1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conservation law using Killing vector

  1. Jul 8, 2008 #1
    In Hartle's GR book (p. 177), there is a derivation of [tex]\xi \cdot u = constant[/tex], where [tex]\xi[/tex] is a Killing vector, [tex]u[/tex] is four-velocity along a geodesic in an arbitrary metric, and

    [tex]L = (-g_{\alpha\beta}\frac{dx^\alpha}{d\sigma}\frac{dx^\beta}{d\sigma})^\frac{1}{2}[/tex]

    The derivation goes:

    [tex] \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 [/tex]

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

    [tex]\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}[/tex]

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

    Last edited: Jul 8, 2008
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: Conservation law using Killing vector
  1. Null Killing vectors (Replies: 3)