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  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 $c=1$.


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

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)

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

where $v$ is the isntantaneous velocity of $\mathcal{P2$}$ measured by $\mathcal{P1}$.

Since $\tau = \int_{\tau _1}^{\tau _2} d\tau$ and $\tau ' = \int_{\tau _1}^{\tau _2} \sqrt{1 - v^2} d\tau$ and $\sqrt{1-v^2} \le 1$ , we conclude that $\tau > \tau '$.


I can use the same argument as above, by considering there are two infinitesimally separated events in the rest frame $K$ of $\mathcal{P1}$ such that the spacetime interval measured is $ds^2 = d\tau^2$ this interval measured by $K'$, the instantaneous rest-frame of $\mathcal{P2}$ gives the opposite relation

d\tau^2 = d\tau' ^2 - dx' ^2

giving me the opposite result. But this is inconsistent since all particles agree that $\mathcal{P2}$ is moving because he's accelerated. Where's my mistake. How to correct this inconsistence?

asked Feb 7, 2020 in Theoretical Physics by Lil' Gravity (0 points) [ no revision ]

1 Answer

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Consider a particle moving in an inertial system between two points. The proper time of the particle for its movement between the two points is lesser equal to the time observed in the inertial system for the particle's movement between the two points.

In your example, \(K\) is a possible choice of inertial system, and \(P_1\)is the observer. But it could be any other inertial system, and the observer need not move between the two points. (And you don't need the system \(S\) for your argument.)

With the roles of the particles reversed (I mean the second part of your argument), the trajectories of the particles are relevant, however. Again you show that the proper time of the particle, now \(P_1\), is shorter than the time in another system.

Your problem is:
In the first part of your argument, \(d\tau\) is the time, as measured in \(K\), it takes \(P_2\), not \(P_1\), to travel between the two points. In the second part of your argument, \(d\tau\) is the time, as measured in \(K\), it takes \(P_1\) to travel between the two points.

Thus in the two parts of the argument \(d\tau\) has different meanings, hence the seeming contradiction.

There is, if viewed from a different perspective, also a problem with your approach to the events A and B and the treatment of "infinitesimal" segments along the trajectories (vectors tangent to the trajectories). It appears you want both particles to meet at A and then to meet again at B.

In the first part of your argument, you consider a tangent vector in the direction of the proper time in \(K'\), whereas in the second part you consider a tangent vector in the direction of the proper time in \(K\). Now these two vectors cannot be parallel, unless \(dx'=0\), in which case, for this tangent vector or trajectory segment, \(d\tau=d\tau'\). Except for this special case the two vectors certainly are different. Expecting equality you find a contradiction.

answered Feb 7, 2020 by anonymous [ no revision ]

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