Px148 Notes

University of Warwick PX148 Classical Mechanics and Special Relativity

Mike Allen Based heavily on the material of Prof. T. R. Marsh

Updated: December 4, 2013

2 Foreword These notes mostly show the essentials of the lectures, i.e. what I write on the visualizer and what students should know. Occasionally I include information that I may have said but not written during lectures, when I think it would help you follow the notes. Really I may put little boxes like this and you prompts in the important equations and ideas appear in margin too. should try to learn and remember them. The notes are terse, and brief to the point of grammatical inaccuracy. This is because they are notes and are not intended to replace the book (University Physics, 13th ed., by Young & Freedman). I make them available in case you had to miss a lecture or find it difficult to make notes during lectures, but if you rely on these notes alone and do not read books as well, you will struggle.

Lecture 1

Introduction to PX148 Objectives • To introduce the course.

1.1

Classical Mechanics

Until ∼ 1600, Aristotle’s view of the Universe, in which Earth defined a “state of rest”, was widely believed in Europe. “Common sense”: eventually pretty much everything falls to the ground and stops moving. “Common sense is the collection

Galileo’s experiments – real and “thought” – overturned these ideas. of prejudices Newton, born shortly after Galileo died in 1642, put Galileo’s ideas acquired by age eighteen.”, into a rigorous mathematical framework in “Principia” (1687). A.Einstein

Newton’s work showed that, starting from a few experimentally-based principles, a huge range of physical phenomena could be understood quantitatively. The physics of these phenomena – e.g. the motions of stars, forces holding up a bridge – are termed Classical Mechanics. Classical mechanics is the bedrock of physics and engineering, and is capable of great precision for motion at everyday speeds. One can’t appreciate the rest of physics without a firm grasp of classical mechanics, so this is where we start.

3

LECTURE 1. INTRODUCTION TO PX148

1.2 1.2.1

4

Newton’s Laws Newton’s First Law – N1

Every body continues in its state of rest or uniform motion unless acted on by a force.

This comes straight from Galileo.

More compactly: A body moves at constant velocity unless acted upon by a force. uniform motion:

in a straight line at constant speed.

constant velocity: the same; constant direction and speed at rest:

a special case of constant velocity

Aristotle believed that the velocity of an object was proportional to the force acting on it. Newton realized that it is the acceleration (rate of change of velocity) that is proportional to force. The property of objects that resists changes in velocity is known as “inertia”; we measure it as mass (sometimes “inertial mass”). There is an implicit assumption here. N1 only holds in special frames of reference known as “inertial frames”: non-accelerating, non-rotating. Otherwise we would measure accelerations even in the absence of forces. What defines an “inertial frame” (e.g. what do they not accelerate or rotate relative to?) is a deep and as-yet-unresolved question, but if you are in one, you feel no force acting upon you!

1.2.2

Newton’s Second Law – N2

The rate of change of momentum of a body is equal to the total force acting upon it. or, for short:

Newton, of course, invented calculus. Or was it Leibniz?

LECTURE 1. INTRODUCTION TO PX148

5

Force equals rate of change of momentum. momentum:

mass × velocity, i.e. p = mv Really vectors!

According to N2 therefore F =

dp , dt

where t is the time. If the mass m is constant this can also be written F =m

dv = ma, dt

where a is the acceleration of the body, its rate of change of velocity. Example 1.1. An oil tanker has mass m = 500 000 t = 5 × 108 kg (1 t is a metric tonne, 1000 kg). Its engine and propeller can exert a maximum force of F = 5 × 106 N. How long does it take to reach a speed of 5 m s−1 , starting from rest? Starting from 5 m s−1 , what is the tanker’s “stopping distance”? [Ignore water resistance.] Answer.

F 5 × 106 = = 10−2 m s−2 . 8 m 5 × 10 dv/dt = a, so it will take 500 s to reach v = 5 m s−1 . a=

Using v 2 = u2 + 2as, with initial speed u = 5 m s−1 , final speed v = 0 m s−1 , acceleration a = −0.01 m s−2 , then the stopping distance s is given by v 2 − u2 −25 s= = = 1250 m. 2a 2 × −0.01

1.2.3

Newton’s Third Law – N3

To every action there is an equal and opposite reaction.

LECTURE 1. INTRODUCTION TO PX148 action:

in more modern parlance, “force”

reaction:

again, just means ‘force”

6

Lecture 2

Forces Keypoints from Lecture 1 • Classical mechanics is the bedrock of physics and engineering. • Newton’s Laws: physical principles, not mathematical axioms. • Newton’s Laws: consistent with experimental observations; further experiments could prove them wrong.

Objectives • Meet the Forces: reaction, friction, springs.

2.1

Types of Force

In an elementary analysis of a bridge or building, we are concerned with forces in balance with no resultant acceleration (the next step would be a dynamic analysis to understand a structure’s ability to withstand earthquakes perhaps). Here we look at some different forces.

7

LECTURE 2. FORCES

2.1.1

8

Reaction forces

N

N

Reaction forces are communicated through inter-atomic forces. Try to squeeze nuclei closer together, electron “clouds” and nuclei repel ~ , for each other through electrostatic forces. Often denoted N , or N “normal” to surface.

2.1.2

Static Friction

Almost no surface is completely smooth at the atomic level. For example, 0.1 µm = 10−7 m roughness is a very smooth surface, but still means peaks and valleys ∼ 1000 atoms in height or depth!

Large N

Small N FS

FS

It requires a certain minimum sideways force to make two surfaces start to slide past each other. Experimentally it is found that this force required is linearly proportional to the normal reaction force N . FS ≤ µS N.

No slippage: normal reaction force:

N

LECTURE 2. FORCES

9

static frictional force:

FS

coefficient of static friction:

µS

A few values steel–steel teflon–teflon rubber–rubber

µS 0.3 0.04 1.2

NB It is possible to have µS > 1.

The friction acts parallel to the surfaces. Consider a stationary block on a slope: Draw diagrams for all but the most trivial problems!

N

Fs

θ W=mg

θ

1. Only the forces on the block are shown. 2. Equal and opposite forces to FS and N act on the slope, but we are not interested in them here, so we don’t show them. 3. If static, the total force on the block is zero. 4. All forces are drawn acting through the same point because we are not interested in rotation (for now at least). 5. FS must act up the slope. 6. Force is a vector and so must balance separately along all independent directions or “degrees of freedom” (two in this case). 7. Must therefore resolve forces in this number (two) of directions. 8. The choice of directions is up to you: try to make things easy.

See UP ch4 on “free body” diagrams. I don’t personally like their use of wiggly lines to scrub forces that they also plot resolved into components. I recommend drawing each force once only.

LECTURE 2. FORCES

10

Here, perpendicular and parallel to the inclined plane are good choices: N − W cos θ = 0, FS − W sin θ = 0. Hence, FS = N tan θ ≤ µS N , for no slippage, i.e. tan θ ≤ µS . So, measuring the critical θ, at which slipping occurs, gives µS .

2.1.3

Kinetic friction

Once slippage occurs, friction still occurs, and is still found to be proportional to the normal force, but the coefficient changes. The force then is Slippage: FK = µK N. kinetic frictional force:

FK

coefficient of kinetic or dynamic friction:

µK

Note that the speed does not appear. In general µK < µS , so once slippage starts, it continues, but µK is similar in size to µS .

2.1.4

Spring forces

Springs are resistant to being stretched or compressed. For small displacements from equilibrium, x, they offer a force that is in proportion to x: F = −kx, force constant:

k (characteristic of the spring)

minus sign:

force always opposes direction of displacement

Example 2.1. A mass m is attached to a spring with constant k and dragged along a horizontal table with static coefficient of friction µS . How far must the spring stretch to get it moving?

LECTURE 2. FORCES

11

Answer. Normal reaction supports weight so N = mg. Force needed to make mass move therefore equals F = µS mg. Therefore the spring stretches by x=

F µS mg = . k k

Lecture 3

Equilibrium Keypoints from Lecture 2 • At a microscopic level, reaction and frictional forces are a manifestation of the electromagnetic forces between atoms, in combination with quantum mechanics. • Maximum static frictional force is proportional to the normal force N : FS ≤ µS N. • Kinetic frictional force is proportional to the normal force and (almost) independent of the sliding velocity: FK = µK N. • µK < µS ; also, friction coefficients can exceed 1. • Spring forces act to restore an equilibrium length and for small extension x, are given by F = −kx.

Objectives • How to set up and solve problems in statics 12

LECTURE 3. EQUILIBRIUM

3.1

13

Forces in balance

When nothing moves, all forces must balance. This applies to individual components and the whole system. In all cases forces must sum to zero. Consider:

2

θ

1

Cylindrical logs at equilibrium in a box. Assuming that there is no friction then the forces acting appear as follows:

N2

N2

R N1

R

N1

W2

F W1

F

Forces on each log and the box. (We take the box to be fixed!). All contact forces have a normal reaction component only (no friction). ~ 1 = m1~g and N3 has been imposed on the action/reaction pairs. W ~ 2 = m2~g are the weights (assumed known). We also assume that W the angle θ is known (fixed by the geometry). There are four unknown forces. The zero force condition allows us to derive relations between them, e.g. on log 1: ~ 1+F ~ +W ~ 1+R ~ = ~0. N

LECTURE 3. EQUILIBRIUM

14

This is really two equations in one. N1 − R cos θ = 0, F − W1 − R sin θ = 0 .

horizontally: vertically:

Something similar applies to log 2. Equally we could consider the two logs together: ~ 1+F ~ +W ~ 1+W ~ 2+N ~ 2 = ~0, N which becomes N1 − N2 = 0, F − W1 − W2 = 0.

horizontally: vertically:

Four unknowns, four equations, giving answers in terms of W1 , W2 , and θ. F = W1 + W2 ;

R=

W2 ; sin θ

N1 = N2 =

W2 cos θ sin θ

Example 3.1. A 1 kg mass is placed at the mid-point of a nearhorizontal, inextensible, light rope 2 m long. At equilibrium, the midpoint of the rope lies 1 cm below its two ends. What is the tension in the rope? Answer. Translating the words into a figure: T

T

l θ

θ

h

mg

where ` = 1 m, h = 1 cm and m = 1 kg Resolving forces vertically:

Therefore

2T sin θ = mg. h Geometry: sin θ = . ` mg` 1 × 9.81 × 1 T = = = 490.5 N. 2h 2 × 0.01

LECTURE 3. EQUILIBRIUM

3.2

15

Reflections on Newton’s Laws

Here are some left-overs from the first two lectures.

3.2.1

N1 is a special case of N2

If F = 0, N2 says dv/dt = 0, i.e. the velocity v remains constant in the absence of force, which is N1 .

3.2.2

N2 is not just a definition of force

N2 has not merely introduced a new abstract concept “force” to represent the rate of change of momentum, dp/dt. It is more than this: can devise ways to exert a constant force (e.g. a spring, mass on a string over a pulley) which can be applied to different masses so N2 is experimentally testable.

3.2.3

N3 is needed for consistency with N1 & N2

Consider an object at rest. Imagine it to have two halves, which may exert forces on each other. Unless these forces are equal and opposite, the object would accelerate on its own, violating N1 . Hence N3 is required by N1 and N2 .

3.2.4

Scalars vs Vectors

Physical quantities seen so far such as m, v, p, a, F , and t fall into two classes. Some (such as mass m and time t) have a size or “scale” or “magnitude” but no direction. These are sometimes called “scalar” quantities, and can be represented mathematically by real numbers.

LECTURE 3. EQUILIBRIUM

16

Others (velocity v, momentum p, force F ) have a magnitude but also a direction. Mathematically we associate them with vectors. Written properly, N2 in the form “F = ma” becomes ~ = m~a. F We will see many such vector relations. A scalar cannot equal a vector; the following are therefore both invalid: r = 2~v , 7 m + ~v = ~p. 7 For simple one-dimensional problems we can stick to scalars. There wasn’t time in the lecture to do the following example, but you might like to look at it anyway. Example 3.2. Estimate the force exerted by a hammer on a nail. Answer. Must guess reasonable values. Let’s take the mass of the hammer’s head to be m = 0.5 kg, its speed at contact with the nail v = 4 m s−1 , and the nail to be driven d = 1 cm in to the wood by each blow. Assume the force F to be constant, for t seconds, until the hammer comes to rest. Then from N2 the change in momentum ∆p is given by ∆p = −mv = −F t, so

mv . t The speed drops uniformly from v to 0, so the mean speed = v/2, and thus d 2d t= = , v/2 v therefore mv 2 0.5 × 42 F = = = 400 N. 2d 2 × 0.01 F =

Both this example and the second part of the first lecture’s example could be worked out from KE = work done = force × distance.

In general try to think in terms of vectors.

LECTURE 3. EQUILIBRIUM

17

A suddenly acting force is sometimes called an “impulse”. The impulse is defined as force × time = F t, or equally well as the change in momentum ∆p.

Lecture 4

Newton’s Law of Gravity Keypoints from Lecture 3 • Newton’s Laws are self-consistent; N3 is required for N1 and N2 to be true. • Physical quantities require a variety of mathematical objects to represent them. We will use scalars and vectors (in later modules you will meet objects called “tensors” which extend this further). • It is important to combine these correctly in equations. • N1 applies in situations of static equilibrium. All forces must then be in balance. This condition and geometry then allow one to work out the forces.

Objectives • To introduce and discuss gravitational attraction.

18

LECTURE 4. NEWTON’S LAW OF GRAVITY

4.1

19

Newton’s Law of Gravity r

1 r^12

F 12

2

The gravitational force between two point masses m1 and m2 a distance r apart is given by Gm1 m2 F = , r2 where G = 6.67 × 10−11 N m2 kg−2 is the gravitational constant. It is an attractive force that acts along the line joining the two masses. These statements can be combined into a single vector expression: ~ 12 = − Gm1 m2 rˆ 12 , F r2 ~ 12 is the force exerted by 1 on 2 and rˆ 12 is a unit vector where F pointing from 1 to 2. The minus sign ⇒ attraction. Pedantic note: in the often-seen equation F =−

Gm1 m2 , r2

the minus sign has no clear meaning because it is not vectorial. Newton’s shell theorem: the same equation applies to two spheres if r is the distance between their centres.

4.2

Inertial vs Gravitational mass

For a moment, distinguish the inertial mass mi which appears in N2 from the gravitational mass mg which appears in the above equations.

I assume the shell theorem without proof; it took Newton years to prove (he had to develop calculus). It is not hard to prove, but would cost a full lecture. I encourage you to look it up.

LECTURE 4. NEWTON’S LAW OF GRAVITY

20

Why should they be the same? Applying N2 to mass 2: F = hence

mi2 a2

Gmg1 mg2 , = r2

Gmg1 a2 = 2 r



mg2 mi2

 .

If mg1 = ME , the mass of Earth, and r = RE , the radius of Earth, then this says   GME mg2 g2 = , 2 RE mi2 hence the acceleration due to gravity varies as mg /mi . Galileo’s famous experiment from the leaning tower of Pisa implies that mg /mi is constant. Modern experiments show mg /mi constant to < 1 part in 1013 . From now on we do not distinguish between “gravitational” and “inertial” mass, so we drop the superscripts. The acceleration due to gravity is the same for all objects: g=

4.2.1

GME = 9.81 m s−2 . 2 RE

Unclear whether he did so in reality as opposed to thought. Very odd coincidence in Newtonian mechanics; a key principle of General Relativity.

N3 and gravity.

~ 21 of particle 2 on particle 1? What about the gravitational force F Swap the labels 1 and 2, and note that rˆ 21 = −ˆ r 12 with r unchanged: ~ 21 = − Gm2 m1 rˆ 21 = Gm1 m2 rˆ 12 = −F ~ 12 . F 2 2 r r So Newton’s Law of gravity is consistent with N3 . Example 4.1. Three identical iron spheres of radius 5 m are placed in contact as shown below. Calculate the gravitational force on sphere A due to B and C.

Would this be the case if F = Gm21 m2 /r2 ? What about G(m1 + m2 )/r2 ?

LECTURE 4. NEWTON’S LAW OF GRAVITY

21

B A R

C

Answer. The force on A just due to B has magnitude FBA =

GmB mA . r2

Setting r = 2R, and mA = mB =

4π 3 R ρ, 3

where R = 5 m is the radius of the spheres and ρ = 7870 kg m−3 is the density of iron, then FBA

4Gπ 2 4 2 16Gπ 2 R6 ρ2 = Rρ. = 9 × 4R2 9

This gives FBA =


2 4 × 6.67 × 10−11 × π 2 × 54 × 7.87 × 103 = 11.33 N. 9

The total force is not twice this since the forces from B and C upon A are not parallel. Instead only the component towards the centre of mass of the three spheres matters, giving √ Ftot = 2FBA cos(30◦ ) = FBA 3 = 19.62 N. Example 4.2. A person of mass m = 71.4 kg weighs 700 N standing on the ground. How much less would she weigh at the top of the Eiffel tower (height h = 320 m)? [Radius of Earth RE = 6370 km. Hint: use h/RE  1].

LECTURE 4. NEWTON’S LAW OF GRAVITY

22

Answer. Weight on ground W0 =

GME m = mg. 2 RE

Weight at height h 2 GME m mg RE mg Wh = = = , (RE + h)2 (RE + h)2 (1 + h/RE )2

Since h/RE  1 (the symbol  means “much less than”) we can expand using (1 + x)−2 = 1 − 2x + · · ·   2h Wh ≈ mg 1 − R   E 2h Wh − W0 ≈ − mg = −0.07 N, RE since 2h/RE ≈ 10−4 and mg = 700 N.

