Four-velocity in a static spacetime

In summary, 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. The relevant Christoffel symbols have been calculated, and the geodesic equation reduces to \dot{V}^0 = 0. The goal is to prove that V^0 should be coordinate-time-independent.
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scottie_000
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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 [tex] V^a = (V^0,\textbf{0}) [/tex], where [tex] V^0 = V^0(\textbf{x}) [/tex] is a function of only spatial position

Homework 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]


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

Related to Four-velocity in a static spacetime

What is four-velocity in a static spacetime?

Four-velocity in a static spacetime refers to the four-dimensional vector that describes the velocity of an object in a static or stationary spacetime. It takes into account both the time and space components of an object's velocity.

How is four-velocity calculated in a static spacetime?

In a static spacetime, four-velocity is calculated by taking the derivative of an object's position with respect to proper time. This means that it takes into account the object's motion through both time and space.

What is the significance of four-velocity in a static spacetime?

Four-velocity in a static spacetime is significant because it helps us understand the motion and behavior of objects in a stationary environment. It allows us to accurately describe the velocity of objects and how they move through spacetime.

How does four-velocity differ from regular velocity?

Four-velocity differs from regular velocity in that it takes into account the time component of an object's motion. Regular velocity only considers the spatial component, while four-velocity combines both time and space.

Can four-velocity be negative in a static spacetime?

Yes, four-velocity can be negative in a static spacetime. This indicates that an object is moving in the opposite direction of the time component, which is possible in a stationary environment. However, the magnitude of four-velocity is always positive.

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