Derivation of equation for geodesic deviation

Question

I am trying to figure out the calculation which leads to the geodesic deviation on this site. So far I understood all steps until (14.7) and managed to show that (14.6) = (14.7), namely

$$ \ddot\xi^\alpha + \left( \partial_\delta \Gamma^\alpha_{\beta\gamma} \right) \dot x^\beta \dot x^\gamma \xi^\delta + \Gamma^\alpha_{\beta\gamma} \dot x^\beta \xi^\gamma + \Gamma^\alpha_{\beta\gamma} \dot x^\gamma \xi^\beta \\ = \frac{d}{d\tau} \left( \dot \xi^\alpha + \Gamma^\alpha_{\beta\gamma} \dot x^\gamma \xi^\beta \right) - \left( \partial_\delta \Gamma^\alpha_{\beta\gamma} \right) \xi^\beta \dot x^\gamma \dot x^\delta - \Gamma^\alpha_{\beta\gamma} \xi^\beta \ddot x^\gamma + \left( \partial_\delta \Gamma^\alpha_{\beta\gamma} \right) \dot x^\beta \dot x^\gamma \xi^\delta + \Gamma^\alpha_{\beta\gamma} \dot x^\beta \dot \xi^\gamma $$

So far so good. After "rearranging the terms and with the dummy indices suitably relabeled we get the following"

$$ \frac{D^2 \xi^\alpha}{D\tau^2} + \left( \partial_\gamma \Gamma^\alpha_{\beta\delta} - \partial_\delta \Gamma^\alpha_{\beta\gamma} + \Gamma^\epsilon_{\beta\delta} \Gamma^\alpha_{\epsilon\gamma} - \Gamma^\epsilon_{\beta\gamma} \Gamma^\alpha_{\epsilon\delta} \right) \dot x^\beta \xi^\gamma \dot x^\delta $$

I couldn't show the equality of this and (14.7). I've used the geodesic equation to get rid of the $\ddot x$ and used

$$ \frac{D^2 \xi^\alpha}{D\tau^2} = \frac{D}{D\tau} \left( \dot x^\sigma \nabla_\sigma \xi^\alpha \right) = \dot x^\rho \nabla_\rho \left( \dot x^\sigma \nabla_\sigma \xi^\alpha \right) \\ = \ddot \xi^\alpha + \Gamma^\alpha_{\rho\delta} \dot x^\rho \dot \xi^\delta + \left( \partial_\rho \Gamma^\alpha_{\sigma\gamma} \right) \dot x^\sigma \xi^\gamma \dot x^\rho + \Gamma^\alpha_{\sigma\rho} \dot x^\sigma \dot x^\rho + \Gamma^\alpha_{\rho\delta} \Gamma^\delta_{\sigma\gamma} \dot x^\sigma \xi^\gamma \dot x^\rho$$

with $$ \nabla_\sigma \xi^\alpha = \frac{d\xi^\alpha}{d x^\sigma} + \Gamma^\alpha_{\sigma\gamma} \xi^\gamma$$

but no matter what I did it didn't work out. I must be doing a major mistake somewhere along the lines but I just can't find it. I would greatly appreciate some help.

Answer 1

First, note that equation (14.6) in the linked page has a mistake. On the second line of equation (14.6), the quantities $\xi$ should be $\dot\xi$. Equation (14.7) is correct, but (14.6) is missing the dots. Every term in those equations should have two derivatives with respect to the world-line parameter.

Here's a way to organize the calculation that makes the details easier to track — because it avoids introducing details until they're needed. Start with the equation for a single geodesic. $$ \ddot x^a=-\Gamma^a_{bc}\dot x^b\dot x^c. \tag{1} $$ If we write $\delta x^a$ (instead of $\xi^a$) for a variation of the geodesic, then taking the variation of equation (1) gives $$ \delta \ddot x^a =-2\Gamma^a_{bc}\dot x^b\delta \dot x^c -(\delta \Gamma^a_{bc})\dot x^b\dot x^c. \tag{2} $$ assuming that $\Gamma^a_{bc}=\Gamma^a_{cb}$. Equation (2) is a corrected version of equation (14.6) in the linked page. Now, set that aside for a moment, and consider the quantity $$ \frac{D^2}{Du^2}\delta x^a, \tag{3} $$ where $u$ is the worldline parameter (so the dots denote derivatives with respect to $u$) and where $$ \frac{D}{Du}v^a \equiv \dot x^b\nabla_b v^a = \dot v^a+\Gamma^a_{bc}\dot x^b v^c \tag{4} $$ for any vector $v$. Expand (3) just far enough to expose the $\delta\ddot x$ term, like this: \begin{align} \frac{D^2}{Du^2}\delta x^a &= \frac{d}{du}\left(\frac{D}{Du}\delta x^a\right) + \Gamma^a_{bc}\dot x^b\left(\frac{D}{Du}\delta x^c\right) \\ &= \delta\ddot x^a + \frac{d}{du}\left(\Gamma^a_{bc}\dot x^b\delta x^c\right) + \Gamma^a_{bc}\dot x^b\left(\frac{D}{Du}\delta x^c\right) \tag{5} \end{align} Substitute (2) into (5) to get $$ \frac{D^2}{Du^2}\delta x^a = \frac{d}{du}\left(\Gamma^a_{bc}\dot x^b\delta x^c\right) + \Gamma^a_{bc}\dot x^b\left(\frac{D}{Du}\delta x^c\right) -2\Gamma^a_{bc}\dot x^b\delta \dot x^c -(\delta \Gamma^a_{bc})\dot x^b\dot x^c. \tag{6} $$ Before expanding things any further, remember that we expect the terms involving $\delta\dot x$ to cancel. Just by inspecting equation (6), we can verify that those terms do cancel. After discarding those terms, equation (6) becomes $$ \frac{D^2}{Du^2}\delta x^a = \frac{d}{du}\left(\Gamma^a_{bc}\dot x^b\right)\delta x^c + \Gamma^a_{bc}\dot x^b\left(\Gamma^c_{de}\dot x^d\delta x^e\right) -(\delta \Gamma^a_{bc})\dot x^b\dot x^c. \tag{7} $$ Now use $\delta \Gamma^a_{bc}=\partial_d\Gamma^a_{bc}\delta x^d$ to get $$ \frac{D^2}{Du^2}\delta x^a = \frac{d}{du}\left(\Gamma^a_{bc}\dot x^b\right)\delta x^c + \Gamma^a_{bc}\dot x^b\left(\Gamma^c_{de}\dot x^d\delta x^e\right) -(\partial_d\Gamma^a_{bc}\delta x^d)\dot x^b\dot x^c. \tag{9} $$ The next step is to rewrite the right-hand side using a single factor of $\delta x$, which requires re-labeling a couple of indices. From there, showing that the right-hand side can be written in terms of the curvature tensor should be relatively straightforward, because most of the other details have already been eliminated.