An Introduction to Tensor Calculus Relativity and Cosmology 3rd Ed~tqw~_darksiderg

An Introduction to Tensor Calculus, Relativity and Cosmology Third Edition

,Y

D. F. Lawden

Department ol Mathematics The Unil'ersity (f Aston in Birminyham

• 'I

~

"f.( "

1807~19ll2 fillr, e.g. l'A u /1" )' This contraction is also referred to as t with respect to the index j and we shall write divergence of A,j (12

13. Pseudotensors '.!I,; is a pseudo tensor if, when the coordinates arc subjected to the tramformati (X. I), its components transform according to the law

'.ITij

= IAI(/"(/JI '.!l"r

(13

I A 1 being the determinant of the transformation matrix A. Since for orthogol transformations IAI = ± I (equation (8.10)). relative to rectangular Cartesi fmmes, tensors and pseudotensors are identical except that, for certain changes axes, all the components of a pscudotem.or will be reversed in sign. For examl if in Iff 3 a change is made from the right-handed system of axes to a left-hane system, the determmant of the transformation will be - 1 and the cornponent~ a pseudotensor will then be subject to this additional sign change. Let l'u. • be a pseudotem.or of the Nth rank which is skew-symmetric.", respect to every pair of indices. Then all its components are zero, except those which the indices i, j, ... , n are all different and form a permutation of

30 numbers 1,2, ... , N. The effect of transposing any pair of indices in cij ... n is to change its sign. It follows that if the arrangement i,j, . .. , n can be obtained from I, 2, ... , N by an even number of transpositions, then cij n = + 1"2. N' whereas if it can be obtained by an odd number cij .. n = - 1'12 N' Relative to the x;-axes let c l2 N = 1. Then, in this frame, cij .. n is 0 if i,j, '.' . , n is not a permutation 01 1, 2, ... , N, is + I if it is an even permutation and is - I Ilit is an odd permutation. Transforming to the .x',-axes, we tlnd that

( 13.2) But cij • is also skew-symmetric with respect to all its indices, since this is a property preserved by the transformation. Its components are also 0, ± I therefore and cij .. • is a pseudotensor with the same components in all frames. It is called the fA'"i Cir'ila pseudotensor. It may be shown without dilliculty that: (i) the sum or difference of two pseudotensors of the same rank is a pseudotensor. (ii) the product of a tensor and a pseudotensor is a pseudo tensor. (iii) the product of two pseudotensors is a tensor. (iv) the partial derivative of a pseudotensor with respect to Xi is a pseudotensor. (v) a contracted pseudotensor is a pseudotensor. Thus, to prove (iii), let '.!Ii , 'B i be two pseudovectors. Then

'lI, 'Bj

=

I

A 12 ai,ajl '.!l" 'B1

= (li,a jl

'.!l" 'B1

(13.3)

The method is clearly quite general. The remaining results will be left as exercises for the reader.

14. Vector products. Curl Throughout this section we shall be assuming that N = 3, i.e. the space will be ordinary Euclidean space. Let A;, B; be two vectors. Then cu,A,B, is a pseudotensor of rank 5. Contracting twice, we get the pseudovector (14.1)

whose components are

I) at rest disintegrates into three fragments (each or rest mass m) which move apart in directions making equal angles with each other. Show that, in a rrame or rererence in which one or the rragments is at rest, the angle between the directions or motion or the other two rragments is 2 cot 1 (J3.l.). 41. A positron travelling with velocity 3c/5 is annihilated in a collision with a stationary electron, yielding two photons which emerge in opposite dire;.;tions along the track or the incoming particle. II' m is the rest mass or the electron and positron, show that the photons have energies 3mc 2 /4 and 3mc 2 /2. 42. A positron having momentum p collides with a stationary electron. Both particles are annihilated and two photons are generated whose lines or motion make equal angles 0: on opposite sides of the line or motion orthe positron. Prove

67 that p siniX taniX "" 2mc where m is the rest mass of both the positron and electron, If iX = 60", calculate the velocity of the positron. (Ans. 4c/5.) 43. A particle of rest mass mo collides elastically with an identical stationary particle and, as a result, its motion is deflected through an angle O. If T is its KE before the collision and T' is its KE afterwards, show that

m.

44. A particle of rest mass collides elastically with a stationary particle of rest mass m 2 « m.) and, as a result, is deflected through an angle iI. If E, E' are the total energies of the particle m I before and after collision respectively, prove tha t cosO

