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When I was studying special relativity, I've learned that the relativistic action for a free particle is defined as $$ S = \lambda \int_{\tau_0}^{\tau_1} d\tau $$ Where $\lambda$ is a constant that is fixed a posteriori and $d\tau$ is the Lorentz invariant line element, or the proper time of the particle. It's defined this way since it should be invariant, so it will the equations of motion.

My question is: Doesn't it overkill the property of invariance? The action in Newtonian Mechanics is not invariant under Galilean transformations, but we use it to obtain Galilean invariant results. It would be possible to have other suitable actions for a free particle?

EDIT I made a mistake about saying that the action in Newtonian mechanics is not Galilean invariant. In fact, the Lagrangean is not, as explained in the answer below.

Ruffolo
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The usual non-relativistic action is Galilean invariant: $$ S=\frac m2\int v^2dt $$ After all, it is the non relativistic limit of a Lorentz invariant action. You can check it directly too. The subtlety is that the Lagrangian itself is not invariant. A Galilean transformation adds a boundary term which does not change the Euler Lagrange equations. Back to relativity, you can similarly look for Lorentz invariant actions without restricting to Lorentz invariant Lagrangians, but you will not find new solutions.

LPZ
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