22.6
22.6 a) (Hartle's answer) The answer is 22.27
b) For a = 0 of 22.31 we have ^{¶e}/_{¶t} + Ñ·S = 0. The LHS is
e_{0} E · 
¶E
¶t

+ m_{0} B 
×

= 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 e_{0} [ E div E  (E·Ñ) E + ^{1}/_{2} grad(E^{2}) ] + (similar in H) (of course, div E = 0).
22.10 (a) The energy density measured by an observer with velocity V is
g^{2} [A + B (V^{x})^{2} + C (V^{y})^{2} + D (V^{z})^{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¶x^{a}

= G^{b}_{ab} = 
1
2

g^{mn} 
¶g_{mn}
¶x^{b}


 (*) 
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 g_{a} ) = ^{1}/_{2} ln ( 
Õ
a

g_{a} ) hence f = ( 
Õ
a

g_{a} )^{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 d^{3}x is conserved.
2213 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) = ¶^{2}G+ G¶G = g¶^{3}g + ¶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.