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.

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