22.6 22.6 a) (Hartle's answer) The answer is 22.27

b) For a = 0 of 22.31 we have e/t + ·S = 0. The LHS is
e0 E ·  E

t
+ m0 B

 B

t
 
= E·(×H) - H ·(×E)
By a vector identity you recognize this as - ·(E×H).

The case a = i is handled similarly. It helps to recognize Tij/xj as the ith component of the vector e0 [- E div E - (E·) E + 1/2 grad(E2) ] + (similar in H) (of course, div E = 0).

22.10 (a) The energy density measured by an observer with velocity V is
g2 [A + B (Vx)2 + C (Vy)2 + D (Vz)2]
Since each component of V can range between -1 and 1, the conditions are
A+B 0, A+C 0, A+D 0     or, for the usual negative B, C, D,     A max(|B|, |C|, |D|)

(b) no

22.12 The point and logic of this problem is to show that ·J = 0, which is often called a "conservation law" yields actual (integrated) conserved quantities, and that one can get such a conserved J from T if there is a spacetime symmetry.

(a) For arbitrary J, f must satisfy
 f

fxa
= Gbab =  1

2
gmn  gmn

xb
(*)
It's obviously easiest if you can guess the solution. Otherwise, let the diagonal elements of the metric be ga, with ga = 1/ga and integrate,
ln f = 1/2

a 
(ln ga ) = 1/2 ln (

a 
ga )     hence     f = (

a 
ga )½ = ABCD.
(Of course, if would be enough to show that a solution to (*) exists.)

(b) The two terms in ·(xT) both vanish, one because ·T = 0 generally, the other because of Killing's equation and the symmetry of T. Thus xb Ttb ABCD d3x is conserved.

22-13 a) In the LIF the G's vanish at one point. We must therefore differentiate R first and then use G = 0. Symbolically, R = G+ GG, hence the terms in the Bianchi identity are of the type
(G+ GG) = 2G+ GG = g3g + g 2 g + GG
Now we may put G = 0 = g to get the required result.

b) Hartle says: In the LIF there are (sic) a total of eight terms. By relabeling the dummy indices and using the fact that partial derivatives commute they can all be seen to cancel.