Lecture 5

Systems of Particles Keypoints from Lecture 4 • Newton’s Law of Gravity: ~ 12 = − Gm1 m2 rˆ 12 . F r2 • Note: both sides are vectors, as they must be. • By experiment it is found that the masses that appear in Newton’s Law of Gravity are the same as the masses required for N2 . • Newton’s Law of Gravity is consistent with N3 .

Objectives • Discuss how to handle composite objects • Define centre of mass and mention centre of gravity

5.1

Many Particles

How do Newton’s Laws apply to composite objects, with multiple accelerations and forces? 23

LECTURE 5. SYSTEMS OF PARTICLES

24

Consider a set of N particles with both inter-particle and external forces acting:

1

F1

3

F31 F13

4 F2

5

2 6

~ ji and the external Let the force of particle j acting on i be called F ~ i. force on particle i be simply F Then the total force acting upon particle i is given by X ~ ~ ji = mi~ai . Fi + F j6=i

Summing over all particles, the total force acting on the N particles is XX X X ~i+ ~ ji = F F mi~ai . i

i

i

j6=i

N3 ⇒ the second term = ~0 since for every Fji , say, there is a cancelling ~ =P F ~ i , so Fij . The first term is the total external “applied” force F i X ~ = F mi~ai . i

~ = m~acm , where m = P mi is the total mass, We can write this as F i if we define P mi~ai ~acm = Pi . m i i Remembering that ~ai =

d2~r i , dt2

LECTURE 5. SYSTEMS OF PARTICLES then ~acm

d2 = 2 dt

25

P  d2~r cm mi~r i i P = , dt2 i mi

where ~r cm

P mi~r i = Pi . m i i

This is a mass-weighted average of the position vectors of the particles. This is the definition of the “centre of mass”. Hence we have the important result: Newton’s second law applies to the total mass and the acceleration of the centre of mass of composite bodies.

5.2

Centre of Mass of Continuous Bodies

If N → ∞ while mi → 0, then the centre-of-mass sums become integrals: R ~r dm ~r cm = R . dm This applies to each component one-by-one, e.g. R x dm xcm = R . dm “dm” is an infinitesimal element of mass. Example 5.1. A bar of length L has an increasing density per unit length of ρ = αx where α is a constant and x is the distance from one end. What is the position of its centre of mass? Answer. By definition, dm = ρdx. The total mass of the bar is Z Z L Z L M = dm = ρ dx = αx dx = 21 αL2 . 0

0

Similarly, Z

L

Z x dm =

Z xρ dx =

0

0

L

αx2 dx = 13 αL3 .

LECTURE 5. SYSTEMS OF PARTICLES Hence xcm

26

R x dm 2 = R = 3 L. dm

Example 5.2. Find the centre of mass of a two-dimensional uniformly filled semi-circle of radius R. Answer. Orient the semi-circle with its straight edge on the yaxis: y

x

dx

By symmetry, ycm = 0. We seek xcm . Assume a density per unit area of σ. Semicircle area is 12 πR2 . Z M = dm = 12 πσR2 . Slice the semi-circle up vertically, so the area of each slice is 2ydx, so dm = 2yσ dx, and x2 + y 2 = R2 , so Z Z R Z R
1/2 x dm = x 2yσ dx = 2σ x R 2 − x2 dx 0 0 h
iR 2 3 2 2 3/2 1 = 3 σR . = 2σ − 3 R − x 0

Therefore, finally xcm

2σR3 /3 4 = = R ≈ 0.424R. πσR2 /2 3π

LECTURE 5. SYSTEMS OF PARTICLES

5.3

27

Centre of Gravity

In the lecture I demonstrated the location of the centre of gravity but did not explicitly write down the equation. The total external gravitational force on a set of particles X ~ FG = mi~g (~r i ), i

can be reduced to a single force acting through one point, the centre of gravity. If ~g is constant then the centre of gravity and the centre of mass are one and the same. I hope to return and give a proper definition of the centre of gravity later.

Lecture 6

Force and acceleration Keypoints from Lecture 5 ~ as the total externally• N2 applies to composite objects with F applied force, m as the total mass and ~a as the acceleration of the centre of mass. • The centre-of-mass is defined by P mi~r i ~r cm = Pi . i mi • In the “continuum limit” ~r cm

R ~r dm . = R dm

• Gravity acts through the centre of gravity. It coincides with the centre of mass, if ~g is constant, but otherwise in general it will not be the same.

Objectives • How to set up and solve dynamical, non-equilibrium, problems.

28

LECTURE 6. FORCE AND ACCELERATION

6.1

29

Forces out of balance

When forces are out of balance, accelerations occur. Consider the block on a slope again:

N

FK

θ

a W=mg

θ

Again, only forces on block are shown; the friction force is now the kinetic friction FK = µK N . The acceleration is indicated with a double arrow: it is not a force, but the result of the forces. Resolving perpendicular to and parallel to (down) the slope: N − mg cos θ = 0, mg sin θ − FK = ma. Therefore, as before, N = mg cos θ, and so FK = µK mg cos θ, and so a = g(sin θ − µK cos θ). Example 6.1. Consider the block and slope again, but let’s assume that the slope is a wedge that can move, and that there is no friction. What will the accelerations of block and wedge be? Answer. Draw diagrams with forces on each component separately: N m a2

a2 M

θ

M

a1 mg

Forces on block

θ

W’

m N N’

θ

Forces on wedge

Here it is convenient to measure the acceleration of the block ~a1

LECTURE 6. FORCE AND ACCELERATION

30

relative to the wedge. The true acceleration of the block (i.e. relative to an inertial frame) is ~aB = ~a1 + ~a2 , and this is what we must use in N2 . We resolve forces and accelerations. Wedge, horizontal: N sin θ = M a2 .

(6.1)

Wedge, vertical: not needed, as we are not interested in N 0 . Block, perpendicular to slope: N − mg cos θ = −ma2 sin θ.

(6.2)

N.B. ~a1 has no component perpendicular to the slope, so the acceleration on the right is all from ~a2 . Block, parallel to slope: mg sin θ = m(a1 − a2 cos θ).

(6.3)

Equally well we could have resolved horizontally and vertically for the block, which would give two equations different from, but exactly equivalent to, Eqs. (6.2), (6.3). Use Eq. (6.2) to eliminate N from Eq. (6.1): m(g cos θ − a2 sin θ) sin θ = M a2 , so, rearranging, mg cos θ sin θ . M + m sin2 θ Eq. (6.3) and Eq. (6.4) then tell us a1 : a2 =

mg cos2 θ sin θ a1 = g sin θ + , M + m sin2 θ which reduces to a1 =

(M + m)g sin θ . M + m sin2 θ

Check: dimensions

4;

horizontal slope

for θ = 0, a1 = 0, a2 = 0 4;

vertical slope

for θ = 90◦ , a1 = g, a2 = 0 4;

heavy wedge

for m/M → 0, a1 → g sin θ, a2 → 0 4.

(6.4)

LECTURE 6. FORCE AND ACCELERATION

6.2

31

Physics jargon

Physics problems can seem rather artificial, but the point is to try to isolate the physics of interest without the clutter of multiple effects. To this end one needs to know the meaning of a few code words, some of which we have had already, but now seems a good point to list them all. smooth:

zero friction.

light:

as in “light pulley” – zero mass, no force needed to accelerate it.

rigid:

does not bend at all

flexible:

totally bendy. Strings are almost always “flexible”.

ideal:

a string is ideal if it is light, flexible and does not stretch at all; an “ideal pulley” is light, frictionless and rigid.

uniform:

e.g. “uniform sphere” – constant density throughout.

elastic:

“elastic collision” – conserves kinetic energy (“inelastic” kinetic energy lost to heat, sound etc)

6.3

Strings and pulleys

Illustrate with an example. Example 6.2. Consider the following arrangement. The pulleys are light and frictionless; the string is ideal; the table is smooth. Table and wall are fixed. Find the accelerations of the two blocks and the tension in the string.

LECTURE 6. FORCE AND ACCELERATION

32

Wall m2

m1

Table

Answer. No force needed to turn the pulleys or accelerate the pulleys or the string ⇒ tension in string is constant: T a1

T T

m1

m2

a2

m1 g

Therefore F = ma on block 1 (downwards): m1 g − T = m1 a1 ,

(6.5)

while block 2 (leftwards): 2T = m2 a2 .

(6.6)

Two conditions, but three unknowns: need one more condition. String does not stretch ⇒ m1 moves twice as fast as m2 : a1 = 2a2 .

(6.7)

Eqs (6.5), (6.6) and (6.7) then imply: 4m1 g , m2 + 4m1 2m1 g a2 = , m2 + 4m1 m1 m2 g T = . m2 + 4m1 a1 =

Added after the lecture: the diagram below should make it clear why the bottom of the string moves twice as fast as the moving pulley. The

LECTURE 6. FORCE AND ACCELERATION numbers are just the lengths of the straight sections of string. 6

4

2 5

3

4

33

Lecture 7

Equations of motion Keypoints from Lecture 6 • If forces on an object don’t balance, it accelerates. • Always be clear about what quantities you do and don’t know in a problem: you will need one condition for every “unknown”. • Code words: smooth, light, rigid, flexible, ideal, uniform, elastic, inelastic. • Tension in ideal strings is constant. With pulleys can double, triple, etc, forces.

Objectives • How to solve equations of motion

7.1

Constant acceleration – “SUVAT”

“Equations of motion” are expressions for ~r , ~v and ~a as functions of time.

34

LECTURE 7. EQUATIONS OF MOTION

35

Consider motion under constant acceleration, a, in 1D. We have dv = a. dt Integrating: v

Z

Z dv =

u

t

Z

t

a dt = a 0

dt, 0

so dx = u + at. (7.1) dt Here u is the initial speed at t = 0 and v is the speed after t seconds. Integrating again Z Z v=

x

t

dx = x0

(u + at) dt, 0

so s = x − x0 = ut + 12 at2 ,

(7.2)

where s is the distance travelled in t seconds. If we square both sides of Eq. (7.1) we can simplify using Eq. (7.2): v 2 = u2 + 2u at + a2 t2
= u2 + 2a ut + 12 at2 | {z } s

so finally v 2 = u2 + 2as.

(7.3)

Example 7.1. In the UK’s driving theory test, the “braking distance” at 70 miles per hour is said to be 75 m. What is the acceleration? Answer. One mile = 1600 m, so u=

1600 × 70 = 31 m s−1 3600

The final speed v = 0, so using Eq. (7.3), a= or −0.65g.

v 2 − u2 0 − 312 = = −6.4 m s−2 , 2s 2 × 75

LECTURE 7. EQUATIONS OF MOTION

7.2

36

Time-dependent acceleration

Suppose now that the acceleration is a function of time, a = a(t), then we must keep a inside the integral: Z v Z t dv = v − u = a(t) dt. u

7.3

0

Position-dependent acceleration

Suppose a = a(x), i.e. it is a function of position. Then dv = a(x). dt Integrating Z

Z dv =

a(x) dt,

but we don’t know x = x(t) – that’s what we are trying to determine! Use the chain rule dv dv dx dv = = v = a(x), dt dx dt dx so we can write

Z

v

x

Z v dv =

u

or

1 2 (v − u2 ) = 2

a(x) dx, 0

Z

x

a(x) dx. 0

RHS now calculable. If we are lucky, we might be able to work out the integral, and integrate again to get x. Incidentally we can (again) relate this to change in energy equals work done, if we multiply both sides by m. See later.

LECTURE 7. EQUATIONS OF MOTION

7.4

37

Velocity-dependent acceleration

Now suppose a = a(v), then dv = a(v), dt which can be integrated as follows: Z t Z v dv = dt = t, 0 u a(v) to relate v and t. Re-arrange to give v = v(t), integrate to get x = x(t). Example 7.2. Starting from rest at t = 0, an object of mass m falls through air under gravity, experiencing a drag force F =−

mv , τ

where τ is a constant with dimensions of time. Calculate its speed, and the distance it has dropped, by time t. Answer. N2 becomes: mg −

mv dv =m . τ dt

Divide out m, rearrange, and integrate Z t Z v h iv dv dt = t = = −τ ln(g − v/τ ) , 0 0 0 g − v/τ
= −τ ln(g − v/τ ) − ln g ,   g − v/τ = −τ ln . g Re-arranging g−

v = ge−t/τ , τ


or v = gτ 1 − e−t/τ .

LECTURE 7. EQUATIONS OF MOTION

38

Tends to a maximum value of gτ , the “terminal velocity”. Setting v = dx/dt, we can integrate again Z t   h it −t/τ −t/τ x= gτ 1 − e dt = gτ t + τ e , 0 0 h i −t/τ = gτ t − τ (1 − e ) . x ∝ t2 at short times, and x ∝ t at long times.

x gτ v

0

2

1

t/τ

3

Lecture 8

Work and Energy Keypoints from Lecture 7 • Starting from an expression for the acceleration ~a, and given initial values of position ~r and ~v , it is possible, in principle, to determine ~r (t), ~v (t) and ~a(t), the “equations of motion”. • “SUVAT” is the simple case of constant a. • More generally there are three cases a = a(t), a = a(x), a = a(v), each of which requires its own handling. • (More generally still ⇒ differential equations.)

Objectives • Work, kinetic energy, conservative forces, potential energy

8.1

Work and Kinetic Energy

Consider a 1D, position-dependent force, F (x) acting on a mass m: F (x) = m 39

dv . dt

LECTURE 8. WORK AND ENERGY

40

As last lecture, we can use the chain rule dv dx dv dv = = v. dt dx dt dx Now let the mass move from position A to B, and integrate:  B Z B Z B 1 2 F (x) dx = mv dv = mv . 2 A A A

(8.1)

Defining the kinetic energy T by T = 12 mv 2 , and the work done ∆W by Z

B

∆W =

F (x) dx, A

Eq. (8.1) says that The work done equals the change in kinetic energy. If F (x) = F , a constant, then Z ∆W = F

B

dx = F (xB − xA ),

A

i.e. “work done = force × distance”, but not true in general. Work ≈ force × distance if the step is so small that F does not vary much. As distance → 0, an infinitesimal step dx, it becomes exact: dW = F dx.

8.2

3D

Now consider a general position dependent force acting on a mass m: ~ (~r ) = m d~v . F dt

LECTURE 8. WORK AND ENERGY

41

Cannot simply extend previous approach because in d~v d~v d~r = , dt d~r dt the first derivative is undefined. Instead we generalise dW = F dx. A natural extension of this to 3D is ~ · d~r . dW = Fx dx + Fy dy + Fz dz = F Using N2 d~v d~v · d~r = m · ~v dt, dt dt = m~v · d~v ,
= d 12 m~v · ~v .

dW = m

Therefore generalising the definition of kinetic energy to
T = 12 m~v · ~v = 12 mv 2 = 12 m vx2 + vy2 + vz2 , we recover work done = change in kinetic energy, but now Z B ~ · d~r . F ∆W = A

Physical interpretation: F A

θ dr

B

~ and d~r : Defining θ as the angle between F
~ · d~r = F cos θ dr, dW = F so dW = the component of force in the direction of movement times the (infinitesimal) distance moved. As a result the term on the right in Z B ~ · d~r . ∆W = F A

is called a line integral. Simply think of this as the sum of the infinitesimal contributions from each step d~r .

LECTURE 8. WORK AND ENERGY

8.3

42

Conservative forces, potential energy

The relation between work and kinetic energy is most useful in conservative force fields. “conservative” A conservative force is one for which the work done moving from one point to another is independent of the path. P

B P’

A

Z

B

A

or Z

B

A

as in energy conserving.

Z ~ · d~r = F P

Z ~ · d~r + F P

A

B

B

A

~ · d~r F

~ · d~r , F 0 P

I =

~ · d~r = 0. F

P0

~ (~r ) allows one to define a potential energy U (~r ) A conservative force F because the work done by it in moving a particle from A to B is only a function of the end points, not the path taken: Z B
~ · d~r = − U (B) − U (A) = −∆U, F ∆W = A

where U is a function to be determined. The minus sign is a convention to ensure that ∆W > 0 when moving from a high to a low potential. Since ∆T = ∆W , for conservative forces ∆T = −∆U , and so if we define the total energy E = T + U , ∆E = 0 and thus E = T + U = constant. This is the conservation of mechanical energy, which was later generalised to the conservation of energy when heat was recognised as another form of energy.

Note reversed limits on second integral. Circle on final integral indicates that the path is a closed loop.

LECTURE 8. WORK AND ENERGY

8.4

43

Deriving a force from a potential

If a potential can be defined then the corresponding force can be obtained as follows: dU , in 1D dx ˆ = −∇U, ~ ~ = − ∂U ˆı − ∂U ˆ − ∂U k F ∂x ∂y ∂z F =−

~ = where ∇

in 3D

∂ ∂ ∂
, , is the gradient operator. ∂x ∂y ∂z

~ (~r ) is conservative, i.e. not every force field Not every force field F is derived from a potential. For example, frictional forces are not conservative.

Lecture 9

More energy Keypoints from Lecture 8 • A force acting on a moving particle performs “work”, measured in Joules [J]. The work done moving from A to B is defined as Z B ~ · d~r . F ∆W = A

• ∆W = ∆T , the change in kinetic energy, where T = 21 mv 2 . • If the work done from A to B is independent of the path taken, ~ is said to be conservative and can be represented in the form F ~ = −∇U ~ . of a (scalar) potential function U , where F • In this case T + U = E, the total mechanical energy, is constant.

Objectives • Energy conservation, power.

44

LECTURE 9. MORE ENERGY

9.1

45

The work you do

Last lecture discussed work done by a conservative force, which ends up in kinetic energy of a particle. If you move a particle slowly from A to B in the same force field, by N3 you apply an equal but opposite force, so where last time we had ∆U = −∆W, when expressed in terms of the work you do, ∆WY , ∆U = ∆WY . i.e. you must do positive work to move the particle to a place of higher potential; the force does work moving the particle to a lower potential.

