Understanding Relativistic Ptcl Lagrangian: S1 vs. S2

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SUMMARY

The discussion focuses on the relationship between two relativistic particle actions, S1 and S2, defined as S1 = ∫(-1/2 mg_{μν}ẋ^μẋ^ν - U) dτ and S2 = ∫(-m√(g_{μν}ẋ^μẋ^ν) - U) dτ. Both actions yield the correct geodesic equations under the Euler-Lagrange framework, provided τ is affine. However, S1 is more versatile for null curves by treating mass as energy, while S2's derivation from the action S = -m∫ds fails for massless particles. The stress-energy tensor relations reveal that S1 and S2, despite having the same extrema, exhibit different behaviors in relation to the Einstein-Hilbert action.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with the Einstein-Hilbert action
  • Knowledge of stress-energy tensor concepts
  • Basic principles of general relativity
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  • Study the derivation of the Einstein-Hilbert action in detail
  • Explore the implications of mass-energy equivalence in relativistic mechanics
  • Investigate the properties of geodesics in curved spacetime
  • Learn about the role of the stress-energy tensor in general relativity
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The discussion is beneficial for theoretical physicists, graduate students in physics, and researchers focusing on general relativity and Lagrangian mechanics.

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.
 
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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