=

(E +m1(1)E' -m l c 2 E -mic 4 --

. . .. - .. ~ .---.-.- " ' -

[(e -mic 4 )(E'l

-mic4 )]1/ 2

45. A pion having rest mass mo is moving along the x-axis of an inertial frame Oxyz with speed 4c/5, when it disintegrates into a muon having rest mass 2m o/3 and a neutrino. The neutrino moves parallel to the y-axis. Prove that the angle made by the muon's velocity with the x-axis is tan - I (l/8)and calculate the energy of the neutrino. (Ans. m oe l /6.) 46. A nucleus having rest mass mo disintegrates when at rest into a pair of identical fragments of rest mass (A < I). Show that the speed of one particle relative to the other is 2 -A 2 )C/(2 _A l ). If A is small, show that this speed is less than c by a fraction A4 /8. 47. A positron collides with a stationary electron and the two particles are annihilated. Two photons are genera ted, the lines of motion of which make angles of 30" and 9tr with the original line of motion of the positron. Calculate the original velocity of the positron and show that the energy of one of the photons i~ equal to the in ternal energy of the electron. (The rest masses of an electron and 2 positron are equal.) (Ans. J3c/2.) 48, A moving positron collides with a stationary electron, Both particles an annihilated and two photons are generated, the lines of motion of which bot~ make angles of 60" with the original line of mOl ion of the positron, Prove that the total energy of each photon is 4m oc 2 13, where mo is the rest mass of the positron and of the electron. 49. A nucleus having rest mass mo is moving with velocity 4('/5 when it emits a photon having energy m oe l /3 in a direction making an angle of 60" with the line of motion of the nucleus. Show that the subsequent direction of motion of the nucleus makes an angle tan' I (J3/7) with its initial direction of motion anJ calculate the new rest mass of the nucleus, Show that, relative to an inertial frame in which the nucleus was initially at rest, the line of motion of the photon makes an angle of 120' with thc original direction of motion of the nucleus. (Am. -J 13m o/6.)

J(I

pm o

68 50. A pion has rest mass mo and momentum P when it disintegrates into a pair of photons having energies E and E'. The directions of motion of the photons are perpendicular and that of the photon having energy E makes an angle rx with the original direction of motion of the pion. Prove that

P = moc(sin 2rx) -

J d cot rx),

E = moc l

If],

£' = moe J(~tanrx). l

51. A particle having rest mass mo is moving with an unknown velocity when it absorbs a neutrino whose energy is 3m oc l /2. The angle between the paths of the particle and neutrino is rx, where cosrx = 1j3_ After absorption, the rest mass of the particle is 2m o- Calculate the original velocity of the particle and show that its path is deflected through an angle fI, where tanli = 4 J2/5, as a result of the encounter_ (AIlS_ 3c/5.) 52. A particle whose rest mass is mo moves along the x-axis of an inertial frame under the action of a lorce , 2m oc l u j = (~-=-~l AI time t = 0, the particle is at rest at the origin 0, Show that the time Iaken for the particle to move from 0 to a point x ( < 0) is given by ( =

[XJ' /]

I 3c (/

(x

+ 3u)

53. Oxr are rectangular axes of an inertial frame. A particle having rest mass m o is projected from the origin with momentum Po along Ox. II is acted upon by a constant force/parallel to Or. Show that its path is the catenary y = Wo - ," cosh ( 'Ix - I .f CPo

)

where w5 = m5c4 + p{,c 2 54, A particle having rest mass mo moves along a straight line under Ihe action ofa frictional force of magnitude mou/k opposing its motion; I' is the speed of the particle and k is a constant. Show that the time which elapses whilst the particle's velocity IS reduced from 41'/5 to 3c/5 is [log (3/2) + 5/12Jk. 55. A particle having rest mass mo moves on the x-axis under an attractive force to the origin of magnitude 2m oc l /x l . Initially It is at rest at x = 2. Show that its motion is simple harmonic with period 471:/(. 56. A space ship, with its motors closed down. is moving at high velocity (J lhrough stationary interstellar gas which causes a retardation as measured by the crew of magnitude rxl'], Show that the distance it moves through the gas whih;t its velocity is reduced from V to U is I rx

II _! x

log 1+ x I -x

I

V U

57. A particle has rest mass mo and 4-momentum P. An observer has 4-velocit) Y in the same frame. Show that, for this observer, the particle's: (i) energy is - p. Y; (ii) momentum is of magnitude J[P 2 + (p.y)1/c2]; (iii) velocity is of magnitude J[I +e 2p 2/(p.y)2J. (HilI(: All these expressions are invariant.) 5R. A particle has rest massm u and moves along the x-axis under the action of~ force given at any point having coordinate x by ,

f

m uc 1w 2x

= -- i;T=- (,l~]+ ~/ ~])3/]

wand a being constants. It is projected from the origin with velocity wa. Shoy that its velocity at any later time is given by 1,2 = (li(a 2 -- x 2 ). What does thi, imply for the particle's motion? 59. A particle of rest mass mu moves along the x-axis of an inertial frame undethe action of a force m oe 2

,

f

=

2(1

+ 2X[/2),1/2

At time 1 = 0, the particle is at rest at the origin. Show that, at any later time coordinate is given by x

=

I,

it,

2 + el ~ 2 J (I + et)

60. A particle of rest mass /flo moves under the action of a central f01'ce. (r, l:) are its polar coordinates in its plane of motion relative to the force centre as poll. VIr) is its potential energy when at a distance r from the centre. Obtain Lagrange's equations for the motion in the form d. )2 (yr)~yrl d1

+ ..I

/flo

= 0,

V'

d 2. ., (yr 0) dl

=0

where 1'= [1_(r 2 +r 2()2)/e 2] [/2 Write down the energy equation for the motion and obtain the differential equation lor the orbit in the form h

22(d II

2 tl

dlf + U

)

=

C~V, m~(:2 V

where u = I/r and It, C are constants. In the inverse square law case when V ~ ~ p/r, deduce that the polar equation of the orbit can be written

lu = 1+ /,COSI/O 2 where /12 = I _p2 /m~h2 e . If p/molJ( is small, show that t he orbit is approximately an ellipse whose major axis rotates through an angle Trp2 /m~h2 (2 pu revolution. 61. A particle, having rest mass mo, is at rest a; the origin of the x-axis at time 1 = O. It is acted upon by a force f: directed along the posItive x-axis, whOle

