objects. We also discuss the effect of Earth’s daily and annual motion on the positions of these objects. Finally, we explain how time is measured in astronomy. In the next unit, you will learn about various techniques and instruments used to make astronomical measurements. Study Guide In this unit, you will encounter the coordinate systems used in astronomy for the first time. In order to understand them, it would do you good to draw each diagram yourself. Try to visualise the fundamental circles, reference points and coordinates as you make each drawing. You could use clay/plastecene balls or spherical potatoes to model various planes on the celestial sphere. You could also mark the meridians and the coordinates on them for a visual understanding of the coordinates we describe in this unit.

29

Basics of Astronomy

Objectives After studying this unit, you should be able to: •

identify the fundamental great circles such as horizon, celestial equator, ecliptic, observer’s/local meridian;

•

assign the horizon and equatorial coordinates of a celestial object; and

•

calculate the apparent solar time at a given mean solar time on any day.

2.2

CELESTIAL SPHERE

When we look at the clear sky at night, the stars appear to be distributed on the inside surface of a vast sphere centred on the observer. This sphere is called the celestial sphere (Fig. 2.1).

Fig.2.1: The celestial sphere centred on the observer. Note that the celestial sphere is extremely large compared with the Earth; for all practical purposes, the observer and the centre of the Earth coincide

Actually, the stars are at different distances from us. But since the distances of the stars are very large, they appear to us as if they are at the same distance from us. So it is sufficient if only the directions of stars are defined on the sphere. Since the distance is not involved, it is usual to take the radius of the celestial sphere as unity. You know that on the surface of a sphere, we need just two coordinates to describe the position of a point. We shall study below some ways used to specify these two coordinates for stars and other celestial objects. But before that you should familiarise yourself with the geometry of a sphere. 30

2.2.1

Basic Concepts of Positional Astronomy

Geometry of a Sphere

Let us consider a sphere of radius R = 1 unit (Fig. 2.2). We know that a plane intersects the sphere in a circle. If the plane passes through the centre of the sphere we get a great circle, and if it does not pass through the centre then it is a small circle.

Fig.2.2: Great and small circles on a sphere. A great circle is the intersection of a sphere and a plane passing through its centre.

In Fig.2.3, FG is a great circle whose plane passes through the centre O, and EH is a small circle. Points P and Q are called the poles of the great circle FG. Notice that PQ is the diameter of the sphere perpendicular to the plane of the great circle FG. The distance between two points on a sphere is measured by the length of the arc of the great circle passing through them. It is the shortest path on the surface of the sphere between these points. Thus, in Fig. 2.3, the distance between points C and D on the great circle FG is the arc length CD. It is equal to R (=OD) times the ∠COD in radians. Since R = 1, arc CD = ∠COD. In astronomy, it is convenient to denote distances on a sphere in angles (radians or degrees) in this way. An angle between two great circles is called a spherical angle. It is equal to the angle between the tangents to the great circles at the points of their intersection. It is also the angle between the planes of the two great circles. In Fig. 2.3, PACQ and PBDQ are two great circles intersecting at the points P and Q. ∠APB and ∠CPD are spherical angles between these great circles. These angles are equal. They are also equal to ∠COD, since the arc length CD is common to both great circles. P

E

H A

F

B G

O C

D

Q Fig.2.3: Spherical angle is the angle between two great circles

31

Basics of Astronomy

2.2.2

Spherical Triangle

A closed figure formed by three mutually intersecting great circles on a sphere is called a spherical triangle. For example, in Fig. 2.3, PCD is a spherical triangle. All the 6 elements of the triangle, namely, 3 sides and 3 angles are expressed in angular units. The elements of a spherical triangle ABC in Fig. 2.4 are the 3 spherical angles denoted by A, B and C and the 3 sides denoted by a, b and c.

B a c C b A

Fig.2.4: Spherical triangle is bounded by three arcs of great circles: AB, BC and CA. The spherical angles are A, B, C and the corresponding sides are a, b and c

Note that a spherical triangle is not just any three-cornered figure lying on a sphere. Its sides must be arcs of great circles. The essential properties of a spherical triangle are: i) The sum of the three angles is greater than 180°, ii) Any angle or side is less than 180°, iii) Sum of any two sides/angles is greater than the third side/angle. Note: Remember that the sides are being measured in degrees or radians. The four basic formulae relating the angles and sides of a spherical triangle (Fig. 2.4) are given below. These are for reference only. You need not memorise them. Sine formula sin A sin B sin C = = sin a sin b sin c

(2.1)

Cosine formula cos a cos b cos c

= = =

cos b cos c + sin b sin c cos A cos c cos a + sin c sin a cos B cos a cos b + sin a sin b cos C

(2.2)

Analogous cosine formulas are:

32

sin a cos B = cos b sin c − sin b cos c cos A sin a cos C = cos c sin b − sin c cos b cos A

(2.3)

Basic Concepts of Positional Astronomy

Four parts formula cos a cos B = sin a cot c − sin B cot C

(2.4)

Now that you have become familiar with the geometry of the celestial sphere, you can learn about the different coordinate systems used to locate a celestial object.

2.3

ASTRONOMICAL COORDINATE SYSTEMS

The method of fixing the position of a star or celestial object on the celestial sphere is the same as that of fixing the position of a place on the surface of the Earth. We shall, therefore, first consider geographical coordinates of a place on the surface of the Earth.

2.3.1

Geographical Coordinates

As you know, any place on the Earth is specified by two coordinates: geographical latitude and geographical longitude (see Fig.2.5).

P

Longitude Prime meridian Latitude Fig.2.5: The geographical coordinates (longitude and latitude) uniquely define any position on the Earth’s surface

Note that the line joining the poles is always perpendicular to the equator. The circles parallel to the equator are the circles of latitude. The great circles drawn through the north and South Poles are called the circles of longitude. How do we specify the coordinates of the place P on the Earth’s surface? The great circle passing through P is called its meridian. By international agreement, the geographical coordinates (longitude and latitude) of the place P are specified with respect to the equator and the meridian through Greenwich, which is also called the prime meridian. 33

Basics of Astronomy

The general definitions are given in the box ahead.

Definitions of longitude and latitude The longitude of a place is defined by the angle between Greenwich meridian and meridian of the place. It is measured from 0 to 180°° east or west of Greenwich. The latitude is defined by the angular distance of the place from the equator. It is measured from 0 to 90°° along the meridian of the place north or south of the equator.

In Table 2.1, we give the longitudes and latitudes of some cities in India. Table 2.1: Longitudes and latitudes of some cities in India Cities

Longitude

Latitude

Allahabad

81° 50′E

25° 26′N

Ahmedabad

72° 44′E

23° 08′N

Chennai

80° 10′E

13° 06′N

Kanyakumari

77° 32′E

8° 05′N

Kolkata

88° 21′E

22° 34′N

Mumbai

72° 51′E

19° 07′N

New Delhi

77° 12′E

28° 46′N

Shillong

91° 52′E

25° 34′N

Srinagar

74° 48′E

34° 05′N

Ujjain

75° 46′E

23° 10′N

We shall now extend the concept of geographical coordinates on the Earth to define the coordinates of a celestial body on the celestial sphere. But before that you may like to work out a problem. Spend 3 min.

SAQ 1 Using Table 2.1, find the difference between a) the longitudes of Ahmedabad and Shillong, and b) the latitudes of Srinagar and Kanyakumari. The longitude-latitude system illustrates the principle of an astronomical coordinate system. What we require for reference is: i)

A fundamental great circle (such as the equator), and

ii) A reference point, or origin on the chosen great circle. Depending on the choice of these two references, we describe the following coordinate systems, which are most often used in astronomy: 1. Horizon system 2. Equatorial system 34

2.3.2

Basic Concepts of Positional Astronomy

Horizon System

Suppose the observer is at the point O at latitude α (see Fig. 2.6). The point vertically overhead the observer is called the zenith. It is denoted by Z in Fig. 2.6. The point vertically below the observer is called the nadir. It is denoted by Z′. The great circle on the celestial sphere perpendicular to the vertical line ZO Z′ is called the celestial horizon or just the horizon. This is the fundamental great circle chosen for reference in the horizon system.

Note that the term horizon used here does not have its common meaning as the line where the sky meets the land.

The great circle ZNZ′SZ containing the zenith of the observer is called the observer’s meridian or the local meridian. N and S are the points of intersection of the horizon and the observer’s meridian. These points indicate geographical north and south directions. Either N or S can be taken as the reference point in this system.

Zenith Z Observer′′s meridian

X α O N

S horizon

Celestial equator

Nadir Z′′

Fig.2.6: The celestial horizon and the observer’s meridian

Let X be the celestial body, say a star. You can now fix the coordinates of X in this system. Draw the great circle ZXZ′ through the star X in Fig.2.6 (see Fig.2.7a). Let it intersect the horizon at Y. The position of X in the horizon system is defined with respect to the horizon and the reference point N or S. One coordinate of X is the arc YX, called the altitude (a). Mark it on Fig. 2.6. To fix the other coordinate, choose N as the origin. Then the arc length NY is the second coordinate, called the azimuth (A). Mark it on the above figure. The position of X from the zenith, the arc ZX = 90−a, is called the zenith distance and is denoted by z. Mark z on Fig. 2.6. Thus, in the horizon system the position of X can be specified by the coordinates (A, a) or (A, z).

35

Basics of Astronomy

Compare your drawing with Fig. 2.7a.

Zenith Z

zenith distance z

X

X

altitude

W horizon

S

W horizon

a altitude

O

A

azimuth

NS

a O

Y

E

N

A

E

Y

azimuth

Nadir Z′′ (a)

(b) Fig.2.7: a) Horizon coordinate system; b) the azimuth gives the direction in which to look for X and the altitude gives the angle by which a telescope must be raised from the horizon

The altitude (a) varies from 0° to 90° from the horizon to the zenith or 0° to −90° from the horizon to the nadir. The azimuth (A) of the star is measured from N eastward or westward from 0 to 180°. Thus, in Fig.2.7a, the great circle arc NY along the horizon is the azimuth (east) of X. If S is chosen as origin, the azimuth is measured from 0 to 360° through west.

Horizon coordinate system Horizon Coordinates: (A, a) or (A, z) The azimuth, A, is the arc length NY along the horizon if N is taken as the origin. The altitude, a, is the arc length YX along the great circle ZXYZ′ containing the zenith and the star. The zenith distance, z, is the arc length ZX from the zenith to the star on the great circle ZXYZ′. Fundamental great circle: Horizon Reference point: The points of intersection N or S of the horizon and the observer’s meridian.

36

The horizon system is a very convenient system and most small telescopes use it to locate a celestial object. The azimuth gives the direction in which to look for an object. The altitude then gives the angle by which the telescope must be raised from the horizon to locate the object (Fig. 2.7b).

Basic Concepts of Positional Astronomy

The pole star is situated in the geographical north direction. Let us locate its coordinates in the horizon system. Example 1: Coordinates of the pole star in the horizon system

RECALL

What are the horizon coordinates of the pole star? Solution Study Fig. 2.8. The pole star is at P. Notice that the angle α (∠BCO) is the latitude of the observer. Simple geometry shows that this angle is equal to ∠PON. Do you recognise that the arc PN or ∠PON is the altitude of P?

P O

λ

Zenith Z Celestial sphere

P 90°° − α

D W α S

The latitude of a point P on the surface of the Earth is given by the angle between the line OP and a line in the plane of the equator. Here O is the centre of the Earth.

α

B αO C Earth

N North Pole Equator

Nadir Z′′ Fig.2.8: Horizon coordinates of the pole star. BC and OD are parallel. Hence ∠ZOD = α, ∠POD = 90°° and ∠ZON = 90°°. ∠POZ = 90°° − α and ∠PON = α

Thus, the altitude of the pole star is equal to the latitude of the observer. What is its azimuth? With respect to N, it is 0° and with respect to S, it is 180° west. Thus the horizon coordinates of the pole star are (0°,α).

The longitude tells us our east-west position. It is measured from the 0º line of the prime meridian of Greenwich. Lines of longitude run from the North Pole to the South Pole.

You may now like to attempt a problem. SAQ 2 Suppose you wish to point a small telescope to a star whose azimuth is 30° and altitude is 45°. Describe the procedure you will adopt. The horizon system, though simple, has two shortcomings. Firstly, different observers have different horizons. Therefore, at a given time, the horizon coordinates of an object will be different for different observers. Secondly, as the Earth rotates from west to east, celestial objects move from east to west across the sky. Looking from far above the North Pole, say from a spacecraft, the motion of the Earth is anticlockwise. Thus, even for the same observer, the horizon coordinates keep changing with time. So, the horizon system of coordinates is not very useful for practical purposes. These

Spend 3 min.

37

Basics of Astronomy

drawbacks are removed in the equatorial system, a system that gives the same coordinates for an object for all observers on the Earth.

2.3.3

Equatorial System

Consider Fig. 2.9 showing the celestial sphere for an observer O. The great circle whose plane is parallel to the equatorial plane of the Earth, and contains the centre O of the celestial sphere is called the celestial equator. P and Q are the poles of the celestial equator: P is the north celestial pole and Q, the south celestial pole. These poles are directly above the north and south terrestrial poles. As you know, the point P also points to the pole star. In Fig. 2.9, the great circle NESW is the observer’s horizon with zenith Z as its pole and RWTE is the celestial equator for which P and Q are poles. The celestial equator and the observer’s horizon intersect in two points, E and W, called the East and West points. Notice that the observer’s meridian is a full great circle, unlike the meridian.

Any semi-great circle through north and south celestial poles P and Q is called a meridian. But as you have learnt in Sec. 2.3.2, the full great circle through the observer’s zenith (PZRQT) is called the observer’s meridian or the local meridian. Zenith Z

U

R

P

Celestial Equator

Celestial sphere

Diurnal circle

X

W O S

horizon

C D

E

N V Observer’s meridian

Q

T Nadir Z′′

REMEMBER THE DIURNAL CIRCLE OF A STAR IS ALWAYS PARALLEL TO THE CELESTIAL EQUATOR.

38

Fig.2.9: Schematic diagram of the celestial equator and diurnal circle UVX, upper (U) and lower (V) transits. In the segment DUC the star is above the horizon. It is below the horizon in the segment CVD

As a result of the Earth’s rotation from west to east, the celestial bodies appear to move across the sky in the opposite direction, i.e., from east to west. This daily or diurnal motion is common to all celestial objects. The path of a star X will be along the small circle XUV parallel to the celestial equator. It is called the diurnal circle (Fig. 2.9). It intersects the observer’s meridian at the points U and V. The star completes one circuit of its diurnal circle in 24 hours, which is the Earth’s rotation period. As you can see from the Fig. 2.9, the star will cross (transit) the observer’s meridian at the points U and V during its diurnal motion. These are called the upper and lower transits. A star is said to rise when it rises above the observer’s

horizon. It is said to set when it goes below the horizon. In Fig. 2.9, in the segment DUC, the star is above the horizon. It is below the horizon in the segment CVD.

Basic Concepts of Positional Astronomy

Now, suppose we want to define the equatorial coordinates of a star X (see Fig. 2.10). The celestial equator (RWTE) is the fundamental great circle in this system. The reference point is R, the point of intersection of the observer’s meridian (ZRQP) and the celestial equator above the horizon. The semi-great circle PXQ is known as the hour circle of the star X. It is called hour circle because it indicates the time elapsed since the star was on the upper transit on the observer’s meridian. Let the hour circle of X intersect the celestial equator at J (Fig. 2.10).

Z

U

R

P

Diurnal circle

X

J W S

Celestial Equator

O

C

N

D

E

V

Q

T Z′′

Fig.2.10: Schematic diagram showing local equatorial coordinates (H = RJ, δ = JX). Note that as the diurnal circle is parallel to the celestial equator, JX does not change due to the star’s diurnal motion. But RJ changes as the star moves in the diurnal circle

One coordinate of the star X is given by the great circle arc RJ along the celestial equator. It is called the hour angle (H) of the star. It is measured towards west from 0 to 24 hours. The other coordinate required to define the position of the star is the circle arc JX from the equator along the star’s hour circle. It is called declination (δ). As the path of the diurnal motion of the star is parallel to the celestial equator, the declination, δ, of a given star does not change during the diurnal motion. The hour angle and the declination specify the position of the star in the local equatorial system. In astronomy, angles such as hour angle are frequently expressed in time units according to the relation: 24 hours = 360°, lh =15°, 1m = 15′ and 1s = 15″. 39

Basics of Astronomy

The declination of a star varies from 0 to 90° from the celestial equator to the celestial poles. It is taken as positive if the star is situated north of the equator and negative if it is south of the equator. Notice that in this system, one of the coordinates, the declination, does not change during the diurnal motion of the star while the other coordinate, the hour angle, varies from 0 to 24 hours due to the rotation of the star. That is why this system is called the local equatorial system. Local Equatorial System Local Equatorial Coordinates: (H, δ) The hour angle, H, is the arc length RJ along the celestial equator. The declination, δ, is the arc length JX along the star’s hour circle. Fundamental great circle: Celestial equator. Reference point: Intersection of observer’s meridian and celestial equator.

We shall now define a coordinate system in which both the coordinates of a star remain unchanged during its diurnal motion. For this, we choose the celestial equator as the fundamental great circle and the vernal equinox (ϒ) as the reference point. Let us understand what ϒ is. Vernal equinox You know that the Earth moves around the Sun in a nearly circular orbit. The plane of the Earth’s motion around the Sun intersects the celestial sphere in a great circle called the ecliptic (Fig. 2.11a). As seen from the Earth, the Sun appears to move around the Earth in this plane. Thus the ecliptic is the annual path of the Sun against the background stars. In Fig. 2.11a, R ϒ T is the celestial equator and M ϒ M′ is the ecliptic. The ecliptic is inclined to the equator at an angle of 23°27'. Points K and K′ are the poles of the ecliptic. The ecliptic and the celestial equator intersect at two points called vernal equinox (ϒ) and autumnal equinox (Ω). During the annual apparent motion, the Sun is north of the equator on ϒ M Ω and south of the equator on Ω M′ ϒ. Thus, the Sun’s declination changes from south to north at vernal equinox (ϒ) and from north to south at autumnal equinox (Ω). We can define the position of a star X with respect to the celestial equator and the vernal equinox (Fig. 2.11b). Let the hour circle of X intersect the celestial equator at J. One coordinate of X is the arc length ϒJ along the celestial equator measured from vernal equinox eastward. It is known as the right ascension of the star and is denoted by α. The other coordinate is the declination δ (arc JX ) as already defined. As the Earth rotates, the points ϒ and J on the celestial equator rotate together. Thus, the separation between ϒ and J does not change and the right ascension of the star remains fixed. This system of coordinates (α, δ) is called the universal equatorial system. The α and δ of stars do not change appreciably for centuries. 40

Basic Concepts of Positional Astronomy

North celestial pole K P

R

Autumnal equinox

Sun

Ω

Celestial equator

δ

Earth M'

M

Ecliptic

ϒ Vernal equinox

α ϒ

T

Ecliptic (b)

Q

South celestial pole

K'

(a) Fig.2.11: a) The ecliptic, vernal equinox and autumnal equinox; b) universal equatorial coordinates (α α, δ)

Universal Equatorial System Universal Equatorial Coordinates: (α, δ) The right ascension, α, is the arc length ϒJ eastward along the celestial equator. The declination, δ, is the arc JX along the star’s hour circle. Fundamental great circle: Celestial equator. Reference point: Vernal equinox.

Table 2.2 gives the universal equatorial coordinates of some prominent objects. Table 2.2: Universal equatorial coordinates of prominent astronomical objects Object

Universal equatorial coordinates

Crab nebula

RA 5: 35 (h,m)

Declination 22°:01m

Andromeda

RA 00:42.7 (h,m)

Declination 41°:16m

Ring nebula

RA 18:53:35.16 (h,m,s) Declination 33°:01m

Orion nebula

RA 05:35.4 (h,m)

Declination −05°:27m 41

Basics of Astronomy

Let us now find the coordinates of a star. Example 2: Equatorial Coordinates Calculate at latitude of 50°N, the zenith distance of a star of declination 20°N when it is on the observer’s meridian. Solution Look at Fig. 2.12.

Z

20°° N X

40°° P

Celestial equator

R

20°° S X'

50°° O

S

N

horizon

T Q

Fig.2.12

Since the latitude of the place is 50°N, the altitude of the pole star is also 50°. So, arc PN = 50°. Therefore, arc PZ = 40°, since arc NZ is 90°. As the declination of X is 20°N, RX = 20°. Thus, arc ZX, which is the zenith distance of X, is 30°, since RP = 90°. If the declination of the star were 20°S, it would have been at X′, then the zenith distance ZX′ would have been 50° + 20° = 70°. You should now practice calculating the equatorial coordinates of a few celestial objects. Spend 10 min.

SAQ 3 a) Determine the declination (δ) of (i) celestial North Pole, (ii) celestial South Pole (iii) zenith. (Hint: Look at Fig. 2.12 again). b) Determine the right ascension (α) and declination (δ) of (i) vernal equinox, (ii) ecliptic north pole, (iii) ecliptic South Pole. (Hint: Look at Fig. 2.13. Notice that arc ϒR = 6h.)

42

Basic Concepts of Positional Astronomy

K P

δ

R

Sun

Celestial equator Earth M'

M

Ecliptic

ϒ Vernal equinox

T

δ Q

K' Fig.2.13

We shall now briefly describe the diurnal motion of the stars and the Sun as seen from different places on the surface of the Earth.

2.3.4

Diurnal Motion of the Stars

See Fig. 2.14, drawn for an observer in northern latitudes. The diurnal circles of stars north of the equator are more than half above the horizon (arc JKL). The diurnal circles for stars south of the equator are less than half above the horizon (arc MCM′). Z K

R

Celestial equator C S

Celestial equator

P

Z

W J

O

M

N L

M' E

Dec. 21

Horizon

S

Q

T

Mar. 21 Sept. 22

N

June 21

Z' (a)

(b)

Fig.2.14: a) Diurnal circles for stars north and south of the equator; b) diurnal circles on particular days

43

Basics of Astronomy

Thus, an observer in northern latitudes will see more northern stars for longer duration above the horizon compared to the southern stars. For an observer on the equator, the zenith is on the celestial equator. Hence, both northern and southern stars are equally visible for 12 hours as their diurnal circles are perpendicular to the horizon (Fig. 2.15). Celestial equator

Z

Horizon Q

P

O

δmax = 90°°

δmax = 90°°

Z' Fig.2.15: Diurnal circles of stars seen from the equator

If we observe stars to the north, we find that some stars never set, i.e., they never go below the horizon. Instead they trace complete circles above it (Fig. 2.16). Such stars are called circumpolar stars. (The Egyptians called such stars as the ones that know no destruction.) Such stars are also there near the South Pole. Zenith Z

R

P

Celestial Equator

Celestial sphere

X Star never sets

W O

Horizon

S

N C D

E

Q

Observer’s meridian

Star never rises

T Nadir Z′′

Fig.2.16: Circumpolar stars

44

For an observer at latitude φ in the northern hemisphere, the declination of circumpolar stars is δ ≥ (90 − φ) (Fig. 2.17). The southern stars with δ < (φ − 90) are never seen above the horizon. They never rise. The situation is exactly the opposite in the southern hemisphere. For an observer on the equator, φ = 0, and the maximum value of δ is 90° at the north and South Poles. Hence, there are no circumpolar stars, as the declination of stars can never exceed 90° (see Fig. 2.15).

Basic Concepts of Positional Astronomy

Z

R

P

W X

Celestial equator

S

O

N φ

E

90°°− φ

α

δ

J Q

T

Z′′ Fig.2.17: Declination of circumpolar stars in the northern hemisphere for an observer at latitude φ in the northern hemisphere

At North Pole (φ = 90°), all the stars lying north of the equator, that is, δ ≥ 0 are circumpolar, while the southern stars (δ< 0) are below the horizon. These will be invisible (Fig. 2.18). Z, P

Celestial equator

Horizon O

Z', Q Fig.2.18: Circumpolar stars at the North Pole

Motion of the Sun You have studied in Sec. 2.3.3 that because of the yearly revolution of the Earth round the Sun, the Sun appears to move around the Earth along the ecliptic. It moves at a

45

Basics of Astronomy

rate of about 1° per day. As a result, the right ascension and declination of the Sun change continuously. The Sun is at ϒ (Fig. 2.19) on or around March 21 every year. At that time its coordinates are α = 0, δ = 0. At M, around June 21, the Sun is at α = 6h, δ = +23.5°. At Ω around September 23, α = 12h, δ = 0° for the Sun. At M' around December 22, the Sun is at α = 18h, δ = −23.5° (see Table 2.3). The point M with δ = δmax = + 23.5° is the summer solstice which marks the beginning of Dakshinayan, the south-ward motion of the Sun. The opposite point M' with δ = δmin = − 23.5° is the winter solstice which is the beginning of Uttarayan, the north-ward journey of the Sun. Note that the arc lengths ϒM and ϒM′ are equal even though it may not seem so in Fig. 2.19.

Table 2.3 ~ Day

(α, δ)

March 21

(0, 0)

June 21

(6 , + 23.5°)

September 23

(12 , 0°)

December 22

(18 , -23.5°)

North celestial pole

h

K P

h h

Celestial equator

Ω (Sept. 23) ( α = 12h, δ = 0°°

R

Sun

Earth

(Dec. 22) M' α = 18h, δ = −23.5°°

(June 21) α = 6h, δ = +23.5°°

M

Ecliptic

ϒ (Mar 21) α = 0, δ = 0

T

Q South celestial pole K' Fig.2.19: Motion of the Sun

The Sun as a Circumpolar Star You have learnt that for an observer at latitude φ, all stars with δ ≥ 90° − φ are circumpolar. Thus, at North Pole φ = 90° and for a star to be circumpolar, δ ≥ 0. You have also learnt that the declination of the Sun varies from −23.5° to +23.5° annually. Hence, for an observer on the North Pole, the Sun is circumpolar from March 21 to September 23 when the Sun is north of the equator since δ ≥ 0 for the Sun during this period. Further, from September 23 to March 21 the Sun is invisible at the North Pole, since its declination is negative during this period. On March 21, the Sun appears above the horizon and continues to remain above the horizon for 6 months till September 23. 46

Thus, it is seen that the Sun is above the horizon for six months and below the horizon for 6 months for observers at the North Pole.

Basic Concepts of Positional Astronomy

Can you say how an observer at the South Pole sees the Sun? The Sun is visible from September 23 to March 21 at the South Pole and invisible for the next six months. During the ‘day’ of 6 months, that is, from March 21 to September 23 at the North Pole, the Sun moves around the sky in small circles parallel to the celestial equator, which coincides with the horizon (see Fig. 2.20). Z, P

Celestial equator Horizon O

Z', Q Fig.2.20: Sun’s motion for an observer on the North Pole for whom the celestial equator coincides with the horizon.

On a given date, the Sun is circumpolar at all those places whose latitude φ ≥ (90 − δ), where δ is the declination of the Sun on the date under consideration. On June 21/22, for example, when the Sun's declination is +23.5°, the Sun is circumpolar at all the places north of 66.5° latitude. You may now like to stop for a while and concretise these ideas. Try this SAQ. SAQ 4 a)

Find the approximate zenith distance of the Sun on June 22 and December 23 at Delhi (φ = 28°46′N).

Spend 10 min.

b) At what latitudes is the star Procyon (δ = 05°18′S) circumpolar?

2.3.5

Conversion of Coordinates

We now state the relevant formulae for conversion of coordinates from local equatorial system to horizon system. Their mathematical derivation can be found in any standard book dealing with positional astronomy such as Smart (1960). You need not memorise these formulae. 47

Basics of Astronomy

Local Equatorial and Horizon System: Conversion of (H, δ) to (A, z) sin z sin A = cos δ sin H sin z cos A = sin δ cos φ − cos δ sin φ cos H cos z = sin δ sin φ + cos δ cos H cos φ

(2.5)

Conversion of (A, z) to (H, δ): cos δ sin H = sin z sin A cos δ cos H = cos z cos φ − sin z sin φ cos A sin δ = cos z sin φ + sin z cos φ cos A

(2.6)

So far we have discussed how to locate the position of a celestial object in horizon, local equatorial and universal equatorial coordinate systems. However, you know that along with the space coordinate, we also need to specify the time to be able to completely describe the motion of an object. Let us now learn about how time is measured in astronomy.

2.4

MEASUREMENT OF TIME

The rotating Earth is our natural clock. The ancient people were familiar with rotation and revolution of the Earth around the Sun and the revolution of the moon around the Earth. It was natural for them to base time measurement on these motions. Their basic unit of time was from one Sunrise to the next. They called it the day. Revolution of the Earth around the Sun provided another unit called the year. A third unit, the month, is approximately the period of revolution of moon around the Earth. The basic unit of time, the day, is defined as the time required for the Earth to rotate once on its axis. If this rotation is considered with respect to the Sun then the time for one rotation relative to the Sun is the solar day, that is, the time from one Sunrise to the next Sunrise. On the other hand, rotation period of the Earth relative to the stars is called a sidereal day. The sidereal day is different from a solar day. This difference arises due to the Earth’s revolution around the Sun (see Fig. 2.21).

C F

B

F

F

F'

Fig.2.21: The difference between one sidereal day and one solar day

In Fig. 2.21, the Earth is initially at A in its orbit around the Sun. Suppose at a given place F, the Sun and a distant star are overhead. Thus, it is noon for an observer at F (the foot of the solid arrow) and midnight at F′ (the foot of the dotted arrow). One sidereal day later (say at time t), the Earth is at B but now the solid arrow does not point to the Sun, though it would still point towards the distant star as it is very far 48

away compared to the Sun. So at time t, the Sun is not overhead at F, and it is not noon at F. The solid arrow points towards the Sun only a little later at C, when the Earth has turned upon its axis and moved a bit in its orbit.

Basic Concepts of Positional Astronomy

The time from A to C is equal to one solar day, while the time from A to B is equal to one sidereal day. The Earth has to rotate by about 1° more to bring the Sun overhead and complete the solar day. It takes roughly 4 minutes for the Earth to rotate through 1° since it rotates through 360° in 24 hours. Thus, the sidereal day is shorter than the average solar day by about 4 minutes. In a month (30 days), the sidereal time falls behind the mean solar time by two hours, and in six months the difference becomes 12 hours. Then at solar midnight we have sidereal noon. Since the Sun regulates human activity on the Earth, we use solar time in our daily life. The sidereal time is not useful for civil purposes but is used in astronomical observations.

2.4.1

Sidereal Time

Due to the rotation of the Earth, the vernal equinox ϒ, moves along the equator once in 24 hours like any other star. The hour angle of ϒ for an observer is called the sidereal time for that observer or the local sidereal time (LST).

Z P X R J W S

N ϒ E

Q Z' Fig.2.22: Sidereal Time

In Fig. 2.22, NWSE is the horizon, P and Z are the north celestial pole and the zenith, respectively. The hour angle (HA) of vernal equinox is the great circle arc Rϒ measured towards the west from the observer’s meridian. When ϒ is at R, at upper transit on the observer’s meridian, its HA is 0h and consequently the local sidereal time (LST) is 0h at this instant. The interval of time between two successive transits of vernal equinox over the same meridian is defined as one sidereal day. One sidereal day is divided into 24 sidereal hours, one sidereal hour into 60 sidereal minutes and one sidereal minute into 60 sidereal seconds. Let X be the position of a star in Fig. 2.22. The arc RJ (towards west) is the HA of X (HAX) and the arc ϒJ (towards east) is the right ascension of X (RAX). We have Rϒ = RJ + ϒJ. Hence 49

Basics of Astronomy

LST = HAX + RAX

(2.7)

This is an important relation in the measurement of time. If X is the Sun, denoted by , then we have the relation LST = HA

2.4.2

+ RA

(2.8)

Apparent Solar Time

The hour angle of the Sun at any instant is the apparent solar time. The interval between two consecutive transits of the Sun over the same meridian is called one apparent solar day (see Fig. 2.23). Local celestial meridian

Celestial Equator (a)

Ecliptic

EAST

SOUTH

WEST

Local celestial meridian

Celestial Equator (b)

Ecliptic

EAST

SOUTH

WEST

Fig.2.23: Apparent solar day

In Fig. 2.23a, we show the Sun at noon on March 21, when the Sun and the vernal equinox are on the local meridian. The celestial sphere rotates and some time during the next day, the vernal equinox is again on the meridian. That interval of time is one sidereal day; a fixed point on the celestial sphere has gone around once. But in that time the Sun has moved eastward along the ecliptic (see Fig. 2.23b). But the Sun is not yet on the meridian and it is not yet noon. The celestial sphere has to rotate a bit more to bring the Sun up to the meridian. This time interval represents the apparent solar day.

50

The apparent solar day is found to vary throughout the year. Thus, the Sun is an irregular timekeeper and consequently the time based on the motion of the Sun is not uniform. The non-uniformity in the solar time is due to two reasons: the first is that the path of apparent motion of the Sun is elliptical and not circular. This means that the Sun moves at a non-uniform rate. The second reason is that the Sun's apparent motion is in the plane of the ecliptic and not in the plane of the equator; and the measurement of time, that is, the measurement of hour angle is made along the equator.

2.4.3

Basic Concepts of Positional Astronomy

Mean Solar Time

In order to have a uniform solar time, a fictitious body called the mean Sun is introduced. The mean Sun is supposed to move uniformly (that is, in a circular path) along the equator completing one revolution in the same time as the actual Sun does round the ecliptic. The hour angle of the mean Sun at any instant is defined as the Mean Solar Time (MST), or simply, the Mean Time. The time interval between the successive transits of the mean Sun over the same meridian is defined as the mean solar day. The mean solar day is subdivided into hours, minutes and seconds. The civil day begins at mean midnight for reasons of convenience. Our clocks are designed to display the mean solar time. By international agreement, the time of the Greenwich meridian is called the Greenwich mean time (GMT). It can be related to the mean time of any other place through the longitude of that place. Since the Earth completes one rotation on its axis, that is 360 degrees of longitude in 24 hours, each degree of longitude corresponds to 4 minutes of time. Therefore, GMT at a given instant is related to the mean time of any other place by a simple relation: Local Mean Time = GMT + λ (in units of time)

(2.9)

where λ is the geographical longitude of the place expressed in units of time. In the above relation, λ is to be taken as positive if the longitude of the place is east of Greenwich, and negative if it is west of Greenwich. The above relationship holds good for both the sidereal as well as the solar time. It is obvious that if each town and city in a country were to use its own local time, the life of citizens would become very difficult. To overcome this problem, a whole country uses the local time of a place as its standard time. For example, the Indian Standard Time is the local time of a place on longitude 82.5° east. The IST is, therefore, ahead of the GMT by 5½ hours. Very large countries such as United States have more than one standard time. Example 3 What will the local time of Shillong be when the local time of Ahmedabad is 6 p.m.? Solution Refer to Table 2.1. The difference between longitudes of Ahmedabad and Shillong is approximately 19°. Since each degree of longitude is equal to 4 minutes of time, the difference in the local time of these two places is 19 × 4 = 76 minutes. Since Shillong is east of Ahmedabad, local time of Shillong will be ahead of the local time of Ahmedabad by one hour and sixteen minutes. And now you can attempt an exercise. SAQ 5

Spend 5 min.

Refer to Table 2.1. If the local time of Mumbai is 8:00 p.m., what would be the local time at Kolkata at that time? We will now discuss the Equation of Time, which relates the apparent solar time and the mean solar time. 51

2.4.4

Basics of Astronomy

NOTE

Equation of Time

As we have seen in Eq. (2.8)

HA is the symbol of hour angle for the Sun.

LST = HA

RA is the symbol of right ascension of the Sun.

+ RA

(2.10)

Using Eq. (2.10) for the Sun and the mean Sun, we can write RAMS − RA

HAMS is the symbol

= HA

− HAMS

(2.11)

The difference in right ascensions of the mean Sun and true Sun at a given instant (see Fig. 2.24a) is called the Equation of Time (ET).

of hour angle for the mean Sun.

Celestial equator

RAMS is the symbol

non-uniform rate

of right ascension of the mean Sun. Ecliptic uniform rate

uniform rate (a)

Real Sun

time of perihelion

time of equinox

Fictitious Sun Equation of time RAMS − RA

Not to scale

Sundial-clock time

(b)

18 16 14 12 10 8 6 4 2 0 −2 −4 −6 −8 −10 −12 −14 −16

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Fig.2.24: Equation of time

52

Mean Sun

Sep

Oct

Nov

Dec

Basic Concepts of Positional Astronomy

Thus, we have ET = HA

− HAMS

(2.12)

We see that the ET at any instant is the difference between the apparent solar time given by HA and the mean solar time given by HAMS. In Fig. 2.24b, you can see that during the year the ET varies between −14 1/4 to + 16 1/4 minutes and it vanishes 4 times during the year on or about April 16, June 14, September 1 and December 25. Any discussion on time in astronomy would perhaps be incomplete if we do not mention the calendar.

2.4.5

Calendar

The civil year contains an integral number of 365 mean solar days. As the actual period of Earth’s revolution, called a tropical year contains 365.2422 mean solar days, a fraction 0.2422 of a day is omitted each year. This resulted in the loss of a number of days over several centuries and the civil year would get out of step with the seasons. To overcome this confusion, Julius Caesar introduced Julian calendar, named after him, in which the year was made of 365.25 mean solar days. Consequently, every 4th year was made to contain 366 mean solar days. This was called a leap year. The extra day was added in February. Thus, in this calendar, a year, which is divisible by 4, is a leap year in which the month of February has 29 days. In Julian calendar, the tropical year was assumed to be of 365.25 mean solar days, but its actual length is 365.2422 mean solar days. The small difference between assumed and actual lengths of the year, created serious problem by the 16th century and the civil year was out of step in relation to the seasons. To overcome this problem, Pope Gregory, introduced in 1582 the calendar known as Gregorian calendar, which we are using now. In this calendar, a leap year is defined as before, with the exception that when a year ends in two zeros, it will be a leap year only if it is divisible by 400. Thus in a cycle of 400 years, there are 100 leap years according to Julian Calendar while there are only 97 in Gregorian Calendar, since years 100, 200 & 300 are not leap years. This makes the average civil year to consist of 365.2425 mean solar days, which is very close to the true length of the tropical year. No serious discrepancy can arise in the Gregorian calendar for many centuries. With this we come to the end of Unit 2. In this unit, we have studied various aspects of positional astronomy. We now summarise the main points of the unit.

2.5

SUMMARY

•

In positional astronomy, we are concerned with the relative directions of the heavenly bodies, which appear to be distributed on the surface of a vast sphere called the celestial sphere.

•

Any plane passing through the centre of the sphere intersects it in a circle called the great circle. The extremities of the diameter of the sphere perpendicular to the plane of the great circle are called poles of the great circle. Arcs of three great circles form a spherical triangle in which any angle / side is less than 180°. The sum of the three angles is greater than 180°.

•

Due to rotation of the Earth on its axis, the heavenly bodies appear to move across the sky from east to west in small circles parallel to the equator.

53

Basics of Astronomy

•

In the horizon coordinate system, the altitude and azimuth specify a star’s position. The altitude of the north celestial pole at any place is equal to the latitude of the place.

•

In the local equatorial system, the coordinates of a heavenly body are given by the hour angle and the declination. In the universal equatorial system, the declination and the right ascension specify the position of a celestial object.

•

Circumpolar stars are those, which never set, i.e., they are always above the horizon. At a given place of latitude φ, a star becomes circumpolar if its declination δ ≥ (90 − φ). It follows that at geographical North Pole (where φ = 90°) the Sun is circumpolar for 6 months from March 21 to September 23. Thus, at poles the day will last for 6 months and night for 6 months during the year. At equator (φ = 0) there is no circumpolar star and all the stars move in circles perpendicular to the horizon. The day and night are of 12-hour duration each.

•

Hour angle of vernal equinox is defined as the sidereal time. The sidereal day is a measure of the rotation period of the Earth with respect to the distant stars.

•

The hour angle of the mean Sun is defined as mean solar time and hour angle of the Sun is defined as apparent solar time.

•

The difference between the apparent solar time and the mean solar time is known as the equation of time. The difference between the two systems of time varies between −14 1/4 to +16 1/4 minutes and becomes zero 4 times during the year.

2.6

TERMINAL QUESTIONS

Spend 30 min.

1.

What is the latitude of the place at which celestial horizon and equator coincide? At what latitude is the celestial equator perpendicular to the horizon?

2.

Show that a star attains its maximum altitude when it is on the observer’s meridian.

3. A and B are two places in north latitude on the surface of the Earth; their latitudes are 24° 18′N and 36° 47′N, respectively; and their longitudes are 133° 36′E and 125° 24′W, respectively. Find the difference in their local time. 4. Show that at a place of latitude φ (North) at upper transit, the zenith distance of a star equals φ − δ .

2.7

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1. a) Difference in the longitudes of Ahmedabad and Shillong = 91° 52′ − 72° 44′ = 19° 08′ b) Difference in latitude between Srinagar and Kanyakumari = 34° 05′ − 8° 05′ = 26° 00′ 2. Turn the telescope 30° degrees from north to west, keeping it horizontal. Then raise it so that it makes an angle of 45° with the horizontal. 3. a) See Fig. 2.25. δ of celestial north pole = 90°. δ of celestial South Pole = − 90°. δ of zenith = φ, latitude of the place. 54

Z

20°° N X

Basic Concepts of Positional Astronomy

40°° P

Celestial equator

R

20°° S X'

50°° O

S

N

horizon

T Q

Fig.2.25

b) See Fig. 2.26. (α,δ) of vernal equinox = (0,0). (α,δ) of ecliptic north pole = (18h, 66½°) since ϒJ is measured eastward. (α,δ) of ecliptic South Pole = (6h, − 66½°).

K P

R

Sun

Celestial equator Earth M'

M

Ecliptic

ϒ Vernal equinox

T

Q

K' Fig.2.26

55

Basics of Astronomy

4. a) At Delhi, zenith makes an angle of 28° 46′ with the equator. The Sun on June 22 and December 22 makes an angle of 23° 27′ with the equator. Therefore, zenith distance of the Sun on that day = 28° 46′ − 23° 27′ = 5° 19′ b) φ ≥ 90 − δ = 84° 42′ 5. The difference in longitudes of Mumbai and Kolkata is around 15 ½°. Thus, the time difference is 62 minutes. Since Kolkata is to the east, the local time there will be 2 minutes past 9 p.m. when it is 8 p.m. in Mumbai. Terminal Questions 1. φ = 90°, φ = 0°. 2. See Fig. 2.27. At the transit point M on the diurnal circle, the altitude is maximum. Z

M Celestial equator

R

P W O

S

N

E

Q

T Z' Fig.2.27

3. Difference in longitude = 133° 36′ E − 125° 24′ W = 259°. ∴

Time at A will be ahead by 259° × 4 min = 1036 min = 17 h 16 min.

4. See Fig. 2.28. The star is at U. In general, z = φ − δ. φ

z

δ

Z

U

Horizon S

O

N Celestial equator

Z' Fig.2.28

56

UNIT 3 ASTRONOMICAL TECHNIQUES

Astronomical Techniques

Structure 3.1

Introduction

3.2

Basic Optical Definitions for Astronomy

Objectives Magnification Light Gathering Power Resolving Power and Diffraction Limit Atmospheric Windows

3.3

Optical Telescopes Types of Reflecting Telescopes Telescope Mountings Space Telescopes

3.4

Detectors and Their Use with Telescopes Types of Detectors Detection Limits with Telescopes

3.5 3.6 3.7

Summary Terminal Questions Solutions and Answers

3.1

INTRODUCTION

In Unit 1 you have learnt about various measurable astronomical quantities and the methods of measuring some of them such as distance, size and mass. You have also studied about the radiant flux or brightness of different astronomical sources like the planets, the sun, stars and galaxies. A great deal of information in astronomy is gathered with the help of ground-based instruments. Therefore, in this unit we describe some instruments as well as the techniques used to measure radiant flux and to analyse the radiation emitted by astronomical objects. We shall first revisit the basic optical definitions, viz. magnification, light gathering power and resolving power, relevant to astronomy. You may recall having studied them in Unit 11 of the Physics elective PHE-09 entitled Optics. In Sections 3.3 and 3.4, we shall describe instruments like optical telescopes and detectors that help us measure radiant flux from stars and other objects. A telescope is like a camera. It is a light-focusing instrument, while a detector, like a photographic film, is the medium on which the photons leave an impression, which can be measured later. You know the poem ‘Twinkle-twinkle little star’ from your early days. In this unit, you will study how the Earth’s atmosphere, that makes the stars twinkle, affects the observations made with telescopes on the ground. Finally, you will learn about techniques to detect faint astronomical light, which originates from very distant objects. In the next unit we discuss the basic principles of physics applicable in astronomy and astrophysics. Objectives After studying this unit, you should be able to: •

define angular magnification, light gathering power and resolving power of a telescope;

•

describe different types of reflecting telescopes and telescope mountings;

57

Basics of Astronomy

•

explain how atmosphere affects the incoming starlight;

•

describe the different detectors and techniques used for observations; and

•

estimate detectability limit of a telescope.

3.2

BASIC OPTICAL DEFINITIONS FOR ASTRONOMY

The telescope is the most common instrument used in ground-based astronomy. Most people think of a telescope only in terms of its magnifying power. However, magnification is only one of three powers of a telescope, the other two being its resolving power and light gathering power. Before we describe various telescopes, we would like to discuss these concepts as applicable in astronomy.

3.2.1

Magnification

As you know from the course in Optics, the magnifying power of a telescope refers to its ability to make the image appear bigger. If the angle subtended by a distant object at the objective of a telescope is β and that subtended by the virtual image at the eye is α (Fig. 3.1), then the angular magnification (A.M.) of the telescope is given by A.M. =

α

(3.1)

β

fe

fo

α Fo

Fe

β Fe

To top’ point of virtual image at infinity Fig.3.1: The image of a distant object is formed at the focal plane of the objective. The rays entering the eye are rendered parallel by the eye-piece. The eye sees the image at infinity

We calculate the A.M. of a telescope by dividing the objective’s focal length by the focal length of the eyepiece. f A.M. = o fe

(3.2)

For example, if a telescope has an objective with a focal length of 60 cm and an eyepiece of focal length 0.5 cm, its angular magnification is 60/0.5, or 120 times. We say that the magnification is 120 X.

58

Further, the A.M. is also equal to the ratio of entrance pupil (e.g., the objective lens diameter) to the human eye’s pupil diameter. If we assume the diameter of a normal human eye pupil to be about 8 mm for the eye adapted to perfect darkness, then the lowest possible magnification will be equal to the ratio of objective diameter (in mm) to 8. Similarly, if we assume a normal pupil diameter of 3 mm, the highest possible magnification will be the diameter of the objective divided by 3.

Astronomical Techniques

The highest possible magnification of a telescope is limited by its optics which includes the quality of lenses and mirrors and the thermal insulation of the telescope tube so that the exchange of heat does not disturb the air inside the tube. The magnification is also limited by the disturbance of light rays suffered in the Earth’s atmosphere. Also, the higher the magnification, the smaller is the field of view, i.e., the area of the sky which can be observed by the telescope becomes smaller (see Fig. 3.2).

(a)

(c)

(b)

Fig.3.2: Increasing the magnification decreases the field of view and makes the image dimmer. Pictures show simulated views of the Orion Nebula at magnifications of approximately 50x, 80x, and 120x

Notice that the images in Figs. 3.2 b and c are dimmer compared with the one in Fig. 3.2a. Is there a way to make it brighter? We can do so by gathering more light from the object. This brings us to the concept of light gathering power of a telescope.

3.2.2

Light Gathering Power

The light gathering power of a telescope refers to its ability to collect light from an object. Most interesting celestial objects are faint sources of light and in order to get an image we need to capture as much light as possible from them. This is somewhat similar to catching rainwater in a bucket, the bigger the bucket, the more rainwater it catches. Similarly, the light gathering power of a telescope is proportional to the area (i.e., diameter squared) of the telescope objective. Now, the area of a circular lens or mirror 2 of diameter D is π (D/2) . Thus, the ratio of light gathering powers of two telescopes is given by LGP1 LGP2

=

D1 D2

2

(3.3)

For example, a telescope of 100mm diameter can gather (100/8)2 = 156.25 times more light than a human eye with a typical pupil diameter of 8 mm. Similarly, if we compare telescopes of 24 cm and 4 cm diameters, the former gathers (24/4)2 = 36 times more light than the latter. Thus, even a small increase in telescopes diameter produces a large increase in its light gathering power and allows astronomers to study much fainter objects. The ability of a telescope to reveal fine detail of an object is determined by its resolving power, which we now discuss.

59

Basics of Astronomy

3.2.3

Resolving Power and Diffraction Limit

You have learnt in Sec. 11.2 of Unit 11 of Optics that the image of a point object through an optical instrument is not a sharp point-like image but a bright circular disc called the Airy disc. The disc is surrounded by a number of alternate bright and dark fringes produced due to diffraction. The central bright disc represents the image of the object. Now suppose we wish to observe two close stars that appear equally bright. We should be able to see two Airy discs. However, whether we see them as distinct discs or overlapping each other, depends on the resolving power of the telescope. Fig. 3.3 shows three situations where two equally bright stars are closely placed. The pattern of rings seen at each star image is called the Airy pattern. The two stars are said to be just resolved when we can just infer their images as two distinct Airy discs (Fig. 3.3b). To resolve these Airy discs, we use the Rayleigh criterion.

Rayleigh criterion Two equally bright stars are said to be resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.

(a)

(b)

(c)

Fig.3.3: Resolving a pair of two equally bright stars. In a) the stars are easily resolved; in b) the stars are JUST resolved, and in c) the stars are too close and not resolved.

You may like to ask: What factors determine a telescope’s resolving power? Naturally, the quality of lenses is a major factor. But even with perfect optics, the resolving power of a telescope is limited by diffraction. The diffraction limit of resolution (R) of a telescope is defined as R (in radians) = (1.22 λ /D),

(3.4)

where λ is the wavelength of light and D is the telescope diameter. Both λ and D have to be expressed in the same units. [Note that 1 degree = 60 arc-minutes (60′) = 3600 arc-seconds (3600″)]. R can be expressed in arc-seconds as R (arc-sec) =

1.22λ (206265) D

For human eye 60

R ~ 1′ for absolute sharp vision,

(3.5)

Astronomical Techniques

R ~ 2′ for clear vision and R ~ 4′ for comfortable vision. Remember that the smaller the value of R, the better is the instrument’s ability to resolve nearby objects. Example 1: Diffraction limit of resolution of telescopes of different sizes for a given wavelength Table 3.1 shows the diffraction limit of resolution of telescopes of different sizes for λ = 457 nm. Table 3.1: Diffraction limit of resolution of various telescopes Telescope Diameter (mm)

R (″″)

50

2.3

100

1.15

200

0.58

400

0.29

500

0.23

You may like to calculate R for a telescope. SAQ 1 Calculate the diffraction limit of resolution of Mount Palomar telescope of 200 inch diameter for λ = 457 nm. Compare its light gathering power with a telescope of 200 mm diameter.

Spend 10 min.

Based on size alone, the largest telescopes should have large resolving powers. But the resolution of large telescopes is limited by the passage of light through the Earth’s atmosphere. When we look through a telescope, we are looking through several kilometres of turbulent air, which blurs the image. The Earth’s atmosphere does not allow ground-based telescopes to resolve better than 1-2 arc-seconds in the sky (for even the best astronomical sites). The major limitation for ground-based astronomy is the Earth’s atmosphere and it can affect the observations in many ways. Moreover, all wavelengths cannot pass through the atmosphere. This brings us to the concept of atmospheric windows.

3.2.4

Atmospheric Windows

Some of the effects of the Earth’s atmosphere on electromagnetic radiation are: Absorption, scintillation, scattering and turbulence. Atmospheric molecules such as carbon dioxide and water vapour give rise to absorption. Thus, only certain bands of frequencies in the electromagnetic spectrum pass through the atmosphere. These regions of the electromagnetic spectrum are called atmospheric windows. Fig. 3.4 shows the absorption properties of the Earth’s atmosphere. On the x-axis is the wavelength in cm and on the y-axis is the altitude in km at which the intensity of the radiation entering the atmosphere is reduced to half. 61

Basics of Astronomy

Infrared and microwaves are blocked by the atmosphere.

Gamma rays, X-rays and UV rays are blocked by the atmosphere. Gamma ray −12

10

UV

X ray 10

10

7

4 × 10− m

Transparent

MicroInfrared wave UHF VHF FM

−8

−10

Opaque

Radio window

Transparency of Earth’s atmosphere

Altitude

Visual window

−4

Visual

10

−2

10 7

7 × 10− m

AM 2

4 1 10 10 Wavelength (metres)

Fig.3.4: The spectrum of electromagnetic waves from gamma rays to long radio waves showing the optical and radio ‘windows’ and the regions of atmospheric transparency

The atmosphere allows only visible radiation and radio waves to come through to the surface of the Earth. For observations at other wavelengths we have to fly detecting instruments to altitudes at which these wavelengths are not completely absorbed. Before the advent of artificial satellites, the instruments were flown in balloons and rockets. Now-a-days observations are carried at various wavelengths by instruments on board the space satellites. It is important to remember that astronomers like to observe objects in as many wavelengths as possible. This helps them to understand astronomical objects better. Spend 5 min.

SAQ 2 For the same diameter, compare the resolving power of an optical telescope operating at λ 457 nm and a radio telescope operating at λ 1 cm. You have learnt in SAQ 2 that the resolving power of a radio telescope is quite poor. Therefore, radio telescopes need to have very large apertures. So far we have discussed some optical definitions relevant to astronomy. We now turn our attention to the instruments and techniques used for making astronomical observations from the Earth. Let us now learn about optical telescopes.

3.3

OPTICAL TELESCOPES

Telescopes have undergone tremendous evolution from Galileo’s times when only refracting telescopes were used. The most recent telescopes have mirrors that can be actively shaped according to the observer’s need. The refracting telescopes became outdated very soon because i) it was difficult to make good large lenses required for large gathering area, ii) large lenses were very heavy and balancing the telescope became difficult, and iii) lenses suffered from optical aberrations. Reflecting telescopes with mirrors (parabolic or hyperbolic) have been around for more than a century. In the last few decades, advanced technologies have been developed which allow the observer to effectively control the optical system so that it can be adapted to the needs of the observations. Such active and adaptive optics have completely changed this basic tool of observational astronomy. We now describe various types of reflecting telescopes. 62

3.3.1

Astronomical Techniques

Types of Reflecting Telescopes

James Gregory of Scotland (1638-75) proposed the first design for a reflecting telescope in which he used a paraboloidal mirror as the objective (primary mirror) to minimize chromatic and spherical aberrations (Fig. 3.5). In this type of telescope, light from a distant object hits the primary mirror and is reflected to a secondary mirror which reflects it down to the telescope tube again to a secondary focus. The light emerges from the telescope through a small central hole in the primary mirror and is observed with an eyepiece or a detector.

Parallel light rays from the star

Focal point

Secondary mirror

Paraboloidal primary mirror

Fig.3.5: Gregorian reflecting telescope with equatorial mounting; ray diagram for the telescope

The performance of the Gregorian telescope was not satisfactory and Newton (16421727) built a working reflector. A Newtonian telescope (Fig. 3.6) uses a parabolic primary but a flat mirror as a secondary, which is set at 45° to the axis of the tube. The light is brought to a focus at the side of the telescope where the eyepiece is placed. This system is widely used even now.

Secondary mirror

Eye-piece

Primary mirror

Fig.3.6: Newtonian reflecting telescope

A little later, a French optician, G. Cassegrain (who lived at the time of Newton), designed another reflector (Fig. 3.7). Now-a-days the most commonly used telescope is the Cassegrain type with a central hole in the primary mirror which allows the light to come out for placing an eyepiece or any other general detector or instruments. This design also allows the folding of the focussed beam and thus provides a very compact telescope. 63

Basics of Astronomy

Fig.3.7: Cassegrain reflecting telescope

All telescopes need to be pointed at the desired part of the sky, and then they have to follow or track the objects in the sky as their direction changes due to Earth’s rotation. For example, suppose you are viewing Jupiter and its moons with a telescope. Due to the Earth’s rotation, you will see Jupiter moving across your view until it is gone. You will have to move your telescope to follow Jupiter if you want to keep viewing it. However, to track it, you will need a mounting for the telescope so that it can be turned in the desired direction. We will now briefly describe telescope mountings.

3.3.2

Telescope Mountings

Most of the small telescopes (less than 1 metre diameter) use the equatorial mount (Fig. 3.8). In an equatorial mounting, the pier or the base on which the telescope is mounted is set so that its axis points to the North Pole. This is done by raising the axis by an angle equal to the latitude of the place. This axis is called the polar axis. A rotation about the polar axis is used for adjustment in right ascension. The telescope is also provided with motion about an axis perpendicular to the polar axis, called the declination axis, for adjustment in declination. Thus, a combination of these two motions allows the telescope to point to any object whose equatorial coordinates are known.

Fig.3.8: The equatorial mounting

Since the telescope is fixed on the Earth, it moves with the Earth. In order that the object remains in the field of the telescope, the mounting is made to rotate in the direction opposite to that of the Earth with the same speed as that of the Earth. All 64

these adjustments make this type of mounting heavier than the altitude-azimuth mounting described ahead. That is why it is not used with very big telescopes.

Astronomical Techniques

All modern telescopes (larger that 2 m diameter) use Alt-Azimuth type mount shown in Fig. 3.9. You have learnt in the last unit that altitudes and azimuths of objects change with the location of the observer and with time for the same observer. This was a major limitation of this type of mounting in earlier times but with advances in the technology using computers in the past few decades, it is no more an issue.

AZIMUTH AXIS

ALTITUDE AXIS

Fig.3.9: The Alt-Azimuth type mount

In Fig. 3.10a we show the largest telescope in India set up by the Indian Institute of Astrophysics at Kavalur Observatory in Karnataka.

(a)

(b)

Fig.3.10: a) The Vainu Bappu telescope in Kavalur Observatory; b) domes housing twin Keck telescopes

At present, the largest operating telescopes are the twin Keck 10 m telescopes at Mauna Kea, in Hawaii, placed at about 4200 m height above sea-level (Fig. 3.10b). There are plans now for making very large aperture (up to 20 − 30 m) telescopes!

65

Basics of Astronomy

3.3.3

Space Telescopes

You have learnt in Sec. 3.2 that the main advantages of a telescope over direct observation with the human eye, are magnification, light collection and resolution. For light collection, one could fabricate telescopes of increasingly large diameters. But these will be limited by the effect of the Earth’s atmosphere. However, if we could put a telescope in space (high above the Earth’s atmosphere, in a balloon or a satellite), then the Earth’s atmosphere would not be a limiting factor, and we could achieve diffraction-limited images. In such situations the size of the telescopes that can be built for such platforms is the only limitation due to the costs involved and limitations of available technology. The Hubble Telescope of about 2 m diameter is the best example of this type of telescope (Fig. 3.11). Over the past decade it has made very high quality observations of stars, nebulae, galaxies, supernovae and other objects. Some of these objects belong to a very early phase of the universe. These observations have led to improved understanding of these objects.