9.2

Gravitational potential energy r

M

dr ds

If a mass m moves d~s under gravitational attraction from a mass M , then ~ · d~s = − GM m rˆ · d~s = − GM m dr. dW = F r2 r2 As usual, infinitesimal terms of higher order are dropped. Integrating: Z

B

∆W = −GM m A

  1 1 dr = GM m − . r2 r(B) r(A)

This is only a function of the value of r at the end points, hence gravity is conservative with potential energy U (r) = −

GM m + constant. r

I use d~s because d~r causes confusion here because of the use of dr to represent the change in distance between the masses; i.e. dr 6= |d~s|.

LECTURE 9. MORE ENERGY

46

The constant is not uniquely defined, but usually we take U (∞) = 0, so the constant = 0 and U (r) = −

GM m . r

Often we use the “gravitational potential” to mean the potential energy per unit mass GM U (r) =− . u(r) = m r Sometimes this is written φ(r).

9.3

Using energy conservation

Energy conservation is useful when one needs to know the speed at a particular place but not time. Example 9.1. An object is released from rest at a height h = 3000 km above the surface of Earth. Ignoring air resistance, at what speed will it hit the ground? Compare your answer with assuming potential = mgh. Answer. Gravity is conservative so T + U = E = constant, 1 GME m T = mv 2 and U = − , 2 r

with

where r is the distance of the object from the centre of the Earth. Conservation of energy means: −

GME m 1 2 GME m = mv − , RE + h 2 RE

(T = 0 at the start). Therefore   1 1 2 v = 2GME − , RE RE + h 2GME h . = RE (RE + h)

LECTURE 9. MORE ENERGY

47

Putting G = 6.67 × 10−11 N m2 kg−2 , ME = 5.97 × 1024 kg, RE = 6.37 × 106 m and h = 3 × 106 m, find v = 6.3 km s−1 . Using 21 mv 2 = mgh (inaccurate, since h is not small compared with RE )   2GM E h v 2 = 2gh = 2 RE one finds v = 7.7 km s−1 . mgh over-estimates the force and hence the kinetic energy. Example 9.2. Escape velocity: What speed is needed to escape the gravitational pull of an object of mass M and radius R? Answer. The speed must be enough for T > 0 at r = ∞ where U = 0, i.e. E > 0. Therefore, by conservation, E = T + U > 0 at the start as well. r GM m 2GM 2 1 , or v > . mv > 2 R R Escape velocity from Earth = 11.2 km s−1 , from the Sun = 618 km s−1 . This is a fundamental constraint upon planetary atmospheres.

9.4

Power

The amount of work performed per unit time is the power, and is measured in Joules per second or Watts. ~ · d~r , then Since dW = F P =

dW ~ · d~r = F ~ · ~v . =F dt dt

Power is a scalar Example 9.3. The “Lorentz force” is the force that acts on a charged particle in an electromagnetic field, and is given by ~ = q(E ~ + ~v × B), ~ F

LECTURE 9. MORE ENERGY

48

~ and B ~ where q is the particle’s electric charge, ~v is its velocity and E are the electric and magnetic fields (and × is the vector, or “cross”, product). What is the rate of work on the particle? Answer. ~ · ~v = q(E ~ · ~v + ~v × B ~ · ~v ) = q E ~ · ~v , F i.e. only the electric field performs work on the particle. There was not enough time in the lecture to go through the next example, but you may still find it interesting. Example 9.4. The Dinorwig pumped storage power station in Wales stores water in one lake 500 m higher than another in order to provide a source of rapidly accessible power. When running at its full power of 1.7 GW, what is the flow rate of water? Answer. If the water flows at a rate of m ˙ = dm/dt, measured in mass per unit time, then the energy released per unit time (= power) falling through a height h is given by P = mgh. ˙ Therefore 1.7 × 109 P m ˙ = = = 3.47 × 105 kg s−1 , gh 9.81 × 500 or 347 tonnes/sec. In reality it is higher still – 390 tonnes/second because it is not 100% efficient.

Lecture 10

Rockets and circular motion Keypoints from Lecture 9 • Gravity is a conservative force. The gravitational potential energy of two masses m1 and m2 separated by a distance r is given by Gm1 m2 U =− , r a scalar equation. • If position but not time is defined in a problem, then energy conservation might be useful. • The speed needed to give an object a positive total energy E = T + U > 0 in a gravitational field is known as its “escape velocity”. • Power, measured in Watts, is the rate of performing work.

Objectives • The mechanics of rockets (changing mass!) • Making a start on circular motion. 49

LECTURE 10. ROCKETS AND CIRCULAR MOTION

10.1

50

Rockets

Rockets burn fuel which is ejected at high speed in one direction while the rocket is pushed in the opposite direction.

10.1.1

Rockets in space

Assuming that the rocket and its exhaust form an isolated system, then there is no overall change in momentum. Consider a time t when the mass of a rocket is m and its speed v, followed by t + dt when its mass is m + dm, its speed v + dv having expelled a mass −dm at speed −ve relative to the rocket. ve is the exhaust speed, determined by the combustion energy of the fuel. Convention! dm Conserving momentum: mv = (m + dm)(v + dv) + (−dm)(v − ve ).

will, of course, be a negative quantity.

Expanding the terms in brackets mv = mv + m dv + v dm + dm dv − v dm + ve dm. Cancelling terms and dropping the dm dv second-order term we have m dv = −ve dm. Integrating: Z

v

Z

m

dv = −ve v0

m0

and thus v − v0 = ∆v = ve ln

dm , m m  0

m

,

where m0 is the initial mass.

10.1.2

Infinitesimals again: dm → 0 and dv → 0.

Rockets at launch

When launched from Earth, the rocket and its exhaust are acted upon by gravity. During time dt, the force of gravity mg changes the overall momentum by −mg dt and thus m dv = −ve dm − mg dt.

This is a classic equation of rocketry.

LECTURE 10. ROCKETS AND CIRCULAR MOTION

51

Integrating as before (taking v0 = 0 and assuming g is constant) m  0 − gt. v = ve ln m The rocket’s acceleration at launch is a=

dv ve dm =− − g. dt m dt

The total thrust is F = ma + mg = −ve

dm . dt

Example 10.1. At launch the Saturn V rocket of the Apollo missions had a mass of 3000 t. Burning 15 t of fuel per second, the thrust was 35 MN. Calculate the exhaust speed and, assuming that both it and fuel burn rate remain constant, the “G-force” on the astronauts as a function of time, (a + g)/g, and the rocket speed, at t = 100 s. Answer. We know that m0 = 3.0 × 106 kg, dm = −1.5 × 104 kg s−1 , dt F = 3.5 × 107 N. Therefore F 3.5 × 107 ve = = = 2333 m s−1 . 4 −dm/dt 1.5 × 10 The G-force experienced by the astronauts is a+g F 3.5 × 107 /9.81 1.19 = = = = , g mg 3.0 × 106 − (1.5 × 104 t) 1 − t/200 with t measured in seconds. This looks like:

LECTURE 10. ROCKETS AND CIRCULAR MOTION

52

5

G-force

4 3 2 1 0 0

50

100

150

t / sec

The speed after 100 s is m  0 v = ve ln − gt, m   3000 = 2333 ln − 9.81 × 100, 3000 − 15 × 100 = 636 m s−1 .

10.2

Circular motion

Motion at constant speed in a circle is the simplest case of acceleration as the result of a change in direction rather than speed. Consider a particle at position ~r , rotating about an axis perpendicular to ~r .

r (t+δ t) O

δθ r (t)

δr

The central engine was cut at 150 secs to prevent too high an acceleration. Almost twice the speed of sound.

LECTURE 10. ROCKETS AND CIRCULAR MOTION

53

In time δt, vector changes by δ~r , while staying constant in magnitude: r = |~r (t)| = |~r (t + δt)|. The velocity is given by δ~r . δt→0 δt

~v = lim The magnitude of δ~r is



δθ |δ~r | = 2r sin 2

 .

For small x  1, x3 x5 + − . . . ≈ x, sin x = x − 3! 5! so |δ~r | ≈ r δθ, and therefore

dθ = rω, dt where ω = dθ/dt is the angular velocity measured in radians per second. For circular motion, ~v is always perpendicular to ~r . v = |~v | = r

NB The above argument can be applied in a similar way to any vector ~ rotating about a perpendicular axis. For example, for the rate of A change of velocity ~v , we have a = vω, where the acceleration vector ~a is directed perpendicular to ~v , inwards towards the centre of the circle.

NB In this case |δ~r | = 6 δr because the magnitude of ~r is constant.

Lecture 11

More circular motion Keypoints from Lecture 10 • If a rocket ejects mass −dm at exhaust speed ve , conservation of momentum gives a change in velocity dv m  0 m dv = −ve dm ⇒ v − v0 = ∆v = ve ln m ~ of constant magnitude A = A ~ , and per• A rotating vector A pendicular to the axis of rotation, changes with magnitude dA ~ = ωA, dt ω=

dθ is the “angular frequency” or “angular velocity”. dt

~ ~ itself. • The rate of change dA/dt is perpendicular to A • For the position vector ~r we get the important relations v = ωr for the tangential speed and a = ωv for the acceleration directed in towards the centre.

Objectives • Centripetal acceleration, circular orbits 54

LECTURE 11. MORE CIRCULAR MOTION

11.1

55

Centripetal acceleration

Just like the radius vector, the velocity vector of a particle moving in a circle at constant speed sweeps out a circle: 3 2 2

1

1

δv

3

r O

δθ v(t+δ t) δθ

v(t)

Thus just as v = ωr, so a = ωv, and ~a is perpendicular to ~v . It points towards O, the centre of the circle. We have three equivalent formulae: v2 a = ωv = ω r = . r 2

These give the centripetal acceleration of an object moving in a circle of radius r, at speed v or, equivalently, with angular velocity ω. Example 11.1. Draw the forces and acceleration on a car driving at constant speed in a circle. Obtain an expression for the maximum speed with no skidding.

Answer. The car moves in a circle because of the centripetal force provided by friction between tyre and track: Top view

v F a

Rear view a F mg

F = ma, so mv 2 F = < µS mg, r

LECTURE 11. MORE CIRCULAR MOTION

56

where µS = coefficient of static friction, so √ v < µS gr. e.g. µS = 0.8, R = 50 m, then v < 19.9 m s−1 ≈ 45 mph. Example 11.2. Calculate the centripetal acceleration of a point on Earth’s equator. Answer. Here a = ω 2 r is the easiest to use because we know Earth’s spin period.Using ω = 2π/P , P = 1 day, and r = 6370 km we have 2  2π × 6.37 × 106 = 0.0337 m s−2 = 0.0034 g. a = ω2r = 24 × 3600

11.2

NB No “centrifugal” outwardspointing force!

Angular velocity vector

~. The equation v = ωr can be made vectorial by defining ω ~ is a vector of length ω pointing The angular velocity vector ω along the axis of rotation. ~ is set by the right-hand rule: if the fingers of your The direction of ω right-hand point in the direction of rotation, your thumb points in the ~. direction of ω Q: what is the direction of ω for the Earth?

With this definition ~v = ω ~ × ~r , where ~r is the position vector measured from a point on the axis.

r sin φ ω

P

r φ

LECTURE 11. MORE CIRCULAR MOTION

57

~ at speed Let’s check. The point P spins round the axis marked by ω ωr sin φ into the page, so as claimed ~v = ω ~ × ~r .

11.3

Circular orbits

Centripetal acceleration is key to understanding orbits. Consider a satellite in orbit around Earth: V a F R

Earth

The force acting is gravity which has magnitude F =

GME m , R2

which causes centripetal acceleration of the satellite of a = V 2 /R so GME m V2 F = =m . R2 R Therefore the speed is given by V2 =

GME . R

Using V = ωR, we also have ω2 =

GME 4π 2 = , P2 R3

LECTURE 11. MORE CIRCULAR MOTION

58

where P is the orbital period. So, P 2 ∝ R3 : we will see later that this is Kepler’s third law. For low Earth orbit, R = RE + 200 km = 6570 km ME = 5.97 × 1024 kg G = 6.67 × 10−11 N m2 kg−2 ⇒ V = 7.8 km s−1 ⇒ P = 88.4 min Example 11.3. In a classical model of a hydrogen atom, the electron (mass me = 9.109 × 10−31 kg) orbits the nucleus (a proton) in a circle of radius r = 5.29 × 10−11 m, attracted by the electric “Coulomb” force which is given in this case by F =

e2 , 4π0 r2

where e = 1.60 × 10−19 C is the magnitude of the charge on the electron, and 0 = 8.85 × 10−12 F m−1 is the “permittivity of free space”. Calculate the orbital speed of the electron. Answer. Using N2 and a = v 2 /r, we have s 2 2 v e e2 F = me = , so v = . r 4π0 r2 4π0 me r Putting in the numbers gives v = 2.19 × 106 m s−1 < 0.01 c. This is a non-relativistic speed. However for heavier atoms, when the nuclear charge is of order 100, e2 → 100e2 , relativistic effects start to appear.

We should mention that this model is incorrect! Quantum physics is needed

Lecture 12

Torque and Angular Momentum Keypoints from Lecture 11 • Just like the position vector ~r , the velocity vector ~v of a particle moving at constant speed in a circle sweeps around at angular velocity ω. • The resultant centripetal acceleration a is given by a = ωv = ω 2 r =

v2 . r

• Circular motion can be described vectorially using the angular ~ which rotates around an axis ~ . Any vector A velocity vector ω in this way satisfies ~ dA ~ ~ × A. =ω dt • Gravitational force causes the centripetal acceleration of satellites and planets in circular orbits whereby V2 =

59

GM . R

LECTURE 12. TORQUE AND ANGULAR MOMENTUM

60

Objectives • Torque and angular momentum

12.1

Moments

Consider a simple (light) lever in balance: W1 + W2 r1

r2 P

W1

W2

y x

As Archimedes proposed, to find the correct ratio of W1 and W2 we need to balance moments: r1 W1 = r2 W2 . Is this a new principle in addition to N1 , N2 and N3 ? In fact no. To see this, first generalise “moments” by defining the ~ acting at position vector ~r to be ~ exerted by a force F torque τ ~. ~ = ~r × F τ Physical interpretation:

O

F θ

r θ

r sin θ

LECTURE 12. TORQUE AND ANGULAR MOMENTUM

61

The moment of F around O has magnitude = F r sin θ. (The closest approach distance r sin θ is sometimes known as the “lever arm”.) ~ is the direction The only sensible direction to associate with ~r and F ~. ~ = ~r × F perpendicular to both. All properties are satisfied by τ NB Torques must always be measured with respect to a specific point – choose a different point, get a different torque. From N2 ~ = d~p = m d~v , F dt dt

so

~ = m~r × τ Now

d~v . dt

d d~r d~v d~v (~r × ~v ) = × ~v +~r × = ~r × . dt dt dt dt | {z } zero

Therefore we can write d~v d d = m (~r × ~v ) = (~r × ~p) . dt dt dt The quantity ~r × ~p is called the “moment of momentum”, or more usually the angular momentum. Defining ~ = m~r × τ

~ = ~r × ~p, L we have ~ = τ

~ dL . dt

Torque equals the rate of change of angular momentum.

12.2

Systems of particles

As before, consider a system of particles with internal and external forces. The force on particle i was given by X ~ ~ ji = d~pi . Fi + F dt j6=i

Since d~r /dt = ~v , and the cross product of a vector with itself is zero.

LECTURE 12. TORQUE AND ANGULAR MOMENTUM

62

The torque around a point O is ~i+ ~r i × F

X j6=i

~ ~ ji = dLi . ~r i × F dt

The total torque on the particles is therefore ~ = τ

X

~i+ ~r i × F

i

XX i

~ ji = ~r i × F

X dL ~i i

j6=i

dt

.

For the internal forces, the double sum contains both ij and ji terms:
~ ji + ~r j × F ~ ij = ~r i − ~r j × F ~ ji because of N3 . ~r i × F ~ ji , which act along the vector between particles For “central” forces F ~r i − ~r j , this is zero. So this just leaves the external forces. Again, the

1

F 21

F 12

2

vectors are parallel. See diagram

O so ~ = τ

X i

~i= d ~r i × F dt

! X i

~i L

=

~ dL . dt

Thus The total torque from external forces on a composite body equals the rate of change of its total angular momentum.

12.3

Back to levers

A body in equilibrium has zero torque about any point:

It is your choice: make things as simple as possible. In this case P looks good.

LECTURE 12. TORQUE AND ANGULAR MOMENTUM

63

W1 + W2 r1

r2 P

W1

W2

y x

In vectors

~r 1 = −r1ˆı, ~r 2 = +r2ˆı, ~ 1 = −W1ˆ, F ~ 2 = −W2ˆ, F

so the total torque is X ~ i = (r1 W1 − r2 W2 ) ˆı × ˆ = ~0. ~ = ~r i × F τ i

ˆ is not zero, and therefore Now ˆı × ˆ = k r1 W1 = r2 W2 , the standard equation for balancing a lever, obtained by “taking moments about P ”. Conclude: Archimedes’ lever equation is contained within Newton’s ~,L ~ = ~r × ~p, and τ ~ ~ = ~r × F ~ = dL/dt, Laws, and in τ we have a much more powerful version of it.

