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 $\mathcal{P2}$?
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$\begingroup$ Related post by OP: physics.stackexchange.com/q/529574/2451 $\endgroup$– Qmechanic ♦Commented Feb 9, 2020 at 11:11
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2 Answers
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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 $B$:
$t_{A-A_1}' = \gamma_1 (t_1 - v_1 \Delta x_1)$
$t_{A_1-A_2}' = \gamma_2 (t_2 - v_2 \Delta x_2)$
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$t_{A_{n-1}-B}' = \gamma_n (t_n - v_n \Delta x_n)$
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.
answered Feb 8, 2020 at 21:55
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$\begingroup$ I understood your approach, I like it. But can I use the same argument of plugging $N$ inertial frames to the opposite case, i.e, if I discretize the time measured by $non-accelerated \ particle$ by $N$ inertial frames? $\endgroup$ Commented Feb 8, 2020 at 22:17
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$\begingroup$ I think I can't because if i do that, i obtain the opposite andswer. Can you argue why I can't use the opposite analysis? $\endgroup$ Commented Feb 8, 2020 at 22:17
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1$\begingroup$ Any inertial frame is equivalent in special relativity. But if one is chosen, all the maths must be calculated in that frame. $\endgroup$ Commented Feb 8, 2020 at 22:24
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$\begingroup$ So your approach is correct, because all the analysis are are made on the same frame, the rest frame of the free-particle. So is my attempt wrong because I'm doing the analysis from the point of view of many different inertial reference frames? $\endgroup$ Commented Feb 8, 2020 at 22:43
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1$\begingroup$ Yes. It is like the twin paradox, that uses 2 steps. You can choose the earth frame, the traveller twin going away frame, or the traveller twin coming back frame. But there is not a "traveller twin frame", because there are two of them, completely different. $\endgroup$ Commented Feb 8, 2020 at 23:50
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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.
