We know that the Lagrangian of a relativistic particle is as follows: $$L = -mc^2\sqrt{1-(v/c)^2},$$ with the action being the integral of this Lagrangian with respect to time in the reference frame of the particle. My question is, does the action rely on a particular choice of the inertial frame?
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No, the action $S=-E_0\tau$ is minus the rest energy $E_0=m_0c^2$ times the lapse $\tau$ of proper time of the massive particle. Both factors $E_0$ and $\tau$ are invariants, so the action $S$ is independent of the choice of coordinate system. See also this related Phys.SE post.
Qmechanic
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I do not understand how τ is invariant. Surely the lorentz transforms state otherwise? – Rich Piana Oct 26 '20 at 20:02
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$\tau$ is invariant under Lorentz transformations. Can you see why? – Qmechanic Oct 27 '20 at 08:22