The action of a relativistic free particle is
$$\mathcal{S}=\int^{t_{1}}_{t_{0}} L dt\tag{1},$$
for
$$L=-\frac{mc^{2}}{\gamma}.\tag{2}$$
I understand that a particle will follow the trajectory of stationary action. I've read that from this Lagrangian, one can deduce that the trajectory of a particle is one which minimises the Minkowski distance (i.e. a geodesic). I'm not entirely sure how we can deduce that here. Classically I can see this, since for a free particle, with velocity in the $x$-direction, we can say that $L=\frac{1}{2}m\dot{x}^{2}$. Therefore
$$\begin{split}
\mathcal{S} &=\int^{t_{1}}_{t_{0}}\frac{1}{2}m\dot{x}^{2}dt \\ &=\int^{x_{1}}_{x_{0}}\frac{1}{2}m(\frac{dx}{dt})^{2}\frac{dt}{dx}dx \\&=\int^{x_{1}}_{x_{0}}\frac{1}{2}mvdx \\&=\frac{1}{2}mv(x_{1}-x_{0})
\end{split} \tag{3}$$
Hence, classically, minimising the distance travelled, $x_{1}-x_{0}$, will minimise the action. Unfortunately, it's not clear to me why relativistically a particle will travel the minimum minkowski distance. Also, it's not clear to me whether or not the Minkowski distance is minimised for the trajectory of a general particle, or just that of a free particle. Any help is really appreciated 🙂
Best Answer
It is straightforward to derive that OP's action (1) is $$S~=~ - m_0c ~ \Delta s, $$ where $$\Delta s~=~c\Delta\tau $$ is the spacetime distance $$\Delta s ~=~\int \!ds$$ in the $(+,-,-,-)$ sign convention, $$(ds)^2~=~g_{\mu\nu}dx^{\mu} dx^{\nu},$$ cf. e.g. this Phys.SE post.
Therefore a minimum for the action $S$ is the same as a maximum for the spacetime distance $\Delta s$.
The fact that the straight line of a free particle maximizes (rather than minimizes) the spacetime distance $\Delta s$ follows from the reversal of the triangle inequality in Minkowski spacetime.