# Linearized Gravity Wald

1. Jan 28, 2014

### center o bass

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

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