homework 6 9-8 a) Since Schwarzschild time = clock time at infinity, we can directly use W = df/dt = (M/r3)½, so the period is
P = 2p
W
= 2p73/2 M ~ 116 M
b) Note this involves both "gravitational" and "Doppler" shift, both of which are contained in ut = dt/dt. This can be obtained from
u·u = -1 = -

1- 2M
r


ut2 + r2uf2 =

1- 2M
r
-r2W2

ut2 =

1- 3M
r


ut2
This gives Pspaceship = (4/7)½P = 28pM ~ 88 M

9-12 As one of you wrote:
l
e
= b     rmin = R     W = df
dt
= b
R2


1- 2M
R


    dr
dt
= 0


ds2 = -dtshell2 + drshell2 + r2(dq2 + sin2qdf2)


dtshell
dt
=   _______
1 - 2M/r
 
       drshell
dr
= 1
  ______
1-2M/r
       Wshell = df
dt
dt
dtshell
= b
R2
  ______
1-2M/R
 
       drshell/dtshell = 0


V = RWshell = b
R
  
 


1- 2M
R
 
.

9-14 The answer depends on how one interprets "the area swept out by an orbit". Since this takes place in spacetime, one might think of an area in spacetime, swept out by the pure radial direction (which is well-defined from the spacetime's symmetries). However, Hartle had in mind the projection of the radial direction on a surface t=const, where the area element is dA = (1-2M/r)-½rdrdf. One has to integrate this from 2M to r(s) to get dA/ds in terms of df/ds, where s is either Schwarzschild time (where df/ds = l2/r2) or proper time. For neither does one get a constant. (Of course, r2 df/dt is constant, but that is an area rate is a fictitious flat 3D geometry, and includes r < 2M ...).

9-18 The easy way is to note that this metric is "conformally flat" (flat metric × overall factor), hence any null curve in this metric corresponds (via the same formula in terms of coordinates) to a null curve in flat space. Hence the light bending is also the same - there is none. (One ought to show that the correspondence also works between null geodesics; for example, from the variational principle, roughly d(f ds2) = f d(ds2) when ds2 = 0.)
The hard way is to follow the steps to (9.78), finding
df/dr = 1/r2(1/b2 - 1/r2)½
independent of M, hence the same as for M=0 (no deflection).

9-22 a) For a sufficiently small impact parameter b, a light ray can be deflected by df = p, hence returns in the direction it came from.

b) The detector subtends and angle d/R at the black hole. Think of b as a function of f, then the range of impact parameters that will hit the detector is db = (db/df)·2(d/R), corresponding to a ring of area d(pb2) = 2pbdb, so we find for the detected
number/sec = 4p|b db/df|f = p (d/R) f*.

Numerical integration gives b(p) = 5.35 M,     db/df = 0.17 M so
Number/sec = 11.4 f* M2 (d/R).

c) The size of a 1015 g black hole is ~ 10-15 m, much smaller than the retroreflector on the moon, which returns only a few photons/s to the telescope. So it will take an impossibly long time to detect backscattered photons from such a much smaller black hole




File translated from TEX by TTH, version 2.79.
On 2 Dec 2003, 22:10.