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 ¶T^ij/¶x^j as the i^th 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 g_a, with g^a = 1/g_a 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 òx_b T^tb 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.