12.4

Centre of Gravity

We return to discuss centre of gravity. Consider a rigid body composed of several masses mi at positions ~r i under the influence of gravity. The ~ i = wi gˆ , where the unit vector gˆ points force on each mass will be f “down” and wi = mi g is the weight. Consider the torque due to all these weights about some arbitrary point (which, for convenience, we choose to be the origin here). The position of the centre of gravity,

LECTURE 12. TORQUE AND ANGULAR MOMENTUM

64

~r cg , is defined such that, if the individual forces were replaced by the
~ =P f ~ i = P wi gˆ acting at ~r cg , then the torque on total force f i i the body would be the same. Thus X ~i ~ = ~r i × f ~r cg × f X i

i  X  wi ~r cg × gˆ = wi~r i × gˆ . i

This equation should apply no matter what the orientation of the body, or equivalently, whatever the orientation of gˆ , so P X  X wi~r i wi ~r cg = wi~r i ⇒ ~r cg = Pi . w i i i i If g is the same everywhere, so wi = mi g, the factors of g cancel and this is the same equation as the one defining the centre of mass ~r cm . If the magnitude of g varies from place to place, ~r cg and ~r cm will not coincide. For example, g varies slightly with height above the earth, as a result of which (“University Physics”, p347) the centre of gravity of the Petronas towers in Kuala Lumpur lies about 2 cm below the centre of mass. In the general case of variable gravity, the problem has no consistent solution. It is relatively easy to show that the standard method for locating the centre of gravity of a body by suspending it from several points in turn, is consistent with the above definition.

Actually, if this is the case, we can’t simply drop the dependence on gˆ ; things get a little more complicated.

Lecture 13

Moments of Inertia Keypoints from Lecture 12 ~ around a point O is called the torque • The moment of a force F ~, ~ = ~r × F τ where ~r is the position vector of the point at which the force acts relative to O. • The moment of a particle’s momentum ~p about O is called its ~ and is given by angular momentum L ~ = ~r × ~p. L • Rate of change of angular momentum equals applied torque ~ dL . dt This applies to particles and composite bodies. ~ = τ

• If the weight of a body is concentrated at the centre of gravity, the torque will be unchanged.

Objectives • Moments of inertia 65

LECTURE 13. MOMENTS OF INERTIA

13.1

66

Angular momentum of a particle moving in a circle

The angular momentum about a point O of a particle of momentum ~p and position vector ~r relative to O is given by ~ = ~r × ~p. L This is general. Now consider the specific case of circular motion around O at angular velocity ω: v m r O

~ around O is a vector parallel With ~p = m~v , the angular momentum L ~ (i.e. along the axis of rotation, out of the page) with magnitude NB This holds to ω 2

only about O.

L = mrv = mr ω. The kinetic energy of the particle 1 1 1 T = mv 2 = m(rω)2 = mr2 ω 2 . 2 2 2 Defining a new quantity I = mr2 , the “moment of inertia”, we have L = Iω, 1 T = Iω 2 . 2

13.2

Rigid-body rotation

The equation for kinetic energy is easily generalised to the rotation of rigid bodies. However, everything must be defined relative to the axis of rotation: Here r (or ri ) is the perpendicular distance of each mass element (or particle) from the axis.

I is the angular counterpart of mass m, but unlike m, I is usually direction dependent.

LECTURE 13. MOMENTS OF INERTIA

67

ω

r dm

Summing over all particles we then have T =

1X 2

mi vi2 =

i

1 2

! X

mi ri2 ω 2 ,

i

so we still have T = 21 Iω 2 if I=

X

mi ri2 =

Z

r2 dm =

Z

r2 ρ dV,

i

where we take the continuum limit on the right (ρ is the density per unit volume). This comes with a warning: if the axis is chosen in a different direction, in general the value of I will be different. This complicates things, and we will not go into those complications in this module. I is not really a ~ and ω ~ is more compliFor these reasons, the connection between L cated, in general, than the analogous link between ~p and ~v . In general ~ and ω ~ are not parallel. In this module we will only discuss cases for L ~ ~ are parallel: which L and ω ~ = Iω ~, L with I given as before. This is true for all directions through spheres or cubes, symmetry axes of cuboids, cylinders and discs. In the cases we shall encounter ~ dL d~ ω ~ = τ =I , dt dt where d~ ω /dt is the angular acceleration. This is the angular analogue ~ = m~a. of F

scalar: it is a second-rank tensor. This will become clear in later years.

LECTURE 13. MOMENTS OF INERTIA

13.3

68

Example moments of inertia

Example 13.1. Calculate the moment of inertia of a thin ring of radius R, mass M about an axis through its centre, perpendicular to the plane of the ring Answer. Looking along the rotation axis: R δθ θ

With all elements of mass at the same distance R, the answer is obviously I = M R2 , but let’s do it more formally as well. Assume a mass per unit length along the ring of σ, then a small section dθ has mass dm = σR dθ, and we have Z Z 2π I = r2 dm = σR3 dθ = 2πσR3 . 0

Then, since M = 2πRσ, we have I = M R2 . Example 13.2. Calculate the moment of inertia of a uniform circular disc of radius R, mass M about an axis through its centre, perpendicular to its face. e.g. a coin. Answer. Split into thin rings, integrate. R

r+dr r

Assume mass per unit area σ, then mass of a thin ring radius r, radial thickness dr is dm = 2πrσ dr.

LECTURE 13. MOMENTS OF INERTIA

69

It is r from the axis at all points, hence Z Z R 1 I = r2 dm = 2πσr3 dr = 2πσR4 . 4 0 The mass of the disc M = πR2 σ, and therefore 1 I = M R2 . 2 Can calculate the following moments of inertia: • uniform sphere, radius R, mass M , about an axis through its centre: I = 25 M R2 ; • thin rod, length L, mass M , about a perpendicular axis through 1 its centre: I = 12 M L2 ; • the same rod, but about a perpendicular axis through one end: I = 13 M L2 . Test your understanding by doing these yourself!

Lecture 14

Angular momentum in use Keypoints from Lecture 13 • The kinetic energy of an object in rigid-body rotation is 1 T = Iω 2 , 2 where I, the moment of inertia, is given by Z I = r2 dm, where r is the perpendicular distance of element dm from the rotation axis. • The angular momentum of an object in rigid-body rotation can (often, and always in this module) be written as ~ = Iω ~. L • Some example moments of inertia were given.

Objectives • Examples of angular momentum 70

LECTURE 14. ANGULAR MOMENTUM IN USE

14.1

71

Cylinder rolling down a slope ω N

FS

v a mg θ

A cylinder of radius r, mass m and moment of inertia I rolls without slipping down a slope which makes an angle θ to the horizontal. What is its acceleration a? Because of torques, must take care over where the forces act. The weight acts through the centre of gravity, equivalent to the centre of mass in a uniform gravitational field. ~ =m~a”: Unknowns: N , FS , a, dω/dt, [v, ω]. First, “F perpendicular to slope: N − mg cos θ = 0 It soon becomes obvious that we don’t need this equation, since N doesn’t appear anywhere else. parallel to slope: mg sin θ − FS = ma. ~ Next, “~τ = dL/dt”, around the axis of the cylinder with L = Iω: rFS = I

dω . dt

[NB Both N and mg act through axis of cylinder and add no torque.] The final relation comes from the no-slip condition v = rω, or a=r

dω . dt

These last two equations combine to give FS =

I a. r2

LECTURE 14. ANGULAR MOMENTUM IN USE

72

Inserting into the equation for acceleration down the slope: mg sin θ − so a=

I a = ma, r2

g sin θ . 1 + I/mr2

Clearly, the ratio I/mr2 plays a major role. mass concentrated on axis: I = 0, a = g sin θ, like frictionless mass. uniform cylinder: I = 12 mr2 , a = 23 g sin θ, not so fast. hollow tube:

14.2

I = mr2 , a = 21 g sin θ, slower still.

Conservation of angular momentum

In some circumstances, no external torques act and then ~ dL = 0, dt ~ is constant and so L In the absence of external torques, angular momentum is conserved. Example 14.1. During a Type II supernova the Earth-sized core (R = 6370 km) of a star collapses to a neutron star of radius R = 10 km in a few seconds. If the initial spin period of the core was P = 1000 s, calculate the final spin period. Answer. Assuming uniform spheres, L = 25 M R2 ω. If L is conserved during the collapse, then since ω = 2π/P , we must have R2 /P = constant, so P ∝ R2 . Hence  Pf =

10 6370

2

× 1000 = 2.5 × 10−3 s.

LECTURE 14. ANGULAR MOMENTUM IN USE

73

Conservation of angular momentum is important in formation of the solar system, the Galaxy, ice skaters in a spin, divers, cats falling, spacecraft, helicopters, . . .

14.3

This is why “pulsars” spin so rapidly – the current record-holder spins 716 times/second!

Gyroscopic Precession (approx.) L

Ω

dL=(Ω dt)L sin θ Ω dt L sin θ

ω

top view

side view

Ω

r θ

mg θ

L

O r sin θ Consider a tilted gyroscope, spinning rapidly about its axis at angular velocity ω. The centre of the wheel is at ~r from the base. It is subject to a torque from gravity. The torque ~r × (m~g ) points into the page, ~ ~ so the so dL/dt also points into the page. This is perpendicular to L, ~ changes. direction (not the magnitude) of L dL ~ dL ~ ~ = m~r × ~g ⇒ = mgr sin θ =τ dt dt ~ the gyroscope “precesses” around a vertical Given the direction of L, ~ sweeping out a cone. Only the horizontal component of L ~ axis, with L is changing, the vertical component remains constant. If it precesses ~ then at an angular velocity Ω dL ~ dL ~ ~ ~ = Ω × L ⇒ = ΩL sin θ dt dt

You need to imagine the tip ~ tracing a of L circle about the vertical axis. Since it lies at angle θ to the vertical, the radius of the circle is L sin θ.

LECTURE 14. ANGULAR MOMENTUM IN USE

74

If the gyroscope has moment of inertia I, then L = Iω where ω is its spin rate. Hence mgr sin θ = Ω Iω sin θ. Therefore

mgr . Iω NB Independent of θ. The derivation becomes inaccurate at lower spin ~ to point precisely along the axis rates, when we can no longer take L of the gyroscope. Precession rate: Ω =

The Earth’s axis precesses once every 26,000 years because of torques from the Sun and Moon. Called “precession of the equinoxes” and visible through a slow alteration in the path the Sun traces out through the constellations during the year.

Lecture 15

Angular work, power Keypoints from Lecture 14 • Torques as well as forces must be taken into account when angular acceleration occurs. • In the absence of external torque, angular momentum is conserved. • A spinning object under a torque of fixed orientation relative to its spin axis exhibits precession.

Objectives • Angular work and power

15.1

Angular work

Consider applying a force to spin up a rigid body, rotating around a fixed axis:

75

LECTURE 15. ANGULAR WORK, POWER

76

F dθ r

Work done dW = F (r dθ) = (F r) dθ = τ dθ. Then using τ = dL/dt and L = Iω:   dω dω dL 1 2 dθ = I dθ = I ωdt = Iω dω = d Iω . dW = dt dt dt 2 So dW = dT , where now the rotational kinetic energy 1 T = Iω 2 . 2

15.2

General work on a rigid body

The work done by external forces on a rigid body is given by X ~ i · d~r i . F dW = i

For rigid bodies, all internal work terms cancel. Also for rigid bodies d~r i can be written as the sum of the change in the position of the centre of mass plus a rotation around an axis through the centre of ~ dt. mass. Define the vector d~θ = ω d~r i = d~r cm + d~θ × (~r i − ~r cm ). Therefore X X   ~ ~ i · d~θ × (~r i − ~r cm ) dW = F i · d~r cm + F i

~ · d~r cm + =F

i

X

  ~ i (scalar triple product) d~θ · (~r i − ~r cm ) × F

i

~ · d~r cm + τ ~ cm · d~θ. =F

LECTURE 15. ANGULAR WORK, POWER

77

Then we can set ~ = m d~v cm F dt

~ cm = I and τ

d~ ω cm dt

~ cm is the angular where ~v cm is the velocity of the centre of mass and ω velocity about an axis through the centre of mass. Using these we can show as before that dW = dT where 1 2 1 2 T = mvcm + Iωcm , 2 2 Hence the kinetic energy of a rigid body splits into a translational part and a rotational part. Recall cylinder problem of last lecture.

15.3

Angular power

Power generalises to P =

dW ~ · ~v cm + τ ~ cm · ω ~ cm , =F dt

the second term is the angular part. Example 15.1. Due to tidal torques from the Moon, the length of Earth’s day is getting longer at a rate of about 1 ms per century. Calculate the torque acting on Earth, the power being lost from Earth’s spin, and the power going into the Moon’s orbit. Answer. We have Earth’s rotation period P = 24 h = 86 400 s and so ω = 2π/P = 7.27 × 10−5 s−1 . 0.001 dP = = 3.2 × 10−13 dt 100 × 365.25 × 86400 and

1 dω 1 dP dω ω dP =− ⇒ =− ω dt P dt dt P dt and we will ignore the negative sign. τ=

dL dω =I dt dt

and I = 25 M R2

LECTURE 15. ANGULAR WORK, POWER

78

with M = 5.97 × 1024 kg, R = 6.37 × 106 m. Putting it together τ=

2 5

7.27 × 10−5 × 3.2 × 10−13 86400 16 or 2.6 × 10 N m.

× 5.97 × 1024 × 6.372 × 1012 ×

= 2.6 × 1016 kg m2 s−2

Power lost from Earth = τ ω = 1.9 × 1012 W. For the power gained by the Moon, we multiply the same torque (angular momentum conservation) by the angular frequency of the Moon’s orbit, which is roughly Ω = ω/29 if we take the period to be 29 days. Thus Power gained by Moon = τ Ω = 6.5 × 1010 W. The discrepancy in power is lost as heat in the oceans.

About 20× UK’s generating capacity

Lecture 16

Collisions Keypoints from Lecture 15 • The kinetic energy of a rigid object has components from both movement of, and spin around, the centre of mass: 1 2 1 2 T = mvcm + Iωcm . 2 2 ~ acting on an object spinning with angular velocity • A torque τ ~ does work at a rate τ ~ ·ω ~ (just τ ω if parallel). ω

Objectives • Collisions in 1D and 3D • Centre of mass frame.

16.1

Collisions in 1D

Particle collisions are the fundamental way in which we probe matter, e.g. Rutherford’s discovery of the atomic nucleus. Details depend upon the forces involved, but momentum and energy conservation lead to 79

LECTURE 16. COLLISIONS

80

important governing principles. This makes collision problems easier to solve, generally, than equations of motion. Consider a particle of mass m1 and speed u1 , which collides from the left with a particle of mass m2 , speed u2 < u1 . What are their final speeds, v1 and v2 ?

Before

During

m1

m2

u1

u2

m1 m2

After m1

m2

v1

v2

Assumptions: 1. During the collision, total momentum is unchanged. This may either be because there are no external forces, or because the collision occurs so quickly that any impulses due to the external forces are vanishingly small. Remember, impulse = force × time = change in momentum. 2. Potential energy is irrelevant. This may be because the interactions are short-ranged, and only apply “during” collisions, not before or after. In an elastic collision, kinetic energy is conserved. Assuming that the collision is elastic, then: After Before 2 2 2 × Kinetic energy: m1 v1 + m2 v2 = m1 u21 + m2 u22 Momentum: m1 v1 + m2 v2 = m1 u1 + m2 u2 Here is one way to solve these equations. Multiply the energy equation by m1 + m2 and square both sides of the momentum equation: energy → m21 v12 + m1 m2 v12 + m1 m2 v22 + m22 v22 = . . . momentum → m21 v12 + m22 v22 + 2m1 m2 v1 v2 = . . . (for the right hand sides, just replace v by u). Subtract, and note that all the surviving terms have a factor m1 m2 which can be cancelled out v12 + v22 − 2v1 v2 = u21 + u22 − 2u1 u2 ⇒ (v2 − v1 )2 = (u2 − u1 )2

LECTURE 16. COLLISIONS

81

Discounting the uninteresting (no-collision) case which leaves the relative velocity of the two particles unchanged, we end up with (v2 − v1 ) = −(u2 − u1 ) meaning that the relative velocity is reversed by the collision. It is then simple to solve for v1 and v2 , using the momentum equation. 

   m1 − m2 2m2 v1 = u1 + u2 m1 + m2 m1 + m2     m2 − m1 2m1 u1 + u2 v2 = m1 + m2 m1 + m2 Suppose that particle 2 is a target, initially at rest, u2 = 0. Then just the first terms on the right are non-zero. equal masses: m1 = m2 , giving v1 = 0 and v2 = u1 , so the particles simply exchange velocities. heavy target: (m2 /m1 ) → ∞, giving v1 = −u1 , v2 = 0, incident particle rebounds. heavy projectile: (m2 /m1 ) → 0, giving v1 = u1 , v2 = 2u1 , incident particle unperturbed, target particle goes faster. Suppose instead that the centre of mass is at rest. Then both the initial and final momenta are zero, so m1 u1 + m2 u2 = 0,

m1 v1 + m2 v2 = 0,

u1 + v1 = u2 + v2

and from this it is easy to see m1 v1 = −m1 u1 = m2 u2 = −m2 v2 i.e. all the particle momenta in this frame have equal magnitude, just the signs are altered on collision. We can always transform the problem into this frame by subtracting ucm =

m1 v1 + m2 v2 m1 u1 + m2 u2 = = vcm m1 + m2 m1 + m2

from the initial velocities, doing the calculation, and then adding it back on again.

LECTURE 16. COLLISIONS

16.2

82

Collisions in 3D

Collisions in 3D are handled similarly, using vectors.