70 magnitude when the particle's velocity is I' is given byf = m o ke 2 Ill. Show that at time I( > 0), I' = e sinO, where 0 is positive acute and sec/! = I + kl. Deduce that, at t he same time, the coordinate of t he particle is given by x = e(tan/!- O)/k. 62. If v is the 3-velocity ofa particle and {i = (1- ,,2I( 2 )- [/2, prove that v'v = I,i, and V'

d (flv) = (h'D dl

If mo is t he particle's rest mass. define the 3-force f acting upon it and deduce from the above result that v'f= rile 2 , where m is the inertial mass. 63. A particle having rest mass mo moves along the x-axis under a force of attraction towards the origin - mooix. It is initi>llly at rest at the point x = a.. Show that the velocity with which it passes through the origin is wae

J (4e 2 + oill 2)

~--2e2+ (~2 ~2-

64. If the force f always acts along a normal to a particle's path, show that the sreed I' of the panicle is constant. Write down the equation of motion of the rarticle and deduce Ihat till" curvature of Ihe rath is given by K =.Ilml,2 If the particle moves in a eircle 01 radius a under a COnSlal1l radial fon;e.l; show that its speed I' is given by

where A =.IiJl2moe2 and mo is the particle's rest mass. 65. A nucleus is moving along a straight hIll' when it ejects an electron. As measured by a stationary observer, the speed of the electron is !c and the angle between the lines of motion of the nucleus and electron is 60". If the speed of the electron relative to the nli(!cus is also calculate the speed of the nucleus. (Ans.

ie,

8cI17.) 66. A particle having rest mass mo. initially at reSl at the origin of an inertial frame. moves along its x-axis under t he act ion of a variable force fdirected along the axis and given by the formula f = m oe l l2 (I + X). Show that the particle's velocity I' is given by I' = Xl /2 el(l + X)I /2. Putting x = sinh 20, if ( is the time and ( = 0 at O. prove that c( = /I + sinhllcoshll. 67. A particle having rest mass mo is moving with speed when it is subjected to a retarding force. When the particle's inertial mass is m. the magnitude of the retarding force is IXm 2 (IX is constant). Show that the time needed by the force to bring the particle to rest is rcc/6m olX. 68. If T u is the energy--momentum tensor for an elastic fluid and V, is its 4velocity of flow, by verifying the equation T u Vi = -C 2 /1 00 V; in a frame in which the fluid is momentarily at rest, prove it in any frame. Deduce the equations

J

ie

II. = (/loov. + r.pt'pl( 2)1 (! _v 2 Ie]),

/l =

1100

+ 1I.1'.le]

71 Hence derive the following formulae for the clements of l~j:

T,/I = lion V, VII

+ r,p + T" V, V/de2 + T,/I Vp V4 /e 2

7:'4 = T4 , = II"" V, V.

1~4 = /1"0 V4 V4 -

V, Vp /e

T,p

2

69. A perfed fluid is streaming radially outwards across the surface of asp here with radius H and centre O. If the motion is steady and there IS no exterml force field, show that equation (21.20) leads to the equations

d dr

(r

3

('A)

-I r

2 dp

d). r'-r

= 0,

dr

+ 3A

u

= 0

where r is radial distance from 0, p is the pressure, v is the speed of flow and A = (/loo+p/e 2 )r/r(l-t,2/e2). If p vanishes over the sphere r = Rand p= Pat great distances, and if /10" is constant outsidc the sphere, show that in thi\ region

p = (P

+ e 2 /1(0) J (I

r 11'(I-t,2/e 2 )

=

\/2

--

(,2 /(

2

) - e21/oo

R 2 J(p2 +2Pe 2/1o,,)/e/l,,0

70. A straight rod has cross-sectional area A and mass m per unit length. It lies along the x-axis of an inertial frame in a state of tcnsion F. Show Ihat the encrgy momcntum tensor has components whIch art: Ihe clements of a 4 x 4 diagonal matrix. with diagonal elements ( - F/ A, 0, 0, - me 2 / A). Deducr that ar observer moving along the x-axis with speed u, "ees the inertial mass per uni, length of the rod to be

m --

FU 2/c4

-- I .:.: ~i/~i 2

Deduce that F cannot exceed me • 71. Assuming that the energy momentum tensor T i} is a tensor with respect t, a general Lorentz transforma tion Xi = ai}x} + hi, write down the transfclrmatio equations for T u in the special case where a4, = a,4 = 0, a44 = I. Deduce th2 /I, g" g,P are 3-tensors with respect to a simple rotation of the frame Ox,x 2 -, without relative motion. 72. Relative to a frame S, a fluid has flow velocity (u, 0, 0) at a certain point. I the frame SO rela tive to which the fluid is s ta tionary at the point, the stress tenSe has components I~pand the fluid density is /100' Show that the energy- mDmentul tensor in the frame S has components

T

_ 1 I --

o

1'1

I

+l/ooU

=_ u2 /(:2

2 '

('2/100

T

+ I? \

iu

- ------- '-l_u 2 /e 2 e

\4-

T 22

= I~2'

T 13

=

r~_"

72

e

2

T44 = _.-

/IO O

T 31

= (I _u 2(e 2 )- 1!2'~1'

T\4

= (I

2 -u (e 2

r 1!2'~3iu/e.

