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Linearized Gravity Wald

  1. Jan 28, 2014 #1
    Due to the diffeomorphism invariance of GR ##(M, g_{ab})## and ##(M, \phi_*g_ab)## represents the same physical situation. (##\phi_*## denotes the pushforward, Wald uses another convention and denotes the pushforward by ##\phi^*##.) In linearized gravity one imagines a one-parameter family of solutions ##g_{ab} = g_{ab}(\lambda)## and one taylor expands this about a know solution ##g_{ab}(0)## such that to first order ##g_{ab} = g_{ab}(0) + \left. \tfrac{d g_{ab}}{d\lambda}\right|_{\lambda = 0} \lambda## where ##\gamma_{ab} := \left. \tfrac{d g_{ab}}{d\lambda}\right|_{\lambda=0}## is called the perturbation of ##g_{ab}(0)##.

    Now due to diffeomorphism invariance ##\gamma_{ab} = \left. \tfrac{d g_{ab}}{d\lambda}\right|_{\lambda = 0}## and ##\gamma'_{ab} = \left. \tfrac{d (\phi_{\lambda})_* g_{ab}}{d\lambda}\right|_{\lambda = 0}## represents the same physical perturbation. Now Wald states that it is not hard to see that

    ##\gamma'_{ab} - \gamma_{ab} = - \mathcal{L}_v g_{ab}##

    I have a question regarding this derivation. The standard definition of the lie derivative of a tensor ##T## where v is the tangent vector to the one-parameter group of diffeomorphisms ##\phi_{\lambda}##.

    $$(\mathcal{L}_v T)_p = \lim_{\lambda \to 0} \frac{(\phi_{-\lambda})_* T_{\phi(p)} - T_p}{\lambda}$$

    Let drop the abstract index notation, then we get

    $$\gamma'_p- \gamma_p = \left. \frac{d}{d\lambda}\right|_{\lambda=0}( (\phi_\lambda)_* g_{\phi_{-\lambda}(p)}(\lambda) - g_p(\lambda)) = \lim_{\lambda\to 0} \frac{(\phi_\lambda)_* g_{\phi_{-\lambda}(p)}(\lambda) - g_p(\lambda) -( (\phi_0)_* g_{\phi_{0}(p)}(0) - g_p(0))}{\lambda} = \lim_{\lambda\to 0} \frac{(\phi_\lambda)_* g_{\phi_{-\lambda}(p)}(\lambda) - g_p(\lambda) }{\lambda}\\
    = \lim_{-\lambda\to 0} \frac{(\phi_{-\lambda})_* g_{\phi_{\lambda}(p)}(-\lambda) - g_p(-\lambda) }{-\lambda} = - \lim_{\lambda\to 0} \frac{(\phi_{-\lambda})_* g_{\phi_{\lambda}(p)}(-\lambda) - g_p(-\lambda) }{\lambda}.$$

    Which is close to the definition but the terms in the numerator still depends on ##\lambda##. Im tempted to taylor expand these about ##\lambda = 0##, but the problem is that the pushforward on the first term also depends on ##\lambda##, so that destroys the Lie derivative. So how do I convince myself that this limit is indeed

    $$- \lim_{\lambda\to 0} \frac{(\phi_{-\lambda})_* g_{\phi_{\lambda}(p)}(0) - g_p(0) }{\lambda}?$$
     
  2. jcsd
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