Here is the system of interest, S is considered at rest and S' is moving wrt to S as shown:

...a............v.............a

.------>------------------><------.

S................................S'

The magnitude of constant acceleration is given by a, and the magnitude of constant velocity is given by v. S' undergoes constant acceleration from the rest position in the frame S. At the end of the journey, the S' frame is once again at rest in the S frame. Will the time shown on the clock of S' be less than, greater than, or equal to the time shown on the clock of S when S' finishes it's journey. If it cannot be determined, then why not?

EDIT: (This is a clarification of the diagram as described in a post below)

S is an inertial reference frame, all accelerations/velocities of S' are measured wrt to S. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has reached a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, -a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0.

...a............v.............a

.------>------------------><------.

S................................S'

The magnitude of constant acceleration is given by a, and the magnitude of constant velocity is given by v. S' undergoes constant acceleration from the rest position in the frame S. At the end of the journey, the S' frame is once again at rest in the S frame. Will the time shown on the clock of S' be less than, greater than, or equal to the time shown on the clock of S when S' finishes it's journey. If it cannot be determined, then why not?

EDIT: (This is a clarification of the diagram as described in a post below)

S is an inertial reference frame, all accelerations/velocities of S' are measured wrt to S. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has reached a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, -a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0.

Last edited: