Consider two wordlines $x^\mu(\sigma)$ and $x^\mu(\sigma)+\xi^\mu(\sigma)$, where $|\xi|<|x|$. The two wordlines fulfill the geodesic equation
$$\frac{\text{d}x^\mu}{\text{d}\sigma^2}+\Gamma^\mu_{~\nu\rho}(x)\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}x^\rho}{\text{d}\sigma}=0$$
and
$$\frac{\text{d}\left(x^\mu+\xi^\mu\right)}{\text{d}\sigma^2}+\Gamma^\mu_{~\nu\rho}(x+\xi)\frac{\text{d}\left(x^\nu+\xi^\nu\right)}{\text{d}\sigma}\frac{\text{d}\left(x^\rho+\xi^\rho\right)}{\text{d}\sigma}=0$$
Expanding the second equation around $x$ and subtracting it from the first equation yields:
$$\frac{\text{d}^2\xi^\mu}{\text{d}\sigma}+2\Gamma^{\mu}_{~\nu\rho}(x)\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}\xi^\rho}{\text{d}\sigma}+\xi^\sigma\frac{\partial}{\partial x^\sigma}\Gamma^{\mu}_{\nu\rho}\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}x^\rho}{\text{d}\sigma}=0 \tag{1}$$
My book now states that equation (1) can be written as
$$\frac{D^2\xi^\mu}{D\sigma^2}=-R^{\mu}_{\;\,\nu\rho\sigma}\xi^\rho\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}x^\sigma}{\text{d}\sigma} \tag{2}$$
where the covariant derivative of a vector $V^\mu$ along a worldline is defined as:
$$\frac{DV^\mu}{D\sigma}=\frac{\text{d}V^\mu}{\text{d}\sigma}+\Gamma^{\mu}_{\alpha\beta}\frac{\text{d}x^\alpha}{\text{d}\sigma} V^\beta$$
I want to show that from (2) follows from (1). The left term in equation (2) is:
$$\frac{D^2\xi^\mu}{D\sigma^2}=\frac{D}{D\sigma}\left[\frac{\text{d}\xi^\mu}{\text{d}\sigma}+\Gamma^{\mu}_{\alpha\beta}\frac{\text{d}x^\alpha}{\text{d}\sigma} \xi^\beta\right] \\ =\frac{\partial}{\partial \sigma}\left[\frac{\text{d}\xi^\mu}{\text{d}\sigma}+\Gamma^{\mu}_{\alpha\beta}\frac{\text{d}x^\alpha}{\text{d}\sigma} \xi^\beta\right]+\Gamma^{\mu}_{\kappa\delta}\frac{\text{d}x^\delta}{\text{d}\sigma}\left(\frac{\text{d}\xi^\kappa}{\text{d}\sigma}+\Gamma^{\kappa}_{\nu\rho}\xi^\nu \frac{\text{d}x^\rho}{\text{d}\sigma}\right)$$
The other term that appears in (2) is
$$R^{\mu}_{\;\,\nu\rho\sigma}\xi^\rho\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}x^\sigma}{\text{d}\sigma}=\left(\partial_\rho\Gamma^\mu_{\nu\sigma}-\partial_\sigma\Gamma^\mu_{\nu\rho}+\Gamma^\mu_{\alpha\rho}\Gamma^\alpha_{\nu\sigma}-\Gamma^\mu_{\alpha\sigma}\Gamma^\alpha_{\nu\rho}\right)\xi^\rho\frac{\text{d}x^\nu}{\text{d}\sigma}\frac{\text{d}x^\sigma}{\text{d}\sigma}$$
When I write everything out I get a lot of terms and I don't see why a lot of them should cancel. Maybe you could give me a hint if I am on the right track?
Best Answer
You are on the right track! All of these terms indeed do cancel. In your equation $$\frac{D^2\xi^\mu}{D\sigma^2}=\frac{\partial}{\partial \sigma}\left[\frac{\text{d}\xi^\mu}{\text{d}\sigma}+\Gamma^{\mu}_{\alpha\beta}\frac{\text{d}x^\alpha}{\text{d}\sigma} \xi^\beta\right]+\Gamma^{\mu}_{\kappa\delta}\frac{\text{d}x^\delta}{\text{d}\sigma}\left(\frac{\text{d}\xi^\kappa}{\text{d}\sigma}+\Gamma^{\kappa}_{\nu\rho}\xi^\nu \frac{\text{d}x^\rho}{\text{d}\sigma}\right),$$
you should have written $\frac{\text{d}}{\text{d}\sigma}$ instead of the partial derivative. With this, and remembering that
$$\frac{\text{d}}{\text{d}\sigma} = \frac{\text{d}x^\mu}{\text{d}\sigma}\frac{\partial}{\partial x^\mu},$$ you should recover the terms that involve the derivatives of the Christoffel symbols contracted twice with the four-velocity and $\xi$. A simple relabeling will get you the result after that.