Fig.3.11: The Hubble Space Telescope

Both the types of telescope − ground-based and space-based − have somewhat different roles to play and complement each other. That is why there is always a demand to build large diameter ground-based telescopes. The active and adaptive telescopes are the latest instruments in use today. In the active telescopes, the shape of the primary mirror can be changed (although it retains its parabolic form) according to the observer’s need. This is done with the help of small peizo driven pistons placed at the back of the mirror support. This is now routinely done for large mirror based telescopes. In the adaptive telescope the changes in the shape of the primary mirror are done in a controlled manner depending upon the changes in the Earth’s atmosphere during night. In some sense, the adaptive telescopes can beat the effect of the atmosphere and render images as good as the space telescopes for some duration in the night! Telescopes must use detectors to help us obtain useful information about the universe. We now discuss various detectors and how these are used with telescopes.

3.4

66

DETECTORS AND THEIR USE WITH TELESCOPES

Detectors are used for measuring the light output from a telescope and play a major role in obtaining information about the stars, galaxies, etc. The actual light available from an astronomical object is very small as will be clear from the following example.

Astronomical Techniques

Example 2: Estimate of Available Light from a Star Vega is one of the bright stars in the sky and is about 26 light years away. The radiation emitted in visible band at the star’s surface is 175,000 Wcm−2. This reduces by 10−16 times on its arrival outside Earth’s atmosphere. Another 20% is lost in the atmosphere due to absorption and scattering. About 30% is lost in the telescope optics. Thus, if we use a 25 cm (10″) telescope to collect this light, then only about 0.5 × 10−9 W is available at its Cassegrain focus. Eventually only a fraction of this is actually detected since the detector is never an ideal one and has an efficiency of at best 80%. It is amazing that even with such a small amount of light detected, we can study many things about a star, e.g., its light variation, brightness, composition. We can even take a spectrum (which actually divides this small available light into smaller bits of wavelengths)!

Detectors are used with telescopes in the following two modes of operation: •

Imaging: This involves taking direct pictures of star fields and extended objects like gas clouds or galaxies. Since sharp images are required over a wide field which may extend up to several square degrees, careful optical design is a natural requirement.

•

Photometry: This involves measuring total brightness, spectrum etc. of single objects. Compared to imaging mode, poorer images are acceptable in this case but the stellar image has still to be small enough to enter an aperture or slit of a spectrograph.

We now briefly describe various types of detectors.

3.4.1

Types of Detectors

Detectors used in the imaging mode are mainly 2-dimensional (2D type) since we are trying to form images of objects in a given area. Examples of such detectors are the photographic emulsion, human eye and the most modern detector, the chargecoupled device (CCD). Detectors used for photometry of single objects are 1D type (one dimensional), since they receive photons from one object only. The photometer is a 1D detector. Photometer Before the advent of CCDs, the measurements of light intensity and colour were made using a photometer, a highly sensitive light meter attached to a telescope (Fig. 3.12a). A photometer is still used in the photometry of single stars. It is used more commonly for stars whose light output varies with time, called variable stars. The most important component of a photometer is a photomultiplier tube that is based on the photoelectric effect about which you have studied at +2 level. A photon when incident on a photocathode emits an electron. The electric current thus generated is amplified further and can be measured directly. The calibration of intensity or colour is done by observing a comparison star. Today, however, most photometric measurements are made on CCD images. Charge-coupled device A charge-coupled device (CCD) is a special computer chip of the size of a postage stamp (Fig. 3.12b). It contains a large number (~ millions) of microscopic light detectors arranged in an array. A CCD can be used like a small photographic plate, though it is much more sensitive. CCDs detect both bright and faint objects in a single exposure. The image from a CCD is stored in a digitised form in a computer. Therefore, brightness and colour can be measured to high precision. Moreover, it is

67

Basics of Astronomy

easy to manipulate the image to bring out details. At present, the only major drawback of CCD is that its maximum size is limited (about 70 mm square) as compared to the most basic 2D detector, i.e., photographic plates which can be as large as 300 mm square. This disadvantage of CCD is also being overcome by combining a large number of CCDs.

(a)

(b)

Fig.3.12: a) Photometer attached to a telescope; b) a CCD

Efficiency of a Detector A basic parameter which defines the efficiency of any detector is its Quantum Efficiency (Q.E.). It is the ratio of number of photons actually detected (or recorded) by it to the number of photons recorded by an ideal and perfect detector. Since ideal detector by definition would detect all photons incident on it with 100% efficiency, this ratio is nothing but the ratio of actually detected photons by the detector versus the number of photons incident on it. Fig. 3.13 gives the efficiency curves for various detectors.

QUANTUM EFFICIENCY (PERCENT)

100

CCD

10

PHOTOMULTIPLIER TUBE

PHOTOGRAPH

1

EYE

0.1 300

500

700

900

1100

λ (nm) Fig.3.13: Quantum efficiency of various detectors in terms of wavelength of light

68

Note that the human eye and photographic emulsion are detectors with the lowest sensitivity and photomultiplier tubes are only marginally better. The CCD works over a large wavelength region in the visible band with a Q.E. of the order of 60-80%. You must remember that the y-axis in Fig. 3.13 is in log scale. As you have read in Unit 1, the limiting magnitude of the naked eye is about +6. This means that the faintest object visible under normal observing conditions, with an eye adapted to darkness has a magnitude of +6. It takes 25-30 minutes for the eye to get completely dark-adapted, otherwise we can see only up to fourth or fifth magnitude. Experiments have, however, shown that a dark-adapted eye, looking at a patch of dark sky through a modest telescope can see stars as faint as mv = +8.5. This corresponds to a flux of about 200 photons per second. We would also like to find out how far the limits of detection are extended when we use a telescope with various detectors.

3.4.2

Detection Limits with Telescopes

Modern detectors like CCDs can detect individual photons, and the limiting magnitude is normally governed by the background noise. For a typical photographic plate, only about 0.1% of the incident photons are recorded depending on the type of grains on the film. On such a plate, an image is detectable only after about 5 × 104 photons have been received.

Astronomical Techniques

In spectroscopy or spectrophotometry, we analyze the light in great detail by using a spectrograph which spreads the light into a spectrum according to wavelength. You are familiar with a prism dispersing white light into its component colours. Nearly all modern spectrographs use a grating instead of a prism and the spectrum is recorded directly on a CCD camera. Since we understand how light is emitted, scattered, or absorbed by matter, a spectrum carries a lot of information about the source of light and the medium through which it has passed. We shall discuss this in detail in Unit 7.

At 200 photons per second, the photographic plate can match eye’s limit with an exposure of about 4 minutes. Obviously, the plate can detect much fainter objects with longer duration exposures. The limit to the flux of visible radiation which is detectable is given by F

vis

lim

∝

1

D 2t

(3.6)

where t is the exposure time and D is the telescope aperture. For a telescope, the magnitude limit in the visible range, in a very approximate manner without considering any efficiency factors etc., is given by mvlim ~ 2 + 5 log10 D

(3.7)

where D is in mm. Thus using the human eye as the detector along with a telescope we can have the limiting magnitudes as mv ~ 12.9 for a 6 inch and mv ~ 15.3 for an 18 inch telescope. Spend 5 min.

SAQ 3 Find the magnitude of the faintest object that the 3.5 metre New Technology Telescope at the European Southern Observatory in Chile can detect. With this we come to the end of the unit. We now summarise its contents.

3.5 •

SUMMARY If a telescope has an objective of local length f0 and the eye-piece has a focal length fe, then angular magnification is equal to the ratio f0/fe. 69

Basics of Astronomy

•

The light gathering power of a telescope refers to its ability to collect light from an object.

•

The resolving power of a telescope is its ability to reveal fine detail. The resolving power of a telescope is given by α=

11.6 D

where D is the diameter of the telescope’s objective. A point light source observed through a telescope would not appear as a point source. Diffraction would cause the image to appear as a round disk of light, called Airy’s disk. The diffraction limit of resolution in radians is given by R = 1.22

λ D

where λ is the wavelength of light. Both λ and D must have the same units. •

At the Earth’s surface, electromagnetic radiation can be detected only in the radio, infrared and optical windows.

•

Telescopes can be of refracting or reflecting types.

•

All modern telescopes are reflecting type. There are three kinds of reflectors: Gregorian, Newtonian and Cassegrain. The Cassegrain reflectors are the most popular.

•

Telescopes can be operated in imaging or photometry mode.

•

The larger is the diameter of the objective, the fainter is the source that a telescope can detect.

•

The human eye, photographic emulsion, photometer and charge-coupled devices are various types of detectors.

•

Of these the CCDs are used most by modern astronomers. These are used for recording images, measuring brightness and colour of celestial objects.

3.6

TERMINAL QUESTIONS

Spend 30 min.

1. Why can infrared observations be made from high mountains while X-ray observations can be made only from space? 2. Nocturnal animals have large pupils in their eyes. Can you relate that to astronomical telescopes? 3. The moon has no atmosphere. If we had an observatory on the moon, at what wavelengths could we observe the astronomical objects? 4. What is the resolving power of a 20 cm telescope if observations are made at λ 550nm? 5. What do two stars 1.5 arc-seconds apart, look like through a 25 cm telescope ? 6. Compare the light gathering powers of the 5 m telescope and a 0.5 m telescope. 70

3.7

Astronomical Techniques

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1.

R=

=

1.22λ D

× 206265 arc - sec

1.22.457 × 2.06265 2 × 2.54

× 10 − 2 arc - sec

= .023 arc - sec

2 LGPMP 200 × 2.54 × 10 = = (2.54)2 × 10 2 = 645.2 LGP200mm 200

2. Ratio of resolving powers =

1.22λ1 D

× 206265 ×

D 1.22λ 2

×

1 206265

λ 4.57 ×10 − 9 × 10 2 = 1 = λ2 1 = 4.57 × 10 − 7

The resolving power of the radio telescope is 4.57 × 10 − 7 times the resolving power of an optical telescope operating at 4.57 nm. 3. Limiting magnitude mν = 2 + 5 log10 D = 2 + 5 log10 (3.5 × 100 × 10 ) = 2 + 15 + 5 log10 (3.5)

= 19.7 Terminal Questions 1. At the top of a high mountain, we can receive only IR. X-rays are available only much above the top of the highest mountains. Therefore, to observe in X-ray band, we will need to observe from a space-based telescope. 2. The larger the diameter of the pupil, larger is the light gathering power. Nocturnal animals are adapted to night vision by virtue of this. Astronomical telescopes also require bigger and bigger objectives to see fainter and fainter objects. 3. At all wavelengths, provided we have detectors sensitive in those regions. 4. 0.58 arc-seconds. 5. Resolved. 6. The 5 m telescope has 100 times the light gathering power of a 0.5 m telescope.

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Basics of Astronomy

UNIT 4 PHYSICAL PRINCIPLES Structure 4.1

Introduction

4.2

Gravitation in Astrophysics

Objectives Virial Theorem Newton versus Einstein

4.3 4.4

Systems in Thermodynamic Equilibrium Theory of Radiative Transfer Radiation Field Radiative Transfer Equation Optical Depth; Solution of Radiative Transfer Equation Local Thermodynamic Equilibrium

4.5 4.6 4.7

Summary Terminal Questions Solutions and Answers

4.1

INTRODUCTION

So far in this block we have provided the basic information which is useful in astronomy. You have learnt about astronomical quantities of interest, various coordinate systems, astronomical instruments and techniques. We now turn our attention to astrophysics. The aim of astrophysics is to apply principles of physics to understand and explain the behaviour of various astronomical systems. There are certain physical principles and concepts which are used in astrophysics so universally that it is worthwhile to discuss them before we begin studying specific astronomical systems. We have decided to put together some such basic physical principles in this Unit. You must have already learnt many of these principles in your other physics courses. Now you will learn how these can be applied to astrophysical systems. Let us consider some issues which are of universal concern in astrophysics. We know that gravitation is the dominant force in virtually any astrophysical setting. Since gravitation is always attractive, it must be balanced in some way in a system which is not shrinking. We shall discuss a very general and powerful principle called the virial theorem, which helps us understand how gravitation is balanced in astrophysical systems. Another topic of importance in astrophysics is the interaction of radiation with matter. Astrophysics is a very special science in which we cannot do experiments with our systems (stars, galaxies, etc.) in our laboratories. Virtually everything we know about these systems is learnt by analysing the radiation reaching us from these systems. If we want to make inferences about the systems which emitted the radiation or through which the radiation passed, then we need to understand how matter and radiation interact with each other. Many astronomical systems like stars emit radiation simply because they are hot. So we often need to apply various principles of thermal physics to understand how matter in these systems behaves. Therefore, we plan to recapitulate some of the important results of systems in thermodynamic equilibrium and then develop the theory of transfer of radiation through matter. Objectives After studying this unit, you should be able to: 72

•

apply virial theorem to simple astrophysical systems;

•

identify the situations in astrophysics to which Newton’s theory of gravitation or general theory of relativity can be applied;

•

determine the specific intensity, energy density, radiant flux and radiation, pressure for a given radiation field; and

•

solve the radiative transfer equation for simple cases and interpret the results.

Physical Principles

Study Guide In this unit, we will be using certain concepts discussed in various units of the physics electives PHE-01 entitled ‘Elementary Mechanics’ (Unit 10), PHE-06 entitled ‘Thermodynamics and Statistical Mechanics (Unit 9), and PHE-11 entitled ‘Modern Physics’ (Unit 9). Please keep these units handy for ready reference.

4.2

GRAVITATION IN ASTROPHYSICS

In your elementary physics courses, you must have learnt about two important longrange forces, whose range of influence extends to infinity. These are the gravitational force and the electromagnetic force. The electromagnetic force can be attractive or repulsive depending on the nature of charges. So, if a system has equal amounts of positive and negative charges, and if there are no relative motions between these two types of charges, then electromagnetic forces are screened off. That is, the system does not produce a large-scale electromagnetic field. On the other hand, gravitation is always attractive and cannot be screened off. So it is the dominant force acting over the entire universe. Since gravitation is always attractive, a natural question to ask is: Why do the celestial objects not shrink in size? Obviously, it has to be balanced in a system which is not shrinking in size (such as a star or a galaxy). We shall now use Newtonian theory of gravitation to discuss an important theorem (virial theorem) which tells us how this balancing takes place. Then we shall briefly consider whether Newtonian theory of gravitation is adequate in astrophysics or whether we have to apply a more complete theory of gravitation − the general theory of relativity due to Einstein.

4.2.1

Virial Theorem

Why is our solar system not shrinking in size? We shall analyse this problem in the non-inertial frame of reference attached to a planet. From Unit 10 of the course PHE-01 entitled ‘Elementary Mechanics’, you know the answer: the Sun’s gravitational attraction on a planet is balanced by the centrifugal force due to the orbital motion of the planet. Let M and m be the masses of the Sun and a planet in the solar system. For simplicity, if we assume the planet to go in a circular orbit of radius r with speed v, then the force balance equation in the given non-inertial frame of reference is mv 2 GMm = r r2

We can rewrite this equation in the form 2

1 GMm =0 mv 2 − 2 r

(4.1)

Now note that the gravitational potential energy of the system (the Sun and the planet) is GMm EG = − , r 73

Basics of Astronomy

whereas the kinetic energy of the system is EK =

1 2 mv 2

if the Sun is assumed to be at rest. We can now rewrite Eq. (4.1) in the following form 2 E K + EG = 0

(4.2)

This is the virial theorem for a planet going around the Sun. What is the significance of this result? This tells us that the gravitational potential energy and the kinetic energy of a system will have to be of the same order if gravitation is to be balanced by motion. We have proved virial theorem for the simple case of a planet going around the Sun in a circular orbit. However, Eq. (4.2) can be proved quite generally for a system in which gravitation is balanced by motions such that the system is not shrinking in size. The motions needs not be circular, but can be of any type. For example, inside a star, gravity is balanced by thermal motions of its particles (atoms, electrons, ions). Even in this situation, Eq. (4.2) can be shown to hold provided we take the total kinetic energy of all the particles in the star for EK. If a galaxy or a star cluster is not shrinking in size, the total kinetic energy EK of the stars in it should be related to the total gravitational potential energy EG by Eq. (4.2).

Fig.4.1: Spiral galaxies

In a type of galaxy known as spiral galaxy, stars seem to be moving in nearly circular orbits. However, in typical star clusters and in galaxies known as elliptical galaxies, stars move randomly. Due to these random motions, the stars do not fall to the centre due to gravitation. Do you feel puzzled by the idea that random motions can balance gravitation? To understand this concept, consider the air around you. Earth’s gravity is pulling all the molecules of the air. Then why are not all molecules settling on the floor of the room due to this attraction? It is the random motion of the molecules which prevents this from happening. Deriving the virial theorem Eq. (4.2) for a completely general situation is a very mathematically involved problem. It is beyond the scope of this elementary course. So let us apply the virial theorem in a simple situation so that you feel comfortable with it. Example 1: Estimating the average temperature in stellar interior Fig.4.2: The Sun

A star (we have the Sun in mind) has a mass of 1033g and a radius of 1011cm. Make an order-of-magnitude estimate of the average temperature in the interior of the star. Solution

74

Since this is an order of magnitude estimate, we shall take the virial theorem to imply that the potential and the kinetic energies are approximately equal. In c.g.s. units, we

Physical Principles

take the following approximate values of various physical constants: gravitational constant G ≈ 10−7, Boltzmann constant kB ≈ 10−16, mass of hydrogen atom mH ≈ 10−24. The gravitational potential at the surface of the star has the magnitude GM/R. The approximate gravitational potential energy of the star is given by multiplying this by M, which is

( )2 ≈ 1048 erg

GM 2 10− 7 × 1033 = R 1011

Now we need to equal the total kinetic energy to this. Now, suppose that the star is made up of hydrogen. Then, the star consists of about M/mH particles. Each of them has kinetic energy of order kBT. Therefore, the total kinetic energy of the star is M 1033 × 10 −16 T k BT ≈ ≈ 1041T − 24 mH 10

If this is equated to 1048 erg, then we obtain the temperature of the star as T ≈ 107 K Note that detailed calculations suggest a temperature of about 15 million degrees at the centre of the Sun. The above order of magnitude estimate thus provides a fairly good estimate of the average temperature inside a star.

You may now like to solve a problem to fix these ideas. SAQ 1 A globular cluster of stars has about a million stars. The stars inside such a cluster of radius 1020 cm have random velocities of order 106 cm s−1. Estimate the mass of the star cluster and the number of stars in it. Take the mass of a star to be about 1033g.

Spend 10 min.

Hint: The mass of the cluster is Nm, where N is the number of stars in the cluster. Now, equate the total K.E. of the cluster to its P.E. After Newton formulated his theory of universal gravitation, for more than two centuries it was regarded as a supreme example of a successful physical theory. However, in 1915, Einstein showed that this theory was incomplete and formulated his new theory of gravitation known as general relativity. In this section, we investigate the situations in which Newton’s theory may have to be replaced by the general theory of relativity.

4.2.2

Newton versus Einstein

We now know that Newton’s theory is only an approximation. But it is such an exceptionally good approximation in most circumstances that we do not need general relativity at all. Only when the gravitational field is sufficiently strong, we have to apply general relativity. Although we shall not discuss general relativity in this elementary course, we would like to point out when you can safely use Newtonian theory and when general relativity is needed. Even people without any technical knowledge of general relativity now-a-days have heard of black holes. These are objects with gravitational fields so strong that even light cannot escape. Let us try to find out when this happens. Newtonian theory does not tell us how to calculate the effect of gravitation on light. So let us figure out when a particle moving with speed of light c will get trapped, according to Newtonian theory. Suppose we have a spherical mass M of radius r and a particle of mass m is

75

Basics of Astronomy

ejected from its surface with speed c. The gravitational potential energy of the particle is −

GMm r

If we use the non-relativistic expression for kinetic energy for a crude estimate (we should actually use special relativity for a particle moving with c!), then the total energy of the particle is E=

1 2 GMm mc − 2 r

Newtonian theory tells us that the particle will escape from the gravitational field if E is positive and will get trapped if E is negative. In other words, the condition of trapping is 1 2 GMm mc − 1

(4.3)

It turns out that more accurate calculations using general relativity give exactly the same condition (4.3) for light trapping, which we have obtained here by crude assumptions. General relativity is needed when the factor f =

2GM c2r

(4.4)

is of the order unity. Newtonian theory is quite adequate if f is much smaller than 1.

Let us investigate the case of the Sun. Example 2 Let us examine if the Newtonian theory is adequate for the Sun. Solution The Sun has mass 1.99 × 1033 g and radius 6.96 × 1010 cm. Substituting these values in Eq. (4.4), we get f = 4.24 × 10−6 T2 > T1

0.5 T2

T1

0 0

2,000

4,000

6,000

8,000 10,000

12,000 14,000

16,000 18,000 20,000

wavelength (angstrom) Fig.4.3: Blackbody radiation curve

We can now use these results to understand the interaction of matter with radiation.

4.4

THEORY OF RADIATIVE TRANSFER

Matter can both emit and absorb radiation. It is possible to use quantum mechanics to calculate the rates at which atoms emit or absorb energy. Here, however, we shall not do it. We would study the processes of emission and absorption by matter, by introducing suitable coefficients of emission and absorption. Radiative transfer is the name of the subject in which we study the interaction of radiation with matter having prescribed emission and absorption coefficients. Let us first consider how we can provide the mathematical description of radiation at a given point in space.

4.4.1

Radiation Field

You know that it is particularly easy to give a mathematical description of blackbody radiation, which is homogeneous and isotropic inside a container. Specifying the energy density uν associated with the frequency ν, which is given by Planck’s law (Eq. 4.9), more or less provides us complete information about blackbody radiation. In general, however, the radiation is not isotropic. When we have sunlight streaming into a room, we obviously have a non-isotropic situation involving the flow of radiation from a preferred direction. We now define the radiation field for a non-isotropic situation. Let us consider a small area dA at a point in space (Fig. 4.4). Let dEν dν be the energy of radiation passing through this area in time dt from the solid angle dΩ centred at θ and lying in the frequency range ν, ν + dν. The energy dEνdν is proportional to the area dA cosθ projected perpendicular to the direction of radiation, time interval dt, solid angle dΩ and frequency range dν. Hence we can write dE d = I ( r , t , nˆ )cos dA dt d dν

(4.10)

79

Basics of Astronomy

where nˆ is the unit vector indicating the direction from which the radiation is coming ˆ is the unit vector normal to the area dA. The quantity I (r , t , nˆ ) is called the and N specific intensity. As you can see, it is a function of position r, time t and direction nˆ .

nˆ

dΩ θ dA

Fig.4.4: Illustration of specific intensity

Radiation field If I (r , t , nˆ ) is specified for all directions at every point of a region at a time, then we say that the radiation field in that region is completely specified.

In this elementary treatment, we shall restrict ourselves only to radiation fields which are independent of time. If we know the radiation field at a point in space we can calculate various quantities like radiant flux, energy density and pressure of radiation. For example, radiant flux is simply the total energy of radiation coming from all directions at a point per unit area per unit time. Hence, we simply have to divide Eq. (4.10) by dA dt and integrate over all solid angles to get the flux. Thus, we can define the radiant flux associated with frequency ν, and the total radiant flux as follows: Radiant flux in terms of specific intensity The radiant flux associated with frequency ν is given by F = I cos dΩ

(4.11)

The total radiant flux is obtained by integrating over all frequencies F = F dν

(4.12)

The pressure of the radiation field over a surface is given by the momentum exchanged per unit area per unit time, or momentum flux, perpendicular to that surface. Let us obtain an expression for momentum flux. You know from Unit 3 of PHE-11 that the momentum associated with a photon of energy dEν is dEν /c. Its component normal to the surface dA is dEν cosθ/c. On dividing this by dA dt, we get the momentum flux associated with dEν ; 80

Momentum flux =

Physical Principles

dE cos θ c dA dt

Using Eq. (4.10), we get the expression for momentum flux in terms of specific intensity: dE cos θ 1 I = cos 2 d c dAdt c

(4.13)

The radiation pressure pν is obtained by integrating the momentum flux over all directions. Radiation pressure p =

1 I cos 2 d . c

(4.14)

If the radiation field is isotropic, i.e., it is independent of θ and φ, then dΩ = sinθ d dϕ and d

p =

I c

cos2 d

=

=

4 . Hence, we get,

4 I 3 c

(4.15)

We now apply these results to calculate energy density and radiation pressure. Spend 5 min.

SAQ 5 Perform the integration in Eq. (4.15) and verify the result. Example 3: Calculating energy density, specific intensity and radiation pressure Calculate the energy density uν of a radiation field at a point and use that expression of energy density to write down the specific intensity of a blackbody radiation. Show that the pressure due to isotropic radiation is given by 1/3 of the energy density. Solution Let us consider energy dEν of radiation associated with frequency ν as given by Eq. (4.10). This energy passes through area dA in time dt in the direction nˆ . Since the radiation traverses a distance cdt in time dt, we expect this radiation dEν to fill up a cylinder with base dA and length cdt in the direction nˆ during this time. Now the volume of this cylinder is cosθ dAcdt. Therefore, from Eq. (4.10), the energy density of this radiation in the solid angle dΩ is dE I = d cosθ dAcdt c

To get the total energy density of radiation at a point associated with frequency ν, we have to integrate over all directions, so that u =

I dΩ c

For isotropic radiation u =

4 Iν c 81

Basics of Astronomy

Since blackbody radiation is isotropic, the specific intensity of blackbody radiation usually denoted by Bν (T) should be independent of direction. Hence the energy density of blackbody radiation is simply given by u =

B (T ) dΩ c

Since Bv(T) is independent of direction, integration over all solid angles gives 4π. ∴

u =

4πB (T ) c

Therefore, making use of Eq. (4.9), we now get the specific intensity of blackbody radiation as B (T ) =

2h 3 c

2

1 h −1 exp k BT

Using Eq. (4.15), we get the radiation pressure for isotropic radiation as 1 p = u 3

For black body radiation p =

4 B (T ) . 3 c

In astrophysics, we need to understand the interaction of matter and radiation to explain spectra of objects such as stars, interstellar gas clouds and galaxies. We now discuss the effect of matter on radiation field.

4.4.2

Radiative Transfer Equation

If matter is present, then in general the specific intensity of the radiation field keeps changing as we move along a ray path. Before we consider the effect of matter, first let us find out what happens to specific intensity in empty space as we move along a ray path. See Fig. 4.5. Let dA1 and dA2 be two area elements separated by a distance R and perpendicular to a ray path. Let Iν1 and Iν2 be the specific intensities of radiation in the direction of the ray path at dA1 and dA2. R

dA1

dA2

Fig.4.5: Two area elements perpendicular to a ray path

82

We want to determine the amount of radiation passing through both dA1 and dA2 in time dt in the frequency range ν, ν + dν. If dΩ2 is the solid angle subtended by dA1 at

dA2, then according to Eq. (4.10), the radiation falling on dA2 in time dt after passing through dA1 is

Physical Principles

I 2dA2 dtd 2 d

From considerations of symmetry, this should also be equal to I 1dA1dtd 1d

where dΩ1 is the solid angle subtended by dA2 at dA1. Equating these two expressions and noting that dΩ1 =

dA1 , d Ω = 2 R2 R2

dA2

we get Iν1 = Iν2

(4.16)

In other words, in empty space the specific intensity along a ray path does not change. If s is the distance measured along the ray path, then we can write dI =0 ds

(4.17)

in empty space. At first sight, this may appear like a surprising result. We know that the intensity falls off as we move further and further away from a source of radiation. Can the specific intensity remain constant? The mystery is cleared when we keep in mind that the specific intensity due to a source is essentially its intensity divided by the solid angle it subtends, a quantity called the surface brightness of an object. This means that the specific intensity is a measure of the surface brightness. As we move further away from a source of radiation, both its intensity and angular size falls as (distance)2. Hence the surface brightness, which is the ratio of these two, does not change. Let us now consider what happens if matter is present along the ray path. If matter emits, we expect that it will add to the specific intensity. This can be taken care of by adding an emission coefficient jν on the right hand side of Eq. (4.17). On the other hand, absorption by matter would lead to a diminution of specific intensity and the diminution rate must be proportional to the specific intensity itself. In other words, the stronger the beam, the more energy there is for absorption. Hence the absorption term on the right hand side of Eq. (4.17) should be negative and proportional to Iν. Thus, we obtain the radiative transfer equation which gives the value of specific intensity in the presence of matter: Radiative transfer equation dI = j − ds

I

(4.18)

where αν is the absorption coefficient The radiative transfer equation provides the basis for our understanding of interaction between radiation and matter. It is fairly trivial to solve this equation if either the emission coefficient or the absorption coefficient is zero. Let us consider the case of jν = 0, i.e., matter is assumed to absorb only but not to emit. Then Eq. (4.18) becomes

83

Basics of Astronomy

dI =− ds

(4.19)

I

On integrating this equation over the ray path from s0 to s, we get s

I ( s ) = I ( s0 ) exp −

(s′) ds′

(4.20)

s0

We will discuss below more general solutions of the radiative transfer equation. These solutions will provide us answers to questions such as: Why is the radiation emitted from nebula usually in spectral lines? Why do we see absorption lines in stellar spectra?

4.4.3

Optical depth; Solution of Radiative Transfer Equation

To obtain a general solution of the radiative transfer equation, we need to define two quantities, namely, the optical depth and the source function. Let us first define the optical depth τν through the following relation: d

=

ds

(4.21)

such that the optical depth along the ray path between s0 and s becomes s

=

(s′) ds′

(4.22)

s0

If matter does not emit radiation, i.e., jν = 0, it follows from Eqs. (4.20) to (4.22) that the specific intensity along the ray path falls as I ( τ ) = I ( 0) e −

(4.23)

Based on the values of τν we can define objects as optically thick or optically thin. Optically thick and optically thin objects If the optical depth τν >> 1 along a ray path through an object, then the object is known as optically thick. An object is known as optically thin if τν 1 , the system is optically thick.

91

UNIT 5 THE SUN

The Sun

Structure 5.1

Introduction Objectives

5.2 5.3 5.4

Solar Parameters Solar Photosphere Solar Atmosphere Chromosphere Corona

5.5 5.6 5.7 5.8 5.9 5.10

Solar Activity Basics of Solar Magnetohydrodynamics Helioseismology Summary Terminal Questions Solutions and Answers

5.1

INTRODUCTION

From Unit 4, you know that the principles of different branches of physics such as mechanics, thermodynamics and quantum mechanics are used in astronomy and astrophysics. In the present and subsequent Units, you will use these principles to investigate the behaviour and properties of the universe and its constituents. On the cosmic scale, the Sun is just another star; there are bigger and brighter stars in the universe. The Sun is, however, very important to us because i) it is the nearest and the only star in our planetary system and ii) it provides almost all of our energy. Do you know that a slight variation in the energy received from the Sun can threaten life on the earth! Further, the Sun being the nearest star, we can study its structure, atmosphere, and other physical characteristics in greater detail. The information/data so obtained can be used to test the theories of stellar structure and evolution. In this way we can improve our theories and have a better understanding of other stars. In the present Unit, you will study about the Sun. Due to the efforts of astronomers, today we have detailed information regarding the Sun. In Sec. 5.2, you will learn to arrive at the estimates of the basic solar parameters such as mass, radius and effective surface temperature. As far as we are concerned, all the visible radiation from the Sun comes from its surface layer called the photosphere. Above the photosphere is the atmosphere of the Sun consisting of two distinct layers namely the chromosphere and the corona. In Sec. 5.3, you will study the characteristic features of these layers. Interaction of the Sun’s magnetic field with highly mobile charged particles in it gives rise to a variety of observable events. These events, collectively known as solar activity, have been discussed in Sec. 5.4. The theoretical analysis of the interaction of magnetic field with conducting matter in motion is known as magnetohydrodynamics and it provides a basis to understand solar activity and related features of the Sun. Solar magnetohydrodynamics has been discussed in Sec. 5.5. In Sec. 5.6, you will learn, in brief, about helioseismology which provides valuable information about the Sun’s internal structure. Objectives After studying this unit, you should be able to: •

estimate values of the basic parameters of the Sun; 5

The Solar System and Stars

•

describe different layers of the Sun’s atmosphere;

•

describe some of the observed features such as sunspot, prominence and solar flare associated with solar activity;

•

explain the role of the Sun’s magnetic field in solar activity;

•

derive the basic results of solar magnetohydrodynamics; and

•

explain the seismology of the Sun.

5.2

SOLAR PARAMETERS

The basic parameters of the Sun are its mass (MΘ), radius (RΘ), luminosity (LΘ) and effective surface temperature (Teff). In the following discussion, you will learn to estimate these parameters. Mass: You know that the mean distance of the Sun from the Earth is 1.5 × 1011m. It is called the mean solar distance, a. In astronomy we measure mean distances in terms of a and it defines the Astronomical Unit (1 AU = mean solar distance). To obtain an expression for the mass of the Sun, we may use Kepler’s third law under the assumption that the mass of the planet can be neglected in comparison with the mass of the Sun. This assumption is valid for the Sun-Earth system and we can write: 4π 2 a 3 You may recall from Unit 6 of the physics elective course PHE-01 entitled Elementary Mechanics that Kepler’s third law is given by: T2 =

4π2 a 3 GM

where T, a, G and M are respectively the time period, semi-major axis, gravitational constant and mass of the Sun.

Spend 5 min. In astronomy, we encounter typically small angles (of the order of minute (′) and second (″)), for which simple approximations are used for trigonometric functions. Thus, for small values of the angle θ, tanθ = θ = sinθ

P2

= GM Θ

(5.1)

where a, P, G and MΘ are the mean solar distance, orbital period (~ 365 days) of the Earth, gravitational constant and mass of the Sun, respectively. With the values of the orbital period and the mean solar distance available at present, the value of GMΘ is estimated to be 132712438 × 1012 m3s−2. Since the laboratory measurements for G gives a value equal to 6.672 × 10−11 m3kg−1s−2, we obtain: MΘ ≈ 2 × 1030kg. This value is taken as the mass of the Sun as it exists today. In fact, solar mass decreases continuously since the Sun continuously emits radiation and particles which carry with them some mass. However, the total mass loss during the Sun’s estimated life time (~ 1010 yrs) is found to be less than 1027kg. This value is much less than the error in measurement of the solar mass and is, therefore, negligible. This method can also be used to estimate the masses of the satellites/Moons of the planets in our solar system. How about solving an SAQ of this nature? SAQ 1 One of the four Galilean satellites of the planet Jupiter is Io. Its orbital period is 1.77 days. The semi-major axis of its orbit is 4.22 × 1010cm. Calculate the mass of Jupiter under the assumption that the Jupiter is too massive in comparison to Io. Radius: The radius of the Sun can be estimated if we know the values of its angular diameter, θ and the mean solar distance, a (Fig. 5.1). The angular diameter of the Sun is 32′ and mean solar distance is 1.5 × 1011m. Thus, with the help of Fig. 5.1, we obtain the value of the solar radius RΘ as:

and cosθ = 1 However, as you know, angles must be measured in radian if these formulae are to be valid.

6

RΘ =

1 2

[ (1.5 × 10

11

8

= 6.7 × 10 m.

m) × ( 32 × 2.9 × 10 − 4 rad)

]

Astronomical observations indicate that the solar radius is not constant; rather, its value changes slowly. Over a period of ~ 109 years, the average change is about 2.4 cm per yr. Further, radius of the Earth is 6.4 × 106m. Thus, the Sun’s radius is

1 AU

The Sun

Sun 32´

Earth

Fig.5.1: When viewed from the Earth, the angular diameter of the Sun is approximately 32′′

almost 100 times larger than that of the Earth. To get an idea of the relative sizes of the Sun and the Earth, refer to Fig. 5.2. Sun

Moon’s orbit

Earth

Fig.5.2: The size of the Sun is so big that it can contain the Earth as well as the orbit of the Moon!

Luminosity: The solar luminosity, LΘ, is defined as the total energy radiated by the Sun per unit time in the form of electromagnetic radiation. To estimate the value of luminosity of the Sun, let us imagine a sphere with the Sun at its centre (Fig. 5.3). The radius of this imaginary sphere is a, the mean distance between the Sun and the Earth. Now, each unit area A of the sphere receives energy equal to S, called the solar constant. Therefore, luminosity can be expressed as: LΘ = 4πa2S

(5.2)

a

1m Earth

Sun

A

1m

1 AU

Fig.5.3: Imaginary sphere of radius a surrounding the Sun where a is its mean solar distance from the Earth

7

The Solar System and Stars

Since the solar radiation is absorbed in the Earth’s atmosphere, it is obvious that S should be measured above the atmosphere. S has now been measured accurately using satellites and its value is 1370 Wm−2. Substituting the value of S and the mean solar distance, a = 1.5 × 1011m in Eq. (5.2), we get: LΘ = 3.86 × 1026 W. Temperature: The temperature of the Sun at its surface and its interior regions are different. The surface temperature can be estimated using Stephan-Boltzmann law which you studied in our course on Thermodynamics and Statistical Mechanics (PHE-06). We leave this as an exercise for you in the form of an SAQ. Spend 5 min.

SAQ 2 Assume that the Sun radiates like a black body at temperature T. Calculate T using −8 −2 −4 Stephan-Boltzmann law. Take Stephan constant σ = 5.67 × 10 Wm K . On solving SAQ 2, you would have found that the temperature of the Sun is 6000 K. This estimated temperature is called the effective surface temperature because it is the temperature of a black body whose surface emits the same flux as the Sun. This is the temperature of the surface layer of the Sun called the photosphere from which all the radiation is emitted. To appreciate the validity of the approximation that the Sun radiates like a black body, refer to Fig. 5.4. It shows the observed solar radiation in the ultraviolet, visible and infrared regions of electromagnetic spectrum. Also plotted in this figure is the energy curve of a black body at 6000 K. In view of the similarity of the two curves, it is fair to assume that the Sun radiates like a black body at temperature 6000 K.

You learnt that the Sun radiates like a black body at approximately 6000 K. You may think that if all the energy is radiated outward from the Sun, its surface should gradually cool down. This does not happen because there is a constant flow of energy from the interior of the Sun towards its surface. Fig.5.4: Solar energy curve

The Sun is a hot, bright gaseous ball and it does not have a well defined surface like the Earth. The visible surface of the Sun is called photosphere. Let us learn about it now.

5.3

SOLAR PHOTOSPHERE

The photosphere (Fig. 5.5) is the visible surface of the Sun. All the light received from the Sun, in fact, comes from the photosphere. You may ask: Why do not we receive the radiation in the same form as generated in the interior of the Sun? At the centre of the Sun, the energy is generated in the form of high energy photons called γrays. As these photons travel outwards, they collide with particles of matter and lose Fig.5.5: The Sun’s photosphere energy continuously. By the time these photons reach the surface − the photosphere − they are reduced to photons of visible region of electromagnetic spectrum. So, visible radiation is emitted from the photosphere. 8

The Sun

The density of photosphere is 3400 times less than the density of the air we breathe. The thickness of the photosphere is about 500 km and the temperature at its base is ~ 6500 K. The temperature decreases upward and reaches a minimum value of ~ 4400 K at the top. This assumption is corroborated by the Sun’s absorption spectrum which indicates that the light we receive must be passing through a cool gas in which photons get absorbed. The photosphere is not a quiet region (see Fig. 5.6). It shows a granular structure. If you look at Fig. 5.6a carefully, you can see that the photosphere consists of bright and irregularly shaped granules; each granule surrounded by dark edges. It has been found that these granules are very hot and their typical size is ~ 1500 km. The hot gas in the granules rises up with a speed of the order of 500 ms−1 and bursts apart by releasing energy. The cool material subsequently sinks downward along the dark edges or lanes between granules. The rising hot granules are seen only for a very short time (~ 10 minutes) before they dissolve.

Granule boundary Granule

(a)

Granule

(b)

Fig.5.6: a) Photograph of the photosphere showing granular structure; and b) schematic diagram showing granules and their boundaries

The question is: What causes granulation of the photosphere? It is caused due to convection (a mode of energy transport by matter, about which you will study in Unit 8). The granulation can be visualised (Fig. 5.6b) as the top layer of a region where, due to convection, hot gas from below the photosphere moves upward. Thus, the centre of the granule is hotter and it emits more radiation and looks brighter in comparison to the edges which are relatively cooler and emit less radiation. Convection based explanation seems valid because the spectra of granules indicate that their centres are much hotter than the edges. Further, the solar granulation provides observable evidence supporting the idea that there exists a convection zone below the photosphere. You may now like to know: What is the chemical composition of the photosphere? It consists of 79 percent hydrogen and the remaining 21 percent consists of nearly 60 other chemical elements. Interestingly, all the elements of the photosphere are known elements and their proportion in the earth is more or less the same as that in the photosphere. This similarity in the chemical compositions of the photosphere and the earth is of utmost importance for understanding the formation of the solar system. Though the photographs of the Sun give the impression that it has a clear edge, such clear and distinct edge does not exist. Outside the apparent edge are the Sun’s outer layers, collectively known as the Sun’s atmosphere. These layers can be seen and probed and valuable information about their physical characteristics can be obtained. Let us learn about the various layers of the solar atmosphere.

As they look inwards into the solar atmosphere, astronomers have discovered that, within a distance of about 500 km, the solar atmosphere changes from being optically thin to optically thick. This distance is only ~ 0.07% of the Sun’s radius. This region gives the impression that the Sun has a sharp edge as viewed from the earth.

As you have learnt in Unit 4, optically thin/thick medium refers to the medium characterised by low/high absorption of electromagnetic radiation.

9

The Solar System and Stars

5.4

SOLAR ATMOSPHERE

The Sun’s atmosphere is divided into two layers namely, the chromosphere and the corona. A schematic diagram of these layers is shown in Fig. 5.7. Corona 2600 km Transition zone 2300 km

Chromosphere

500 km Photosphere 0 km

Sun’s interior

Fig.5.7: Schematic diagram showing the layers of solar atmosphere

5.4.1

Fig.5.8: Chromosphere just before a total solar eclipse

You may be aware that each chemical element emits/absorbs electromagnetic radiations of characteristic wavelengths. These radiations of different wavelengths (also called lines) constitute the spectrum of the element. Balmer lines, which fall in the visible region, are the spectral lines of hydrogen atom. The emission/absorption of the spectral lines is predominantly dependent on the temperature and density of the material. You will study about the origin of spectral lines in Unit 7.

10

Chromosphere

Chromosphere lies above the photosphere (Fig. 5.7) and extends up to ~ 2000 km. This layer of the solar atmosphere is normally not visible from the Earth because of its faintness. However, it can be seen during a solar eclipse. The name chromosphere is derived from the fact that a few seconds before and after a total solar eclipse, a bright, pink flash appears above the photosphere (Fig. 5.8). The spectrum obtained at that time is called a flash spectrum (Fig. 5.9).

Fig.5.9: Emission lines observed at the time of total solar eclipse

The appearance of pink colour is due to the emission of the first Balmer line (Hα) which occurs in the red region. The temperature, density and pressure in the chromosphere determine the intensities of various emission lines. In the chromosphere, the density decreases by a factor of ~ 104 from that of the photosphere

while the temperature rises to ~ 25000 K within a short distance of ~ 2000 km. Therefore, the spectral lines that are not produced at relatively higher density and lower temperature of the photosphere are formed in the chromosphere as emission lines (see margin remarks).

The Sun

At this point, it is logical to ask: Why does the temperature in the chromosphere increase with height? The clue to the answer of this question lies in observing the chromosphere just before the total solar eclipse. Hot gas, in the form of jets called spicules, is observed throughout the chromosphere (Fig. 5.10). These spicules extend upward in the chromosphere up to a height of ~ 10000 km and last for as long as 15 minutes. This implies that the lower part of the chromosphere is highly turbulent and the spicules transport energy and matter from the photosphere to the chromosphere. This causes heating of the chromosphere.

Fig.5.10: Spicules in the Sun’s chromosphere

Now, your next logical question could be: What causes spicules? The origin of spicules is not yet understood completely. However, it appears to be caused by the Sun’s magnetic field. Further, just above the chromosphere, there exists a transition region extending up to ~ 3000 km. In this region, the temperature rises sharply to ~ 106 K (Fig. 5.11). The transition region links the chromosphere with corona, the outermost part of the solar atmosphere.

1014

107

1012 1010

105

108

104

106

103 10

2

10

4

10

6

Density (cm 3)

Temperature (K)

106

104

Height from the Sun's surface (km)

Fig.5.11: The variation of temperature and density in the Sun’s atmosphere with distance

11

5.4.2

The Solar System and Stars

Corona

Temperature (103 K)

40000 20000 10000 200 40

Chromosphere

Corona, the outermost layer of the Sun’s atmosphere is named after the Greek word for Crown. Like the chromosphere, the corona can be observed only during total solar eclipse − when the Moon completely covers the solar disc (Fig. 5.12). You may wonder why we cannot see corona at normal times! The fact is that the density of matter in both the chromosphere and corona is very low (see Fig. 5.11). They emit very little light and, as a result, they are very faint. In the bright light of the photosphere, they are not visible.

Corona Si X

Si VII

Mg X O VI OV O II Si IV O II Si II

10

1 2 4 10 20 40 Height above the surface(103 km) Fig.5.13: Height versus temperature plot showing emission lines of different ionised atoms

Spend 5 min.

Fig.5.12: Two photographs of the solar corona

The spectrum of corona consists of bright lines superimposed on a continuous spectrum. When these lines were first discovered, they were thought to be due to a new element, coronium, not found on the earth. Later, it was realised that these lines were due to highly ionised atoms and not due to the so-called ‘new’ element coronium. Fig. 5.13 shows the temperature and height in the corona at which emission lines of various ionised elements are formed. You may note here that to excite the emission lines from highly ionised elements, say spectral line of SiX ( read margin remarks), a temperature greater than 2 × 107K is required. The observed emission lines of highly ionised atoms of iron, nickel, neon, calcium etc., in the spectrum of corona clearly indicate that the temperature prevailing in corona is very high (more 6 than 10 K). Now, before proceeding further, how about testing yourself? SAQ 3 a) The temperature of chromosphere and corona is very, very high in comparison to that of the photosphere. Still, we observe that the photosphere is the brightest of the three. Why? b) Calculate the temperature at which a particle will have sufficient energy to ionise a hydrogen atom.

In astronomy, it is common to denote a neutral atom, such as silicon, as SiI. By this convention, singly ionised silicon is denoted by SiII. Hence, SiX denotes a silicon atom whose nine electrons have been removed due to repeated ionisation.

12

Due to high temperature, electrons in the corona region have high energies. These electrons interact with ionised atoms and give rise to emission of X-rays. The coronal X-ray emission is much larger than that of the photosphere. Remember that the temperature of the photosphere is only 6000 K. So, it emits very little energy in the X-ray region. The Sun, as observed in X-rays is shown in Fig. 5.14 which clearly indicates the existence of very high (~ 106 K or more) temperature in the corona. You have already learnt that the temperature of the photosphere is lower than that of the chromosphere and as one goes further up in the corona, temperature rises to more than a million degree K. This gives rise to a very simple but important question: Despite being closer to solar interior, why is the photosphere far cooler than the corona? You know that the second law of thermodynamics precludes such a scenario

as heat cannot flow from a cooler region to a hotter region on its own. We also know that the radiation from the photosphere passes through corona almost freely because of its (corona’s) low density. Since hardly any absorption of radiation takes place in the corona, the existence of such high (~ million degree) temperature in the corona presents a paradoxical situation. Several mechanisms have been proposed to resolve the paradox. It is now generally believed that the magnetic field of the Sun might, in some way, be responsible for coronal heating. You will study the basics of this mechanism in Sec. 5.5 of this Unit. The observed overlapping of regions of intense Xray emission and strong magnetic fields lend support to this idea. Solar Wind Unlike its visual appearance, the solar corona extends much beyond into the space. The outer layer of the solar atmosphere, in fact, continuously emits charged particles which fill the entire solar system. This emission is called the solar wind. It comprises streams of charged particles (mainly protons and electrons) and causes continuous loss of mass from the Sun. The phenomenon of solar wind was predicted much before its detection. Its characteristics can be investigated using rockets and satellites. For instance, the solar wind velocities range from 200-700 km s−1 at the distance of the Earth from the Sun. The number density of the solar wind at this distance is ~ 7 particles per cm3. You may like to know: What gives rise to the solar wind? In view of the high temperature prevailing in the corona, the gas contained therein exerts tremendous pressure outward. In fact, the pressure is much higher than the inward pressure due to the Sun’s gravity. The gas, therefore, streams outward from the Sun and fills the interplanetary space. In 1962, Mariner II spacecraft detected the solar wind by its onboard instruments.

Outer radiation belt

The Sun

Fig.5.14: X-ray picture of the Sun

One of the manifestations of solar wind is observed in the shape of comets. You know that the tail of a comet points away from the Sun. It is because the solar wind sweeps along the material of the comet.

Magnetic axis of the Earth

Inner radiation belt

Rotational axis of the Earth

Fig.5.15: van Allen radiation belts

The electrically charged particles carried by the solar wind cannot cross the lines of force of the Earth’s magnetic field. These particles are deflected by the Earth’s magnetic field, spiral around the field lines and move back and forth between the magnetic poles of the Earth. As a result, two doughnut-shaped zones of highly energetic charged particles are created around the Earth and they are collectively called the van Allen radiation belts. These radiation belts are shown in Fig. 5.15. Astronomers have observed a variety of short-lived events, collectively known as solar activity, occurring on or near the surface of the Sun. The root cause of all these

You know that when a charged particle passes through a magnetic field, it experiences a force which changes its direction of motion. The force experienced by a moving charged particle is known as Lorentz force.

As you know, satellites and space missions comprise very sophisticated integrated circuits, solar cells and other electronic gadgets. Therefore, care is taken to minimise the damaging effects of the radiation belts on the satellites and space missions.

13

The Solar System and Stars

activities is the existence of strong and localised magnetic field in the photosphere. Studies of these events/activities provide valuable information about the Sun and the nature of its magnetic field. You will now learn about some of these short-lived events.

5.5

SOLAR ACTIVITY

Sunspots If you look at the photographs (Fig. 5.16) of the Sun, you see dark spots on its visible surface. These dark spots are called sunspots. Sunspots can be seen sometimes even with unaided eye at sunrise or sunset. (But you should not attempt to see the Sun with unaided eye as it may cause irrepairable damage to your eyes because of its intense brightness.) Naked eye observations of sunspots date back to ~ 2000 years in China. It was in the seventeenth century that Galileo, using the telescope which he himself had fabricated, observed sunspots and found that these dark spots were in motion. This led him to suggest that the Sun was spinning in space. Galileo also observed that the sizes and shapes of the sunspots kept changing as they rotated with the Sun.

Fig.5.16: Photographs of a sunspot group

The sunspot temperature is ~ 4000 K. With such high temperature, you may wonder, why they appear dark! Sunspots appear darker because they are cooler than their surrounding areas in the photosphere that have an effective temperature of 6000 K. A typical white light picture of a large sunspot is shown in Fig. 5.17. Note that it consists of a dark central region, called umbra, surrounded by a less dark region, called penumbra. We do not see such details in the picture of smaller sunspots.

Fig.5.17: Sunspot structure

14

At this stage, a logical question is: Why is the temperature of the sunspots lower than their surroundings? It is due to the existence of strong magnetic fields in the sunspots. In the presence of a magnetic field, a spectral line emitted by an atom at a

single wavelength is split into three lines. This is called the Zeeman Effect. Such Zeeman splitting is observed in the spectrum of sunspots. Since the line separation, ∆λ is proportional to the applied magnetic field, a magnetic field up to ~ 3000 Gauss has been estimated in sunspots. Fig. 5.18 shows the mechanism of Zeeman splitting of a spectral line and the Zeeman splitting of a spectral line of a sunspot.

The Sun

Energy levels Transition

Spectrum

(a)

(b)

Fig.5.18: Zeeman splitting of a) a spectral line; and b) a spectral line of the sunspot

The presence of strong magnetic fields in sunspots restrains the flow of hot material from layers below the photosphere. Therefore, within a sunspot, less heat comes up and they (sunspots) are cooler/darker than the surrounding region. Within a sunspot, the umbral magnetic field is quite intense ~ 3000 Gauss. It spreads like an umbrella and weakens in the penumbral region. The field strength in the penumbra is estimated to be ~ 1000 Gauss. Sunspots can last for weeks. The question is: How do these cooler regions survive for so long amidst the hotter regions? This could happen due to the magnetic field. You know that magnetic field exerts pressure (equal to B2/2µ) across the lines of force. This pressure, along with the pressure of matter inside a sunspot balances the material pressure outside and the sunspots can exist in equilibrium. Sunspot Cycle The observed motion of the sunspots indicates that the Sun is spinning in space. In 1843, Heinrich Schwabe, a German who observed the sky for fun, discovered a periodic variation in the numbers of visible sunspots. He found an interval of 5.5 yrs between the time when maximum number of sunspots (sunspot maxima) were observed and the time when the minimum number of sunspots (sunspot minima) were observed. Over the last two centuries, sunspot observations clearly suggest a periodic variation of about 11 years between two successive sunspot maxima (Fig. 5.19a). Another important observation pertaining to sunspot is that the sunspot zones migrate along solar latitude. It is observed that the first sunspot zone appears at latitude of ~ 35° in, say, the northern hemisphere and it migrates to lower latitudes. It lasts till it reaches a latitude of ~ 10°. The latitude migration of sunspot zones is shown in Fig. 5.19b. This is the famous butterfly diagram which shows a period of ~ 11 years between the successive occurrences of a sunspot at a given latitude. It is believed that the sunspot cycle is caused due to differential rotation of the Sun; it rotates faster at the equator compared to higher latitudes. 15

The Solar System and Stars

Number of Sunspots

Year (a)

Year (b) Fig.5.19: a) Sunspot cycle; and b) butterfly diagram which shows the migration of sunspots from higher to lower latitudes

Solar Prominences Refer to Fig. 5.20 which depicts loop like structures surging up into the corona when the Sun is viewed along the edge of the solar disc. These structures are called prominences which are intimately connected with and formed due to the Sun’s magnetic field. The material in prominences comprises hot ionised gases trapped in magnetic fields associated with the active regions. Since these gases came from deeper layers of the solar atmosphere, they are cooler and therefore denser than the coronal gas. It is for this reason that prominences appear as bright structures. However, when viewed against the photosphere (solar disk), prominences appear as dark snake like objects, called filament.

Fig.5.20: Solar prominences

Solar Flares 16

Yet another form of solar activity is called solar flare. Solar flares are sudden eruptive events which occur on the Sun (Fig. 5.21). Each event may involve energy in the

The Sun

Approximate size of the Earth Fig.5.21: Solar flare

range of 1022 to 1025 Joules. Usually the flares last anywhere between a few minutes to more than an hour. A large flare may have linear dimension as large as 105 km and may be seen as a short-lived storm on the Sun. Such energetic eruptions are usually linked to sunspots because these quite often occur at the top of magnetic loops that have their feet in sunspots. Thus, the most likely places of occurrence of solar flare are the regions of closely packed sunspots. The tremendous amount of energy carried in solar flare is released in the form of X-rays, ultraviolet and visible radiation, high speed electrons and protons. You may ask: What is the source of energy in solar flares? This question can be answered on the basis of a model for solar flare shown in Fig. 5.22.

Fig.5.22: A solar flare model

17

A Model of Solar Flare

The Solar System and Stars

All kinds of solar activities i.e., the sunspots, prominences, flares etc., are possibly linked to the release of stored magnetic energy. It is believed that the energetic solar eruptions are caused due to coming together and merging of magnetic fields in the active regions (the phenomenon is known as magnetic field reconnection) and thereby releasing the stored magnetic energy. To appreciate this phenomenon, magnetic lines of forces can be considered as stretched springs with certain amount of energy associated with each unit length. If the length of the lines of forces gets reduced by certain mechanism, energy is released. This is what happens when lines of force pointing in opposite directions meet and merge with each other (Fig. 5.22). In order to fix these ideas, you should answer the following SAQ. Spend SAQ 4 3 min. a) What is the basis to conclude that the Sun is rotating in space? b) What is the difference between spicules and solar prominences? So far, you have studied about the photosphere, solar atmosphere and solar activity. You must have noted that the Sun’s magnetic field plays an important role in solar activity. Further, gaseous matter in the Sun is in the ionised form, that is, it is a conducting matter. We will now try to understand the nature of interaction between the Sun’s magnetic field and the conducting matter (fluid) in motion. This understanding is of utmost importance in astronomy because, everywhere in the universe, we find conducting matter moving in the presence of magnetic fields. The study of the motion of conducting fluid in the presence of magnetic field is called magnetohydrodynamics. Let us now turn to this subject.

5.6

BASICS OF SOLAR MAGNETOHYDRODYNAMICS

We begin our discussion of solar magnetohydrodynamics with Maxwell’s equations. You may recall from the physics course entitled Electric and Magnetic Phenomena (PHE-07) that three of the Maxwell’s equations can be written as: ∇ . B = 0,

(5.3)

∇ × B = µj,

(5.4)

and You are familiar with the Ohm’s law expressed mathematically as: j=σE However, when a conducting fluid is moving with velocity v in a magnetic field B, there is an induced electric field given as v × B. Thus, in the presence of two kinds of electric fields − one due to charged particles and another due to their motion − the Ohm’s law takes the form: J = σ (E + v × B)

18

∇×E= −

∂B ∂t

(5.5)

where µ, B, E and j are the magnetic permeability of the medium, magnetic field intensity, electric field intensity and electric current density respectively. Note that in Eq. (5.4), we have not written the displacement current term. This is because, magneto-hydrodynamic phenomena in the Sun are usually slow whereas the displacement current gives rise to fast phenomena such as electromagnetic radiation. Further, if the fluid (conducting matter) velocity is v and its electrical conductivity is σ, then Ohm’s law gives (see the margin remark): j = σ (E + v × B)

(5.6)

The Sun

where (v × B) term represents the electric field induced due to the motion of conducting fluid in the presence of magnetic field. Using Eqs. (5.3) to (5.6), you can readily obtain: ∂B ∂t

= ∇ × ( v × B ) + η∇ 2 B

(5.7)

where η = (µσ)−1 is called the magnetic diffusivity. Its value is generally constant in solar conditions. Spend 3 min.