+ T~ 1112("2

-1= utleY

-

and dcdul:c that TI I

r\) 1-

T 21 =

(l _u2/(2)112T~I.

T 21

= T~2'

'2.\

T.n = (I _U2/('2)112 and A respectively, of the field. It is proved in textbooks devoted to the classical theory (Coulson and Boyd, 1979) that A satisfies the equations . I iltj> dlv A +~---

(,2ilt

=0

V2A-~ il2~ = ('1

(ltl

-

(25.1)

Jloj

(25.2)

and tj> satisfies the equation (25.3)

75 where ('2 = 1/ J10/;0' We now define a 4-vector potential 0 in any inertial frane S by the equation 125.4) 0= (A, iNc) It is easily verified that equations (25.2), (25.3) are together equivalent equation

~)

the

25.5) where the operator [J2 is defined by [25.6)

The space components or equation (25.5) yield equation (25.2) and the time component, equation (25.3). If Q" J i are the components or nand J respecively, equation (25.5) can be written [25.7) in which form it i~ clearly covariant with respect to Lorentz transformllions provided n is a vector. This confirms that equation (25.4) does, in fact, ddine a quantity with the transformation properties of a vector in space· time. Next, it is necessary to show that equation (25.1) IS also covariant with nspeel to orthogonal transformations in space time. It is clearly equivalent 1D the equation (25.R) divO=Q'.i=O which is in the required form. J heing given, n is now determined hy equations (25.7) and (25.R).

26. The field tensor When A and 4> are known in an inertial frame, the electric and magnetic intelsities E and B respectively at any point in the electromagnetic field follow from the equations E = - grad

4) - ?A ..

(26.1)

(It

(26.2)

B = curl A

Making use of equations (4.4) and (25.4), these equations are easily shown to be equivalent to the set

= ~!.?4

_ iE C

x

(lX,

i. (lQ - E = -.4 -

DQ 1 (lX 4 (lQ 2

.

(lX2

(lX 4

i.

(lQ4

DQ,

1'['"

=. (lX-,

('

y

(lX

4

(26.3)

76 (1Q

(lQ

B, = -1--' B .1' =

(Ix]

(lQ,

i' Q .1

(l

i"'\ I

XJ

(lQ

B,

2

( X2

=.

).

(1.\ I

(26.4)

(lQ\ (1\2

Equations (26.3) and (26.4) indicate that the six components of the vector~ i E/,., B with respect to the rectangular Cartesian inertial frame S are the six distInct non-J:ero components in space time of the skew-symmetric tensor n J . , -- n,. i' We have proved, therefore, that equations (26.1), (26.2) are valid in all inertial frames if

B,

-- B v

0

Bx

BI'

- Bx

0

-iE),.

iE),.

if),.

iE),.

()

-- B, 0 (I'i;! = (

. IE'/') -

if~·\,j,.

(265)

is assumed to transform as a tensor with respect to Or! hogonal transforma tion~ in space time. The equations (26.3) and (26.4) can then be summarized in the tensor equation (266) F,j is called the electromugrwti,.jidd II'fJ.\or_ The close relationship between the electric and magnetic aspects of an electromagnetic field is now revealed as being due to their both contributing as (':Jmponents to the field tensor which serves to unite them. Consider now equations (3.2) and (3.3). Employing the field tensor defined by equation (26.5)and theeurrent density given by equation (24.7),and recalling that B = J10 H, D = Do E, these equations are seen to be equivalent to

(26.7)

or, in short, (26.8)

an equation which is covariant with respect to Lorentz transformatIOns. Finally, consider equations (3.1) and (3.4). These can be written (~FI4

42 + ?F 2,1 =0 (lX 4 (~FI 1 (11"'4+ . + . =() +

(lF

41

tl \

(IF

(lx..,

(1.\2

(1

.,

(' I- 12

+ i'I' 24 (lX

(1'\4 (lF

\4-

(2t

('XI

+

t

("F

41 =0

(lX 2

(IF (IF\! +. '. + , 12 =0

B

(lX

;"":Xl

2

(·X.\

These equations arc summari)'ed thus: (26. lfany pair from i, i, J" arc equal, since F'j IS skew-symmetric, the left-hatld meml of this equation is identically )'efll and the equation is trivial. The fOIl' possi cases when i, i, J,; arc dIstinct are the equations (26.9). Equation (26.10)is a len: equal ion and is therefore also covariant With respect to Lorentz transformatio To sum lip, Maxwell's equations in 4-dimensional covariant form are: F,j.

F'j. ,

j

= It"J,

+ Fj" , + F". j = 0

l

(26.

.r

Given J, at all pomts in space time, these equations determine the iield tensor The solution can be found in terms of a vector potential Q, which satisfies following eq ua tions:

Q", =0 } /loJ,

(26.

Q,.)j = -

n, being determined,

F'j follows from the equation F'j =

Qj., -Q"

27. Lorentz transformations of electric and

(26.

j

ma~netic

vectors

Since F'j is a tensor, relative to the special Lorentz transformation zero components transform thus:

(~.1)

its n,

F2 ., = 1-'2.1 F.11

= F.1lcoslX + FJ4sin

IX

(2

F I2 = Fl 2coslX+F 42 sinlX

FI4

=

1-'14

(24= -F 2 ,sinlX+F 14 coslX P.14

= -

F.11

sin IX + F.14COS IX

(2

78

Substituting for the components of Fij from equation (26.5) and for sin IX, cos IX from equations (5.7), the above eqnations (27. f) yield the special LorenlJ: transformation equations for B, viz. (27.3) Similarly, equations (27.2) yield the transformation equations for E, viz.

