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I Relativistic point particle Lagrangian

  1. Jan 29, 2017 #1
    Can someone help me understand how the following two actions are related?
    [itex]S_1 = \int \left(-\dfrac{1}{2}mg_{\mu\nu}\dot{x}^\mu\dot{x}^\nu - U\right) d\tau[/itex]
    [itex]S_2 = \int \left(-m\sqrt{g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu} - U\right) d\tau[/itex]
    Both of them lead to the correct geodesic equation as the Euler-Lagrange equations, as long as τ is affine (S2 requires some additional assumptions to get rid of an additional factor that pops up), but I can't for the life of me explain the factor of 2 difference. I've seen both of them used in various places, but I can't seem to find any explanation of the difference between them. I also can't see any transformation (for a particle with fixed mass m under a potential U) that would show they're equivalent...

    S2 has a straightforward 'derivation' from the action [itex]S=-m\int{ds}[/itex] for proper time s, but it seems to breakdown for massless particles. S1 on the other hand can be made to easily work for null curves, by treating m as the particles energy rather than its rest mass, but doesn't have any derivation I'm aware of.

    *note* for more context, the problem I'm seeing is that these two actions are related very differently to the stress-energy tensor. I find that after solving for the stress energy tensor, [itex]S_1=-\frac{1}{2}T^\mu_\mu[/itex] but [itex]S_2 = -T[/itex]. That means that for S1, the Einstein-Hilbert action exactly cancels out the particle's, which is behavior I would expect (since [itex]2L_{EH}=R^\mu_\mu=T^\mu_\mu[/itex]). For S2 though, the two do not cancel out...
  2. jcsd
  3. Jan 30, 2017 #2


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    They have the same extrema. So from the EOM they are equivalent. You can compare this to e.g. a function f(x) and ln[f(x)], which also have the same extrema.
  4. Jan 30, 2017 #3
    So you're saying there's no difference? You can just square any piece of the Lagrangian and it won't change anything?? I understand that you could square the entire Lagrangian, but there are other parts here (e.g. the Einstein-Hilbert Lagrangian, the potential term, and terms for whatever else may be going on in the universe). And if they're truly equivalent, why would anyone ever use S2?? Surely S1 is much simpler to work with...
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