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Show that geometry has local inertial frames

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex] ds^2 = g_{tt} dt^2 + g_{tx} (dt dx + dx dt) [/itex]
    with [itex] g_{tt} = -x [/itex] and [itex] g_{tx} = 3 [/itex]

    "Show that this is indeed a spacetime, in the sense that at every point, in any coordinates, the matrix [itex] g_{\mu \nu} [/itex] can be diagonalized with one positive and one negative entry. Hint: You can diagonalize the matrix, but that is the hard way. There is a one-line answer."

    2. Relevant equations
    Christoffel Symbols:
    [itex] g_{\alpha \delta} \Gamma^{\delta}_{\beta \gamma} = \frac{1}{2} \left( \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} + \frac{\partial g_{\alpha \gamma}}{\partial x ^{\beta}} - \frac{\partial g_{\beta \gamma}}{\partial x^{\alpha}} \right) [/itex]

    [itex] ds^2 = g_{\mu \nu}(p) x^{\mu} x^{\nu} [/itex]

    3. The attempt at a solution
    I think I have all the components for the solution but I can't put the pieces together. I think I'm supposed to show that the geometry defined in 1. has local inertial frames. Can I do this by showing that the Christoffel symbols vanish (I don't think they all do)? And how do I show that this is valid for any coordinates?
    I've also considered expanding the metric as described here: http://mathpages.com/rr/s5-07/5-07.htm :
    [itex] g_{\mu \nu}(p + x) = g_{\mu \nu}(p) + g_{\mu \nu,\alpha} x^{\alpha} + \frac{1}{2} g_{\mu \nu,\alpha \beta} x^{\alpha} x^{\beta} + \cdots [/itex]
    with
    [itex] g_{\mu \nu, \alpha} = \left( \frac{\partial g_{\mu \nu}(x)}{\partial x^{\alpha}} \right)_p [/itex]
    and
    [itex] g_{\mu \nu, \alpha \beta} = \left( \frac{\partial^2 g_{\mu \nu}(x)}{\partial x^{\alpha} \partial x^{\beta}} \right)_p [/itex]
    and then take [itex]x[/itex] to be vanishingly small so that we only use the first term in
    [itex] ds^2 = g_{\mu \nu}(p) x^{\mu} x^{\nu} [/itex]
    In that case [itex] g_{\mu \nu} [/itex] is just a matrix with numbers which can be diagonalized.

    The one-line answer probably has something to do with the fact that some specific derivatives vanish but I don't understand why this is the answer.
     
  2. jcsd
  3. May 1, 2013 #2
    Maybe det g can tell you something useful?
     
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