During

m1

m1 m2

v1

Before u1 m1

u2 m2

n^

m2 v2 After

~ 1 + m2 u ~ 2 = m1~v 1 + m2~v 2 Momentum conservation: m1 u Energy conservation: m1 u21 + m2 u22 = m1 v12 + m2 v22 . ~ 1, u ~ 2, Together, these make four equations. However, if we are given u we have six unknowns: ~v 1 , ~v 2 . For a complete solution, we need two more pieces of information. Typically these come from the geometry of the collision. For instance, if the particles are spherical, we can take the force between them (and hence the resulting impulse) to lie along ˆ but not the the line joining their centres. This fixes the direction n, magnitude, of the change in momentum of each particle. In any case a similar analysis to the 1D case results in 2 ~v 1 − ~v 2 2 = u ~1 −u ~ 2 The relative speed of the two particles is not changed by the collision, just the direction. We can take this a little further by realizing that the components of ˆ are not affected by the collision, while the velocity perpendicular to n ˆ obey essentially the same collision dynamics components parallel to n as the 1D case analyzed in the last section. In the centre-of-mass frame, things are again simpler. The total momentum is zero, so ~ 1 = −m2 u ~2 m1 u

m1~v 1 = −m2~v 2

and the magnitudes of the momenta are all equal m1 u1 = m2 u2 = m1 v1 = m2 v2 .

LECTURE 16. COLLISIONS

83

This implies that each particle has the same speed before and after the collision, in this frame, but the directions alter. The incoming momenta are equal and opposite; the outgoing momenta are also equal and opposite; the momenta are simply rotated through the same angle.

m2v 2 m2u2 m1u1 m1v1

Lecture 17

Orbits under central forces Keypoints from Lecture 16 • Can always assume momentum conservation in collisions. • Only kinetic energy needs to be considered once particles are well separated; in elastic collisions it is conserved. • The effect of collisions is sometimes easiest to understand in the centre of mass (CoM) frame which moves with velocity ~ cm = u

m1 u~1 + m2 u~2 . m1 + m2

• In the CoM frame, the speeds of each particle are the same before and after elastic collisions (directions change in general).

Objectives • Rutherford scattering and inelastic collisions • The effective potential and orbits.

84

LECTURE 17. ORBITS UNDER CENTRAL FORCES

17.1

More on Collisions

17.1.1

Rutherford scattering

85

Rutherford scattered α-particles (helium nuclei, m1 = 4 u) through gold foil (gold nuclei, m2 = 197 u). One in 8000 suffered almost 180◦ deflection. Rutherford deduced the existence of small “nuclei” containing most of the atomic mass. m1 v1 alpha particle

θ

u1

nucleus

m2 v2

Suppose we can measure u1 , v1 and the deflection angle θ, and that ~ 2 = 0. We would like to know: the gold nucleus is initially at rest, u what is m2 ? In the laboratory frame, momentum conservation gives 2 ~ 1 − ~v 1 m2~v 2 = m1 (~ u1 − ~v 1 ) ⇒ m22 v22 = m21 u
Combining this with energy conservation m2 v22 = m1 u21 − v12 gives 2 u ~ 1 − ~v 1 m2 u21 + v12 − 2u1 v1 cos θ = = m1 u21 − v12 u21 − v12 Really large deflection angles (with cos θ → −1) tend to be associated with a large mass ratio m2 /m1 , which, at the time, was a big surprise.

17.1.2

Inelastic collisions

So far we have assumed that the kinetic energy is conserved, but this is not always the case. The kind of analysis seen above can be repeated, keeping the momentum conservation law, but allowing a certain fraction of the available kinetic energy to be lost (e.g. dissipated as heat).

LECTURE 17. ORBITS UNDER CENTRAL FORCES

86

Not all of the energy can be lost. The energy associated with centre-ofmass motion must still be present after the collision. Only the energy associated with relative motion can be dissipated. The most extreme example is the completely inelastic collision. Viewed in the centre-of-mass frame, two particles with equal and opposite momenta collide, lose all kinetic energy, and end as a single stationary particle. In the laboratory frame, the two particles simply stick together, and continue moving at a common final velocity given by momentum conservation ~ 1 + m2 u ~ 2 = (m1 + m2 )~ m1 u ucm = (m1 + m2 )~v cm , Looked at either way, it is easy to see that the kinetic energy lost is 2 1 2 m1 u1

+ 21 m2 u22 − 12 (m1 + m2 )u2cm

General inelastic collisions lie between the two extremes.

17.2

The effective potential

Both gravity and the Coulomb (electrostatic) force are central, conservative forces. Central:

force acts along the line of centres ⇒ conserves angular momentum.

Conservative: a potential energy can be defined. Because the force is central U = U (r). We will work in 2D polar coordinates (r, θ).

LECTURE 17. ORBITS UNDER CENTRAL FORCES

87

The velocity can be written v

ˆ ~v = vr rˆ + vθ θ,

^

vθ θ

ˆ are unit vectors in the where rˆ and θ radial and tangential directions and the components are given by vr =

dr , dt

vθ = r

r^ vr

r θ O

dθ . dt

ˆ are perpendicular to one another, the speed can be Since rˆ and θ found from Pythagoras and the kinetic energy is  2  2
dr dθ T = 12 mv 2 = 12 m vr2 + vθ2 = 12 m + 12 mr2 . dt dt If U (r) is the potential energy, then E = T + U is conserved. Only the tangential component of velocity contributes to the angular momentum dθ L = rmvθ = mr2 = constant, dt dθ L = and the energy equation becomes dt mr2  2 dr L2 1 + + U (r) = E, 2m 2 dt 2mr | {z }

This allows us to write

effective potential

so this takes the form of the energy of a one-dimensional system  2 1 dr m + Ue (r) = E = constant (17.1) 2 dt where the effective potential is given by Ue (r) = U (r) +

L2 . 2mr2

Notice that the “kinetic energy” is the radial part only; the tangential kinetic energy of the system is hidden away in the effective potential.

LECTURE 17. ORBITS UNDER CENTRAL FORCES

17.3

88

Orbits under gravity

For gravity, U (r) = −GmM/r, so L2 GM m Ue (r) = − , 2mr2 r which looks like:

E>0

D

Ue

0

B

A

E 0:

has a minimum radius (e.g. D above), but no maximum, hence it is “unbound”. An example would be a star passing by another star. For 1/r2 forces, these orbits are hyperbolas.

LECTURE 17. ORBITS UNDER CENTRAL FORCES E = 0:

89

another special case. The minimum energy that allows escape; for 1/r2 forces, these orbits are parabolas.

Circles, ellipses, parabolas and hyperbolas are all curves generated by the intersection of planes with cones, or “conic sections”.

17.4

Circular orbits

The radius of a circular orbit is defined by the minimum of Ue dUe (r) = 0, dr and thus

dU L2 − = 0. dr mr3 dθ Remembering that L = mr2 = mr2 ω, we find dt mrω 2 =

dU . dr

This is a balance between centripetal acceleration and the radial force, as seen before in circular motion. For gravity with U = −GM m/r ω2 =

17.5

GM . r3

Centrifugal barrier

The L2 /2mr2 term in Ue (r) prevents the particle reaching small radii, hence the term “centrifugal barrier”. It comes from the conservation of angular momentum. If Ue (r) contains a negative term of a functional form steeper than 1/r2 at short distances (e.g. −1/r3 ) the centrifugal barrier may be beaten and the particle could reach r = 0. This happens with black holes.

Lecture 18

Kepler’s Laws Keypoints from Lecture 17 • In the CoM frame, the speeds of each particle are reduced by inelastic collisions. • Rutherford scattering: elastic collisions of α-particles off heavy nuclei. • By conserving angular momentum, one can reduce the 2D problem of orbits in a central potential to an equivalent 1D problem, describing the radial motion only, with a modified form of the potential, known as the “effective potential”. • If the potential is U (r) and the angular momentum is L, then the effective potential is given by: Ue (r) = U (r) +

L2 . 2mr2

• The effective potential allows one to classify orbits in terms of dr turning points when Ue (r) = E, implying that = 0. dt

Objectives • Kepler’s laws, elliptical orbits 90

LECTURE 18. KEPLER’S LAWS

18.1

91

Kepler’s Laws

Using observations of Tycho Brahe, Johannes Kepler formulated three laws of planetary motion: Far from easy K1 : The orbits of the planets are ellipses, with the Sun at one focus. K2 : A line from a planet to the Sun sweeps out equal areas in equal times.

given that we see the planets from another planet, Earth.

K3 : The square of a planet’s orbital period P is proportional to the cube of the length of the semi-major axis a of its orbit. See figure. The two hatched areas are the same and indicate what is meant by “sweeping” an area. By K2 , these take the same time.

Polar equation of an ellipse: r=

l , 1 + e cos θ

where l is the “semi-latus rectum” [half perpendicular length].

I will take this as the definition of an ellipse

LECTURE 18. KEPLER’S LAWS

92

Distances from Sun closest furthest perihelion aphelion θ=0 θ=π l l rP = rA = 1+e 1−e The semi-major axis is a=

1 2 (rP

1 + rA ) = 2



l l + 1+e 1−e

 =

l . 1 − e2

Kepler’s Laws are not as fundamental as Newton’s Laws, and indeed can be derived from them, as we shall see.

18.2

Kepler’s Second Law K2

K2 has a beautifully simple explanation.

B o

r dθ

dθ r

A

Area of triangle OAB = 12 base × height = 12 (r dθ)r = 12 r2 dθ, dθ L Rate of sweeping out area = 21 r2 = , dt 2m where L is the angular momentum. Thus K2 is just conservation of angular momentum.

18.3

Kepler’s First Law K1

To describe the shape of an orbit, we need an equation linking r and θ. We can get this from  2 dr L2 GM m 1 m + − = E. 2 dt 2mr2 r

Note the small extra part from the change in r is second order ((r dθ)dr) and can be neglected in the limit.

LECTURE 18. KEPLER’S LAWS

93

by using the chain rule dr dθ L dr dr = = , dt dθ dt mr2 dθ which gives  2 L2 GM m dr L2 − = E. + 2mr4 dθ 2mr2 r This looks complex, but is simplified by a neat trick: set r = 1/u, then  2  2 1 du dr dr 1 du =− 2 , = 4 , dθ u dθ dθ u dθ and we get  2 L2 du L2 2 u − GM mu = E. + 2m dθ 2m Multiply through by m/L2 and call GM m2 /L2 = C  2 du 1 + 12 u2 − Cu = Em/L2 = constant. 2 dθ | {z } | {z } “potential” “kinetic”

This looks very much like the energy (kinetic plus potential) of a particle whose position is u, and whose mass is unity. Of course, we have to pretend that θ is the time, but still we can use this analogy to write down a version of N2 , with the force derived from the “potential energy” U † (u) = 12 u2 − Cu: d2 u dU † = − = C − u. dθ2 du Fortunately this is easy to solve u = C(1 + e cos θ), where e is a constant of integration, and therefore using r = 1/u, l , 1 + e cos θ the polar equation of an ellipse, with r measured from the focus, with r=

1 L2 l= = . C GM m2 This establishes K1 .

check by substitution

LECTURE 18. KEPLER’S LAWS

18.4

94

Kepler’s Third Law K3

The area of an ellipse is A=

πl2 (1 − e2 )3/2

This is not obvious, but feel free to try to prove it

.

√ The area is swept out at rate 21 (L/m) and L/m = GM l as shown in the last section. So, the time taken to sweep out the whole area is 2A 2πl2 2π l3/2 2π 3/2 √ √ P = = = = a , L/m (GM l)1/2 (1 − e2 )3/2 GM (1 − e2 )3/2 GM or 4π 2 GM = 3 . 2 P a This establishes K3 , P 2 ∝ a3 , and generalises the relation for circular orbits by replacing the radius r with the semi-major axis a.

18.5

Orbital energy

The orbital energy is  2 1 dr L2 GM m E= m + − . 2 dt 2mr2 r dr l Now, = 0 at the closest approach (perihelion) distance rP = , dt 1+e so L2 GM m 2 E= (1 + e) − (1 + e). 2ml2 l Substituting L2 = GM m2 l, E=

 GM m  GM m (1 + e)2 − 2(1 + e) = − (1 − e2 ), 2l 2l

or E=−

GM m . 2a

The same result follows from using the aphelion distance rA =

l . 1−e

Lecture 19

Introduction to Special Relativity Keypoints from Lecture 18 • Kepler devised three empirical rules to explain Tycho’s precise observations. Explaining these mathematically was probably Newton’s greatest feat. • Kepler’s second law is conservation of angular momentum. • Kepler’s third law is the period relation for circular orbits with the semi-major axis a replacing the circular orbit radius: 4π 2 GM = . P2 a3

Objectives • Galilean invariance and transformations • Einstein’s postulates of special relativity

95

LECTURE 19. INTRODUCTION TO SPECIAL RELATIVITY 96

19.1

Galilean Invariance

Galileo argued that it was not possible to tell whether one was at rest or uniformly moving. e.g. water poured from a flask on a ship lands on the deck underneath the underneath the flask, whether the ship is moving or at rest. How else could This is summarised in Galileo’s “principle of relativity”: The laws of mechanics are the same in all uniformly moving reference frames. Uniformly moving: the “inertial frames” we encountered before, in which N1 and N2 apply All velocity is relative; there is no “preferred rest frame”, no “absolute reference frame”.

19.1.1

The Galilean Transformations

Einstein found the need to question the relationship between position and time between relatively moving inertial frames.

S

S’ u

y

y’

x=0 z

x’ = 0

z’ u

x x’

Standard configuration of two inertial frames S and S 0 . Consider two inertial frames labelled S and S 0 , oriented such that S 0 moves at speed u in the positive x direction relative to S, and with

drinks be served on an aircraft?

LECTURE 19. INTRODUCTION TO SPECIAL RELATIVITY 97 their respective axes, x, y, z and x0 , y 0 , z 0 aligned. We will be using this set-up again and again, so make sure you understand it fully. We will refer to this setup as the “standard configuration”. Define the time t = 0 as the time when the coordinate origins coincide, i.e. x = x0 = 0. Then at time t, S 0 has moved by ut relative to S, and so we have x0 = x − ut. Including all coordinates we have the “Galilean transformation” relating Einstein’s terminology; coordinates of the same “event” in two different frames: 0

t x0 y0 z0

0

= t, = x − ut, = y, = z,

t=t, x = x0 + ut0 , y = y0, z = z0.

Galileo never wrote them down as such.

Galilean addition of velocities Suppose an object moves at speed v 0 in the x0 -direction as measured in frame S 0 . What speed v does it move relative to S? May seem Speed v in the x direction is rate of change of x with time t, so: dx , dt d(x0 + ut0 ) = , dt d(x0 + ut0 ) = , dt0 dx0 d(ut0 ) = 0 + , dt dt0 = v 0 + u.

v=

This is what we would have expected from everyday experience. Examples: swimming upstream compared with swimming downstream; throwing objects from a moving vehicle. We will apply the same method but get a different answer in Relativity.

“obvious” but we will soon see examples where it isn’t.

LECTURE 19. INTRODUCTION TO SPECIAL RELATIVITY 98

19.2

Galilean invariance and Light

The unification of electricity, magnetism and light (Maxwell 1865) was a great triumph of 19th century physics. Maxwell showed that disturbances in electromagnetic fields travel at √ a unique speed, c = 1/ µ0 0 ≈ 3.00 × 108 m s−1 , the speed of light. c is predicted to be the same in all directions. According to Galilean velocity addition, someone moving at speed u in the ∓x-direction must measure the speed of light moving along the +xaxis to be c ± u. This means that its speed should vary with direction, and they must therefore determine a different set of equations from Maxwell’s. Do we just happen to be in the one-and-only inertial frame where Maxwell’s equations hold true? Seems contrary to the principle of relativity.

19.3

The luminiferous aether

Perhaps, like sound in air or waves on water, there is some substance which carries light, and is almost stationary relative to Earth, so that we see the same speed of light in all directions. This hypothetical stuff was called the “luminiferous aether”. The speed of sound is different in different media: air 350 m s−1 water 1500 m s−1 steel 6000 m s−1 Theory of waves relates these speeds to the stiffness or elastic coefficients of the medium of propagation. The aether needs odd properties: incredible stiffness to support the high speed of light, yet unable to support longitudinal waves like sound; present in glass, air and a vacuum, but can’t be felt.

LECTURE 19. INTRODUCTION TO SPECIAL RELATIVITY 99 By the time of Einstein, there was no evidence for the existence of the aether, which is now consigned to the dustbin of defunct theories.

19.4

Postulates of Special Relativity

In 1905 Einstein published his paper on special relativity, which avoided introducing the aether. Brief word about Michelson-Morley experiments dating from 1880’s. They did not determine the speed of light, rather they compared round-trip times for light in two different directions, which was further evidence against the aether. Bit more detail later. Einstein’s 1905 paper is based on two “postulates”:

1. The laws of physics are the same in all inertial frames. 2. The speed of light in empty space is the same in all inertial frames.

Postulate 1, the relativity principle RP , is just Galilean invariance, generalized from “mechanics” to “physics”. Postulate 2 “solves” the problem with Maxwell’s equations: we are not in a special frame: all frames measure c. No aether is required. Postulate 2 is manifestly inconsistent with Galilean velocity addition: consider observers A and B who measure the same pulse of light, but B travels towards the light source at 0.5c. According to Einstein they both measure the same speed c! We cannot prove these postulates, but we can explore their consequences and test their predictions experimentally.

