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.

  • $\begingroup$ What happened to the Christoffel symbols? I think you're missing some terms in your expressions for the covariant derivatives $\endgroup$ Nov 24, 2018 at 17:54
  • $\begingroup$ @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 $\lambda=\phi$. $\endgroup$
    – Stig
    Nov 24, 2018 at 18:42

2 Answers 2


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.


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