# Killing vector tangents to geodesics

1. Jun 13, 2015

### Andre' Quanta

Suppose to have a killing vector that its norm is null, so at the same time is also a null geodesic.
Does the metric have special propierty? What can i say about the Killing vector and its proprierties?

2. Jun 13, 2015

### Staff: Mentor

Why does the Killing vector being null imply that it's a null geodesic? Not all null curves are null geodesics.

3. Jun 13, 2015

### WannabeNewton

Because $\xi^{\nu}\nabla_{\nu}\xi^{\mu} = -\xi^{\nu}\nabla^{\mu}\xi_{\nu} = - \frac{1}{2}\nabla^{\mu}\xi^2 = 0$ if $\xi^{\mu}$ is a null Killing field so the integral curves of $\xi^{\mu}$ are null geodesics.

To the OP, you need to be more specific. There is a myriad properties that such a space-time possesses, which you can find coherently interspersed throughout the text "Exact Solutions of Einstein's Field Equations" by Stephani et al.

As for the null Killing vector, the most important property is that $\nabla_{[\mu}\omega_{\nu]} = 0 \Leftrightarrow \omega^{\mu} = 0$ where $\omega^{\mu}$ is the twist of the null Killing field. I typed up a short calculation demonstrating the non-trivial direction; see the attached image.

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