Lecture 20

The Lorentz Transformations Keypoints from Lecture 19 • Maxwell’s equations predict a unique velocity for light. • Combined with Galilean velocity addition v = v 0 + u, this suggests that we are in a special inertial frame because in any other frame we should see c ± u, depending upon direction. • The “standard configuration” of inertial frames S and S 0

S

u

• The Galilean transformations: 0

t x0 y0 z0

= t, = x − ut, = y, = z,

S’

y

y’

0

t=t, x = x0 + ut0 y = y0, z = z0.

x=0 z

x’ = 0

z’ u

x x’

• Postulates of Special Relativity 1. The laws of physics are the same in all inertial frames. 2. The speed of light in empty space is the same in all inertial frames. 100

LECTURE 20. THE LORENTZ TRANSFORMATIONS

101

Objectives • To discuss what we mean by “simultaneous events”. • To derive the Lorentz transformations

20.1

Measuring Distance and Synchronizing Clocks

We need to be a little more careful than usual in defining even fairly basic things. An event is specified by its position and the time at which it occurs. Both of these can be measured in different reference frames (e.g. S and S 0 ). We need to have a way of measuring distances, and synchronizing clocks at remote locations. Within the framework of special relativity, we can use light for both these purposes, since c is constant. For a round trip, the time taken for the outward trip must equal the time taken for the return trip. Example: A light signal is sent from O to A, and is reflected back, arriving again at O, 2 seconds later. Therefore, A is (or was) 1 lightsecond away (3 × 105 km) and A received the signal 1 s after it was sent. In other words, the reflection event occurred at x = 3 × 105 km, t = 1 s. (We assume that we take the origin of spatial coordinates x = 0 at O, and the time origin t = 0 when the signal is sent).

20.2

Relativity of Simultaneity

Consider three equally spaced observers, A, O and B, all at rest in frame S 0 .

A

c

O

c

B

O sends a light signal in both directions at time t0 = 0: at velocity −c towards A and at velocity +c towards B.

LECTURE 20. THE LORENTZ TRANSFORMATIONS

102

• Event EA (light reaches A) . • Event EB (light reaches B). • These events occur at the same time t0 . • In frame S 0 , EA and EB are simultaneous. Plotting a “space-time” diagram for S 0 (ct0 versus x0 ):

x’ = const

ct’ EB

EA

t’ = const

O

A

B

x’

Space-time diagram for frame S 0 Light travels at speed c so the light rays satisfy x0 = ±ct0 , and appear as ±45◦ lines on the diagram. Now consider the view from S, in which A, O and B all move at speed u in the +x direction, and so follow “worldlines” obeying x = ut + const:

x’ = const

ct tB

EB

t

ons ’=c

t

tA EA A

O

B

x

Space-time diagram for frame S

Green vertical lines are “worldlines” through space-time

LECTURE 20. THE LORENTZ TRANSFORMATIONS

103

Because of the second postulate, on the S-frame diagram the light paths are still at ±45◦ on the diagram. Events, EA and EB , which are simultaneous in S 0 , cannot be simultaneous in S. Hence t0 6= t !!!!! No “absolute time”

20.3

The Lorentz Transformations

If t0 = t is wrong, we need to re-think the Galilean transforms. Focus on x and t (y and z will turn out to be unchanged). We define the origins O and O0 so that when x = x0 = 0 they set their clocks to make t = t0 = 0. The connection between x and x0 must be linear x0 = γ(x − ut),

x = γ(x0 + ut0 ),

where the new quantity γ is to be determined, and will most likely depend on u. If γ → 1, we regain the Galilean transforms. Why? Any nonlinear terms, like x2 , sin x, would imply that inertial motion in S (such as x = x0 + vt) would become non-inertial in S 0 . The above choice of origins ensures that there are no constant terms. The dependence on x − ut ensures that the origin of S 0 , defined by x0 = 0, will appear as the line x = ut in S. The second equation above is obtained by simply relabelling S → S 0 , S 0 → S and u → −u. It ensures that the origin of S, defined by x = 0, appears as the line x0 = −ut0 in S 0 . We can obtain γ as follows. The equations must describe the coordinates of a light signal leaving the origin x = x0 = 0 at t = t0 = 0: x = ct x0 = ct0

⇒

ct0 = γ(ct − ut) = γ(c − u)t ct = γ(ct0 + ut0 ) = γ(c + u)t0

Multiplying these two equations together gives
c2 tt0 = γ 2 c2 − u2 tt0 s r 2 c 1 1 p γ= = = c2 − u2 1 − u2 /c2 1 − u2 /c2

LECTURE 20. THE LORENTZ TRANSFORMATIONS

104

This quantity is very common in special relativity and is called the Lorentz factor. Finally, we can obtain the transformation equations for time, by inserting our equation for x0 into the equation for x:   x = γ(x0 + ut0 ) = γ γ(x − ut) + ut0 = γ 2 x − γ 2 ut + γut0   2 γ − 1 x t0 = γ t − γ2 u But

γ 2 − 1 u2 = 2 and so we get γ2 c ux  t =γ t− 2 , c 0



ux0 t=γ t + 2 c 

0



where the second equation follows in a similar way.

t0 x0 y0 z0 where

The Lorentz transformations    0 ux  ux =γ t− 2 , t = γ t0 + 2 , c c 0 = γ(x − ut), x = γ(x + ut0 ), = y, y = y0, = z, z = z0. 1 γ=p 1 − u2 /c2

You should remember these! We will be using them a lot.

Lecture 21

Length contraction and time dilation Keypoints from Lecture 20 • Simultaneity is relative

x’ = const x’ = const

ct tB

ct’ EB

EA

t’ = const

EB

onst ’=c

t

tA EA

O

A

B

x’

A

O

• The Lorentz transformation equations  ux  t =γ t− 2 , c x0 = γ(x − ut), y 0 = y, z 0 = z, 0

  ux0 0 t=γ t + 2 , c x = γ(x0 + ut0 ), y = y0, z = z0. 1

γ=p 1 − u2 /c2 105

B

x

LECTURE 21. LENGTH CONTRACTION AND TIME DILATION106

Objectives • Explaining length contraction and time dilation

21.1

The Lorentz Factor
−1/2 1 = 1 − u2 /c2 . γ = γ(u) = p 1 − u2 /c2

At everyday speeds γ is close to 1, but γ → ∞ as u → c: Description Speed of sound Low Earth orbit electrons in old TVs LHC protons, 7 TeV Ultra high energy cosmic ray protons

Speed, u 340 m s−1 7.8 km s−1 1.2 × 108 m s−1 0.99999999c ≈c

γ 1 + 6.4 × 10−13 1 + 3.4 × 10−10 1.078 ∼ 7500 ∼ 5 × 1010

10 8

γ

6 4 2 0

0

0.2

0.4

0.6

0.8

speed u/c

The Lorentz factor γ versus u/c.

1

LECTURE 21. LENGTH CONTRACTION AND TIME DILATION107

21.2

Length contraction

The length of something is defined in terms of the distance between two points measured at the same time. However, the same time in one frame is not the same time in another. Consider a bar at rest in S 0 , lying on the x0 -axis with ends at (x0A , 0, 0) and (x0B , 0, 0). In S 0 the length is L0 = x0B − x0A . In S the bar is moving with speed u, so we must take care to measure the position of each end at the same time t in S. Use the Lorentz transformation from S → S 0 because it involves t rather than t0 : Can use S 0 → S but takes more x0A = γ(xA − utA ), effort x0B = γ(xB − utB ), subtract: x0B − x0A = γ(xB − xA ), since tA = tB L0 = γL where we define L as the length measured in the S frame. So L = L0 /γ. An object moving relative to an observer appears to be shortened by a factor γ = (1 − u2 /c2 )−1/2 in its direction of travel. This is known as “length contraction” or “Lorentz contraction”. L0 is called the “rest length” or “proper length”. Lorentz contraction is reciprocal: S measures objects in S 0 to be contracted, but, equally, S 0 measures objects in S to be contracted. Each thinks the other’s metre rule to be shorter than their own. Example: the Galaxy is ∼ 100 000 ly = 9.5 × 1020 m across (ly means light-years). To an ultra-high energy cosmic ray with γ = 5 × 1010 9.5 × 1020 = 1.9 × 1010 m, about 10% of the it appears to be just 10 5 × 10 distance between the Earth and Sun.

21.3

Time dilation

Consider a time interval ∆t0 = t0B − t0A measured by a clock at rest in S 0 . Use the Lorentz transformation S 0 → S because it involves x0 rather than x: Can use S → S 0 but takes more effort

LECTURE 21. LENGTH CONTRACTION AND TIME DILATION108 tA = γ(t0A + ux0A /c2 ), tB = γ(t0B + ux0B /c2 ), subtract: tB − tA = γ(t0B − t0A ) since x0A = x0B ∆t = γ∆t0 We have set x0A = x0B because the clock is at rest in S 0 . So ∆t = γ∆t0 . In frame S, in which the clock is moving, the intervals appear longer than they do in S 0 , in which it is at rest. Time is “dilated”. ∆t0 , the time interval measured in the frame in which the clock is at rest, is called the “proper time”. It is less than the time interval measured in any other frame. Example: consider again the ultra-high energy cosmic rays with γ = 5 × 1010 . These travel at very near c, and hence to us they take 100 000 yr to cross the Galaxy, which is ∼ 100 000 ly across. In their rest frame they take ∆t0 =

∆t 100 000 yr = = 63 s. γ 5 × 1010

Note that this is consistent with the contracted width of the Galaxy in the frame of the cosmic rays, as per previous example.

21.4

Atmospheric muons and time dilation

Rossi and Hall performed a classic experiment on time dilation in 1941. Cosmic rays hitting the upper atmosphere produce particles called muons at a height of about 15 000 m. The muons travel mostly downwards towards the ground, many of them at speeds approaching c. Muons decay, e.g. µ− → e− + ν¯e + νµ , with their number when at rest obeying the relation N (t) = N (0)e−t/τ with a mean lifetime τ = 2.2 µs.

muons

mountain

LECTURE 21. LENGTH CONTRACTION AND TIME DILATION109 Rossi and Hall measured the muon flux Φ(0) at ground level and Φ(h) at a height h = 1600 m up a mountain in Colorado. They specifically selected high-energy muons travelling near c, so travel time t = 1600/c. One therefore might expect that a fraction   Φ(0) N (t) 1600/c = = exp − = 0.088, Φ(h) N (0) 2.2 × 10−6 Φ(0) = 0.7. Φ(h) Simply inserting this ratio on the right of the above equation suggests that the muons are travelling at ≈ 7c instead of c. should survive over this distance. Rossi and Hall measured

Explanation. The muon’s decay lifetime is dilated in our frame. The measured survival rate requires a dilation factor γ ≈ 7. s r   u 2 −1/2 2 γ −1 48 u γ = 1− = = ≈ 0.99 ⇒ 2 c c γ 49 In the muon frame, the decay lifetime is not dilated, but the distance to be covered between the top and bottom of the mountain is Lorentzcontracted, and hence the time taken to travel this distance is shorter by a factor γ ≈ 7. So again, the muon population only decreases to 0.7× its initial value. The Global Positioning System (GPS) provides a famous example of time dilation at work. If ignored, GPS positions would be more than 10 km off after just a day! [Both gravitational effects and the dilation due to special relativity have to be allowed for.]

Lecture 22

Michelson-Morley Experiment Keypoints from Lecture 21 An observer in frame S measures the length of a moving object by recording coordinates of each end of the object simultaneously in his or her frame, i.e. S. Observes lengths to be contracted relative to their values at rest: L=

L0 . γ

An observer in S “measures” the duration of intervals on a clock at rest in S 0 to be dilated: ∆t = γ∆t0 . “Moving clocks run slow.”

Objectives • The Michelson-Morley Experiment

110

LECTURE 22. MICHELSON-MORLEY EXPERIMENT

22.1

111

The Michelson-Morley Experiment

As promised, we go back to explain some of the experimental background to Einstein’s bold assertion that the speed of light is the same in any inertial frame. As we shall see, the experiments do not measure the speed of light (much later, Michelson did do this, but it is not crucial to the theory of relativity). What they do is directly compare the speed of light, or rather, its round-trip time, in different directions, and by obtaining a null (negative) result, fail to support the idea of the aether. Actually there Earth orbits the Sun at ∼ 30 km s−1 , always changing direction. The aether surely doesn’t keep step with Earth’s orbit does it? Can we measure an aether “wind” blowing by? Look for change in speed of light with direction. Light travels fast; cannot simply time it. Instead consider comparing the times taken on two round trips, one parallel to the wind, the other perpendicular to it. B

u LB

Aether wind

O

LA

A

Light is sent from O to A and back, and O to B and back with the aether passing by at speed u as indicated. Attempt to analyze using Galilean addition of velocities.

were suggestions of a partial “dragging” of the aether by Earth, but we still expect some “wind”.

LECTURE 22. MICHELSON-MORLEY EXPERIMENT

112

First consider the O → A → O trip. This takes LA LA + , c −u c + u  (c + u) + (c − u) 2LA c = LA = 2 2 2 − u2   c −u  c  2LA 1 2LA = = γ2 2 2 c 1 − u /c c p introducing γ = 1/ 1 − u2 /c2 for neatness (it has no special relativity implications here). Trip up and back in direction of wind takes a little longer (since γ > 1) than if no wind was blowing (2LA /c). tA =

Path O → B → O is a little trickier; consider two views of it: u B

u

B

c

c

LB

v u O

O

u O

Aether frame

Earth frame

This path is best viewed in the aether frame since it is by definition the one in which light has speed c. √ By Pythagoras, the “cross-wind” speed is v = c2 − u2 , therefore     2LB 2LB 1 2LB p = tB = √ = γ. c c c2 − u2 1 − u2 /c2 Cross-wind trip also takes longer, but not by so much. Time difference:  2 2 γ LA − γLB c Still not useful – we would need to know LA and LB incredibly precisely to make any use of this. Instead, rotate the paths by 90◦ , then ∆t = tA − tB =

∆t90 =

 2 γLA − γ 2 LB c

The idea is to look for change in ∆t with rotation. We then don’t need to know LA and LB so well. This can be seen from the form of the expression for δt which does not involve a difference between LA and LB , unlike ∆t.

LECTURE 22. MICHELSON-MORLEY EXPERIMENT

113

The difference between the two times is δt = ∆t − ∆t90 =

  2 (LA + LB ) γ 2 − γ c

Since u  c, then u2  c2 , and we can approximate this using the binomial theorem (1 + x)n = 1 + nx + · · · , valid for x  1 :    2 2 −1 u u γ2 = 1 − 2 =1+ + ··· c c2   2 −1/2 u2 u 1 γ = 1− 2 =1+ 2 + ··· c c2  2    2 u L + L u A B 1 and δt ≈ . Therefore γ 2 − γ ≈ 2 c2 c c2

22.1.1

Experimental realisation

Take LA = LB ≈ 15 m, u = 30 km s−1 , then δt ≈ 1 × 10−15 s! Measuring this seems hopeless, but in a truly beautiful experiment, Michelson (1881) and Michelson and Morley (1887) managed better than this using “interferometry”. B

LB

A S O

LA

D

A light source at S produces light that is split at O by a half-silvered mirror. The two beams produced are reflected off mirrors at A and B,

LECTURE 22. MICHELSON-MORLEY EXPERIMENT

114

recombined at O and detected (by eye) at D. The time difference is seen as a change in position of bright and dark interference fringes. They rotated their apparatus by floating it in a bath of mercury (!). The change expected = (cδt)/λ fringes, where λ is the wavelength of light. Michelson and Morley used LA = LB = 11 m, λ = 589 nm. For u = 30 km s−1 , they expected 0.4 of a fringe shift.

Don’t try this at home. Mercury is highly toxic.

Found: no shift within uncertainty ±0.01 fringes. There is no aether drift! ≡ ±2 × 10−17 s!

22.2

Lorentz-Fitzgerald Contraction

After Michelson and Morley, Fitzgerald (1889) and Lorentz (1892) kept the aether hypothesis alive by suggesting that all objects contract L → L/γ along the direction of the aether wind. This would affect the LA length in the formula ∆t = tA − tB =

 2 2 γ LA − γLB c

changing it to  2 γLA − γLB . c If LA = LB then the contraction gives ∆t = 0. ∆t = tA − tB =

This contraction is similar to the Lorentz contraction formula. However, here, the contraction is the result of some unspecified interaction between the aether and all objects and is the same for all observers, whereas in Einstein’s it is an outcome of different frames of reference. Einstein’s explanation is far more satisfying.

Lecture 23

Doppler Effect Keypoints from Lecture 22 • Michelson and Morley’s interferometer. • Expect sometimes to move relative to the aether as the Earth switches direction in its 30 km s−1 motion around the Sun. S • Should give rise to changes in round-trip time for light. • Detect nothing.

B

LB

A

O

LA

D

Objectives • Doppler effect

23.1

“Seeing” vs “Observing”: the Relativistic Doppler Effect

The use of the term “observer” in SR can be misleading. The times that enter the Lorentz Transformation equations exclude light travel115

LECTURE 23. DOPPLER EFFECT

116

time. They are not the times that a single observer would see. This is well illustrated by the Doppler effect: the change in frequency of a signal due to the motion of source or observer. This effect is important in astronomy and cosmology: by observing lines in a spectrum, we can deduce how quickly remote objects are receding from us. Good news: the equations for the relativistic Doppler effect on light signals are simpler than for the familiar case of sound waves travelling in air. Consider a source of light B moving away from an astronomer A with speed u. It emits pulses of light at regular intervals ∆t0 in its own reference frame. This corresponds to a “proper” frequency f0 = 1/∆t0 . What frequency f does A detect?

B

c A

u At position •, B is just about to emit another pulse; the previous pulse was emitted at position ◦. The proper time between two consecutive pulses ∆t0 = 1/f0 . A will “observe” this to be dilated to γ ∆t = γ∆t0 = . f0 In A’s frame, in the time taken to emit two pulses, B moves away by an additional u∆t. This delays the arrival of the second pulse by an additional u∆t , c so A sees the pulses separated by   u u u ∆tA = ∆t + ∆t = 1 + ∆t = γ 1 + ∆t0 . c c c

LECTURE 23. DOPPLER EFFECT

117

Hence the frequency seen by A, f = 1/∆tA is given by f 1 = = f0 γ(1 + u/c)

s

1 − u/c . 1 + u/c

The first form has the advantage that the two effects are clearly separated: the γ factor arises from time dilation, while the term in parentheses arises from the light travel time. The second form follows from the fact that 1 1 =p γ=p 1 − u2 /c2 (1 − u/c)(1 + u/c) and may be easier to remember. If B travels towards A, similar reasoning gives 1 f = = f0 γ(1 − u/c)

s

1 + u/c . 1 − u/c

Since wavelength λ = c/f then the equivalent relations are (  λ u + sign moving away =γ 1± with λ0 c − sign moving towards • When B moves away from A, the signal received at A has a longer wavelength, lower frequency: it is “red-shifted”. • When B moves towards A, the signal received at A has a shorter wavelength, higher frequency: it is “blue-shifted”. The nomenclature comes from the assumption that the original signal lies in the visible spectrum. Example 23.1. A star called SS 433 throws out two jets of material at 0.26c in opposite directions. At what wavelengths would a spectral line of rest wavelength λ0 = 656.2 nm emitted by the material in the jets be seen by an astronomer in line with them?

