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' q)( _ * 0.For the above choice of f and g, determine as a function of time the invariant distance betweenparticles that move on x = 0, y = 0, z = a, and similarly the distance between particles that move onx = 0, y = a, z = 0.Interpret your results physically.8. The Robertson-Walker metric may be writtendr22222+r(d+sind).ds2 = dt2 + a21 Kr 2Explain the significance of the quantities a and K, and of the world-lines (r, , ) = constant.Show that photons can travel down curves ( = constant, = constant).Given that a = (t/t0 )2/3 for K = 0, find the distance now (t0 ) in the case K = 0 between us and agalaxy from which we are currently receiving photons emitted at t1 .Suppose the Universe is closed with the Earth at the point r = 0. A distant galaxy of radius R iscurrently distance D from us with its centre on the line = 0. Show that its rim is at angular coordinate(1 + z)R K=.sin(D K)where z is the galaxys redshift. Simplify this formula for the case z 1 and discuss the differencebetween the general result and this case.9. Show that for any two vectors u, v we have(u )v (v )u = [u, v] ,where the vector [u, v] is defined by[u, v] u v v u .

Classical Fields III 3For each fixed , x (, ) defines a geodesic, with the affine parameter. Show that

dx dx,d d

Show further that(x )(x )

= 0.

dxdx= R x x ,dd

where x dx/d and the curvature tensor R can be taken to be defined by

(u )(v ) (v )(u ) [u, v] w = R u v w ,

with u, v and w arbitrary vectors.Two masses are dropped from points a small height apart. Show that just after they are released,the separation x between them satisfiesD2 x= c2 R 00 x ,Dt2where x0 ct. Hence show that the gravitational field at the Earths surface has the curvature componentRz 00z = 2g/(c2 R)where z is an upwards directed coordinate, g is the usual acceleration due to gravity, and R is theEarths radius.

4th

Prof J.J. Binney

year: Option C6

Classical Fields III: Solutions1. Let e1 = ei1 , e2 = ei2 with 1 , 2 real, then = u + iv = 1 e1 + 2 e2 = ( 1 cos 1 + 2 cos 2 ) + i( 1 sin 1 + 2 sin 2 )So we require1 =

sin 2 u cos 2 vsin(2 1 )

and similarly for v. Given that 1 6= 2 the i can be be determined.The general covariant derivative in this case is a = a + a1 1 + a2 2 which coincideswith D = + i(q/h)A if we adopt 1 = 0 in the limit 1.The speed of transverse waves on the rope iscs =so F/A < c2

s

tension=mass/length

s

FA

cs < c.

6. Extremizing the Lagrangian c2 t2 + z 2 + r02 (2 + sin2 2 ) we findd =0(2c2 t)dd 2r 2 sin cos 2 = 0(2r02 )0d 21 sin 2 2 = 0

d(2z) =0dd =0(2r02 sin2 )d + 2 cot = 0

sot = 0Now

z = 0

= 12 sin 2

R = +

= cot

= cot + = csc2 + cot2 = 1

c2F/A

Classical Fields III: Solutionsand

3

R = +

= cot ( 12 sin 2) + +

= cos 2 + cot ( 21 sin 2) = sin2

So

R = diag(0, 0, 1, sin2 )1/c20 0 000000 0R = 00 0 10200 0 0 sin

00100 1/r0200

and R = 2/r02 . Finally

000 0=00201/(r02 sin )

0000

001/r020

00 01/r02

G = diag(0, 0, 1, sin2 ) + r02 diag(c2 , 1, r02 , r02 sin2 )8G= diag(c2 /r02 , 1/r02 , 0, 0) = 4 diag(T00 , Tzz , 0, 0)cso Tzz = c4 /(8Gr02 ). The tension isZ 2Z mr 2 c4c4 (1 cos m )F = r02d Tzz = 2(1 cos m ) 0 2 =d sin 8Gr04G007. The Minkowski metric is dudv + dy 2 + dz 2 .Extremizing the Lagrangian u v + f 2 y 2 + z 2 we obtaindu=0d

;

dv 2f f 0 y 2 2gg 0 z 2 = 0;d

d 2(f y)d= f 2 y + 2f f 0 u y

0=

;

d 2(g z)d= g 2 z + 2gg 0 u z

0=

so the non-vanishing Christoffel symbols arevyy = 2f f 0

vzz = 2gg 0

yuy = f 0 /f

zuz = g 0 /g

From eqs of motion above can see that y = 0 y = 0 and similarly for z, so y = const, z = constare solutions. Also then u = v are required, so u and v are linear in . If x is constant c 2 d 2 = dudv,which is consistent with this linearity. Thus constant x, y, z defines geodesics.Lf 0 = + u0 , Lf 00 = 20 + u00 and Lg 0 = 20 u00 sog 0020 + u0020 + u00f 00+=fgL + uL uwhich always vanishes because the numerators vanish unless u = 0, and when u = 0 the denominatorsare equal so the two terms cancel.Dz =

Z

a

dz g =a

Z

aa

dz [1

uu(u)] = 2a[1 (u)]LL

uDy = 2a[1 + (u)]LZa

Dx =

dx = 2a

a

Thus

n2a(1 ct/L) for 0 < ct ; D = 2a(1 + ct/L) for 0 < cty2aotherwise2aotherwiseso at t = 0 particles are impelled towards each other along z and away from each other along y by adisturbance that propagates along x. This is a gravitational shock wave.Dz =

n

Classical Fields III: Solutions

4

8. The eqns of motion are

d a2 r sin 2 2(2a2 r2 )dd0=(2a2 r2 sin2 )dso r 2 sin2 = const. If this const is zero and = /2, then d(r 2 )/d= 0, which is satisfied by = 0at all .0=

The current distance is obtained by integrating ds = a(t0 )dr from zero to the coordinate rg of thegalaxy, and we have D = a(t0 )rg = rg because currently a = 1.RtRt2/3 1/3Since photon propagates radially, dt = a(t)dr, and rg = t10 dt/a = t10 dt/(t/t0 )2/3 = 3t0 (t0 1/3

2/3

1/3

1/3

t1 ). Hence D = 3t0 (t0 t1 ).We have K > 0 because the universe is closed, so the distance to the galaxy is.Z rgdrD = a(t0 )1 Kr 20Z ga(t0 )d where sin K r= K 0Hence sin( KD/a(t0 )) = Krg .

At t1 let the edge of the galaxy be at angular coordinate m , so R = a(t1 )rg m and(1 + z) KRKRR==m =a(t1 )rga(t1 ) sin( KD/a(t0 ))sin( KD)because a(t1 ) = (1 + z)1 .9.u v v u = u v v u + u v v u = [u, v]

by the symmetry of .dxdxdxdxd dxd dxdx dx=,==0d dddddd dd d

In the given definition of R we put u = w = dx /d and v = dx /d and havedxdxdxx x x = R x x ddd

The second term in the brackets on the left vanishes because x( ) is geodesic. Moreover,dxdx x = x because [, ] = 0ddso we can rewrite the first term and then have the equation of geodesic deviation:dxdx(x )(x )= R x x .ddDropped masses have geodesic paths x(, ) with x 0 ' c. Since x = d/d , when we multiplythe equation of geodesic deviation by a small number we getd2x ' c2 R 00 x .d 2But from elementary mechanics z = GM/R2 , where M is the Earths mass and R is the particlesdistance from the centre of the Earth. Thus varying z we have2GMzz=R3Comparing with the z component of the equation of geodesic deviation and setting g = GM/R 2 weobtain Rz 00z = 2g/(c2 R).