Solving the geodesic equation.

Posted on March 4, 2011 by johannesvalks

In this post, I will write about the geodesic equation. I suggest a simplified form that is simpler to solve.

The geodesic equation is given by

\displaystyle \frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\kappa\lambda} \frac{x^\kappa}{ds} \frac{x^\lambda}{ds} = 0,

where \displaystyle \Gamma^\mu_{\kappa\lambda} = \frac{1}{2} g^{\mu\alpha} \left\{ \partial_\kappa g_{\lambda\alpha} + \partial_\lambda g_{\alpha\kappa} - \partial_\alpha g_{\kappa\lambda} \right\}. We have used the convention that \displaystyle \partial_\kappa g_{\lambda\alpha} = \frac{\partial g_\lambda\alpha}{\partial x^\kappa}. Then we may write

\displaystyle \frac{d}{ds} \left( g_{\alpha\mu} \frac{x^\mu}{ds} \right) = \frac{1}{2} \left( \partial_\alpha g_{\kappa\lambda} \right) \frac{x^\kappa}{ds} \frac{x^\lambda}{ds}

Then it is clear that IF the metric g_{\mu\nu} does not depend on x_\alpha we obtain the equation

\displaystyle \frac{d}{ds} \left(g_{\alpha\mu} \frac{dx^\mu}{ds} \right) = 0,

hus \displaystyle g_{\alpha\mu} \frac{dx^\mu}{ds} = \textit{constant}.

Consider for example the Schwarzschild metric, given by

\displaystyle ds^2 = \psi(r) c^2 dt^2 - \frac{1}{\psi(r)} dr^2 - r^2 d\theta^2 - r^2 \sin^2(\theta) d\phi^2.

where \displaystyle \psi(r) = 1 - \frac{2GM}{rc^2}.

Then for \theta we obtain the equation

\displaystyle \frac{d}{ds} \left( g_{\theta\theta} \frac{d\theta}{ds} \right) = - r^2 \sin(\theta) \cos(\theta) \left(\frac{d\phi}{ds}\right)^2,

where \theta = \pi/2 is a solution. Next we consider the equation

\displaystyle \frac{d}{ds} \left( g_{rr} \frac{dr}{ds} \right) = \left( \psi'(r) + \frac{\psi'(r)}{\psi^2(r)} \dot{r}^2 - 2r \dot{\phi}^2 \right) \left(\frac{dt}{ds} \right)^2.

Say we consider circular orbits – thus dr=0 then we obtain

\displaystyle 0 = \left( \psi'(r) c^2 - 2r \dot{\phi}^2 \right) \left(\frac{dt}{ds} \right)^2,

thus we obtain

\displaystyle \frac{GM}{r^2} - r \dot{\phi}^2 = 0,

so

\displaystyle \dot{\phi}^2 = \frac{GM}{r^3},

which is know as the Third Law of Kepler.

We have here derived the Third law of Kepler for the Schwarzschild metric, without calculating the affine connection \Gamma^\mu_{\kappa\lambda}.

Have fun!

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