I am trying to derive the geodesic equation using variational principle. My Lagrangian is $$ L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$ Using the Euler-Lagrange equation, I have got this. $$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

How do I prove the right hand side to be zero to get the geodesic equation?

Basically, I can rewrite the above mentioned equation as $$ \frac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{dt} \frac{dx^\lambda}{dt} = \frac{dx^\mu}{dt} \frac{d\ \log(L)}{dt}$$

I need to know why is the following true? $$\frac{d\ \log(L)}{dt} =0$$

I know that the derivation might be simpler using a different Lagrangian but I want to do it using this one.

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    $\begingroup$ Duplicate of physics.stackexchange.com/q/94348/50583 $\endgroup$ – ACuriousMind Feb 10 '16 at 16:02
  • $\begingroup$ @ACuriousMind I have seen the accepted answer to that question. I didn't quite understand the need to change the parameter. $\endgroup$ – Yogesh Yadav Feb 10 '16 at 16:08
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    $\begingroup$ $g_{rs}\dot x^r\dot x^s=\text{const.}$ if you pick a nice parameterization... $\endgroup$ – Ryan Unger Feb 10 '16 at 16:11
  • $\begingroup$ I have edited the question. $\endgroup$ – Yogesh Yadav Feb 10 '16 at 16:13

The geodesic equation for a general parameterization takes the form $$ \frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\rho\sigma} \frac{dx^\rho}{d\tau} \frac{dx^\sigma}{d\tau} = K(x(\tau))\frac{dx^\mu}{d\tau} $$ The RHS of the above is only zero when $\tau$ is chosen to be an affine parameter along the geodesic.

If you choose the affine parameter then $$ g_{\mu\nu}(x(\tau)) \frac{dx^\mu}{d\tau} \frac{ d x^\nu}{d\tau} = \varepsilon $$ where $\varepsilon$ is a constant. In this case, your equation takes the form required.

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  • $\begingroup$ small typo in the second equation: there should be no free index $\endgroup$ – Bort Feb 10 '16 at 16:45

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