Four-velocity in a static spacetime

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In a static spacetime with a time-independent metric, the four-velocity for an observer at rest is expressed as V^a = (V^0, 0), where V^0 is a function of spatial position. The relevant line element is ds^2 = -e^{2\phi} dt^2 + h_{ij}dx^i dx^j, and the calculated Christoffel symbols indicate that the spatial components of the four-velocity are zero. The geodesic equation simplifies to \dot{V}^0 = 0, suggesting that V^0 is constant along the observer's trajectory. The challenge remains to demonstrate that V^0 depends solely on spatial variables, as the dot represents proper time rather than coordinate time.
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Homework Statement


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

Homework 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


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?
 
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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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