Lagrangian formulation of free massive point particle in special relativity

Question

I wonder if there is a way to reproduce the 4-force generalization for Newton's equation for a free particle i.e.

$$ m\frac{d^2x^\mu}{d \tau^2} = 0, \qquad \text{ for } \, \mu =0,1,2,3, \tag{1} $$

with just calculus of variations? If I start out with the action

$$ S = -mc^2\int\frac{dt}{\gamma}\tag{2} $$

with

$$ L =- \frac{mc^2}{\gamma},\qquad \gamma:= \frac{1}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\tag{3} $$

and put this in Euler-Lagrange equation I obtain

$$ \frac{d}{dt}\left( \frac{\partial L}{\partial v^i} \right) - \left(\frac{\partial L}{\partial x^i}\right) = 0 $$ $$\qquad\Leftrightarrow\qquad \frac{d}{dt}\left( \gamma m v^i\right)= \frac{d}{dt}p^i= 0 \qquad \text{ for } \, i =1,2,3. \tag{4} $$

I believe this is done right? But what about the case

$$ m\frac{d^2x^0}{d \tau^2} = 0,\tag{5} $$

i.e. the case for the temporal index $\mu = 0$?

Answer 1

The resolution to OP's question seems to be that OP's action (2) and Lagrangian (3) correspond to the static gauge $$t~\equiv~x^0~=~\tau \tag{A}$$ of a generally covariant action principle$^1$

$$S[x]~=~\int\! d\tau ~L, \qquad L~:=~ -m\sqrt{-\dot{x}^2}, $$ $$ \dot{x}^2~:=~g_{\mu\nu}~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\tau}, \tag{B}$$

where $\tau$ is an arbitrary world-line parameter (which does not have to be the proper time). See also e.g. this, this, and this Phys.SE posts.

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$^1$ In this answer we work in units where the speed-of-light $c=1$ is one, and we use the Minkowski sign convention $(−,+,+,+)$.