Ex=f;"

E,.={I(Ev-uB,),

[,={I([-;,+uB v )

(274)

The inverse equations can he written down hy exchanging 'harred' and 'unbarred' symhols and replacing 1/ by -II. As an example of the usc to which these transformation formulae may he put, consider the electromagnetic field due to an infinitely long, uniformly charged wire lying at rest a long the .x-axis of the inertial frame ,'i. I I' lJ is the charge per unit length, it is well known that the electric intensity is everywhere perpendicular to the wire and is of magnitude i//(2m: oi'), where r is the perpendicular distance from the wire. Thus, at the point (.x,)'. z), the components of E are given by

Ex = 0,

(27.5)

The magnetic induction vanishes, The electromagnetic field observed from the parallel inertial frame S (relative to which S has velocity (u, 0, 0)) is given by the inverses of equations (27.3) and (27.4) to have components B x = 0,

By =

{Iuliz 2~;~C2 (y2+ Z2) ,

(276)

at the point (x, y, z) (having used the transformation equations y = }'. z = z). A segment of the wire having unit length in S' will appear in S to have length J (1 - 11 2/(.2); however, the charge on the segment must be the same in both frames, viz. q. It follows that the charge per unit length as observed in S is if = {Ii]. Thus, the charge which flows past a fixed point on the x-axis of S in unit time will be {lUi] = i; i therefore measures the current flowing along this axis. Since c 2 = l/p",;", equations (27.6) can now be written 13, = 0,

By

=

Poiz 2IT(y2

+

Z2)'

(27.7)

These equations imply that the magnetic induction is of magnitude p oi/2lTr and

that the B-lines are circles with centres on Ox and planes parallel to Ovz. Th s result f(lr a long straight current is a well-known one in the classical theory. Tic electric intensity is of magnitude qj2nfor and is directed radially from the wire; however, in the case ofa current due to the flow of negatively charged electrons n a stationary wire, this field IS cancelled hy the contrary field due to the positi 1c charges on the atomic nuclei.

28. The Lorentz force We shall now calculate the force exerted upon a point charge e in motion in is the longit ude of any point on the Earth's surface, the distancc d,Y between the points (Ii, (Pl, (II + dll, C,o;;S

continuous

98 and specifies an affine connection between the points of Y' N' A space which is affinely connected possesses sutlicient structure to permit the operations of tensor analysis to be carried out within it. For we can now write < - iJA'd X 1 _ ['kijA d X 1 d Ai - vA ik 0

•

ox)

(lA,

= ( '~j (~x

k

fijA k

)

dx l

.

(33.9)

But. as we have already explained. the left-hand member of this equation is a vector for arbitrary dx 1 and hence it follows that (33.10) is a covariant tensor, the covariant derivative of Ai' It will be observed from equation (33.10) that, if the components of the atlinity all vanish over some region of .'I' N. the covariant and partial derivatives are identical over this region. However, this will only be the case in the particular reference frame being employed. In any other frame the components of the affinity will, in general, be non-zero and the distinction between the two derivatives will be maintained. In tensor equations which are to be valid in every frame. therefore, only covariant derivatives may appear, even if it is possible to find a frame relative to which the affinity vanishes. We have stated earlier that. when defining an affine connection, the components ofan aflinity may be chosen arbitrarily. To be precise, a coordinate frame must first be selected in .'I'N and the choice of the components of the affinity is then arbitrary within this frame. However, when these have been determined, the components of the affinity with respect to any other frame are, as for tensors, completely fixed by a transformation law. We now proceed to obtain this transformation law for affinities.

34. Transformation of an affinity The manner in which each of the quantities occurring in equation (33.10) transforms is known, with the exception of the aflinity 1. The transformation law for this affinity can accordingly be deduced by transformation of this equation. Relative to the x-frame, the equation is written

n

-

Ai;]

=

(lAi

75ii

r~lAk

(34.1)

Since Ai, A,;j are tensors, (34.2)

ox' ox'

~.

A=----A ••J O.Xi oi] ,.t

(34.3)

Substituting in equation (34.1). we obtain (34.4) Employing equation (D.IO) to substitute for A and cancelling a pair of identical terms from the two members of equation (34'.4), this equation reduces to (34.5) Since A, is an arbitrary vector, we can equate coefficients of A, from the two members of this equation to obtain

Dx'

(Ix' Dx'

iJ2 x'

r~ ilik - (Sii ii:p r :, + ;j.iiiJ~j

(34.6)

Multiplying both sides of this equation by Dx'jiJx' and using the result ilY{ ax' i"--x' ;ii k

(34.7)

yields finally (348) which is the transformation law for nn affinity. It should be noted that, were it not for the presence of the second term in the right-hnnd member of equation (34.R), I'~ would transform as a tensor of the third rank having the covarinnt and contravariant characteristics suggested by the positions of its indices. Thus, the transformation law is linear in the components of an atJinity but is not homogeneous like a tensor transformation law. This has the consequence that, if all the components of an affinity are zero relative to one frame, they are not necess;lrily zero relative to another frame. However, in general, there will be no frame in which the components of an affinity vanish over a region of.'!' N, though it will be proved that, provided the affinity is symmetric, it is always possible to find a frame in which the components all vanish at some particular point (see section 39). Suppose r~, r~* are two allinities defined over a region of .'I'N' Writing down their transformation laws and subtracting one from the other, it is immediate that -k

