20-18 The covariant components are

xa = gab xb = ga1
Substitute this into the formula (20.68) for the covariant derivative to find
 Ñb xa = 1 2 æè ¶ga1 ¶xb - ¶gb1 ¶xa + ¶gab ¶x1 öø .
The first term is antisymmetric in (a, b), and the last term vanishes because of x1-independence. Thus the symmetrized Ñb xa vanishes - that's Killing's equation. (This does not, of course, prove that every vectorfield generating a symmetry of the metric satisfies Killing's equation - but it is so :)

20-20
a) In rectangular coordinates the Christoffel symbols vanish, so Killing's equation becomes xa,b + xb,a = 0. The first two have constant components, whose derivatives vanish, for the third, for example, the (1, 2) component is (-y),y + x,x = -1 + 1 = 0.

b) If you verify this by starting with /f, you will be doing problem 8-8 over again; else, similarly but "the other way round", express /x in terms of /r and /f.

c) Let the symmetry origin be at (X, Y), so coordinates centered there are x¢ = x - X,     y¢ = y - Y. The desired K. V. is
 h¢ = - y¢ ¶ ¶x¢ + x¢ ¶ ¶y¢ = h - X ¶ ¶y + Y ¶ ¶x .

21-2

21-11 Hartle's official answer is: "Running the Mathematica program Curvature and the Einstein Equations for the metric given in Problem 8.12 gives R = -2."