Maximum proper time in Minkowski Spacetime for free particles

Question

Consider two events $\mathcal{A}$ and $\mathcal{B}$ corresponding to the beggining and the ending of trajectories of two massive particles, respectively. The particle named $\mathcal{P1}$ is in free motion, and the other particle $\mathcal{P2}$ is in accelerated motion. Both particles measure events $\mathcal{A}$ and $\mathcal{B}$ as events that occurs at same place, in both rest-frames, so both particles also measure their respective proper times elapsed between these events. How can I prove that the proper time of the free particle $\mathcal{P1}$ is bigger than proper time of the accelerated particle $\mathcal{P2}$?

Answer 1

If we approximate the accelerated path from $A$ to $B$, by $n$ "inertial" steps:
$A$ to $A_1$, $A_1$ to $A_2$,..., $A_{n-1}$ to $B$:

$t_{A-A_1}' = \gamma_1 (t_1 - v_1 \Delta x_1)$
$t_{A_1-A_2}' = \gamma_2 (t_2 - v_2 \Delta x_2)$
.
.
.
$t_{A_{n-1}-B}' = \gamma_n (t_n - v_n \Delta x_n)$

where $t_k$ is the time from $A_{k-1}$ to $A_{k}$ measured by the inertial frame, $\Delta x_k$ is the coordinate difference $x_{k} - x_{k-1}$ , measured also by the inertial frame. And $v_k$ is the speed of each step.

Adding the times:

$t_{A-B}' = \Sigma \gamma_k t_k - \Sigma \gamma_k v_k \Delta x_k$

But: $\Delta x_k = v_kt_k$

$t_{A-B}' = \Sigma \gamma_k t_k - \Sigma \gamma_k v_k^2 t_k = \Sigma \gamma_k t_k (1 - v_k^2) = \Sigma \frac {t_k}{\gamma_k}$

As $t_{A-B} = \Sigma t_k\implies t_{A-B} > \Sigma \frac {t_k}{\gamma_k}$

$t_{A-B} > t_{A-B}'$

The accelerated path is the limit when the time of each step go to zero and the number of steps go to infinity.

Answer 2

It is quite obvious from action for massive particle: $$ S = -m c^2 \int d\tau $$ where $\tau$ is proper time of particle.

Note, that due to minus sign in action, particle move on trajectory with maximal proper time, and this trajectory is free motion, as consequence of absence of force. So any other trajectory will have bigger value of action and smaller proper time.