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Timelike killing vektors

  1. May 16, 2014 #1
    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?
     
  2. jcsd
  3. May 16, 2014 #2

    WannabeNewton

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    Consider the vector field ##\xi^{\mu}## which in the coordinates for which ##\partial_0 g_{\mu\nu} = 0## takes the form ##\xi^{\mu} = \delta^{\mu}_0##.

    Then ##\mathcal{L}_{\xi}g_{\mu\nu} = \partial_0 g_{\mu\nu} = 0## so ##\xi^{\mu}## is a time-like Killing field for this metric.

    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##.
     
  4. May 16, 2014 #3

    Bill_K

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    Why is this the wrong way to think of it?? It seems to me a perfectly reasonable way to think of it.

    Someone hands you an unfamiliar metric. What's the first thing you do -- write out Killing's Equation and try to solve it?

    Or do you say, aha, I see it doesn't depend on one of the coordinates ζ, so there's an obvious Killing vector.
     
  5. May 16, 2014 #4

    PeterDonis

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    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.
     
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