Variational principle for a particle with varied mass I can't quite understand how the variational principle works. For a relativistic free particle we can find the extremum of the action
$$
S=-m\int ds
$$
and find the equations of motion. But what if the mass isn't constant and instead we have
$$
S=-\int m(s)ds.
$$
How does the variation work in this case?
 A: To apply the variational principle to the action you've written down, you need to consider the particle's spacetime position $x^\mu$ as a function of the parameter $s$ along the particle's path:
$$
S = - \int m \sqrt{ g_{\mu \nu}[x^\rho] \dot{x}^\mu \dot{x}^\nu} \, ds
$$
where we now view $x^\mu(s)$ as a function of the parameter $s$, with $\dot{x}^\mu$ as its first derivative.  We then apply the Euler-Lagrange equations to get a set of equations of motion for the world-line coordinate functions $x^\mu(s)$.  If we choose our parameter $s$ to be the proper time along the worldline, then these EOMs are equivalent to the geodesic equation.
With this in mind, it is not a big stretch to apply the Euler-Lagrange equations to a modified action of the form
$$
S = - \int m(s) \sqrt{ g_{\mu \nu}[x^\rho] \dot{x}^\mu \dot{x}^\nu} \, ds
$$
The process would be exactly the same:  apply the Euler-Lagrange equations to this action to get a set of equations of motion for $x^\mu(s)$.  The only difference is that this "action" now explicitly depends on $s$;  this is analogous to a classical particle Lagrangian that explicitly depends on $t$.
The resulting equations of motion will not have the same "nice" properties that the geodesic equations do, and (as noted in the comments) it may be difficult to give them a physical interpretation;  but you're free to write them out and try to solve them.
