# Show that killing vector field satisfies...

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1. May 11, 2016

### Augbrah

I'm trying to do past exam papers in GR but there are some things I don't yet feel comfortable with, so even though I can do some parts of the question I would be very happy if you could check my solution. Thank you!

1. The problem statement, all variables and given/known data

Spacetime is stationary := there exists a coord chart with a timelike coordinate $x^0$ such that metric components $\partial_0 g_{\mu \nu} = 0$.
Spacetime is static := there exists a coord chart such that $\partial_0 g_{\mu \nu} = 0$ and $g_{0i} = 0$

i) [Done] Show that spacetime is stationary if and only if there exists a timelike Killing vector field V.
ii) [Done] Show that if spacetime is static, there exists a timelike Killing vector field satisfying $V_{[\alpha}\nabla_\mu V_{v]}=0$
iii) Let V be timelike Killing vector field with $V_{[\alpha}\nabla_\mu V_{v]}=0$. Show that this condition implies $$\nabla_\mu(|V|^n V_\nu)-\nabla_\nu(|V|^n V_\mu)=0$$ Where $|V|^2=V_a V^a$ and $n$ is integer which should be determined.

2. Relevant equations
Killing vector field V satisfies $\nabla_\mu V_\nu + \nabla_\nu V_\mu = 0$

3. The attempt at a solution
i) I have done this part. For timelike $V^\mu$ we can find an inertial frame s.t. $V^\mu = \delta^\mu_0$. Then $V_\mu = g_{\mu 0}$. Hence $\nabla_\mu V_\nu + \nabla_\nu V_\mu =...=\partial_0 g_{\mu\nu}$, by expressing Christoffel symbols as partial derivatives of a metric other derivatives nicely cancel.

So we proved that timelike Killing V ⇔ $\partial_0 g_{\mu\nu}=0$.

ii) If spacetime is static, it is also stationary, so we already know from i) there must exist timelike Killing V, just need to show it satisfied the equation (right?). By expanding antisymmetrization we have
$$V_{\alpha}(\nabla_\mu V_{\nu} - \nabla_\nu V_{\mu}) + \text{(other 2 cyclicly permuted pairs)}=0$$ Studying the first term and using same identification $V_\mu = g_{\mu 0}$ and as before remembering that for static spacetime $g_{0i} = 0$ as well as $\partial_0 g_{\mu \nu} = 0$ we get:
$$V_{\alpha}(\nabla_\mu V_{\nu} - \nabla_\nu V_{\mu}) = g_{\alpha 0} (\partial_\mu g_{\nu 0} - \partial_\nu g_{\mu 0}) = 0$$

Same procedure for other two terms and total sum is 0.

iii) Besides product rule I did not got far. Any suggestions?

2. May 11, 2016

### strangerep

You should probably how far you got with part (iii).

3. May 28, 2016

### Rudyard

I might be missing something but I don't think what you wrote at the end of part ii) is correct. If you put μ=i and ν=0 this is not zero but equal to $g_{\alpha 0}\partial_i g_{00}$ which is not necessarily zero
You really need to antisymmetrise in order to get zero identically

Last edited: May 28, 2016