## Saturday 12 October 2019

### Symmetries and Killing vectors

 Sean Carroll, my guide and nemesis
I'm now reading section 3.8 on symmetries and Killing vectors. It's not too hard to follow but there are a few stumbling blocks.

After equation 3.161 for the geodesic in terms of 4 momentum $p^\lambda\nabla_\lambda p^\mu=0$  Carroll says that by metric compatibility we are free to lower the index $\mu$. Metric compatibility means that $\nabla_\rho g_{\mu\nu}=\nabla_\rho g^{\mu\nu}=0$ so I tried to show that, given that, $\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$. Here was my first attempt:
Lower the index with the metric, use the Leibnitz rule, use metric compatibility$$\nabla_\lambda p^\mu=\nabla_\lambda g^{\mu\nu}p_\nu=p_\nu\nabla_\lambda g^{\mu\nu}+g^{\mu\nu}\nabla_\lambda p_\nu=0+\nabla_\lambda p^\mu$$Then I tried painfully expanding $\nabla_\lambda g^{\mu\nu}p_\nu$ and got the same result. So then I asked on Physics Forums: Why does metric compatibility imply  $\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$? I got my wrist slapped by martinbn who pointed out that  $\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$ made no sense because there are different types of tensors on each side of the equation. (The $\mu$ is up on one side and down on the other). I was embarrassed😡.

What Carroll is really saying is that metric compatibility means that$$\nabla_\lambda p^\mu=0\Rightarrow\nabla_\lambda p_\mu=0$$which is quite different and easy to show:$$\nabla_\lambda p^\mu=0$$$$\Rightarrow g^{\mu\nu}\nabla_\lambda p_\nu=0$$$$\Rightarrow g_{\rho\mu}g^{\mu\nu}\nabla_\lambda p_\nu=0$$$$\Rightarrow\delta_\rho^\nu\nabla_\lambda p_\nu=0$$$$\Rightarrow\nabla_\lambda p_\rho=0$$I posted something very like those steps and there was silence which usually means they are correct. The first step uses, $\nabla_\lambda p^\mu=g^{\mu\nu}\nabla_\lambda p_\nu$, which can be done in several ways
1) $\nabla_\lambda p^\mu$ is a tensor so you can lower (or raise) an index with the metric as usual.
2) $\nabla_\lambda p^\mu=\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)=p_\nu\nabla_\lambda g^{\mu\nu}+g^{\mu\nu}\nabla_\lambda p_\nu=g^{\mu\nu}\nabla_\lambda p_\nu$ as in (1) use the Leibnitz rue and metric compatibility
3) $\nabla_\lambda p^\mu=\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)=g^{\mu\nu}\nabla_\lambda p_\nu$ using Carroll's third rule for covariant derivatives: That they commutes with contractions.

The Leibnitz rule was the second rule of covariant derivatives and I discussed all four in Commentary 3.2 Christoffel Symbol. The third caused angst and another question on PF. I now think that the third rule is just saying that because the covariant derivative is a tensor you can raise and lower indexes on it. 3 and 1 above are really the same. I have suitably amended Commentary 3.2 Christoffel Symbol.

Sometimes I hate Carroll!

There was also another post on the thread ahead of the first two which referred to a similar question on Stack Exchange. MathematicalPhysicist was asked to show that $$U^\alpha\nabla_\alpha V^\beta=W^\beta\Rightarrow U^\alpha\nabla_\alpha W_\beta=W_\beta$$The proof for this is very similar to the above:$$U^\alpha\nabla_\alpha V^\beta=W^\beta$$$$\Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}W_\gamma$$$$\Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha V_\gamma=g_{\mu\beta}g^{\beta\gamma}W_\gamma$$$$\Rightarrow U^\alpha\delta_\mu^\gamma\nabla_\alpha V_\gamma=\delta_\mu^\gamma W_\gamma$$$$\Rightarrow U^\alpha\nabla_\alpha V_\mu=W_\mu$$Once again there are three ways to do the first step. Metric compatibility is not essential.

See Commentary 3.8 Symmetries and Killing vectors.pdf first two pages. Then I run into another problem with Killing.