Q:

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.

A:

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.