SAQ 5 Derive Eq. (5.7). The first term on the right hand side of Eq. (5.7) gives the change in B due to the fluid motion. The second term represents the change in B due to conductivity σ. It is generally called the Ohmic decay of the field. Note that for v = 0, Eq. (5.7) reduces to:

∂B = η∇ 2 B ∂t

(5.8)

Eq. (5.8) is the diffusion equation. It gives the rate at which magnetic field diffuses out due to conductivity. In the limit of infinite conductivity, we have η → 0, and Eq. (5.7) becomes ∂B ∂t

= ∇ × ( v × B)

(5.9)

Without going into the mathematical details, it is possible to understand the relative importance of the two terms on the right hand side of Eq. (5.7). To do so, let us obtain the orders of magnitude, of the values of the two terms. The order of magnitude of the VB , where V, B and L are the typical values of the fluid velocity, first term is L magnetic field and the dimension of the system. Similarly, the magnitude of the ηB . The ratio of the two terms is called magnetic Reynold number, second term is L2 Rm, and is given by: Rm =

VL η

(5.10)

If Rm >> 1, the second term in Eq. (5.7), i.e. the diffusion term, is negligible. The condition Rm >>1 is obtained in two situations: when the conductivity is very high because η appears in the denominator, and secondly when the dimension of the system, L is very large. In astrophysical systems, we have Rm >> 1 because of the second situation as their dimensions are very large. In any case, Rm >> 1 implies that there is no decay of the magnetic field as if the conductivity of the medium is infinite. Actually, the conductivity is finite, but the large dimensions ensure that Rm >> 1 and so there is no decay of the field. If we drop the diffusion term in Eq. (5.7), the remaining equation (Eq. (5.9)), implies that the magnetic flux linked to a cross-section of the fluid remains unchanged as the fluid moves about. In other words, the magnetic field is frozen in the fluid. The idea of frozen field means that the magnetic flux is transported along with the material motion. We can show this formally in the following manner:

19

Let us consider a cross-sectional area A placed in a magnetic field B (Fig. 5.23). The magnetic flux linked with area A may be written as:

The Solar System and Stars

l

Φ=

B.d a

(5.11)

A

dl

A

v × dldt

A'

Fig.5.23: Motion of a closed curve l with fluid with velocity v

where da is an element of area on the surface A. Let l be the curve enclosing the area A. Let us further assume that, in the time interval dt the area changes from A to A′ as the fluid moves around. The magnetic flux Φ′ linked with A′ may be different from Φ because (i) magnetic field B may have changed and/or (ii) some flux might have been exchanged through the surface of the volume generated between A and A′. Now, the rate of change of magnetic field is ∂B/∂t. The area of the curved surface surrounding the volume between A and A′ is v × dl dt where dl is an element of length of the contour of A. Therefore, the difference ∆Φ between fluxes through A′ and A is given by: ∂B

∆Φ = A

You may recall that area is a vector quantity having direction along the normal to the area, whose sense is defined by the right hand rule.

∂t

.da + o B.. ( v × dl ) dt. l

(5.12)

Further, using Stoke’s theorem, we can write the second term on the right hand side of Eq. (5.12) as: o B.. ( v × dl) = − ∇ × ( v × B) . da l

(5.13)

A

So, we can write the rate of change of flux (Eq. (5.12)) as: dΦ = dt

A

∂B − ∇ × ( v × B ) . da ∂t

(5.14)

∂B = ∇ × ( v × B) . Thus, from Eq. (5.14), we find that the ∂t rate of change of magnetic flux is zero. It implies that Φ remains constant. We may, therefore, conclude that in astrophysical systems the lines of forces are completely attached or glued to the moving fluid when Rm >> 1.

From Eq. (5.9), we have

Now, let us pause for a moment and think about the significance of the above conclusion for solar activity. Recall that the active regions consisting of sunspots have strong magnetic fields and solar activities such as prominence and flare occur in these regions. The structures associated with solar prominences and solar flares are very similar to the magnetic lines of force. It is, therefore, believed that these activities are caused due to frozen magnetic lines of force in the conducting fluid.

20

You have learnt earlier in this unit that, in almost all the events associated with the Sun’s magnetic field, energy is also transported. The energy transported by the magnetic field glued to the conducting matter is responsible for heating the chromosphere and corona. Would not you like to know how it happens? To understand this process, we begin with the fact that an electric current exists in the solar atmosphere due to the drift of electrons with respect to ions carrying opposite charges. If a magnetic field B is also present in the plasma, then a volume force, also 1 called the Lorentz force (= j × B), acts on the material. Since j = (∇ × B ) (from µ Eq. (5.4)) we may write the expression for Lorentz force as:

The Sun

j× B =

=

1 (∇ × B ) × B µ 2 1 (B.∇ )B − ∇ B µ 2µ

(5.15)

The first term on the RHS of Eq. (5.15) denotes magnetic tension and the second term denotes the gradient of magnetic pressure. Therefore, in the presence of a magnetic field, a lateral pressure (due to the second term in Eq. (5.15)) acts on a conducting gas. To maintain equilibrium, the lateral pressure due to magnetic field is balanced by the gas pressure. The magnetic tension term is similar to the tension of a stretched string. When a stretched string is distorted, its tension provides the restoring force. Therefore, when the magnetic field lines (frozen in the conducting fluid) are disturbed we have a similar restoring force (due to the magnetic tension). Thus, a disturbance may propagate as a transverse wave along the magnetic field lines. Such waves are called Alfven waves. The speed of these waves is given by: vm =

B 4πρ

.

where ρ is the density of the plasma. Since Alfven waves propagate along the magnetic field lines, it is possible to transport energy outward along magnetic field lines which are threading the outer solar atmosphere. This energy is believed to be responsible for heating the chromosphere and corona. Spend 3 min.

SAQ 6 A pressure of 103Pa (Pascal) prevails in the solar atmosphere. What should be the strength of the magnetic field required to balance such a pressure? Now, before we close our discussion about the Sun, we will briefly discuss a new area of solar research called helioseismology.

5.7

HELIOSEISMOLOGY

In 1962, Leighton, Noyes and Simon noticed wiggling back and forth of some of the absorption lines in the solar spectrum with periods ~ 5 min. They conjectured that the movement of the surface of the Sun is responsible for such observations. Subsequently, careful observations of the solar surface confirmed the idea of waves rising up and down on the surface of the Sun (Fig. 5.24). Astronomers now use these waves to probe the solar interior much the same way as seismologists probe the earth’s interior using vibrations caused by an earthquake. So far, millions of different vibrational modes also called acoustic modes or p-modes have been observed on the Sun’s surface. All of them have different frequencies and surface patterns. Thus, the Sun appears to have a rhythmic surface motion similar to that of a beating heart. The observed rise and fall of the surface of the Sun is, in fact, due to superposition of many different acoustic modes. These modes are now believed to be driven by irregular motions in the convective envelope under the solar surface.

The name helioseismology derives from the fact that, in Greek, helios means the Sun and the word seismos is used for an earthquake. As you know, seismology refers to the study of seismic waves moving inside the Earth. Their arrival at various points on the Earth’s surface enables us to find the point of origin of these waves. This way we are able to construct the internal structure of the Earth.

21

The Solar System and Stars Disk of Sun

Position

5 min

Time

Fig.5.24: Waves on the surface of the Sun

The combined effect, or the superposition of millions of these acoustic waves, results in the observed up and down motion of the photosphere with a period of the order of 5 minutes. The extent or size of a wave is called its horizontal wavelength. The relation between sizes and periods of waves, obtained theoretically, suggested that only specific combination of periods and sizes can resonate inside the Sun. Fossat and Grec observed the solar oscillation from the South pole for around 120 hours. The analysis of this continuous record showed that the entire surface of the Sun is ringing like a bell with periods in the range of 5 minutes and the vibrations may last for days and weeks.

Since we cannot go inside a star, we build theoretical models which tabulate pressure, temperature and density at various depths inside the star.

The natural frequencies of oscillations can be computed for any solar model. In view of the precision now possible for determining the frequencies, we may compare these with those computed for a given solar model. In case there is lack of agreement between the predicted and observed frequencies, the model is slightly modified to improve agreement. The improvement in the model brings it closer to reality. It has been found that the depth of the solar convection zone is at a radius of 71.3 percent of radius of the Sun. It has also been confirmed now that the proportion of helium in the Sun lies between 0.23 − 0.26. This is quite consistent with the value of 0.25 for helium believed to have been formed in the early phase of the universe after the bigbang. Now, let us summarise what you have learnt in this unit.

5.8

22

SUMMARY 30

•

The mass, radius and effective surface temperature of the Sun are 2 × 10 kg, 7 × 108m and 6000 K, respectively.

•

The Sun’s atmosphere comprises of the photosphere, chromosphere and corona.

•

The photosphere is the visible surface of the Sun and all the radiation we receive from the Sun is emitted by this layer.

•

The photosphere has a granular structure which is caused by convection of hot gas from below the photosphere.

•

The chromosphere, which lies above the photosphere, extends up to ~ 2000 km and is normally not observable from the earth except during total solar eclipse. The temperature of the chromosphere is much higher than the photosphere.

•

Relatively higher temperature of the chromosphere is perhaps caused due to spicules − jets of hot gas − which extend upward in the chromosphere.

•

The outermost layer of the solar atmosphere, corona is also not visible at normal times due to its low material density. The temperature of the corona is of the order of 106K, much more higher than that of the photosphere.

•

The sunspots are the dark spots on the solar disk. They appear dark because they are cooler than their surrounding areas in the photosphere. The movement of sunspots indicate that the Sun is spinning in space.

•

Solar prominences are the loop like structures surging up into the corona. They are caused due to the Sun’s magnetic field.

•

Solar flares are sudden eruptive events involving energy in the range of 1022 to 1025 joules. The most likely places of occurrence of solar flares are the regions of closely placed sunspots and they are possible caused due to the release of stoned magnetic energy.

•

The study of the motion of conducting matter in the presence of magnetic field is called magnetohydrodynamics.

•

The magnetic Reynold number is given by Rm =

The Sun

VL η

which indicates that for infinite conductivity (i.e. η → 0), there is no decay of the magnetic field. In other words, magnetic field is frozen in the conducting matter in motion. •

Transportation of magnetic flux with conducting matter explains some of the solar activities like prominence and solar flare.

•

Helioseismology has its origin in the observed wiggling back and forth of some of the absorption lines in the solar spectrum. The movement of the surface of the Sun, causing these back and forth motion of absorption lines, gives rise to waves on it. Investigation of the nature of these waves provides valuable information regarding the internal structure of the Sun.

5.9

TERMINAL QUESTIONS

Spend 20 min.

1.

In a sunspot, magnetic diffusivity, linear dimension and velocity of conducting fluid respectively is 103m2s−1, 104km, and 103ms−1. Estimate the magnetic Reynold number, Rm. Is it possible to assume that the conductivity in the sunspot is virtually infinite?

2.

The temperature inside a sunspot is 4000 K and that of its surface is 6000 K. Calculate the strength of the magnetic field inside the sunspot which will balance the pressure inside and outside. [Hint: Remember that the magnetic pressure is B2/2µ where µ is magnetic permeability of the medium and its value for the present case can be taken as 4π × 10−7 NA−2].

3.

The number density of particles (assume hydrogen) in the photosphere is 1020 particles per cm−3 and the strength of the magnetic field of the Sun is 1 G. Calculate the velocity of the Alfven waves in the photosphere.

23

The Solar System and Stars

5.10 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1. Since the mass of the planet Jupiter is very large compared to its satellite Io, we can use Kepler’s third law for Jupiter and Io system. Thus, we can write: 4π 2 a 3 P2

= GM J

(i)

where MJ is the mass of Jupiter. From the problem, we have 10

8

a = 4.22 × 10 cm = 4.22 × 10 m P = 1.77 days = 1.77 × 24 × 60 × 60 s G = 6.672 × 10−11 m3 kg−1 s−2 Substituting these values in Eq. (i) above, we get MJ = =

4π 2 a 3 GP 2

4 × (3.14) 2 × (4.22 × 10 8 m) 3 (6.672 × 10 −11 m 3 kg −1s −2 ) × (1.77 × 24 × 60 × 60 s) 2

= 1.97 × 1027 kg 2. According to the Stephan-Boltzmann law, the amount of energy radiated by a black-body per unit time per unit area at temperature T is given by E=σT4

(i)

where σ is Stephan constant. The energy radiated by the Sun can also be expressed in terms of its luminosity LΘ as E=

LΘ

(ii)

4πR 2

where R is the radius of the Sun. Comparing the above two expressions, we can write LΘ

4πR

2

= σT 4

Rearranging the terms and substituting the values of R (= 6.7 × 108m), −8 −2 −4 26 σ (= 5.67 × 10 W m K ) and L = (3.86 × 10 W), we get,

T= 24

LΘ

4πR 2 σ

1/ 4

=

=

1/ 4

3.86 × 10 26 W

The Sun

4 × (3.14) × (6.7 × 10 8 m) 2 × (5.67 × 10 −8 Wm − 2 K − 4 ) 1/ 4

3.86 × 10 2 4 × 3.14 × (6.7) 2 × 5.67

× 1016 K 4

= 0.5895 × 104K ≈ 6000 K. 3. a) It is because the density of matter in photosphere is much higher than the density of matter in chromosphere and corona. b) We know that the ionisation energy of the hydrogen atom is 13.6 eV. Further, energy acquired by a particle at temperature T is kBT. If this energy is equal to the ionisation energy of the hydrogen atom, it will be ionised. Thus, we must have k BT = (13.6 × 1.6 × 10 −19 ) J T=

(13.6 × 1.6 × 10 −19 J) (1.38 × 10 −23 JK −1 ) 4

≈ 13.6 × 10 K. 4. a) Motion of sunspots. b) Spicules are the jet like structures comprising hot gas and is observed throughout the chromosphere. Prominences look like structures surging up into the corona. 5. Ohm’s law can be written as: J = σ (E + v × B) Taking curl of both sides, we get: ∇ × J = σ (∇ × E + ∇ × (v × B)) Using Eqs. (5.4) and Eq. (5.5) we can write: 1 σµ 1 σµ

(∇ × ∇ × B) = −

∂B ∂t

+ ∇ × ( v × B)

[∇(∇. B) − (∇.∇) B] = −

∂B ∂t

+ ∇ × ( v × B)

From Eq. (5.3), we have ∇ . B = 0. Thus, we get ∂B ∂t

= ∇ × ( v × B) +

1 σµ

∇2 B

25

The Solar System and Stars

= ∇ × ( v × B) + η∇ 2 B

which is Eq. (5.7). 2

6. We know that the pressure generated by magnetic field B is equal to B /2µ. Thus, for generating pressure equal to 103 Pa, we must have

B2 ≈ 103 Pa 2µ Substituting µ = 4π × 10−7 NA−2 in the above expression, we get:

B ≈ 5 × 10 −2 T = 5 × 10 2 G. Terminal Questions 1. The magnetic Reynold number is given by (Eq. (5.10)):

Rm =

VL η

Substituting the values of V, L and η from the problem, we get Rm =

(10 3 ms −1 ) × (10 7 m ) 10 3 m 2 s −1

≈ 107 Yes, conductivity can be taken to be virtually infinite because Rm >> 1. 2. On the basis of the equation of state for the sunspot and using the fact that magnetic pressure is equal to B2/2µ, we can write: B2 = Nk B (T2 − T1 ) 2µ where N is number density. Since µ = 4π × 10−7 NA−2 , T2 = 6000 K and T1 = 4000 K, we can write:

B2 = 2 × (4π × 10−7 NA−2) × (1023 m−3) × (1.38 × 10−23J K−1) × (2000 K) B = 0.08 T = 800 G 3. Velocity of Alfven waves is given by vm =

B 4πρ

where ρ is density. Substituting the values of B and ρ, we get: 1G v = 20 m 4π × (10 × 1.6 × 10 − 24 g cm −3 ) 25 cm s−1. 26

UNIT 6 THE SOLAR FAMILY

The Solar Family

Structure 6.1

Introduction Objectives

6.2 6.3 6.4 6.5 6.6 6.7 6.8

Solar System: Facts and Figures Origin of the Solar System: The Nebular Model Tidal Forces and Planetary Rings Extra-Solar Planets Summary Terminal Questions Solutions and Answers

6.1

INTRODUCTION

In Unit 5, you have learnt about various features of the Sun such as solar atmosphere, solar activity and energy transportation. You know that the Sun is the only star in our solar system. It is the most massive object in the system and all the planets revolve around it. You have already studied about the solar system in your school science course as well as in Foundation in Science and Technology (FST-1) course. Therefore, you know most of the facts and figures such as mass, density, distance from the Sun, and surface temperature of all the nine planets of the solar system. However, being a student of physics and astronomy, you may not be satisfied only with the facts and figures and would like to know: How did the solar system come into being? Is it possible to explain the observations pertaining to the solar system on the basis of the principles of physics? Do all the stars have planetary systems similar to our solar system? Does life exist on any other planet? In the present unit, we shall address some of these issues. In Sec. 6.2, you will recapitulate some facts and figures about the solar system. This is necessary because any model for the formation of the solar system must be consistent with these observations. In Sec. 6.3, you will learn the nebular model which is at the core of all the contemporary theories of formation and evolution of the solar system. In this section you will also discover that the nebular model explains most of the dynamic properties of the solar system. Further, it has been argued that the gravitational force gives rise to tides in the oceans on the Earth and planetary rings around the outer planets such as Jupiter and Saturn. The genesis of tidal forces and planetary rings has been discussed in Sec. 6.4. And, in Sec. 6.5, you will learn about the efforts made by astronomers to investigate the existence of extra-solar planets. Objectives After studying this unit, you should be able to: •

describe the planets of the solar system;

•

understand how the terrestrial planets differ from the jovian planets;

•

explain the nebular model of the solar system;

•

discuss the role of gravitational forces in generating tides and in the formation of Earth’s tidal bulge and planetary rings; and

•

discuss the possible existence of extra-solar planets.

27

The Solar System and Stars

In a recent development, the International Astronomical Union (IAU) has decided to remove Pluto from the list of planets in the solar system. So, the solar system now has only eight planets. Pluto has now been categorised as an object of the Kuiper belt, found in the outer region of the solar system. Kuiper belt contains many objects of the size of Pluto. In fact, it was the realisation that many more objects of the size of Pluto could be discovered in future that prompted IAU not to consider Pluto as planet.

6.2

SOLAR SYSTEM: FACTS AND FIGURES

The solar system consists of the Sun, nine planets (see the margin remarks), satellites of planets, asteroids and comets. The nine planets, arranged according to their increasing distances from the Sun, are: Mercury (Buddha), Venus (Shukra), Earth (Prithvi), Mars (Mangal), Jupiter (Brihaspati), Saturn (Shani), Uranus (Arun), Neptune (Varun) and Pluto (Yama). The sizes of these planets with respect to the Earth are shown in Fig. (6.1). Note in Fig. 6.1a that the sizes of the first four planets are similar to that of the Earth and they are called terrestrial planets. On the other hand, from Fig. 6.1b, it is obvious that the sizes of the next four planets are bigger than the Earth. They are called jovian planets. The status of the ninth planet, Pluto, is somewhere in-between.

Relative sizes of the terrestrial planets (Earth = 1)

Relative sizes of the jovian planets (Earth = 1) R = 11.3

R = 0.38

R = 4.0

Mercury

Uranus R = 1.0

R = 0.95

R = 1.0

Jupiter

Earth R = 9.4 Venus Earth

R = 3.9

R = 0.53 R = 0.26

Moon

Neptune

Saturn R = 0.18

Mars Pluto (a)

(b)

Fig.6.1: Sizes of a) terrestrial; and b) jovian planets relative to the Earth

The properties of the terrestrial and jovian planets are different. For example, terrestrial planets are mainly made of rocks and metals having an average density of 4 or 5 g cm−3 whereas jovian planets consist mainly of gas and ice with an average density of 1 or 2 g cm−3. All these planets revolve around the Sun in elliptical orbits. The planetary orbits are almost in the same plane except that of Pluto which is inclined at an angle of ~ 17° to the common plane (Fig.6.2). Asteroids are believed to be the captured objects which were wandering in the solar system. Their orbits are mostly located in between the orbits of Mars and Jupiter. Since all the planets revolve round the Sun, it is considered the ‘head’ of the solar family. In addition, the Sun contains almost 99.87% of the total mass of the solar system. Among the planets, Jupiter and Saturn are the most massive, accounting for 92% of the mass of all the planets. 28

The Solar Family

Earth Mercury Sun

Venus

Mars

(a)

Jupiter Uranus Saturn

Neptune

Pluto

(b) Fig.6.2: Orbits of the a) terrestrial; and b) jovian planets in the solar system

Table 6.1 summarizes some basic data about these planets and the Moon. Many of these features could directly be attributed to the distance of a planet from the Sun. Table 6.1: Some basic data of planets and the Moon Planets

Mass [kg]

Mercury

3.30×10

23

4.87×10

24

5.97×10

24

5.5

6.42×10

23

3.9

1.90×10

27

5.69×10

26

1.1

Venus Earth Mars Jupiter Saturn

Density [g cm−3]

5.4 5.2

1.3 0.7

Mean distance from the Sun (km)

Rotation Period (hours or days)

Revolution Period (days)

Inclination of orbit to the plane of the Solar System (deg)

Percentage of light reflected

5.8×10

7

58 d

88

7.0

7

1.1×10

8

243 d

245

3.4

76

1.5×10

8

23 h 56 min

365.25

0.0

39

2.3×10

8

24 h 37 min

687

1.8

18

7.7×10

8

9 h 55 min

4333

1.3

51

1.4×10

9

10 h 30 min

10743

2.5

50

8.70×10

25

2.8×10

9

17 h 14 min

30700

0.8

66

Neptune

1.03×10

26

1.7

4.5×10

9

18 h

60280

1.8

62

Pluto

1.00×10

22

2.1

5.9×10

9

6 d 09 h 17 min

90130

17.1

15

5

27.33 d

27.33

5.0

Uranus

22

Moon 3.3 7.35×10 3.8×10 * * This is Moon’s mean distance from the Earth.

SAQ 1 Using the data given in Table 6.1, verify Kepler’s third law: T2 r3

Spend 5 min.

= Const .

where T is the orbital period of a planet and r is its mean distance from the Sun. Further, as mentioned above, the division of planets into two groups namely, terrestrial and jovian, is based on the similarity of some major characteristics of the planets in a particular group. Table 6.2 lists these characteristics of terrestrial and

29

The Solar System and Stars

jovian planets. Such classification helps in developing theoretical models for their formation and evolution. Table 6.2: Characteristics of the terrestrial and jovian planets Characteristics Basic form

Terrestrial

Jovian

Rocky

Gas/Liquid/Ice

Mean orbital distance (AU)

0.39−1.52

5.2−30.1

Mean surface temperature (K)

200−700

75−170

Mass (relative to the Earth)

0.055−1.0

14.5−318

Equatorial radius (relative to the Earth)

0.38−1.0

3.88−11.2

Mean density (g cm−3)

3.96−5.42

0.68−2.30

Sidereal rotation period (equator)

23.9 h − 243 h

9.8 h − 19.2 h

Number of known Moons

0−2

8−20

Ring systems

No

Yes

We present some salient features of the individual planets in Table 6.3. Table 6.3a: Salient features of the terrestrial planets Mercury • It can be seen near the horizon at sunset or sunrise with unaided eyes. • Like the Moon, it has no atmosphere and its surface is full of craters. • It is, at the same time, the coldest and the hottest planet because its periods of revolution and rotation are almost equal which keeps its same surface face the Sun all the time. • Its surface temperature varies between + 340°C to − 270°C.

30

Venus • It appears as the third brightest object in the sky after the Sun and the Moon, as it is nearest to the Earth. • Its surface is dry, hot and volcanic. Its atmosphere contains about 96 percent carbon dioxide, 3.5 percent nitrogen and remaining half percent is water vapours, argon, sulphuric acid, hydrochloric acid etc. • The planet is covered by a thick cloud mainly consisting of sulphuric acid droplets. • Its surface temperature is very high (~ 470°C) which is perhaps caused due to the greenhouse effect: the infrared radiation emitted by the planet is not allowed to escape due to the presence of carbon dioxide in its atmosphere thus causing the heat received from the Sun to be trapped and raise its temperature.

Earth

Mars

• Its crust, extending to 10 km deep under the oceans and up to 40 km under the continents, consists mainly of silicon (27.7%), and oxygen (47.3%). Elements like aluminium, iron, calcium, sodium etc. make the bulk of its matter and less than 2% is made of all other elements.

• Though half in size of the Earth, this planet has various similarities with the Earth

• Its rotation axis is tipped by ~ 23° causing various seasons and polar caps. • Its atmosphere consists of distinct layers called troposphere, stratosphere and ionosphere. Troposphere comprises mainly of 78% nitrogen, 21% oxygen; stratosphere contains ozone which absorbs the harmful ultraviolet radiation from the Sun.

i) A day on Mars is 24h 40 minutes and a year lasts for 1.88 Earth year, ii) its rotation axis is tipped at 25° and thus has seasons and polar caps. • This is the most extensively probed planet and a few automated laboratories have also been landed. The atmosphere and geology of this planet has many features similar to the Earth. • Martian surface has craters of all sizes and enormous volcanoes. • Martian soil, like the Earth, is mostly made of silicates. However, due to the presence of 16 percent iron oxide in its soil, it has the characteristic red colour. It is also known as the Red Planet.

The Solar Family

Table 6.3b: Salient features of the Jovian planets Jupiter

Saturn

Uranus

Neptune

Pluto

• It is the largest (having diameter 11.2 times that of the Earth) and the most massive planet (contains 71% of all the planetary mass).

• It is the second largest planet exceeded in size and mass only by Jupiter. Like Jupiter, it consists mainly of hydrogen and helium.

• This planet is smaller than Jupiter and four times farther from the Sun. In a telescope, it appears as a green disc with vague markings.

• This planet was discovered in 1846 and it is so far away that, seen from this planet, the Sun would look like a bright spot!

• This planet was predicted to exist theoretically to account for the observed irregularities in the orbits of Uranus and Neptune.

• It is the last (sixth) planet visible from the Earth. It has beautiful rings (which can be seen through a telescope). The rings are, in fact, a thousand tightly packed individual ringlets.

• Its axis of rotation is tipped 97.9° from the perpendicular to its orbit. This causes its poles to nearly point towards the Sun and it would seem that the planet is rolling along the orbit like a wheel.

• It is like a spinning ball of gas and liquids with no solid surface. In this regard, it is similar to the Sun. It has a large number of satellites. • It is covered by a turbulent, gaseous atmosphere comprising of hydrogen, helium and small traces of water vapour, ammonia, methane etc. • It has a Great Red Spot on its surface which has an oval shape. It is big enough to accommodate two Earths! It is presumed to be due to huge cyclonic disturbance in its atmosphere. • In view of its composition, size and the number of Moons, Jupiter looks like a star having its own ‘solar system’! • It has rings and it emits radio waves.

• The temperature at its cloud tops is − 180°C. It is colder than Jupiter. • This planet is less dense than water and being mostly liquid and rotating rapidly, its shape is flattened. • It has a large number of satellites orbiting at the edge of the rings. These satellites (Moons) are composed of rock and ice and have craters.

• The rings of this planet were discovered as late as 1977 (Voyager 2) and they comprise of very dark material, as black as coal. • It has a large number of Moons.

• Its colour is faint blue which is caused due to larger percentage of methane present in it. • Its cloud temperature is about − 237°C. • Like other jovian planets, this planet also has rings. • Two Moons of this planet are visible from the Earth. • Triton is the largest Moon of this planet which is orbiting it in the clockwise direction i.e. opposite to the rotation of the planet. Triton also has atmosphere of its own comprising of nitrogen and methane.

• It is a very small (about one-fifth the size of the Earth), cold and dark planet. • Unlike most planetary orbits, Pluto’s orbit is quite elliptical and therefore it can come closer to the Sun than Neptune. • Being so far away from the Sun, this planet is cold enough to freeze most compounds. • Its mass is only 0.002 times the Earth mass!

• Its main satellite, Titan, is very large (diameter 5800 km) and has atmosphere of its own as dense as ours.

31

At this stage, you may pause for a moment and think about what have you learnt till now. You have basically recapitulated what you learnt in school about the solar system and have acquainted yourself with the characteristic features of each of its planets. In the next section, you will learn about the origin of the solar system. A model for the origin of the solar system must explain its characteristic features listed below:

The Solar System and Stars

i)

Most of the mass of the solar system is contained in the Sun.

ii) Except for Mercury and Pluto, the orbital planes of all the planets are more or less in the same plane. iii) When viewed from above, the planets are found to revolve around the Sun in the anticlockwise direction; the direction of rotation of the Sun is also the same. iv) Except for Venus, Uranus and Pluto, the direction of rotation of planets is the same as their direction of revolution. v)

The direction of revolution of the satellites of each planet is the same as the direction of rotation of the planet itself.

vi) Total angular momentum of all the planets is more than the angular momentum of the Sun. vii) Terrestrial planets comprise mainly of rocky material whereas jovian planets comprise mainly of gaseous material. Spend 3 min.

SAQ 2 List some common features of terrestrial and jovian planets.

6.3

ORIGIN OF THE SOLAR SYSTEM: THE NEBULAR MODEL

The formation of the Earth or, in fact, the entire solar system has been of considerable interest to human being for ages. A variety of speculative ideas were proposed which gave rise to two kinds of theories: catastrophic and gradualistic. According to the catastrophic theories, the planets were formed out of the material ejected from the Sun when a giant comet collided with it or the planets came into being due to ripping-off of material from the Sun caused by the tides generated by a close by passing star. However, these theories do not explain many features of the solar system. Further, during the 20th century, astronomical evidences have been found which support the gradualistic theories. According to these theories, formation of planets is a gradual process and is a natural by-product of the formation of stars like the Sun. Contemporary gradualistic theories of the solar system are based on the nebular hypothesis. According to this hypothesis, the Sun as well as the planets formed from an interstellar cloud of gas and dust. A model of the formation of the solar system based on this hypothesis is called a nebular model. The basic principle of physics underlying the nebular model is the Newton’s law of gravitation. According to the nebular model, formation of stars including the Sun, begins when the interstellar cloud with enough mass and low internal pressure, contracts due to its own gravity.

32

Can you guess the consequences of gravitational contraction of an interstellar cloud? You are on the right track if you think that it leads to increase in density of the cloud. Simultaneously, the kinetic energy of the particles in the cloud increases which

causes increase in its temperature. As a result, the internal pressure of the collapsing cloud increases. Eventually, the gravitational contraction is balanced by the internal pressure of the cloud. The contracting cloud is called a protostar and when its internal temperature is high enough to initiate thermonuclear reactions, a full-fledged star (such as the Sun) is born. This is the general picture of the formation of a star according to the nebular model.

The Solar Family

Well, you may ask: How did our solar system come into existence? According to the nebular model, our solar system formed due to gravitational contraction of the rotating interstellar cloud called the solar nebula. Due to rotation, the solar nebula takes the form of a disc (Fig. 6.3). When the Sun formed at the centre of this disc and became luminous enough, the remaining gas and dust was pushed away due to the Sun’s radiation pressure. The blown away gas and dust condensed into various planets orbiting the Sun. One of the natural consequences of the solar nebula model is that most stars in the galaxies should have planetary systems!

(a)

(b)

(d)

Venus Jupiter

(e) Uranus

Saturn Asteroids Pluto

Neptune (c)

Mars

Mercury Earth

(f)

Fig.6.3: Schematic diagram showing different stages (a to f) of the formation of the solar system

Formation of Solar Nebular Disk You must have noted that one of the significant features of the solar nebular model is the formation of a disk of interstellar cloud around the Sun. You may ask: Is there any evidence supporting the formation of the disk? Two facts support this assumption. Firstly, as you know, the orbits of planets, except that of Mercury and Pluto, lie in nearly the same plane. Such coplanar planetary orbits are consistent with the formation of nebular disk. Secondly, the rotation of the Sun and revolution of the planets is along the same direction and their equatorial planes are very close to the plane of the solar system. Thus, the motion of the Sun and the planets are consistent with the disk hypothesis. Your next logical question could be: Can we explain the formation of nebular disk on the basis of physical principles? It can indeed be done provided we assume that the interstellar cloud, giving rise to the solar system, had some initial rotational motion. Having assumed this, we can invoke the principle of conservation of angular momentum to explain the formation of a disk. As the cloud contracted due to gravity, each gas and dust particle would come closer to the axis of rotation. To conserve the angular momentum during the contraction, the particles coming nearer to the axis of rotation must revolve faster. At some point, due to the increased speed of revolution,

33

the centripetal acceleration of the particles of the cloud balance the gravitational contraction and equilibrium is attained.

The Solar System and Stars

To understand the flattening of the rotating cloud, refer to Fig. 6.4. Particles at points A (near the pole) and B (near the equator) have the same angular momentum because points A and B are equidistant from the axis of rotation. Thus, it is energetically more favourable that particle at A falls under gravity along the line AB and reaches the point B (closer to the centre) than say a particle at point C to come to point B. It is so because, in the first situation, no change in angular momentum is involved. Thus, rotating cloud near the poles contracts more than those near the equator giving rise to the formation of disk.

(a)

(b)

(c)

Fig.6.4: a) Interstellar cloud with initial rotational motion; b) due to gravitational attraction, material at the poles contracts along the rotation axis; and c) the final disk shaped solar nebula

The principle of the conservation of angular momentum also explains why most of the planets rotate along the same direction. It is because each planet has retained some of the angular momentum of the solar nebula. Let us discuss the angular momentum of the solar system to appreciate this point better. Angular Momentum of the Solar System The present distribution of angular momentum in the solar system seems inconsistent with the basics of the nebular model. The argument goes like this. Since the Sun and the planets are formed from the same spinning interstellar cloud, the angular momentum per unit mass of each of them must be same. The facts are otherwise: the Sun possesses approximately 99.9% of the total mass of the solar system but only 1% of its total angular momentum; orbital angular momentum of Jupiter exceeds the rotational angular momentum of the Sun by a factor of 20; among the planets having 99% of the total angular momentum of the solar system, Jupiter possesses the most of it! To appreciate these facts and figures, you should solve an SAQ. Spend 5 min.

SAQ 3 Calculate the total angular momentum of the Sun-Jupiter system assuming that Jupiter has a circular orbit of radius 5.2 AU, and its orbital period is 11.86 yr. Assume that the Sun interacts only with Jupiter.

34

You may ask: How does the nebular model explain the inconsistency in the distribution of angular momentum? It is proposed that during the formation of the solar system, the angular momentum is transported from the central part of the nebula

to the outer regions. Two possible mechanisms for such a transfer have been suggested. According to the first mechanism, the interaction of the charged particles and the magnetic field of the evolving Sun causes transfer of angular momentum. You know that charged particles spiral along the magnetic field. Thus, the charged particles created by ionisation of solar nebula by the Sun are dragged along by the magnetic field of the rotating Sun (Fig. 6.5). The magnetic field links the outer nebular matter (charged particles) with Sun’s rotation. As a result, the nebular material in the outer region gains angular momentum at the expense of the Sun. The other mechanism suggested for transfer of angular momentum is based on viscosity. You know that due to viscosity, motion of fluid in one part is affected by the motion of the fluid in the adjoining part. Thus, it is quite possible that the slow moving particles located at outer edges of the nebula gain in velocity due to their interaction with the fast moving particles in the smaller orbits of the nebula and vice-versa. This may cause transfer of angular momentum from the Sun to the outer planets of the solar system.

The Solar Family

Now, you will learn about formation and evolution of planets according to the nebular model. Formation of Planets You know that the solar nebula consisted of gas and dust. As the solar nebula contracted, it became hot enough and most of the dust particles evaporated. So, the solar nebula consisted mainly of gaseous matter. The question is: How did the planets form from the solar nebula? The formation of planets is a two-stage process: firstly, small solid particles are formed from the gaseous matter and then these particles stick together and grow into planets.

Fig.6.5: Magnetic lines of force of the rotating Sun

Depending upon the temperature of the nebular region, the nebular gas condensed into solid matter of different types. In the inner region, the temperature was very high and materials with very high melting points were formed. The sequence of condensation of gas into different types of materials, from the centre of the nebula (the Sun) to its periphery, is called the condensation sequence and is given in Table 6.4. Table 6.4: The condensation sequence Temperature (K)

Material(s) formed

Planet (Temperature of formation)

1500

Metal oxides

1300 1200

Metallic Iron and Nickel Silicates

1000

Feldspars

Venus (900 K)

680

Troilites (FeS)

Earth (600 K) Mars (450 K)

175

H2O, Ice

Jovian (175 K)

Argon-neon Ice

Pluto (65 K)

65

Mercury (1400 K)

From Table 6.4, it is evident that terrestrial planets formed from high density materials and jovian planets formed from low density materials.

With the help of computers, it is possible to study the behaviour of a large number of particles when they interact in the manner as in the process of accretion. Such computer experiments are called simulation. The results of such simulation studies suggest that there could be as many as 100 planetesimals of the size of the Moon along with a large number of smaller objects in the region of inner planets. In addition, there could be about 10 bodies with masses comparable with the planet Mercury and many other objects of the size of the planet Mars.

35

The Solar System and Stars

Having discussed the general chemical composition of planets, we now focus on the evolution of planets. The evolution of planets involves three stages: a) The first stage involves the growth of macroscopic grains of solid matter from the interstellar cloud. The size of these grains range from a few cm to a few km and they are called planetesimals. Planetesimals can grow through two processes: condensation and accretion. In the condensation process, grains grow by adding one atom at a time to a ‘nucleus’ atom, from the surrounding gaseous cloud. This is similar to the growth of snowflakes in the Earth’s atmosphere. In the accretion process, solid particles stick together. Further, the planetesimals would tend to rotate in the plane of the solar nebula. b) In the second stage, planetesimals coalesce and form protoplanets − objects having planetary sizes and masses. You may ask: How do the protoplanets form? Since all the planetesimals are moving along the same direction in the nebula, they collide with each other at a low relative velocity and stick together to form protoplanets. Further, growth of protoplanets is helped by gravity because the nebular matter is attracted by the protoplanets. c) At the third stage, when a protoplanet grows into a stable planet, a large amount of heat is generated in its core due to the decay of short-lived radioactive elements. Heat is also generated due to collision of these planets with other objects. Due to high temperature, the planets melt and facilitate the process of gravitational separation in which materials in the planet segregate themselves according to their density. Therefore, the inner regions of the planets hold heavier elements and compounds and lighter elements are pushed to the surface.

The creation of a massive object, like Jupiter, influences the orbits of nearby planetesimals. Most of the objects present in the asteroid belt today had their orbits changed gradually into more eccentric orbits till they were sucked in by Jupiter. It is also likely that they either left the solar system or crashed into the Sun. This process might in fact have resulted in the smaller mass left in the asteroid belt and also a smaller planet Mars.

As we go further from Jupiter and beyond in the nebula, the material density becomes very low. So, the accretion process for the formation of a planet-like object takes much longer time. Saturn perhaps took twice as long to form as Jupiter, while the planet Uranus took an even longer time. The planet Neptune is estimated to have taken twice the time it took to form Uranus.

36

This, in the nutshell, is the ‘story’ of planet building! You must have noted that there are many unanswered questions and many ifs and buts in the mechanism described above. To understand the details of formation of planets is still an active area of research in astronomy and astrophysics. We will, however, not go into those details and close our discussion with a few words about the success of the nebular model. The solar nebular model successfully accounts for the following three important features of the solar system: i)

Disk shape of the solar system: The model suggests that the rotating solar nebula ultimately evolves into disk shape due to gravitational contraction and conservation of angular momentum.

ii) The orbits of most of the planets are coplanar: As per the model, this situation exists because of the disk shape of the solar nebula from which planets are formed. iii) The direction of rotation of the Sun and the directions of revolution of most of the planets are the same: It is so because the Sun and the planets formed from the same rotating nebula. You have also learnt that the condensation sequence of the solar nebula explains why terrestrial planets comprise of compounds having high melting point and jovian planets comprise of ices and gases. The nebular model has some very obvious limitations. You know that the Moon’s surface is not smooth; it has craters of varied sizes, small and large. Also, its surface composition is extremely poor in hydrogen, helium etc. These observations suggest the continued collisions of bigger planetesimals with the Moon’s surface even after its formation. It is quite likely that other objects (planets) in the solar system experienced similar collisions with planetesimals. Thus, the theory of solar system formation must account for massive encounters endured by planets in the early stages of their

The Solar Family

formation. These massive collisions or encounters are possibly responsible for different orientation of the spin axis of the planets in the solar system. It is now known that Venus, Uranus and Pluto have retrograde rotations. The cause for tipping of rotation axis, retrograde rotation etc. of planets cannot be understood as such by gradualistic models like the nebular model. We need to consider catastrophic events. During the formation of the solar system, the wandering planetesimals did perhaps collide with other planets such as Mercury, Venus, Earth etc. The collision with Mercury ripped off its low density mantle while such a collision with Venus flipped its rotation axis. A collision with Earth led to the formation of the Earth-Moon system. Further, massive planetesimals crashed into Mars as well as other outer planets changing orientation of their rotation axes. It is also possible that some of these planetesimals were captured by planets as their Moons. Now, before you proceed further, how about answering an SAQ? SAQ 4

Spend 7 min.

a) List two evidences supporting the assumption that a disk shaped solar nebula existed during the evolution of the solar system. b) Why do terrestrial planets comprise mainly of materials having high melting points? c) What is the difference between the condensation and accretion processes?

6.4

TIDAL FORCES AND PLANETARY RINGS

You must be aware of tides occurring on the Earth’s surface. In fact, those of you who live in the coastal areas must be quite familiar with tides. Two high tides occur once in every ~ 24 hours depending on the local features of the coastal area and its latitude. However, generally people are not familiar with the tidal bulge of the Earth along the equator which measures around 10 cm. Do you know what causes tides or tidal bulge of the Earth? It is caused due to the difference between gravitational force of the Moon at different locations, say points A and B, on the Earth (Fig. 6.6). Due to similar reasons, tidal bulge (as large as 20 m) caused by the Earth has also been observed on the Moon. The larger value of the bulge of the Moon is because the Earth is more massive than the Moon. Tidal effects play an important role in astronomy, particularly in understanding the creation of large number of satellites and ring systems in the jovian planets. We now derive an expression for the tidal force for the Earth-Moon system. Then we shall use this concept for a qualitative explanation of the formation of planetary rings.

A Moon

B Earth

Fig.6.6: Tidal bulge of the Earth is due to the difference in gravitational force of the Moon experienced by mass elements at, say points A and B, on the Earth.

37

The Solar System and Stars

Let a mass m on the Earth be located at a distance r from the centre of mass of the Moon (Fig. 6.7). If the mass of the Moon be M, the magnitude of the gravitational force acting on m, in the direction shown, is given by: F =G

Mm

(6.1)

,

r2

where G is the gravitational constant.

dr

r M

m

m

Moon Earth

Fig.6.7: Gravitational force on an element of mass m located on the Earth at a distance r from the centre of mass of the Moon

Now consider a similar mass located at a distance dr from the earlier mass element along the same line. The difference between the gravitational forces experienced by these two mass elements can be obtained by differentiating F in Eq. (6.1):

dF = −2G

Mm dr r3

(6.2)

Note that the difference in the gravitational forces on the two equal masses separated by a distance dr will cause them to move with different accelerations. The differential force given by Eq. (6.2) is called the tidal force. The r3 term in the denominator of Eq. (6.2) clearly shows that the tidal force has a stronger dependence on distance compared to the gravitational force (which varies as 1/r2). Further, in case of the Earth-Moon system, the tidal force becomes more pronounced if the mass element is closer to the Moon. Now to appreciate the effect of the tidal force of the Moon on mass elements located at different points on the Earth, e.g., at the equator and at the poles, let us obtain a general expression for the tidal force considering the Earth as a two-dimensional object. Let us consider an element of mass m located at the centre (C) of the Earth and another element of mass m located at an arbitrary point (P) at latitude φ on the surface of the Earth (Fig. 6.8). By assuming that the Moon lies along the x-direction, the components of gravitational force at points C and P can be written as: FC , x = G

Mm r2

,

FC , y = 0,

and FP, x = G

Mm s2

FP , y = −G 38

cos θ ,

Mm s2

sin θ.

(6.3)

The Solar Family

where s is the distance between points P and the centre of mass of the Moon.

y

x P R

s

φ

r

Moon

C

Earth

Fig.6.8: The schematic diagram of the tidal force on an arbitrary point on the Earth due to Moon

Let the unit vectors along the x- and y-directions be ˆi and ˆj , respectively. Thus, from Eq. (6.3), the difference between the magnitudes of gravitational forces at the points P and C can be written as: ∆F = FP − FC

= GMm

cos θ 1 ˆ sin θ ˆ − 2 i − GMm j 2 s r s2

(6.4)

Further, s can be expressed in terms of r, R and φ as:

s 2 = (r − R cos φ) 2 + R 2 sin 2 φ ≈ r 2 1−

2R cos φ r

2

Substituting for s in Eq. (6.4) using the small angle approximations, cosθ ≈ 1 and R sinθ ≈ sin φ, and using binomial expansion we obtain: r

∆F ≈

GMmR (2 cos φ ˆi − sin φ ˆj) r3

(6.5)

SAQ 5

Spend 6 min.

Derive Eq. (6.5). You may ask: What does Eq. (6.5) signify physically? This expression clearly indicates that the magnitude of the tidal force is dependent on the latitude. For mass element located at the equator of the Earth, that is, for φ = 0, the magnitude of tidal force is maximum. And, at the poles, where φ = π/2, the magnitude of the tidal force is minimum. This is the cause of tidal bulge around the equator and also causes tides in the oceans.

39

The Solar System and Stars

In the above discussion, we have considered the Earth-Moon system to be an isolated one. In fact, we must also consider the tidal forces due to the Sun. When the Sun, Earth and the Moon are all aligned in a straight line, the differential forces discussed above add up to produce large tides on the Earth. These tides are called spring tides. When the Sun, Earth and the Moon form a right angle, the differential forces due to the Sun and the Moon are directed opposite to each other and the tides on the Earth (known as neap tides) are very small. Yet another effect of tidal force is that the Earth’s speed of rotation is slowed down. This results in longer days at present compared to many years ago. Further, just as the Moon causes tides on the Earth, the Earth also gives rise to tides on the Moon. You know that on the Earth we see the same side of the Moon which implies that its rotation and revolution periods are the same. It is quite likely that earlier the Moon’s rotation period was shorter than its orbital period. As time progressed, because of the tidal friction, the rotation period of the Moon increased and has become equal to its orbital period (also called one-to-one synchronous rotation). We find such synchronous motion to be quite a common phenomenon in the solar system. For instance, the two Moons of Mars and four Moons of Jupiter and majority of Saturn’s Moons are in synchronous rotation. Many moons of the outer planets behave in a similar manner. You know that most outer planets of the solar system have rings around them −5 comprising of small (~ 10 µm to 10 m) particles. The concept of tidal forces can be invoked to understand these ring systems. There are two possible scenarios. Since the jovian planets are massive, tidal gravitational force of the planets on their satellites must be very strong. Thus, if a satellite comes very close to, say, Jupiter, it experiences a strong tidal force and breaks up into pieces. The particles formed during such a process revolve around the planet in ring formation. According to the second scenario, during the formation of jovian planets, the tidal forces restrained the particles from condensing into a satellite. You may argue: The tidal forces must ultimately disrupt the ring system; what maintains the ring system? It is suggested that the Moons associated with these planets play an important role in preserving the rings. The particles in the rings are restrained from moving out of their orbit due to the combined gravitational forces of these Moons. Human beings have always wondered about the existence of planets around other stars. Only in recent times there has been a confirmation that planets do exist outside the solar system. These planets are called extra-solar planets. You will learn about them now.

6.5

Fig.6.9: Photograph of an extra solar planet taken by Hubble telescope

40

EXTRA SOLAR PLANETS

On August 4, 1997, the Hubble Space telescope took the first image of an extra-solar planet around another star (Fig. 6.9). The picture shows a double star located at about 450 light years away towards the constellation of Taurus. It was the first direct look at a planet outside our solar system. It is now believed that this planet is a “runaway” object, thrown out of the binary system, as indicated by the filament tracing. The planet is located at about 1500 times the Earth-Sun distance from its parent. The binary system is believed to be 300,000 years old and the planet seems to be quite similar to Jupiter with a mass of around 2-3 times that of Jupiter. This discovery led to careful investigation by several groups of astrophysicists and by now more than 100 such extra solar planets have been found. Young planets in the new systems are difficult to detect as the parent stars overshadow their feeble glow. Since it is difficult to detect a planet orbiting a distant

star, we look for alternative methods of detection, such as the influence of the planets on their parent stars. You may like to know: What are these influences? It is easy to think of gravitational influence. As the planet orbits the star, it will tug at it from different sides causing it to wobble back and forth. The gravitational influence gives rise to the following methods for detection of the planet:

The Solar Family

a) Astrometric Detection b) Radial Velocity Detection The first method is based on measuring the position of a star relative to its background stars. If the star is accompanied by an orbiting planet, it gets a tug from the planet and its position changes a little. This change in the position of the star is measured which would show a periodic change (back and forth) indicating the presence of an orbiting object. The second method is based on Doppler effect. As an orbiting planet tugs on to its companion star, the light from the star experiences a Doppler shift. If the planet pulls the star slightly away from us, the light emitted by star would shift towards red end of the electromagnetic spectrum while if it tends to pull the star towards the Earth, the light would be shifted towards the blue end of the spectrum. To measure the Doppler shift, we choose a particular spectral line and observe its shift from red to blue and back. In 1995, Michel Mayor and Didier Qucloz of the Geneva Observatory observed that the Sun-like star, 51 Pegasi is wobbling back and forth at the rate of 56 ms−1. The only valid explanation for this observation was the presence of a planet like object orbiting the parent star. The mass of the planet was estimated to be half the mass of Jupiter with a radius of about 0.05 AU. Subsequently, Mayor also detected a planet of about 0.16 times the mass of Jupiter orbiting the star HD 83443 in the constellation of Vela about 141 light years away from the Earth. Recently David Charbonneau, Timothy Brown and Robert Gilliland used the Hubble spectrometer and made the first direct detection and chemical analysis of the atmosphere of the planet HD 209458b orbiting a yellow, Sun-like star, HD 209458, in the constellation of Pegasus. The mass of the planet is estimated to be 70% of the mass of Jupiter. It passes in front of its star every 3.5 days and contains sodium in its outer layers. The extremely short period of the planet suggests its very close proximity to the star and therefore its atmosphere getting heated to around 1100 degrees Celsius. Many new results are pouring in and are compelling astronomers to have a re-look on the theories of solar system formation. The search for extra-solar planets is going hand in hand with the search for life in the Universe. Exploration of the solar system has so far not revealed any signs of even primitive life on any of the planets or their satellites. Scientists strongly believe that there must be a large number of extra-solar planets where conditions are suitable for the existence of life. Indeed, some of these planets could be hosting life more advanced than our own. It is possible that some of these intelligent beings are trying to contact us just as we are looking for them. It must be remembered that any message from any of these beings would be coded in radio waves. That is why SETI, an organisation searching for Extra Terrestrial Intelligence, is looking for such signals in the radiation at radio frequencies coming from outside the solar system. Scientists are hopeful that someday they would be able to detect extra-solar intelligence. Let us now summarise what you have learnt in this Unit:

41

The Solar System and Stars

6.6

SUMMARY

•

The first four planets of the solar system namely Mercury, Venus, Earth and Mars are called terrestrial planets and they mainly comprise of solid matter. The next five planets, namely Jupiter, Saturn, Uranus, Neptune and Pluto are called jovian planets and they mainly comprise of ices and gases.

•

One of the significant features of jovian planets is the ring system.

•

Some of the characteristic properties of the solar system are: i)

Most of the mass of the solar system is contained in the Sun.

ii) Orbital planes of all the planets except Mercury and Pluto are coplanar. iii) The direction of rotation of the Sun and direction of revolution of the planets is the same. iv) Total angular momentum of all the planets is more than the angular momentum of the Sun. •

According to the nebular hypothesis of the origin of solar system, the Sun as well as the planets formed from an interstellar cloud of gas and dust. A model based on this hypothesis is called nebular model.

•

According to the nebular model, formation of stars including the Sun takes place when the interstellar cloud contracts due to its own gravity.

•

The contracting interstellar cloud takes a disk shape due to its rotational motion. Formation of nebular disk causes coplanar planetary orbits.

•

The present distribution of angular momentum in the solar system seems inconsistent with the basics of the nebular model.

•

The angular momentum problem is addressed by proposing that there is a transfer of angular momentum from the inner to outer regions of the solar system during its evolution.

•

The formation of planets results due to condensation of nebular gas as per the condensation sequence.

•

The formation of planets involves three stages: i) Formation of planetesimals, ii) Formation of protoplanets, and iii) Stabilisation of the planet.

•

Tidal forces result due to the difference in gravitational force at two points on the Earth. For the Earth, tidal forces due to the Moon have a significant effect. Tidal force at the equator is maximum and at the poles, its value is minimum. The general expression for the tidal force is given by ∆F ≅

•

6.7 42

GMmR r3

(2 cos φ ˆi − sin φ ˆj)

In the recent past, a few extra-solar planets have been detected.

TERMINAL QUESTIONS

Spend 20 min.

1. Explain, in your own words, the theory of the solar system formation based on nebular hypothesis. 2. Explain how angular momentum can be transferred from the Sun to the outer planets of the solar system.

The Solar Family

3. Is it possible that the Earth also suffered collisions with other bodies of the solar system and its surface also had craters like those on the surface of Mercury? Can you guess what happened to those craters? 4. Explain how extra-solar planets can be detected. Why can they not be seen directly?

6.8

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1.

To verify Kepler’s third law, let us calculate (T 2 r 3 ) for three representative planets, say the Mercury, the Earth and the Saturn. From Table 6.1, we have

TMercury = 88 days = 88 × 24 × 60 × 60 s = 7.6 × 106s and rMercury = 5.8 ×10 7 km = 5.8 ×1010 m

So, (TMercury ) 2

=

(rMercury ) 3

(7.6 ×10 6 s) 2 (5.8 ×1010 m) 3

= 2.9 ×1019 m − 3s 2 Similarly, using data from Table 6.1, you can easily show that: (TEarth ) 2 (rEarth )

3

= 2.9 ×1019 m − 3 s 2

and (TSaturn ) 2 (rSaturn )

3

= 2.9 ×1019 m − 3s 2

Since the ratio (T 2 r 3 ) for all the three planets has the same value 2.9 ×1019 m − 3s 2 , Kepler’s third law is true.

2.

See text.

3. For two bodies of mass M1 and M2 moving around their centre of mass, the reduced mass is given by: µ=

M 1M 2 M1 + M 2

So, for the Sun-Jupiter system, we can write: 43

The Solar System and Stars

µ=

MΘ M J MΘ + M J

≈ MJ Using Kepler’s third law, it can be readily shown that the angular momentum L and the period T of the orbiting mass are related as: L=

2πabµ

T

.

where a, and b respectively are the semi-major and semi-minor axes of the elliptical orbit of the planet. As per the problem, the orbit of Jupiter is to be considered circular. Thus, we can write: a = b = 5.2 AU

= 5.2 × 2.279 × 1011m Substituting the values of MJ = 2 × 1027 kg and T = 11.86 yr, we get:

L=

=

(

2 × 3.14 × (5.2 × 2.279 × 1011 m) 2 × 2 × 10 27 kg (11.86 × 365 × 24 × 3600 s)

)

17.63 × 10 51 kg m 2 4.44 × 10 9 s

= 3.9 × 10 42 kg m 2 s −1 4. See text. 5. Refer to Fig. 6.8. From Eq. (6.4), we have the difference between the magnitudes of gravitational force at points P and C as:

∆ F = FP − FC = GMm

cos θ s2

−

1 r2

ˆi − GMm sin θ ˆj s2

s 2 = r 2 + R′ 2 = r 2 + R 2 sin 2 φ = (r − R cos φ) 2 + R 2 sin 2 φ 2 2 2 2 2 ≅ r − 2r R cos φ + R cos φ + R sin φ = r 2 − 2r R cos φ + R 2 2R (neglecting higher order terms in R) ≈ r 2 1− cos φ r Substituting for s2 and making small angle approximations, cos θ ≈ 1 and R sin θ ≈ sin φ, we can write: r 44

The Solar Family

R 1

∆ F = GMm r

=

2

1−

2R r

− cos φ

GMm

r

2

r − 2 R cos φ

r

1 ˆ i − GMm r2

r r

2

R sin φ

− 1 ˆi −

r − 2 R cos φ

1−

sin φ 2R r

ˆj cos φ

ˆj

2 R cos φ ˆ R sin φ ˆj i− r − 2 R cos φ r − 2 R cos φ

=

GMm r2

≅

GMm R 2 cos φ ˆi − sin φ ˆj r3

(

)

(because r>> 2 R cos φ).

Terminal Questions

1. 2. 3. 4.

See text. See text. See text. See text.

45

The Solar System and Stars

UNIT 7 STELLAR SPECTRA AND CLASSIFICATION Structure 7.1

Introduction Objectives

7.2 7.3

Atomic Spectra Revisited Stellar Spectra Spectral Types and Their Temperature Dependence Black Body Approximation

7.4 7.5 7.6 7.7 7.8

H-R Diagram Luminosity Classification Summary Terminal Questions Solutions and Answers

7.1

INTRODUCTION

In Unit 5, you have studied about the Sun and in Unit 6, you have learnt about the solar system. You also learnt the characteristic features of the solar atmosphere and solar activity. You now know that the Sun is the nearest and the only star in our solar system. All the stars, including the Sun are located at very large distances from the Earth. Thus, a logical question is: How do we obtain information about the stars? All the information about stars is obtained by analysing their spectra and this is the subject matter of the present Unit. You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course that the radiation emitted by an object at a given temperature covers a range (spectrum) of wavelengths with a characteristic peak wavelength. The value of the characteristic wavelength depends on temperature of the object. Therefore, by a careful analysis of the radiation emitted by a star, we can estimate its temperature. In addition, we can also obtain useful information regarding composition, pressure, density, age etc. of a star on the basis of its spectrum. In Sec. 7.2, we briefly recapitulate the atomic origin of emission and absorption spectra and explain how the temperatures and luminosities of stars can be inferred from their spectra. One of the earliest uses of stellar spectra was to classify stars on the basis of strength of certain spectral lines. Later on, it was discovered that the relative strengths of spectral lines depend basically on the star’s temperature. The spectral classification has been discussed in Sec. 7.3. The most comprehensive classification of stars was done on the basis of the correlation between their luminosities (an observable parameter of stars) and temperatures. This classification gave rise to HertzsprungRussell (H-R) diagram about which you will learn in Sec. 7.4. This diagram is of utmost importance in astronomy. In Sec. 7.5, you will learn yet another classification of stars, called luminosity classification, which tells us about the size of a star. Objectives After studying this unit, you should be able to: • •

46

• • • •

explain the atomic origin of emission and absorption spectra; discuss the correlation between the strength of spectral lines of an element and the temperature of the region containing the elements; list the spectral types of stars and their characteristic features; discuss the salient features of different groups of stars on the H-R diagram; describe the need for luminosity classification of stars; and estimate the size of a star on the basis of its luminosity class.

