Deriving Geodesic Equation from Lagrangian

  • #1
Hi,

If I have a massive particle constrained to the surface of a Riemannian manifold (the metric tensor is positive definite) with kinetic energy $$T=\dfrac 12mg_{\mu\nu} \dfrac{\text dx^{\mu}}{\text dt} \dfrac{\text dx^{\nu}}{\text dt}$$ then I believe I should be able to derive the geodesic equations for this manifold by applying the Euler-Lagrange equations to the Lagrangian $$L:=g_{\mu \nu}\dfrac{\text dx^{\mu}}{\text dt} \dfrac{\text dx^{\nu}}{\text dt}.$$ However, when I go to do this, here's what I find: $$\dfrac{\partial L}{\partial x^{\sigma}} = \dfrac{\partial g_{\mu\nu}}{\partial x^{\sigma}} \dfrac{\text dx^{\mu}}{\text dt}\dfrac{\text dx^{\nu}}{\text dt}.$$ Moreover, $$\dfrac{\text d}{\text dt}\left(\dfrac{\partial L}{\partial (\text dx^{\sigma}/\text dt)}\right)=\dfrac{\text d}{\text dt}\left(2g_{\sigma\mu} \dfrac{\text dx^{\mu}}{\text dt}\right)=2g_{\sigma\mu} \dfrac{\text d^2x^{\mu}}{\text dt^2}.$$ Setting these expressions equal and multiplying by the inverse metric, I obtain $$\dfrac{\text d^2x^{\tau}}{\text dt^2} - \dfrac 12 g^{\tau\sigma}\dfrac{\partial g_{\mu\nu}}{\partial x^{\sigma}} \dfrac{\text dx^{\mu}}{\text dt} \dfrac{\text dx^{\nu}}{\text dt} = 0.$$ This looks similar to the geodesic equation, but something is off about the "Christoffel Symbols" of this equation.

What's wrong with my derivation? Any help is appreciated. Thanks.
 

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  • #2
stevendaryl
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You're using ##t## for proper time, not coordinate time, right?

If you take the derivative of ##2g_{\sigma \mu} \frac{dx^\mu}{dt}## you get ##2 \frac{d g_{\sigma \mu}}{dt} \frac{dx^\mu}{dt} + 2 g_{\sigma \mu} \frac{d^2 x^\mu}{dt^2}##. I know that ##g_{\sigma \mu}## is not explicitly a function of ##t##, but it is a function of ##x^\lambda##. You use the chain rule (or whatever it's called):

##\frac{d g_{\sigma \mu}}{dt} = \frac{\partial g_{\sigma \mu}}{\partial x^\lambda} \frac{dx^\lambda}{dt}##
 
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  • #3
Orodruin
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Why is this in the relativity forum? It is purely classical mechanics.

You're using ##t## for proper time, not coordinate time, right?
It is a massive particle constrained to move freely on a manifold. Think spherical pendulum with g being the metric on the sphere. It has nothing to do with proper time.
 
  • #4
Yes, this is purely classical mechanics. I posted in here because I was going to have some follow up questions regarding relativity, but I've resolved them now. Thanks.
 
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Orodruin
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Yes, this is purely classical mechanics. I posted in here because I was going to have some follow up questions regarding relativity, but I've resolved them now. Thanks.
I would suggest you do not do things this way. You managed to confuse one poster (who anyway was able to help you, but it could have been worse). Instead, I suggest you post questions where they belong and if you have follow-ups or spin-offs more suited for a different part of the forum they can be posted in a new thread.
 
  • #7
stevendaryl
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Why is this in the relativity forum? It is purely classical mechanics.



It is a massive particle constrained to move freely on a manifold. Think spherical pendulum with g being the metric on the sphere. It has nothing to do with proper time.
Well, without specifying what the metric is, it works just as well for classical or relativistic physics.
 
  • #8
Orodruin
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Well, without specifying what the metric is, it works just as well for classical or relativistic physics.
Of course, it is just the derivation of the geodesic equations, I am not arguing that. Just saying that it may be confusing to the OP to start talking about proper time for a classical mechanics problem (g is specified to be positive definite in the OP and it explicitly talks about a particle constrained to a manifold).
 

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