LECTURE 23. DOPPLER EFFECT

118

Answer. The wavelength 656.2 nm lies in the red region of the spectrum, but this is not what we see. The line from the receding jet will be seen at: r r 1 + 0.26 1.26 λ= λ0 = × 656.2 = 856.3 nm. 1 − 0.26 0.74 This is infra-red (not visible to naked eye). The line from the approaching jet will be seen at: r r 1 − 0.26 0.74 λ= λ0 = × 656.2 = 502.9 nm. 1 + 0.26 1.26 This is green. Example 23.2. A clock is stationary at x0 = 0 in the S 0 frame, which moves at u = 53 c, γ = 54 , relative to the S frame. The clock counts down the time from ct0 = −4 to ct0 = 0 as it approaches the origin of S. Use a space-time diagram to explain the difference between the times ct, in S, at which these clock ticks occur, and what is seen by an observer at x = 0. Answer. At x0 = 0, ct0 = −4 we have, from the Lorentz Transformations,

x

x = γ(x0 + ut0 ) = γut0 = −3,
ct = γ ct0 + (u/c)x0 = γ ct0 = −5

0 −1

−1 −2

All the ticks of the clock ct0 = −4, −3, −2, −1

−2 −3

−3

ct 0

happen at x = 0, therefore in the S frame they occur at ct = γct0 = −5, −3.75, −2.5, −1.25.

−4

−4 −5

−3 −2 −1 0

The world-line of the clock is shown as the red line, with the events indicated using a clock symbol. The times ct can be read off the vertical axis. This is what we mean by the (loose) phrase “Moving

LECTURE 23. DOPPLER EFFECT

119

clocks run slow”: time dilation applies, ∆t = γ∆t0 . However, the light signals, going from the ticks of the clock to the observer at x = 0, take time to get there. The paths of these signals are indicated by green lines at 45◦ . The observer at the origin of S sees the clock ticks at the times when they reach x = 0, ctA = −2, −1.5, −1, −0.5. s 1 − u/c 0 This fits the Doppler formula: ∆t = ∆t = 12 ∆t0 . We 1 + u/c could say “approaching clocks appear to run fast (departing clocks appear to run slow)”. Take care to be clear what you mean!

Lecture 24

The Twin Paradox Keypoints from Lecture 23 When one “measures” the time of an event, in a certain frame, we imagine using a clock located at the event, which has been previously synchronized with all the other clocks in that frame. If one sees a clock at a remote location, the light-travel time between the location and the observer must be included. This is exemplified by the Doppler effect: ( f 1 + for receding light source = f0 γ(1 ± u/c) − for approaching light source where f0 is the proper frequency emitted by an object at rest in the S 0 frame, the 1/γ term comes from measuring the frequency in the S frame, and the term in parentheses comes from the light travel time. An equivalent formula is s ( f 1 ∓ u/c upper signs for receding light source = f0 1 ± u/c lower signs for approaching light source

Objectives • The Twin Paradox 120

LECTURE 24. THE TWIN PARADOX

24.1

121

Reciprocity of Time Dilation

As with length contraction, time dilation is reciprocal. That is, just as an interval of time ∆t0 measured in S 0 is dilated to ∆t = γ∆t0 in S, so an interval of time ∆t measured in S is dilated to ∆t0 = γ∆t in S 0 . There is no contradiction since the time interval in the first instance applies to events at the same x0 , whereas it applies to events at the same x in the second instance.

24.2

The Twin Paradox

This immediately leads to the most well-known “paradox” in relativity: the paradox of the travelling twin. It should be noted that there is no “paradox”: Special Relativity provides a perfectly consistent description of this as it should. A theory with an elementary internal inconsistency would not be much of a theory. The problem comes from a gut feeling “but surely . . . ”. This is strong enough that the twin paradox has caused more controversy than any other aspect of Special Relativity. The “paradox” is this. Alice and Bob are twins. Alice takes a relativistic journey to a star 30 light-years away and back, travelling so fast that, to her, the trip only takes one year. On her return she finds Bob 60 years older. In the 1960s film, “The

But from Alice’s point of view, wasn’t it Bob who went away and came Planet of the Apes”, Charlton back . . . shouldn’t he be the younger? Heston returns

Alice and Bob can’t both be younger than each other so which of them, to Earth 2000 years after he if either, is right? Solution: identify inertial frames and events. To avoid having to deal with acceleration, assume Alice instantaneously accelerates to cruising speed. We need 3 inertial frames to describe the problem: S: Bob’s rest frame (Earth) S 0 : Alice’s rest frame, outbound, +u relative to Bob

left it, but only one year has elapsed for him.

LECTURE 24. THE TWIN PARADOX

122

S 00 : Alice’s rest frame, inbound, −u relative to Bob. In “standard configuration” there are three events: 1. Alice leaves Earth, x = x0 = x00 = 0, t = t0 = t00 = 0. 2. Alice reaches star and turns round, x = L, t = L/u. 3. Alice returns to Earth, x = 0, t = 2L/u.

Use the Lorentz Transformations for time to compute t0 and t00 at these events, remembering γ(u) = γ(−u) = γ:  ux  0 t =γ t− 2 , c   ux t00 = γ t + 2 . c

Event 1

2

3

x 0

L

0

2L u   L uL 2L 0 γ − 2 γ u c u   2L L uL γ 0 γ + 2 u c u

t 0 t0 t00

L u

Total time elapsed for Bob: ∆tB = (t2 − t1 ) + (t3 − t2 ) = t3 − t1 =

2L . u

Total time elapsed for Alice needs more care: ∆tA = (t02 − t01 ) + (t003 − t002 ),     2L L uL L uL − 2 −0+γ −γ + 2 =γ u c u u c   2 u 2L =γ 1− 2 , c u   1 2L = , γ u ∆tB = . γ Conclusion: Alice will be younger than Bob, having aged by a factor γ less than Bob during her trip.

LECTURE 24. THE TWIN PARADOX

123

There is no symmetry between Alice and Bob: Alice switches between inertial frames; Bob does not. NB You may sometimes see it said that one needs gravitation or general relativity to discuss the twin paradox. One can, but it is not required.

24.3

The twin paradox by Doppler shift

Imagine that Alice and Bob had a pair of identical clocks and both used telescopes to look at each others’ clocks during the trip to count the ticks. This is equivalent to having them send signal pulses to each other at regular intervals. The ticks define a unit of time in the local frame; for example we can have one tick per second or one per year. If the rest frame tick rate of each clock is f0 , then we apply the Doppler equation to find what is seen. outward f0 f− = γ(1 + u/c)

return f+ =

f0 . γ(1 − u/c)

Bob’s view. Bob is watching Alice and her clock. Alice turns round at t = L/u, but Bob does not immediately see this because of the finite speed of light. Instead he sees this event at a time L/c later. So we can count the total number of ticks that he sees from Alice’s clock:     L L L L NB←A = + f− + − f+ u c u c  L = (1 + u/c)f− + (1 − u/c)f+ , u   2L = f0 . γu Bob sees Alice age by 2L/γu units of time. Alice’s view. Alice is watching Bob and his clock. When Alice turns round she immediately sees the frequency shift. Since each leg for her

Half the ticks come in each leg, but the durations of each leg are different.

Alice measures each leg to be of equal duration, but different numbers of pulses arrive, because the frequencies are different

LECTURE 24. THE TWIN PARADOX

124

is length-contracted to L/γ, each only takes a time L/γu. Thus L (f− + f+ ), γu     L 1 L 1 2 = 2 + f0 = 2 f0 , γ u 1 + u/c 1 − u/c γ u 1 − u2 /c2   2L = f0 . u

NA←B =

Alice sees Bob age by 2L/u units of time. There is no “paradox”, and both these calculations give the same answers as in the previous section. Example 24.1. Illustrate the exchange of signals between the twins for the case u/c = 53 , γ = 54 , L = 3 ly. The round trip time for Alice is ∆tA = 8 yr, while ∆tB = 10 yr passes for Bob. Let the interval between signals be 1 yr, or f0 = 1 yr−1 . The Doppler factors for frequency, at this speed, are 12 (separating) and 2 (approaching). Answer. The red angled lines show Alice’s world line. The light signals are shown by the 45◦ green lines. Bob sees 4 signals in ∆t = 8 yr from the first leg, and the other 4 signals in just ∆t = 2 yr from the second leg (including Alice’s “Hello” at the moment of her return). Alice receives 10 signals from Bob: just 2 arrive in ∆t0 = 4 yr during the first leg, and the remaining 8 come in ∆t00 = 4 yr during the second leg (including Bob saying “Hello” as she returns).

10

ct

ct

10

8

8

6

6

4

4

2

2 x

0

x 0

0

1

2

3

0

1

2

3

Times of the events (1=departure, 2=turnround, 3=return) are: t1 = 0 t01 = 0 t001 = 0

t2 = 5 t02 = 4 t002 = 8.5

t3 = 10 t03 = 12.5 t003 = 12.5

Lecture 25

The Transformation of Velocities Keypoints from Lecture 24 • Twin “paradox”: at first one thinks that, based on the view of each twin, the other one ages more. • There is no paradox: the travelling twin genuinely ages less than the stay-at-home one. • The travelling twin has to switch inertial frames during their trip; there is no symmetry between the two.

Objectives • Addition of velocities

25.1

Relativistic addition of velocities

According to Einstein, light emitted from an object travelling at 0.5c towards you, approaches you at c, not 1.5c. What about the relative speed of two objects approaching each other, both at 0.8c? 125

LECTURE 25. THE TRANSFORMATION OF VELOCITIES 126 As usual, we start with S and S 0 in standard configuration, with S 0 moving at speed u in the +ve x-direction relative to S. Consider an object moving at speed v 0 relative to S 0 . At what speed v does it move in S? The speed in S is given by v = dx/dt, while Recall that the v 0 = dx0 /dt0 . To express v in terms of v 0 , we need to express x and t as Galilean0 answer functions of x0 and t0 , i.e. we need the Lorentz Transformations from is v = v + u. S 0 to S: t = γ(t0 + ux0 /c2 ), x = γ(x0 + ut0 ). We then use the chain rule to re-express v:  −1 dx dt dx dx dt0 = 0 = 0 v= . dt dt dt dt dt0 From the Lorentz Transformations  0  dx dx =γ + u = γ (v 0 + u) , 0 0 dt dt (NB γ = γ(u) is fixed since u is constant.) Similarly  
u dx0 dt 0 2 = γ 1 + = γ 1 + uv /c . dt0 c2 dt0 Hence we arrive at Einstein’s relation for the addition of velocities: v0 + u v= . 1 + uv 0 /c2 For the inverse, we swap v ↔ v 0 and u → −u: v0 =

v−u . 1 − uv/c2

Example 25.1. Two particles travel directly towards each other, each at speed 0.8c relative to the same observer. What is the speed of one particle in the rest frame of the other? Answer. Assume that the particles are located on the x-axis. Take the particle travelling from right to left as being at the origin of frame S. Then the observer is in frame S 0 moving at speed 0.8c in the +ve x-direction. The other particle is moving from left to right at speed 0.8c relative to the S 0 frame. We can apply

LECTURE 25. THE TRANSFORMATION OF VELOCITIES 127 the velocity addition formula to find that the speed of the other particle in S is v0 + u 0.8c + 0.8c v= = = 0.9756c. 1 + uv 0 /c2 1 + (0.8c)(0.8c)/c2 Example 25.2. Two aeroplanes approach each other directly, each travelling at 300 m s−1 . Calculate the speed each measures the other to be travelling at, in terms of a difference relative to the Galilean result. Answer. We are interested in v0 + u − (v 0 + u) = δv = 0 2 1 + uv /c



 1 − 1 (v 0 + u). 0 2 1 + uv /c

Since u, v 0  c, expand the denominator using (1 + x)−1 ≈ 1 − x: uv 0 0 δv ≈ − 2 (v + u) c Plugging in the figures  2 300 δv = − × 600 = −6 × 10−10 m s−1 . 8 3 × 10 Newtonian physics is pretty good in its regime of application!

25.1.1

More general motion

We must consider relativistic effects on the other velocity components too. The Lorentz transformation for position gives y = y 0 and z = z 0 , but the time transform also comes in. If an object moves in S at velocity (vx , vy , vz ), what would be measured in S 0 ? As before dy 0 dy 0 vy0 = 0 = dt dt



dt0 dt

−1 .

LECTURE 25. THE TRANSFORMATION OF VELOCITIES 128 Since y 0 = y, the first term is just vy . From the Lorentz Transformation t0 = γ(t − ux/c2 )    u dx uvx  dt0 =γ 1− 2 =γ 1− 2 , dt c dt c So vy0 =

vy . γ(1 − uvx /c2 )

vz0 =

vz . γ(1 − uvx /c2 )

Similarly

25.1.2

Aberration of light

Example 25.3. A light-beam in S travels down the y-axis towards the origin. What would S 0 measure? Answer. Therefore

The light-beam has velocity components (0, −c, 0). 0−u = −u, 1 − u × 0/c2 −c c vy0 = = − , γ(1 − u × 0/c2 ) γ vz0 = 0.

vx0 =

So S 0 would see the light-travelling at angle θ to the y 0 -axis where tan θ =

γu . c

For u  c, γ ≈ 1, θ  1 so tan θ ≈ θ. Thus θ≈

u radians. c

As Earth goes around the Sun at 30 km s−1 it reverses its motion every 6 months, and the above relation causes the positions of stars to vary by 30 = ±10−4 rad = ±2100 . θ=± 5 3 × 10 where the arcsecond 00 is (1/3600) of a degree ◦ .

LECTURE 25. THE TRANSFORMATION OF VELOCITIES 129 This was first observed by Bradley in 1729 who explained it in terms of motion with respect to the aether. The usual analogy is the way rain appears to come towards you when driving in a car. It is known as the “aberration of light”. It was one of the reasons why Michelson and Morley’s failure to detect motion with respect to the aether was so puzzling. Relativity explains it and the Michelson-Morley experiment with ease.

Lecture 26

Relativistic Mass and Momentum Keypoints from Lecture 25 Relativistic addition of velocities. An object with velocity (vx0 , vy0 , vz0 ) in S 0 is measured in S to have velocity u + vx0 vx = , 1 + uvx0 /c2 vy0 vy = , γ(1 + uvx0 /c2 ) vz0 vz = . γ(1 + uvx0 /c2 ) Examples: stellar aberration, “Fresnel drag” (speed of light in water is partially affected by the water speed).

Objectives • Relativistic momentum

130

LECTURE 26. RELATIVISTIC MASS AND MOMENTUM

26.1

131

Momentum 6= mv

Consider two particles of equal mass m, speeds ±v, which collide to form a single stationary particle of mass M , in frame S. (This is a completely inelastic collision.):

v

v

v’

Before m

m After

m

m v

M M S

S’

Completely inelastic collision, seen from a frame S in which the end product is stationary and S 0 in which one of the original particles is stationary. Galilean case conserves mass and momentum if p = mv and M = 2m. Now consider a frame S 0 moving at u = −v in which the right-hand particle is stationary. In S 0 the left-hand particle moves at From last lecture v−u 2v 0 v = = , 1 − uv/c2 1 + v 2 /c2 while the final product moves at v. In S 0 , then, 2mv initial momentum = mv 0 = 1 + v 2 /c2 final momentum = M v = 2mv. These are not equal. Conclusion: one or both of the Newtonian relations for mass and momentum are wrong!

26.2

Relativistic momentum

Suppose that mass is a function of speed m = f (v) m0 where m0 is the “rest mass” and f (v) a function: f → 1 as v → 0. Suppose that momentum is still given by p = mv = f (v) m0 v.

LECTURE 26. RELATIVISTIC MASS AND MOMENTUM

132

To work out f (v), consider the same inelastic collision as before. In S, the centre of mass frame, momentum conservation gives f (v) m0 v + f (−v) m0 × (−v) = 0. From this it immediately follows that f (v) = f (−v). Mass conservation gives (using this result): f (v) m0 + f (−v)m0 = 2f (v)m0 = M0 , where M0 is the rest mass of the combined particle (stationary in S). Let’s abbreviate f = f (v) and f 0 = f (v 0 ). In S 0 , we have: momentum conservation: f 0 m0 v 0 = f M0 v = 2f 2 m0 v, mass conservation: f 0 m0 + m0 = f M0 = 2f 2 m0 , where we eliminate M0 using the previous result. Dividing out m0 gives us two equations f 0 v 0 = 2f 2 v,

f 0 + 1 = 2f 2 .

Eliminating f 0 :


2f 2 v = 2f 2 − 1 v 0 2v . or v 0 = 2 − 1/f 2 We can compare this with the velocity transformation formula: v0 =

2v 1 + v 2 /c2

to give v 2 /c2 = 1 − 1/f 2 1 ⇒ f2 = 1 − v 2 /c2

⇒

1/f 2 = 1 − v 2 /c2 1 ⇒ f=p = γ(v). 1 − v 2 /c2

NB The derivation of f (v) is a plausibility argument, not a proof, because we had to assume that the same mass function applied to both momentum and mass conservation. It must be tested by experiment, which it is, every day, in particle accelerators around the world.