7.2

Stellar Spectra and Classification

ATOMIC SPECTRA REVISITED

To appreciate the relation between the physical parameters of an object and the radiation emitted by it, let us imagine heating an iron bar. First, the iron bar begins to glow with a dull red colour. As the bar is heated further, the dull red colour changes to bright red and then to yellowish-white. If we can prevent the bar from melting and vaporising with further increase of temperature, it would glow with a brilliant bluish colour. This simple experiment reveals how the intensity and colour of light emitted by a hot object varies with temperature. Similar is the situation with stars. You must have noticed, while looking at night sky, that all stars do not look the same in terms of their brightness or colour: the differences are determined by their surface temperatures.

If you happen to look at Orion nebula (Fig. 7.1), it has not only blue stars but also some red stars. Similarly the brightest visible star to the naked eye − Sirius (Vyadh) − looks whitish; Canopus (Bramhahrudaya) is yellowish and Aldebran (Rohini) is reddish.

Fig.7.1: The Orion nebula

From Unit 3, you know that spectroscopy refers to the analysis of light in terms of its wavelength and it is extensively used to obtain information regarding temperature, composition etc. of stars. Such analyses provide valuable information. But, the question is: How do we analyse the light from a star? To answer this, we need to recapitulate the atomic origin of emission and absorption of light. Recall from the Modern Physics (PHE-11) course that: Electrons in an atom exist in certain allowed energy states. Each element has a characteristic set of energy levels (see Fig. 7.2).

i)

0 −0.85 eV −1.50 eV

n=4 n=3

−3.40 eV

n=2

−13.60 eV

n=1

0

0

−25eV

−25eV

(a)

(b)

Fig.7.2: Energy level diagram of a) hydrogen atom; and b) helium atom

47

The Solar System and Stars

ii)

When an electron makes a transition from a higher energy level to a lower energy level, radiation is emitted. An electron goes from a lower energy level to a higher energy level when it absorbs energy and vice-versa (Fig. 7.3). Such transitions follow certain selection rules and are always accompanied by the emission/absorption of radiation. You know from Unit 9 of the PHE-11 course that the wavelength (or frequency) of the emitted/absorbed radiation is determined by the difference in the energies of the two atomic energy levels: ∆ E = E2 − E1 = hv

(7.1)

where E1 and E2 are the energies of the levels involved in transition, h is Planck’s constant , v is the frequency of the emitted/absorbed radiation and ∆ E is the energy of the corresponding photon.

E4 E3

Energy

E2

etc… Level 4 Level 3

Level 2

Bright emission line E1

Level 1 (a)

E4 E3

Energy

E2

E1

etc… Level 4 Level 3

Level 2

Dark absorption line

Level 1 (b)

Fig.7.3: The atomic electron a) emits radiation when it makes a transition from a higher energy level to a lower energy level giving rise to emission spectrum; and b) makes a transition to a higher energy level on absorbing radiation giving rise to absorption spectrum

To understand the origin of atomic spectra, refer once again to Fig. 7.3 which shows the energy level diagram and transition of electron between these levels for a hydrogen atom. In Fig. 7.3(a), a hydrogen atom makes a transition from the 2nd energy level to the 1st, giving off light with an energy equal to the difference of energy between levels 2 and 1. This energy corresponds to a specific colour or wavelength of light -- and thus we see a bright line at that exact wavelength! This is an emission spectrum. On the other hand, what would happen if we tried to reverse this process? That is, what would happen if a photon of the same energy was incident on a hydrogen atom in the ground state? The atom could absorb this photon and make a transition from the ground state to a higher energy level. This process gives rise to a dark absorption line in the spectrum as shown in the figure. This is an absorption spectrum. 48

Stellar Spectra and Classification

If light from a star with a continuous spectrum is incident upon the atoms in its surrounding atmosphere, the wavelengths corresponding to possible energy transitions within the atoms are absorbed. Thus, an observer will see an absorption spectrum. −0.54eV

n=∞

−0.85eV −1.51eV

n=4 n=3

−3.39eV

n=2

0 eV

Lyman series

−13.6eV

1215

n=1

972

1025

950

937

λ (Å) (a)

n=4

n=3

n=2

Balmer series

4000

5000

6000

7000

λ (Å)

n=1 (b) Fig.7.4: Emission of radiation by hydrogen atoms corresponding to Lyman and Balmer series: a) Transitions to level n=1 gives Lyman series; and b) Transitions to n=2 level gives Balmer series

On the basis of above, we can understand the genesis of atomic spectra. Let us take the simplest example of the spectrum of hydrogen atom. In Fig. 7.4 we show the various spectral series for the hydrogen atom. The transition of electrons from higher energy levels to the lowest (n = 1) energy level gives rise to emission of a series of characteristic wavelengths known as Lyman series. Similarly, transition of electrons from higher energy levels to the first excited (n = 2) energy level results into the emission of another series of characteristic wavelengths called the Balmer series. You may note that some of the spectral lines in Balmer series fall in the visible region of the electromagnetic spectrum (Fig. 7.4b). This makes them very useful spectral lines for spectroscopic analysis. Further, transitions to the second excited (n = 3) energy level give rise to the Paschen series. Emission Spectrum Now, let us consider the nature of emission of radiation by hydrogen gas (a collection of many hydrogen atoms) kept at a high temperature. Each atom in the gas can

49

The Solar System and Stars

absorb thermal energy as well as the energy due to collisions with other atoms. As a result, each atom would absorb a different amount of energy. Therefore, the electrons in the lowest energy level of various atoms get excited to different higher energy levels (Fig. 7.5). The question is: What kind of emission spectrum would we obtain in such a situation? etc… Level 4

etc… Level 4

Level 3

Level 3

Level 2

Energy

Energy

Level 1

Level 2

Level 1

Fig.7.5: At high temperatures, electrons in hydrogen atoms get excited to various higher energy levels; transitions to levels 2 and 3 are shown here as examples

To answer this, let us consider a hydrogen atom in this hot gas whose electron has been excited to the second excited energy level (Fig. 7.6). This electron can return to the lowest energy level either directly (Fig. 7.6a) or in steps. For example, it can go to the first excited energy level and then to the lowest level (Fig.7.6b). In the two cases, the energies of the emitted photons would be different. Thus, in a gas of hydrogen atoms at high temperature, electrons can be excited to many possible levels and can make transitions to any of the lower levels emitting radiations of the corresponding frequencies. This implies that photons of different frequencies will be emitted and we will observe more than one emission line in the emission spectrum of a hot gas. etc… Level 4

etc… Level 4

Level 3

Level 3

Level 2

Energy

Energy

Level 1 (a)

Level 2

Level 1 (b)

Fig.7.6: Different ways of transition of electron in a hydrogen atom from an excited energy level to the lowest energy level

However, you must remember that the spectrum of hydrogen gas is not continuous, that is, not all frequencies will be emitted; radiations of only those frequencies will be emitted whose energies are equal to the difference in energy between the allowed energy levels. Therefore, the emission spectrum of hydrogen gas contains bright lines of certain definite frequencies.

50

The frequencies contained in the emission spectra of each element are unique: no two elements have the same type of emission spectra because each one of them has a characteristic set of allowed energy levels.

Stellar Spectra and Classification

Absorption Spectrum Let us assume now that the hydrogen gas is not at a high enough temperature to excite the electrons to higher energy levels. In such a situation, there is no emission of radiation. If white light (continuous spectrum) is passed through this cool hydrogen gas, it will absorb light of only those frequencies which give electrons just enough energy to make a transition to one of the excited energy levels (see Fig. 7.3b). Therefore, just as photons of certain energies are emitted by hydrogen gas at a high temperature, photons of the same energies are absorbed by the cool hydrogen gas. You can argue that atoms absorbing photons and giving rise to absorption spectrum will eventually re-emit photons of the same frequencies. Thus, it should cancel out the effect of absorption and we could not observe any absorption spectrum! The question, therefore, is: How to explain the observed absorption spectrum? The re-emitted photons are repeatedly absorbed by the atoms of the cool gas before they finally escape it. These re-emitted photons escaping from the gas travel along different directions. Thus, the photons of certain frequencies, which were originally coming towards the spectroscope as a part of white light, are scattered by the atoms of the cool gas. As a result, the intensities of the light corresponding to these frequencies are very low and we observe an absorption spectrum. Thus when white light passes through a cool hydrogen gas, we obtain a series of dark lines and such a spectrum is called the absorption spectrum. Do remember that for any element, the dark lines of absorption spectrum occur at exactly the same frequencies (or wavelengths) where we observe bright lines in its emission spectrum. Now, before proceeding further, you should answer an SAQ to fix these ideas. SAQ 1 a) Why do the spectra of different elements have different sets of lines? b) What do you understand by a continuous spectrum? Under what condition do we observe it? Let us pause for a moment and think: How does the knowledge about the origin of spectral lines help in analysing stellar spectra? Firstly, you have learnt that each element produces a unique set of spectral lines. Therefore, observing the spectrum of a star, we can tell which element(s) are present in its atmosphere. Secondly, the nature of spectrum  emission or absorption  depends on the temperature prevailing in the atmosphere through which star light passes. Thus, the nature of spectrum gives us an idea about the temperature of the star’s atmosphere. In addition, stellar spectra also provide information about the density of the star’s atmosphere and the motion of stars with respect to the Earth. Whether a star is moving towards or away from the Earth can be inferred on the basis of Doppler effect:

Spend 5 min.

Strictly speaking, the stellar spectra tell us about the composition of the star’s atmosphere. However, it is generally believed that the composition of the atmosphere reflects the composition of the star itself.

According to the Doppler effect, there is an apparent change in the wavelength of radiation emitted by a star due to its (star’s) motion; if the star is moving away from the Earth, the wavelengths are shifted towards longer wavelengths (the phenomenon is known as red shift) and if the star is moving towards the Earth, the wavelengths are shifted towards shorter wavelengths (the phenomenon is known as blue shift). With this background information, you are now ready to study stellar spectra. In this context, it is useful to remember a set of empirical rules of spectroscopic analysis given below:

51

The Solar System and Stars

i)

A hot and opaque solid, liquid or highly dense gas emits a continuous spectrum; Fig. 7.7a (you have learnt about optically thin and thick media in Unit 4). ii) A hot gas produces emission spectrum and the number and position of the emission lines depends on the composition of the gas (Fig. 7.7b). iii) If light having continuous spectrum is passed through a gas at low temperature, an absorption spectrum consisting of dark lines is produced (Fig. 7.7c). Again, the position and number of dark lines are characteristic of the elements present in the gas. iv) When light with continuous spectrum passes through a very hot, transparent gas, a continuous spectrum with additional bright lines is produced. Fig. 7.7 shows some of these situations.

Hot light source

Spectroscope

Cloud of gas

Continuous spectrum with dark lines Spectroscope

Spectroscope

(c)

Bright-line spectrum

Continuous spectrum (a)

(b) Fig.7.7: Diagrammatic representation of empirical rules for spectroscopic analysis of stellar spectra

7.3

STELLAR SPECTRA

Before discussing stellar spectra, you may like to know how they are obtained using a spectroscope. Refer to Fig. 7.8 which shows how to obtain a spectrum using a spectroscope. Light from the telescope (pointing towards the star) is passed through a slit, a collimating lens, through the prism of the spectroscope to obtain the spectrum. The resulting spectrum is focussed by another lens and can either be viewed directly or photographed.

Prism Red Light from telescope

Source slit

Collimating lenses

Violet

Fig.7.8: Schematic diagram of a spectroscope

52

For a casual observer, all stars look alike. But, if observed carefully, some stars appear bright and some faint. It is usually difficult to make out the colours of stars if they are fainter than the second magnitude. This is due to poor colour sensitivity of the human

eye to fainter light. Stellar spectra, however, show that individual stars differ widely from one another in brightness and spectral details. While spectra of some stars contain lines due to gases like hydrogen and helium, others show lines produced by metals. Some stars have spectra dominated by broad bands of molecules such as titanium oxide. A typical spectrum of a star consists of a continuum, on which are superposed dark absorption lines. Sometimes emission lines are also present. An example of a typical stellar spectrum is the solar spectrum (Fig. 7.9). Solar spectrum can be easily obtained by passing a narrow beam of sun light through a prism. It consists of a continuum background superposed by dark lines.

Stellar Spectra and Classification In earlier times, the spectrum was exposed on a photographic film. The film was then developed in the usual way and scanned for intensity at each point of the film. The data so obtained was plotted to obtain the spectrum. Now-adays, with the advent of modern detectors like the CCDs (about which you have studied in Unit 3), the spectrum can be directly stored in a computer and plotted.

Fig.7.9: Solar spectrum

The dark lines in the solar spectrum are called Fraunhofer lines. You know that the intensities of spectral lines indicate the abundance of various elements to which the lines belong. The important lines in stellar spectra are due to hydrogen, helium, carbon, oxygen, neutral and ionised metal atoms. Bands in the spectra are caused by the molecules such as titanium oxide, zirconium oxide, CH, CN, C3 and SiS2.

You have learnt in Unit 1 that the brightness of stars is expressed in magnitudes.

Similarities in stellar spectra provided the basis for classification of stars into certain categories. The earliest classification was done by Annie J. Cannon. She classified more than 2, 50,000 stars by observing the strength of absorption lines, particularly, the hydrogen Balmer lines. In this way stars have been classified into seven major spectral types, namely, O, B, A, F, G, K and M. For greater precision, astronomers have divided each of the main spectral types into 10 sub-spectral types. For example, spectral type A consists of sub-spectral types A0, A1, A2.... A8, A9. Next come F0, F1, and so on. Thus there are 70 sub-spectral types possible. However, in practice, all 70 types have not been observed. The advantage of the finer division is to estimate the star's temperature to accuracy within about 5 percent. The Sun, for example, is not just a G star, but a G2 star, with a surface temperature of about 5800 K.

7.3.1

Spectral Types and Their Temperature Dependence

Fig. 7.10 shows a set of spectra of seven main types of stars: O, B, A, F, G, K and M (hottest O-type to coolest M-type). It was shown by M.N. Saha, an Indian scientist, that Cannon’s spectral classification can be explained primarily on the basis of the temperatures of stars because the intensities of spectral lines depend on the surface temperature of a star. To elaborate this argument, let us take the case of spectral lines of hydrogen Balmer series as an example. You know that 50 to 80 percent of the mass of a typical star is made of hydrogen. However, only a small fraction of the stars show Balmer lines in their spectra! It is so because most of the stars do not have sufficient surface temperature to raise the electrons in hydrogen atoms to the second and higher energy levels. And, as you know, for production of Balmer lines, electrons in the hydrogen atoms make transitions from the higher energy levels to the first excited energy level. For example, a M type star whose surface temperature is about 3000 K, does not have enough energy for production of Balmer lines. Therefore, even though there is an abundance of hydrogen in the stars, Balmer lines may not be observed in the spectra of all of them: the surface temperature prohibits the formation of Balmer lines of hydrogen.

M.N. Saha’s conclusion that Cannon’s spectral classification is based on surface temperature of stars is based on the fact that ionisation of atoms is a process similar to chemical reaction. This means that at a given temperature, relative number of atoms in various stages of ionisation which are in equilibrium, is fixed. Further, the intensity of an absorption line is a function of the number of atoms which can absorb radiation corresponding to this line and this number itself is a function of temperature.

53

The Solar System and Stars

Hydrogen Balmer series

H lines

H

H

H

O

B A F G K M 350 nm

400 nm KH lonized Ca

Various 450 nm metals Calcium

500 nm

Fig.7.10: Examples of spectra of various types of stars (O to M type)

On the other hand, very hot stars with surface temperatures higher than 25000 K also do not contain Balmer lines in their spectra. Can you guess why is it so? It is because such stars are so hot that electrons of the hydrogen atoms are ripped off and the atoms are ionised. Such ionised atoms cannot produce spectral lines. It is only when the surface temperature of a star is between 7500 K and 11000 K that the conditions are favorable for the production of Balmer lines. Therefore, surface temperature of a star determines which spectral lines would be formed and what their intensities would be. After hydrogen, the second most abundant element in stars is helium. A helium atom has two electrons which are held very tightly together in their lowest states. Hence, helium absorption lines are seen in the spectra of relatively hot stars with surface temperatures in the range 11000 K to 25000 K. In stars hotter than 25000 K, one of the two electrons in helium atoms is torn away. The question is: Are the spectral lines produced by singly ionised helium atoms similar to those produced by unionised helium atom? The spectral lines produced by singly ionised helium are different from those produced by neutral or un-ionised helium. Further, stars with temperatures in excess of 40000 K are so hot that helium is completely ionized which cannot produce any spectral lines. In some stars, conditions are favourable for a molecule to produce spectral lines. Very cool stars with temperatures less than 3500 K show very strong, broad bands of titanium oxide (TiO). Table 7.1 shows the values of the temperature and some other important parameters of seven main spectral types. Table 7.1: Spectral types and their parameters Spectral Approx. class temp. (K)

54

O B

40,000 20,000

Hydrogen Balmer lines Weak Medium

A F G

10,000 7,500 5,500

Strong Medium Weak

K

4,500

Very weak

M

3,000

Very weak

Other spectral features

Naked-eye example (Star)

Color

Ionised He Ionised and Neutral He Ionised Ca weak Ionised Ca weak Ionised Ca medium Ionised Ca strong TiO strong

Meissa (O8) Achenar (B3)

Blue Blue/White

Sirius (A1) Canopus (F0) Sun (G2)

White Yellow/White Yellow

Arcturus (K2)

Orange

Betelgeus (M2)

Orange/Red

There is an alternative method to determine the temperature of a star on the basis of its spectrum. This method is based on the principle of black body radiation and you will learn it now.

7.3.2

Stellar Spectra and Classification

Black Body Approximation

As you know, the colour and brightness of a star is different from other stars. You also know that these parameters depend on temperature. Does it, therefore, mean that we can estimate the temperature of stars on the basis of their observed colour and brightness? It can indeed be done if we consider a star as an ideal object called black body. Refer to Fig. 7.11 which shows a set of black body spectra at various temperatures. Note that hotter bodies radiate most of their total energy in the shorter wavelength part of the spectrum. On the other hand, the cooler bodies have the peak of their radiation at the longer wavelength side of the spectrum and the total energy radiated by them is relatively low. It has been observed that the outer envelope of a star’s spectrum is quite similar to a black body spectrum at a certain temperature. Thus, a stellar spectrum can be approximated to a black body spectrum. You should, however, note that the absorption features of stellar spectra distinguish it from the black body spectrum. Therefore, the main spectral types, namely O to M, can be considered as referring to different temperatures. In other words, we can say that spectral type directly refers to the effective temperature of the star (see Table 7.1).

2

log10 F

0

-2

-4

-5

-4.5

-4

-3.5

log10 (cm) Fig.7.11: Black body radiation curves for various temperatures

To estimate the temperature of an astronomical object such as a star on the basis of black body approximation, go through the following Example carefully. 55

The Solar System and Stars

Example 1 An astronomical object named Cygnus X-1, a strong X-ray source, is found to radiate like a black body with peak wavelength at 1.45 nm. Calculate its temperature. Assume that the constant for Wien’s displacement law is equal to 2.9×10−3 mK. Solution You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course that the Wein’s displacement law is given by: λpeak T = constant where λpeak is the value of wavelength corresponding to the peak of the black body spectrum and T is the temperature of the black body. For Cygnus X-1, we have λpeak = 1.45 × 10−9m and value of the constant = 2.9 × 10−3mK. Thus, temperature of Cygnus X-1, T=

2.9 × 10 −3 mK 1.4 × 10

−9

≅ 2 × 10 6 K .

m

As you know, the temperature of a star is not a directly measured quantity; rather, we infer this parameter on the basis of black body approximation. Would it not be better to classify stars on the basis of a parameter, say luminosity, which is more easily measurable? This is what was attempted by Ejnar Hertzsprung and Henry Norris Russell and their classification of stars resulted in a graph known as H-R diagram. You will study about it now.

7.4

H-R DIAGRAM

H-R diagram is a graph which enables us to simultaneously classify stars on the basis of their temperatures and their luminosities. Since H-R diagram involves parameters such as temperature, luminosity and radius, it would be advisable to first recapitulate these terms and their interdependence in the context of stars. The luminosity (L) of a star is defined as the total energy radiated by it, in one second, consisting of radiations of all wavelengths. On the other hand, you may recall from Unit 1 that the absolute visual magnitude refers only to the visible range of the electromagnetic radiation. Therefore, corrections need to be applied for the radiations emitted at other wavelengths. The required extent of correction depends upon the temperature of the star. This correction is called Bolometric Correction (B.C.) and for medium temperature stars like our Sun, its value is small. After making necessary correction, we obtain the absolute bolometric magnitude: for the Sun it is + 4.7 and for Arcturus, it is − 0.3. Thus, the expressions for the apparent magnitude as well as the absolute magnitudes of a star are written as: mbol = mv + B.C.

(7.2a)

M bol = M v + B . C .

(7.2b)

and

56

where mbol, mv, M bol and M v are respectively the apparent bolometric magnitude apparent visual magnitude, absolute bolometric magnitude and absolute visual magnitude of a star. Further, the absolute bolometric magnitude of a star can be calculated from the following relation:

bol M star − M bol = −2.5 log

Lstar L

Stellar Spectra and Classification

(7.3)

You have already learnt that a difference of 5 in absolute magnitude implies a ratio of 100 in luminosity. So, the luminosity of Arcturus is 100 LΘ . The most luminous stars can have luminosities of the order of 105 LΘ , whereas the least luminous ones have luminosities ~ 10−4 LΘ . Thus, if we can measure the parallax of a star, we can find its distance, calculate its absolute visual magnitude, calculate the bolometric correction and obtain absolute bolometric magnitude. As a result, we can estimate the luminosity of a star in terms of the Sun’s luminosity. How about solving an SAQ to check your understanding of the terms discussed above? SAQ 2 The absolute visual magnitude of a star is 8.7 and for its temperature, the bolometric correction is − 0.5. Calculate the absolute bolometric magnitude and the luminosity of the star.

You have learnt in Unit 1 how the parallax of a star is measured.

Spend 5 min.

You know that one of the fundamental parameters of a star is its diameter and it is related to the star’s luminosity and temperature. To appreciate the relation amongst these parameters, consider a normal candle flame which has a low surface area. The candle flame, despite being very hot, cannot radiate much heat and its luminosity is low. However, if the candle flame were 25 cm long, it would have larger surface area. In this case, despite being at the same temperature, it would radiate more heat and would have high value of luminosity. In the same way, a star’s luminosity is affected by its surface area and temperature. To obtain a relation between the luminosity and diameter of a star, we can assume the star to be spherical in shape having surface area 4πR2, where R is the radius. Further, from basic thermodynamics (PHE-06 course), you know that, the total energy radiated by a black body per second per unit area is σT 4 (Stefan’s law). Assuming that the star radiates like a black body, we can express its luminosity (L) as the product of its surface area and the energy radiated by it per unit area per second, i.e., L = 4πR 2σT 4

(7.4)

Thus, by comparing the luminosity L of a star, with LΘ , the luminosity of the Sun, we get:

L R = LΘ RΘ

2

T TΘ

4

(7.5)

where RΘ and TΘ are the radius and temperature of the Sun. Let us now refer to Fig. 7.12 which shows a H-R diagram for all known stars in our solar neighbourhood. Note that the H-R diagram is a plot between absolute magnitude or luminosity (along y-axis) and temperature (along x-axis). The H-R diagram contains quite a lot of information about stars. Since the absolute magnitude or luminosity refers to the intrinsic brightness of a star, H-R diagram relates the intrinsic brightness of a star with its temperature. Moreover, the H-R diagram separates the effects of temperature and surface area on the luminosity of the stars because the brightness of two stars at the same temperature is proportional to their radii. This feature enables us to classify stars in terms of their diameters. Further, you must remember that the location of a star on the H-R diagram is in no way related to its location in space: a star located near the bottom of the diagram simply means that its luminosity is low and similarly, a star in the right indicates that its temperature is low (because the temperature decreases away from the origin) and so on. Another interesting feature of the H-R diagram is that the position of a star on

57

The Solar System and Stars

it changes with time. This implies that the star’s luminosity and temperature change with its age. Again, this change in position of a star has nothing to do with the star’s actual motion. The stars have been divided into different types/groups on the basis of their location on the H-R diagram. In Fig. 7.12, you may note that the distribution of stars follows a pattern such that a majority of stars fall along a central diagonal called the main sequence. The main sequence stars account for nearly 90 per cent of all stars. The other types of stars such as giants, supergiants, white dwarfs populate other regions. The giant stars (named so because of their big size) located at the top right of the H-R diagram have low temperature but high luminosity. O

B

A

Spectral class F G

K

M

Supergiants -20

17

Absolute magnitude

6 -10

Red giants

3.2

Main sequence

1.8 1.5 1.3

(a) 1

5

0.7 White dwarfs

0.5 0.3

10

20000

−10

O

10000 7000 6000 Temperature (K) B

Spectral Class A

F

5000

G

3000

K

M

−−−5

Absolute Magnitude

0

(b)

+5

+10

+15

+20

20,000

14,000

10,000

7000

5000

3500

2500

Temperature (K) Fig.7.12: a) A schematic H-R diagram (Note that on the main sequence, masses of the stars are indicated in units of solar mass.); and b) magnified version of the H-R diagram

58

Stellar Spectra and Classification

When we move further up in the H-R diagram (Fig. 7.12), we find super giants. These high luminosity and low temperature stars must be extraordinarily big in size. It has been estimated that the diameters of super giants are roughly 100 to 1000 times the diameter of the Sun! Further, when we come to the bottom left (below the main sequence), we come across stars known as white dwarfs  stars which are very hot but their luminosity is very low. Obviously, white dwarf stars must be very small in size compared to the Sun to have such low luminosities. Example 2 A bright star in Orion constellation, Betelgeuse, has a surface temperature of 3500K and is 105 times more luminous than the Sun. Calculate its radius in terms of RΘ , the radius of the Sun. What kind of star could it be on the basis of the H-R diagram? Solution We have TB = 3500K;

TΘ = 5800K;

LB = 105 LΘ

From Eq. (7.5), we have R B2 TB4 LB = 2 4 LΘ RΘ TΘ

or, 10 5 = [3500 5800]4 (R B /RΘ )2

or, R B ≈ 1000RΘ

The luminosity and surface area of Betelgeuse is much higher than the Sun whereas its temperature is lower than the Sun. These characteristics, as per the H-R diagram, indicate that Betelgeuse is a super giant. To fix the ideas expressed above, you should answer the following SAQ. SAQ 3

Spend 8 min.

a) Suppose that the surface temperature of two stars A and B is the same and the luminosity of A is higher than B. Which of the two stars is bigger in size? Why? b) Choose a typical giant star in Fig. 7.12 and estimate its radius. Although the basic H-R diagram depicts stars in terms of their temperatures and absolute magnitudes, we can have its many other representations as well depending upon the extent of information we wish to incorporate in it. Fig. 7.13 shows a composite H-R diagram incorporating several parameters such as luminosity, absolute magnitude, temperature, and spectral type.

59

The Solar System and Stars

Fig.7.13: A composite H-R diagram showing various parameters of the stars

Spend 5min.

SAQ 4 a) In which part of the main sequence are the less massive stars located? b) A main sequence star has a luminosity of 400 LΘ. What is its spectral type? Thus, on the basis of above discussion, you will agree that the H-R diagram provides information about the following parameters of stars: a) b) c) d) e)

size luminosity mass spectral type, and absolute magnitude

It is because of this reason that the H-R diagram is so important in astronomy. So far, you have studied about stellar spectra and classification of stars on the basis of their temperature. You have also learnt that, on the basis of H-R diagram, one can get an idea about the size of a star. However, to get a better idea about the size of a star, we can use the correlation between the sharpness of spectral lines and luminosity. This gives rise to luminosity classification. You will learn about it now.

7.5

LUMINOSITY CLASSIFICATION

We have discussed earlier in this unit that the luminosity of a star predominantly depends on its size: the bigger the star is, the higher is the value of its luminosity. Detailed study of the stellar spectra helps in ascertaining the density and size of a star which give a fairly good idea of its luminosity. You may ask: Which features of the stellar spectra provide such information? Recall that the spectrum of an isolated atom differs from that of the gas of such atoms. The difference is manifested in the form of broadening of spectral lines of the atom in the latter case due to collision between atoms. The next logical question is: Under what conditions can collisions readily take place? For frequent collisions, a denser gas provides a better environment than a rarer gas. Therefore, if the spectral lines in stellar spectra are broadened, we can safely conclude that the density of the star is high. Examples of such stars are the main sequence stars. On the other hand, giant stars have very low densities. As a result, their spectral lines are fairly narrow compared to the main sequence stars. 60

Thus, sharp lines in the stellar spectra clearly indicate that the size of the star is large and hence its luminosity is high. The converse is also true. Thus, looking at the width of spectral (absorption) lines, we can obtain a fairly good idea about its luminosity. As such, stars have more or less continuous range of luminosities. Still, on the basis of the width of their spectral lines, they are categorized into various luminosity classes. Table 7.2 gives the various luminosity classes denoted by Roman numerals I to V with the supergiants further subdivided into classes Ia and Ib. The star Rigel (β Orionis) is a bright supergiant (class Ia) and Polaris, the North star is a regular supergiant (class Ib).The star Adhara ( Canis Majoris) is a bright giant (Class II); Capella (α Aurigae) is a Giant (III) and Altair (α Aquilae) is a subgiant (IV). The Sun is a main-sequence star (V). Thus, the complete spectral classification for the Sun is G2V. This complete classification is also called the spectro-luminosity classification. The spectro-luminosity class of star Vega is A0V.

Stellar Spectra and Classification Table 7.2: Luminosity classes

Ia

Bright supergiant

Ib

Supergiant

II

Bright giant

III

Giant

IV

Subgiant

V

Main-sequence star

Refer to Fig. 7.14 which shows the position of luminosity classes on the H-R diagram. You must remember that the lines corresponding to each class is just an approximation; star of a particular class may lie just above or below the line corresponding to that class.

Fig.7.14: H-R diagram depicting the location of luminosity classes

Now, let us sum up what you have learnt in this unit.

7.6

SUMMARY

•

Emission and absorption of radiation is caused due to transition of electrons in atoms between allowed energy levels. Analysis of stellar spectra provides information about the temperature and size of a star.

•

On the basis of the strength of spectral lines, particularly Balmer lines, in stellar spectra, stars were classified into seven main spectral types namely O, B, A, F, G, K and M. It was shown by M N Saha that this classification essentially refers to the temperatures of stars.

•

H-R diagram enables us to classify stars on the basis of their temperatures and luminosities. 61

The Solar System and Stars

•

Luminosity of a star is defined as the total energy rediated by it in one second consisting of radiations of all wavelengths. The relation between luminosity, radius and temperature of a star is given by: L LΘ

=

R RΘ

2

T

4

TΘ

•

On the basis of H-R diagram, stars are grouped into four categories namely main sequence, giants, super giants and white dwarfs. Stars of each group have characteristic temperatures, sizes and luminosities.

•

The density of a star affects the sharpness of its spectral lines. This fact is used for luminosity classification of stars.

•

According to the luminosity classification, stars are classified into five classes: I, II, III, IV and V with class I further sub-divided into Ia and Ib.

7.7

TERMINAL QUESTIONS

Spend 25 min.

1. Assign spectral class to each of the objects whose characteristics are given below: a) b) c) d) e)

Temperature ~ 40,000K Weak Balmer lines but moderately strong Ca lines Strongest hydrogen lines Molecular bands of TiO Neutral helium lines

bol 2. A star has luminosity ~ 100 LΘ and apparent bolometric magnitude, m star = 9.7. bol If Sun has M = + 4.7, calculate the distance of the star.

3. Calculate the radius of a star which has the same effective temperature as the Sun but luminosity 10,000 times larger. 4. In the following table, which star is a) the brightest; b) the most luminous; c) the largest; and d) the smallest? Star 1. 2. 3. 4. 5.

7.8

Spectro-luminosity type G2V B1V G2Ib M5III White Dwarf

mv 5 8 10 19 15

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1. a) Spectra of different elements have different lines because their atomic energy levels are different. b) A continuous spectrum contains all wavelengths. We get such a spectrum when the density of a system is high, as in a solid. 2.

From Eq. (7.2b), we know that M bol = M v + B . C .

Substituting the values of M v and B.C., we get, M bol = 8.7 − 0.5 = 8.2 62

Stellar Spectra and Classification

Further, from Eq. (7.3), we have:

Lstar (M bol − M bol ) = 10 − 4 star Θ LΘ = 10 −0.4 (8.2−4.7 ) =

1 25

Thus, we get, Lstar =

LΘ . 25

3. a)

Since LA > LB, the surface area of A is greater than the surface area of B. So, A is bigger.

b)

In Fig. 7.12, we find that the temperature of a typical giant star can be taken as 5000 K. So, we can write for a typical giant, T = 5000 K. Further, the luminosity of a typical giant can be written as, L = 100 LΘ Thus, using Eq. (7.5), we get: L R = LΘ RΘ

2

T TΘ

4

or, R RΘ

2

=

L LΘ

TΘ T

4

= 100 × (6 / 5) 4

or, R = 14.4 RΘ . 4. a) Lower part of the main sequence. b) From Eq. (7.3), we have bol

Lstar LΘ

bol

M star − M Θ = − 2.5 log

= − 2.5 log 400 =

6.5

Thus, bol M star =

6.5 + 4.7 =

1.8

If we look at the H.R. diagram to identify the class of the star having this value of absolute magnitude, we find that its class should be B7, approximately. Terminal Questions 1. (a) O (b) G (c) A (d) M (e) B 2. We have from Eq. (7.3): bol M star − M bol = −2.5 log

Lstar LΘ

or, bol M star = 4.7 − 2.5 log (100) = − 0.3

63

The Solar System and Stars

Further, we have: bol bol m star − M star = 5 log

d 10

9.7 − (− 0.3) = 5 log (d/ 10 )

or, d = 1000 pc .

3. From Eq. (7.5), we have: L1 L2 = R12T14 R22T24

or, 10 4 = R12T14 R 2 T14 ;

or,

R1 = R

10 4 = 100 R

4. [Hint: Use Fig. 7.12 to determine the absolute magnitudes.] The answers are: a) 1; b) 3; c) 3; d) 5.

64

UNIT 8 STELLAR STRUCTURE

Stellar Structure

Structure 8.1

Introduction Objectives

8.2 8.3 8.4 8.5 8.6

Hydrostatic Equilibrium of a Star Some Insight into a Star: Virial Theorem Sources of Stellar Energy Modes of Energy Transport Simple Stellar Model

8.7 8.8 8.9

Summary Terminal Questions Solutions and Answers

8.1

INTRODUCTION

Polytropic Stellar Model

In Unit 7, you have learnt about the classification of stars on the basis of their spectra. You know that the H-R diagram is obtained on the basis of the luminosity, and effective temperature of stars and enables us to classify them in the most comprehensive manner. A careful look at this diagram reveals that there are gaps between families of stars, e.g., between the main sequence and giants. You may wonder why such gaps should exist when we have such a large number of observable stars! Further, you may like to know: What causes some ordinary stars to become giants and others to become dwarfs? These and similar other questions cannot be answered on the basis of observations alone. We require the knowledge of the physical conditions in the interior of the stars. In other words, we need to know: How the temperature, pressure and density of a star vary in its interior? In the present unit, you will study the physical principles which form the basis for understanding the internal structure of stars. You know that the Sun is emitting radiation at a constant rate and its diameter shows no significant variation with time. This implies that the Sun as well as other stars are in mechanical and thermal equilibria. In Sec. 8.2, you will study about hydrostatic equilibrium and its consequences for the variation of density and pressure inside a star. In Sec. 8.3, you will learn how to estimate the internal temperature of a star on the basis of the virial theorem: statement of relation between the kinetic and potential energies of a system in equilibrium. In addition to the considerations of hydrostatic and thermal equilibria, the mechanism of energy generation and transport play an important role in deciding stellar structures. In Sec. 8.4, you will learn why only the energy generated due to nuclear reaction needs to be considered as the source of stellar energy. You will also learn some important mechanisms of nuclear energy generation in stars. In Sec. 8.5, various modes of transportation of energy from the interior to the surface of stars have been discussed. You will discover the conditions for the formations of convective and radiative zones in the stellar interior. Finally, in Sec. 8.6, we discuss the computation of a simple stellar model. We also discuss how the results of this model compare with the observations. Objectives After studying this unit, you should be able to: •

list the basic assumptions for the theoretical study of stellar structure; 65

The Solar System and Stars

•

show that the values of interior pressure and temperature of a star are higher than the values at the surface by several order of magnitudes;

•

explain that the nuclear energy generation is the only important energy generation process in stars;

•

predict when a radiative or a convective zone will be formed in the stellar interior;

•

compute polytropic stellar model and compare the theoretical results with observations; and

•

solve numerical problems based on these concepts.

8.2

HYDROSTATIC EQUILIBRIUM OF A STAR

You know that stars, including the Sun, are made of hot gas. We cannot probe the interior of stars to determine their physical parameters and their variation with time and distance because of their high temperatures and enormous distances from the Earth. The question, therefore, is: How do we determine the internal structure of a star? Astrophysicists construct theoretical models of stars and compare their predictions with observations. To keep the theoretical analysis simple, the following assumptions are made: i)

The star is spherically symmetric: You know that stars have rotational motion which alters their spherical shape. Since the rotational motion is slow in most cases, it does not have appreciable effect on the shape of the stars. Spherical symmetry is, therefore, a valid assumption.

ii) The star is in dynamic equilibrium: Dynamic equilibrium means that the energy radiated by a star is equal to the energy supplied from its core. This assumption seems valid because luminosities of stars have been observed to be constant over a considerable period of time. iii) The star is in thermally steady state: This implies that the temperature at each point within a star is constant over a considerable period of time. Note that this assumption does not mean that the entire interior of a star is at the same temperature. Under these assumptions, the theoretical understanding of stellar structure is based on four equations based on certain fundamental principles of physics. First of all, let us consider the principle of hydrostatic equilibrium. You know that the observable stellar parameters such as luminosity change very slowly. We can, therefore, safely assume that a star is in hydrostatic equilibrium, that is, it is neither expanding nor contracting (at least not very rapidly). This equilibrium is maintained by a balance between the force of gravity acting inwards and that due to the gradient of pressure of the gas acting outwards. To find out the consequences of the hydrostatic equilibrium, let us consider an element of volume dV at a point A inside a star at a distance r from its centre (Fig. 8.1). If ρ(r) is the density of matter inside dV, that is, at a distance r from the centre, the mass enclosed in the volume element dV is ρ(r) dV. Further, if M(r) is the mass inside the sphere of radius r, the gravitational force acting on the mass inside dV is given by: GM (r ) r2

66

ρ(r )dV

(8.1)

Stellar Structure

dV (=dA.dr) dP.dA r

dr R

Fig.8.1: An element of volume dV at a distance r inside a star in hydrostatic equilibrium

Now, in view of the spherical symmetry of the star, the pressure, density and temperature may be taken as identical at all points over the spherical surface of radius r. Therefore, the net hydrostatic force acting on the volume element dV and pushing it outward can be written as: dP . dA,

(8.2)

where dA is the area of the volume element dV perpendicular to r and dP is the pressure difference between two sides of the volume element along the radius. For hydrostatic equilibrium, the gravitational force must be equal and opposite to the hydrostatic force. Thus, on the basis of Eqs. (8.1) and (8.2), we can write: GM (r ) r2

ρ(r )dV = −dP.dA

=−

dP .dV dr

This yields the equation of hydrostatic equilibrium. Equation of Hydrostatic Equilibrium dP dr

=−

GM (r ) r2

ρ(r )

(8.3)

We can also write Eq. (8.3) as dP dr

= − ρ(r ) g (r )

where g(r) is the acceleration due to gravity given by g(r) =

(8.4) GM (r )

. From Eq. (8.3), r2 it is obvious that it is the pressure gradient that supports the star and not the pressure. If we denote the mass of the whole star by M and its radius by R, the mean density, < ρ > of the whole star can be expressed as: =

M 4 3

πR

(8.5) 3

67

The Solar System and Stars

Let us now consider a spherical shell of the star between radii r and r + dr. The volume of the matter enclosed in this shell is 4πr2dr. Since ρ(r) is the density of stellar matter at distance r, the mass of this spherical shell is:

dM (r ) = 4π r 2 ρ(r ) dr Thus, we can express the total mass inside the sphere of radius r as:

M (r ) =

r 0

ρ (r ) 4 π r 2 dr

Differentiating both sides of the above equation with respect to r, we get the mass continuity equation for the star: Mass continuity equation dM (r ) dr

= 4πr 2 ρ(r )

(8.6)

Eqs. (8.3) and (8.6) constitute two basic equations of stellar structure. Further, the state of hydrostatic equilibrium in a star enables us to obtain a relation between its gravitational potential energy and the kinetic energy of its constituent particles. This relation is known as the virial theorem. You will learn about it now.

8.3

SOME INSIGHT INTO A STAR: VIRIAL THEOREM

You may be aware that the virial theorem is applicable for a system of perfect gas particles. However, this theorem can also be applied to a star because it (star) can be considered as a system of free particles. To obtain the relation between the potential and kinetic energies of a star, let us consider the equation of hydrostatic equilibrium 3 (Eq. (8.3)). On multiplying Eq. (8.3) by 4πr and integrating over the radius of the star, we get: R 0

dP dr

R

4πr 3 dr = −

ρ( r )

GM (r ) r

0

2

4πr 3 dr

(8.7)

On integrating by parts, the left hand side of Eq. (8.7) gives: Actually, the pressure on the surface of a star is not zero. Its value is, however, much smaller than the pressure in the interior. Therefore, it can be assumed to be zero. This is also the case with density.

R 0

dP dr

R 3

4πr dr = P 4πr

3

R

− 0

3P 4πr 2 dr

0

Since P = 0 at r = R, the first integral on the right hand side vanishes and the above equation reduces to: R

dP

0

dr

4πr 3 dr = −3

R

P 4πr 2 dr

0

Thus, Eq. (8.7) can be written as: R

−3 0

68

P 4πr 2 dr = −

R

ρ(r ) 0

GM (r ) r

4πr 2 dr

(8.8)

Now, assuming that the star comprises of monoatomic gas, its total thermal (or internal) energy U can be written as:

R

2U = 3

Stellar Structure

P 4πr 2 dr

(8.9)

0

Spend 3 min.

SAQ 1 Derive Eq. (8.9). Substituting Eq. (8.9) in Eq. (8.8), we obtain: R

2U =

ρ( r )

GM (r ) r

0

4π r 2 dr

(8.10)

The right hand side of Eq. (8.10) can be expressed in terms of the gravitational potential energy. The gravitational potential due to the mass M(r) inside the sphere of GM (r ) radius r is − . Therefore, total potential energy due to all mass elements dM (= r 4πr2ρ(r) dr) of the star is: M

Ω=−

GM (r ) r

0

R

dM = −

GM (r ) r

0

4πr 2 ρ(r )dr

(8.11)

Thus, on the basis of Eqs. (8.10) and (8.11) we get the virial theorem. Virial theorem 2U + Ω = 0

(8.12)

You can see from Eq. (8.11) that the gravitational potential energy Ω of the whole star can be determined only if we know the variation of density ρ(r) inside. Assuming that ρ(r) ≈ < ρ > , the mean density of stellar matter, we may write: M (r ) ≈

4 3

πr 3 < ρ >

(8.13)

Substituting < ρ > for ρ(r ) and M(r) from Eq. (8.13) in Eq. (8.11) and integrating, we get: Ω≈−

3 GM 2 5 R

(8.14)

Now, to find an expression for the internal energy of a star, we make use of the equation of state of a gaseous system: M (r ) =

ρ P= k BT µm −23

dM =

where kB (= 1.38 × 10 J K ) is the Boltzman constant, T is the temperature and µm is the mean mass of a gas particle. For pure hydrogen gas, µ = 1. Substituting Eq. (8.15) in Eq. (8.9), we get: 3 R 2 0

P 4πr 2 dr =

3 M k BT 2 0

m

4

2

π . 3r < ρ > dr

3 dM

U=

3

πr < ρ >

3

(8.15) −1

4

2

= 4 π r dr

dM 69

since 4πr 2 dr = dV and ρdV = dM . If we define the mean temperature of the star, 1 M = TdM , then the above expression can be written as: M 0

The Solar System and Stars

U=

3 kB M 2 µm

(8.16)

Substituting the expressions for potential energy (Eq. (8.14)) and internal energy (Eq. (8.16)) in the virial theorem (Eq. (8.12)), we get:

< T >=

1 µmG M 5 kB R

Thus

< T > ∝ M 2 / 3 < ρ > 1/ 3

(8.17)

where R has been expressed in terms of the mean density. Spend 10 min.

SAQ 2 a) Verify the results contained in Eqs. (8.14) and (8.17). b) Assume that the Sun is made of pure hydrogen (µ = 1). Show that the mean temperature of the Sun is < T > ≅ 4 × 106K. On solving SAQ 2, you must have appreciated that the internal temperature of the Sun can be estimated without making any detailed calculations. Further, it is obvious from Eq. (8.17) that if two stars have the same mass, the denser one will be hotter. For a sun-like star, the effective surface temperature is Te ≈ 5780 K. And from the solution 6 of SAQ 2, we find that the mean solar temperature, < T > ≈ 4 × 10 K. This means that the internal temperature must be much higher. You may ask: What causes such high internal temperature in stars? To answer this, we must investigate the sources of energy generation in stars. The stars can have possibly three kinds of energy sources: gravitational, chemical and nuclear. You will study about them now.

8.4

SOURCES OF STELLAR ENERGY

Let us first consider the gravitational energy as the source of energy for stars. From Eq. (8.14), note that when the whole matter of the star is scattered at infinity, its gravitational potential energy is zero. When the stellar matter is assembled to make a 3 GM 2 . This means that during the 5 R gravitational contraction of a star, that is, during the formation of a star, an energy

star of radius R, its potential energy becomes −

equal to

3 GM 2 is released. 5 R

It can be shown (TQ 2) that the Sun’s energy would last only for about 107 yrs if gravitational energy was its only source. But, the results obtained on the basis of radioactive dating of different types of meteorites, deep terrestrial oceanic sediments and lunar rocks suggest an age of ≈ 5 × 109 years for the Sun. Thus, the gravitational potential energy cannot be the source for solar luminosity. We must look for some other source of energy for stars like the Sun. 70

We are now left with two other possible processes, namely, chemical and nuclear.

The possibility of a chemical process as the source of energy in stellar interior is also ruled out on the basis of results obtained in the following example.

Stellar Structure

Example 1 Assume that the Sun consists of hydrogen and oxygen and the proportion of these elements is such that the entire solar material could be burned and transformed to water vapour. Show that the total energy available from this process would last only 4 for ~ 10 yr given that 10 eV is liberated in the formation of each water molecule. Solution The molecular weight of water (H2O) is 18 u = 18 × 1.6 × 10−27 kg = 28.8 × 10−27 kg. The mass of the Sun is 2 × 1030 kg. Therefore, the total number of water molecules present in the Sun is given by: 2 × 10 30 kg 28.8 × 10

− 27

≈ 6 × 10 55.

kg

Thus, total energy liberated due to the formation of water vapours in the Sun is: 6 × 1055 × 10 eV × 1.6 × 10−19 J (eV)−1 ≈ 9.6 × 1037 J. Since the Sun radiates energy at the rate ~ 4 × 1026 Js−1, the duration over which the Sun would radiate all its energy generated due to formation of water molecules is: ≈

9.6 × 10 37 J 4 × 10

26

Js

−1

≈ 2.4 × 1011 s ~ 10 4 yr .

In view of the fact that the Sun has an estimated age of ~ 5 × 109 yr, the result of the above example clearly shows that chemical process cannot be responsible for generation of energy in stars. Let us now look at the possibility of nuclear processes for generating energy in a sunlike star. You may recall from your school physics that nuclear reactions are of two types: fission reactions and fusion reactions. In nuclear fission, large unstable nuclei like 238U break into smaller nuclei and energy is released. Since the abundance of such nuclei is negligible in stars, such a process is also ruled out as a source of stellar energy. We are, therefore, left with only fusion process to be considered as possible energy source. In nuclear fusion process, two lighter nuclei combine and form a new nucleus and energy is released. The amount of energy released depends on the binding energy per nucleon of the elements involved in the reaction. You may recall that the binding energy of a nucleus is the energy required to separate its constituent nucleons by a large distance. Refer to Fig. 8.2 which depicts binding energy curve. Note that as the mass number increases from zero, the value of binding energy per nucleon increases. This implies that if two or more, lighter nuclei (such as hydrogen) are fused together 3 4 to create a relatively heavier nuclei (such as He or He), we will have surplus of energy. This is precisely what happens in fusion reactions: enormous amount of energy would be released due to fusion of hydrogen nuclei and consequent production of helium. Further, Fig. 8.2 also shows that the value of binding energy per nucleon saturates at around mass number 50 and shows a very slow decrease beyond. The nature of the binding energy curve, therefore, indicates that small-mass as well as large-mass nuclei are less tightly bound than medium mass (such as Fe) nuclei.

71

Before proceeding further, you may like to convince yourself whether or not the energy generated due to nuclear fusion can account for the observed luminosity of the Sun. To do so, solve the following SAQ.

The Solar System and Stars

Fig.8.2: Variation of the binding energy per nucleon with mass number

Spend 5 min.

SAQ 3 Assume that originally the Sun comprised only of hydrogen and that the inner 10 percent of the Sun’s mass could be converted into helium. For how long would the 26 −1 Sun be able to radiate at the rate given by L = 4×10 Js . The mass of the Sun is 30 MΘ = 2×10 kg.

To know the amount of energy released due to fusion of four hydrogen nuclei and formation of a helium nucleus, note that the total mass of four hydrogen atoms is 4.031280 u while that of a helium atom is 4.002603 u. The mass defect is 0.02867u and in accordance with the 2 Einstein equation, E = mc , the energy released due to fusion of four hydrogen nuclei into a helium nucleus is equal to 2 (0.02867 u) × c where c is the velocity of light.

72

On solving SAQ 3 you find that the Sun would indeed be able to radiate energy at the present rate for another 5 billion years by fusing hydrogen into helium. Now the next logical question is: How is the fusion of hydrogen nuclei into helium nuclei made possible? In other words, you may like to know what the pre-conditions for nuclear fusion to take place are and how these conditions are obtained in the stellar interiors. Nuclear fusion involves fusion of two positively charged nuclei against Coulomb repulsion. The possibility of such a process can be understood on the basis of the potential energy curve that an atomic nucleus would experience when it approaches another atomic nucleus (Fig. 8.3). Note that the potential energy curve consists of two distinct regions. Region I corresponds to the situation when separation between the two nuclei is such that their potential energy is due to Coulomb repulsion. Region II represents the situation when two nuclei are very close to each other and the curve is in the form of a potential well. The potential well illustrates the strong nuclear forces that bind the nuclei. From Fig. 8.3, it is obvious that the two nuclei can fuse only when the approaching nuclei are able to overcome the repulsive Coulomb barrier. This means that the approaching nuclei have sufficient energy to overcome the barrier. The question is: At what temperature, the approaching nuclei will have sufficient energy? Using classical dynamics, we find that the temperature required to provide sufficient energy to two nuclei so that they overcome Coulomb’s barrier is much 7 higher than the core temperature Tc (~ 1.5 × 10 K) of the Sun. Thus, classical physics

fails to explain the possibility of fusion reactions in stars. You must convince yourself about the inadequacy of classical physics to explain nuclear fusion by solving the following SAQ.

Stellar Structure

Coulomb barrier

Energy

← Ekin ∼ kT

H Distance

Nuclear potential well (binding energy)

He

C O Si Fe Region II

Region I

Fig.8.3: Schematic representation of the potential energy barrier experienced by an atomic nucleus approaching another atomic nucleus

Spend 5 min.

SAQ 4 Suppose that two nuclei have charges Z1e and Z2e and in order to interact, they must be separated by a distance ~ 10−13m. Calculate their mutual potential energy. If their relative kinetic energy is 3 kBT, calculate the temperature required by two hydrogen nuclei to overcome this potential barrier. The temperature in the core of the Sun is ~ 1.5 × 107K. At this temperature, only a few proton-proton fusion can take place. There is, however, a finite probability that particles with insufficient energy can tunnel thorough the potential barrier and react. Now, let us look at some thermonuclear fusion reactions through which lighter nuclei fuse and release energy. In stars like the Sun, one of the prominent thermonuclear reactions is the so-called proton-proton chain or p − p chain. This reaction proceeds sequentially through three steps as given below:

H 1 + H 1 → H 2 + e + v, (1.19 MeV) H 1 + H 2 → He 3 + γ, (5.49 MeV) 73

He 3 + He 3 → He 4 + H 1 + H 1 , (12.85 MeV)

The Solar System and Stars

Note that in these reactions, four hydrogen nuclei combine to form a helium nucleus. The amount of energy liberated at each step is given within brackets. The p − p chain produces most of the energy in Sun and other such stars. Further, a relatively smaller amount of energy in the Sun is also generated by the thermonuclear reaction known as carbon-nitrogen-oxygen (CNO) cycle as given below:

H 1 + C12 → N 13 + γ, (1.45 MeV) N 13 → C13 + e + v, (2.22 MeV) H 1 + C13 → N 14 + γ, (7.54 MeV) H 1 + N14 → O15 + γ, (7.35 MeV) O15 → N 15 + e + v, (2.71 MeV) H 1 + N15 → C12 + He 4 , (4.96 MeV). You may note that in the CNO cycle, the carbon destroyed in the first step gets regenerated in the last step. Further, similar to the p − p chain, CNO cycle also produces a helium nucleus from four hydrogen nuclei. A CNO cycle, however, cannot begin unless carbon is present. The thermonuclear energy generated in the stars depends on the abundance of fusionable matter and the interior temperature of the stars. This provides us a basis to link the luminosity of stars with its mass. Let ε denote the rate of energy generated per unit mass in thermonuclear reactions. Then, the luminosity dL caused by an element of mass dM can be written as: dL = ε dM

(8.18)

Since dM = 4πr2ρ(r) dr, Eq. (8.18) can be expressed as:

dL = ε 4πr 2 ρ( r ) dr

(8.19)

Eq. (8.19) is one of the basic equations of stellar structure. The energy generated at the core of a star must flow towards its surface because the temperature of the core is very high compared to the star’s surface. Further, since the star’s surface continuously radiates energy, it will cool off unless the radiated energy is replaced. Energy transport in the star has important consequences for its structure because the transport process determines the temperature (and pressure) of different layers of the star’s interior. Would you not like to know what the different mechanisms of energy transport in a star are? This is the subject matter of the next section.

