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Homework Help: Four-velocity in a static spacetime

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data
    I am given a static spacetime line element which has the property that the metric is time independent. I am asked to calculate some of the Christoffel symbols, which I have done.

    The question asks to show that for an observer at rest, the four-velocity is given by [tex] V^a = (V^0,\textbf{0}) [/tex], where [tex] V^0 = V^0(\textbf{x}) [/tex] is a function of only spatial position

    2. Relevant equations
    Line element [tex] ds^2 = -e^{2\phi} dt^2 + h_{ij}dx^i dx^j [/tex]
    Relevant Christoffel symbols (as calculated)
    [tex] \Gamma^0_{00} = 0 [/tex]
    [tex] \Gamma^0_{0i} = \frac{\partial \phi}{\partial x^i} [/tex]
    [tex] \Gamma^0_{ij} = 0 [/tex]
    Four-velocity [tex] V^a = \frac{dx^a}{d\tau} [/tex]
    Geodesic equation:
    [tex] \dot{V}^0 + 2\frac{\partial \phi}{\partial x^i}V^0 V^i = 0 [/tex]


    3. The attempt at a solution
    I am willing to believe that the spatial part of [tex] V^a [/tex] is 0, since I am told the observer is at rest. Is this correct?
    Given this, I think the geodesic equation should become just
    [tex] \dot{V}^0 =0[/tex]
    but I don't see how this shows that [tex] V^0 [/tex] should be a function of just spatial variables, since the dot represents proper time, not coordinate time.
    Is there any way to prove that it ought to be coordinate-time-independent?
     
  2. jcsd
  3. Feb 10, 2010 #2
    Anyone have any ideas? I'm sure it's very simple, but I can't think how to actually prove it
     
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