# Four-velocity in a static spacetime

by scottie_000
Tags: four vector, general relativity, geodesic, rest, spacetime
 P: 49 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 $$V^a = (V^0,\textbf{0})$$, where $$V^0 = V^0(\textbf{x})$$ is a function of only spatial position 2. Relevant equations Line element $$ds^2 = -e^{2\phi} dt^2 + h_{ij}dx^i dx^j$$ Relevant Christoffel symbols (as calculated) $$\Gamma^0_{00} = 0$$ $$\Gamma^0_{0i} = \frac{\partial \phi}{\partial x^i}$$ $$\Gamma^0_{ij} = 0$$ Four-velocity $$V^a = \frac{dx^a}{d\tau}$$ Geodesic equation: $$\dot{V}^0 + 2\frac{\partial \phi}{\partial x^i}V^0 V^i = 0$$ 3. The attempt at a solution I am willing to believe that the spatial part of $$V^a$$ is 0, since I am told the observer is at rest. Is this correct? Given this, I think the geodesic equation should become just $$\dot{V}^0 =0$$ but I don't see how this shows that $$V^0$$ 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?
 P: 49 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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