Can someone help me understand how the following two actions are related?(adsbygoogle = window.adsbygoogle || []).push({});

[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 (S_{2}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...

S_{2}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. S_{1}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 S_{1}, 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 S_{2}though, the two do not cancel out...

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# I Relativistic point particle Lagrangian

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