8.5

MODES OF ENERGY TRANSPORT

There are three basic energy transport processes: conduction, radiation and convection. Fig. 8.4 shows a schematic diagram for these three processes. 74

Stellar Structure

Conduction

Convection

Radiation

Fig.8.4: Three modes of energy transport: conduction, radiation and convection

Conduction is the most familiar form of heat flow and this process works through vibrating atoms. The energy of the vibrating atoms is transferred to the nearby cooler atoms by collisions and energy transport through conduction works better in solids and not in gases. You will learn later in this Unit that conduction is responsible for energy transfer in stars like white dwarfs whose interior is in a crystallised form having density ~ 106g cm−3. For ordinary gaseous stars, conduction process is not important. Radiation is the next most familiar mode of energy transport and is responsible for energy transport in some layers of the interiors of almost all the stars. The process of energy generation in the central region of a star produces very high energy photons which are γ-rays. As these photons travel outwards, they collide with matter. At each collision, γ-ray photon loses energy and when it reaches the surface, its frequency lies in the visible range. The progress of photons is extremely slow as they travel outward and this, in fact, regulates the solar luminosity at the level of ~ 1026 Js−1. The absorption of the energy of γ-rays by the stellar gas, is characterized by the absorption coefficient, also known as the opacity, kλ of the gas. The subscript λ indicates that the absorption depends on the wavelength. The important sources of opacity at high temperatures inside a star are: i) Electron scattering: The scattering of photons by free electrons. ii) Photoionisation: The energy of photon is used for successive ionisation of atoms/ions. We can obtain an expression for the opacity on the basis of qualitative arguments. Consider a slab of the stellar gas of thickness dx. Let Fλ denote the flux of radiation that strikes at one end. If the mass density of the gas is ρ, the amount of radiation absorbed by the slab is proportional to i) the density of matter in the slab ii) the incident flux and iii) the thickness of the slab. Thus, the amount of flux absorbed is: dFλ ∝ ρFλ dx

= −k λ ρFλ dx,

(8.20)

where the minus sign indicates absorption. Photons generated inside the Sun do not reach the solar surface directly; they are scattered by the electrons and nuclei. This scattering is isotropic and thus their forward and backward scattering is equally likely. The travel of photons inside a star is, therefore, like that of a drunken person. It is also called the random walk. It takes thousands of years for these photons to reach the star’s surface. To have an idea about the time taken by a photon to reach star’s surface from the core, go through the following example carefully. 75

The Solar System and Stars

Example 2 Suppose that the energy transport due to radiation process is analogous to the random walk. Compute the time taken by a photon, generated in the core of the Sun, to reach the solar surface. Given that for the Sun, the mean free path is ~ 0.5 cm for photon at an average density and temperature of 1.4 g cm−3 and 4.5 × 106 K, respectively. Solution You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course that, according to the theory of random walk, we can write the mean square distance, < D 2 > moved in N steps as: < D2 > = N 2

This gives N =

< D2 > 2

, the number of steps required to travel a distance < D 2 > in

steps of size in one dimension. For three dimensional space, N ~ 3

< D2 > 2

. Now,

for the photon at the core which has to reach the surface, we have < D 2 >= R 2 . Hence, the time taken for the photon to reach the solar surface is given by t=

3R 2 c.

since, in each step, the time taken is /c where c is the velocity of light. Substituting the values of R, c and , we get r

t ≅ 30,000 yr.

r + dr

Fig.8.5: Spherical shell of thickness dr around a surface of radius r

Now, you may like to know: How does the radiative transport of energy give rise to temperature gradient in stars? To find out, let us consider a thin spherical shell around a spherical surface of radius r as shown in Fig. 8.5. Let the temperature of the sphere be T. Since the spherical region within radius r acts as a source of black body radiation, the radiative flux incident on the inner side of the shell can be expressed by (Stefan Law): 4

F(r) = σT ,

(8.21)

where σ is the Stefan’s constant. Similarly, the radiative flux emerging outward from the shell surface at r + dr is: F (r + dr) = σ (T + dT )4

(8.22)

where T + dT is the temperature at the surface of radius r + dr. It is important to mention here that dT is negative because the surface at r + dr is cooler than the inner surface of the shell located at r. Since dT is very small compared to T, we can expand (T + dT)4 using binomial expansion. Doing so, we get: F (r + dr) = σ T 4 + 4σ T 3 dT 76

Therefore, the flux absorbed by the shell can be written as:

Stellar Structure

dF = F (r + dr) − F(r) = 4σT 3 dT

(8.23)

Combining Eqs. (8.20) and (8.23), we get: 4σT 3dT = − k(r) ρ(r) F(r) dr

(8.24)

Since luminosity, L(r) = 4πr2F(r), we can rewrite Eq. (8.24) as: k (r )ρ(r ) dT =− F (r ) dr 4σT 3 =−

k (r )ρ(r ) acT

3

L(r )

(8.25)

4πr 2

ac , where a is a constant and c is the velocity of light. Eq. (8.25) gives the 4 temperature gradient within a star due to radiative transport of energy. This expression 3 needs to be multiplied by an extra factor of on the right hand side so that it 4 becomes consistent with the one obtained by incorporating the details of such a process. We, therefore, write the temperature gradient as:

since σ =

dT 3 k (r ) ρ(r ) =− dr 4 acT 3

L(r )

(8.26)

4πr 2

Eq. (8.26) is yet another basic equation of stellar structure. Now, let us consider the third mode of energy transport, namely, convection which plays an important role in stars. Convection refers to the process in which heat energy is transported by mass motion, i.e., by transport of the hot/cool matter itself. You are familiar with convection currents or bubbles moving up and down when water in a beaker is gradually heated from below. Such a motion also takes place in certain regions in stars with hot fluid masses rising outward releasing their heat energy and the cooler matter sinking downward to receive more energy. Fig. 8.6 depicts the photospheric granulation which is a strong evidence supporting convective transport of energy at the base of Sun’s photosphere. Convection causes mixing of the stellar

Convective zone

Radiative zone

Fig.8.6: Photospheric granulation of the Sun which is caused due to convective transport of energy in which the hot matter comes out to the solar surface from layers below the surface

constituents in certain regions inside a star and produces homogeneity of chemical composition by transferring heavy elements from interior to the surface. You may ask:

77

The Solar System and Stars

Under what condition(s) does convection become the dominant mode of energy transport? The process is dominant when the temperature gradient becomes too steep. Steep temperature gradients are generally created in regions with high opacity which restrains the flow of energy through radiative process. To determine the temperature gradient in a convective region of a star, we consider a situation where hot bubbles of gas rise up and expand adiabatically. After rising through a characteristic distance, the bubbles lose extra heat and get mixed up with the surroundings. For such a process, the bubble’s adiabatic temperature gradient is given by: dT γ − 1 T dP = dr ad P dr γ

(8.27)

Eq. (8.27) can be obtained by using the adiabatic relation P = K ργ and the equation of k state of the gas, P = B ρT . Also, from the equation of state, we have P = NkBT, µm where N is the number of particles per unit volume. In astrophysics, it is usual to take the mass of a particle as µm, where µ is called the mean molecular weight and m is the dP we use Eq. (8.3) in Eq. (8.27). With these mass of a proton. Further, for dr substitutions, it is possible to write Eq. (8.27) as: dT γ − 1 µm GM (r ) =− dr ad kB γ r2

(8.28)

Eq. (8.28) is one of the basic equations of stellar structure. You should note that only if the actual temperature gradient in a star is steeper than the adiabatic gradient given by Eq. (8.27), convective transport of energy can take place. We call the actual temperature gradient in such a case as superadiabatic. Therefore, for convection to take place, we must have: dT dr actual

>

dT dr ad

(8.29)

In fact, it can be shown that convection dominates the radiative transport of energy in dT a region if is slightly superadiabatic. In any case, the actual mode of dr actual energy transport in a region inside a star depends on the temperature gradient existing there.

78

Now, you should pause for a moment and think what you have learnt so far. You have learnt to derive certain equations on the basis of the principles of physics connecting various parameters of a star. These equations are known as the basic equations of stellar structure. You may ask: Why did we do all this? How do equations of stellar structure help enhance our understanding of stars? These equations are used to develop theoretical models of stars. If the predictions of these models are in agreement with observations, we can conclude that the assumptions made about the parameters of stellar interior are valid. In case of disagreement between theoretical prediction and observations, the models are ‘fine-tuned’ by modifying the initial assumptions. You will indeed appreciate that this is the only way to investigate the interior of stars because we simply cannot look into those interiors. In the next section, you will learn to develop a stellar model on the basis of equations of stellar structure.

8.6

Stellar Structure

SIMPLE STELLAR MODEL

Developing a stellar model essentially involves solving the equations of stellar structure for a star. As such, it is a very complex task because large numbers of equations with several unknowns need to be solved. This does not mean that we cannot get a physical picture of a star. We find that, with some valid approximations, simple stellar models are easier to calculate. Such models help understand the basis of some of the empirical laws, such as, mass luminosity relation. Before discussing any stellar model, let us first list the basic equations of stellar structure. Basic Equations of Stellar Structure In the previous sections, we used the following basic physical principles to obtain the equations of stellar structures: • • • •

Hydrostatic equilibrium, Equation of state for stellar matter, Mechanism of stellar energy generation, and Modes of energy transport in stellar interior.

These basic equations are used to compute theoretical stellar models. This is equivalent to “constructing a theoretical star”! Once different models are computed, their location in H-R diagram is found out. It is so because the H-R diagram sets a detailed standard to be met by any theory of stellar structure and evolution. A theoretical model is wrong if the physical characteristics of the computed “star” are such that its location falls in the gap or empty regions of the H-R diagram. The physical parameters of a star at any point in its interior are temperature, T(r), pressure P(r), density ρ(r), and luminosity L(r). The basic equations of stellar structure are: Hydrostatic equilibrium:

dP GM (r ) =− ρ(r ) dr r2

(8.30)

Mass continuity :

dM = 4πr 2ρ(r ) dr

(8.31)

dT 3 kρ L(r ) =− dr 4ac T 3 4πr 2

(8.32)

Energy transport: (radiative) Energy transport: (convective) Energy generation: (Thermal equilibrium) Equation of state:

dT 1 m GM (r ) = − 1− dr r2 γ kB

(8.33)

dL = 4πr 2ρ(r )ε(r ) dr

(8.34)

P (r ) = Rρ(r )T (r )

(8.35)

In the above equations, ε(r) is the thermonuclear energy production rate per unit mass. You have learnt about it earlier in relation with luminosity. Further, the opacity k occurring in these equations depends on temperature and density of the gas. In fact, the exact form of the opacity relation depends on the process responsible for it. Computation of stellar opacity is, however, a complex process and is usually approximated by Kramer’s opacity relation given as 79

The Solar System and Stars

k = const.Z (1 + X )

ρ

(8.36)

T 3.5

where X denotes the amount of hydrogen in a gram of stellar matter (it is also called the abundance of hydrogen), Z denotes the abundance of heavier elements. (In astrophysics, elements heavier than helium are called heavier elements.) The opacity relation given by Eq. (8.36) is valid for stars on the main sequence. To obtain the values of physical parameters by integrating the stellar structure equations, we invoke the following boundary conditions: M(r) = 0 and

L(r) = 0 at r = 0 (the centre of star),

(8.37)

and M(r) = M;

L(r) = L and

T(r) = Teff at r = R (surface of a star) (8.38)

With this background information, we are now in a position to discuss a stellar model.

8.6.1

Polytropic Stellar Model

In such a model, there is no need to know the actual source of energy generation in the star. Further, we assume that any change in the equilibrium structure of a star takes place in such a way that the specific heat remains constant, i.e., dQ = C = constant dT

(8.39)

where C denotes the heat capacity when neither pressure (P) nor volume (V) is constant. We call such a change as a polytropic change. An adiabatic or an isothermal change, therefore, represents a polytropic change of zero and infinite heat capacities, respectively. Instead of the adiabatic relation dQ = 0, we now have dQ = CdT. In such a situation, the first law of thermodynamics dQ = CV dT + PdV takes the form CdT = CV dT + PdV Now, using the equation of state for a perfect gas, PV = RT and the fact that R = CP − CV, we can write the above expression as:

(C − CV ) dT = (C P − CV ) dV T

V

(8.40)

where CP and CV are the specific heats at constant pressure and constant volume, respectively. Let us now define an exponent γ′ similar to the adiabatic exponent γ: γ′ =

CP − C CV − C

(8.41)

Thus, we can write from Eq. (8.40) that 80

TV γ′−1 = const,

PV γ′ = const .

(8.42)

Spend 5 min.

SAQ 5

Stellar Structure

Derive Eq. (8.42). It is usual to express the physical variables, e.g., density, pressure and temperature in terms of the polytropic index n defined as: n=

1 γ′ − 1

(8.43)

′

Since PV γ −1 = const. , we can write: 1+

P = Kρ

1 n

(8.44)

where K is a constant. Further, the density is expressed in terms of a non-dimensional parameter θ defined in terms of the central density ρc as ρ = ρc θ n.

(8.45)

We, therefore, get the following expressions for pressure and temperature: P = Pc θ n+1

(8.46a)

T = Tc θ

(8.46b)

n +1

where Pc = Kρ c n and Tc =

µm 1/ n Kρ c . We shall see below that Pc and Tc are the kB

central pressure and temperature. SAQ 6

Spend 5 min.

Derive Eq. (8.46a) and (8.46b). With this formal introduction to polytropic changes, let us consider the following stellar structure equations: Hydrostatic Equilibrium:

dP

=−

dr dM

Mass continuity:

dr

GM ( r ) r2

ρ( r )

= 4πr 2 ρ(r )

(8.47)

(8.48)

Substituting Eq. (8.48) in Eq. (8.47) and rearranging terms we can write: 1 d

r 2 dP

r 2 dr

ρ dr

= − 4π G ρ(r )

(8.49)

Substituting for P and ρ from Eqs. (8.44) and (8.45) in Eq. (8.49), we get: (n + 1) K 4πG

1

ρn c

−1

1 d r

2

dr

r2

dP dr

= − θn

(8.50) 81

To write Eq. (8.50) in a simpler form, let us introduce a dimensionless variable ξ as follows:

The Solar System and Stars

r = αξ;

(8.51) 1

where α =

(n + 1) K 4πG

1

ρn c

−1 2

. Substituting Eq. (8.51) in Eq. (8.50) and rearranging

terms we get:

Lane-Emden equation 1 d 2

ξ dξ

ξ2

dθ

= − θn

dξ

(8.52)

Eq. (8.52) is known as Lane-Emden equation. Solution of this equation, for a given n, provides the density and pressure profile inside a star. The boundary conditions under which this equation must be solved are: θ = 1 and

dθ =0 dξ

at

ξ = 0.

(8.53)

θ is also known as Lane-Emden’s function. Analytical solutions of Eq. (8.52) with the specified boundary conditions are possible only for n = 0, 1 and 5. The analytical expressions for the Lane-Emden functions for these values of n are: n = 0; θ 0 = 1 − n = 1; θ1 =

ξ2 6

sin (ξ) ξ

n = 5; θ 5 = 1 +

(8.54)

1 2 −2 ξ

3

Fig.8.7 shows the density profile inside a polytrope for n = 1.5 and 3. Spend 10 min.

SAQ 7 Verify that the Lane-Emden equation (Eq. (8.52)) is satisfied by the solutions given by Eq. (8.54). The general solution of the Lane-Emden equation is in the form of a series for θn as given below: θn = 1 −

82

ξ2 n 4 + ξ − ... 6 120

(8.55)

Stellar Structure 1 Polytropes 0.8

0.6 ρ /ρ ρc

n = 1.5

0.4

n=3

0.2

0 0

0.2

0.4

0.6

1

0.8

r/R Fig.8.7: Density profile of a polytropic star

Current stellar models calculated for the Sun that fit the observations indicate that a large fraction of hydrogen at the centre of the Sun has been converted to helium (estimated 40 percent hydrogen and 60 percent helium). Energy transport is still radiative in the solar interior upto ~ 0.73 RΘ and beyond this distance, the temperature gradient reaches a value which sets up convection. Fig. 8.8 illustrates the internal structure of the present Sun. Photosphere Chromosphere 2000 km Convection Zone 330 km Radiative Zone 1.6 × 107 K 160 × 103 kg m−3 1/4R 0.80R

Corona

R

Core 8 × 104 5 × 105

20 × 103

106 Temperature (K) 4 × 10−4 6 × 103 K 8 × 10−5 Density (kg m−3)

Fig.8.8: The internal structure of the Sun as per a stellar model which shows the main regions of the Sun and values of its important physical parameters

83

The Solar System and Stars

Computation of actual stellar models involves complex mathematics and needs extensive computer resources. It is primarily due to the complexities of the equations of stellar structure. Now, let us summarise what you have learnt in this Unit.

8.7

SUMMARY

•

To understand the internal structure of stars, theoretical models of stars are developed and the predictions of these models are compared with observations.

•

The principle of hydrostatic equilibrium is one of the fundamental principles of physics invoked for investigating stellar structure.

•

The equation of hydrostatic equilibrium is given as: GM (r ) dP =− ρ( r ) dr r2

•

The mass continuity equation is given as: dM (r ) = 4πr 2 ρ( r ). dr

•

The relation between total thermal energy and the gravitational potential energy of a system of perfect gas particles such as a star is given by the so called virial theorem: 2U + Ω = 0.

•

Of the three possible sources namely gravitational, chemical and nuclear of energy generation in stars, only nuclear processes can give rise to such high internal temperatures. Energy is generated in stars due to fusion reactions, particularly due to fusion of hydrogen nuclei and consequent formation of helium.

•

One of the basic equations of stellar structure, as given below, relates the luminosity (an observable parameter) of a star with its density: dL = ε 4πr 2 ρ( r ) dr

where ε is the rate of energy generated per unit mass in thermonuclear reactions. •

Energy generated at the core of a star is transported to its surface through one or more than one of the three basic energy transport processes namely conduction, radiation and convection.

•

The temperature gradient in a star is produced due to radiative transport of energy and is given as: dT 3 k (r ) ρ(r ) =− dr 4 acT 3

• 84

L(r ) 4πr 2

Equations of stellar structure are used to develop theoretical models of stars. If the predictions of a model are in agreement with observations, we conclude that the assumptions made about the parameters of stellar interior are valid.

•

Developing a stellar model basically involves solving equations of stellar structure. This is quite a complex process because a large number of equations with several unknowns are to be solved.

•

In polytropic stellar model – a relatively simple stellar model – we do not need to know the source of energy generation. In this model, it is assumed that any change in the equilibrium structure of a star do not alter its heat capacity:

Stellar Structure

dQ = C = constant dT

•

The Lane-Emden equation is given as: 1 d

ξ2

2

ξ dξ

dθ dξ

= − θn

Solution of this equation provides the density and pressure profile inside a star.

8.8

TERMINAL QUESTIONS

Spend 25 min.

1. Show that when four protons combine to form helium, the energy released is ~ 26.7 MeV. 2. At present, the Sun radiates energy at the rate of ~ 4 × 1026W. Assuming that the gravitational contraction is the only source of the Sun’s radiant energy, how long, since its creation, it would have radiated energy at the present rate? Take M Θ = 2 × 10 30 kg and RΘ = 7 × 10 8 m . 3. Show that Eq. (8.55) is consistent with Eq. (8.54).

8.9

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1.

The left hand side of Eq. (8.8) is: R

3

P 4πr 2 dr = 3

0

R

PdV = 2 0

3

R

2 0

PdV = 2U

since the internal or thermal energy can be expressed as 2.a)

3 PdV . 2

From Eq. (8.11), we can write the total potential energy as: R GM ( r )

−Ω=

0

r

4πr 2 ρ(r ) dr

Assuming that ρ(r) ≈ < ρ > , the mean density of stellar matter, we can write: −Ω=

R GM ( r ) 0

r

4πr 2 < ρ > dr =

RG

4π 3 r < ρ > 4πr 2 < ρ > dr 0 r 3 85

The Solar System and Stars

R M2 4 π 4π 4 r < ρ > 2 dr = 3 G.r 4 dr 3 3 R6 0

R

=

3G 0

(using Eq. (8.13)) =

3GM

2

R

R6

r 4 dr

0

or, 3 GM 2 5 R

Ω=−

This is Eq. (8.14). Further, from the virial theorem (Eq. (8.12)), we can write: U =−

=

Ω 2

3 GM 2 10 R

(using Eq. (8.14))

Also, from Eq. (8.16), we have: U=

3 kB

M

2 µm

Comparing the above expressions for U, we can write: 3 GM 2 10

R

=

3 kB M 2 µm

or,

< T >=

1 GM µ m 5 R kB

Further, M=

4 πR 3ρ 3

or, 1 3

M R∝ ρ

Substituting this value of R, we get

∝M

2

1

3

3

which is Eq. (8.17). b) Eq. (8.17) can be written as: = 86

1 GM µ m 5 R kB

Substituting the values of G, M, R and kB, we get:

Stellar Structure

1 (6.7 × 10 −8 cm 3 g −1s − 2 ) × (2 × 10 33 g) × (1.6 × 10 −24 g) = . 5 (1.4 × 10 −16 erg K −1 ) × (7 × 1010 cm)

≈ 4 × 106 K. 3.

Mass of four hydrogen atoms = 4.031280 u Mass of a helium atom = 4.002603 u So, the mass defect = 0.02867 u Thus, the energy released per gram when four protons combine to form helium is =

(0.02867 u) × (9 × 10 20 cm 2 s − 2 ) 4.0328

≈ 6 × 1018 erg. Thus, we can write: lifetime of the Sun =

(2 × 10 33 g) × (6 × 1018 erg g −1 ) ( 4 × 10 33 erg s −1 ) = 3 × 1017 s

≈ 1010 yr. 4.

We can write the potential energy of two nuclei Z1e and Z1e separated by a distance r as: Z1 Z 2 e 2 r

P.E. =

=

e2 r

if Z1 = Z2 = 1

Since the potential barrier will be overcome by the relative kinetic energy of the two nuclei, we can write: 3k B T =

e2 r

or, T=

=

e2 3k B r 4.8 × 4.8 × 10 −20 3 × 1.4 × 10 −16 × 10 −11

≈ 5 × 107 K.

87

The Solar System and Stars

5.

From Eq. (8.40), we have:

(C − CV ) dT = (C P − CV ) dV T

V

If we define an exponent γ ′ as γ′ =

Cp −C Cv − C

we can write: CP − C −1 CV − C

γ′ − 1 =

=

C P − C − CV + C CV − C

=

C P − CV CV − C

Further, Eq. (8.40) can be written as:

(CV

− C)

dT dV + (C P − CV ) =0 T V

or, dT dV + ( γ ′ − 1) =0 T V ′

TV γ −1 = Const.

Further, using the equation of state of a perfect gas: PV = RT we can write: dP dV dT . + = P V T

Substituting for

dT , we get: T

dP dV + γ′ =0 P V

or, ′

PV γ = Const.

6.

From Eq. (8.44), we have:

P = Kρ 88

n +1 n

(

= K ρc θ n +1 n

= Kρ c

n

)

n +1 n

Stellar Structure

(using Eq. (8.45))

θ n +1

= Pc θ n +1 n +1 n

where Pc = Kρ c

Further, we can write pressure as: P=

kB ρT µm

T=

µm P kB ρ

or,

n +1

µm ρ n = K kB ρ

(substituting Eq. (8.44))

=

µm µm K (ρ c θ n ) 1 / n Kρ1 / n = kB kB

=

µm Kρ1c/ n θ kB

= Tc θ where Tc =

7.

(substituting Eq. (8.45))

µm Kρ1c/ n kB

Lane-Emden equation (Eq. (8.52)) is: 1 ξ

2

d dξ

ξ2

dθ = − θn dξ

For n = 0, this equation reduces to: d dθ ξ2 = −1 dξ ξ 2 dξ

1

And, for n = 0, we have θ = 1 −

ξ2 6

Substituting this value of θ in the left hand side of the above equation, we get: d ξ ξ2 − 2 dξ 3 ξ

1

=

d ξ3 − 3 ξ 2 dξ

1

=

1 ξ2

−

3ξ 2 3

= −1.

89

The Solar System and Stars

Further, for n = 1, we have θ =

sin ξ from Eq. (8.54). So, we get: ξ

dθ cos ξ sin ξ = − dξ ξ ξ2 dθ in the left hand side of Lane-Emden dξ

Again, substituting this value of equation, we get: 1 ξ

d (ξ . cos ξ − sin ξ ) dξ

2

1

=

ξ

2

[− ξ sin ξ + cos ξ − cos ξ] = −

sin ξ = −θ ξ

as required. ξ2 n = 5, we have, θ = 1 + 3

For

−1 / 2

or, dθ ξ2 1 =− 1+ dξ 2 3

Substituting the value of

−3 / 2

.

2ξ 3

dθ in the left hand side of Lane-Emden equation, we dξ

get: −3 / 2 1 d ξ3 ξ2 1+ − 3 3 ξ 2 dξ

−3 / 2 −5 / 2 ξ3 3 ξ2 ξ2 3ξ 2 2ξ + = − 1+ . 1+ . 2 3 3 3 2 3 3 ξ

1

ξ2 = − 1+ 3 ξ2 = − 1+ 3

−3 / 2

ξ2 ξ2 1+ + 3 3

−5 / 2

ξ2 = − 1+ 3

−5 / 2

1+

ξ2 ξ2 − 3 3

−5 / 2

= − θ5 , as required.

Terminal Questions 1. When four protons combine to form a helium atom, the mass defect is 0.02867 u. Thus, energy released in this process: 90

E = mc 2

Stellar Structure

= (0.02867 ×1.6 ×10 − 27 kg) × (9 ×1016 m 2 s − 2 ) = 4.128 ×10 −12 J

=

4.128 ×10 −12 1.6 ×10 −19

eV

= 26 MeV. 2. Assuming the Sun to be a sphere, its gravitational potential energy can be written as: 3 GM 2 5 R

Ω=

So, the time τ for which the Sun will radiate with its luminosity, L can be written as: τ=

3 GM 2 1 . 5 R L

=

3 GM 2 1 1 . . yr . 5 R L 3 × 10 7

=

3 (6.7 × 10 −11 m 3 kg −1 s −1 ) × (4 × 10 60 kg 2 ) 1 × yr . 5 (7 × 10 8 m) × (4 × 10 26 kg m 2 s −1 ) (3 × 10 7 )

= 10 8 ×

1 yr 5

≈ 2 × 10 7 yr .

3. Eq. (8.55) is: θn = 1 −

ξ2 n 4 + ξ − ... 6 120

For n = 0, this equation reduces to: θ0 = 1 −

for

ξ2 6

→ a constant

n = 1,

θ1 =

1 1 ξ3 ξ5 sin ξ = ξ− + − ... ξ ξ 3! 5!

91

The Solar System and Stars

=

1 ξ2 ξ4 ξ3 ξ5 ξ− + − ... = 1 − + 6 120 6 120 ξ

=1−

for

ξ2 n 4 + ξ − ... 6 120

n = 5, ξ2 θ5 = 1 + 3

−1 / 2

=1−

1 ξ2 − 1 − 3 1 ξ4 + . . . . 2 3 2 2 2 9

=1−

ξ2 ξ4 ξ2 5 4 + =1− + ξ − ... 6 24 6 120

=1−

ξ2 n 4 + ξ − ... 6 120

Thus, we find that Eq. (8.55) is consistent with Eq. (8.54).

92

UNIT 9 STAR FORMATION

Star Formation

Structure 9.1

Introduction

9.2

Basic Composition of Interstellar Medium

Objectives Interstellar Gas Interstellar Dust

9.3

Formation of Protostar Jeans Criterion Fragmentation of Collapsing Clouds

9.4

From Protostar to Pre-Main Sequence

9.5 9.6 9.7

Summary Terminal Questions Solutions and Answers

9.1

INTRODUCTION

Hayashi Line

In Unit 8, you have learnt the basic physical principles governing the processes in the interior of stars. The existence of stars was taken for granted and their coming into being was not discussed. It is, however, logical to ask: Where do the stars come from? How are the stars formed? In general terms, you did study about the formation of the solar system including the Sun (the only star in our solar system) in Unit 6. But the focus of that unit was to understand the formation and characteristics of the planets. In the present unit, you will learn about the formation of stars. You must have observed stars in the night sky. On a clear dark night away from city lights, you can observe that the space surrounding the bright stars is fairly luminous. This is due to the scattering of light from the star by the gas and dust in its surroundings. Further, astronomical observations suggest that the more luminous and massive stars are younger and have been formed recently. Since these stars are completely surrounded by gas and dust, it is believed that the stars are formed from it. The gas and dust in the interstellar space is called the Inter Stellar Medium (ISM) and it is the major constituent of our galaxy − the Milky Way Galaxy. You will learn the basic composition of ISM and methods of its detection in Sec. 9.2. Under suitable conditions, a cloud of gas and dust condenses or collapses due to its own gravity and forms protostars. In Sec. 9.3, you will learn the Jeans criterion which is a rough measure of the mass and size of the interstellar cloud which may collapse and give birth to a star. You will discover that the interstellar cloud must fragment repeatedly to form stars. In Sec. 9.4, you will learn how a protostar evolves into a full fledged star and becomes a member of the main sequence (in the H-R diagram discussed in Unit 7).

It is usual to write our galaxy with the upper case G, that is, as Galaxy.

The various stages from the initial collapse of a cloud of ISM upto the pre-main sequence are collectively considered as birth of a star, that is, the process of its formation. You may ask: What happens afterwards? This issue is addressed in the next Unit on Nucleosynthesis and Stellar Evolution where we discuss the life of stars. Objectives After studying this unit, you should be able to: • • • •

describe the basic composition of ISM; derive Jeans criterion for the stability of a gas cloud; explain the necessity of repeated fragmentation of a collapsing cloud; and discuss the evolution of an interstellar cloud into a pre-main sequence star.

5

From Stars to Our Galaxy

9.2

BASIC COMPOSITION OF INTERSTELLAR MEDIUM

The interstellar medium (ISM) makes up only 10 to 15 percent of the visible mass of the Milky Way Galaxy. It comprises matter in the form of gas and dust (very tiny solid particles). About 99 percent of ISM is gas and the rest is dust. You may like to know: Which elements are present in the ISM? An important clue for investigating the composition of the ISM is the fact that the birth and death of stars is a cyclic process. This is so because a star is born out of ISM, and during its life, much of the material of the star is returned back to the ISM by the process of stellar wind (in case of the Sun, it is solar wind; see Unit 5) and other explosive events such as Nova and Supernova. The material thrown back into the ISM may form the constituents of the next generation of stars and so on. To know the basic composition of ISM, astronomers use photographs and spectra. In the following discussion, we shall confine ourselves to the composition of the ISM of the Milky Way Galaxy in the neighbourhood of the Sun. Further, for simplicity, we will first discuss the composition of interstellar gas and then come to interstellar dust.

9.2.1

Interstellar Gas

Hydrogen and helium are the two major constituents of interstellar gas; hydrogen constitutes about 70 percent and the rest is helium. Indeed, other elements are also present but in very small quantities. The analysis of the radiation received from ISM has enabled astronomers to classify the gaseous matter filling the interstellar space into the following four types: i) ii) iii) iv)

H II region, H I region, Inter-cloud medium, and Molecular cloud.

We now briefly describe these regions and their possible roles in the formation of stars. i) H II region: As the name suggests, these regions of ISM mainly consist of (singly) ionised hydrogen. In addition, these regions contain ions of other elements such as oxygen and nitrogen and free electrons. Since the ionisation energy of hydrogen atom is very high, such regions can exist only in the vicinity of very hot stars. H II region can be viewed even with naked eye in the constellation of Orion. If you look carefully at Orion’s sword, you will see that one of the objects is a hazy cloud of gas (Fig. 9.1). This bright cloud is called Orion nebula. The spectrum of this nebula shows lines of hydrogen and some other elements in its bright line spectrum. Such bright nebulae are also called emission nebulae. It has been discovered that emission nebulae do not shine by their own light. They absorb high energy ultraviolet photons from hot stars. These photons ionise the gas in the nebulae and subsequently, low energy photons are radiated. Since the spectrum of the nebula consists of many emission lines of hydrogen, it indicates that the light must have been emitted by a low-density gas. Further, the emission lines of hydrogen are very strong and the red, blue and violet Balmer lines blend together resulting in the characteristic pink-red colour of the nebula (Fig. 9.1).

6

Star Formation

Fig.9.1: Orion nebula − a typical bright and diffuse nebula

At this stage, you may ask: Why are the H II regions always found in the emission nebulae? Note that only those photons which have wavelengths shorter than 91.2 nm have sufficient energy to ionise hydrogen. Such high energy photons can be produced in sufficiently large numbers by hot (~ 25,000 K) stars only. And stars (such as hot O or B star) having temperature of this order are located in or near the emission nebulae. Further, H II regions have very low density (~ 109 particles m−3). They provide observable evidence supporting the existence of matter in interstellar space. ii) H I region: Although it has been generally believed by astronomers that hydrogen atoms populate interstellar space, they could not detect H I gas till 1951. The reasons are obvious: it is not possible for the neutral hydrogen in ISM to produce emission line as it is in the ground state. However, with the development of radio telescopes, it is now possible to detect H I region. The detection of H I in the ISM is based on the detection of a unique radiation of wavelength 21 cm. SAQ 1

Spend 5 min.

Calculate the energy of electromagnetic radiation having wavelength 21-cm. On solving SAQ 1, you must have found that the value of the energy of radiation corresponding to wavelength 21-cm is very small. You may ask: What kind of transition produces such a low energy photon? In a hydrogen atom, an electron revolves around the proton. Since the proton and the electron possess spin, there are two possible ways for their spins to align with respect to each other. The two spins may be parallel (aligned) or anti-parallel (anti-aligned) (Fig. 9.2). It is known that the parallel spin state of hydrogen atom has slightly more energy than the anti-parallel spin state. Therefore, if there is a flip from the parallel to the antiparallel state, there is a loss in energy of the hydrogen atom and it results in the

7

From Stars to Our Galaxy

emission of a photon. The frequency and the wavelength of such emissions are 1420 MHz and 21 cm, respectively. 21-cm radiation

Proton

Electron

Parallel spins

Anti-parallel spins

Fig.9.2: Parallel and anti-parallel spin alignments of the electron and proton in a hydrogen atom

Further, the question is: Can we obtain 21-cm radiation in laboratory conditions? It is not possible because the best vacuum that can be produced in a laboratory has a much higher density of atoms than that found in the ISM. So, in laboratory conditions, practically all the hydrogen atoms get de-excited due to higher collision rate and we cannot obtain 21-cm line. Thus, the only place where favourable conditions for emission of such radiations can exist is outer space. The 21-cm line was first detected in the year 1951 using a radio telescope even though it was predicted as early as in the 1940s. Since its detection, it has become a common tool in astronomy to map the location and densities of H I regions. This helps in determining the structure of galaxies including our own. Investigations of diffuse interstellar H I clouds suggest that their temperatures are in the range of 30 – 80 K, masses in the range of 1 – 100 solar masses and the number densities in the range of 108 – 109 cm−3. iii) Inter-cloud Medium: Having read about clouds of ionised and neutral hydrogen in ISM, you would like to know: Is the space between the interstellar clouds empty? It is not so; the inter-cloud space consists of a) neutral hydrogen gas with a density of 105 atoms m−3, and 4 −3 b) hot (~ 8000 K) ionised gas with very low density (~ 10 ions m ). You would further like to know: Is there any interaction between the H I clouds and the inter-cloud medium? The H I clouds are very cool and have high densities whereas inter-cloud medium has very low density and high temperature. The pressure of a region, being a function of its density and temperature, in the H I clouds and in the inter-cloud medium is about the same and they are in equilibrium.

Hydrogen molecules have been detected in the ISM by infrared spectroscopy.

8

iv) Molecular Cloud: Analysis of optical spectra of interstellar medium reveals that matter exists in molecular form in ISM. Since hydrogen is the most abundant matter in ISM, it mainly consists of the hydrogen molecules (H2). However, molecules of hydrogen do not emit photons of radio wavelength and vast clouds of hydrogen molecules remain undetected by radio spectroscopy. Other molecules, such as CO (carbon monoxide), capable of emitting in radio wavelength, can indeed be detected. In fact, nearly 100 such molecules have been detected. But the basic question is: How do these molecules form in ISM? It is believed that the atoms come in the vicinity of each other and bond to form molecules on the surfaces of the dust grains (about which you will learn later in this Section). These molecules are very weakly bonded and can be easily broken by high energy photons. Thus, they can exist only deep inside dense clouds. Also, efficient release of energy by molecules makes these dense clouds very cool. These dense, cold clouds are called molecular clouds.

In our Galaxy, the largest of these cool, dense molecular clouds are called giant molecular clouds (GMC). They are 15 to 60 pc across and may contain 100 to 106 solar masses! The internal temperature of GMC is very, very low (~ 10 K). The question is: Do the giant molecular clouds have any role in the formation of stars? You know that young stars are surrounded by H II regions. And H II regions are invariably found near giant molecular clouds. This proximity indicates that the GMC plays an important role in the formation of stars. Thousands of GMC exist in the spiral arm of our Galaxy. The association of O and B main sequence stars with GMC suggests that star formation takes place in these regions. We will talk more about it later in this Section.

9.2.2

Star Formation

Interstellar Dust

Interstellar medium also contains dust grains, which constitute about one percent of the interstellar mass. Although the temperature of interstellar dust region is very low (~ 100 K), it can be detected by infrared telescopes. A typical dust grain is made of thin, highly flattened flakes of silicates and graphites coated with ice. Fig. 9.3 shows a typical dust grain, which is of the size of the wavelength of blue light. Mantle (ice)

Core (Silicates, Graphite)

0.1 m

Fig.9.3: A typical dust grain

Now, you may ask: What is the evidence supporting the existence of dust in ISM? The two observable effects due to dust are extinction and reddening. Refer to Fig. 9.4. Note that, besides the brighter gas and dust regions surrounding the stars, there are darker regions as well. You may think that these regions are devoid of stars. It is not so. In fact, these darker regions are so dense that they completely stop the light emitted by stars behind them and therefore no light is able to pass through. This phenomenon is known as interstellar extinction. The extent to which light is scattered or absorbed in a dust cloud depends on the number density of particles and on its thickness.

9

From Stars to Our Galaxy

Dark region

Fig.9.4: Dark interstellar space caused due to extinction

To obtain an expression for the apparent magnitude of a star, located behind the dense cloud of dust, recall (from Unit 1) that the relation between the apparent magnitude (mλ), the absolute magnitude (Mλ) and the distance d in parsec of a star at wavelength is written as: m = M + 5 log10 d − 5

(9.1)

since the absorption and scattering of light is dependent on the wavelength. For stars suffering extinction, we can write Eq. (9.1) as: mλ = M λ + 5 log10 d − 5 + aλ

(9.2)

where aλ is the magnitude of light scattered or absorbed along the line of sight. Eq. (9.2) indicates that absorption increases the magnitude of a star. A star which would be visible to the naked eye, for instance, may be invisible due to the large extinction i.e., sufficiently large aλ. Further, the extinction may be expressed in terms of the optical depth as: aλ = 1.086 τλ Spend 10 min.

(9.3)

SAQ 2 The relation between optical depth, τλ and intensity Iλ for a star is given by I = I 0e − τ .

Show that aλ = 1.086 τλ. (Hint: Remember that apparent magnitude may be written as m = K− 2.5 log I, where K is a constant.) To appreciate the fact that extinction depends on the density of dust grains, we can express the optical depth τλ in terms of the number density of particles, n and scattering cross section σλ as: s

10

τλ =

n ( s) σ λ ds. 0

(9.4)

Assuming that the scattering cross-section σλ is constant along the line of sight, we obtain from Eq. (9.4): τλ = σλ Nd

Star Formation

(9.5)

where Nd is the column density of particles, i.e., the number of particles in a cylinder of unit cross section stretching from the star to the observer. Eq. (9.5) shows that extinction depends on the amount of interstellar dust present in the path of light from the star. BD + 56524

a /aν

HD 48099

Herschel 36

0

2

4

6

8

l/ ( m-1)

Fig.9.5: Schematic diagram showing the variation of

Fig. 9.5 shows the variation of

aλ aν

with

1 λ

a 1 . Here, aν is the amount of extinction in with aν

the visual band of wavelengths centered at 5500 A . In Fig. 9.5 we observe a peak in the ultraviolet region which indicates that radiations of corresponding wavelengths are strongly absorbed by ISM. Such a peak, therefore, provides a basis for determining the composition of ISM. It is now known that graphite interacts strongly with electromagnetic radiation of wavelength around 2175 A . Therefore, the occurrence of peak at λ = 2175 A in Fig. 9.5 suggests the presence of graphite as one of the constituents of ISM. Further, the presence of absorption bands at 9.7 µm and 18 µm in the observed spectrum (not shown in Fig. 9.5) indicate the presence of silicate grains in the ISM. Yet another manifestation of interstellar dust is in the form of interstellar reddening. You know that an O star should be blue in colour. But, it has been observed that some stars with the spectrum of an O star look much redder. This is caused due to scattering of light from stars by interstellar dust.

11

From Stars to Our Galaxy

Blue light Red light

Star Telescope Interstellar dust

Blue light

Fig.9.6: Interstellar reddening caused due to the scattering of blue light coming from star

Refer to Fig. 9.6. The light coming from a star behind the dust cloud is scattered. Since the typical size of dust grains is of the order of the wavelength of blue light, the blue light from star is scattered more than the red light. As a result, some of the blue light from the star is lost and it appears redder. You might wonder as to why we discussed ISM in such detail. It is because astronomers suspect that stars are formed from ISM. The most important observation supporting this hypothesis is the association between young stars and clouds of gas: wherever we find the youngest group of stars, we also find large clouds of gas illuminated by the hottest new stars. In the following section, you will learn about the processes and principles involved in the formation of stars from ISM.

9.3

FORMATION OF PROTOSTAR

In the previous Section, you learnt that the giant molecular clouds (GMCs) are the likely sites where star formation can take place. However, GMCs are very unlike stars: their typical diameter is 50 pc, typical mass exceeds 105 solar masses and their density is 1020 times less than a typical star. Thus, the question is: How do stars form from such molecular clouds? The simplest answer to this question is that the stars form due to gravity: a cloud of gas contracts due to self-gravity. This contraction increases density and temperature of the core, initiating generation of nuclear energy. You may ask : How do we reconcile the typical mass of a star with the mass of a GMC (~ 105 solar mass)? This problem was addressed by Jeans who proposed a criterion for the mass of a contracting cloud which may evolve into a star.

9.3.1

Jeans Criterion

Jeans proposed that there are two competing processes in the gravitational collapse of a molecular cloud. On the one hand, the gravitational contraction increases the internal pressure of the cloud which tends to expand the cloud. On the other hand, gravity acts on the cloud and tends to further contract it. Which of these two processes will dominate is determined by the mass of the cloud. If the internal pressure is more than the gravitational force, the cloud will break up. A clump of cloud must have a minimum mass to continue collapsing and give birth to a star. This minimum mass is called the Jeans mass. It is a function of density and temperature. To obtain an expression for the Jeans mass, we make the following simplifying assumptions:

12

i) the cloud is uniform and non-rotating; ii) the cloud is non-magnetic; and iii) the gas and dust is confined to a certain region of space by the gravitational force and is in hydrostatic equilibrium.

For such a system, we may write the relation between kinetic and potential energies as (see virial theorem, Unit 8): 2U + Ω = 0,

Star Formation

(9.6)

where U is the internal kinetic energy and Ω is the gravitational potential energy of the cloud. If M and R are the mass and radius of the cloud, respectively, the potential energy of the system can be written as (Eq. (8.14), Unit 8): Ω=−

3 GM 2 . 5 R

(9.7)

If the number of particles in the cloud is N and its temperature is T, the internal kinetic energy of the cloud can be written as: U=

3 Nk BT , 2

(9.8)

M , m where µ m is the mean molecular weight and m is the mass of a hydrogen atom. If the total internal kinetic energy is less than the gravitational potential energy, the cloud will collapse. This condition reduces Eq. (9.6) into

where kB is the Boltzmann constant. Further, the number of particles N =

2U < Ω

Substituting for Ω and U from Eqs. (9.7) and (9.8), respectively, we get: 3k BTM 3 GM 2 < . m 5 R

or, M>

5k BTR mG

(9.9) 1

3M 3 On substituting R = in Eq. (9.9), we find that the minimum mass that will 4πρ initiate a collapse is given by:

5 k BT M ≈ MJ = mG

3 2

1

3 2 4πρ

(9.10)

Here MJ is called the Jeans mass and Eq. (9.9) is known as Jeans criterion. The Jeans mass is the minimum mass needed for a cloud to balance its internal pressure with self-gravity; clouds with greater mass will collapse. Jeans criterion can also be expressed in terms of the Jeans length, RJ given by:

RJ ≈

15k B T 4π G m

1 2

(9.11)

13

From Stars to Our Galaxy

which can be obtained by putting M =

4π 3 R ρ in Eq. (9.9). For pure hydrogen, 3

µ = 1. Spend 5 min.

SAQ 3 A collapsing cloud is made of neutral hydrogen (H I) only. If the temperature of the 5 −3 cloud is 50 K and its number density is 10 m , calculate its Jeans mass. In a situation where Jeans criterion is satisfied, the cloud must collapse gravitationally. When the collapsing could attain high density, it becomes gravitationally unstable and breaks up into smaller pieces. Thus, the general picture is that stars form in groups due to fragmentation. You will learn more about it later in this section. At this stage, you may ask: How long does it take for a cloud to collapse? It is a good idea to estimate the minimum time which will be taken if we assume that the cloud collapses only under the influence of self-gravity and there is no other process taking place to slow down the collapse. This assumption is called free-fall collapse, and it implies that the pressure gradient in the interior of the clump is negligibly small. To obtain a rough estimate of the free-fall time, you may recall that a particle on the surface of a star of mass M and radius R experiences an acceleration g given by: g=

GM

.

R2

(9.12)

If the time taken by this particle to fall through a distance R is tff we can write: R=

1 g t ff 2 2

or 2R t ff = g

1 2

(9.13)

Substituting Eq. (9.12) in Eq. (9.13), we get: 3

tff =

=

2R GM

1 2

2R 3 4π 3 G R ρ 3

1 2

1

=

14

3 2 2πGρ

(9.14)

Eq. (9.14) shows that the free-fall time, tff is a function of the cloud’s initial density and it does not depend on the initial radius or the initial mass of the cloud. If some part of a collapsing cloud, say in the central region, is denser than its surroundings, the collapse of such a region is likely to be faster than that of the surrounding region.

Spend 5 min.

SAQ 4

Star Formation

Calculate the free-fall time for a molecular cloud whose initial density is 10−17 g cm−3. Now you know that the larger gas clouds collapse if their masses exceed the Jeans mass. In a typical situation of an H I cloud with T = 100 K, ρ = 10−24 g cm−3 and µ = 1, we find that for gravitational collapse, the mass of the cloud must be greater than 105 times the solar mass. So, you may be led to believe that the stars could be formed with masses of this order. However, observations suggest that the stars are formed in groups and the masses of the stars are in the range of 0.1 − 120 MΘ. Thus, the range of stellar mass is much smaller than 105 MΘ. This has led astronomers to propose the fragmentation of interstellar clouds during collapse. You will now learn about it.

9.3.2

Fragmentation of Collapsing Clouds

As mentioned earlier, Jeans criterion provides the theoretical justification for fragmentation of a collapsing cloud. Refer to Fig. 9.7 which shows the fragmentation of interstellar cloud of mass M and Jeans mass MJ such that M > MJ. Since M is greater than MJ, the interstellar cloud collapses. This results into the increase of the density and temperature of the cloud which may, in turn, change the Jeans mass for the cloud. Let the changed Jeans mass be M ′J (≠ MJ ). Interstellar cloud M > MJ

Collapse

M'J

Stage II

M'J > MJ

M, M'J M > M'J

MJ

Stage I

M > MJ > M'J

M, M'J M < M'J Stable

M'J < M1

Stage III

M'J < M2

M'J < M3

Stage IV Fragmentation Stage V M'J < M1

M'J < M2 M'J < M3

Fig.9.7: Fragmentation of interstellar cloud

15

From Stars to Our Galaxy

Further, from Eq. (9.10) it is evident that, if the cooling of this cloud is not efficient, the Jeans mass will increase (stage II in Fig. 9.7). If the mass of the cloud is less than the new (increased) value of the Jeans mass, the collapse will stop and if M > M ′J , the cloud will collapse further. On the other hand, if the cooling of the collapsing cloud is efficient, the temperature of the cloud will fall and the Jeans mass would decrease (stage III in Fig. 9.7). In such a situation, it is possible that the mass in certain regions of the cloud is more than the reduced Jeans mass (stage IV in Fig. 9.7). This may trigger further collapse of such regions resulting into fragmentation of the cloud (stage V in Fig.9.7). Fragmentation of clouds continues until clouds of still smaller masses, which are gravitationally stable, are created. These fragments of molecular clouds are the birth places of the stars in the mass range of 0.1 − 120 MΘ. You must note that for the scenario discussed above to be true, the Jeans mass should not be a constant during collapse. It must decrease with increase in density in a local region during collapse. This would lead to further gravitational instabilities which may lead to separate individual collapsing regions. Further, while discussing the collapse of interstellar clouds, we assumed that the collapse process is isothermal. This can be considered to be a valid assumption as far as the initial stages of collapse are concerned. As the collapse begins, the cloud is likely to be optically thin due to its low density. Therefore, the gravitational energy released during collapse is radiated away completely, keeping the temperature of the cloud unchanged. You may ask: What happens if the collapse is adiabatic instead of isothermal? In that case, the energy released during the collapse is used up in raising the internal energy of the cloud and its temperature increases. The increase in temperature thus affects the Jeans mass. Let us now obtain an expression for the Jeans mass for adiabatic collapse. You may recall from the course entitled Thermodynamics and Statistical Mechanics (PHE-06) that, for an adiabatic process, the relation between temperature T and density ρ of a system is given by: T = K1 ργ−1

(9.15)

where K1 is a constant and γ is the ratio of heat capacities. Using this relation in the expression for the Jeans mass (Eq. (9.10)) we can write: 5k BT MJ = µmG

3 2

1

3 2 4πρ 3

5k B K1 2 = µmG

1

3 2 ( 3 γ − 4) / 2 ρ 4π

Thus M J ∝ ρ ( 3 γ − 4) / 2

(9.16)

For the cloud comprising only atomic hydrogen, we have γ = 5/3. Therefore, for adiabatic collapse of the cloud, we get: M J ∝ ρ1/ 2 16

(9.17)

Eq. (9.17) shows that Jeans mass increases with increase in density. However, the relation between the Jeans mass and density for the isothermal collapse is given as MJ ∝ ρ−1/2. Comparison of these two results indicates that in a switchover from isothermal to adiabatic collapse, the Jeans mass reaches a minimum for the fragments of molecular clouds.

Star Formation

The gravitational collapse is a complex process. It is difficult to ascertain exactly when a switchover from isothermal to adiabatic collapse takes place. Moreover, the transition is neither instantaneous nor complete. The minimum Jeans mass has been estimated to be ~ 0.5 MΘ. This is the right order of magnitude for the mass of the fragments of clouds which become stars. The fragmentation of collapsing clouds leads to the formation of protostars − prestellar objects which are hot enough to radiate infrared radiation but not hot enough to generate energy by nuclear fusion. Due to gravitational collapse, the fragmented clumps of cloud contract and at the core of each of them, high-density region develops which is surrounded by a low-density envelope. Matter continues to flow inward from the outer parts of the clumps. The protostar begins to take shape deep inside the enveloping cloud of cold, dusty gas. These clouds are called cocoons because they hide the forming protostar from our view. So, these are the initial stages of the formation of stars. Let us now discuss this further.

9.4

FROM PROTOSTAR TO PRE-MAIN SEQUENCE

We have seen above that initially a cloud collapses, perhaps isothermally, and with the passage of time the cloud collapse tends to become adiabatic. As a result, increased pressure inside tends to slow the time scale of collapse particularly near the centre of the cloud. Further, fragmentation of collapsing cloud leads to a stage when a protostar is formed and its interior attains hydrostatic equilibrium. The interior of the protostar at this stage becomes almost convective, that is, the temperature gradient inside becomes larger than the adiabatic gradient. The stellar object so formed is called a pre-main sequence star. To understand the evolution of a pre-main sequence star, its position on the HR diagram and changes in its interior during evolution, you must study the concept of a Hayashi line.

9.4.1

Hayashi Line

In the HR diagram, Hayashi line or track runs almost vertically in the temperature range of 3000 to 5000 K as shown in Fig. 9.8. This line is important in the discussion of pre-main sequence evolution of stars because of the following features: i)

When a protostar is formed, the interior of the cloud attains hydrostatic equilibrium. At this stage, it is fully convective. The position of such an object must fall on this line.

ii) For a given mass and chemical composition, this line represents a boundary in the HR diagram. It separates the HR diagram into allowed and forbidden regions. The forbidden region occurs to the right of this line and the stars in this region cannot attain hydrostatic equilibrium. For stars falling to the left of this line, energy transport due to convection/radiation or both is possible.

17

Luminosity (Log scale)

From Stars to Our Galaxy

Hayashi line

Main Sequence 3.8

3.7 3.6 Log Teff

3.5

Fig.9.8: The Hayashi line

The internal temperature of pre-main sequence stars is quite low and cannot ignite thermonuclear reactions. To compensate for the loss of energy through radiation, a protostar must contract and this leads to increase in its thermal energy. Because of increase in temperature and pressure in the interior of the protostar, its collapse slows down. In other words, we expect longer evolution time in the pre-main sequence stage for smaller mass protostars. For more massive stars, the pre-main sequence evolution is faster and their pre-main sequence lifetime is shorter. Now, let us sum up what you have learnt in this unit.

9.5 • •

• •

•

SUMMARY The space between the stars, called the interstellar medium (ISM), is not empty. It contains gas and dust, and the gas is mostly hydrogen. The interstellar gas is not uniformly distributed. At places, it is highly concentrated. These regions are called gas clouds. It is in these gas clouds that new stars are formed. The gaseous matter in the interstellar space is classified into four types, namely, HI region, H II region, inter-cloud medium and molecular clouds. The interstellar dust, which constitutes about one percent of the interstellar mass, gives rise to extinction and reddening. Extinction depends on the density of dust grains and reddening is caused due to scattering of light from stars by interstellar dust. Jeans proposed that a molecular cloud must have certain minimum mass, called Jeans mass, for its collapse due to self-gravity. The expression for the Jeans mass is:

MJ =

• 18

5k BT µ mG

3 2

3 4πρ

1 2

In terms of the size, the Jeans criterion implies that if a gas cloud becomes larger than a certain size, called Jeans length, it collapses under the gravitational force. The expression for the Jeans length is:

15k B T RJ = 4πGµ mρ

•

9.6

Star Formation

The expression for the free-fall time taken by a cloud to collapse under the assumption of free-fall collapse is:

t ff

•

1 2

3 = 2πGρ

1 2

Repeated fragmentation of collapsing clouds leads to the formation of protostars. A protostar contracts due to gravitational force and its temperature and density increases to such an extent that the nuclear reaction at its core becomes feasible. A star is then said to be born. Such stars are called pre-main sequence stars.

TERMINAL QUESTIONS

Spend 25 min.

1.

What is the origin of the 21 cm line of hydrogen? Why can we not obtain this line in a terrestrial laboratory? Explain the importance of this line in astronomy.

2.

What is interstellar reddening? What does it tell us about the composition of interstellar matter?

3.

Derive Eq. (9.11). For the data given in SAQ 3, calculate the Jeans length.

9.7

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1.

You know that the energy of electromagnetic radiation of frequency ν and wavelength λ is given by: E=h =h

c λ

where h is the Planck’s constant and c is the velocity of light. Substituting the values of h, c and λwe can write.

E=

(6.63 × 10 −34 Js) (3 × 10 8 ms −1 ) (0.21m) =

6.63 × 3 ×10 − 24 J 21

≈ 10 −24 ×

1 1.6 × 10 −19

eV

≈ 6.25 ×10 −6 eV 19

From Stars to Our Galaxy

2.

As per the problem, I λ = I λ 0 e − τλ

Taking logarithm on both sides and multiplying by 2.5, we get : 2.5 log I λ = 2.5 log I λ 0 − 2.5 τ λ log 10 e

If m0 is the original magnitude and m is the increased magnitude because of extinction, we have: K − m = K − m0 − 2.5 × 0.4343τ λ

or, m − m0 = 1.086τ

because we can write the apparent magnitude as: m = K − 2.5 log I .

Further, the magnitude of light scattered or absorbed along the line of sight can be written as: a λ = m − m0 = 1.086 τ λ

3.

From Eq. (9.10), we have the expression for Jeans mass: 3 2

5k BT MJ = µ mG

3 4 πρ

1 2

For pure hydrogen, µ = 1 . So, we have:

(

)

5 × 1.38 × 10 − 23 JK −1 × (50 K ) MJ = 1.67 × 10 − 27 kg × 6.67 × 10 −11 m 3 kg −1s − 2

(

) (

3 × 5 4 π × 10 ×1.67 ×10 − 27 kgm −3

(

= 10 24

5 × 1.38 × 5 1.67 × 6.67

3 2

× 1010

3 2

)

1 2

)

300 4 π ×1.67

1 2

kg

10 34 × 5.45 × 3.75 × 10 3 = MΘ 2 × 10 33 ≈ 10 5 M Θ 4.

From Eq. (9.14), we have the expression for the free-fall time for a collapsing cloud as:

t ff = 20

3 2πGρ

1 2

Substituting the values of G and ρ, we get:

Star Formation 1 2

t ff =

=

≈

3 2π × 6.67 ×10 −11 m 3 kg −1s − 2 × 10 −14 kgm −3 1 2

30 2π × 6.67 1012 3 × 10

7

×1012 s

≈ 3 × 10 4 yr

yr

Terminal Questions

1.

See text.

2.

See text.

3.

The expression for Jeans mass is given by (Eq. (9.10)): 5k B T MJ = µmG

3 2

3 4π ρ

1 2

Now, for a spherical cloud of mass M and radius R, we have: M=

4 π R3 × ρ 3

So, we can write: MJ =

4 π R J3 × ρ 3

Substituting the above expression for MJ in Eq. (9.10), we can write: 5k B T 4 π R J3 × ρ = µ mG 3

3 2

3 4πρ

1 2

or, 15k B T RJ = 4πG µ m ρ

1 2

As per the data given in SAQ 3, we have T = 50K and the number density 105 m−3. Further, for pure hydrogen, µ = 1 . Substituting these values in the expression for RJ, we get: 21

From Stars to Our Galaxy

15 × (1.38 ×10 − 23 JK −1 ) × (50 K) RJ = 4π × (6.67 × 10 −11 m 3 kg −1s − 2 ) × (10 5 × 1.67 × 10 − 27 kg m −3 ) × (1.67 × 10 − 27 kg m −3 )

≈ 1019 m ~ 300 pc

22

1 2

UNIT 10 NUCLEOSYNTHESIS AND STELLAR EVOLUTION

Nucleosynthesis and Stellar Evolution

Structure 10.1

Introduction Objectives

10.2 10.3 10.4

Cosmic Abundances Stellar Nucleosynthesis Evolution of Stars Evolution on the Main Sequence Evolution beyond the Main Sequence

10.5 10.6 10.7 10.8

Supernovae Summary Terminal Questions Solutions and Answers

10.1 INTRODUCTION In Unit 9, you have learnt that the gravitational collapse of interstellar cloud leads to the formation of stars. You may recall that a stable star is formed when an equilibrium between gravitational force in the collapsing cloud and the radiation pressure due to nuclear energy generated in its core is attained. In this regard, some logical questions which might come to your mind are: Since nuclear fuel burning at the core of a star like the Sun is finite, what happens when the fuel is exhausted? Does the same type of nuclear energy generation process take place in all stars? What happens to the material produced in the nuclear reactions in the stars? If all the new born stars find place on the main-sequence of the H-R diagram, how do we explain the existence of red giants, supergiants and white dwarfs? To answer these questions, you need to learn about the life of a star (that is, stellar evolution) after it has been formed and found its place on the main-sequence. This is the subject matter of the present Unit. An interesting approach to understand stellar evolution is to ask ourselves: How did the ninety-three natural (chemical) elements, which are the building blocks of all living and non-living matter around us, come into existence? All the elements, except hydrogen and most of helium, were made inside stars through the process of nucleosynthesis. The stars make elements during their life-time. Interestingly, all the stars produce the same elements which we find on the Earth. Further, we find the same elements everywhere in the universe and more or less in the same proportion. The relative proportion of these elements in the Universe is called cosmic abundances about which you will learn in Sec. 10.2. In Sec. 10.3, you will learn about various nucleosynthesis processes taking place inside the stars at different stages of their life. You will discover that the conditions for different processes are different. This provides a basis to track the evolution of stars along the main-sequence and afterwards. The low-mass and high mass stars evolve differently leading to different end products such as white dwarfs, neutron stars and black holes. You will learn about this in Sec. 10.4. In Sec. 10.5, you will learn that, when the entire nuclear fuel at the core of a star is exhausted, rapid gravitational collapse takes place which may result in a violent explosion, called supernova, of the stellar envelope. Objectives After studying this unit, you should be able to: • •

list the various methods of determining cosmic abundances; list and explain various nucleosynthesis processes in the context of stellar evolution;

23

From Stars to Our Galaxy

• • • • •

describe where and how chemical elements are formed; explain the significance of stellar mass in respect of evolution of stars; estimate the lifetime of stars on the main sequence; describe the conditions for the formation of red giants, supergiants and white dwarfs; and discuss conditions for the supernova explosions and the fate of left over stellar cores.

10.2 COSMIC ABUNDANCES You know that the atom is the smallest unit of an element and atoms of each element are unique. Atoms may be said to be the basic building blocks of matter. Molecules are formed when atoms combine due to electrical forces between them. More complex substances are formed when there is chemical reaction between elements.