LECTURE 26. RELATIVISTIC MASS AND MOMENTUM

133

Thus conservation of mass and momentum continue to work if the mass and momentum of a particle of rest mass m0 and velocity ~v are given by: m = γm0 , ~p = γm0~v . with γ = γ(v). Note that these reduce to the Newtonian m = m0 , ~p = m~v for v  c.

26.2.1

Nothing travels faster than light

As v → c, γ(v) → ∞, so p → ∞. Very different from the Newtonian formula where p → mc remains finite as v → c. Since force equals rate of change of momentum, no amount of force can accelerate a particle of non-zero rest mass to v = c, let alone v > c.

26.2.2

Mass

In the derivation we found that M0 = 2γm0 . Two particles of rest mass m0 , each moving at speed v, collide and produce a single particle of rest mass 2γm0 > 2m0 . We seem to have created mass from nothing! We will return to this in the next lecture. Example 26.1. How fast does a particle have to travel for its momentum to be twice the Newtonian estimate? Answer. We need v such that γ(v)m0 v = 2m0 v, or γ = 2. Since 1 γ=p , 1 − v 2 /c2 then

v2 1 1 − 2 = 2, c γ

LECTURE 26. RELATIVISTIC MASS AND MOMENTUM so v = c

134

r √ r 1 1 3 1− 2 = 1− = = 0.866. γ 4 2

Example 26.2. A particle of rest mass m0 travelling at v = 45 c collides and merges with an identical but stationary particle. What are the speed V and rest mass M0 of the combined particle? Answer. For v = 54 c, γ = 53 . Initial mass = γ(v)m0 + m0 = 83 m0 . Therefore the final mass M = γ(V )M0 = 83 m0 . Initial momentum p = γ(v)m0 v = the final momentum is

5 3

× m0 × 45 c = 34 m0 c. Therefore

P = γ(V )M0 V = 43 m0 c. Dividing, P = 12 c, M 1 2 so γ(V ) = √ = √ = 1.155 1 − 0.25 3 M 8/3 4 and hence M0 = = √ m0 = √ m0 = 2.309 m0 . γ(V ) 2/ 3 3 V =

[Newtonian answers: V = 25 c, M0 = 2m0 .]

Lecture 27

Relativistic Energy Keypoints from Lecture 26 The addition of velocities means that the Newtonian formula for momentum, ~p = m~v , with m constant, cannot be right. We presented a plausibility argument based on a head-on, symmetric, inelastic collision of two particles, that m = γm0

and

~p = γm0~v ,

p where m0 is the rest mass and γ = 1/ 1 − v 2 /c2 . The merged particle has more rest mass than the combined rest mass of the two original particles: rest mass is not conserved! p → ∞ as u → c: no object with mass can travel as fast as light.

Objectives • E = mc2

135

LECTURE 27. RELATIVISTIC ENERGY

27.1

136

Force

Force equals rate of change of momentum along with ~p = γm0~v leads to: ~ = d~p , F dt d~v dγ = γm0 + m0~v , dt dt dγ = γm0~a + m0~v . dt γ is changing with time because v is changing with time. ~ = ~ is not even necessarily parallel to ~a. Not only do we see F 6 m~a, F Furthermore, force is not invariant between frames and does not have simple transformation properties. Focus instead on momentum and energy.

27.2

Kinetic energy

We first encountered kinetic energy by considering a force F doing work: dp dp dW = F dx = dx = v dt = v dp. dt dt
In classical mechanics, p = mv, so dW = mv dv = d 12 mv 2 , and the total work is obtained by integrating (from rest), Z v W = v dp = 12 mv 2 , 0

which we call the kinetic energy. In relativity, kinetic energy is simply the total energy minus the rest energy. We define it in a similar way, but must take account of the variation of m with v, through m = γm0 .   dW = v dp = v d(γm0 v) = m0 v d(γv) = m0 v γ dv + v dγ . Now we need dγ: −1/2  −3/2 (−2v) dγ d 3v = 1 − v 2 /c2 = − 21 1 − v 2 /c2 = γ × dv dv c2 c2 2 v v so dγ = γ 3 2 dv and v dγ = γ 3 2 dv. c c

LECTURE 27. RELATIVISTIC ENERGY

137

Hence we get  dW = m0 v γ + γ

3



v2 c2



 dv = m0 γv 1 + γ

2



v2 c2

 dv.

The expression in parentheses simplifies dramatically  2 v 2 /c2 1 2 v = 1 + = = γ2 1+γ 2 2 2 2 2 c 1 − v /c 1 − v /c So now we can write dW = m0 γ 3 v dv Looking back, we notice that γ 3 v dv = c2 dγ, so this result can be written even more nicely as dW = m0 c2 dγ = dm c2 where, finally, we have written m = γm0 , so dm = dγ m0 . This is a fascinating result. It means that the effect of applying a force and doing work is to increase γ, or equivalently the (velocitydependent) mass, rather than simply increase the velocity. We write the kinetic energy as the integral of dW , recalling that, at zero velocity, γ = 1, or m = m0 . Denoting it as K here, We need to Z distinguish it 2 2 from two other K = dW = (γ − 1)m0 c = (m − m0 )c . energies.

27.3

Newtonian limit

By the binomial theorem (1 + x)n = 1 + nx + · · · , in the limit v  c,
−1/2 v2 = 1 + 2 + ··· γ = 1 − v 2 /c2 2c   2 v Therefore K = 1 + 2 + · · · − 1 m0 c2 = 12 m0 v 2 + · · · 2c

27.4

as expected.

Equivalence of mass and energy

In the inelastic collision considered in the last lecture, two particles of rest mass m0 and speed v merged to make a single one of mass 2γm0 .

LECTURE 27. RELATIVISTIC ENERGY

138

The increase in rest mass is ∆m0 = 2(γ − 1)m0 =

∆K , c2

where ∆K is the total kinetic energy input. This suggests that energy and mass are inter-convertible, according to ∆E = ∆m0 c2 . Perhaps all mass is potentially convertible, and we can define a rest energy E0 = m0 c2 . In that case K = (γ − 1)m0 c2 = (m − m0 )c2 = E − E0 , where the total energy E is given by E = E0 + K = γm0 c2 = mc2 . E includes the rest mass energy E0 = m0 c2 plus kinetic energy gained through work. In relativity if one says “the energy of a particle” one means the total energy, E. The equivalence of mass and energy has been verified in countless particle physics experiments when high kinetic energy particles collide to create other particles. Example 27.1. The Sun radiates 3.8 × 1026 W of power. At what rate does its mass change as a result? Answer. As the Sun loses energy, it must change mass at rate dM 3.8 × 1026 =− = −4.2 × 109 kg s−1 . 8 2 dt (3 × 10 ) Example 27.2. The explosion of 1 t of TNT releases 4.2 × 109 J. What is the TNT equivalent locked up in 1 kg of matter? Answer. Equivalent TNT mass =

1 × (3 × 108 )2 = 21.4 × 106 t. 9 4.2 × 10

Example 27.3. Calculate the speed of an electron which has kinetic energy K = 30 keV.

LECTURE 27. RELATIVISTIC ENERGY

139

Answer. The rest energy of an electron E0 = m0 c2 = 511 keV, so K = (γ − 1)m0 c2

⇒

γ =1+

30 = 1.059 511

and, then, as usual v = c

 1/2 1 1− 2 = 0.328. γ

Lecture 28

Mass-EnergyMomentum Keypoints from Lecture 27 Total energy (often just “energy”): E = mc2 = γm0 c2 . Rest mass energy: E0 = m0 c2 . Kinetic energy: K = (γ − 1)m0 c2 . Total energy = rest mass energy + kinetic energy: E = E0 + K.

Objectives: • The close relationship of mass, energy and momentum in relativity

140

LECTURE 28. MASS-ENERGY-MOMENTUM

28.1

141

Mass, Energy and Momentum

In Newtonian mechanics, mass, energy and momentum are separately conserved, and independent quantities. In Special Relativity, energy and mass blur into one. Example 28.1. A kilogram of water is heated from 20 ◦C to 80 ◦C. By how much does its mass increase? Answer. Assuming that the specific heat capacity of water is 4200 J K−1 , the total energy input = 4200 × 60 = 252 kJ, so ∆m =

2.52 × 105 ∆E = = 2.8 × 10−12 kg. 2 16 c 9 × 10

This is partly stored in higher speeds of the molecules as kinetic energy, but one can equally regard this as an increase in rest mass. The key relations to remember are: ~p = γm0~v , E = γm0 c2 . These can be combined in a couple of ways. Firstly, dividing: pc v = c E which can be used to get the velocity from the energy and momentum. Secondly,
E 2 − p2 c2 = γ 2 m20 c4 − c2 v 2 ,   v2 2 2 4 = γ m0 c 1 − 2 , c = m20 c4 . Thus E 2 − p2 c2 = m20 c4 = E02 . m0 , the rest mass, is frame independent, therefore E 2 − p2 c2 is as well: it is “Lorentz invariant”. NB. This equation relates the squares of things with the dimensions of energy. It is not the same as saying that energy equals rest energy plus kinetic energy.

LECTURE 28. MASS-ENERGY-MOMENTUM

142

This relation applies to multiple particles, but it needs careful interpretation: • E=

P

the total energy in a given frame;

• ~p =

P

the total momentum in that frame;

i Ei ,

pi , i~

• E0 is the total energy in a frame for which ~p = ~0. Very useful relation for particle collisions.

28.2

Massless particles

The above relation allows us to discuss particles with zero rest mass. E 2 − p2 c2 = m20 c4 → 0, so massless particles obey the relation E = pc, (strictly |p| in one dimension, or |~p| in 3D). The one certain example of a massless particle is the particle of light, the photon. As a corollary, since v/c = pc/E, we see that massless particles travel at the speed of light. Example 28.2. A stationary spacecraft, equipped with a “photon drive”, annihilates half of its mass (stored as a matter/anti-matter mix), converting it into photons, and directing the resulting radiation in a narrow beam. What is its speed afterwards? Answer. Let the spaceship rest mass be m0 before conversion, and 21 m0 afterwards. After the conversion, let the spaceship have energy ES and momentum ~pS , while the photons have energy EP and momentum ~pP . Momentum conservation implies that ~pS = −~pP , so we can set p = |~pS | = |~pP |. The photon energy obeys EP = pc, and so energy conservation

Until fairly recently, neutrinos were also believed to be massless, but this is no longer the case.

LECTURE 28. MASS-ENERGY-MOMENTUM

143

implies that m0 c2 = ES + EP , or pc = m0 c2 − ES . The spaceship energy obeys the relation
2 4 1 m c = ES2 − p2 c2 0 2 = (ES − pc)(ES + pc) = (2ES − m0 c2 )m0 c2 and so we can write 2ES − m0 c2 = 41 m0 c2 2ES = 45 m0 c2 ES = 45 ( 12 m0 c2 ) so γ = 54 . This gives β = v/c = 35 .

28.3

Particle Accelerators

Particle accelerators use kinetic energy to create new particles. The Lorentz invariant E 2 − p2 c2 = E02 places important constraints on accelerators. This is both invariant between inertial frames and during collisions since E and p are conserved. Example 28.3. A particle of rest mass m0 = 1 GeV/c2 is accelerated to a (total) energy of E = 100 GeV and collided with an identical, but stationary particle, to create a single particle. Calculate the rest mass of the final particle. Answer. Let E0 = m0 c2 be the rest energy of both the particles, and E the energy, and p the momentum, of the moving particle. Then these quantities will satisfy E 2 − p2 c2 = E02 . Conserving energy and momentum: Ef = E0 + E,

and pf = p,

NB Most of the energy is lost in the photons.

LECTURE 28. MASS-ENERGY-MOMENTUM

144

where Ef and pf are the energy and momentum of the final particle. If the rest mass of this particle is M0 , M02 c4 = Ef2 − p2f c2 , = (E0 + E)2 − p2 c2 , = E02 + 2E0 E + (E 2 − p2 c2 ), = 2E02 + 2E0 E. Putting the numbers in p 2 M0 c = 2 × 12 + 2 × 1 × 100 = 14.2 GeV, so M0 = 14.2 GeV/c2 . Now consider a head-on collision of two such particles, each with E = 100 GeV. Now all the input energy = 200 GeV can go into the product. The first method wastes energy in the motion of the final particle, required to conserve momentum. Example 28.4. Show that even if a photon has enough energy, it cannot convert spontaneously into an electron-positron pair. Answer. Useful trick: move to the “centre of mass” frame of the e− e+ pair in which p = 0, but E > 2me c2 . But photons are massless, i.e. E = pc. It is impossible to satisfy these conditions whilst conserving E and p, QED. Photons can produce e− e+ pairs if another particle is present to conserve momentum (e.g. a nucleus).

Lecture 29

Relativistic Geometry Keypoints from Lecture 28 Mass and energy are interconvertible. Total energy E and momentum p lead to a Lorentz invariant: E 2 − p2 c2 = m20 c4 . For massless particles (photons): E = pc. Dual-beam particle accelerators can put much more of their energy into new particles than single beam plus stationary targets.

Objectives • The geometry of spacetime

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LECTURE 29. RELATIVISTIC GEOMETRY

29.1

146

Relativistic intervals

A familiar analogy: rotation of coory’ dinate axes about the origin ! ! ! x’ = 0 0 x cos θ sin θ x = 0 y − sin θ cos θ y

y x=0

The quantity r2 = x2 +y 2 is invariant to such a rotation: x2 + y 2 = x02 + y 02 .

y’ = 0 x’ 1

y=0 x

Similarly, the distance between any two points ∆r2 = ∆x2 + ∆y 2 is invariant. Rotations leave lengths unchanged. The x- and y-axes are defined by y = 0 and x = 0; the circle uses the preservation of length to establish the scale on the rotated axes. The equivalent quantity for Lorentz Transformations is the “interval” s where (omitting y and z as they are unchanged) s2 = c2 t2 − x2 . s2 is invariant between frames: c2 t2 − x2 = c2 t02 − x02 . Similarly, the interval, ∆s2 = c2 ∆t2 − ∆x2 , between any two events, is also Lorentz invariant. We seek a similar description of LTs using spacetime diagrams which are plots of ct vs x. The geometric nature of Special Relativity was first appreciated by Minkowski.

See Problem Sheet 7

LECTURE 29. RELATIVISTIC GEOMETRY The Lorentz Transformations can be written, with β = u/c, as ! ! ! 0 ct 1 −β ct = γ x −β 1 x0

ct

See Problem Sheet 9

ct’

x=0 x’ = 0

The x0 - and ct0 -axes, equivalent to ct0 = 0 and x0 = 0, appear in S as: ct = βx,

147

ct’ = 0

x’

ct = 0 x

ct = β −1 x.

This fixes the directions of a grid of constant values of t0 and x0 ; how about the spacing? In other words, how do we set the units on each axis? Consider

2

2 2

2

2 02

02

s = c t − x = c t − x = −1 or x2 − c2 t2 = x02 − c2 t02 = 1 This is a hyperbola passing through x = 1, ct = 0, and x0 = 1, ct0 = 0. Its intersection with the line ct0 = 0 defines unit length on the x0 axis. A similar hyperbola, s2 = +1, passing through x = 0, ct = 1, and x0 = 0, ct0 = 1, may be used to define the units on ct0 . • T.D. is time dilation: a second in S 0 appears longer in S. • L.C. is length contraction: a metre in S 0 appears shorter in S.

NB: s2 can be negative!

LECTURE 29. RELATIVISTIC GEOMETRY

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148

Past, Future, Causality

The relativistic space-time interval helps us classify pairs of events. FUTURE

x’ x D B

ELSEWHERE

ELSEWHERE

ct ct’

C

A

PAST

time-like intervals:

∆s2 > 0, c2 ∆t2 > ∆x2 (in any inertial frame). There exists a frame S 0 in which the events occur in the same place and are separated only in time. On the diagram, A—B is such an interval. In other words, for some choice S 0 , the ct0 axis (x0 = constant) can be made parallel to the line A—B. Also, since c2 ∆t2 = ∆s2 + ∆x2 , that particular frame is the one in which ∆t2 is a minimum.

space-like intervals:

∆s2 < 0, c2 ∆t2 < ∆x2 (in any inertial frame). There exists a frame S 0 in which the events occur at the same time and are separated only in space. On the diagram, C—D is such an interval. In other words, for some choice S 0 , the x0 axis (ct0 = constant) can be made parallel to C—D. Also, since ∆x2 = c2 ∆t2 + (−∆s2 ), that particular frame is the one in which ∆x2 is a minimum.

For a time-like interval, a signal (travelling no faster than c) could in principle connect the two events in any frame. This also means that, if ∆t > 0 in one frame, then ∆t0 > 0 in any other frame. (And likewise, ∆t < 0 in one frame, then ∆t0 < 0 in any other frame). This, in turn, means that we can make a causal connection between the events, i.e.

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one of them (the earlier one) could, in principle, have caused the other (the later one) without any inconsistency between frames. For a space-like interval, a signal connecting the two events would have to travel faster than c (in any frame). This also means that there exist some frames in which ∆t and ∆t0 have opposite signs. In this case we cannot make a causal connection between the events, because observers in different frames would disagree on which came first. For an observer at the origin, space-time is divided by the lines x = ±ct into 1. “future” and “past” time-like events, causally connected to the observer by possible signals travelling at, or less than, the speed of light; 2. space-like events called “elsewhere”, not causally connected to the observer, because the connecting signals would have to travel faster than light. In two or three spatial dimensions, a “light cone” defined by c2 t2 = x2 + y 2 + z 2 divides space-time into these two classes of events.