Fourvelocity in a static spacetimeby scottie_000 Tags: four vector, general relativity, geodesic, rest, spacetime 

#1
Feb910, 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 fourvelocity 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] Fourvelocity [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 coordinatetimeindependent? 



#2
Feb1010, 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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