# Prooving that free-particles measure larger proper time than accelerated oned in Minkowski Spacetime

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Let $S$ be an Inertial Frame. Consider two timelike separated events in S: $\mathcal{A}$ and $\mathcal{B}$. Consider two massive moving particles in $S$: $\mathcal{P1}$ and $\mathcal{P2}$ wich both have trajectories beggining at $\mathcal{A}$ and ending in $\mathcal{B}$. $\mathcal{P1}$  have a non-accelerated motion, with a _constant_ 3-velocity  relative to $S$. $\mathcal{P2}$ have an accelerated trajectory with instantaneous 3-velocity relative to $S$. Inertial Observers can disagree if particle $\mathcal{P1}$  is moving or not, but all inertial observers will agree that particle $\mathcal{P1}$  is in movement.

Consider that the clocks moving with the particles and the clock o $S$ starts at 0, and the particles start to move in theis respective trajectories. Since $\mathcal{P1}$  measures a proper time $\tau$ and $\mathcal{P1}$  measure another proper time $\tau '$ between $\mathcal{A}$  and $\mathcal{B}$, I'm trying to prove that all observers agree that $\tau > \tau '$. I'm using the metric signature $(+,-,-,-)$. Also I'll assume one-dimensional spacial motion without loss of generality and using $c=1$.

_ATTEMPT OF A PROOF:_

Let  $K$ be the rest-frame of $\mathcal{P1}$ an $K'$ be some instantaneous rest-frame of $\mathcal{P1}$

Consider two events in $K'$ infinitesimally separeted such that the spacetime interval in $K'$ is given by $ds^2 = d\tau ' ^2$. These infinitesimally separated events in $K$ have spacetime interval given by $ds^2 = d\tau' ^2 = d\tau^2 - dx^2$, where $dx$ is the spacial displacement of those events measured by $K$. Then we have that

\begin{align}
d\tau ' ^2 &= d\tau ^2 - dx^2 \\ &= d\tau^2\left[ 1 - \left( \frac{dx}{d\tau}\right)^2\right] \\ &= d\tau ^2 (1 - v^2)
\end{align}

\begin{equation}
\Longrightarrow d\tau' = d\tau \sqrt{1 - v^2} \Longrightarrow d\tau' < d\tau
\end{equation}