1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook