Why Is the Norm of the Tangent Vector Constant in Geodesic Equations?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 3K views
Nabigh R
Messages
11
Reaction score
0
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?
 
Physics news on Phys.org
Nabigh R said:
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?

Well, [itex]s[/itex] 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

[itex]ds = \sqrt{|dx^\mu dx_\mu|}[/itex]

For this choice, [itex]|\frac{dx^\mu}{ds} \frac{dx_\mu}{ds}| = 1[/itex]
 
  • Like
Likes   Reactions: 1 person
Nabigh R said:
$$ \ddot{x}^\tau + \Gamma^\tau_{\mu \nu} \dot{x}^\mu \dot{x}^\nu = \frac{1}{2} \dot{x}^\tau \frac{d}{ds} \ln \left| \dot{x}_\nu \dot{x}^\nu \right| $$
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?
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.
 
Thanks Bill, Wikipedia too says a similar thing, but what exactly is an affine parameter?