- #1

George Keeling

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- I think I now realise stuff about Killing vectors. Am I correct?

I'm on to section 5.4 of Carroll's book on Schwarzschild geodesics and he says stuff in it which, I think, enlightens me on the use of Killing vectors. I had to go back to section 3.8 on Symmetries and Killing vectors. I now understand the following:

A Killing vector satisfies $$

\nabla_{(\mu}K_{\nu)}=0

$$and that implies that ##K_\nu p^\nu## is conserved along a geodesic. ##p^\nu## is a 4-momentum.

Carroll says "In fact, if a vector ##K^\mu## satisfies Killing's equation, it will always be possible to find a coordinate system in which ##K=\partial_{\sigma\ast}##". ##x^{\sigma\ast}## was the coordinate of which the metric was independent. That is ##\partial_{\sigma\ast}g_{\mu\nu}=0##. So in GR if the coordinate was ##t## then we would have a ##K^\mu=\left(1,0,0,0\right)##. If the metric equation does not contain ##t## it must have such a Killing vector. It must then satisfy the equation ##\nabla_{(\mu}K_{\nu)}=0## and that implies that that ##K_\nu p^\nu## is conserved along a geodesic.

A few lines further on he says

"As we investigate in Appendix B, Killing vector fields on a manifold are in on-to-one correspondence with continuous symmetries of the metric [isometries] on that manifold".

From the second statement (slightly stretched) we infer that a time isometry must imply a Killing vector ##K=\left(1,0,0,0\right)## and with a Schwarzschild metric a ##\phi## isometry implies a Killing vector ##R=\left(0,0,0,1\right)##.

A Killing vector satisfies $$

\nabla_{(\mu}K_{\nu)}=0

$$and that implies that ##K_\nu p^\nu## is conserved along a geodesic. ##p^\nu## is a 4-momentum.

Carroll says "In fact, if a vector ##K^\mu## satisfies Killing's equation, it will always be possible to find a coordinate system in which ##K=\partial_{\sigma\ast}##". ##x^{\sigma\ast}## was the coordinate of which the metric was independent. That is ##\partial_{\sigma\ast}g_{\mu\nu}=0##. So in GR if the coordinate was ##t## then we would have a ##K^\mu=\left(1,0,0,0\right)##. If the metric equation does not contain ##t## it must have such a Killing vector. It must then satisfy the equation ##\nabla_{(\mu}K_{\nu)}=0## and that implies that that ##K_\nu p^\nu## is conserved along a geodesic.

A few lines further on he says

"As we investigate in Appendix B, Killing vector fields on a manifold are in on-to-one correspondence with continuous symmetries of the metric [isometries] on that manifold".

From the second statement (slightly stretched) we infer that a time isometry must imply a Killing vector ##K=\left(1,0,0,0\right)## and with a Schwarzschild metric a ##\phi## isometry implies a Killing vector ##R=\left(0,0,0,1\right)##.

**Is that all correct?**I want to check before I proceed, thanks.