Relativistic action is a constant?

Question

Say that you want to find the equations of motion of a free relativistic massive point particle by minimizing the action

$$S=-mintmathrm{d}tau,sqrt{eta_{munu}frac{mathrm{d}x^mu}{mathrm{d}tau}frac{mathrm{d}x^nu}{mathrm{d}tau}}.tag{1} $$

But I'm very confused, because it seems to me that $$ sqrt{eta_{munu}frac{mathrm{d}x^mu}{mathrm{d}tau}frac{mathrm{d}x^nu}{mathrm{d}tau}}=sqrt{u_nu u^nu}=c,tag{2}$$

where $u_nu$ is the 4-velocity, so the Lagrangian is just a constant and all the derivatives are 0. There must be a giant fault in my reasoning but I just can't see it.

Qmechanic · Accepted Answer

Well, this is a case of mistaken identity. OP is probably thinking that the letter $tau$ stands for proper time, but in the off-shell action $S[x]$ the integration variable $tau$ is just a parameter that parametrizes the world-line (WL) of the massive point-particle. It would probably have been more pedagogical to use another letter for the integration variable, say $lambda$. See also this & this related Phys.SE posts.

Although the off-shell action $S[x]$ is actually reparametrization invariant, it is not possible to choose the integration parameter to be proper time for all virtual paths as it would be incompatible with the boundary conditions of the variational problem, cf. e.g. my Phys.SE answers here & here. But we can of course choose the classical path itself to be parametrized by proper time if we want to.

Relativistic action is a constant?

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