homework 6 21-19 You have to read through the whole problem, else you don't realize that it's only up to coordinate transformation (that is, there exists some coordinate system, in which ...and at least one such coordinate system is a Lorentz-gauged one), and assuming asymptotic flatness, that you are to find the "most general" (would "general" be enough? Can you be more general than general??) solution of the Einstein equations.

(a) static here means invariant under t -t. (That's not quite the same thing as time-independent source, for example the Tmn for a rotating sphere would be time-independent, but not time-reversal invariant). Considered as a coordinate transformation (which it is) it would change the sign of the hit, so invariance requires that those vanish.
Note that a coordinate choice is implied here. In a general (even the most general!) coordinate system the hit would not vanish. Have we shown that the coordinates we desire exist? I think not - we have assumed they exist in order to even formulate time-reversal invariance.
So there is a way to do this properly by using an appropriate definition of "static": a metric is static if it admits a surface-orthogonal timelike Killing vector field. The orthogonal surfaces that exist, by this definition, through each point are unique (given a unique Killing vector), because locally there is only one (spacelike) tangent hypersurface orthogonal to a (timelike) vector. Therefore the surfaces also cannot cross. If the space is time-orientable (another assumption!) you can then associate a number with each surface - that's t. And you can choose coordinates on one surface, and let the integral curves of the Killing vector (which always exist) carry those coordinates to the other surfaces. Do you get to all of them? That has to do with whether the one surface you chose is a Cauchy surface. Do you get back to your original surface, to a different point and therefore to an inconsistency of your coordinates? You have to make more assumptions, else even by cutting and pasting Minkowski space you can get counterexamples.

(b) Hartle tells you to impose the Lorentz gauge condition (for later convenience, so the Einstein equations become just the Laplace equation), but this does not entirely fix the gauge, you can still change h by (21.53) with xa = (0, xi) with xi independent of t without changing what you achieved in (a), hit = 0. "These three functions can be used to make the three off-diagonal components of hij vanish."
Geometrically this means that you have that one spacelike surface, that's a 3D manifold on which you can choose 3 coordinates so the coordinate lines are mutually perpendicular everywhere. Hmm. Forgetting about global problems again ...

(c) Now the Lorentz gauge condition shows that the metric has the desired form, for example Vx = 0 says
x[-hxx + hyy + hzz + htt] = 0
hence -hxx + hyy + hzz + htt is independent of x, and since it's zero at x , it vanishes everywhere. From Vy = 0 = Vz one gets similar equationsa cyclically permuted, solveing simultaneously gives the form (21.25).

21-24 Let Xi be the displacement between the "observer" and one of the test particles before the wave passes, and let xi be the change in Xi due to the wave. The geodsic equation is for this case
 d2xi

dt2
= - Ritjt Xj.
(1)
Let A, B range over the transverse directions, then
RAtBt = -  1

2
 2hCB

t2
dCA
When this is substituted in (1) it can be integrated (since X is constant in time) for a wave of form (16.2) to get essentially (16.13), xA = 1/2XA f(t), xz = 0.

16-8 (a) Since the same force, in the x-direction, is acting on all charges (assuming equal e/m), the ring will oscillate as a whole in the x-direction, not changing its shape.

(b) A gravitational wave of the form hAB = d1A d1B could produce this motion, but no combination of + and × polarizations will do this.

16-13 (Hartle's solution)
As in Section 16.16.5, in geometrized units,
f (energy flux) = (energy density) =  w2 a2

32p
with dimension L-2. The dimensionless amplitude a is comparable to dL/L. To convert the geometrized units cm-2 for flux back to erg/cm2·sec, note
(f in cm2) =  G

c5
(f in erg/cm2·sec).
Recognizing also that (w in cm-1) = (w in Hz/c) one has
f ~  c3

32pG

 w

1 Hz

2

 

 dL

L

2

 
   erg

cm2·s
Taking w ~ 200 Hz, dL/L ~ 10-21 gives

f ~ 2×10-1 erg/cm2·s

A star a distance r away with the luminosity of the sun, LS ~ 4×1033 erg/s, would give a flux
f =  LS

4pr2
.
A star of the sun's luminosity would give the same flux in electromagnetic radiation at a distance of ~ 0.01 pc.