Sorry for making a new thread on what I've posted elsewhere, but it seems all the relativity experts have ignored my question.


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?

The time on S' will be less than the time on S.

Provided that the S frame remains inertial. That needs to be specified.

This thread was initially meant to go into a more complex discussion but the crash of sciforums prevented. You may lock this thread.

No. Please tell us what you have in mind. It might be interesting. Although the original question wasn't a very good one. I suppose it was asked in order to get definite statements from users.

No. Please tell us what you have in mind. It might be interesting. Although the original question wasn't a very good one. I suppose it was asked in order to get definite statements from users.
I do not think any such discussion will be productive. However if you wish, the following discussion would be much more interesting:

I assume you are familiar with the muon experiment and that length contraction is used to explain the muons reaching the earth surface when created in the upper atmosphere. Can you answer the following question: What is the length of the known universe according to a muon moving at .998c through the atmosphere? You can assume the atmosphere to be a vaccum (The result wouldn't be too much different). I'll also give you that the length of the known universe in the Earth frame is almost 156 billion light years.

If you're here, I'll point out that I've answered this in the MacM AR thread, where you originally asked the question to me.