Four-velocity in a static spacetime


by scottie_000
Tags: four vector, general relativity, geodesic, rest, spacetime
scottie_000
scottie_000 is offline
#1
Feb9-10, 04:34 PM
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 [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?
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scottie_000
scottie_000 is offline
#2
Feb10-10, 03:51 AM
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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