Confused about relativistic point particle action in MTW In MTW page 179 exercise 7.2 they give the following action for an particle in an electromagnetic potential.
$$I = -\frac{1}{16 \pi} \int F_{\mu \nu} F^{\mu \nu} d^4x + \frac{1}{2}m \int \frac{dz^{\mu}}{d\tau}\frac{dz_{\mu}}{d\tau}\,d\tau + e \int \frac{dz^{\mu}}{d\tau} A_{\mu}(z)\,d\tau. \tag{7.5}$$
However, I am puzzled by the middle term.
$$\frac{1}{2}m \int \frac{dz^{\mu}}{d\tau}\frac{dz_{\mu}}{d\tau}\,d\tau.$$
Wouldn't the relativistic term here would be roughly,
$$mc^2 \int d\tau~?$$
This looks like a non-relativistic term, which seems strange given the use of four-vectors here.
Can someone clear up this confusion for me?
 A: The action you have in your question is for a relativistic massive charged particle in an EM field, and the term $$\frac{1}{2}m \int \frac{dz^{\mu}}{d\tau}\frac{dz_{\mu}}{d\tau}d\tau$$ is the kinetic energy of the action. But in your last comment above, you state that the particle is free. The action you also refer to $m\int d \tau$ is that for a free relativistic particle in a curved spacetime.
How you get that term is by noting that the standard Lagrangian for  such (an on-shell) particle is $$\mathcal{L}(\dot{z}) = -mc\sqrt{g_{\mu\nu}\dot{z}^\mu \dot{z}^\nu}$$ for a particle that has a worldline described by $z(λ)$, parametrized by an affine parameter $\lambda$ with a spacetime metric $g_{\mu\nu}$.
If $\lambda=\tau$  (the proper time), then $\sqrt{g_{\mu\nu}\dot{z}^\mu \dot{z}^\nu} = \sqrt{c^2}=c$ which means the Lagrangian reduces to $$\mathcal{L}(\dot{z}) = -mc^2$$ which is a constant. As you seem to be working in units where $c=1$ this implies the action is given by $$I=\int\mathcal{L}(\dot{z})\ d\tau=-m\int d\tau$$
A: TL;DR: OP's has a point. It is inconsistent to use proper time $\tau$ as a world-line parameter before performing the variation in the stationary action because not all virtual paths are parametrized by their proper time. Only after the variation has been performed, and the Euler-Lagrange (EL) equations obtained, it is consistent to use proper time $\tau$ (of the solution curve) as a world-line parameter, cf. e.g. my Phys.SE answer here.
More details: In this answer we focus on the middle term in eq. (7.5) since the first & last terms are only spectators to OP's question. MTW explains on the previous page 178 that $\lambda$ denotes an arbitrary world-line (WL) parameter, while $\tau$ denotes proper time.
More generally, the "master" action for a relativistic point particle is
$$\begin{align}I[z,e]~=~&\int_{\lambda_i}^{\lambda_f} \! d\lambda~L,\cr
L~=~&\frac{\dot{z}^2}{2e}-\frac{e(mc)^2}{2}, \cr 
\dot{z}^2~:=~&g_{\mu\nu}(z)~ \dot{z}^{\mu}\dot{z}^{\nu}~<~0,  \cr 
\dot{z}^{\mu}~:=~& \frac{dz^{\mu}}{d\lambda},\end{align}\tag{A} $$
where $e(\lambda)>0$ is an einbein field, cf. e.g. this related Phys.SE post. This action (A) has WL parametrization invariance.
The usual square-root action [that OP implicitly mentions in their last expression] can be recovered by integrating out the einbein field $e$, cf. e.g. this Phys.SE post.
If we instead fix a gauge
$$e(\lambda)~=~e_0~:=~\text{some constant},\tag{B}$$
then the action (A) becomes proportional to MTW's middle term (7.5) [up to an irrelevant constant term $-\frac{e_0(mc)^2}{2}$ in the Lagrangian]. After the variation, one may relate the WL parameter
$$\lambda~=~\frac{\tau}{me_0} + \text{constant}\tag{C}$$
to proper time.
