Four-velocity in a static spacetime

[scottie_000]

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?

 

[scottie_000]

Anyone have any ideas? I'm sure it's very simple, but I can't think how to actually prove it