-k

(l:e (lX'

iJx'

r - r *rJ= -i)x" - -(lxi · t:x.) ··(f'-P*) Sf $1 I)

(34.9)

i.e. the difference of two aflinities is a tensor. However, the sum of two allinities is neither a tensor nor an aflinity. It is left as an exercise for the reader to show,

100 similarly. thai the sum or an atlinity r~i and a tensor A~J is an aflinity. If j is symmetric with respect to its subscripts in one frame, it is symmetric in every frame. For, from equation (34.8),

n

r~ )'

k (lX' ~x' P ()x" P.x) (lx' ,~t

= (1x

llx k (lX' itx' i).x'

=

(lX'

(l.x)

+ (lX k

(12 X'

Px' trx.1iJ.x i (1 2 x'

(lX k

r;,+ ....

(lX' (1;(';(;,\';

r~j

(34.10)

r;,

where, at the first step, we have put

= r~,.

35. Covariant derivatives of tensors In this section, we shall extend the process of covariant dilferentiation to tensors of all ranks and types. Consider first an invariant field V. When V suffers parallel displacement from P to P', its value will be taken to be unaltered, i.e. ii V = 0 m all frames. Hence 135.1)

is the counterpart for an invariant of equation (.BA). It follows that

V,= Vi

(35.2)

i.e. the covariant derivative of an invariant is identical with its partial derivative or gradient. Now let Bi be a contravariant vector field and Ai an arbllrary covariant vector, Then Ai B' is an invariant and, when parallel displaced from P to P', remains unchanged in value. Thus o(A,B i ) = 0 or

oA i Hi

+ A,bB'

=

0

and hence, by equation (33.7), AkDB k =

-

r'~ Akdx j 11'

But, since the A k are arbitrary, their coeflieients equation can be equated to yield

111

(35.3)

the two members of this (35A)

This equation defines the parallel displacement of a contravariant vector. The covariant derivative of the vector is now deduced as before: thus (35.5)

101 and since d x' is an arbitrary vector and dB k _liB' is therl known to be a vector, I'B k

B'=

P,:1 +rkB'

I

(35.6)

I)

is a temor called the covarialll derivative of B k . Similarly, if A~ is a tensor field, we consider the parallel displacement of tht invariant A~B,Ci. where B"C' arc arbItrary vectors. Then, from /)(A~B,C') = ()

(35.7)

and equations (3.17) and (,Vi.4), we deduce Ihat IIA~ ~ r,~A:dxk ..

r,: A~ d x'

(35.HI

It now follows that (35.9)

is the covariant derivative required. The rule for linding the covartant derivative of any tensor will now be plain from examination ofequatiorl D5.9), ViI., the appropriate partial derivative is lirst written down and this is therl followed by 'aflinity terms; the 'aflinity terms arc obtained by writing down an lI1r1er product of the allilllty and the tensor with respect to each of its indices in turn, prefixll1g a positive sign when the index i. contravariant and a negatIve sign when it is covariant. Applying this rule to the tcnsor field whose components at every point arc thos1 of the fun,lamerltal tensor II;. It will be found that (35 IG)

Thus. the fundamental tensor behaves like a constant in covariant differentiatior" FInally. In tlw, section, we shall dcmonSlrate that the ordinary rules for th: diflcrenllation of sums and products apply to the process of covariant dlll"crentiallon. The rlght-han,1 member of equatwn (35'» bcing linear 111 the tensor A~, it follows immedIately that If Ci=A~j-B~

C;,

then

=

A;, + 13;,

(35.11) (,151 =) (35.1 ~)

Now suppose that Then .

I~Ci

C', = " k + r:kC' ('.x-

,"

(1 k (AiB')+ r:kAjB' X

102

(lA;.

)

.

(llBj

.).

= ( 2.x:,,+f;,Aj-f;,A; B'+ ~ii.x:' +f:,B' A; . . = A;,Bj + Bj,Aj

.

(35.14)

which is the ordinary rule for the difTerentiatlOn of a product.

36. The Riemann Christoffel curvature tensor [fa rectangular Cartesian coordinate frame is chosen in a Euclidean space $ Nand if A' are the components ofa vector defined at a point Q with respect to this frame, then 6A; = 0 for an arhitrary small parallel displacement of the vector from Q. This being true for arbitrary A', it follows from equation (35.4) Ihat fi, = 0 with respect to this frame at every point of $ N' Suppose C is a closed curve passing through Q and that A' makes one complete circuit of C. being parallel displaced over each clement of Ihe palh. Then its components remain unchanged throughout the motion and hence, if A' + i\A i denotes the vector upon its return to Q, (36.1) Since i\A i is the difference between two \e..:tors both defined at Q, it is itself a vector and equation (36.1) will therefore be a vector equation true in all frames. Thus, in t.' N. parallel displacement ofa vector through one circuit ofa closed curve leaves the vector unchanged. If. however, Ai is defined at a point Q in an amnely connected Spilce .'I'N, not necessarily Euclidean, it will no longer be possible, in general, to choose a coordinate frame for which the components of the aflinity vanish at every point. As a consequence, if A i is parallcl displaced around C. its components will vary and it is no longer permissible to suppose that upon its retum to Q it will be unchanged, i.e. 1\.4 itO. We shall now calculate i\A' when Ai is parallel displaced around a small circlIlt C endosing the poinr P having coordinates Xi (Fig. 7) at which it is initially defined.

