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

t
+ m0 B
×
 ¶B ¶t

= E·(Ñ×H) - H ·(Ñ×E)
By a vector identity you recognize this as - Ñ·(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·Ñ) 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 ³ 0, A+C ³ 0, A+D ³ 0     or, for the usual negative B, C, D,     A ³ max(|B|, |C|, |D|)

(b) no

22.12 The point and logic of this problem is to show that Ñ·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 f¶xa = 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 å a (ln ga ) = 1/2 ln ( Õ a ga )     hence     f = ( Õ a ga )½ = ABCD.
(Of course, if would be enough to show that a solution to (*) exists.)

(b) The two terms in Ñ·(xT) both vanish, one because Ñ·T = 0 generally, the other because of Killing's equation and the symmetry of T. Thus òxb 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+ G¶G = g¶3g + ¶g ¶2 g + G¶G
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.