1. Jan 9, 2008

### Wannabeagenius

Hi All,

I was recently trying to understand the Special Theory of Relativity and my understanding of the resolution of the twin paradox is that the person who has accelerated away from the observer and accelerates again to join the observer is the person that grows older. I understand the reasoning behind this.

However, what if I accelerated with respect to a stationary observer and then, after several years, the stationary observer accelerated to get back to me. Or vice-versa.

What is the explanation under those circumstances?

Bob Guercio

2. Jan 9, 2008

### Fredrik

Staff Emeritus
If you accelerate the same way, you will be the same age when you meet, regardless of when you started the acceleration.

3. Jan 9, 2008

### JesseM

When do you accelerate? If you are initially at the same position and then you accelerate briefly to build up a velocity relative to the other observer, but after that you move inertially and it's the other observer who accelerates to return to you later, then your initial acceleration doesn't affect the results much (or at all, if your initial acceleration is treated as instantaneously brief)--it's the one who accelerates once a significant distance has built up between you that mainly determines which one will be younger when the two of you reunite.

4. Jan 9, 2008

### Ich

No offence, but I doubt that you understand the reasoning, because there is none. The argument concerning acceleration is given merely to point out that the situation is not symmetric. Acceleration does not make clocks run slower.
That's indeed the best counterexample to the widespread notion that acceleration somehow causes time dilation.
If you draw the example in a space-time-diagram, you see immediately that it is the usual twin paradox, but this time as seen in the frame where the outbound twin is at rest. The "stationary" observer is the "moving" one in this case, therefore he ages less.
The only explanation I know of is the geometrical one: The elapsed time is the distance between two events, calculated sqrt( (t2-t1)²-(x2-x1)²), where t and x are the coordinates of the events in an inertial frame.

5. Jan 9, 2008

### RandallB

Bob
The reason your are getting confusing answers is you have not detailed your question very well. I think I understand what you really mean.
First ignore the first “acceleration” as acceleration does not matter at all when working SR problems.
We just have YOU departing earth at 0.5c

Now after some considerable period of time you want the on earth to accelerate in pursuit to catch up with you.
This gives us a problem as we can hardly call your stationary if they are to be accelerated.
Therefore from now on lets call your “stationary observer” the OBSERVER.

We will ignore the acceleration as well by bring it up to speed in a time interval insignificantly small compared the travel times being used. (even GR gravitational time affects are zero where high accelerations are in effect instant due to short time of application no matter how high the "Equivalent Gravity")
Also, the OBSERVER cannot catch up to YOU by following at the same speed so 0.5c relative to Earth won’t do. We send the OBSERVER at twice the speed or 0.8c.
(Note: If you do not already know that 0.5c plus 0.5c = 0.8c spend some time in a chapter or site on relativistic speed addition. If you cannot do relativistic speed addition you cannot do SR twin examples).
The OBSERVER and YOU had seen the separation speed as 0.5c and now both of you will see the closing speed as 0.5c (do the math).

So just how did YOU get back together with the OBSERVER? Certainly not by changing speed WRT earth, meaning YOU are the one that is stationary not the OBSERVER in your new version of this problem.
So the answers is as the one that remained in the stationary reference frame YOU will have aged much more upon reconnecting with the much younger OBSERVER.

Last edited: Jan 9, 2008
6. Jan 9, 2008

### Fredrik

Staff Emeritus
This thread needs space-time diagrams. Too bad I'm too lazy to draw one.

It also needs more input from the OP. Bob, do you feel that your question has been answered?