# Deriving geodesic equation

1. Feb 9, 2014

### Nabigh R

I am trying to derive the geodesic equation by extremising the integral
$$\ell = \int d\tau$$
Now after applying Euler-Lagrange equation, I finally get the following:
$$\ddot{x}^\tau + \Gamma^\tau_{\mu \nu} \dot{x}^\mu \dot{x}^\nu = \frac{1}{2} \dot{x}^\tau \frac{d}{ds} \ln \left| \dot{x}_\nu \dot{x}^\nu \right|$$
where $\dot{x}^\tau \equiv \frac{d x^\tau}{ds}$ and $s$ is a parameter. Now I get the geodesic equation if the right-hand side vanishes, and the only way that happens is if $\dot{x}_\nu \dot{x}^\nu$ is constant. Now the question is why is it constant?

2. Feb 9, 2014

### stevendaryl

Staff Emeritus
Well, $s$ is an arbitrary parameter. We are free to choose any parametrization we like. One particular parametrization, which is possible for slower-than-light geodesics, is to let

$ds = \sqrt{|dx^\mu dx_\mu|}$

For this choice, $|\frac{dx^\mu}{ds} \frac{dx_\mu}{ds}| = 1$

3. Feb 9, 2014

### Bill_K

The geodesic equation takes its usual form only when s is chosen to be an affine parameter. For a timelike geodesic this means s must be a linear function of proper time, s = aτ + b where a and b are constants. If s is not an affine parameter, the geodesic equation has the extra term you mentioned.

4. Feb 9, 2014

### Nabigh R

Thanks Bill, Wikipedia too says a similar thing, but what exactly is an affine parameter?