# Covariant derivative of a vector in the Schwarzschild metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; 8.

Here’s another example of calculating the covariant derivative in the Schwarzschild (S) metric. We’re given a vector with coordinates in the S metric of:

$\displaystyle \mathbf{v}=\left[1-\frac{2GM}{r},0,0,0\right] \ \ \ \ \ (1)$

The covariant derivative is given by

$\displaystyle \nabla_{j}v^{k}\equiv\frac{\partial v^{k}}{\partial x^{j}}+v^{i}\Gamma_{\; ij}^{k} \ \ \ \ \ (2)$

Since the only non-zero component of ${\mathbf{v}}$ is ${v^{t}}$ and it depends only on ${r}$, most of the terms are zero.

$\displaystyle \Gamma_{\; ij}^{t}=\left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (3)$

$\displaystyle \Gamma_{\; ij}^{r}=\left[\begin{array}{cccc} \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right) & 0 & 0 & 0\\ 0 & -\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ 0 & 0 & -r\left(1-\frac{2GM}{r}\right) & 0\\ 0 & 0 & 0 & -r\sin^{2}\theta\left(1-\frac{2GM}{r}\right) \end{array}\right] \ \ \ \ \ (4)$

$\displaystyle \Gamma_{\; ij}^{\theta}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & -\sin\theta\cos\theta \end{array}\right] \ \ \ \ \ (5)$

$\displaystyle \Gamma_{\; ij}^{\phi}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{r}\\ 0 & 0 & 0 & \cot\theta\\ 0 & \frac{1}{r} & \cot\theta & 0 \end{array}\right] \ \ \ \ \ (6)$

The one non-zero derivative is

$\displaystyle \frac{\partial v^{t}}{\partial r}=\frac{2GM}{r^{2}} \ \ \ \ \ (7)$

and the values of the second term in 2 are

 $\displaystyle v^{i}\Gamma_{\; ir}^{t}$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\ \ \ \ \ (8)$ $\displaystyle v^{i}\Gamma_{\; it}^{r}$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{2} \ \ \ \ \ (9)$

with all other terms being zero.

The covariant derivative is then (with ${j}$ the row index and ${k}$ the column index):

$\displaystyle \nabla_{j}v^{k}=\left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{2} & 0 & 0\\ \frac{3GM}{r^{2}} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (10)$