The schwarzschild metric is given by:
$ds^2=-(1-\frac {2GM}{r})dt^2+(1-\frac{2GM}r)^{-1} dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2$
Here is the well known geodesic equation:
$0=\frac d{d\tau}(g_{\mu \nu}\frac{dx^{\nu}}{d\tau})-\frac 12\partial_{\mu}g_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau}$
I was trying to compute the $\theta$ component of the geodesic equation:
Substitute $\mu$ for $\theta$:
$0=\frac{d}{d\tau}g_{\theta\theta}\frac{d\theta}{d\tau}-\frac 12\frac{\partial g_{\phi\phi}}{\partial \theta}(\frac {d\phi}{d\tau})^2$
$0=\frac{d}{d\tau}r^2\frac{d\theta}{d\tau}-\frac 12(2r^2\sin\theta\cos\theta)(\frac {d\phi}{d\tau})^2$
$0=r^2\frac{d^2\theta}{d\tau^2}-r^2\sin\theta\cos\theta(\frac {d\phi}{d\tau})^2$
However, the result given on a book called A General relativity workbook by Thomas A. Moore is
$0=r^2\frac{d^2\theta}{d\tau^2}+2r\frac{dr}{d\tau}\frac{d\theta}{d\tau}-r^2\sin\theta\cos\theta(\frac {d\phi}{d\tau})^2$
i.e. This result has an extra term than mine, namely $2r\frac{dr}{d\tau}\frac{d\theta}{d\tau}$. My question is: Is it me who is wrong, or is the author the one who gets it wrong?
The Schwarzschild metric is given by,
$$ds^2 = \left( 1-\frac{2GM}{r}\right) dt^2 - \left(1-\frac{2GM}{r} \right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta \, d\phi^2$$
The $\theta$ component of the geodesic $x^\theta(s)$ satisfies,
$$\frac{d^2 x^\theta}{ds^2} + \Gamma^\theta_{ab} \frac{dx^a}{ds}\frac{d x^b}{ds} = 0$$
There are only three non-vanishing Christoffel symbols in this case, namely, $\Gamma^\theta_{r\theta} = \Gamma^\theta_{\theta r} = 1/r$ and,
$$\Gamma^\theta_{\phi\phi} = -\sin\theta \cos\theta$$
Plugging these in, we find,
$$\frac{d^2 x^\theta}{ds^2} + \frac{2}{r}\frac{dx^\theta}{ds}\frac{dx^r}{ds} - \sin\theta \cos\theta \left (\frac{dx^\phi}{ds} \right)^2 = 0$$
To get to your expression multiply by $r^2$:
$$r^2\frac{d^2 x^\theta}{ds^2} + 2r\frac{dx^\theta}{ds}\frac{dx^r}{ds} - r^2\sin\theta \cos\theta \left (\frac{dx^\phi}{ds} \right)^2 = 0$$