u v

p

c

FlU. 7

103 Let U be any point on this curve and let Xi + ~i be its coordinates, the ~i being small quantities. V is a point on C ncar to U and having coordinates Xi + ~i + d~i. When A' is displaced from U to V, its components undergo a change (36,2) where r;k and Aj are to be computed at lJ. Considering the small displacement from P to U and employing Taylor's theorem, the value of rik at U is seen to be

:lr i r'jk + {Px'ik ~"

(36.3)

to the first order in the ~/. In this expression, the affinity and its derivative are to be computed at P. Aj in equalion (36.2) represents the vector after its parallel displacement from P to U, i.e. it is (36.4) where A< A' and ['/, arc all to be cakulated at the point P. To the tirst order in ~' therefore, equation (36.2) may be written

)A' =

(

"-l r! AJ + (Aj ,0 [';k - ['/ rJk

(l.'(/;k

j rl

A')~/Jd ~,.

(365)

Integrating around C. it will be found that

~A' ~

-

r i

AJ ld~k +-

Jk:r

(r'

"d~k

f"' _ iWik )Aj l

rk

j/

(lx'

(

:r~

(366)

(

where the dummy indicesj, r have been interchanged in the final term of the righthand member of equation (36.5).

fd~k = ~~k = 0

Now

Also

fd(('~k) = A(~'~k) =

(36.7)

0

(36.H)

c

f~ld~k =

so that

(

- f'kdel ~ ~

(36.9)

(

implying that the left-hand member of this equation is skew-symmetric in I and k. Since ~', d~k arc vectors, it is also a tensor. Denoting it by ex kl , we have N

~

kl _ ,l(dd'k 'kd") - 2j ~ ~ -~ ~

(36.10)

\04

and equation (36.6) then reduces to the form (36.11) Apart from its property of skew-symmetry, ex" is arbitrary. Nonetheless, since it is not completely arbitrary, the quotient the(Hem (section 32) cannot be applied directly to deduce that the contents of the bracket in equation (36.11) constitute a tensor. [n fact, this expression is not a tensor However, it is easy to prove that, if X~, is skew-symmetric with respect to k, 1 ilnd if Y', defined by the equation (36.12) is a vector for arbitrary skew-symmetric tensors 'X , then X~, IS also a tensor. To prove this. let II" be an arbitrary symmetric tensor. Then the components of the tensor kl

y"

= ex kl + II"

(36.1J)

are completely arbitrary, for, assuming k < I,

)," = ex kl + II",

1"k

= -

ex kl + Il kl

(36.14)

and it follows that the values of l', y'k can be chosen arbitrarily and then (3615) i.e., it is only necessary to fix the value~; of ex".ll in the cases" < 1in order that the ykl shall assume any specified values oler the complelc range of its superscripts, with the exception ofthe cases when twn superscripts are equal. I I' the superscripts arc equal, ex kl = 0 and ykl = II". But thesl~ Il kl are also arbitrary lwd hence so again are the y" with equal superscripts. Since {ikl is symmetric and X~, is skew-symmetric, kl

X~,

{ikl

=

0

(36.16)

Adding equations (36.12) and (36.16), we obtain therefore (36.\7) But ).kl is an arbitrary tensor and hence, by the quoticnt theorem, Xii is a tensor. The multiplier ofex kl in equation (36.11) is not skew-symmetric in k, I. However, it can be made so as follows: Interchange the dummy indices k, 1 in this equation to obtain (36.18) Adding equations (36.11) and (36.18) and noting that ex" = - ex'k , it will be found that (36.19)

105

The bracketed expression is now ,kew-symmetric in k,l and hence

(

r rki ["

_ j/

(~ri) 1 r'rl;k r" + ?f' 1 -l' 4' (".xk ('lx' '

(36.20)

is a tensor. Al bcing arbitrary, it follows that W,'I...

11"

= 1",,1", 1 - I" rl r'Jk + (t i '(kjl ...

(36.21 )

is a tensor, [t is the Riemat/t/ Chrisroffel ('Im'(ltllre ret/"or, E4uation (36,19) can now be written ~A'~, ~Hj.,,41():kI

(36,22)

I I' IJ;., is contracted With respect to the indices i and I. the resulting tensor is called the RiClI rell,\or and is delloled hy 1 are eo-Iat itude and longitude respectively on the surface of a sphere of unit radius, ohtain the metric

d.l"2 = d0 2

+ sin 2 /idtj>2

for the surface. Show that tire only non-vanishing three index symbols for this Yf 2 are

Show also that the only non-vanishing components of B iikl are B 1212

= - B 1221 = B 2121 = - B 2112 = sin 2 0

120 and that t he components of the Ricci tensor are given by R 22 =-sin 2 1i.

