Relativistic mechanics: Does the action rely on a particular choice of the inertial frame?

Question

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?

Answer 1

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.