In school chemistry, you must have learnt about elements, atoms and molecules. Elements are the simplest substances of ordinary matter. Scientists recognise 93 elements as natural. Besides these, there are 20 elements which can be manufactured in the laboratory by bombarding nuclei with α-particles or other high energy particles. These 20 artificial elements are not found in nature because they are unstable and live for extremely short times. Almost all the natural elements are found on the Earth. A few elements which are not found on the Earth are found in other bodies of the solar system. You may be surprised to know that the same 93 elements are also found elsewhere in the universe and their proportion is more or less the same as their proportion in the solar system. The chemical composition of the universe refers to the presence of different types of elements and their proportions in the universe. Since the chemical composition of the universe and that of the solar system are similar, it is one and the same whether we talk of the chemical composition of the universe, or the chemical composition of the solar system. Further, the relative proportions of elements in the universe are called cosmic abundances. Determination of Cosmic Abundances Now, the question is: How do we determine abundances in the solar system, that is, how do we determine cosmic abundances? Some of the important methods to obtain information about abundances are as follows: i)

In the solar system, the immediate source for obtaining such information is obviously the Earth. Samples from many locations on the Earth are analysed in the laboratory. Care is taken that these locations are as diverse as possible.

ii) The next obvious source is the Sun. You may recall from Unit 5 that the dark lines in the solar spectrum, called Fraunhofer lines, are actually absorption lines due to elements present in a slightly cooler layer above the photosphere. Each line in the spectrum is checked against the sample spectra of elements and the elements are identified. The intensity of a particular line gives the abundance of the corresponding element. You may also recall from Unit 5 that the higher layers of the solar atmosphere, the chromosphere and the corona, are at relatively higher temperatures than the solar surface. The spectra of these layers show emission lines due to elements present in these layers. Analysis of these lines also helps in determining solar system abundances. iii)

The Sun also emits streams of particles in the form of solar wind. Occasionally, the Sun emits very high energy particles, called the solar cosmic rays. The compositions of the solar wind and the solar cosmic rays are directly determined by instruments on-board many spacecrafts orbiting the Earth.

iv) The spectrum of other objects in the solar system such as the moon and planets are other sources of information about abundances in the solar system. Samples of dust brought from the moon and chemical analysis of the Martian surface has added significantly to this information. 24

v) You must be aware that small rocky pieces wandering in the solar system, called meteors, occasionally enter the atmosphere of the Earth. If meteors are not burnt completely by the heat generated due to atmospheric friction, they reach the Earth. These pieces are called meteorites. Analysis of their composition provides valuable information about abundances. The spectra of comets are yet another source of information about the solar system abundances.

Nucleosynthesis and Stellar Evolution

vi) Outside the solar system, spectra of other stars and interstellar clouds are important sources of information about the cosmic abundances. Cosmic abundances of various elements have been determined using a variety of methods including those discussed above. Refer to Fig. 10.1 which depicts the variation of cosmic abundances with mass number of elements.

Abundance (Log scale)

10

6

2

-2 0

50

100

150

200

Mass number Fig.10.1: Abundances of various elements in the universe as a function of mass number

The same data is shown in greater detail in Fig. 10.2. The abundances have been expressed in terms of a unit in which the abundance of silicon (Si) is exactly 106. This is because the abundance of Si is very close to this number.

Abundance (Log scale)

10

6

2

-2 0

20

40

60

80

Mass number Fig.10.2: A detailed version of Fig. 10.1 for elements up to mass number 80

The salient features of Figs. 10.1 and 10.2 are as follows: 1. Hydrogen (H1) and helium (He4) are the most abundant elements in the universe. About 90% of the particles in the universe are hydrogen atoms. Helium is the next most abundant element, accounting for about 10% of all the particles. 2. Heavier elements constitute less than 1% of the total matter in the universe.

25

From Stars to Our Galaxy In the language of astronomy, any element heavier than He4 is called a heavy element.

3. If we leave out H1 and He4, we observe that abundances generally increase with mass number up to the mass number around 60. This is in the neighbourhood of iron (Fe56). Around this mass number, there is a broad peak. 4. Beyond the mass number 60, the abundances decrease. At first, the decrease is faster and then it gradually tapers off. 5. There are peaks of abundances corresponding to elements with mass numbers 12, 16, …. and so on (multiples of 4). Moreover, elements with mass numbers, 14, 18, … and so on (multiples of 2) are more abundant as compared to those with odd mass numbers. On the basis of these features of cosmic abundance data, we can conclude that: a) The origins of hydrogen and helium are perhaps different from the origin of heavier elements in the universe. b) Peaks of abundances at mass numbers that are multiples of four could involve a particle such as the α-particle or the helium nucleus, which has mass equal to 4 atomic mass unit (amu).

Spend 5 min.

SAQ 1 On the basis of relative number of atoms of hydrogen and helium in the universe, calculate the fractional mass of the matter in the universe contributed by hydrogen and helium. Having learnt about the cosmic abundances, a logical question that may come to your mind is: Where and how are these elements formed? Astronomical studies tell us that all the elements, except hydrogen and helium, have been synthesised in the stars during their evolution. This is also reinforced by the observation that the older stars in our Galaxy contain much less heavier elements than the younger stars. Thus, we can visualise the following roadmap for creation of elements and how the process is related with the evolution of stars: a) Elements are formed inside the stars. Since the birth and death of stars is a continuous process, the formation of elements is also an on-going process. b) The oldest stars in the Galaxy, called Population II stars, were formed from the original matter of the Galaxy which was mostly hydrogen. These stars had to manufacture their own heavy elements. Therefore, they are relatively poor in heavier elements. c) At the end of their life, some of these stars explode and return the heavier elements formed by them to the interstellar medium. d) From this enriched interstellar material, new stars are formed. These relatively younger stars, also called Population I stars, are rich in heavier elements. In addition, they also manufacture elements in their cores which constitute the raw material for the subsequent generations of stars. The hypothesis that elements are made inside the stars gets support from the detection of elements like technetium in the spectra of some stars. This element is not found in the solar system. Where could this element have been formed except in the stars in which it is observed?

26

Now, the question is: What is the origin of the major constituents, namely hydrogen and helium, of interstellar medium? An acceptable theory in astronomy tells us that hydrogen and helium were formed in a different process (see Unit 15). You could question this theory since you have learnt earlier that He4 is formed from H1 in the core of the Sun and the other main-sequence stars. The fact is that if we take account of all the helium that could have been formed in the stars in all the galaxies, it falls much short of the total helium estimated to be present in the universe (about 30% by mass). To appreciate this statement, solve the following SAQ. SAQ 2

Nucleosynthesis and Stellar Evolution

Spend 10 min.

The atomic weights of hydrogen and helium are 1.0079 and 4.0026, respectively. In the fusion reaction converting hydrogen into helium, one gram of hydrogen produces about one gram of helium and approximately 6 × 1018 ergs energy is released. Given that the luminosity of the Sun is 4 × 1033 erg s−1 and its estimated age is 5 × 109 years, show that only about 5% of its mass has been converted into helium. Take the solar 30 mass as 2 × 10 kg. Having solved SAQ 2, you might conclude that only a small fraction of the total helium present in the universe has been manufactured in the stars. It is, therefore, reasonable to believe that light elements like hydrogen, helium, deuterium (D2), He3, 7 and Li did not form inside the stars. You may ask: Do we have any clue about the origin of these elements? According to one theory about the origin and evolution of the universe, these elements were formed in the first minute after the birth of the universe. At that time, the universe was hot and dense and the conditions were suitable for the formation of light elements. (This issue is discussed in detail in unit 15 of this course.) This theory is supported by the fact that the abundances of light elements predicted by it in the early universe agree very well with the observed abundances. The coincidence is considered a very strong evidence supporting the idea that the early universe was very hot and dense and that it was born in a violent event called the Big-Bang. A clue to support the hypothesis that heavier elements are manufactured inside the stars was provided by Bethe (in US) and Weizsacker (in Germany) in 1938.They showed the possibility of converting hydrogen into helium through nuclear reactions which would take place at high temperatures and high densities. Such conditions are readily available in the interior of stars such as the Sun, which also has plenty of hydrogen. The work of Bethe and Weizsacker gave birth to the field of nuclear astrophysics and subsequently, scientists were able to show that other heavier elements could also have been formed in stars through the process of nucleosynthesis involving a variety of nuclear reactions. Would you not like to know about nucleosynthesis? We discuss this in the next section.

10.3 STELLAR NUCLEOSYNTHESIS You may recall from your school physics that nuclear fusion is the process of coming together of two or more light nuclei to form a new nucleus. Since a fusion reaction involves coming together of charged particles, it requires that they have sufficient energy to overcome the Coulomb repulsion. Such energies are readily available in the form of thermal energy in stellar interiors. The nuclear reactions taking place in such environments are called thermonuclear reactions. The process of creation of new elements in such reactions is called nucleosynthesis. A variety of nuclear reactions can take place depending upon the elements present, temperature and density in the interior of stars. The information about these reactions helps us understand stellar evolution better because on the basis of this information,

27

From Stars to Our Galaxy

we can determine the age of the stars as well as their future. Let us now discuss the major processes by which elements are synthesised in the stellar core.

The temperature inside the stars is very high. Therefore, all the atoms are ionised. The nuclear reactions take place between nuclei and the products are also nuclei. In this context, therefore, whenever we talk of atoms or elements, we really mean nuclei.

1) Hydrogen Burning The first stage of nucleosynthesis is the fusion of hydrogen nuclei and consequent formation of helium. You have already learnt in Unit 5 about the chain of reactions, called the pp-chain, which causes fusion of four hydrogen nuclei (protons) to form helium. This is the process by which the Sun and other similar stars generate their energy: H1 + H1 → H 2 + e + + e H 2 + H1 → He 3 + He 3 + He 3 → He 4 + 2 H1

Another chain reaction has helium as its end product and it produces energy in the main sequence stars. It starts with carbon and is called the carbon-nitrogen cycle (CN-cycle). Obviously, for this reaction, it is necessary that the stars have some carbon to begin with. Such stars are called the second generation stars. The stars which start life with only hydrogen and helium are called first generation stars. The temperature required for CN-cycle is higher than the temperature required for the pp-chain. The nuclear chain reactions involved in the CN-cycle are given below: C12 + H1 → N13 + γ N13 → C13 + e+ + νe C13 + H1 → N14 + γ N14 + H1 → O15 + γ 15

O 15

1

15

+

→ N + e + νe 12

N + H → C + He

4

You may note that the end result of the CN-cycle is to combine four hydrogen nuclei to form helium; carbon merely acts as a catalyst for the reaction and is not consumed in the process. You may ask: If we know that CN-cycle is active in a star, what information can we obtain about that star? Recall from Unit 7 that, as we go up in the H-R diagram, the luminosity increases. Since luminosity of a star is proportional to some power (generally 3.5) of its mass, the mass also increases upwards. Further, the internal temperature of a star is generally proportional to its mass. Therefore, the stars in which energy is generated by the CN-cycle are generally found in the upper region of the main sequence. They have high internal temperature and were formed from material enriched in heavy elements. Spend 2 min.

SAQ 3 On the main sequence in the H-R diagram, where would you find stars which have internal temperatures lower than that of the Sun? 2) Helium Burning

28

When all the hydrogen in a stellar core has been used up, pp-reactions and CN-cycle are no longer possible. As a result, the energy generation stops and the

pressure in the core decreases. It is no longer able to match the inward gravitational pull of the particles and the core contracts. In some stars, the contraction continues till the temperature has risen to about 2 × 108 K. When the core temperature of a star attains this value, it is possible for helium nuclei to fuse together and produce carbon.

Nucleosynthesis and Stellar Evolution

If you look at the periodic table carefully, you will find that at mass number eight, there is a gap. This means that the element at this position is unstable. The question is: How do reactions between helium nuclei overcome this gap? The two He4 nuclei (or α-particles) combine to form Be8. Since Be8 is unstable, it disintegrates into two α-particles in a very short time. In the presence of to and fro reactions of this kind, the stellar core becomes a sea of He4 nuclei with a few Be8 12 8 nuclei floating. These floating Be nuclei combine with α-particles to form C 8 12 nuclei. Despite very low population of Be nuclei, C is formed because the 8 reaction between α-particle and Be has a very high probability of occurrence. So, carbon is formed when three α-particles combine according to the following reactions: 4

4

8

He + He → Be + γ Be8 + He4 → C12 + γ The above nuclear reaction is also called the triple-α α reaction. Once carbon is formed in the stellar core, formation of heavier nuclei becomes possible. You may recall that carbon and some other heavier elements are absolutely essential for the origin of life. Thus, it can be argued that we are here and discussing nucleosynthesis today because triple-α α reactions took place in some stellar cores in the distant past! SAQ 4

Spend 5 min.

a) Why are three α-particles needed to initiate helium reactions? b) Explain the importance of triple-α reaction in the formation of heavy nuclei. In what way is triple-α reaction related to the origin of life on the Earth? 3) Burning of Carbon and Heavier Nuclei Once carbon is formed, it can combine with an α-particle and form oxygen (O16). Afterwards, O16 can react with an α-particle to form Ne20. When all the helium has been converted into carbon in the core of the star, helium reactions stop and energy generation is terminated once again. This leads to further gravitational contraction of the core and temperature of the interior of the star increases. At the enhanced temperature, it becomes possible for two carbon nuclei to combine and form Mg24. By now, you must have noted that every time a particular type of nuclear fuel is used up completely, core contraction due to gravitation takes place. As a consequence, there is a rise in the temperature of the core. When the temperature has risen to the required level, yet another nuclear reaction becomes possible. This cycle continues till all the nuclei in the core have become iron nuclei. You may ask: Why does the series end at iron? To answer this question, refer to Fig. 10.3 which depicts the binding energy curve for nuclei. Note that the average binding energy per nucleon increases with mass number till we reach the mass number of iron. This means that with increasing mass number, the nuclei are more tightly bound and are more stable. It also means that iron is the most stable element. It cannot combine with other nuclei to produce still heavier nuclei. The binding energy curve also gives us a clue for understanding why energy can be 29

From Stars to Our Galaxy

derived by fusing light elements as well as by splitting (fission) the nuclei of heavy elements.

Binding energy per nucleon (MeV)

9

8

7

6

5 0

50

100

150

200

250

Mass number Fig.10.3: Average binding energy per nucleon as a function of mass number

It is, therefore, understandable why elements heavier than iron cannot form as a result of thermonuclear fusion reaction. However, the fact of the matter is that elements heavier than iron do exist in nature. The question, therefore, is: How did they form? It is believed that such heavier elements formed in some special types of nuclear reactions called s- and r-processes. 4) s- and r- processes The elements heavier than iron are probably synthesized by the absorption of one neutron at a time. For example, let us consider the nucleus of atomic number Z and mass number A. It is written as (Z, A). When it absorbs one neutron, its mass number increases by one and it becomes (Z, A+1). If the new nucleus absorbs yet another neutron, it becomes (Z, A+2). If this nucleus emits a β-particle before it can absorb another neutron, it becomes (Z+1, A+2). The last one, nucleus (Z+1, A+2), may absorb another neutron to become (Z+1, A+3). In this way, all the elements up to Bi209 are formed. This process of absorption of one neutron at a time is a slow process and is named as s-process. The neutrons required for this process are produced as a by product of reactions of the following types: 4

13

16

16

16

31

He + C → O + n O +O →S +n Well, you can further ask: Why does the s- process stop at Bi209? It is because the elements heavier than Bi209 are unstable and emit β-particles before they can absorb a neutron. However, if neutrons become available in large numbers, then these nuclei can absorb neutrons rapidly. Due to this rapid process, or the r-process, elements right up to uranium are synthesized in stars.

30

So far, you have studied about cosmic abundances and stellar nucleosynthesis. You now know that all the heavy elements are formed in stars due to thermonuclear reactions. The relative proportion of heavier elements in stars tells us whether it is a first or a second generation star. The various types of nuclear reactions are associated with different stages in the life of a star and also with the location of the star on the HR-diagram. You will now learn about the evolution of stars.

Nucleosynthesis and Stellar Evolution

10.4 EVOLUTION OF STARS You may recall from Unit 9 that gravitational collapse of a gas cloud gives birth to stars. A stable star is formed and finds its place on the main sequence only when it attains hydrostatic equilibrium, that is, when the pressure inside the star balances the gravitational force acting inwards. In the following, we shall discuss the life of stars on the main sequence and afterwards.

10.4.1 Evolution on the Main Sequence For most of their lives, stars live on the main sequence on the H-R diagram. We would, therefore, like to know: i) How long does a star live on the main sequence? and ii) What happens in the core of a star while it is on the main sequence and when it departs from the main sequence? When a star arrives on the main sequence, it is said to have been born. That is why the main sequence is called zero-age main sequence (ZAMS). While the star is on the main sequence, it burns hydrogen either through pp-chain or through CN-cycle. There is little change in its luminosity. The best example of the main sequence star is the Sun. It is known that the luminosity of the Sun has not changed much during its lifetime of 5 billion years. You may ask: How do we estimate the life of a star on the main sequence? Fortunately, for the Sun, we have enough data to make an intelligent guess. It is estimated that the core of the Sun has about 10% of its total mass and from SAQ 2, we know that so far it has burnt only about half of this (hydrogen) mass to make helium. Therefore, it is estimated that the Sun would stay on the main sequence for another 5 billion years. Let us now ask ourselves a more general question: How long does a star live on the main sequence? To address this question, we need to look critically at the massluminosity relation of stars. We know that the ultimate source of energy for a star is its mass. When it is burning hydrogen to form helium, it is actually converting its mass into energy. The relation between luminosity (L) and mass (M) of a main sequence star is given by: L ∝ M 3.5

(10.1)

Since luminosity is the energy radiated by a star per second and M is its total mass which can be converted into energy, the time for which it will stay on the main sequence can be written as: τ ≈ M / L.

(10.2)

From Eqs. (10.1) and (10.2), we can write: τ ∝ M −2.5

(10.3)

The constant of proportionality is determined by the expected lifetime of the Sun. Eq. (10.3) shows that the more massive a star is, the shorter is its life on the main sequence. To get a feel of the numerical values of the age of stars on the main sequence, you should solve the following SAQ. SAQ 5

Spend 5 min.

The estimated lifetime of the Sun on the main sequence is ~ 1010 years. Calculate the main sequence lifetime of a star of mass (i) 10 MΘ and (ii) 0.5 MΘ. 31

10

10

10

4

10

2

10

0

Lifetime (yr)

Luminosity

10

8

Lifetime

10

6

10

4

0

5

10 Mass (solar unit)

15

10 20

Luminosity (solar unit)

From Stars to Our Galaxy

-2

Fig.10.4: Lifetime and luminosity of stars on the main sequence as function of mass

Helium in the core

Expanding outer layers

Refer to Fig. 10.4 which shows the lifetime of stars on the main sequence as a function of the stellar mass. You may note that a star of mass 10 MΘ will stay on the main sequence for about 3 × 107 years and a star of mass 0.5 MΘ will stay on the main sequence for about 6 × 1010 years. You may ask: How come a more massive star which has more nuclear fuel has shorter lifetime on the main sequence? Recall that massive stars must burn their fuel at a faster rate to emit more energy from their surfaces per second. Thus, they run out of fuel sooner. By the same token, lower mass stars burn fuel at a slower rate. In this context, it is interesting to note that the estimated age of the universe is ~ 1 to 2 × 1010 years, a time period much shorter than the estimated lifetime of low mass stars on the main sequence! Having got a fairly good idea about lifetimes of stars of different masses on the main sequence, you may like to know: How does the core of a main sequence star evolve?

Hydrogen in the shell fusing to helium Fig.10.5: When hydrogen in the core of a star is exhausted, the core contains pure helium and hydrogen burns in a thin shell around the core

Spend 3 min.

While a star is on the main sequence, its core becomes progressively richer in helium. When the core consists only of helium, no nuclear reaction takes place because the temperature is not high enough for the next stage of nuclear reactions. As a result, the pressure in the core decreases and the core must contract under its own weight. The gravitational energy released due to contraction raises the temperature of hydrogen in a thin shell around the core so much that it starts burning (Fig. 10.5). The helium made in the shell adds to the mass of the core whose contraction is accelerated. The energy produced in the shell and the gravitational energy due to the contracting core push out the envelope of the star due to radiation pressure. As a result, the star expands in size. Its surface becomes cooler but its luminosity increases enormously because of increased surface area. It becomes a giant star. SAQ 6 It is estimated that after its life on the main sequence, the Sun will swell to 200 times its present radius. If, at that time, its surface temperature is half of its present temperature, calculate the luminosity of the Sun in terms of its present luminosity. Let us now look at the events that take place in the life of a star when it leaves the main sequence.

32

10.4.2 Evolution beyond the Main Sequence

Nucleosynthesis and Stellar Evolution

When a star leaves the main sequence, it becomes a giant star because of its increased size. Since the surface temperature of these stars is low, they appear red and are also known as red giant stars. The more massive (> 10 MΘ) stars become so huge because of expansion that they are called supergiant stars. The time taken by a star to travel from the main sequence to the giant or supergiant branch is much shorter than its stay on the main sequence. The Sun will also become a giant star after about 5 billion years. Meanwhile, the cores of these stars continue to contract and their temperatures rise 8 further. When the temperature of the core is about 2 × 10 K, triple α - reactions become possible. The energy released in these reactions builds up the pressure in the core and further contraction of the core is halted. Further details of evolution depend on the mass of the star. Let us discuss some illustrative cases now. Mass of the star is ~ 1 MΘ If the mass of the star is approximately 1 MΘ, the helium burning is rather abrupt and a large amount of energy is suddenly released. This phenomenon is called a helium flash. Several circumstances, some not yet completely understood, combine at this stage to force the star to throw away its outer envelope. The ejected matter surrounds the star. This object is called a planetary nebula (Fig. 10.6). In a small telescope, it appears like a planet, hence the name planetary nebula.

Fig.10.6: A planetary nebula

Within a short time, the gas surrounding the star vanishes due to interaction with the interstellar medium. Simultaneously, the core again begins to contract and becomes very dense and the matter in the core becomes a degenerate gas. Degenerate gas is a particular configuration of a gas composed of fermions (electrons, neutrons, etc.) whose behaviour is regulated by a set of quantum mechanical laws, in particular, Pauli’s exclusion principle (explained in the physics electives PHE-11 entitled Modern Physics). This configuration of a gas is usually reached at high densities. Recall that according to Pauli’s exclusion principle, no more than two fermions (of opposite spin) can occupy the same quantum state of a system. As the density of electrons in the core of a star increases in a fixed volume, these particles progressively fill the lower energy states. The additional electrons are forced to occupy states of higher and higher energy. This process of gradually filling in the higher-energy states increases the pressure of the electron gas. The pressure of the degenerate gas depends only on the density of the gas and is independent of its temperature. Since degeneracy occurs when the density is high, the pressure of the degenerate gas is high. The equation of state of such a gas is independent of temperature unlike the normal gas. In stars of mass ~ 1 MΘ , only the electron component of the matter becomes degenerate. The pressure of degenerate electrons is sufficient to halt the contraction of the star. A state of equilibrium is established. The star is then called a white dwarf star. The maximum mass of a white dwarf star is ~ 1.4 MΘ. This limit on the mass of a white dwarf star was predicted theoretically by S. Chandrasekhar, an Indian astrophysicist. It is, therefore, called the Chandrasekhar limit. In the H-R diagram, the white dwarf stars occupy the left bottom corner, much below the main sequence as shown in Fig. 10.7. The numbers above the lines in Fig. 10.7 indicate the mass of white dwarf stars in terms of solar mass. 33

From Stars to Our Galaxy

Main sequence

Sun

log (L/L Θ)

0

-1.5 0.89

0.51

0.22

-3.0

White dwarfs

-4.5 4.5

4.3

4.1

3.9

3.7

3.5

log T Fig.10.7: Location of white dwarf stars on the H-R diagram

There is no source of nuclear energy inside a white dwarf star. It is a dead star. It just utilises the thermal energy of its particles to radiate from its surface. It can live like this for several billion years and, afterwards, it becomes a cold object. The path of evolution of such stars on the H-R diagram is shown in Fig. 10.8a. Note that, after the star leaves the main sequence, it swells and heads towards the giant region. Subsequently, depending upon its mass, it may become a white dwarf star. 105 Giant region 10000

Formation of planetary nebula 103

10 solar radii

Luminosity (solar unit)

Luminosity (solar units)

1000

10 Main sequence 10-1

White dwarfs

10-3

Horizontal branch

Red giant

100

10 1 solar radius 1 Main sequence

10-5 80000

40000

20000

6000

Surface temperature of star (K) (a)

2000

0.1 4.1

4

3.9

3.8

3.7

Surface Temperature (K; Log scale) (b)

Fig. 10.8: a) Evolutionary track of a general mass star; and b) evolutionary track of a star of a few solar mass

Mass of the star is ~ 5 MΘ

34

After becoming a giant star, a star having mass approximately equal to 5 MΘ traces its path back on the H-R diagram towards the main sequence along a horizontal line, called the horizontal branch (see Fig. 10.8b). This happens because of the commencement of the helium fusion (also called triple alpha process). As helium is exhausted and the next set of nuclear reactions involving carbon starts, the star

3.6

changes its evolutionary course backwards towards the giant branch. It wriggles several times between the giant and the horizontal branch. During this time, it also sheds a lot of mass. Eventually, it also becomes white dwarf after passing through the planetary nebula phase.

Nucleosynthesis and Stellar Evolution

Mass of the star is > 10 MΘ If the initial mass of the star is several times the solar mass, the helium reactions start gradually. When helium in the core is exhausted, the core contracts once again. The helium reactions are now ignited in a thin shell around the core. The temperature of the contracting core rises and becomes high enough for carbon-carbon reactions to occur. The cycle of core contraction and burning of the next heavy element continues till the core consists of iron. No more fusion reactions are possible now. SAQ 7

Spend 3 min.

Why do fusion reactions stop at iron? Let us pause for a moment and think of what we have learnt about the evolution of stars. We have traced the evolution of stars and simultaneous formation of elements. Depending upon its mass, a star can take different evolutionary courses. Well, you can further ask: What happens after all the nuclear reactions have stopped and the stellar core consists of iron only? We have seen from the binding energy curve (Fig. 10.3) that iron has the highest binding energy per nucleon. It cannot, therefore, burn to give off energy. Thus, the core is forced to contract. The gravitational energy heats the core resulting in the disintegration of iron nuclei into nuclei of helium. The break up of iron is an endothermic process, that is, the reaction absorbs energy rather than giving it out. To feed this process, the only source of energy is the gravitational collapse. As a result, the collapse of the core is accelerated. Faster release of gravitational energy due to collapse gives rise to further break up of iron nuclei into helium nuclei. This, in turn, accelerates the collapse. As the density of the core continues to increase, even the helium nuclei cannot remain intact as nuclei. They break up into protons and neutrons. Coming back to the evolution of the collapsing star, let us first consider its envelope. Due to transfer of energy from the core to the envelope, very high temperatures are produced in the envelope and nuclear reactions in various layers of the envelope are ignited. The energy released due to these reactions heats the envelope so much that it becomes prone to explosive disintegration in a matter of seconds. In this short time before explosion, nuclear reactions produce heavy nuclei all the way up to the iron group. In addition, the large numbers of neutrons released in the nuclear reactions participate in the r-process and heavy nuclei beyond Bi209 are produced. Finally, the envelope explodes. The explosion is called a supernova. The elements built inside the star over hundreds of millions of years of its life are thrown into the interstellar medium. The new stars born from this enriched medium contain a small proportion of heavy elements and these stars are called the second generation stars. It has been suggested that the interstellar material from which the solar system was formed contained heavy elements released in a supernova that took place nearby.

In the nuclear reaction taking place in the core of stars, a large number of neutrinos are also produced. You may be aware that the neutrino has no charge and very small mass, if at all. These particles react with matter extremely weakly. So, their mean free path is very large and these were considered to be the carrier of energy from the core to the envelope. It has, however, been found now that at the extremely high densities developed in the core, the mean free path of neutrinos is shorter than the size of the core. So, they cannot transfer energy to the envelope. A different process is now proposed for the transfer of energy from the core. When the density of matter approaches the nuclear density 15 −3 (~ 10 gcm ), the short range nuclear forces come into play. The particles repel one another strongly and rebound. A bounce is said to travel from the core and into the envelope. It is the bounce that is thought to transfer energy to the envelope.

Well, the envelope explodes but what happens to the core? The core keeps collapsing. According to our present knowledge, if the initial mass of the star is up to about 12 MΘ, then the core is left with a mass of about 2 MΘ to 3 MΘ. The matter in such stars is mostly neutrons. Like electrons, neutrons also obey Pauli’s exclusion principle. At the extremely high density which exists in the core, neutrons become degenerate. The pressure exerted by the degenerate neutrons is sufficiently high to halt the collapse of the core. The core stabilises in the form of a neutron star. 35

From Stars to Our Galaxy

If the initial mass of the exploding star is close to 15 MΘ or more, the core that is left behind has a mass more than 3 MΘ. A core of this mass cannot attain equilibrium of any kind. It keeps contracting. Eventually, its gravitational field becomes so strong that even light cannot escape it. It becomes a black hole.

10.5 SUPERNOVAE Supernovae are extremely violent and bright stellar explosions. The energy released in one such explosion is equivalent to the conversion of 1 MΘ into energy, or about 47 10 J! The luminosity of a supernova is of the order of the luminosity of a whole galaxy. In all supernova explosions, the brightness reaches a peak within a few days of the explosion. During the short time of its maximum brightness, it shines as a very bright object. Thereafter, its brightness decreases, first rapidly and then slowly. The variation of the brightness of an object with time is called the light curve of the object.

Brightness

Light curves of supernovae have been studied extensively because of their importance in many areas of astronomy. Based on the type of light curves, supernovae have been classified as type I and type II. Fig. 10.9 shows the light curves of type I and type II supernovae. You may note that the light curve of type I supernova reaches higher brightness but show a rapid decline. On the other hand, the light curve of type II supernova shows a lower maximum brightness but slower decline. The difference in the light curves of the two types of supernovae indicates differences in the stars which explode.

Type I

Type II

0 Time (days) Fig.10.9: Schematic light curves of supernovae of type I and type II

You may argue: Since supernovae are so bright, it must be easy to observe them. Surprisingly, it is not so; observations of only four supernovae in our Galaxy have been recorded in the last two thousand years! The most famous supernova occurred in the year 1054 A.D. and was observed by the Chinese astronomers in the constellation Taurus.

36

Type I supernovae are further subdivided into two classes: Ia and Ib. The light curve for a supernova of type Ia is shown in Fig.10.10. Again, note that the brightness of the supernova is maximum at the time of explosion and it drops drastically afterwards. The light curves of supernovae of type Ia are almost identical. These supernovae are believed to be caused due to the explosion of white dwarf stars. You may ask: How can a white dwarf give rise to a supernova explosion because its mass cannot be greater than 1.4 MΘ? This would be possible if we imagine a white dwarf star in a binary system with a main sequence or a giant star as its companion. Because of its strong gravitational field, the white dwarf star sucks matter from the companion. As

Nucleosynthesis and Stellar Evolution

Brightness

its mass increases beyond 1.4 MΘ, it explodes as a supernova. Since all stars have the same mass at the time of explosion, it is possible that supernovae Ia reach the same brightness or absolute magnitude. Further, the observed apparent magnitude of a supernova can be used to determine its distance using its distance modulus (defined as m − M) because all type Ia supernovae have the same absolute magnitude at the time of the maximum brightness. In recent years, type Ia supernovae have been used successfully to find distances of distant galaxies. This has been very useful in understanding the nature and the ultimate fate of the universe.

Supernova explosion 0

Time (days)

Fig.10.10: Schematic diagram of the light curve of type Ia supernova

Type Ib supernovae are thought to be due to carbon burning in the degenerate core of a star. Carbon burning in this situation is abrupt and very rapid. It is called a carbon flash. It generates so much energy that the star explodes. The stars which explode to become supernovae of type II have usually masses higher than 10 MΘ. Their light curves are all distinct. These supernovae leave behind neutron stars or black holes, which are detected several years after the explosion. You may ask: How do we know all that we have said above about supernovae? At the time of explosion, a star throws a cloud of gas which travels into the interstellar −1 medium with speeds of the order of 10,000 kms . The cloud expands and merges gradually into the interplanetary medium. It seeds the interplanetary medium with heavy elements. The explosion also generates ripples in the surrounding medium which create conditions favourable for the formation of new stars. An example of the what remains behind a supernova explosion is the Crab Nebula. It is the remnant of the supernova of 1054 A.D. A neutron star is located at the centre of the nebula which was formed as a result of the explosion. Now, let us summarise what you have learnt in this unit.

10.6 SUMMARY •

Chemical composition of the universe refers to the presence of different types of elements and their proportions in the universe. The chemical composition of the universe is the same as that of the solar system.

•

Relative proportions of elements in the universe are called cosmic abundances. Observed abundances of elements show that (i) hydrogen and helium are the most abundant elements in the universe, (ii) there is a broad peak of abundances near 37

From Stars to Our Galaxy

A = 56 (iron group of elements), and (iii) there are peaks at mass numbers which are multiples of 4. •

All elements except hydrogen and helium are formed inside the stars due to thermonuclear reaction and creation of new elements in such reactions is called nucleosynthesis.

•

Depending upon the elements present, density and temperature in the interior of stars, nucleosynthesis takes place through one of the three major processes, namely hydrogen burning, helium burning and burning of carbon and heavier nuclei. As long as a star stays on the main sequence, it burns hydrogen.

•

The life span of a star on the main sequence depends on its mass: τ ∝ M −2.5 . Also, the evolution of a star away from the main sequence depends on its mass. Low mass stars become white dwarfs at the end of their lives.

•

Massive stars (> 10 MΘ) explode as supernovae at the end of their lives leaving behind neutron stars or black holes.

10.7 TERMINAL QUESTIONS 1.

Spend 30 min.

List the major processes of formation of elements inside stars. Why can elements beyond iron not be formed by fusion?

2. Suppose that a supernova explosion takes place at the distance of Proxima Centauri (~ 3 pc). If its luminosity equals the luminosity of our galaxy (~1012 LΘ), show that it would appear about as bright as the Sun. Take the absolute magnitude of the Sun as 5 and its apparent magnitude as − 27. 3. For stars having more mass than 10 MΘ, the luminosity is directly proportional to their masses. Show that their lifetime on the main sequence is independent of their masses. 4. Aldebaran (α-Tauri, Rohini in India) is a red giant star of spectral class K5. Its radius is 22 times the radius of the Sun. Its surface temperature is 3800 K. Calculate its luminosity (in units of solar luminosity) and its absolute magnitude. The absolute magnitude of the Sun is 5.

10.8

SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1. 90% hydrogen and 10% helium by number implies that the universe comprises 90 units of mass of hydrogen and 40 units of mass of helium because the mass of a helium atom would be 4 units if mass of a hydrogen atom is taken as 1 unit. So, out of 130 units of mass, 90 units is hydrogen and 40 units is helium. Thus, Percentage of hydrogen =

90 × 100 ~ 70 130

Percentage of helium

40 × 100 ~ 30 130

=

That is, approximately 70 percent of the mass of the matter in the universe is contributed by hydrogen and remaining by helium. 38

2. Total energy radiated by the Sun during its lifetime =

Luminosity of the Sun × Age (in s) of the Sun

=

(4×1033erg s−1) × (5×109×3×107s)

=

6×1050 erg

Nucleosynthesis and Stellar Evolution

[

1 yr = 3×107 s]

Now, this energy was generated in the nuclear reactions within the Sun converting hydrogen into helium. So, to produce this much energy, mass of hydrogen consumed

6 × 10 50 erg

=

18

6 × 10

erg g

−1

= 10 32 g

Therefore, percentage of the total mass of the Sun which has been converted into 10 32 g

helium till date =

2 × 10 33 g = 5%

×100

3. The internal temperature of a star is roughly proportional to its mass. Since luminosity of a star is proportional ~ M 3.5, the stars with lower internal temperature than the Sun have lower luminosities than the Sun. So, these stars will be found in the H-R diagram to the right of the Sun and lower than the Sun. 4.a) Since the nucleus with mass number 8, which nucleosynthesis of two α-particles will produce, is unstable, three α-particles are needed to produce stable nuclei beyond mass number 8. b) If triple - α reaction could not take place, carbon and heavier nuclei would not have been synthesized. Further, since carbon is essential for the origin of life, without triple - α reaction, life on the Earth would not have been possible. 5. The lifetime of a star on the main sequence is given by (Eq. (10.3)): τ ∝ M −2.5

Thus, for the Sun, we have: τ Θ ∝ M Θ −2.5

On the basis of above expressions for τ , we can write: −2.5

τ = τΘ

M M

i) For a star whose mass is 10MΘ, we can write its lifetime on the main sequence as: −2.5 10

τ =10

yr

10 M MΘ

=1010 ×10 −2.5 yr

= 10 7.5 yr = 10 7 10 yr ≅ 3 × 107 yr

39

From Stars to Our Galaxy

ii) For a star whose mass is 0.5MΘ, we can write: M 1 = M 2

Thus, − 2.5 10

=10

1 2

yr

10

≅ 6 × 10 yr 6. We know that the luminosity of a star is given as: L = 4 R2 T 4 So, we can write for the Sun: L = 4π R

2

T4

After the Sun has expanded, we have, as per the problem: R = 200 RΘ and T=

TΘ 2

Thus, the luminosity of the expanded Sun is: 4

T 2

2

L = 4 π (200 RΘ ) σ

=

4 × 4 × 104 R 16

2

T

4

4 × 4 × 10 4 R 2 T 4 L = LΘ 16 × 4 R 2 T 4

L = 2.5 × 103 LΘ 7. Fusion reaction stop at iron because iron is the most stable atom/nucleus. It cannot combine with any other nucleus to produce heavier nuclei. Terminal Questions 1. See text. 2. As per the problem, luminosity of the supernova, L = 1012 LΘ . If the absolute magnitude of supernova is M and that of the Sun is MΘ, we can write:

L = 10 − 0.4 ( M − M Θ ) LΘ 40

1012 = 10 −0.4 ( M − M Θ )

12 = − 0.4 (M − MΘ)

Nucleosynthesis and Stellar Evolution

M = − 30 + 5 = − 25 If m is the apparent magnitude of the supernova, we can write:

m = M + 5 log r − 5 = − 30 + 5 × (log (3))

(r = 3pc)

= −30 + 5 × 0.48 ≈ − 27.5

The value (−27.5) of apparent magnitude of the supernova is approximately same as that of the Sun. So, the supernova will be as bright as the Sun. 3. As per the problem, L∝M Now, the lifetime of a star can be approximated as: ~

or

τ∝

M L M M

= Constant. So, the life time of such stars is independent of their masses. 4. You have seen in SAQ 2 of Unit 5 that, using Stephan - Boltzmann law, we can express the luminosity of a star as

L = 4 R2 T 4 where R is the radius of the star and T is its temperature. We can, therefore, write the luminosity of a star like Aldebaran as: L Ald = 4π R 2 σ T 4

= 4 × (22 R ) 2 (3800 K) 4 And, the luminosity of the Sun can be written as: LΘ = 4 × R 2 (6000 K ) 4

Thus, we have: L Ald 484× (3.8) 4 = L 64 = 95 41

From Stars to Our Galaxy

Further, the ratio of luminosity of Aldebaran and the Sun can be expressed in terms of their absolute magnitude MAld and MΘ, respectively, as: L Ald =10 − 0.4( M Ald − M Θ ) L

Thus, we get: 95 =10 −0.4( M Ald − 5)

because MΘ is 5. Taking logarithm on both sides, we get: log 95 = − 0.4 (MAld − 5) 1.98 = − 0.4 (MAld − 5) or,

M = − 4.95 + 5 = 0.05

42

UNIT 11 COMPACT STARS

Compact Stars

Structure 11.1

Introduction

11.2

Basic Familiarity with Compact Stars

11.3

Theory of White Dwarf

11.4

Neutron Star

Objectives Equation of State and Degenerate Gas of Fermions Chandrasekhar Limit Gravitational Red-shift of Neutron Star Detection of Neutron Stars: Pulsars

11.5 11.6 11.7 11.8

Black Hole Summary Terminal Questions Solutions and Answers

11.1 INTRODUCTION In Unit 10, you have learnt about the evolution of stars, that is, how they live their lives after being formed and take their respective positions on the H-R diagram. You know that the evolution of a star is governed by two competing forces, namely, the gravitational contraction and the radiation pressure due to energy generating thermonuclear fusion reactions. In spite of the huge mass of a star, the amount of its nuclear fuel is finite and it has to end ultimately. When it happens, a star can no longer be prevented from contraction due to gravity. The density of the star increases manifold and it turns into a compact object. In the present Unit, you will study about such compact objects, also called compact stars. There are many interesting questions pertaining to compact stars, such as: Does the gravitational collapse of a star take place uninterrupted? If not, what is the mechanism which balances the force of gravity? Do all stars end their life by becoming compact stars of similar type? If not, what determine(s) the nature of the compact stars? You will discover the answers to these and other related questions as you study this Unit. A compact star can become a white dwarf, a neutron star, or a black hole depending upon its initial mass. Further, the gravitational collapse of compact stars like white dwarf and neutron stars is halted by the degeneracy pressure of fermions − a quantum mechanical phenomenon. You will learn about the degeneracy pressure in Sec. 11.2. The theoretical analysis of the relation between the nature of a compact star and its mass was done by S. Chandrasekhar. This led him to predict a limiting mass for white dwarf stars. You will learn about the theory of white dwarfs in Sec. 11.3. In Sec. 11.4, you will study various characteristics of neutron stars and also understand why it is difficult to detect them optically. All the theoretical predictions about neutron stars could only be put to test when they were observed in the form of pulsars. In Sec. 11.5, you will learn about one of the most interesting objects, called the black hole, which physics has ever predicted. You will discover that the black hole signify the ultimate victory of gravity in the evolution of stars.

43

From Stars to Our Galaxy

Objectives After studying this unit, you should be able to: • • • • • • • •

understand the role of mass in deciding whether a compact star becomes a white dwarf, a neutron star or a black hole; explain the concept of degeneracy pressure of fermions and its role in compact stars; discuss the concept of Chandrasekhar limit and obtain an expression for it; understand the formation of neutron stars and its internal structure; derive an expression for the gravitational red shift of the neutron stars; describe the detection of neutron stars in the form of pulsars and discuss its properties; explain the concept of Schwarzschild radius for black holes; and describe the geometry of space-time around the non-rotating black holes.

11.2 BASIC FAMILIARITY WITH COMPACT STARS You may recall from Unit 10 that the gravitational force is balanced by the outward force due to pressure gradient inside a star. The pressure inside is generated due to thermonuclear reactions. With the passage of time, a star exhausts all its nuclear fuel by ‘burning’, that is, due to nuclear reactions. The star cannot support itself against gravity and begins to collapse. Eventually, the density of the star increases tremendously and the star turns into a very compact object. At this stage, you would like to know: a) In the absence of radiation pressure due to nuclear reactions, what enables compact stars to withstand gravitational squeeze? b) Are all the stars similar to one another after their collapse? To address the first issue, recall from the course entitled Physics of Solids (PHE-13) that when atoms are very close to one another, their quantum energy states overlap and the electrons in those orbitals behave as if they are free from their parent atoms. A similar situation is obtained in compact stars where, due to very high pressure, all the electrons are separated from their parent atoms. This phenomenon is called pressure ionisation. A compact star is, therefore, a collection of nuclei and free electrons. You have studied in Unit 10 how the pressure of the degenerate gas of free electrons restrains the seemingly unstoppable gravitational contraction. This is because a quantum state cannot accommodate more than one fermion (e.g., electron, proton, neutron, etc.). This implies that, when the density of electrons is high, they are forced to occupy quantum states with higher energies because lower states are full. In such a situation, the pressure of the gas depends only on the density and is independent of the temperature. You have learnt that gas of free electrons in such a state is called degenerate electron gas and the pressure exerted by it is called degeneracy pressure. So, the gravitational collapse of compact stars is balanced by the degeneracy pressure of electrons. It is, however, important to note that Pauli’s exclusion principle holds for all fermions. You will learn later in this Unit that degeneracy pressure due to neutrons plays an important role in stabilizing some compact stars.

44

The answer to question (b) raised above is: No. The nature of the remains of a star after death depends on its mass. You may recall from Unit 10 that the mass of a star plays a crucial role in its evolution and determines its luminosity. Similarly, depending upon its mass, a dying star can turn into any one of the three kinds of compact stars, namely a white dwarf, neutron star or black hole. You will study about these compact stars later in the Unit. Here, it should suffice to say that it was genius of

the Indian astrophysicist, S. Chandrasekhar, who first showed that a degenerate star cannot have mass larger than a certain maximum mass. He suggested, on the basis of theoretical calculations, that the degeneracy pressure of electrons will be able to stop further collapse of a star and its mass is less than a certain mass called Chandrasekhar limit. The resulting star is called a white dwarf. If the mass of a collapsing star is more than the Chandrasekhar limit, but less than 3 MΘ then the degeneracy pressure of neutrons can halt the collapse. These stars are known as neutron stars. Further, if the mass of the collapsing star is even higher, there is no way that the collapse can be halted and the collapsing star becomes a black hole.

Compact Stars

Thus, compact stars are simply the end products of ordinary stars and are characterised by smaller sizes and higher densities. To compare and contrast their sizes with that of the Sun, recall that the radius of the Sun is 7 × 1010 cm and its mass is ~ 2 × 1033g. A white dwarf of the same mass as that of the Sun would have a radius about 100 times smaller, that is, around 109 cm. A neutron star of similar mass may have radius of about 10 km only. And a mass equal to that of the Sun is too small for a black hole! Generally, it is believed that the minimum mass of a black hole is three times the mass of the Sun (3MΘ). You may ask: What is the radius of a black hole? Well, that is a somewhat difficult concept. We will talk about it in Sec. 11.5 of this unit.

1016 1014 1012

Neutron stars

Density (g cm-3)

1010 108 White dwarfs

106 104 102

Earth Sun

100 10

Main sequence stars

-2

10-4 101

106

1011

1016

Radius (cm) Fig.11.1: Typical range of densities of celestial objects as a function of their typical radii

To get an idea about the average densities of compact stars vis-à-vis their radii, refer to Fig. 11.1. Note that the main sequence stars, such as the Sun, have density ~ 1 g cm−3. On the other hand, the density of a white dwarf could be ~ 107 to 108 g cm−3 and that of a neutron star ~ 1015g cm−3. A black hole, however, has no defined average density, as you will learn later in this Unit. Also, the radius of a black hole is a theoretical construct which means that if you come within this distance from the centre (where all the mass is concentrated), you cannot escape from the tremendous attraction of its gravitational force. Why only you? Even photons, the fastest particles, cannot escape.

You know from the course entitled Thermodynamics and Statistical Mechanics (PHE-06) that the behaviour of a gaseous system can be understood on the basis of its equation of state.

45

From Stars to Our Galaxy

11.2.1 Equation of State and Degenerate Gas of Fermions In the preceding paragraphs, you have learnt that the densities of white dwarfs and neutron stars are very high compared to the densities of objects we come across in everyday life, or even the densities of ordinary stars like the Sun. Therefore, to know more about the compact stars, you should have some idea about how matter behaves when the density is extraordinarily high. You may ask: Why should matter behave differently? It is because the equation of state of matter at high densities is different from the equation of state at ordinary densities. When the molecules of a gaseous system are few and far between, as in our room, or in the atmosphere, collision is the only way the molecules can interact with one another. Such a gas of molecules is called an ideal gas and its equation of state is given by: PV = NRT,

(11.1)

where P is the pressure, V is the volume, N is the total number of particles, R is the gas constant, and T is the temperature. It is evident from Eq. (11.1) that an equation of state relates different thermodynamic quantities (such as pressure, volume and temperature) of the gas. Such a relation is very useful in investigating the behaviour of gaseous systems. The velocity distribution of the particles of an ideal gas is given by the well known Maxwell’s velocity distribution law which you know already from the kinetic theory of gases.

When the density of the gaseous system becomes as high as in compact stars, the distance between two atoms/molecules becomes comparable to the size of the atoms/molecules or even the size of the nucleons (i.e., protons, neutrons)! At such densities, other forces such as the Coulomb force and nuclear forces begin to influence the dynamical behaviour of the atoms/molecules. As a result, the equation of state of an ideal gas cannot describe the behaviour of high density gases. Now, to get an idea about the equation of state which can describe matter at high densities such as that in compact stars, we note that compact stars consist mainly of degenerate gas of electrons and neutrons. Further, you may recall from the course Thermodynamics and Statistical Mechanics (PHE-06) that we need to use statistics to describe the collective behaviour of a large number of particles (atoms or molecules) of a system. Ordinary gases obey the so-called Maxwell-Boltzmann (M-B) statistics. But, the fermions, which make up the compact stars, obey the so-called Fermi-Dirac (F-D) statistics. So, in the context of the equation of state, our problem reduces to know how F-D statistics influences the physical parameters such as pressure inside a compact star.

You may further recall from this course that the number density =

dN d 3 p d 3x

of

particles in phase space can be written as: dN

g = f d 3 p d 3 x h3 When the particle density is low and temperature is high, F-D as well as B-E (BoseEinstein) distributions become identical to: f (E) = e−(E−µ)/kT. which is known as the Maxwell-Boltzmann distribution.

46

(11.2)

where p and x are the momentum and position variables of the particle, respectively, h is the Planck’s constant and f is the distribution function. In Eq. (11.2), d 3 p d 3 x is the volume element in the phase space, with d 3p containing the momentum elements and d 3x containing the position elements; g is called the statistical weight, i.e., number of states that a single particle can have for a given value of the momentum p. If S denotes the spin of particle, then g = 2S + 1. The distribution function f denotes the average occupation number of a cell in the phase space.

You may also recall that, for an ideal gas in equilibrium at temperature T, the F-D statistics gives the distribution function f as: 1

f (E ) =

Compact Stars

(11.3)

exp [( E − µ) / k BT ] + 1

where kB is the Boltzmann constant, µ is the chemical potential and E is the energy. The energy distribution function given by Eq. (11.3) is valid for fermions such as electron, proton and neutron which have half-integral spins (i.e., 1/2, 3/2 ..). From Eq. (11.3), it is obvious that for T → 0, f (E) becomes a step function (Fig. 11.2), that is, f (E) = 1

when

E< µ

(11.4a)

f (E) = 0

when

E > µ.

(11.4b)

and

Fraction of states occupied

1.0

T=0

0.5

T>0 0.0 0.0

1.0 Energy (E/FF)

Fig.11.2: Fermi distribution function at T = 0 and at temperature, T > 0

When the distribution of fermions is similar to the T = 0 case shown in Fig. 11.2, the gas of fermions is called completely degenerate. Physically, this means that all energy states below a certain energy are fully occupied and the occupancy above this energy is zero. The energy up to which all states are occupied is called the Fermi energy, EF and for fermions at T = 0, EF = µ. Spend 5 min.

SAQ 1 Explain how Eqs. (11.4a) and (11.4b) are satisfied. So, now you know the nature of f for a degenerate gas. The question is: How can we use Eq. (11.2) to arrive at the equation of state for compact stars? Suppose you want to know how many particles are present in a unit volume having all permissible values of momentum. You have to simply integrate the number density in phase space (left hand side of Eq. (11.2)) over all possible momenta, i.e., n=

dN 3

3

d pd x

d3p

(11.5) 47

From Stars to Our Galaxy

Now, if you want to know the energy density ε of particles, you have to integrate over the energies of all the particles: dN

= E d

3

d 3 p,

3

(11.6a)

pd x

where E is the energy of the particle given by:

(

1 2 2 p

)

2 4

E= m c +

(11.6b)

m being the mass of each particle and c being the velocity of light. Similarly, the pressure of the gas defined as the rate of momentum exchanged across an ideal surface of unit area, can be written as: P=

1 3

pv

dN 3

3

d3p

(11.7)

d pd x

Actually, Eq. (11.7) shows that pressure can be expressed as the momentum flux of a gas. Since we are discussing only the isotropic pressure here, the said flux in any one direction (out of three) is one-third of the net momentum flux. That is why a factor of 1/3 has been put before the integral. Assuming that the motion of fermions in compact stars is non-relativistic, it can be shown (we have avoided giving the mathematical details) that Eq. (11.7) reduces to: 5

P = K ρe3 The planetary nebula, so called because of its planetlike appearance, is visually one of the most attractive astronomical objects.

(11.8)

where K is a constant and ρe is the density of electrons. Eq. (11.8) is the equation of state of a degenerate gas of electrons at high density. Note that, unlike the ideal gas, the pressure of a degenerate gas is independent of temperature. This implies that even when temperature of a degenerate gas of fermions is very very low, it can exert tremendous pressure. Let us now discuss white dwarf stars in which only the electrons are degenerate.

11.3 THEORY OF WHITE DWARF After a star has exhausted its nuclear fuel, it begins to collapse due to gravity. If the mass of the star is less than about 8 MΘ, the gravitational contraction is accompanied by expulsion of matter from its outer envelope. The discarded matter forms a ring-like structure around the collapsing star and is called planetary nebula (see Fig. 11.3).

48

Fig. 11.3: A planetary nebula

The core of the star continues to contract till its density attains a value in the range 105 − 108 g cm−3. At this density, a new equilibrium sets in and these stars are called white dwarfs. The evolution of a medium mass star into a white dwarf is shown schematically in Fig. 11.4.

Compact Stars

105 Giant region

Formation of planetary nebula

Luminosity (solar units)

103

10 Main sequence

10-1

White dwarfs

10-3

10-5

80000

40000

20000

6000

2000

Surface temperature of star (K) Fig.11.4: Evoluation of medium mass star leading to the formation of white dwarf

One of the interesting features of white dwarfs is that they have mass almost equal to the mass of the Sun, but their radii are only about five to ten thousand kilometres. The characteristic features of white dwarfs were explained theoretically by S. Chandrasekhar.

11.3.1 Chandrasekhar Limit S Chandrasekhar investigated the effect of the gravitational field on a degenerate gas of electrons to estimate the mass-radius relation of white dwarfs. One of the characteristics of a degenerate gas is that its equation of state does not involve temperature. When particles are non-relativistic, the pressure exerted by such a gas is given by Eq. (11.8). Now, inside a star in hydrostatic equilibrium, the pressure must balance the gravitational pull towards the centre. For a star in hydrostatic equilibrium, (recall from Unit 8 that) the pressure gradient can be expressed as:

1 dPe GM = ρ e dr R2

(11.9)

where M and R are the mass and the radius of the star respectively. If we assume that the density of the star is uniform, we can write:

M =

4 3 πR ρ e 3

(11.10) 49

From Stars to Our Galaxy

Substituting for ρ e from Eq. (11.10) in Eq. (11.9) and integrating from the centre to the surface, we get:

Pe ∝

M2

(11.11)

R4

But, from Eq. (11.8) and Eq. (11.10), we find that:

Pe ∝

M

5 3

(11.12)

R5

Comparing Eqs. (11.11) and (11.12), we obtain:

R ∝

1 M 1/ 3

(11.13a)

We can also write Eq. (11.13a) as

ρ ∝ M2

Eq. (11.13) shows that, as the mass of the white dwarf increases, its radius decreases. The mass-radius relation is plotted in Fig. 11.5. The shrinking of radius is understandable because increase in mass would mean increase in the force of gravitational contraction.

0.02 Radius (solar unit)

S. Chandrasekhar was awarded Nobel Prize in 1983 for his extensive theoretical work on white dwarfs.

(11.13b)

Chandrasekhar limit 0.01

0

0

0.5

1.0

1.5

2.0

2.5

Mass (solar unit) Fig.11.5: Mass-radius plot for white dwarfs

A logical question at this point is: What will happen if we go on adding mass to a white dwarf? Eq. (11.13a) indicates that if enough mass is added to a white dwarf, its radius would ultimately shrink to zero! What is the value of this mass? Chandrasekhar showed that if mass of the white dwarf is about 1.4 times the solar mass, its radius will shrink to zero. This is called the Chandrasekhar limit. If the mass of the white dwarf is more than this limiting mass, its gravitational contraction cannot be balanced by the degeneracy pressure of the electrons. Thus, a star of mass greater than 1.4 solar mass cannot become a stable white dwarf unless it ejects mass in some way. No white dwarf has been discovered which has a mass higher than Chandrasekhar limit. In the absence of energy generating nuclear reactions, there is actually no source of energy left in a white dwarf. Therefore, these stars would go on shining by radiating 50

their thermal energy and in the process, their temperature would decrease. Ultimately, the entire thermal energy will be lost. The star becomes a cold object.

Compact Stars

Would you not like to know how long a white dwarf star takes to cool down? The internal temperature (say, Twd), of a white dwarf is almost constant. Its total thermal energy can, therefore, be written as: U=

=

3 N e k B Twd 2

3 M k B Twd 2 µ mp

(11.14)

where M is the mass of the white dwarf. Eq. (11.14) has been written by taking the energy of an electron at temperature Twd as (3/2)kBTwd, (1/2)kBTwd for each degree of freedom. In Eq. (11.14), µm p is the mean molecular weight of nuclei inside the star. For stars which consist entirely of heavy elements, µ = 2 . If we take M = 1MΘ, and 7 48 Twd = 10 K, the total thermal energy of a white dwarf is ~ 10 ergs! This is a very significant amount of energy. The question is: How long will this amount of thermal energy enable a white dwarf to shine? To know this, solve the following SAQ. SAQ 2 Suppose the luminosity of a white dwarf star of mass 1MΘ is 10−3LΘ. If the luminosity of the Sun, LΘ is 4 × 1026 Js−1, calculate the time for which the white dwarf will keep shining with its present luminosity.

Spend 5 min.

Having solved SAQ 2, you know that the thermal energy of a white dwarf can keep it shining for billions of years! Sirius B, the companion of the ‘Dog Star’ Sirius A, is the earliest white dwarf star to be observed. Fig. 11.6 shows a white dwarf star.

Fig.11.6: A white dwarf star shown by arrow

Fig. 11.7 shows the distribution of the observed white dwarfs on the H-R diagram. Different lines such as 0.89, 0.51 and 0.22 in the Figure indicate the masses of stars in terms of the solar mass. White dwarfs are dimmer compared to the main sequence stars of the same surface temperature because of their smaller size. The difference in absolute magnitudes of white dwarfs compared to main sequence stars is in the range 5 – 10.

51

From Stars to Our Galaxy Main sequence

Sun

log (L/L Θ)

0

-1.5 0.89

0.51

0.22

-3.0

White dwarfs

-4.5 4.5

4.3

4.1

3.9

3.7

3.5

log T Fig. 11.7: Location of white dwarfs in the H-R diagram

As we have mentioned earlier, white dwarfs are compact stars resulting from the death of medium mass stars like the Sun. You may ask: What are the remnant of stars which are much more massive than the Sun? Such stars end their lives as neutron stars. You will learn about it now.

