# Killing vectors - Schwarzschild metric

Given the Schwarzschild metric, $$\mathrm{d}s^{2}=-\left(1-\frac{R_s}{r}\right)\mathrm{d}t^{2}+\left(1-\frac{R_s}{r}\right)^{-1}\mathrm{d}r^{2}+r^{2}\mathrm{d}\theta^{2}+ r^2 \sin^{2}\theta\mathrm{d}\phi^{2},$$ I'm asked to show that $$K^{\mu}=\left(1,0,0,0\right),\; R^{\mu}=(0,0,0,1)$$ are Killing vectors, i.e. they satisfy the Killing equation $$\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0.$$

First of all, using the metric I lower the indices: $$K_\mu=\left(-\left(1-\frac{R_s}{r}\right),0,0,0\right),\; R_\mu=\left(0,0,0,r^2\sin^2\theta\right).$$ Then, for $$R_\mu$$ I tried to compute $$\nabla_\mu R_\nu\equiv\partial_\mu R_\nu - \Gamma^\lambda_{\mu\nu} R_\lambda=\partial_\mu R_\nu - \Gamma^\phi_{\mu\nu} R_\phi;$$ so $$\nabla_r R_\phi=r\sin^2\theta, \; \nabla_\theta R_\phi=r^2\sin\theta\cos\theta.$$ But now it's not clear for me what should I do next.

• What happened to the Christoffel symbols? I think you're missing some terms in your expressions for the covariant derivatives Nov 24, 2018 at 17:54
• @N. Steinle, I have an implicit sum over $\lambda=t,r,\theta,\phi$. Then I realize that the only non-vanishing Christoffel symbol is that one whit $\lambda=\phi$.
– Stig
Nov 24, 2018 at 18:42

Your equation for $$R_\mu$$ reads $$\nabla_\mu R_\nu + \nabla_\nu R_\mu =\left(\partial_\mu R_\nu -\Gamma^\lambda_{\mu\nu} R_\lambda\right) + \left(\partial_\nu R_\mu - \Gamma^\sigma_{\nu\mu}R_\sigma\right)=\partial_\mu R_\nu + \partial_\nu R_\mu -2\Gamma^\phi_{\mu\nu} R_\phi.$$ If you use $$\mu=r$$ and $$\mu=\theta$$ (the only non-vanishing terms), you will find that Killing equation for $$R_\mu$$ it's satisfied.

$$\def\bg{{\mathbf g}}$$ It depends on how Killing vector has been defined. If definition was "a vector field obeying Killing equation", then you are only left to pursue the computation and I can add nothing. But there is an alternative definition, which I prefer:

A Killing field is one whose flow is an isometry for all values of parameter.

This requires to know what is the flow of a vector field and what means "isometry" in a Riemannian manifold. I shortly summarize.

The flow of a vector field $$X$$ is simply the ensemble of its integral curves, each parametrized by a real variable which I will name $$u$$. For each value of $$u$$ the flow of $$X$$ defines a mapping $$\mu$$ of the manifold into itself.

An isometry of a Riemannian manifold is a (differentiable) mapping leaving invariant the metric tensor: $$\mu^*\bg = \bg$$.

Then it can be shown that $$X$$ is a Killing field for the metric $$\bg$$ iff it satisfies the Killing equation (where covariant derivative is defined by Levi-Civita connection of $$\bg$$).

In your problem the flow of $$K$$ is obviously the mapping $$\mu_u: (t,r,\theta,\phi) \mapsto (t+u,r,\theta,\phi)$$ i.e. a time translation by $$u$$. Since $$\bg$$ (i.e. $$ds^2$$) doesn't depend on $$t$$, $$\mu_u$$ is an isometry for all $$u$$.

Analogously, the flow of $$R$$ is $$\nu_u: (t,r,\theta,\phi) \mapsto (t,r,\theta,\phi+u)$$ i.e. a rotation of angle $$u$$. And $$\bg$$ doesn't even depend on $$\phi$$, so $$\nu_u$$ is an isometry as well.