Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Killing Vectors Question

  1. Jan 27, 2014 #1
    See below. I screwed up the edit and the use of tex.
    Last edited: Jan 27, 2014
  2. jcsd
  3. Jan 27, 2014 #2
    EDIT: Used proper tex (hopefully!)

    Hello! I'm working through Weinberg's book Gravitation and Cosmology, and I'm currently in chapter 13, symmetric spaces. I'm trying to follow his derivation of the Killing condition, and I simply cannot, for the life of me, get from equation 13.1.2 to equation 13.1.4. I plugged 13.1.3 into 13.1.2 as he says to, but what I get is very different.
    13.1.2: [tex] g_{\mu\nu} (x) = \frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu}g_{\rho\sigma} (x') [/tex]
    And 13.1.3: [tex] x'^\mu = x^\mu + \epsilon \zeta^\mu (x) [/tex]

    Then, only keep the result of the substitution to first order in epsilon. When I do this, I get:
    [tex] g_{\mu\nu} (x) = \frac{\partial x^\rho}{\partial x^\mu} \frac{\partial x^\sigma}{\partial x^\nu} g_{\rho\sigma} (x') + \epsilon \left [ \frac{\partial \zeta^\sigma (x) }{\partial x^\nu } \frac{\partial x^\rho }{\partial x^\mu } g_{\rho\sigma} (x') + \frac{\partial \zeta^\rho (x)}{\partial x^\mu } \frac{\partial x^\sigma }{\partial x^\nu } g_{\rho\sigma} (x') \right ] [/tex].

    It's supposed to be 13.1.4: [tex] 0 = \frac{\partial \zeta^\mu (x)}{\partial x^\rho} g_{\mu\sigma}(x) + \frac{\partial \zeta^\nu (x)}{\partial x^\sigma} g_{\rho\nu} (x) + \zeta^\mu (x) \frac{\partial g_{\rho\sigma} (x)}{\partial x^\mu} [/tex]

    All of his metrics are functions of x, not x', and he has no epsilon in the equation. That makes it seem to me that the first term on the right hand side of the equation I got has to equal the left hand side, so that they cancel and equal 0. Then the epsilon can divide out. The problem is that then there are only two terms left, as opposed to the three that he has. I'm guessing it has something to do with switching from g(x) to g(x'), but I don't see it. Any help would be greatly, greatly appreciated! Thank you very much!
  4. Jan 27, 2014 #3


    User Avatar
    Science Advisor

    Your expression reduces to ##g_{\mu\nu}(x) = g_{\mu\nu}(x) + \epsilon (\zeta^{\rho}\partial_{\rho}g_{\mu\nu}(x) + g_{\mu\sigma}(x)\partial_{\nu}\zeta^{\sigma} + g_{\rho \nu}\partial_{\mu}\zeta^{\rho}) ## after using the fact that ##g_{\rho\sigma}(x') = g_{\rho\sigma}(x) + \epsilon \zeta^{\gamma}\partial_{\gamma}g_{\rho\sigma}(x) + O(\epsilon^2)## hence ##\zeta^{\mu}\partial_{\mu}g_{\rho\sigma}(x) + g_{\rho\nu}(x)\partial_{\sigma}\zeta^{\nu} + g_{\mu\sigma}\partial_{\rho}\zeta^{\mu} = 0 ## after appropriately relabeling the indices. Don't forget that ##\partial_{\nu}x^{\mu} = \delta^{\mu}_{\nu}##.
  5. Jan 27, 2014 #4
    Thank you very much! I totally forgot about expanding the metric, as well as the delta condition.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook