# Geodesic deviation equation with covariant derivative [closed]

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?

• A solution to that homework question can be found in these lecture notes on p. 87. Aug 3 at 12:27

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.