11.4 NEUTRON STAR To understand the formation of neutron stars, you may recall from Unit 10 that when the nuclear reactions stop, the core of a massive star (~ 10 MΘ to 100 MΘ) collapses and a supernova explosion takes place. The supernova explosion blows away the outer shell of the star. The question is: What is the remnant of a supernova explosion? In 1934, Walter Baade and Fritz Zwicky suggested that the core of a supernova explosion could be a small, high density neutron star. When the core of a star begins to collapse in the absence of nuclear reactions, its density increases and attains a value comparable to the density of white dwarfs. The degeneracy pressure due to electrons can balance the gravitational contraction only when the mass of the object is smaller than the Chandrasekhar limit. If the mass is more than this limiting value, the gravitational collapse continues and the density of the star increases further. When the density is in the range ~ 1014 − 1015 g cm−3, the following two things happen: a) protons and electrons combine to produce neutrons and neutrinos according to the following inverse β-decay reaction: e+p→n+v b) atomic nuclei begin to disintegrate due to pressure ionisation. Since neutrons are fermions, their degeneracy pressure halts the gravitational contraction and a stable neutron star is formed. It has been calculated that the neutron stars have radii of about 10 km and densities of their core is about 1014 to 1015g cm−3. Note that neutron stars are much denser than the white dwarfs.

52

You may ask: Is there any limiting mass for neutron stars similar to the Chandrasekhar limit for white dwarfs? Yes; but the precise value of the limiting mass for the stars is difficult to determine because the behaviour of matter at such high densities is not well understood yet. The estimated limiting mass is in the range 2 – 3 MΘ.

Compact Stars

11.4.1 Gravitational Red-shift of Neutron Stars You have just learnt that the neutron stars are very compact. Their gravitational pull must be very strong indeed. How strong is its gravitational pull? To get an idea, we may estimate the value of gravitational acceleration, g, on a neutron star, say, of mass Mn ~ 1.5MΘ and radius Rn ~ 10 km. The expression for gravitational acceleration on a neutron star can be written as: gn =

GM n Rn2

Substituting the values of Mn and Rn, we get, gn ~ 2 × 1014 cm s−2. The value of gn is ~ 2 × 1011 times higher than the gravitational acceleration on the Earth! That is, the gravitational force on a neutron star is 1011 times stronger than that due to the Earth. Well, to get a feel for the perceptible effect of such a strong gravitational field, let us do a thought experiment. Suppose you are standing on the surface of the star which is

Very often, we do thought experiments: the experiments which cannot be done for practical reasons, but they could be done, in principle.

emitting a yellow light at λ = 5800 A . The light reaches your eyes approximately one metre above the ground. The interesting question is: Will you see the light as yellow? If you think you will, you are wrong! Let us find out the reason. A photon leaving the surface of the neutron star has energy hν and equivalent mass hν/c2. The neutron star will attract this photon and, as a result, the photon has to work ‘hard’ against the star’s gravity to reach your eyes (just as a stone thrown upward from the earth slows down as it goes higher). The work done by the photon to reach hν your eye is, W = g n H , where H ~ 100 cm, is the height of your eyes from your c2 feet. Further, ν can be calculated using the relation λ = c / ν, i.e., ν ~ 5.172413 × 1014 Hz. Now, to arrive at your eyes, the photon works against gravity and hence its energy, that is, frequency will decrease and wavelength will increase. To calculate the change in wavelength or frequency, we can use the energy conservation principle. Let ν′ be the frequency of the light (photons) that you observe. Thus, from the energy conservation principle, we can write: hν ′ = hν − W

= hν −

hν

gnH

c2

= hν 1 −

(11.15)

gnH c2

In terms of the wavelength of photons, Eq. (11.15) can be written as: dλ λ

~

gnH c2

(11.16)

where dλ = λ′ − λ and λ′ = c / ν′. Substituting the values of gn, c and H in Eq. (11.16), we get: dλ λ

~ 2.2 × 10 −5

(11.17) 53

From Stars to Our Galaxy

Thus, yellow light of wavelength λ = 5800 A leaving your feet (the surface of the neutron star) will arrive at your eyes as light of wavelength λ ′ = λ + λ

gnH c2

~ 5800.13 A . Such differences in the wavelengths can

be measured easily these days. The increase in the wavelength of a photon when it comes out of a strong gravitational field is called gravitational red-shift. The significance of this phenomenon lies in the fact that scientists must make necessary corrections in the observed frequency of photon presumably coming from objects such as neutron stars in order to understand physical processes on their surface. It is difficult to observe neutron stars optically because, being very small in size, these are very faint objects. However, with radio telescopes, neutron stars have been detected in the form of pulsars: the stars which emit regular pulses of radiation, very often several times a second.

11.4.2 Detection of Neutron Stars: Pulsars

Number of photons

In 1967, a remarkable discovery was made in the history of astronomy by a student named Jocelyn Bell who was doing her doctoral work under Prof. Anthony Hewish in Cambridge, England. She observed certain periodic pulses of radio waves coming from a certain direction in the sky which were repeated precisely every 1.337s (Fig. 11.8). Very soon, she discovered a few more such objects with different periodicities. These objects are known as pulsars. A year later, it was already clear that these pulses must have been emitted by rotating neutron stars. Periodicities of the pulsars have been found to be between 10−3s and 4s.

Time (a)

(b)

Fig. 11.8: a) Detection of pulses of radio waves, an evidence of the existence of neutron stars; and b) pulsar at the centre of a nebula (Image credit: Chandra X-ray Observatory)

You may ask: How can we be sure that a strong pulse, whose periodicity is about a second, is emitted by a neutron star and not by a white dwarf? When a star rotates, each of its layers experiences centrifugal force directed outward as well as the pointing gravitational force directed inward. If Ω be the angular velocity and R be the radius of the rotating star, the two forces balance in a Keplerian orbit and we can write:

Ω2 R = 54

GM R2

(11.18)

For a white dwarf, we can write

Compact Stars

M ~ ρ ~ 10 7 gcm −3 . So the typical value of Ω is: 3 R

Ω ~ 2.58 s−1.

(11.19)

Therefore, the time period can be written as: T=

2π Ω

~ 2.433s.

(11.20)

From Eq. (11.18), it is clear that for objects with lower densities, the corresponding Ω will also be lower, and hence the time period would be higher! Therefore, white dwarfs and other bigger celestial objects are too big, their densities are too low and they cannot emit pulses with periods as short as those observed in pulsars. In addition, it has been observed that there are pulsars having periods of only a few milliseconds! It is, therefore, almost impossible for white dwarfs to show such periodicities. So, the belief that pulsars must be neutron stars became even stronger. Now, before proceeding further, you should solve an SAQ. SAQ 3 Calculate the gravitational red shift for the yellow light (λ = 5800 A ) on the surface of Sirius B when the photon travels a distance of 1m. Take the mass of Sirius B as M SiB = 1M Θ , and its radius as R SiB = 16000 km.

Spend 5 min.

Emission mechanism and observed profiles Regarding pulsars, a logical question could be: What causes emission of pulses from a neutron star? It is suggested that a neutron star is like a gigantic light house. In a light house, a powerful light-source rotates and ships see it periodically. In exactly the same way, emission of radiation takes place continuously, but due to rotation of the neutron star the emission is detected only periodically by an observer on the Earth (Fig. 11.9). Magnetic axis Radiation

Neutron star

Rotation axis

Radiation

Fig. 11.9: Light-house model of pulsars

55

From Stars to Our Galaxy

You may argue: If emission takes place from the surface of a neutron star, it should have been detected all the time and not periodically! Well, to understand the periodic emission, you need to know that it is linked to the rapid rotation and strong magnetic fields of neutron stars. Suppose a molecular cloud of radius, Rmc ~ 1 pc having a typical magnetic field, Bmc ~ 10−6G collapses into a neutron star of size, Rn ~ 10 km. From Unit 8, you know that if the matter is highly conducting, the magnetic flux linked to it is conserved: B r 2 ~ constant,

(11.21)

Thus, we can write: 2 Bn Rn2 = Bmc Rmc

(11.22)

Substituting the typical values of Rn, Bmc and Rmc, we get: Bn ~ 9.5 × 1018G. Thus, we find that the value of magnetic field associated with a neutron star is very high compared to the Earth’s magnetic field which is only a fraction of a Gauss! The above estimate has been arrived at under the assumption that there is no loss of magnetic field and thus the value indicates the upper limit. In reality, after some dissipation in the process of the formation of a neutron star, the field is smaller, close to 1011 to 1014 Gauss or even less. Generally, this field is bipolar and the magnetic axis is not aligned with the spin axis of the star (Fig. 11.9). As the neutron star spins at great speeds, its magnetic field induces a very strong electric field. As a result, electrons present in the atmosphere of the neutron star are accelerated and attain very high energies. These electrons then gyrate round the magnetic lines of force and emit radiation called synchrotron radiation which is directed along the lines of force. Every time the neutron star rotates, each of the two radiating magnetic poles may point towards us (Fig. 11.9) and we may see two pulses of radiation per rotation of the star. In many objects, you can see these two peaks very distinctly (Fig. 11.8). Thus, the lighthouse model tells us that pulsars are not mysterious objects at all; they are rotating neutron stars. You may further argue: If the neutron star is emitting radiation continuously, it would lose energy; in the absence of nuclear energy, what energy is it losing − the gravitational energy, the rotational energy or the magnetic energy? If it loses gravitational energy, the star would collapse further and the spin period may go down. If it loses magnetic energy, the intensity of radiation will go down with time. However, if it loses rotational energy, the pulsar would slow down and its time period would increase. Observations support the argument that the most dominant component of the energy loss is the loss of rotational energy. So far, you have learnt that stars having mass up to 1.4 MΘ stabilise as white dwarfs and those having mass up to 3 MΘ stabilise as neutron stars. Now, suppose that a star has mass greater than 3 MΘ and all thermonuclear reactions have ceased in it. The question is: What is the fate of such a star? What force will oppose gravity and prevent the complete gravitational collapse of such stars? Obviously, the degeneracy pressures due to electrons and neutrons are insufficient to halt the collapse because the stellar mass is greater than 3 MΘ. In fact, there is no force which can halt the complete gravitational crunch of such a star. This theoretical collapse of a star into a singular point of zero volume and infinite density is called a black hole. You will learn about black holes now. But, before that, how about solving an SAQ? 56

Spend 5 min.

SAQ 4

Compact Stars

Suppose the Sun shrinks to the size of a neutron star of radius 106 cm. Calculate the magnetic field strength at the surface of the neutron star. Take the radius of the Sun to be 1011 cm and the magnetic field at its surface equal to 1 gauss.

11.5 BLACK HOLE When the mass of a star is more than the limiting mass (~3 MΘ) for neutron stars and all nuclear reactions in its core have stopped due to the lack of nuclear fuel, there is no force which can stop its continuous gravitational collapse. Due to such an unhindered collapse, the object collapses to zero radius and infinite density! It is very difficult to visualise such an object physically. However, such an object has an interesting property. Its gravity is so strong that not even light can escape from it. Therefore, such a body is called a black hole. The physics of black hole involves very sophisticated mathematics which is beyond the scope of this course. Nevertheless, we can apply simple principles of physics to conclude that black holes do exist. Let us ask ourselves: For a given mass, what should be the size of a body so that even light (that is, photons - the fastest moving thing) cannot escape from it? The answer to this question has great bearing on our understanding of black holes. The first theoretical attempt to address this question was made by Schwarzschild who used the principles of General Theory of Relativity given by Einstein. The main conclusions of Schwarzschild can also be derived using Newtonian mechanics and the concept of escape velocity. You may recall from school physics that, for a body of mass M and radius R, the escape velocity is given by: v esc =

2GM R

(11.23)

Since we are interested in a black hole and trapping of light by it, we put vesc = c, the velocity of light. Then, Eq. (11.23) reduces to: Rg =

2GM

c2

(11.24)

Rg in Eq. (11.24) is called the Schwarzschild radius. Eq. (11.24) signifies that the size of an object of mass M must shrink to Schwarzschild radius to become a black hole. For example, if the object has mass equal to 1MΘ, its radius must be 3 km if it has to behave like a black hole. In other words, the Sun must shrink to a radius of 3 km to be able to trap light and become a black hole! SAQ 5

Spend 3 min.

Calculate the Schwarzschild radius for the Earth. You may ask: If black holes are point like objects, what is the physical significance of Schwarzschild radius? Schwarzschild radius essentially means that any object (including a photon) which comes within Rg (= 2GMbh /c2) of a black hole is trapped. This limiting value Rg is also called the event horizon, since no event that takes place within Rg from the centre of the black hole can be viewed by any observer at R > Rg. Why is it so? This is because the gravity of the point like object is so strong that even radiation (i.e. photons) cannot escape from the region inside the Schwarzschild radius.

57

From Stars to Our Galaxy

Although the concept of Schwarzschild radius is helpful in visualising a black hole, you have to be careful in not stretching the meaning of escape velocity too far. This is because of Einstein’s General Theory of Relativity (GTR) which treats gravity entirely differently from Newtonian gravity. According to GTR, we cannot talk of escape velocity at all. Einstein proposed that near a gravitating object, the photons move in a curved trajectory as seen by an observer at a large distance. The black hole being compact and massive, its gravitational force is very strong which bends the photon path so much that a photon trying to escape would immediately return back. Bending of light close to a black hole As you know from the well known Einstein mass-energy equation (E = mc2), every form of energy E has an equivalent mass m. So, the photons with energy E = hν will also have a mass, mphoton = hν/c2. This is not the true mass since photons are really massless. Nevertheless, this mass can be attracted by any other massive body such as the Sun. This will bend the path of a photon. In fact, the bending of light was observed by the famous British astronomer Arthur Eddington in the year 1919. He observed a star during the total eclipse whose location in the sky was near the edge of the Sun and found that the apparent location of the star has been changed by a small angle. This convinced him that the light from that star must have been bent by the Sun. Thus, strong bending of light can be taken as an evidence that black holes exist. Types and location of black holes The exact process of the formation of a black hole is not known but there are several possibilities. Unlike the other types of compact stars, black holes do not have any narrow range of masses. There seem to be two populations of black holes: one population has a mass typically 6 to 14 times the mass of the Sun. It is formed due to gravitational collapse after a supernova explosion. These so-called stellar mass black holes could be detected all around the galaxy. Black holes of the second population are very massive, with masses ranging from a few times 106MΘ to a few times 109MΘ. These seem to be located at the centres of galaxies. Detection of a black hole Since radiation cannot escape from a black hole, it cannot be observed / detected directly. However, it can be detected indirectly through its gravitational field. Any matter close-by is strongly attracted by it. Suppose that the black hole is a member of a binary. Then, it sucks matter from its companion. The matter flowing into the black hole gets heated to a very high temperature and emits X-rays. Thus, to search for black holes, we should look for X-ray binaries. Now, let us summarise what you have learnt in this unit.

11.6 SUMMARY

58

•

When the nuclear fuel of a star is exhausted completely, it contracts due to selfgravity and becomes a compact star having density much higher than a main sequence star.

•

In the absence of radiation pressure due to nuclear reactions, the gravitational force in the compact stars is counter-balanced by the degeneracy pressure of fermions like electrons and neutrons.

•

The nature of the remnants of a star after death depends on its mass; the dying star turns into any one of the three kinds of compact stars, namely, white dwarf, neutron star or black hole.

Compact Stars

•

If the mass of a dying star is less than the Chandrasekhar limit (~ 1.4MΘ), it turns into a white dwarf in which the gravitational force is balanced by the degeneracy pressure of electron gas.

•

If the mass of the dying star is more than the Chandrasekhar limit but less than another limiting value (~ 3MΘ), it turns into a neutron star in which the gravitational force is balanced by the degeneracy pressure of neutrons. If the mass of the collapsing star is even higher, the gravitational collapse cannot be halted and the collapsing star becomes a black hole.

•

The degeneracy pressure of a degenerate gas of electrons is given by: 5 3

P = K ρe

•

Chandrasekhar showed that, for white dwarfs, the mass-radius relation is:

1

R∝

M

1 3

and the density-mass relation for such stars is:

ρ∝ M 2 •

Pulsars are stellar objects which emit periodic radio frequency pulses. Pulsars are rotating neutron stars, because white dwarfs cannot emit pulses of this periodicity.

•

Emission of pulses from a rotating neutron star is explained on the basis of lighthouse model.

•

Black holes are objects of zero radius and infinite density. Such an object is difficult to visualise physically.

•

The Schwarzschild radius is given as: Rg =

2GM

c2

and it signifies that the size of an object of mass M must shrink to Rg to become a black hole. •

The Schwarzschild radius is also called the event horizon because an event that takes place within Rg from the centre of the black hole cannot be observed.

•

The physics of black hole cannot be understood on the basis of Newtonian mechanics; we need to invoke Einstein’s General Theory of Relativity (GTR) for this purpose.

11.7 TERMINAL QUESTIONS 1.

Spend 30 min.

For a completely degenerate electron gas, the number density of particles with momenta in the range p and p + dp is given by:

59

From Stars to Our Galaxy

n( p )dp =

8π

h

3

p 2 dp

p≤ p

=0

F

p> pF

where p F is the Fermi momentum (corresponding to Fermi energy). Show that p F increases with n as n

as n 2.

1 3

1 3

. Also show that the average momentum < p > varies

.

The masses and radii of a typical neutron star (NS), a typical white dwarf (WD) and a typical main sequence star (MS) are given below:

NS WD MS

Mass

Radius

1M Θ 1M Θ 1M Θ

10 km 104 km 106 km

Calculate the rotational time periods in all these cases and show that only neutron stars satisfy the pulse time periods observed for pulsars. 3.

Assume that the peak wavelength of X-rays coming from an X-ray binary is 1 A . Calculate the temperature that the falling matter onto a black hole must attain to emit this radiation.

11.8 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1.

At T = 0, the exponent ( E − µ) / k B T in the denominator of Eq. (11.3) goes to infinity. Thus, for E > µ , the denominator of f (E) becomes infinite and we have f (E) = 0. For E < µ , the exponent ( E − µ) / k B T in the denominator of Eq. (11.3) becomes − ∞ at T = 0. Thus, the denominator of f (E) becomes 1 and f (E) = 1.

2.

On the basis of Eq. (11.14), we find that the thermal energy of a star of mass 1M Θ is 10 48 erg. As per the problem, luminosity of the white dwarf star is:

L = 10 −3 LΘ = 10 −3 × ( 4 × 10 26 Js −1 ) = 4 × 10 30 erg . s −1

Thus, the time for which the white dwarf will keep shining can be written as: t = 60

10 48 erg 4 ×10 30 erg . s −1

=

1018 4

1

1 yr = 3 ×10 7 s

yr

3 × 10 7

Compact Stars

10

≅ 10 yr. 3.

On the basis of Eq. (11.16), we can write the expression for the change in wavelength on the surface of white dwarf (Sirius B) as dλ = λ. =

=

c2

λH c

=

g wd H

2

×

GM wd R2

(5800 A ) × (100 cm) (9 ×10 20 cm 2 s − 2 )

×

(6.67 ×10 −8 cm 3 g −1s − 2 ) × (2 ×10 33 g) (1.6 ×10 9 cm) 2

5.8 × 6.67 × 2 ×10 −8 A 9 × 1 .6 × 1 .6

≅ 3.3 ×10 −8 A (Negligible) 4.

From Eq. (11.21), we can write the magnetic field linked to the Sun and to the neutron star (to which the Sun converts) as: BΘ R 2 Θ = B NS R 2 NS = Constant

where BNS and RNS , respectively, are the magnetic field and radius of the neutron star. Thus, B NS = BΘ

R 2Θ R 2 NS

= 1 gauss ×

(1011 cm) 2 (10 6 cm) 2

= 1010 gauss

5.

From Eq. (11.24), we can write the expression for Schwarzschild radius for the Earth as: =

Rg E

=

2GM E

c2 2 × (6.67 × 10 −8 cm 3 g −1s −2 ) × (6 × 10 27 g) 9 × 10 20 cm 2 s − 2

= 0.9 cm

61

From Stars to Our Galaxy

Terminal Questions 1.

As per the problem, 8π

n ( p) dp =

h

n=

8π h

=

pF

3

p 2 dp

3

p 2 dp

0

8π

p3 F

h3

3

Thus, we can write:

pF ∝n

1 3

Now, the expression for average momentum can be written as: pF

1 = n

p n ( p ) dp 0 pF

8π

1 = × n h3

p 3 dp

0

8π 1 p = 3× × F h n 4

4

4

8π 1 n 3 = 3× × h n 4 Thus, we can write:

∝ n 2.

1 3

From Eq. (11.18), we can write the expression for the angular velocity of a rotating star as: GM

Ω=

R3

So, the time period is: T=

2π Ω

= 2π 62

R3 GM

Thus, the time period for the neutron star is:

Compact Stars

1018 cm 3

T NS = 2π

(6.67 ×10 −8 cm 3 g −1s − 2 ) × (2 ×10 33 g)

= 2π × 10 − 4

10 s 6.67 × 2

= 5.4 ×10 − 4 s

Time period for white dwarf is:

TWD = 2π

10 27 cm 3 (6.67 ×10 − 8 cm 3 g −1s − 2 ) × (2 × 10 33 g)

≈ 17 s

And the time period for a main sequence star is:

TMS = 2π

10 33 cm 3 (6.67 ×10 −8 cm 3 g −1s 2 ) × (2 × 10 33 g)

≈ 17 ×10 3 s

Thus, we find that the time period of the pulses emitted by a neutron star is of the same order as of the pulses from pulsars. 3.

We know that, λ m T = 0.3 cm K

As per the problem: λ m = 1 A = 10 −8 cm

Thus, T=

0 .3 10 −8

K

= 3 ×10 7 K

63

From Stars to Our Galaxy

UNIT 12

THE MILKY WAY

Structure 12.1

Introduction

12.2 12.3

Basic Structure and Properties of the Milky Way Nature of Rotation of the Milky Way

Objectives

Differential Rotation of the Galaxy and Oort Constant Rotation Curve of the Galaxy and the Dark Matter Nature of the Spiral Arms

12.4 12.5 12.6 12.7 12.8

Stars and Star Clusters of the Milky Way Properties of and Around the Galactic Nucleus Summary Terminal Questions Solutions and Answers

12.1 INTRODUCTION In Unit 11, we discussed the death of stars and its consequences and thereby completed our discussion about stars. You now know how stars are formed, how they live their lives and how and why they die. If you wish to know more about the Universe, you may ask questions like: What is the structure of the Universe? Is it a single entity comprising unrelated and independent stars or does it consist of substructures in which stars are arranged according to a scheme? If stars constitute communities, what is the nature of such communities and what mechanism brings them into existence? It is now believed that stars arrange themselves into billions of self-contained systems called galaxies. Our star, the Sun, is one of about 200 billion stars in the galaxy called the Milky Way – our home galaxy. In the present Unit, you will learn about the Milky Way galaxy also called the Galaxy. On a clear night sky, you can see a broad white patch running across the sky. It is seen during winter months in the northern hemisphere. The patch is actually made up of hundreds of billions of stars, so close to each other that they cannot be seen individually. This is a part of our home galaxy, the Milky Way. Just as Copernicus discovered that the Earth and the planets revolved around the Sun, astronomers in the early twentieth century discovered that the Milky Way system is indeed a galaxy in which our solar system is situated. In Sec. 12.2, you will learn the basic structure and properties of the Milky Way. It is found that the Sun revolves once in 200 million years around the centre of this galaxy in a nearly circular orbit. In addition, the Galaxy itself is rotating. You will learn the consequences of the rotation of the Galaxy in Sec. 12.3. You will also discover that the Galaxy is a spiral galaxy with several spiral arms. In Sec. 12.4, you will learn about the various constituents of the galaxy, namely the stars, globular clusters, compact stars and the interstellar medium. At the present time, we know a lot about our Galaxy, its shape and its size. Much of the contemporary focus has been to understand the central region of the galaxy and the activities surrounding it. Today, it is believed that the central region contains a massive black hole of mass M ~ 2.6 × 6 10 MΘ. In Sec. 12.5, you will learn about the nature and characteristics of the central region of the Milky Way.

64

Objectives

The Milky Way

After studying this unit, you should be able to:

• • • • •

describe the shape and size of the Milky Way galaxy; explain the consequence of rotation of the Galaxy; describe the major building blocks of the Galaxy including the stars and star clusters; explain the nature and persistence of the spiral arms of the Galaxy; and describe the activities taking place near the centre of the Galaxy particularly the nature of the central compact object.

12.2 BASIC STRUCTURE AND PROPERTIES OF THE MILKY WAY The Milky Way is a highly flattened, disk shaped galaxy comprising about 200 billion (2 × 1011) stars and other objects like molecular clouds, globular clusters etc. It is so huge that, to travel from one edge of the Galaxy to the other, light takes about 100,000 years! Its radius is about 15,000 pc. The solar system is located roughly at a distance of about 8.5 kpc from the centre of the Galaxy. The total mass of Milky Way has been estimated to be about 2 × 1011MΘ. Refer to Fig. 12.1 which shows a schematic diagram of the Milky Way Galaxy. You may note that, broadly speaking, the Galaxy can be divided into three distinct parts: a central bulge (B), the flattened galactic disk (A) and a halo (H) which surrounds the Galaxy.

H B

A

Sun 8500 pc 30,000 pc (a)

(b)

(c) Fig. 12.1: a) A schematic diagram of the Milky Way Galaxy viewed edge on, b) view of the Galaxy from the top; and c) part of the Milky Way visible in the sky

The Milky Way belongs to the Local Group, a group of 3 big and 30 odd small galaxies. After the nearby Andromeda galaxy (M31) shown in Fig. 12.2, ours is the

65

From Stars to Our Galaxy

largest galaxy in the group. It has a few dwarf galaxies as satellites or companions. Prominent among them are Large and Small Magellanic clouds.

(a)

(b)

(c)

Fig. 12.2: a) The Andromeda galaxy; b) the Large Magellanic Cloud; and c) the Small Magellanic Cloud

Let us now discuss the nature of the components of Milky Way.

The Central Bulge The central bulge (Fig. 12.1a) is a more or less spherical cloud of stars. Being located in the disk region of the Galaxy, we cannot see this region in optical wavelengths. It is so because the disk region consists of gas and dust which absorbs optical wavelengths 10 and obstructs our view. The total mass of the bulge is estimated to be about 10 MΘ. Apart from stars, this region consists of gas in the form of molecular clouds and ionised hydrogen. The motion of the stars and the gas near the centre of the bulge suggests that there could be a massive black hole at the centre.

The Disk Component

66

The flattened disk component has a radius of about 15,000 pc. But its thickness is very small. Most of the stars are located along the central plane of the disk and as we move away from this plane, the density of stars decreases.

The most significant feature of the disk component is the existence of spiral arms. Condensation of stars has been observed along the spiral arms. These arms have very young stars called Population I stars, star-forming nebulae, and star clusters. The arms are named after the constellations in the direction of which a large portion of the arm is situated. Our solar system is located on a Local or Orion arm.

The Milky Way

The Halo Component The bulge and the disk components are surrounded by another, not so well defined, and not so well understood spherical component called the halo component. This is mainly made up of gas and older population of stars. These stars exist in very dense clusters; each cluster having 105 to 106 stars. These are called globular clusters. Stars in these clusters are so densely packed that they cannot be resolved, and clusters appear like a circular patch of light (Fig. 12.3).

Fig. 12.3: A globular cluster (NGC 1916) containing millions of stars

The nature of galactic rotation (about which you will learn later in the Unit) suggests that there is a large amount of matter which is governing the motion of the stars in the disk. This matter is not visible in any wavelengths and scientists call it the dark matter. It is believed that the halo contains at least an equal amount of matter as the disk itself, if not more, in the form of dark matter. What could be the nature of this matter? We do not know yet.

12.3 NATURE OF THE ROTATION OF THE MILKY WAY In the early nineteenth century, Jan Oort and Bertil Lindblad studied the motion of a large number of stars located near the Sun. Their study indicated that stars in the Galaxy constitute a gravitational system. Like many other gravitating systems, our galaxy also rotates, though very slowly. The rotational velocity enhances the stability of the Galaxy since the outward centrifugal force can counterbalance the inward pull due to gravity. The assumption that the Galaxy is a gravitating system is at the core of the traditional hypothesis about how the Galaxy came into being. Since the centrifugal force is only along the equatorial plane, there is no obstacle for matter (other than the pressure) to fall along the vertical direction. As a result, the initially spherical distribution of matter has become, after over 10 billion years, the highly flattened galaxy of today. 67

Refer to Fig. 12.4 which shows the initially spherical distribution of matter (Fig. 12.4a) gradually evolving into the present day flattened Galaxy (Fig. 12.4d).

From Stars to Our Galaxy

(a)

(b)

(c)

d)

Fig. 12.4: Evolution of the Galaxy (from a to d) starting from a spherical cloud of gas and settling into the flattened disk shape of today

The rotational velocity of stars in the Galaxy is very slow compared to what we are familiar with in our solar system. For instance, at a distance of roughly 8 kpc away from the centre, the Sun takes about 240 million years to rotate once around the galactic centre. The question is: How do we know about this rotation? How do we measure it? What are the consequences of this rotation on the structure and further evolution of the Galaxy? Let us now learn about these aspects of the Galaxy.

12.3.1 Differential Rotation of the Galaxy and Oort Constants To appreciate the rotation of stars and other objects in the Galaxy, you need to understand the concept of differential rotation. Let us consider the motion of a star in the Galaxy. Suppose, for the sake of argument, that the whole mass of the galaxy is concentrated at its centre and the stars move like planets round the Sun on orbits called the Keplerian orbits. The angular velocity of a star at a distance r from the centre can, therefore, be written as (Eq. (5.1), Unit 5):

ωKep (r ) =

GM Gal r3

1/ 2

.

(12.1)

where MGal is the mass of the Galaxy. We see from Eq. (12.1) that the angular velocity is not a constant. Further, using Eq. (12.1) and the relation v = ωr, we can write the rotational velocity of the star as: vφ (r) ∝ r−1/2.

(12.2)

Now, if stars in the Galaxy are embedded as particles in a rigid body, then the angular velocity, ω of stars would have been constant, independent of its distance from the centre and the rotational velocity vφ (r) of a star in such a system is given by: vφ (r) ∝ r.

(12.3)

Comparison of Eqs. (12.2) and (12.3) clearly shows that the nature of the dependence of rotational velocity on the distance from the centre is different in the two cases – the Keplerian motion and rigid body rotation. When different components of a system rotate independently, the rotation is known as differential rotation. On the basis of observations, we can say that the stars indeed have differential rotation. We do see the signature of this differential rotation in astronomical observations. 68

The Milky Way V0

S

l D

Q

90o +

V R0

R

C

Fig.12.5: Geometry of the differential galactic rotation for stars closer to the Sun

Let us now obtain expression for the velocity of an object in the Galaxy. To do so, refer to Fig. 12.5, which depicts the velocity vectors of the Sun (S) and a star (Q) with respect to the galactic centre (C). Let us assume that the Sun and the star are at a distance of R0 and R , respectively from C and let D be the distance between the Sun and the star. Let V0 be the Sun’s rotational speed and V be the star’s rotational speed; both assumed here to be on circular orbits for simplicity. Let us also assume that l is the angle between the direction of the galactic centre and the direction of the star from the Sun. This is the so-called galactic longitude of the star. (Galactic longitude of the centre is l = 0 by this definition). The Sun-star direction makes an angle α with the velocity vector of the star. If ω0 and ω be the angular velocities of the Sun and the star, respectively, we can write: ω 0 = V0 / R0

(12.4a)

ω =V / R

(12.4b)

and

We are interested in finding the radial speed of the star with respect to the Sun. The radial velocity is the velocity along the line joining the Sun and the star. We can, therefore, write the star’s radial velocity with respect to the Sun, Vr, as: Vr = V cosα − V0 sin l .

(12.5)

where the first term on the right hand side is the component of the star’s space velocity along the Sun-star direction and the second term is the component of the Sun’s own velocity along the Sun-star direction. Thus, radial velocity is essentially the difference between the projected velocities along the Sun-star direction. Substituting Eq. (12.4) in Eq. (12.5), we get: Vr = ω R cos α − ω0R0 sin l

(12.6)

In triangle QSC in Fig. 12.5, the angle SQC is 90 + α. 69

Thus, using the law of sines, we can write:

From Stars to Our Galaxy

sin (90 + α) sin l = R0 R

or, cos α =

R0 sin l R

Substituting this value of cos α in Eq. (12.6), we get: Vr = (ω − ω0) R0 sin l

(12.7)

Similarly, the tangential component of the velocity of star with respect to the Sun can be written as: Vt = V sin α − V0 cos l = (ω − ω0) R0 cos l − ωD.

(12.8)

In the solar neighbourhood, D is the mean square velocity and R

Ω = −G 0

M ( R) dM ( R ) R

(12.19)

Using Eqs. (12.18) and (12.19), we can determine the enclosed mass at a given radius if we know the stellar velocities data.

Enclosed mass (solar unit)

107

106

105 0.001

0.01

0.1 Radius (pc)

1.0

10.0

Fig. 12.10: Graph showing variation of the enclosed mass as a function of the radial distance close to the galactic centre

In Fig. 12.10, we show the enclosed mass as a function of distance from the centre of the Galaxy. It is clear that even though there are no measurements below R < 0.015pc, the enclosed mass seems to have converged to a fixed number. The mass can be read out from the plot itself: it is 2.6 ×106MΘ. The only acceptable solution seems to be that a massive black hole of this mass resides at the centre. For a spiral galaxy this is not very small, but probably in the lower end. Generally, in the centres of spiral galaxies, black holes of mass M ~ 107MΘ are quite common. Spend 5 min.

SAQ 3 What is the mass density of the region around the galactic centre in units of (MΘ pc−3) below r = 0.015 pc? Now, let us summarise what you have learnt in this unit.

12.6 SUMMARY

76

•

The Universe comprises billions of self contained systems called galaxies. The Sun – only star in our solar system – is one of about 200 billion stars in the galaxy called the Milky Way, our home galaxy.

•

The Milky Way is a highly flattened, disk shaped galaxy. Its radius is 15,000 pc and its total mass is estimated to be about 2×1011MΘ.

•

Broadly, the Milky Way can be divided into three distinct portions: a central bulge, the flattened galactic disk, and a halo.

•

The central bulge is a spherical cloud of stars and most of the mass (~1010MΘ) of the Galaxy is contained in it.

•

The disk component consists of spiral arms; most of the stars are located along the central plane of the disk.

•

The halo is made up of gas and older population stars; these stars exist in very dense clusters called globular clusters. The halo contains an equal amount of matter as the disk itself in the form of dark matter.

•

Like many gravitational systems, our galaxy also rotates. The rotational velocity of stars in the Galaxy is very slow compared to what we are familiar with in our solar system. The stars in the Galaxy undergo differential rotation.

•

Rotation curve of a galaxy contains useful information about the distribution of mass in it and it helps us understand whether or not most of the mass of the galaxy is concentrated at the galactic centre.

•

The spiral arms of the Galaxy do not form a single entity that was originally present; rather, they result from the dynamical interaction of the Galaxy with other galaxies and the matter present in the inter-galactic space.

•

The persistence of spiral arms, despite differential rotation, is explained on the basis of density wave model. According to this model, spiral arms are the areas where density of gas is greater than other places. The arms and the space between them contain roughly the same number of stars per unit volume. However, the arms contain larger number of brighter (O and B) stars.

•

Investigations indicate that a massive black hole, having mass 2.6×106MΘ, resides at the centre of the Milky Way.

12.7 TERMINAL QUESTIONS

The Milky Way

Spend 30 min.

1.

Suppose the contribution of the stellar disk to the mass of our galaxy is negligible and the mass of the entire Galaxy is concentrated at the centre located at a distance of 8.5kpc. What would be the mass you need to put at the galactic centre in order that the estimated Oort constants agree roughly with those observed?

2.

Describe the spiral arms of the Galaxy. Explain how they can persist for a long time.

3.

Distinguish between stars of population I and II. Where are they found in the Galaxy?

4.

It is believed that, in its early phase, the universe was very hot. The average energy of particles at that time was ~ 15 GeV. What was the temperature in degree Kelvin?

12.8 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1.

a) From Eq. (12.8), we have the expression for the tangential component of the velocity of star as: Vt = (ω − ω0 ) R0 cos l − ωD

77

From Stars to Our Galaxy

And, from Eq. (12.9) we have dω dR

ω − ω0 =

( R − R0 ) R0

dω dR R 0

= − D cos l

because ( R0 − R) ~ D cos l . Thus, we can write: R0 Vt = − 2 D cos 2l 2

dω − ω0 D dR R 0

because we can approximate, ω ≈ ω0 . Thus, we get R0 2

Vt = D − 2 cos 2l

= D (1 + cos 2 l )

dω − ω0 dR R 0

− R0 2

dω − ω0 dR R0

=D

−

R0 2

R dω − cos 2 l 0 dR R 2 0

dω − ω0 dR R 0

=D

−

R0 2

R dω cos 2 l − 0 dR R 2 0

dω − ω0 dR R 0

= D [ A cos 2 l + B ]

b)

We can write: dω = dR R 0 =

1 R0

d dR

V R

R0

V dV − 0 dR R R0 2 0

where ω and V, respectively, are the angular velocity and radial velocity of a star. Thus, we can write:

A=

V0 − R0

1 2

dV dR R 0

and B=− 78

1 2

V0 + R0

dV dR R0

2.

We can write the time period of the Sun’s rotation as: T=

The Milky Way

2 π R0 V0 9

Since the age of the Sun is 4.6×10 yrs, the number of times (n) it would have revolved around the galactic centre can be written as: n =

=

=

4.6 × 10 9 × 3 × 10 7 s T

(

1 yr = 3 × 107)

(4.6 × 3 × 1016 s) × (250 × 10 3 ms −1 ) 2 π × (8.5 × 103 × 3.1 × 1016 m) 4.6 × 3 × 250 2 π × 8 .5 × 3 .1

≈ 21

3.

Assuming that the mass of the galactic centre is ~ 2.6 ×10 6 M Θ , we can write: Mass density

=

2.6 × 10 6 M Θ 4π × (0.015 pc) 3 3

if we take the mass enclosed within r ≈ 0.015 pc to be 2.6 × 10 6 M Θ . So, we have: mass density =

2.6 ×10 6 M Θ 4π × (0.15 pc) 3 ×10 − 3 3 3 × 2 .6

=

4π × (0.15)

3

×10 9 M Θ pc − 3

≈ 1.8 ×1011 M Θ pc − 3

Terminal Questions 1.

From Eqs. (12.10) and (12.14), we have: V0 = A− B R0

= (15+10) kms−1. kpc−1 −1

= 25 kms .kpc

−1

substituting the values of Oort’s constants A and B. Thus, V0 = 25 kms − 1 . kpc − 1× (8.5 kpc) = 212.5 kms

−1

79

So, for the given values of Oort’s constant A and B, the value of the star’s (the Sun) rotational speed (V0) is 212.5 kms− 1. Now, to determine the mass, MG at the galactic centre whose gravitational force needs to be counter balanced by the centrifugal force experienced by the Sun rotating with speed V0, we can write:

From Stars to Our Galaxy

V 2 = 0 R0 R0 2

G MG

MG =

=

V0 2 R0 G (212.5 ms −1 ) 2 × 10 6 × (8.5 × 10 3 × 3.1 × 1016 m) 6.7 × 10 −11 m 3 kg −1s − 2

1 pc = 3.1×1016 m)

( =

=

( 212.5) 2 × 10 6 × 8.5 × 10 3 × 3.1 × 1016 6.7 × 10 −11 × 2 × 10 30

MΘ

( 2.125) 2 × 10 4 × 10 6 × 8.5 × 10 3 × 3.1 × 1016 6.7 × 10 −11 × 2 × 10 30

MΘ (

= 1010 .

M Θ = 2 ×10 30 kg)

(2.125) 2 × 8.5 × 3.1 MΘ 6 .7 × 2

≈ 1011 M Θ

2.

See text.

3.

See text.

4.

You know from the kinetic theory of gases that, the average energy of the particles at temperature T is given by: E=

3 k BT 2

As per the problem, average energy E = 15 GeV =15 ×10 9 ×1.6 × 1019 J

Thus, we have: 15 × 109 × 1.6 × 10 −19 J =

3 k BT 2

or, T=

(1.5 × 1.6 × 10 −9 × 2 J) 3 × (1.38 × 10 − 23 JK −1 )

≈ 1014 K 80

UNIT 13 GALAXIES

Galaxies

Structure 13.1

Introduction

13.2

Galaxy Morphology

13.3

Elliptical Galaxies

Objectives Hubble’s Classification of Galaxies The Intrinsic Shapes of Ellipticals de Vaucouleurs Law Stars and Gas

13.4

Spiral and Lenticular Galaxies Bulges

Disks Galactic Halo The Milky Way Galaxy

13.5 13.6 13.7 13.8 13.9 13.10

Gas and Dust in the Galaxy Spiral Arms Active Galaxies Summary Terminal Questions Solutions and Answers

13.1 INTRODUCTION On a clear, dark night, you can see the diffuse faint, narrow band of the Milky Way stretching across the sky. It becomes broadened towards the constellation Sagittarius, and is seen to be covered here and there with dark areas. The Italian astronomer and physicist Galileo Galilei was the first to observe the Milky Way through a small telescope. He saw in the faint band an array of stars and clusters of stars, interspersed with dark patches. Following Galileo’s pioneering observations in the 17th century, the Milky Way has been studied extensively with a variety of telescopes of increasing sensitivity and sophistication. We now know that the Milky Way is composed of more than a hundred billion stars, spread in a large, thin disk with a bloated centre, which is known as the bulge. The disk has a diameter of about 30,000 pc, but it is only about 1000 pc thick. There are large spiral features, known as spiral arms in the disk. An object with a structure like that of the Milky Way is called a spiral galaxy. The Sun is an inconspicuous star, situated in the disk of the Milky Way, at a distance of about 8,500 pc from the centre. Our Galaxy is not the only one in existence. Within the observing limits of even a moderately large telescope there are about 10 billion galaxies, covering a wide range of sizes and shapes. While about half the galaxies have shapes like the Milky Way, a large fraction of the rest have the appearance of ellipses. Many galaxies have irregular shapes, while some are mere dwarfs compared to the larger systems. The spiral galaxy NGC 4622, shown in Fig. 13.3, is similar to the Milky Way. If we could move out of our Galaxy, and observe it from a great distance along a line of sight which is perpendicular to the disk, it would show a more or less similar appearance. Some galaxies occur as single objects, while others occur in groups of a small number of galaxies or in large clusters containing thousands of galaxies. It is not unusual to find two galaxies in collision with each other, or interacting with each other from a

5

Galaxies and the Universe

distance. The centres of a small fraction of galaxies contain what is known as an active galactic nucleus. This is a tiny object compared to the whole galaxy, but emits energy which can exceed by far the entire energy output of the rest of the galaxy. Using very large telescopes on the Earth, or the Hubble Space Telescope (HST) which is in orbit around the Earth, it is possible to obtain images of galaxies which are located extremely far away. Light from these galaxies takes a long time to reach us. Light from the galaxies that we detect now, began its journey towards the Earth so long ago that the galaxies, and the Universe itself, were significantly younger at that time. The galaxies at these early epochs are found to be significantly smaller and less well-formed than the galaxies closer to us, which we observe at a much later time in the history of the Universe. From these observations of distant galaxies, some idea is now emerging about the formation of galaxies, and their subsequent evolution to their present state. We shall consider some of these matters in some detail in the following sections. Objectives After studying this unit, you should be able to: •

explain Hubble’s classification of galaxies;

•

describe the properties of elliptical galaxies;

•

state de Vaucouleurs law;

•

describe the properties of lenticular and spiral galaxies; and

•

distinguish between normal and active galaxies.

13.2 GALAXY MORPHOLOGY A galaxy is a system of a very large number of stars, which are bound together by their mutual gravitational attraction. A typical galaxy, like our own Milky Way, contains ~ 1011 stars, spread over a region of size ~ 30 kpc, and has luminosity ~ 1011 LΘ, where LΘ is the luminosity of the Sun (Fig. 13.1).

6

Fig.13.1: The Milky Way as seen in the sky

Galaxies

Fig. 13.2: An artist’s sketch of the Milky Way Galaxy

In spite of their very great energy output, such galaxies appear to be very faint when observed from the Earth, because of their vast distances from us. A few galaxies are visible to the naked eye as faint patches of light on dark nights, and many more are visible when a small telescope is used. Faint, diffuse objects observed in the night sky are called nebulae. The nature of these nebulae was the subject of intense debate in the 1920s. While some of the nebulae are clearly objects in our own Galaxy, like remains of supernovae, it was not clear whether some of the objects were inside our Galaxy, or at great distances outside it. The matter was finally settled by Edwin Hubble who determined that the distance to the Andromeda nebula is about 700 kpc by observing variable stars in it. The modern value of this distance places the nebula far outside the confines of our own galaxy. Hubble’s observation established that the nebula was an independent spiral galaxy; he went on to study and classify many other galaxies.

Fig.13.3: The spiral galaxy NGC 4622. The galaxy is located at a distance of about 70× ×106 pc from us

7

13.2.1 Hubble’s Classification of Galaxies

Galaxies and the Universe

The observed images of galaxies show that they come in a wide variety of brightness, shape, size and structure. Each observed galaxy is different in detail from other galaxies, and when a large number of galaxies are examined, it becomes obvious that there are some basic types into which galaxies can be classified. The first detailed classification system was introduced by Edwin Hubble in 1936. This pioneering work has been followed by other more sophisticated classification systems which take into account the observed properties of galaxies in greater detail, but Hubble’s scheme is the most widely used even at the present time, because of its simplicity and the insight that it provides from observational details which can be easily obtained even with a modest sized telescope. Hubble’s scheme can be illustrated by his tuning fork diagram shown in Fig. 13.4. At the left of the diagram, along the base of the tuning fork, are the elliptical galaxies, which have simple elliptical shapes and appear to be smooth and without any additional structures. Starting with the almost spherical galaxies of type E0, as one moves towards the right, the images of galaxies become increasingly elliptical. The sequence of elliptical galaxies terminates at the point where the two arms of the tuning fork begin. Along the upper arm are the so called normal galaxies, which come in two types: lenticular or S0 galaxies and spiral galaxies. As we move to the right along the upper arm, from spiral type Sa to Sb to Sc, the bulges in the spiral galaxies become less prominent, and the spiral arms appear to be more open. We will describe the various galaxy types in greater detail in the following sections. Normal spiral galaxies

Elliptical galaxies Sa

E0

E4

E7

Sb

Sc

S0 Lenticular galaxy SBa

SBb Barred spiral galaxies

SBc

Fig.13.4: Hubble’s tuning fork diagram

The lower arm of the fork again has lenticular and spiral galaxies, but with a linear central feature called a bar. These barred galaxies constitute about half of all lenticular and spiral galaxies. The distinction between normal and barred galaxies is not absolute, in the sense that most galaxies have some faint bar like features, but a galaxy is called barred only when the bar is very prominent. Every galaxy type along the upper arm of the fork has a barred counterpart along the lower arm. The classification of galaxies as suggested by Hubble is based on very luminous galaxies, with absolute magnitude MB ≤ − 20, which he called giant galaxies. However, when galaxies in our neighbourhood are observed, it is found that the most numerous galaxies are significantly less luminous and more compact than the giants. These dwarf galaxies are designated as dE. 8

Galaxies

Many galaxies have highly irregular shapes, and prominent features like jets, tails and rings. It is believed that these features are often produced because of interactions between galaxies, which can lead to large scale disturbances in the distribution of stars and gas in the galaxies. Many examples of galaxies in on-going close interaction can be seen. Large galaxies can also swallow significantly smaller companions, and this process of cannibalisation can lead to significant changes in the structure of the large galaxy. It is difficult to classify galaxies with highly irregular structures because of their complexity. Spend 5 min.

SAQ 1 A galaxy of absolute magnitude M = − 20 is at a distance of 700 kpc. Would it be visible to the naked eye?

13.3 ELLIPTICAL GALAXIES When elliptical galaxies were first photographed, they were observed to have elliptical shapes and a smooth distribution of light (see Fig. 13.5). They were lacking in features which are very prominent in spiral galaxies, like spiral arms and dark patches and lanes of dust. Later observations with highly sensitive detectors have shown that elliptical galaxies often do have faint features produced by dust and other factors. These faint features turn out to be important indicators of the origin and evolution of elliptical galaxies to their present form.

Fig.13.5: The elliptical galaxy M87

Elliptical galaxies have an enormous range of optical luminosities. The so called giant ellipticals have luminosities L ≥ L*, where L* ≈ 2 × 1010LΘ is a characteristic galaxy luminosity. The number density of galaxies declines sharply for luminosity L > L*. A galaxy with luminosity L ≈ L* has an absolute magnitude of M ≈ − 20. The most luminous elliptical galaxies can have L ≥ 100L*. Elliptical galaxies with L ≤ 3 × 109 LΘ, i.e, M ≥ − 18, are called dwarf ellipticals.

9

Galaxies and the Universe

13.3.1 The Intrinsic Shapes of Ellipticals When the distribution of light in an elliptical galaxy is studied, it is found that the isophotes, or curves of equal light intensity, are elliptical in shape; this in fact gives this type of galaxy its name. In the simplest elliptical galaxies, all the ellipses have the same centre, their major axes are oriented in almost the same direction, and the ellipticities are nearly constant. It was believed at one time that elliptical galaxies acquired their shape due to rotation. The rotation would cause the galaxy to bulge in directions normal to the axis of rotation, because of the centrifugal force. However, observations of luminous elliptical galaxies show that they do not rotate fast enough for the observed flattening to be due to the rotation. It is now known that the shape comes about because of the way the stars in the galaxy move.

13.3.2 de Vaucouleurs Law Most galaxies are so far away that the stars in them cannot be seen individually, and we can only observe the integrated light from stars in different regions of the galaxy. The appearance of an external galaxy is therefore like that of a diffuse object. It is therefore appropriate that we measure the surface brightness of the galaxy, which is the amount of light received per unit angular area of the galaxy, say one square arc second. The surface brightness of light along the isophotes of an elliptical galaxy decreases as we move away from the centre. It was discovered by G. de Vaucouleurs that the surface brightness is a very simple function of the length of the semi-major axis of the isophote. If r is this length, then the surface brightness I(r) is given by 1/ 4

I (r ) = I (0)10 −3.33( r / re )

(13.1)

where re is called the effective radius and I(0) is the surface brightness at r = 0. The total light emitted by the elliptical galaxy is given by ∞

I ( r )2πr dr = 3.37 × 10 −3 πre2 I (0)

LE =

(13.2)

0

Half the total light of the galaxy is emitted from inside re: re

I (r )2πr dr = 0

1 2

LE

(13.3)

de Vaucouleurs’ law takes on a particularly simple form if the intensity is expressed in magnitudes. It can then be written as:

µ ( r ) = µ ( 0) +

10

8.325 1/ 4 r re1/4

(13.4)

Here µ(r) is the surface brightness of the galaxy expressed in magnitudes per square arc second, and µ(0) is the corresponding magnitude at the centre of the galaxy. It follows from this equation that a plot of the surface brightness against r1/ 4 should be a straight line. The surface brightness becomes fainter by 8.325 magnitudes in going from r = 0 to r = re.

Galaxies

The light distribution in most elliptical galaxies does follow de Vaucouleurs law fairly closely. NGC 661 is an excellent example of this; we have shown, in Fig. 13.6, a plot of the surface brightness in magnitude against r1/4. The plot is seen to be a straight line, except in the central bright region where there is significant flattening of the curve. Much of the deviation seen here is due to the effect of the Earth’s atmosphere.

Surface brightness in magnitudes

− 10 −9

NGC 661

−8 −7 −6 −5 −4 −3

1

2 1/4 (Distance from centre)

3

Fig.13.6: The surface brightness distribution of the galaxy NGC661 (centred around 5500 Å 1/4 wavelength). On the x axis is shown r , where r is the major axis distance of the elliptical isophotes from the centre. On the y axis is shown the surface brightness in magnitude, with the origin shifted for convenience

13.3.3 Stars and Gas A study of the colour and spectrum of elliptical galaxies shows that they lack blue stars. These stars are highly luminous, and are much more massive than the Sun. The total lifetime τN of a star, during which the usable nuclear fuel in it is exhausted, depends on its mass, 10

τ N ~ 10

M MΘ

−3.5

yr

(13.5)

The blue stars being more massive than the Sun have a lifetime significantly less than 1010yr, with the most massive stars having a lifetime as short as ~ 107 yr. The lack of such stars means that there has been no star forming activity in ellipticals in relatively recent times. This points to the absence of substantial amounts of cool gas mixed with dust from which stars can be formed. Giant ellipticals contain ≤ 108 − 109 MΘ of cool gas. Observations from X-ray satellites have, however, shown that ellipticals can be highly luminous at X-ray wavelengths, which indicates the presence of significant amounts of hot gas, at temperatures of a few times 107 K. In very bright ellipticals, the mass of the gas can 11 be as high as ~ 10 MΘ, which constitutes ~ 10 − 20% of the visible mass of the galaxy. SAQ 2 Explain in your own words why we expect the gas in elliptical galaxies to be hot.

Spend 5 min.

11

Galaxies and the Universe

13.4 SPIRAL AND LENTICULAR GALAXIES Spiral galaxies are identified by a disk-like structure in which are present the spiral arms. An image of the famous spiral galaxy M31, which is also known as the Andromeda galaxy, is shown in Fig. 13.7. A characteristic of the disk is that it is much extended but rather thin. When a spiral galaxy is viewed face-on, i.e., when the normal to the disk is along the line of sight, the disk appears to be circular, as in the case of NGC 4622 (see Fig.13.3). When there is a non-zero angle between the normal and the line of sight, the disk then appears to be elliptical, as in the case of the Andromeda galaxy. When the disk is viewed edgeon, the disk appears to be rather thin, as in Fig. 13.8. Apart from the disk, spiral galaxies contain a central bulge, which is quite obvious in all the spiral galaxies. They also have a very large but faint halo, whose existence becomes apparent from a detailed study of the distribution and motion of stars. Lenticular galaxies, like the spirals, have a bulge and a disk, but the disk does not contain spiral arms. The bulge and the disk here are of approximately equal prominence. The bulge has properties very similar to elliptical galaxies, except that it contains more gas and dust.

Fig.13.7: The Andromeda galaxy, which is spiral galaxy of type Sb, at a distance of 2.2 million light years from us. Two dwarf galaxies, which are satellites of the Andromeda galaxy are seen in the image

13.4.1 Bulges The bulge of spiral galaxies is a dense system of stars in the inner region of a galaxy, more or less spherical in shape. The bulge of our own galaxy, the Milky Way, can be seen, towards the constellation Sagittarius, and a bulge is clearly visible in the spiral galaxies M31 and NGC891 (see Figs. 13.7 and 13.8). The properties of the bulges are rather similar to the properties of elliptical galaxies.

12

In addition to the systematic motion, the stars in the bulges also have significant random motion, which supports them against gravity. The number density of stars in the bulge is about 104 higher than the number density in the neighbourhood of the

Sun. Bulges have very little gas in them, except near the centre, and so there is not much ongoing star formation. This means that the bulge does not have the short lived, high mass blue stars which are present in the disks, because of the continuous star formation taking place there.

Galaxies

The scale size of the bulges is typically a few kiloparsec (kpc), while the radius of the disk of a spiral galaxy like ours is about 15 kpc. The prominence of the bulge relative to the disk decreases along the Hubble sequence (see Fig. 13.4), with the bulges being most conspicuous in spiral galaxies of type Sa and being almost absent in the galaxies at the end of the Hubble sequence and beyond.

Fig.13.8: The spiral galaxy NGC 891, which is seen edge on. The thin disk and the bulge are clearly seen. Along the plane of the disk is seen a dark band, which is a layer of dust present in the galaxy

13.4.2 Disks The disks in the more luminous (MB ≤ − 20) spiral galaxies extend to ~ 15 kpc or more, while their thickness is only a few hundred parsec, which makes them rather thin. The disks are generally taken to be circular in shape (even though they may appear to be elliptical, due to the projection effect mentioned above). The surface brightness of the disk follows a simple exponential law. At a distance r from the centre, measured along the mid-plane of the disk, it is given by −

r

I d ( r ) = I d (0) e rd

(13.6)

where rd is the disk scale length. This is a few kpc for the typical spiral galaxy. If we move normal to the disk keeping r constant, the surface brightness decreases, and at a distance z from the mid-plane, it is given by

Id (r, z) = Id (r) exp −z/h

(13.7)

where Id (r) is as in Eq. (13.6). The scale length h is a few hundred parsec. A galaxy, whether spiral, elliptical or of any other type, does not have a sharp boundary. The surface brightness, and therefore the number density of stars which produce the light, decreases as one moves away from the centre, and gradually reduces to zero. In such a 13

circumstance, it is best to characterize disk size by scale lengths like rd and h, and the sizes of bulges and ellipticals with the effective radii.

Galaxies and the Universe

The total light emitted by the disk is obtained by integrating Eq. (13.6) over the surface of the disk: ∞

Ld = 0

2πr I d (r ) dr = 2πrd2 I d (0)

(13.8)

If we assume that the distribution of light in the bulge is of the de Vaucouleurs type, then the total light Lb emitted by it is given by Eq. (13.2). The ratio of the emission from the disk to the emission from the bulge is then Ld I ( 0) = 5.94 × 10 2 d Lb I b ( 0)

rd re

2

(13.9)

For a galaxy without a significant disk component, like an elliptical, this ratio is zero. For lenticular galaxies it is ~ 1, and it increases along the Hubble sequence towards the late type spirals. As star formation is an ongoing process in the disk, massive blue stars can be observed in it. Since such stars have a short lifetime, they cannot be found in environments where there is no star formation taking place like bulges of spiral galaxies or elliptical galaxies. The presence of massive stars makes the disk bluer on the whole than the bulge. Stars in the disk are in differential rotation around the centre of the galaxy. You have already read about the differential rotation of stars in our own galaxy, the Milky Way in Block 3. About half of the disks of spiral galaxies, as well as lenticulars, have a linear structure which is known as a bar. We have seen in Section 13.2.1 that the lower arm of Hubble’s tuning fork diagram contains these objects. Spend 5 min.

SAQ 3 Explain in your own words why older galaxies should be redder.

13.4.3 Galactic Halo A halo of stars and globular clusters is found to be surrounding the bulge and disk of our Galaxy. The halo extends to substantial distances beyond the disk, and has the shape of a flattened spheroid (see Fig. 13.9). Such halos are believed to exist in all galaxies to a greater or lesser extent. Globular clusters are gravitationally bound systems of 105 − 106 stars. These clusters are found in the disk plane and close to it, as well as far from the plane. There are about 150 globular clusters associated within our Galaxy, and these are found to be distributed approximately in a sphere around the centre of the galaxy. A characteristic of the globular clusters is that they are very old stars. The clusters far from the plane are very metal poor, the abundance of the heavy elements in them being as small as 1/300 of the solar value. These clusters are estimated to be at least 11 − 12 Gyr old, which makes them the oldest structures in our Galaxy.

