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

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Discussion Overview

The discussion revolves around the derivation of the geodesic equation and the conditions under which the norm of the tangent vector remains constant. Participants explore the implications of different parameterizations and the concept of affine parameters in the context of geodesics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant derives the geodesic equation and notes that the right-hand side vanishes if the norm of the tangent vector, ## \dot{x}_\nu \dot{x}^\nu ##, is constant, questioning why it is constant.
  • Another participant suggests that since the parameter ## s ## is arbitrary, one can choose a specific parameterization for slower-than-light geodesics, leading to the condition that the norm equals one.
  • A third participant emphasizes that the geodesic equation takes its standard form only when ## s ## is an affine parameter, indicating that for timelike geodesics, ## s ## must be a linear function of proper time.
  • A later reply asks for clarification on the concept of an affine parameter, indicating a need for further explanation of this term.

Areas of Agreement / Disagreement

Participants express varying views on the nature of the parameterization and the definition of affine parameters, indicating that there is no consensus on the implications of these concepts for the geodesic equation.

Contextual Notes

The discussion includes assumptions about parameter choices and the nature of geodesics, which may not be fully resolved or defined, particularly regarding the implications of using affine versus non-affine parameters.

Nabigh R
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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?
 
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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, 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
 
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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?
 

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