# Timelike killing vektors

1. May 16, 2014

### befj0001

If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.

How can one prove that?

2. May 16, 2014

### WannabeNewton

Consider the vector field $\xi^{\mu}$ which in the coordinates for which $\partial_0 g_{\mu\nu} = 0$ takes the form $\xi^{\mu} = \delta^{\mu}_0$.

Then $\mathcal{L}_{\xi}g_{\mu\nu} = \partial_0 g_{\mu\nu} = 0$ so $\xi^{\mu}$ is a time-like Killing field for this metric.

But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field $\xi^{\mu}$ satisfying $\nabla_{(\mu}\xi_{\nu)} = 0$. If such a vector field exists then there necessarily exists a coordinate system $\{x^{\mu}\}$ in which $\partial_0 g_{\mu\nu} = 0$ and $\xi^{\mu} = \delta^{\mu}_0$.

3. May 16, 2014

### Bill_K

Why is this the wrong way to think of it?? It seems to me a perfectly reasonable way to think of it.

Someone hands you an unfamiliar metric. What's the first thing you do -- write out Killing's Equation and try to solve it?

Or do you say, aha, I see it doesn't depend on one of the coordinates ζ, so there's an obvious Killing vector.

4. May 16, 2014

### Staff: Mentor

You're assuming that the "0" coordinate is timelike. That's not always the case. Consider, for example, Schwarzschild coordinates on Schwarzschild spacetime inside the horizon; there the "0" (i.e., $t$) coordinate is spacelike, not timelike, but the metric is still independent of $t$, so $\partial / \partial t$ is still a KVF--just not a timelike one.