R 12 =R 2 ,=0, R11=-I,

Prove that the curvature scalar is given by R = - 2. 18. Employing equation (42.9), obtain expressions for V 2 V in cylindrical and spherical polars. 19. [n a certain coordinate system

where 4>, r/J are functions of position. Prove that Btu is a function of r/J only. [I' r/J = -log (o,x') prove that

R Jk

= B~k'=

0

20. [n the :Il 2 whose metric is 2 + r 2 dli 2_ d,·2 = dr __

r2 _ (12

.

r 2 dr 2 -

(--':2-~

(r > (1)

;,2 )2

prove that the differential equation of thc gcodesics may be written

where k 2 is a constant such that k 2 = I if, and only if, the geodesic is null. By putting r dli/dr = tan 4>, show that if the space is mapped on a Euclidcan planc in which r, Ii 2m. Since i141 = C, 1144 = - c 2 (I - 2m/r), equation (53.5) must be replaced by d (dr -,(I .. 2m/I') dU) = 0 dI dI dI

(57.15)

Together with the lirst integral

2 drdu e- dI dI

c"> (I

-·2m/r) (dU)2 --

(57.16)

dI

this leads to the following quadratic f(lr dr/du: r

d ( du

)2 +4mc(I_- R1 Jdrtiu +2/1/C2(1-.2rm)(RI _ 1'1)=0

(57.17)

I

I'

The roots are (57IH)

If r > 2m, one root is positive and onc is negative. However, I' must dccrease iniltally (otherwise the square root in (57.1 H) hec"IlWs imaginary) and so the negative root is taken. I' then decreases steadily to I' n, iU, passage through the Schwarzschild radius being unremarkable. Once inside the Schwarzschild sphere, as already proved, the possibility of escape from the attr;H:tion no longer exists. The world-lines of photom moving radially arc null geodesics governed hy the equation 2cdrdu -e 2(1-2m/r)du 1 = 0 (57.19) There are two families of such geodesics. viz.

~.u = dr

0

and

'

du

21'

dr

r-2m

('

(57.20)

For the first family, equation (57.7) gives

c provided

I'

dl dr

I'

-2m

(57.21)

> 2m. This corresponds to a photon moving towards the centre of

159

attraction. h)r the second family. we find c

dl dr

r

(57.22 1

-2m

in the same region; i.e. a photon moving away from the ccntre of attraetion. A photon hdonging to the first family crosses the Schwarzschild sphere and thell falls into O. Insidc thi, sphere. the photons can he separated into two classes (i) those for which U IS constant along a world-line these photons could have their source ou tSlde the sphere: (ii) thme f()l which the sccond of equations (57.201 is valid and, hence, C1J =

21'

+

4/11

log

(2/11-

r)

+ constant

(57.231

Asr ..... 2m, u ..... - '""j and these photons cannot have had an external source. Since dlJ/dr < 0, these photons also 11111 Into O. It is now clcar that, In the field dc,crrhed hy the metric (57.12), no photon Ot partIcle can cross the Schwarzsclllid sphere in the sense I' increasing. On the othel hand, any photon or partIcle which crosses the sphere in the reverse sense i~ ahsorhed and cannot return to the external world. The conditions inside the sphere 4/(9D 2 ). (b) k = 0,1\ ? 0. (c) k '" - 1.1\ ? 0. In case (i) (a), if 1\ is posllive, show that there is also a sol Ut ion in which S decreases to a nOll-zero minim urn and thereafter steadily increases towards infinity. 8. If k = + I and 1\ = 4/(9D 2 ), show that the radius 5 of the Friedmann model can first increase from zero to a value I/.J 1\, when the cosmos allains the statIC Einstein state (section 66) and thcn, if slightly disturbed from this state, S may either decrease back to zero or continue to increase indefinitely. (This shows that thc statIc Einstein universe is unstable.) 9. A cosmos containing radiation, but no maller, is governed by the equations (68.7) and (68.8). Show that

°

5 2 S2 = e 2 (D - k5 2

+ ~1\54)

where [) is an energy density parameter defined by the equation 3D = KUS 4 10. Sketch the graph l)f S2.~2 against S fllr the universe descrihed in the previous cxercise and dcdlH:e that all the condusions Ii.sted in exercise 7 are valid it 4/(9[)2) is replaced hy 3j(4D). 11. For the universe deserihed in exercise 9, if k ~. 1,1\ = 3j(4D), and 5 = 0 at I = 0, prove that at any later time I, S2 = 2D{ 1 --exp(

If S

=

J (2D) at I

=

-('1/

JD)}

0, prove that thc universe is static hut unsLible.

References

Bateman. H. (1952). Pi/rl/ill !JI//Ncmial FifUlJliolls 0/ Malhm/llfical Physicl. Clmbridgt University Pre". p I H4. Bondi. H. (1960). Co.lnfoloq... Cambridge University Press. Bondi. H .. and (;old. T (194H) Mon. Nol. ROI· . .4.l/r. Soc.. lOll, p. 252. Couboll. C A., and Boyd. T J. M. (1979). Flc('lril'il\', Longman, p. J()9 Hoyle. F (J94H). MOil. NOI. Ro.\'. A.\lr. Soc., lOll. p. J72. Rowan·Robmsoll. M. (1979) CO.ll/1I1' I oml.lco!'e. Oxford University Press.

199

Bibliography Adler. R., Bazin. M .. and Shiffer. M. (1965). Infrodu('/lOf! 10 General Rl'ial;l·iIY. McGrawHill. Aharonl, J. (1959). 'lIw Special I1lCorc o! RcIOli/'iIY, Oxford University Press. Angel. R. B. (19HO). Relalwily: The Theory and il.l· Pllllosopl1\'. Pergamon. Berry, M. ( 1976). Principles o! ('osmolol/l' alld Grarila/loll, Cambndge University Press. Clark, R. W. (197.'). Fin.Hcin. Ihe Li!e and TIIII