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If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.
How can one prove that?
How can one prove that?
Why is this the wrong way to think of it?? It seems to me a perfectly reasonable way to think of it.But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field ##\xi^{\mu}## satisfying ##\nabla_{(\mu}\xi_{\nu)} = 0##. If such a vector field exists then there necessarily exists a coordinate system ##\{x^{\mu}\}## in which ##\partial_0 g_{\mu\nu} = 0## and ##\xi^{\mu} = \delta^{\mu}_0##.
You're assuming that the "0" coordinate is timelike. That's not always the case. Consider, for example, Schwarzschild coordinates on Schwarzschild spacetime inside the horizon; there the "0" (i.e., ##t##) coordinate is spacelike, not timelike, but the metric is still independent of ##t##, so ##\partial / \partial t## is still a KVF--just not a timelike one.If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.