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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?

pxc3110
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1 Answers1

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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 $r^2$:

$$r^2\frac{d^2 x^\theta}{ds^2} + 2r\frac{dx^\theta}{ds}\frac{dx^r}{ds} - r^2\sin\theta \cos\theta \left (\frac{dx^\phi}{ds} \right)^2 = 0$$

JPhy
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