# Conservation law using Killing vector

1. Jul 8, 2008

### stephenmitten

In Hartle's GR book (p. 177), there is a derivation of $$\xi \cdot u = constant$$, where $$\xi$$ is a Killing vector, $$u$$ is four-velocity along a geodesic in an arbitrary metric, and

$$L = (-g_{\alpha\beta}\frac{dx^\alpha}{d\sigma}\frac{dx^\beta}{d\sigma})^\frac{1}{2}$$

The derivation goes:

$$\frac{\partial}{\partial \sigma}\frac{\partial L}{\partial \frac{dx^1}{d\sigma}}} = 0 \\ \Rightarrow \frac{\partial L}{\partial \frac{dx^1}{d\sigma}} = -g_{1\beta}\frac{1}{L}\frac{dx^\beta}{d\sigma} = ... = -\xi \cdot u$$

is conserved along the geodesic. (Here the symmetry associated with $$\xi$$ is in $$x^1$$.) It seems to be saying that

$$\frac{\partial L}{\partial \frac{dx^1}{d\sigma}} = \frac{1}{2L}({-g_{\alpha 1}\frac{1}{L}\frac{dx^\alpha}{d\sigma}-g_{1\beta}\frac{1}{L}\frac{dx^\beta}{d\sigma}) = {-g_{1\beta}\frac{1}{L}\frac{dx^\beta}{d\sigma}$$

but it appears to me that $$\frac{\partial L}{\partial \frac{dx^1}{d\sigma}}$$ has only seven terms, not eight, since $$-g_{11}\frac{dx^1}{d\sigma}}\frac{dx^1}{d\sigma}}$$ appears only once. I'd appreciate it if someone could point out where I went wrong.

Thanks.

Last edited: Jul 8, 2008