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I've been wrestling with this problem for quite a while now, and I can't seem to understand what I am allowed to do or not.

Let us consider the following action : $$S = \int \sqrt{-g_{\mu\nu}\frac{dx^{\nu}}{|ds|}\frac{dx^{\mu}}{|ds|}}|ds|$$

Now, I want to show that this gives the famous geodesic equation if we choose $|ds| = d\tau$, the proper time. I know how the proof goes with the variation of S, but I wanted to prove it with the Euler-Lagrange expression, and that is how I arrive to the root of my problem.

I know that if we choose the parameter as stated before, we must have that $$g_{\mu\nu}\frac{dx^{\nu}}{|ds|}\frac{dx^{\mu}}{|ds|} = -1.$$ Obviously, we can't replace that directly in the initial equation for S, otherwise we will have a constant lagrangian. My question is therefore the following : when can I effectively set $g_{\mu\nu}\frac{dx^{\nu}}{|ds|}\frac{dx^{\mu}}{|ds|}$ to be $-1$, without coming up with a wrong result ? Indeed, let me write the steps for the Euler-Lagrange equation :

$$\frac{d}{d\tau}\left(\frac{\partial}{\partial U^{\alpha}}(\sqrt{-g_{\mu\nu}U^{\nu}U^{\mu}}) \right) = \frac{\partial}{\partial x^\alpha}(\sqrt{-g_{\mu\nu}U^{\nu}U^{\mu}})$$

$$\frac{d}{d\tau}\left(\frac{g_{\alpha \nu}U^{\nu}}{\sqrt{M}} \right) = \frac{\partial_{\alpha}g_{\mu\nu}}{2\sqrt{M}}U^{\mu}U^{\nu}$$

With $$U^{\nu} = \frac{dx^{\nu}}{|ds|}$$ and $$M = -g_{\mu\nu}U^{\nu}U^{\mu}.$$

Now, in the above equation, if I set $M = 1$, then I do find the geodesic equation. Let me rephrase the question : why can I set $M = 1$ now? If I had set $M = 1$ before doing any of the $\frac{\partial}{\partial U^\nu}$ or $\frac{\partial}{\partial x^\nu}$ derivatives, I would have gotten a wrong result. Why can I set $M=1$ inside the $\frac{d}{|ds}$ derivative, but not inside the others? I hope I made myself clear!

Qmechanic
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Frotaur
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1 Answers1

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  1. Because the square root action is reparametrization invariant, the solutions to Euler-Lagrange (EL) equation are geodesics with arbitrary parametrization.

  2. As you already noted, it would be inconsistent to choose $$M~=~{\rm constant}$$ before performing the variation and before doing all the partial differentiations in the EL equation.

  3. By restricting to $$M~=~{\rm constant}$$ after the partial differentiations in the EL equation, you are restricting your parametrized geodesics solutions to only those which are affinely parametrized. (Note that the non-affinely parametrized geodesic equation has an additional term.)

  4. You are allowed to set $$M~=~{\rm constant}$$ before the final total parameter differentiation in the EL equation because this differentiation is along the very same curve rather than, say, a differentiation comparing neighboring curves in a variational process.

  5. See also my related Phys.SE answer here where all of this is explained in more details.

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Qmechanic
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  • Thank you for your answer, could you elaborate also on what do you mean by "reparametrization invariant" ? Also, I understand that this works only for affinely parametrized geodesics (since for them M=1). However, why can't I state that from the start, and put M=1 already inside the partial derivatives ? Is there any reason I can't "say" that we are using affine parametrization before the partial differentiation ? – Frotaur Oct 31 '16 at 22:04
  • This is explained in the linked answer. – Qmechanic Oct 31 '16 at 22:30
  • Thanks again for the precisions, however I still don't understand one last thing. I understand that setting $M=1$ before performing the variation is inconsistent. However, why is the differentiation $\frac{d}{d\tau}$ not considered a "variation" ? – Frotaur Oct 31 '16 at 22:32
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    The final total parameter differentiation is along the very same curve rather than, say, a differentiation comparing neighboring curves. – Qmechanic Oct 31 '16 at 22:45