14

We expect that other galaxies too would have halos like our own galaxy. One way of tracing a halo in external galaxies is through the system of globular clusters, which can be observed to great distances. Globular clusters have been observed around many galaxies, and elliptical galaxies are found to be particularly rich in these objects.

SAQ 4

Spend 5 min.

Galaxies

Explain why metal poor stars are very old.

13.4.4 The Milky Way Galaxy Our Galaxy, called the Milky Way, is a spiral galaxy of type Sbc, which is intermediate to Hubble types Sb and Sc. The bulge of the galaxy is clearly visible to the naked eye in the direction of the constellation Sagittarius. The disk of stars is visible as a diffuse band going across the night sky. A sketch of the structure of the galaxy is shown in Fig. 13.9.

Fig.13.9: A sketch of our Galaxy, showing the bulge, disk and the halo

The visible bulge is about a few kiloparsec in radius, while the extent of the radius of the disk is ~ 15 kpc. The Sun is located about 8 kpc or 28000 lys from the galactic centre, some distance away from the mid-plane of the disk. The luminosity of the disk, i.e., the total amount of energy per second emitted by all the stars in the disk, is ~ 2 × 1011LΘ, where LΘ = 4 × 1033 erg s−1 is the luminosity of the Sun. The total mass of stars in the disk is ~ 6 × 1011 MΘ, where MΘ = 2 × 1033 g is the mass of the Sun.

13.5 GAS AND DUST IN THE GALAXY It was discovered through observations in the early decades of the 20th century that inter-stellar space, i.e., the space between the stars in our Galaxy contains matter in the form of gas and dust. The density of the gas is very low and it is difficult to detect it. The dust usually occurs well mixed with the gas, and constitutes only about one percent of the total material. The dust nevertheless is able to significantly obscure the light coming to us from distant stars, and therefore can be easily detected. The gas and dust together are called the inter-stellar medium (ISM) of the Galaxy. Inter-stellar dust often occurs in the form of clouds, as is evident from the many dark nebulae, like the Horsehead nebula (Fig. 13.10a), which are observed in the Galaxy. The Eagle nebula and the details of the great clouds of dust observed by the Hubble Space telescope are shown in Fig. 13.10b and c. The dust is mostly confined to the disk of the Galaxy, and can be clearly seen in the Milky Way, even with the naked eye, as great dark patches covering the galaxy here and there. It is specially so towards the Galactic centre. 15

Galaxies and the Universe

(a)

(c)

(b) Fig.13.10: a) The Horsehead nebula; b) great columns of dust and gas in the Eagle nebula with c) its details as observed by the Hubble Space Telescope

The distribution of dust in the disk is particularly evident when a spiral galaxy is viewed close to the edge of the disk, as in the case of the galaxy NGC891, which is shown in Fig. 13.8. Dust affects light passing through it by scattering as well as by absorption. Both absorption and scattering by dust remove a fraction of light coming to an observer from a star, and the effect of the two together is termed as extinction. The fraction of light lost due to extinction depends upon the wavelength of light, the size of the dust grains, and the quantity of dust in the path of light. For the dust found in the ISM of our Galaxy, the extinction is approximately proportional to the reciprocal of the wavelength. This means that blue light suffers the extinction most and near-infrared light the least. Because dust is concentrated in the Galactic plane, a star in the plane observed by us is significantly dimmed and reddened.

16

Gas and dust have been observed in other spiral galaxies as well. It was believed for long that elliptical galaxies are free of gas and dust. But X-ray observations have shown that these galaxies contain hot gas. Careful observation in recent years has shown that a surprising number of elliptical galaxies also contain small quantities of dust. This can be distributed in the form of a disk in the central region of the galaxy and sometimes larger disks are also present. Some elliptical galaxies contain a large

Galaxies

quantity of dust, which can be distributed over a region larger than the visible extent of the galaxy. A good example of a giant elliptical galaxy containing dust very prominently is the radio galaxy Centaurus A shown in Fig. 13.11.

Fig.13.11: The elliptical galaxy Centaurus A which is a highly luminous radio source. A prominent dust lane is seen to be running across the face of the galaxy

SAQ 5

Spend 5 min.

Explain why stars towards the centre of our galaxy appear fainter and redder.

13.6 SPIRAL ARMS The disks of spiral galaxies contain spiral arms, which make these galaxies very photogenic. The typical spiral galaxy has two arms, but many galaxies have three or four arms. Spiral arms are found to contain many massive, blue stars. Since such stars can only live for ≤ 10 Myr, the arms must be sites of continuing star formation. The young blue stars are hot, and they emit radiation which can ionize any gas which may be present around them. The gas produces emission lines which can be detected. Such regions ionized by stars are known as HII regions. HII regions are abundant in spiral arms, indicating the presence of large quantities of gas, as well as populations of young massive stars. The arms also contain significant quantities of dust. The disk in a galaxy rotates differentially, that is, stars which are closer to the centre generally rotate with higher angular speeds than those which are further away. This differential motion should lead to tighter winding of spiral arms. It can be shown that for our Galaxy, the arms should have tightened significantly in less than 109 yr. Since the Galaxy was formed more than 10 billion years ago, the arms should have been much more wound up than what is observed. The fact that we do not observe such winding up in our Galaxy and other similar galaxies is explained by the density wave theory of spiral arms. According to this theory, the stars and gas present in the arms are not fixed there. As stars and gas move along their orbits in the disk, at some point they pass through the arms and slow down. This leads to crowding of stars and gas in the arms, and therefore to star formation. The arms can therefore be looked upon as density waves.

17

Galaxies and the Universe

Fig.13.12: Spiral arms in galaxies

13.7 ACTIVE GALAXIES A very small fraction of all galaxies are found to be active, in that they emit very significant quantities of energy which is not produced by the stars. The energy is produced in a very compact region at the centre of the galaxy, and in many active galaxies exceeds by far the energy produced by the stars. The compact region from which the energy is produced is known as the active galactic nucleus (AGN). 11

A “normal” galaxy like the Milky Way contains ≥ 10 stars, each of which emits 1033 erg s−1. The total emission from all the stars is therefore ~ 1044 erg s−1. This energy is produced from a region which extends over a few tens of kiloparsec. An 44 47 −1 AGN, on the other hand, produces energy at the rate of 10 − 10 erg s , from a region which is a fraction of parsec in size. The radiation produced by stars is thermal in character, i.e., the spectrum of the radiation is similar to the spectrum produced by a hot gas. The spectrum of radiation produced by an AGN is quite different, and in simple cases has a power-law form. This means that the intensity varies as some power of the frequency: Iν ∝ νγ

(13.10)

where is a constant. The spectrum of active galaxies is very broad, extending from the radio region to the infra-red, optical, ultraviolet, X-ray and gamma-ray regions. The shape of the spectrum indicates that the physical processes that produce the radiation are different from the processes which produce the spectrum of a hot gas. These processes are called non-thermal. The spectrum from stars has a number of absorption lines in it, which are produced by the absorption of thermal radiation by atoms of specific chemical elements in the cooler region of the stars. While the spectra of AGN do have absorption lines, the spectra also exhibit very prominent emission lines, which are produced by non-thermal processes.

18

There are different kinds of active galaxies. Seyfert galaxies are generally spiral galaxies with an AGN which is moderately luminous. Radio galaxies are always

elliptical, and are associated with highly luminous radio emission. BL-Lacs are active galaxies which are ellipticals. Their characteristic is that, unlike as in the other AGN, emission lines are either completely absent from their spectra, or are weak. The most luminous AGN are the quasars. The luminosity of the AGN here is so high that it outshines the galaxy. The appearance of the object is therefore that of a point source, like a star. Observations by very large telescopes are needed to discern the galaxy associated with a quasar. You will study more about active galaxies in Unit 14.

Galaxies

The source of the immense quantity of energy produced by an AGN is believed to be a black hole, resident at the centre of the galaxy. The black hole is thought to be super massive, i.e., it has mass in the range of ~ 106 − 109 MΘ. Matter from the region surrounding the black hole falls onto it, and gravitational energy is released in the process. This energy is taken up by electrons, which in turn emit the energy through different processes, producing the observed spectrum. The different kinds of AGN observed can be explained in terms of variations of this general theme. While the above picture is more or less accepted, it is not completely proven yet. It has been established that AGN were far more common early in the life of the Universe than at present. Many galaxies which are normal now must therefore have been active in the past. They must therefore contain super-massive black holes which are dormant. Recent observations have shown that a number of normal galaxies may indeed host such black holes, and it is even thought possible that all galaxies have super-massive black holes at their centres. The role of black holes in the formation of galaxies, and whether the black hole or the galaxy came first, is not understood yet and is an area of active investigation. Spend 5 min.

SAQ 6 Explain the difference between thermal and non-thermal radiation. In this unit you have learnt about galaxies. We now summarise its contents.

13.8 SUMMARY •

A galaxy is a system of stars, gas and dust, held together by the mutual gravitational pull of these components. There are billions of galaxies in the universe.

•

Hubble classified these galaxies as elliptical, lenticular and spirals, both normal and barred spirals.

•

The surface brightness of ellipticals varies according to the de Vaucouleur’s law: µ ( r ) = µ ( 0) +

8.325 1/ 4 r re1/4

•

Spiral galaxies are characterised by spiral arms and bulges. The spiral arms contain clouds gas and dust and new stars are continuously being formed there. Most galaxies have halos around them which contain extremely old stars.

•

In the nuclei of many galaxies unusual phenomena take place involving release of huge amounts of energy. These nuclei are called active galactic nuclei. It is believed that this activity is caused by massive black holes sitting in the nuclei. The Milky Way galaxy also has a 106 MΘ black hole at its centre.

19

Galaxies and the Universe

13.9 TERMINAL QUESTIONS

Spend 30 min.

1. Explain Hubble’s scheme of galaxy classification. Why has this scheme proved enduring? What class has been assigned to the Milky Way Galaxy? 2. State de Vaucouleurs law and define the effective radius of an elliptical galaxy. Is this law obeyed by elliptical galaxies? 3. Explain how the surface brightness in the disc of a spiral galaxy varies with the distance from the centre of the galaxy. Show that the total light emitted by the disc is 2πrd2 I d (0) , where rd is the scale length of the disc and Id (0) is the surface brightness at r = 0. What is the meaning of rd? 4. Define an active galaxy. What is the source of its activity?

13.10 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1.

M = m − 5 log r + 5 , where r is in parsec, here 7 × 105 pc. − 20 = m − 5 (5.8451) + 5

2. 3. 4. 5. 6.

m = − 20 + 29.2255 − 5 ≈ 4.2

Since human eye can see up to magnitude 6, this object will be visible to the naked eye. See Text. In older galaxies, there are fewer younger (blue) stars but a large number of red stars as you have studied in Unit 10. See Text. See Text. See Text.

Terminal Questions 1. See Text. 2. See Text. 3. For the first part, see Text. For the second part, ∞

Ld =

0

2πr I d (0) e − r / rd dr

= 2π I d (0) rd2

∞

put r/rd = x

e − x x dx

0

= 2π I d (0) rd2 , since the integral = 1

For the third part, I d (r ) = I d (0) e − r / rd

20

At r = rd, Id becomes 1/e of Id (0), i.e., intensity reduces to 1/e of its initial value. 4. See Text.

UNIT 14

ACTIVE GALAXIES

Active Galaxies

Structure 14.1

Introduction Objectives

14.2

‘Activities’ of Active Galaxies How ‘Active’ are the Active Galaxies? Classification of the Active Galaxies Some Emission Mechanisms Related to the Study of Active Galaxies

14.3

Behaviour of Active Galaxies Quasars and Radio Galaxies Seyferts BL Lac Objects and Optically Violent Variables

14.4

The Nature of the Central Engine Unified Model of the Various Active Galaxies

14.5 14.6 14.7

Summary Terminal Questions Solutions and Answers

14.1 INTRODUCTION In Unit 13, you have become familiar with various types of galaxies. However, there are galaxies which do not fit into the scheme we described in the last unit. They require more detailed understanding. In this Unit, we shall study the nature of these galaxies. These galaxies are ‘active’, namely, there are signs of activity in them. For instance, their intensities may change significantly in a matter of hours to days. The activities may be manifested in the entire bandwidth of the electromagnetic waves, ranging from radio waves to γ-rays. One important point should be clear to you by now and that is: nothing in this Universe is truly inactive. So why should we learn about ‘active’ galaxies in a separate Unit, when all the galaxies are ‘active’? Truly, every galaxy is evolving and therefore changing with time. Only the degree of activity varies. Roughly speaking, we call those galaxies to be active which show properties such as very strong intensity variation in a relatively short period, and show ejection of matter at very high speeds, etc. Since they are compact, the nuclei of these galaxies show more ‘activity’. So, naturally, we are interested in studying these compact regions, which are known as Active Galactic Nuclei or simply AGNs. Several classes of objects may be called ‘AGNs’: These are: Radio Galaxies, Quasars, BL Lac Objects, Optically Violent Variables, Seyfert Galaxies, Star-Burst Galaxies, etc. In this unit, we will try to understand how some of these objects behave and why they behave that way. Objectives After studying this unit, you should be able to: •

describe the nature of activities near the active galactic nuclei;

•

explain how these objects are classified;

•

explain the importance of synchrotron radiation for the study of AGN; and

•

describe the structure of matter very close to the active galactic nuclei.

21

Galaxies and the Universe

14.2 ‘ACTIVITIES’ OF ACTIVE GALAXIES You have learnt in the last Unit that there is something unusual going on in the nuclei of many galaxies. The ‘unusual’ may be the tremendous amount of energy released, or may be the emission of extremely high energy particles in the form of narrow jets. It is obvious that all galaxies are not equally active. Therefore, there is a need to discuss further how active an active galaxy really is.

14.2.1 How ‘Active’ are the Active Galaxies? Active galaxies exhibit evidences of activity in a variety of ways. Their luminosities may be around 1044 erg s−1 (compare this with the luminosity of the Sun, which is 4 × 1033 erg s−1) and can go up to 1048−49 erg s−1. This tremendous energy release is possible only if these objects swallow a few stars per year and convert them into energy. Sometimes the luminosity may vary by tens of percents in a matter of hours! This gives you an idea of the compactness of the size of the nuclei of these galaxies. Recall that light travels at a speed of 3 × 1010cm/s and at this rate in, say, three hours 4 14 time (~ 10 s), it will travel 3 × 10 cm. This is roughly the size of the region from where most of the energy is released. You already know from Unit 11 that a black hole of mass M has a size of 3 × 105 (M/MΘ) cm. Thus, if the activity of the Active Galaxies is governed by a black hole at the centre, it must harbour a black hole of mass of around 109 MΘ or so. Another signature of the activity is that these galaxies emit non-thermal emissions in all the wavelengths, i.e., radio to γ-rays. Compared to a normal galaxy, where the nuclear brightness never masks the spiral arms or the rest of the galaxy, in active galaxies, the nucleus is so bright that the rest of the galactic structure is not seen in most cases. Active galaxies also produce (generally) jet-like outflows of very high energy particles on both sides, normal to the galactic plane. These jets are highly collimated and are continuously ejected. Normal galaxies do not produce these jets. Apart from these, sometimes the emission lines are broad and the state of polarization of the emitted light changes with time in a matter of minutes! Since a tremendous amount of energy is released continuously (rather than a one-time event as in a super-novae), it is natural to assume that the main reason of this is the gravitational energy released by the accretion of matter. What is the source of this matter? What is the nature of the accretion disks? How would the radiation be emitted and at what frequency? Why do the jets form? How are the jets collimated? All these important questions still puzzle astrophysicists today and only some of these questions have been answered satisfactorily.

14.2.2 Classification of the Active Galaxies Active galaxies are classified according to some, often very vague, criteria. Broadly speaking they may be classified in the following way: a) Radio Galaxies: In these galaxies, the emission in radio waves is very strong, larger than 1040 ergs/s. They could be further classified into extended radio galaxies and compact radio galaxies, depending on the size of the emission region. b) Quasars: These objects were called Quasi-Stellar-Objects or Quasars because they ‘looked’ like stars. Basically, in these objects, the nucleus emits most of the radiation. The behaviour in radio is often similar to that of the radio galaxies. In some, radio emission is not prominent and the quasars are called radio quiet quasars. 22

c) Seyfert Galaxies: These are usually spiral galaxies with unusual nuclear brightness. Some, say, 5 to 10% of the Seyfert galaxies may be ellipticals as well. They have very broad emission lines in their spectra which distinguish them from other classes.

Active Galaxies

d) BL Lacertae Objects: These are commonly known as BL Lac objects. They are identified by very rapid variability in radio, infrared and optical emissions. The light is strongly polarized and the polarization varies rapidly. They have no prominent emission lines and the spectrum is dominated by a continuum. e) Optically Violent Variables: These are similar to the BL Lac Objects, but they have weak, broad emission lines. f) Star-Burst Galaxies: In these galaxies the star formation rate is much higher compared to the rates in normal galaxies. The radiation is emitted in the infra-red region. Star formation may be triggered by merger or collision of galaxies.

(a)

(b)

(c)

(d)

Fig.14.1: a) Radio galaxy 3C433 (image courtesy of NRAO/AUI sourced from www.nrao.edu/.../3c433_ 3.5cm_opt_6in_med.jpg; b) Quasar 3C273; c) central region of the Seyfert galaxy NGC 1068; d) Starburst galaxy M82

23

Galaxies and the Universe

Let us now learn about some of these objects which show major activities.

14.2.3 Some Emission Mechanisms Related to the Study of Active Galaxies One of the most important processes that you need to know while studying active Galaxies is the synchrotron radiation. You may also know this radiation as cyclotron radiation, a terminology used for radiation emitted by low energy electrons in the presence of a magnetic field. To emit this radiation two ingredients are required, namely, large number of high energy electrons and a strong magnetic field. Since the magnetic field is present everywhere in AGN, synchrotron radiation is a very common phenomenon. A relativistic electron of rest mass m0, charge e, velocity v (and Lorentz factor γ = (1 − v2/c2)−1/2) in a magnetic field B feels the force, d

e ( γm0 v ) = ( v × B) dt c

(14.1)

The electron moves in a helical path in the magnetic field (Fig. 14.2) with an angular frequency ωg =

eB B = 1.8 × 1011 γm0 c 10 5 G

10 γ

rad s −1

(14.2)

Magnetic field

Electrons in a helical path Fig.14.2: Motion of an electron in a magnetic field and the emission of synchrotron radiation (image credit Gemini Observatory)

The emission is in the shape of a broad spectrum with a peak at critical frequency 2 ωc ~ γ ωg. Because of this continuous radiation, the electron loses energy and starts cooling with time. The cooling time scale is clearly the time in which the whole energy is radiated away: 2

Tcool ~ γm0c /P s. where P is the energy emitted by an electron in one second. 24

(14.3)

To obtain the net emission from the entire gas you need to consider the velocity distribution of the electrons, which may or may not be thermal. In AGN the distribution is non-thermal. For the non-thermal electrons, the emission spectrum looks like a power-law, i.e., the intensity of radiation is expressed as: F( ) = F0

−α

Active Galaxies

(14.4) Flat

Synchrotron radiation is largely linearly polarized. Crudely speaking, an observer viewing the gyrating electron sideways will see an oscillating electron which emits a linearly polarized radiation. From above and below, oppositely directed circular polarization will be seen. These circular polarizations will cancel each other when overall integration is made but the linear polarization will survive and it can be detected easily. Another form of radiation which is very important is the emission line component. Emission lines come from atomic transitions from one bound state to another (Fig. 14.4). etc…

F (ν)

where α is a constant known as the spectral index. When α ≤ 0.4, we call the spectrum a flat spectrum, and when α is larger, we call the spectrum a steep spectrum (Fig. 14.3).

Steep

ν Fig.14.3: Schematic diagram of flat and steep spectra

Level 4 Level 3

Level 2 Energy

Level 1

Fig.14.4: Schematic diagram of an atomic transition from an excited state to a lower state and photon emission

Similarly, when a photon of the ‘right’ frequency ν = (Ef − Ei)/h is incident on an atom, the photon may be absorbed. This absorption line will be very thin if the emitting atom is at rest, but if the atoms are moving around, the line may become broad due to Doppler effect: The line width d will then correspond to the velocity v of the emitting matter: d / ~ v/c, where is the frequency of the radiation emitted in the rest frame and c is the velocity of light. From the broadening of the lines from Seyfert galaxies, sometimes the velocity inferred is anywhere close to 7 × 106 − 2 × 107 m/s. In the next few sections, we shall discuss some of the important classes of the Active Galaxies in more detail. After that, you will learn about the detailed nature of the activities close to the centre, i.e., the nature of the ‘central engines’ which drive these powerful sources in space. But before that, you may like to try an SAQ! SAQ 1

Spend 5 min.

Explain why synchrotron radiation is largely linearly polarized. 25

Galaxies and the Universe

14.3 BEHAVIOUR OF ACTIVE GALAXIES In this section, you will learn about the properties and behaviour of quasars, radio galaxies, Seyferts, BL Lac objects and optically violent variables.

14.3.1 Quasars and Radio Galaxies A quasar (contraction of QUASi-stellAR radio sources) is an astronomical source of electromagnetic energy, including light that dwarfs the energy output of the brightest stars. A quasar may readily release energy in levels equal to the output of dozens of average galaxies combined. The best explanation for quasars is that they are powered by supermassive black holes.Quasars can be observed in many parts of the electromagnetic spectrum including radio, infrared, optical, ultraviolet, X-ray and even gamma rays while most quasars are found to emit in the infrared. The spectra of quasars exhibit characteristic strong emission lines rising above a broad continuum, which are red-shifted due to the high recession velocity of the quasar relative to us. For example, the Lyman line formed by transitions between the n = 1 and n = 2 energy levels in neutral hydrogen normally produces spectral lines with a wavelength of 121.6 nm or 1216 Å which is in the ultraviolet part of the spectrum. In quasar spectra, these are seen at wavelengths in the visible part of the spectrum (see Fig. 14.5).

Fig.14.5: Spectrum of a quasar showing relative intensity on the y axis as a function of wavelength (from www.eso.org)

26

The nuclei of quasars emit UV radiation so strongly that the rest of the galaxy is masked. The radiation that is observed from radio wave to X-ray is totally nonthermal. Synchrotron radiation is the most likely source of this emission. Some quasars could be intense in radio emission while other quasars may have no

significant radio emission. The power emitted in radio waves is usually smaller than that emitted in other wavelengths. In the infra-red region the emitted power is significant.

Active Galaxies

Fig. 14.6: Spectrum of a quasar in the X-ray region showing relative intensity on the y axis (Source: www.estec.esa.nl)

Quasars are found to vary rapidly in a matter of few minutes to weeks, months and even years especially in the optical and radio regions. Since nothing can travel faster than light, the shortest time-scale of appreciable variation gives an idea of the upper limit of the size of the object as well. For instance, if a quasar intensity doubles in a day (tvar ~ 105s), the size of the region emitting most of the radiation must be smaller than the light-crossing time, i.e., Lsize ~ ctvar ≈ 3 × 1015cm. So the central region must be even smaller. This indirectly gives an estimate of the mass of the black hole:

M=

Lsize 3 × 1015

1010 M Θ ,

where Lsize is in units of cm. Emission lines from the quasars are very broad and show characteristic red shifts because these objects are located at very large distances. The red shift indirectly tells you how far the object is situated. The width of the emission lines are determined by large scale motions of the emitting matter. The emitted frequency ν0 is related to the observed frequency ν by, ν = ν0 (1 + v/c). If the emitting atoms follow the Maxwell’s velocity distribution at temperature T, then we can obtain a simple expression for the intensity distribution I(ν) dν.

27

We can write

Galaxies and the Universe

M p c 2 (ν − ν 0 )2 I d (ν) = I 0 exp − 2 k BT ν 02

(14.5)

Here I0 is the maximum intensity which occurs at ν = ν0, and Mp is the mass of the atom of the gas. As mentioned earlier, the width directly gives the information about the velocity, and therefore the temperature of the emitter. If we define half-width of the line as the width at half of maximum intensity, then we can show that half width = 2(ν − ν 0 ) = 2ν 0

Spend 10 min.

2k B T M pc2

ln 2

(14.6)

SAQ 2 What is the half width d of a line of wavelength λ = 6300 A when the temperature of the gas is 105K? Assume O atoms to be the emitters. The quasars which are located far away, i.e., quasars with large red shifts also exhibit a strong degree of absorption features. The red shifts of the absorption lines are almost always found to be less than or equal to the red shift of the emission line itself. This indicates that the absorbing gas is located in between the quasar, or is a part of the quasar itself. Radio Galaxies Those Active Galaxies which emit very large amount of energy in radio waves, in excess of, say, 1040 erg/s, are known as radio galaxies. Some of these could be compact radio galaxies, i.e., the emission is concentrated near the nuclear region. The others could be extended radio galaxies, i.e., the emission is seen in the form of two very large blobs which are well separated and usually co-aligned with the central, optically bright nucleus. Fig. 14.7 shows the nature of the radio emitting regions in a radio galaxy.

28

Fig.14.7: Radio emitting regions in a radio galaxy: jets from the radio galaxy 3C296. The infrared regions are shown in red

43

44

A typical radio lobe in a bright radio galaxy has a luminosity of about 10 or 10 ergs/s. But as we discussed in the last section, this is due to synchrotron radiation and it cools rapidly. In order that the blobs are produced and propagated for millions of years, they must be supplied with energy constantly. Indeed, higher resolution radio images point to an important finding: These radio emitting blobs are supplied with energy by very narrow jets of rapidly moving matter.

Active Galaxies

If one observes these jets at higher and higher resolution, the matter will be found to propagate in the same direction even close to the centre. This indicates that the prime cause of the jet production lies in the activity at the centre. We shall discuss this in Section 14.5.

14.3.2 Seyferts Seyfert galaxies are named after Carl K. Seyfert who, in 1943, described their central regions as having peculiar spectra with notable emission lines. Seyfert galaxies appear to be normal spiral galaxies, but their cores fluctuate in brightness. It is believed that these fluctuations are caused by powerful eruptions in the core of the galaxy. In some Seyferts the nucleus outshines all the stars in the host galaxy.

Relative flux

These active galaxies are distinguished from the others because of the presence of very ‘broad’ and ‘narrow’ (but broader compared to those from the normal galaxies) emission lines. These are known as the Type-I and Type-II Seyferts, respectively. Often in a given galaxy, features of both the types are seen. Fig. 14.8 shows the emission sectrum of a Type-I Seyfert galaxy.

Wavelength Fig.14.8: Emission spectrum of the Type I Seyfert NGC 5548

In the presence of rapidly moving emitters, a line emitted at a fixed frequency ν in the rest frame of the atom, will be observed at a different frequency ′ and as we have discussed before, the shift in frequency is related to the velocity of the atom by dν ν

=

ν′ − ν ν

~ v / c.

(14.7)

Since the sign of v could be both negative and positive, the shift occurs on both the sides of the emitted frequency. Close to the centre of the Seyferts, there are emitting clouds randomly rotating around the nucleus and are responsible for the broad emission lines. These lines show

29

variability in a matter of weeks to months. From this, one can estimate the distance from the centre where the emitting clouds must be located. It turns out that they could be as close as 1016cm. The corresponding velocities could be as high as 7×106 − 2×107 m s−1.

Galaxies and the Universe

Spend 5 min.

SAQ 3 Estimate the distance that an emission cloud must have from the centre of a Seyfert 7 galaxy (having a central compact object of mass 10 MΘ) in order to produce a velocity of 106 m s−1. Use Kepler’s law. Lines emitted from the clouds which are located very far away are narrow and they correspond to velocities of the order of only 200 − 500 km/s. Assuming that the rotation velocities of these clouds are Keplerian, a factor of twenty lower in velocity corresponds to a factor of four hundred larger in distance (for a given central object). Thus, narrow line regions (NLRs) are located very far away.

14.3.3 BL Lac Objects and Optically Violent Variables These types of active galaxies are characterized by a high degree of variability in radio, infra-red and optical wavelengths. For instance there are cases in which the intensity varies by a factor of 10 or more in a matter of weeks. From one night to another, variation up to 10-20% is very common. These objects have no emission lines and the polarization is large and rapidly varying. The emitted radiation is most concentrated at the centre and there is a weak sign, if any, of jets or radio lobes. In optically violent variables, there is a weak broad emission line component. Together these are known as Blazars. Fig. 14.9 shows a typical spectrum from a BL Lac object.

Fig.14.9: Spectrum of the BL Lac object Markarian 421 (Source: www.astr.ua.edu/ keel/agn)

30

Active Galaxies

14.4 THE NATURE OF THE CENTRAL ENGINE So far you have learnt about various classes of active galaxies. But what is really happening close to the centres of these galaxies? We have already mentioned that from the nature of rapid variability, one can infer that the masses of the central objects must be very high and must be concentrated in a compact region. Also, you have learnt that energy is released at a tremendous rate. For instance, a rate of 1047 erg s−1 26 −1 corresponds to a destruction of matter of about 10 g s (using the famous formula of Einstein E = mc2). So in one year, i.e., in 3.15 × 107 s, these galactic centres completely destroy about 3 × 1033 g of matter per year (which is equivalent to 1.5 times the mass of the Sun!). One of the simplest solutions to this energy budget is that the matter can be accreted on to the central compact object and the gravitational energy could be released. This energy is eventually converted into kinetic energy and then to thermal energy and radiation: GMm p r

~

1 2

m p v 2 ~ k B T ~ hν

(14.8)

where, M is the mass of the central compact object, mp is the mass of the particle, T is the temperature of the gas, kB and h are the Boltzmann constant and Planck’s constant, respectively, v is the velocity of matter and is the emission frequency. As matter falls closer and closer to the centre, the velocity is increased. The gas becomes hotter and radiation is emitted at a much higher frequency. Now, what could the nature of the compact object be? In Unit 11, you have seen that only stable compact objects which are also massive are necessarily black holes. So, for the sake of argument, assume that the main cause of the activity is due to matter falling on to a super massive black hole. Does this picture explain everything? Roughly speaking, the answer is ‘yes’!

14.4.1 Unified Model of the Various Active Galaxies So far, you have learnt how matter is accreted on to a black hole and how the radiation is emitted. What we have described so far is what goes on, say, within the inner 1000 rg where rg = 2GMBH /c2 is the Schwarzschild radius of a black hole. When the black hole is super-massive, as in the case of our present discussion, the region farther out emits optical and infra-red radiation since there is supposed to be a large dusty torus. There are smaller clouds of gas, some of which are closer than this torus and move very rapidly. They are responsible for broad line emissions (BLR) since high velocity causes larger broadening of lines. Similarly some of the clouds move around very farther away and move slower. These clouds produce narrower emission lines. Thus, depending on the angle at which the observation is made, the dusty torus may block the BLR only selectively. The possibility that different classes of active galaxies may be actually due to such observational effects has been proved in the past most conclusively. This is because after prolonged observation, objects of a given class have been found to change their class altogether. In Fig. 14.10 we draw the unified model of AGN that is widely adopted in the literature, i.e., all AGN are the same. The differences in lines, jets and spectra may be due to different viewing angles. As the plane of accretion can be randomly oriented in the sky, different lines of sight will result in different kinds of observations. So, the same object may be seen differently and classified differently. It all depends on our line of sight!

31

Galaxies and the Universe

HBLR Seyfert 2

Narrow line clouds

Jet Dusty torus

Cool dust

Broad line clouds Warm dust

Accretion disk Hot dust

Hot dust

Black hole

Warm dust

Cool dust

non-HBLR Seyfert 2

Seyfert 1 Fig.14.10: The unified scheme for understanding AGN

In this unit you have learnt about the active galactic nuclei. Let us now summarise the contents of this unit.

14.5 SUMMARY •

Active galactic nuclei (AGN) are broadly classified as Radio Galaxies, Quasars, BL Lac Objects, Optically Violent Variables, Seyfert Galaxies, Star-Burst Galaxies, etc.

•

The spectra of AGN reveal the nature and extent of activities in them, which serves as the basis of their classification.

•

The emission spectrum of AGN obeys the power law in the radio region: F( ) = F0 v−α where α is a constant known as the spectral index. When α ≤ 0.4, the spectrum is called a flat spectrum, and when α is larger, it is called a steep spectrum.

32

•

One of the most important processes occurring in AGN is the synchrotron radiation.

•

According to the current understanding, all the AGN may actually be the same object seen at different angles.

Active Galaxies

Spend 30 min.

14.6 TERMINAL QUESTIONS

1. Electrons having Lorentz factor γ = 100 are gyrating in a magnetic field of 1010G. At what frequency would its synchrotron emission peak? 2. Derive the expression for the half width (Eq. 14.6) caused by Doppler effect. 3. Explain the origin of broad-line and narrow-line regions in a Seyfert galaxy. 4. Describe briefly the nature of the central engine in an AGN.

14.7 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1. See text. 2.

2k B T

Half width = 2ν 0

=

=

M pc2

ln 2

2

2 × 1.38 × 10 −16 × 10 5

6300 × 10 −8

16 × 1.6 × 10 − 24

2 × 10 6 × 10 6 63

= 1010

ln 2

(in cgs units)

2 × 1.38 × 0.693 × 10 16 × 1.6

200 2 × 1.38 × 0.693 × 10 63 16 × 1.6

= 2.7 × 1010 Hz.

3.

mv 2 R

=

GMm R2

v2 =

GM

=

R

R=

GM v2

6.7 × 10 −8 × 10 7 × 2 × 10 33 1016

(in cgs units)

= 1.34 × 1017 cm = 0.043 pc

Terminal Questions 1.

ω g = 1.8 × 1011

B 10 10

5

γ

rad s −1

ω c = γ 2 ω g = 10 4 ω g = 10 4 × 1.8 × 1011

1010 10 10 5 100

= 1.8 × 1019 rad s - 1 33

Galaxies and the Universe

2.

I d (ν) = I 0 exp −

I max = I 0

at

M p c 2 (ν − ν 0 )2 2 k BT ν 02 ν = ν0

At half the maximum intensity I I= 0, 2

Therefore,

exp −

exp

M p c 2 (ν − ν 0 )2

ν 02

2 k BT

M p c 2 (ν − ν 0 )2

ν 02

2 k BT

M p c 2 (ν − ν 0 ) 2

2k B T

ν 02

=

1 2

=2

= ln 2

Thus, Half width = 2(ν − ν 0 ) = 2ν 0 3. See Text. 4. See Text.

34

2k B T M pc2

ln 2

UNIT 15 LARGE SCALE STRUCTURE AND THE EXPANDING UNIVERSE

Large Scale Structure and the Expanding Universe

Structure 15.1

Introduction Objectives

15.2

Cosmic Distance Ladder An Example from Terrestrial Physics Distance Measurement using Cepheid Variables

15.3

Hubble’s Law

15.4

Clusters of Galaxies

Distance-Velocity Relation The Virial Theorem and Dark Matter

15.5 15.6

Friedmann Equation and its Solutions Early Universe and Nucleosynthesis Cosmic Background Radiation Evolving vs. Steady State Universe

15.7 15.8 15.9

Summary Terminal Questions Solutions and Answers

15.1 INTRODUCTION In previous units you have learnt about stars and galaxies and their properties. You have learnt that there are billions of galaxies in the universe. It is now time to estimate the size of the universe and discuss the current ideas about its major components and its origin. The science which deals with the origin of the universe is called cosmology. In this unit, we discuss some aspects of cosmology. First we learn how distances of distant objects can be estimated so that we can get an idea of the size of the universe. Estimation of distances is a very tricky problem because there is no way of verifying these distances directly. So, we look for internal consistency in various methods, which means that the distance estimates given by them agree with one another. In this unit we discuss the distribution of matter on very large scales also called the large scale structure of the universe. We also look at the kind of matter that may be forming the bulk of the universe. This matter is not visible and is called the ‘dark matter’. It shows itself up only through its gravitational effect. Objectives After studying this unit, you should be able to: •

explain how astronomical distances are measured;

•

state Hubble’s law and explain how Hubble’s constant indicates the age of the universe;

•

explain the need to postulate the existence of dark matter in the universe;

•

derive Friedmann equation and solve it in simple cases; and

•

explain why a hot and dense phase in the early universe is needed to explain the existence of cosmic background radiation and to synthesise light elements. 35

Galaxies and the Universe

15.2 COSMIC DISTANCE LADDER An important step in this direction is to estimate the physical size of distant objects and their distances from us. You know that to specify the position of an object in three-dimensional space, we need 3 coordinates. Since we are observing from the Earth (or its vicinity from satellites), it is best to use spherical polar coordinate system with us as the origin. Two of the coordinates, namely, θ and φ (which are related to the declination and right ascension, respectively) are easily fixed by pointing a telescope in the direction of the object. Fixing the distance to an astronomical object is relatively trickier. In this section, we give an introduction to the basic principles that are used to measure distances. The basic principle involved is to use the properties of the nearby objects and deduce distances of similar objects farther off using these properties. Then we use the latter objects to deduce the distances to objects still farther, and so on. This series of steps which takes us from one step (in terms of distances) to the next step (in terms of distances even farther) is termed the Cosmic Distance Ladder. All this is best understood through an example of estimating distances on the Earth.

15.2.1 An Example from Terrestrial Physics Consider the following situation. You are in a house and there are small plants around the lawn in your house. Outside the house also there are similar plants and in addition there are also some trees. We do not know the height of these trees. Very far away there is the sea and on the beach also there are a number of such trees. The problem is to estimate the distance of the sea from your house. To begin with we do not know about the height of the trees. The obvious thing would be to walk down to the sea (keeping in mind that all the steps should approximately be of the same size) and count the number of steps. The length of the step multiplied by the number of steps gives us the distance to the beach. But you cannot use this method if you are not allowed to, or are unable to come out of the house and go to the beach. The situation is more like this in the context of distance measurement in astronomy. We can make direct measurement only on the Earth’s surface. Since it will take about 1,00,000 years, even to reach the other end of the galaxy by moving at the speed of light, we need to devise a better strategy. Let us come back to our problem of measuring the distance to the sea beach. We will put an extra condition that we are not allowed to come out of the gates of the house. With this constraint we can proceed in the following way. We first measure the heights h of the plants inside the garden of the house. Next we measure the angle θ subtended by similar plants outside which are in the vicinity of h the trees. The distance of these plants from us equals d = . Since the trees are in the vicinity of these plants we know that their distance from usθ is also d. Further, we can measure the angle θ1 subtended by the heights of these trees. The height h1 of these trees is now given by h1 = d θ1. In this way we have achieved the first step in the ladder of finding distances. We next use this information to estimate the distance to the trees near the seaside. Since the trees near the beach are similar to the ones nearby, we assume that their height is also h1. By measuring the angle subtended by these trees we can compute the distance of these trees from us. Because of the physical association of these trees with the sea side we know the distance of the sea from your house. This is the second step of the ladder. Like this we can add more steps and go on to estimate the distances of far off objects. 36

Large Scale Structure and the Expanding Universe

15.2.2 Distance Measurement using Cepheid Variables We will now give an example of using this principle to measure distances in astronomy. In Unit 1 you have learnt the technique of measuring distances by parallax method. In this section we will discuss how this method can be used to calibrate another technique for distance measurements using Cepheid variable stars, which in turn are used to measure even further distances. The brightness of Cepheid variable stars is a periodic function of time (Fig. 15.1b).

Visual Magnitude

3.0 3.5 4.0 4.5 5.0 0

5 Days

10

(b)

(a)

Fig.15.1: a) Cepheid variable stars in NGC 300; b) the brightness of Cepheid stars as a function of time

We have a large number of such stars in our neighbourhood. Their distances can be measured by the parallax method. From these distances and from their observed apparent magnitudes, their absolute magnitudes and luminosities can be calculated. It turns out that their absolute magnitudes are directly proportional to their periods (Fig. 15.2). This is called the period-luminosity relation for the Cepheids. −8

Absolute magnitude

−6

−4

−2

0 0

0.5

1.0

1.5

2.0

2.5

Log period (days) Fig.15.2: Period-luminosity relation for Cepheids

37

Galaxies and the Universe

This relation is firmly established for the local sample. Now we assume that the distant sample of these stars also obeys this relation. So, the observed period of a member of the distant sample is used to find its absolute magnitude. The apparent magnitude can be observed directly. Using the relation between the absolute magnitude and the apparent magnitude: M = m + 5 − 5 log r,

(15.1)

we can find the distance of this star. In this way, the Cepheid variable stars have been used to find distances of nearby galaxies. Used in this manner, Cepheids are called standard candles. We can now use Cepheids to define some other objects, such as supernovae, to act as standard candles to estimate even larger distances. What is the result of these investigations? We find that distant galaxies are rushing away from us with velocities which are proportional to their distances. This is called Hubble’s law. Spend 5 min.

SAQ 1 Explain the concept of a distance ladder.

15.3 HUBBLE’S LAW Hubble’s law is probably the single most important step in our attempt to understand the Universe. This law was discovered by Edwin Hubble and it relates the distances of galaxies with the velocities with which they are receding away from us (Fig. 15.3). 300,000

−1

Velocity (km s− )

250,000

200,000

150,000

100,000

50,000

0

500

1000

1500

2000

2500

3000

3500

4000

Distance (Mpc) Fig.15.3: Hubble’s law

38

The velocity of any object can be split into a component that is along the line-of-sight and another component that is transverse (perpendicular to) the line-of-sight. The lineof-sight component of the velocity can be determined very accurately by Doppler shift of the light that we received from the object. Hence this gives the velocity with which the object is coming towards us or receding away from us. Let us look at this process.

Large Scale Structure and the Expanding Universe

15.3.1 Distance-Velocity Relation Using the methods similar to those mentioned above, Hubble estimated the distances and velocities of a set of galaxies and plotted them. He found that the galaxies in general seem to be receding away from us. This is popularly known as the expanding universe. Further, the velocities with which they are receding away, are directly proportional to their distances from us (Fig. 15.3). This prompted him to propose a law, now known by his name.

Hubble’s law v = Hr,

(15.2)

where v is the line-of-sight velocity of an object, r its distance from us and the proportionality constant H is called the Hubble constant.

The importance of this relation is that, once we know the velocity of a galaxy (by red shift measurement), we can calculate the distance at which it is located. It is important to point out here that Hubble’s law holds even if we were on some other galaxy. Our location in the universe does not have any special importance. Notice that in the above relation H is the slope of the curve shown in Fig. 15.3. Notice also that 1/H has the dimensions of time. In a very simple picture, we can imagine that all the galaxies which are today moving away from one another were at some time in the past together at one point. Some event occurred at that time which triggered the expansion of the universe. This event is usually called the Big Bang. The quantity 1/H measures the time since that even, or the age of the universe. Spend 5 min.

SAQ 2 −1

In astronomy, the velocity is measured in km s . The distance of galaxies is measured in Mpc or million parsec. Find the dimensions of H. Unfortunately the measured value of H has lots of errors. But we know today that it is −1 −1 roughly 70 km s /Mpc . Estimate the age of the universe.

15.4 CLUSTERS OF GALAXIES Galaxies mostly exist as members of large groups called galaxy clusters (see Fig. 15.4). A cluster of galaxies contains about a thousand galaxies. A galaxy cluster consists of a variety of galaxies. It is an observed fact that the gross features of these clusters are very similar. This fact alone has a far reaching implication. We expect that roughly the same kind of physical processes are responsible for their evolution. Further, it is natural to assume that they were created at different times and began to evolve. This is because, in order to initiate the process of creation and subsequent evolution of these clusters one needs a certain combination of astrophysical conditions. These conditions need not be the same at all points at a given time. Let us illustrate this situation in the following manner: Consider a set of systems, A, B, C, … Let the evolution in each of these systems be governed by the same physical processes. If they started to evolve at different times, we would expect that at any time, in particular today, they should be in different stages of evolution. Hence, we would expect they should not show great similarity.

39

Galaxies and the Universe

Fig. 15.4: Cluster of galaxies

Alternatively, if we find that they look similar today, we may naively think that they must have got created at the same time, so that they have had the same time for evolution. For the systems under consideration, namely, the clusters of galaxies, we know that they look similar but at the same time we have reasons to believe that they started their evolution at different times. Our simple-minded reasoning leads to the suggestion that they may have been created at different times but they have reached some kind of steady state today. Such systems have a simple, but important relationship for their kinematic parameters such as their mass and velocity. This relationship is called the virial theorem which we now describe.

15.4.1 The Virial Theorem and Dark Matter The virial theorem says that if a system is bounded and is in equilibrium, then its moment of inertia does not change with time. Such a system will have the following relation between its total kinetic and potential energies: 2T + V = 0,

(15.3)

where T is the total kinetic and V is the total potential energy of the system. Suppose we have a spherical system consisting of N particles that are interacting gravitationally. Let the position of the ith particle be ri and its velocity vi. The total kinetic energy is then T=

1 2

m

N

v2. i =1 i

(15.4)

The total potential energy is obtained by summing over the potential energy of all the pairs. We have V=

40

Gmi m j i> j

ri − r j

(15.5)

For a spherically symmetric distribution, V will come out to be proportional to M (R) R−1, where R is the radius of the system and M the total mass enclosed in a sphere of radius R. The total kinetic energy can be estimated in the following manner. If the velocities are in random directions, some of the particles of the system will contribute a blue shift and some red shift. This will result in broadening of the spectral lines. Hence, from the width of the spectral lines we can estimate the root mean square (rms) velocity. This gives the total kinetic energy. If the system is in equilibrium, we should have T = − V/2

Large Scale Structure and the Expanding Universe

(15.6)

or 2

v rms ∝ R −1 .

(15.7)

When one tries to estimate the mass of the system in the above manner, one finds that there is much more mass in the system as compared to that suggested by the luminous mass alone. Hence, it is postulated that there should be a significant fraction of mass in the form of dark matter. What form this dark matter takes, we do not yet know. One possibility is that it consists of cold, burnt out stars which emit very little radiation. Another possibility is that it is in the form of particles which interact very weakly with normal matter. It is a very active area of research at present.

15.5 FRIEDMANN EQUATION AND ITS SOLUTIONS Cosmology is the study of the overall features of the universe. In this framework, one does not bother about local features of the universe like planets or stars. In cosmology, we set up equations and solve them to obtain the very large scale features of the universe. To set up the basic equations governing the evolution of the universe as a whole, Newtonian theory of gravity is inadequate and is, rigorously speaking, inapplicable. The correct theory to use is Einstein’s General Theory of Relativity. This, however, is outside the scope of this course. All is, however, not lost. It so happens that if we go ahead and apply Newton’s laws (which strictly speaking is not the correct thing to do) the final equations which we get are the same equations that result from the correct theory, namely, the General Theory of Relativity. The aim of this Section is to study the evolution of the Universe using these equations. Hence, we will not be disturbed by the fact that the derivation of these is not rigorous. We will happily go ahead and use these equations since we are aware that the final equations are the correct ones. We know that over a variety of scales the universe is not homogeneous. We have planets, stars, galaxies and clusters of galaxies, which indicate that the universe is far from being homogeneous. However, if we consider the universe as a whole, then at large enough scales the universe seems to be homogeneous and isotropic. This last point needs some explanation. Consider a well maintained lawn. From a distance, the lawn appears uniformly green. But as we start analysing the lawn on smaller scales, it begins to lose its homogeneity. Consider a grass hopper sitting in the lawn. It observes the lawn at a scale which is about the blade of grass. Clearly it will notice that the lawn is not at all homogeneous. This means that the lawn is homogeneously green over distance scales which are much bigger than the size of a blade of grass. In a similar way, we say that over scales of sizes, much bigger than the size of galaxies, the universe is homogeneous. 41

Galaxies and the Universe

The equations that we now derive are the equations that govern the overall evolution of a homogeneous universe. Consider two points A and O. With O as centre, and AO as radius, draw a sphere. The force with which a test particle at A is gravitationally pulled towards O can be calculated by just using the mass enclosed by the sphere. As we know, the mass outside the sphere will not exert any net force on A. Let the density of matter be ρ. Notice that density does not depend upon space. This is because of our condition of homogeneity. However, there is no such restriction on time dependence. So we will include time dependence for generality. The mass of the sphere of radius R = OA is 4π ρ(t ) R 3 (t ), 3

M=

(15.8)

The gravitational potential at A due to this sphere is − GM/R. Further, let v = R be the velocity with which A is moving with respect to O. Depending on the magnitudes of the kinetic energy and the potential energy, the particle at point A may keep moving away from O or may turn back and fall towards O. This is the well known condition for escape velocity. The total energy per unit mass of a test particle at A is E=−

GM R

+

v2

(15.9)

2

or R 2 − 2E =

2GM R

(15.10)

If the energy E is positive, then the distance between A and O will keep increasing and if E is negative, A will attain a maximum distance from O and then begin to fall towards O. This condition on E can alternatively be expressed by replacing −2E by k .2 E ; k = +1 corresponds to the case E < 0. An alternative way of writing the last condition is then R 2 + k .2 E =

2GM R

(15.11)

Dividing both sides by R 2 and expressing the mass M in terms of the density ρ as M=

4 3 πR ρ 3

R2

k .2 E

we get,

R

2

+

R

2

=

8πG ρ 3

Defining a new variable a = R 42

(15.12)

2 E , we can now write the Friedmann equation

governing the evolution of the distance between two particles.

Large Scale Structure and the Expanding Universe

Friedmann equation a2 a2

+

k a2

=

8πGρ

(15.13)

3

We get the same equation from the general theory of relativity. The parameter k then signifies the curvature of space. In the general theory of relativity, the effect of the gravitational field is to make the space curved. The curvature of space is denoted by k, which can take values +1, 0 or −1, depending on the overall density of the universe. The quantity a is called the scale factor and the nature of a as function of t indicates the nature of the expansion of the universe. Fig. 15.5 shows the behaviour of a as a function of t for the three values of k, i.e., +1, 0 and −1. These curves are the solutions of the Friedmann equation. We see that when k = −1 (which according to general theory of relativity implies that the overall density of the universe is less than a certain critical density), or k = 0 (the overall density of the universe equals the critical density), the universe keeps expanding. When k = +1 (the overall density of the universe is greater than the critical density), the universe expands up to a point and then starts contracting. Present observations indicate that the universe will keep expanding, and its expansion will not be followed by contraction. In the early phase of the universe, the curvature must have been small, so it is sufficient to consider the case of k = 0. The solution of the Friedmann equations, of course, depends on the nature of energy density. If ρ ∝ a−n, the equation can be solved for k = 0 and the result comes out to be a ∝ ( n / 2) 2 / n t 2 / n

(15.14)

k0

t Fig.15.5: The variation of a with time

43

Galaxies and the Universe

Spend 5 min.

SAQ 3 Verify that Eq. (15.14) is a solution of Eq. (15.13) when ρ ∝ a− n. Time-temperature Relationship We know from the thermodynamics of radiation that it has pressure and energy density which we denote by prad and ρrad, respectively. They are related to each other through the relation, p rad = ρ rad c 2 3

(15.15)

where c is the speed of light. The distance between points increases as R which, in turn, is proportional to a. Hence, the volume increases as a3 and the energy contained in it as ρ a3. Now, using the first law of thermodynamics, dU + pdV = 0, we have d [ρ c 2 a 3 ] + pd [a 3 ] = 0.

(15.16)

Applying it to radiation, we get ρ rad ∝ a −4

(15.17)

Using this relation in the expression for ρ derived in the last section (with n = 4) we get, a∝ t

(15.18)

At the same time we know that the temperature of radiation is related to its energydensity by ρ rad ∝ T 4

(15.19)

for isotropic and homogeneous radiation field. From the last three equations, we get the relationship of temperature with time as T ∝ 1/ t

(15.20)

We see that t = 0 is both interesting as well as disturbing. This is because at that point of time, the temperature shoots to infinity and so does the energy density. It is this epoch which is termed as the Big Bang.

15.6 EARLY UNIVERSE AND NUCLEOSYNTHESIS We saw that the temperature of radiation varies as inverse of the square root of time. The immediate consequence of this is that the radiation temperature in early times should have been very high. If matter and radiation were in thermal equilibrium, the above statements imply that the temperature of matter was also high in early times.

44

We know from thermodynamics that temperature is a measure of the mean kinetic energy which in turn implies high energy collisions between particles. The earlier the epoch, the higher the energy with which these particles collide with each other. Today we know about the physical phenomena at high energies from experimental investigations using high energy particle accelerators and colliders. We should expect

that the same phenomenon must have taken place in the early universe. In fact, for this reason, early universe is often called the poor man’s laboratory.

Large Scale Structure and the Expanding Universe

An important class of reactions at high energy is those which lead to the synthesis of nuclei of elements. Very early on, there were no complex nuclei. The only ones were the hydrogen nuclei, i.e., protons. At those energies, even if the protons and neutrons combined to form higher atomic number nuclei (e.g., the Helium nucleus), the kinetic energies of the particles were so high that the collisions would have immediately disintegrated them. As the universe cools, a certain temperature is reached when the energies are low enough that this backward reaction (namely disintegration) begins to get suppressed. Hence, stable helium nucleus begins to get formed. As the temperature lowers further, we expect that higher atomic number nuclei will begin to get synthesized. So the question is, “Can we proceed in this way and synthesize all the naturally occurring nuclei”? The answer unfortunately is no! This line of reasoning works for only the elements with first few atomic numbers. As the temperature decreases, we can form Lithium and some Boron. The problem comes up when we need to form Beryllium. In the process of the formation of the stable Beryllium nucleus, one passes through an intermediate stage where, spontaneous disintegration is faster than the fusion. So even before there can be fusion the nucleus which is supposed to participate in the fusion, disintegrates. Hence one cannot form Beryllium by this procedure. Only after the formation of Beryllium, can the nuclei of higher atomic numbers be formed. Hence this is called the “Beryllium Bottleneck”. The universe has to wait for a very long time, namely, till stars form, in order to synthesize elements of atomic number 5 and higher. (Refer to Unit 10 for details of nucleo-synthesis inside the stars).

15.6.1 Cosmic Background Radiation Do we have any signature of the early hot phase of the universe? The answer is yes. In 1965, two scientists at the Bell Telephone Laboratories in America discovered accidentally a radiation at a very low temperature of only 3 K which seemed to come from all directions. It was highly isotropic. It was suggested that the radiation fills the whole universe. Since the wavelength of the peak radiation, ~ 1 mm, falls in the microwave region, it was called the cosmic microwave background radiation (CMBR). This is the relic of the era when the universe was very hot and dense. It is argued that the radiation was once very hot and has been cooled to its present temperature due to the expansion of the universe over billions of years (recall from 4 Eqs. (15.20) and (15.19) that T ∝ 1 t and ρrad ∝ T ). Put in another way, the same energy fills an every increasing volume, so its energy density decreases and so does its temperature. The discovery of the CMBR has great significance for cosmology. (The discoverers of the radiation were awarded the Nobel Prize for their work.) It shows that the universe was once very hot and dense. This lends great support to theories of the universe which maintain that the universe is changing with time, that is, it is evolving. Its existence is a very powerful argument against theories which propose that the universe is steady, that is, it is unchanging. In the theories of the latter type, it is extremely hard to produce such a radiation. CMBR is a topic of intense research today.

15.6.2 Evolving vs. Steady State Universe You have just seen that CMBR points to a phase of the universe when it was very hot and dense. This phase is generally known as the ‘hot Big Bang’. The idea is that some

45

Galaxies and the Universe

violent event took place at that time which sent the universe expanding. You have also seen above that if the early universe had not been hot and dense, it would not have been possible to synthesise light elements, such as H2, H e4 and Lithium. In fact, the prediction of the precise observed abundances of these light elements is a very powerful argument in favour of the universe that changes with time: an evolutionary universe. Yet there is a set of scientists who believe in a steady state universe, a universe which has no beginning and no end, which appears the same at all points in space and at all times. Historically, the steady state theory emerged in the1940s and 1950s, when observational techniques were not much developed, and so the Hubble constant H could not be measured accurately. Recall that 1/H gives a rough time scale of the age of the universe. So, the age of the universe inferred from the value of H at that time turned out to be less than the age of some fossils on the Earth. This was quite embarrassing. To overcome this age problem, the scientists proposed the steady state universe. However, CMBR and synthesis of light elements in the early universe are very powerful arguments against this theory and in favour of the hot Big Bang theory. With this we come to an end of this unit in which you have studied about the large scale structure of the universe. We now present its summary.

15.7

SUMMARY

•

Distances of far-off objects inform us about the large scale structure of the universe. To find distances of other galaxies we employ the concept of the distance ladder. The first step of the ladder is the Cepheid variable stars found in nearly galaxies. Subsequent steps include objects such as supernovae which can be detected even in galaxies which are very far off.

•

The outcome of the exploration of the universe is that the galaxies are rushing away from one another and the universe is expanding.

•

The expansion of the universe is in accordance with Hubble’s law, v = Hr, so that 1/H gives a rough estimate of the age of the universe.

•

The behaviour of the universe is governed by the Friedmann equation: a2 a2

+

k a2

=

8πGρ 3

•

The need for a hot and dense early universe can be explained in connection with the synthesis of light elements and the existence of the cosmic microwave background radiation.

•

The present understanding is that if the universe was once hot and dense, then it must be an evolving universe, and not a steady state universe.

15.8 TERMINAL QUESTIONS

Spend 30 min.

1. Explain how Cepheid variables have been used to measure astronomical distances. 2. State Hubble’s law. How can this be used to get an estimate of the age of the universe? 46

3. Explain why at one time, the steady state theory appeared necessary. What is its status now?

Large Scale Structure and the Expanding Universe

15.9 SOLUTIONS AND ANSWERS Self Assessment Questions (SAQs) 1.

See Section 15.2.

2.

H = 70 km s−1Mpc−1 = 70 × 10 5 cm s −1 / 3 × 1018 × 10 6 cm

=

7 × 10 6 3 × 10

s −1

24

∴ Age of the universe = 3 ×10 24

=

7 ×10 6 × 3 ×10 7

1 H

=

3 × 10 24 7 × 10 6

s

yr

= 1.4 ×1010 yr = 14 billion years.

3.

a2

Eq. (15.13)

a2

=

8πG 3 a2

Since ρ = Ca − n

a

∴

a a

=

8πGC 3

a

−n

2

n−2

∴

2

a 2 da = A dt

=

ρ 8πG 3

a=

n

C a −n

da dt

a 2 ∝t n 2

[C is a constant]

1−

= Aa

n 2

A=

8πGC 3

( 2) 2n . t 2n .

a ∝ n

Hence, proved. Terminal Questions 1.

See Text.

2.

See Text.

3.

See Text.

47