I Understanding Relativistic Ptcl Lagrangian: S1 vs. S2

michael879
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Can someone help me understand how the following two actions are related?
S_1 = \int \left(-\dfrac{1}{2}mg_{\mu\nu}\dot{x}^\mu\dot{x}^\nu - U\right) d\tau
S_2 = \int \left(-m\sqrt{g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu} - U\right) d\tau
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 S=-m\int{ds} 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, S_1=-\frac{1}{2}T^\mu_\mu but S_2 = -T. That means that for S1, the Einstein-Hilbert action exactly cancels out the particle's, which is behavior I would expect (since 2L_{EH}=R^\mu_\mu=T^\mu_\mu). For S2 though, the two do not cancel out...
 
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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.
 
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