# Twin Paradox with no asymmetry

1. Aug 18, 2011

### joeh1971

Here is another variation of the twin paradox.
Suppose we let both twins start their journey from a space station far from any heavenly bodies, so that the whole experiment can be carried out in free space. The twins Jack and John are equipped with identical "twin" shuttles and "twin" clocks synchronized at the departure point O.

The twins travel in opposite directions along the same straight line. Other than the direction, their accelerating and cruising processes are pre-programmed to be identical as measured by their own speedometers, clocks and accelerometers. Thus, they start their journeys with the same accelerations, turn around with the same acceleration and come back to the space station with the same velocity and decelerations. What is the best way to explain that the clocks carried by the twins read the same time when they meet back at the station to compare clock readings or equivalently, that the age of the twins is identical? Shouldn't either twin claim that the other is younger?
I know the resolution of the paradox depends on the fact that there is no asymmetry in this case but I am trying to understand the details of how both end up with the same clock readings since one twin sees the other twin's clock slowing down and vice versa.

2. Aug 18, 2011

### Superstring

Re: Twin "Paradox" with no asymmetry

Why do you assume that?

3. Aug 18, 2011

### PAllen

Re: Twin "Paradox" with no asymmetry

The easiest point to make is to ignore interpretation of what you see with light delays, instead just describe what each literally sees. Then each twin sees the other clock run slow for half the trip and fast for half the trips (pure doppler), such that the result is the same as their own fixed rate clock.

4. Aug 18, 2011

### ghwellsjr

Re: Twin "Paradox" with no asymmetry

During the outbound portion of the trip, both twins will see the other twin's clock as running slow. As soon as they both turn around, they will see the other one's clock immediately run normally. Then near the end of the trip, they will each see the other one's clock run fast. And when they arrive back at the station, their clocks will have the same time on them.

The difference between this and the normal twin paradox is that the images of the others' clocks do not change rates for the same percentages of the trip (and there's no normal interval) and so they end up with different times when they meet.

Last edited: Aug 18, 2011
5. Aug 18, 2011

### Staff: Mentor

Re: Twin "Paradox" with no asymmetry

You don't need to say anything other than "by symmetry". Whatever procedure A uses to calculate B's age B can use to calculate A's age. The details don't matter unless you want to actually calculate the age.

6. Aug 18, 2011

### joeh1971

Re: Twin "Paradox" with no asymmetry

This conclusion is based solely on the time dilation effect. Each twin sees the other moving away with a certain speed and then coming back. So if you have twins A and B. Twin A sees twin B's clock slow down relative to his own clock (i.e. relative to A's clock). Twin B sees twin A's clock slow down relative to his own clock (relative to B's clock). So based only on time dilation, there would be a paradox when they meet.

However, there is another effect. You still have to consider the fact that twin's clocks are separated from each other and so there is a speeding up factor that needs to be considered when the clocks turn around to account for the clocks changing inertial frames (going away frame vs coming back frame). I would llike to show mathematically that both clocks end up with the same time when I calculate that time from Twin A's perspective and then from Twin B's perspective.

7. Aug 18, 2011

### Staff: Mentor

Re: Twin "Paradox" with no asymmetry

The time dilation formula only applies to inertial frames. Neither twin is inertial, so you cannot use it in either twin's frame. It is not applicable to this scenario.

The best way to do this is to pick any arbitrary inertial frame so that it does apply.

8. Aug 18, 2011

### ghwellsjr

Like the one in which the space station is stationary, in which case both twins will experience exactly the same time dilation.

9. Aug 18, 2011

### joeh1971

Re: Twin "Paradox" with no asymmetry

The regular twin paradox is usually solved from the accelerating twins perspective by thinking of the outgoing twin as just switching from one inertial frame into another inertial frame. This introduces a speeding up correction factor for the the clock that remains on earth. The earth twin's clock of course never switches inertial frames and so he doesn't need to add this correction factor to the traveling twin's clock time. The end result is that when the times on each clock are calculated from the earth's perspective and then from the traveling twin's perspective, they both come up with the same answers for both clocks. They both find that more time passes by on the earth twin's clock than on the traveling twin's clock. No Paradox!

However, I am having trouble getting the correct answer when the twin's are moving symmetrically, and both twins change from one inertial frame(outgoing) into a different inertial frame (returning).

Simply saying by symmetry although is correct is not a very satisfying answer.

10. Aug 18, 2011

### Staff: Mentor

Re: Twin "Paradox" with no asymmetry

I have never liked that approach. Dolby and Gull wrote an approach I much prefer.
http://arxiv.org/abs/gr-qc/0104077

I also like these two

Last edited: Aug 18, 2011
11. Aug 18, 2011

### ghwellsjr

Re: Twin "Paradox" with no asymmetry

If you think you have figured out the Twin Paradox by having the traveling twin switch frames at the turn-around point and now you can't figure out why a similar approach doesn't work in this symmetrical scenario, that just points out the fallacy of using multiple frames in the analysis of a scenario. Just use one frame, the easiest one, and you will have no problem with either the Twin Paradox or this Symmetrical Scenario.

Here's the problem when you switch frames: An observer cannot know where the other remote observer is (unless they have a prearranged agreement in which case they don't need to take the trip, they can just calculate the scenario on paper). So when the twins turn around and you know that the other twin is also turning around then you overlook all that time that each twin actually sees each twin as stationary with respect to themself (no Doppler Shift).

That's why I prefer the analysis that depends on what each twin actually sees of the other twin using Relativistic Doppler. That always works without any artificial tweaking of the clocks.

Or just use the space station's rest frame.