# Dumb twin paradox question

1. Feb 7, 2012

### Karl Coryat

[later edit: Sorry for the title -- I am glad to see this was interesting enough to spark discussion.]

Twin A takes off and leaves Twin B behind. Rather than switching on the reverse-thrust, slowing down, and beginning the journey back home (an acceleration that would be distinctly detectable onboard), Twin A has a close encounter with a star, which gravitationally slingshots the ship around the horn and back toward Twin B.

Of course, this acceleration breaks the symmetry of the situation for the twins. But, now Twin A's ship is undergoing a free-fall, weightless acceleration during the turn-around. As I understand it, if the ship were sealed shut, Twin A and his clocks and instruments could just as well think they were in an inertial frame for the entire trip -- only to find themselves back where they started, Twin B having aged more. How do the onboard clocks "know" that they are undergoing an acceleration or are in a gravitational well during the turn-around, so as to tick more slowly than if the star weren't there at all?

I'm sure this has a simple solution, but it isn't coming to me. Thanks for your help.

Last edited: Feb 8, 2012
2. Feb 7, 2012

### ghwellsjr

It's no different than any other "free-fall" scenario such as orbiting the earth which involves a combination of the time-dilation of general relativity and the time-dilation of special relativity due to speed. We usually ignore gravity when explaining the twin paradox because it just complicates things.

3. Feb 7, 2012

### ModusPwnd

The solution is to use the time dilation formula, like usual. Twin A's ship is not in an inertial frame for the entire trip, so you need to integrate up the dilation formula.

4. Feb 7, 2012

### PAllen

In general relativity, you can no longer talk about global inertial frames, with symmetric relationship with each other. It is still true 'sufficiently locally'. In your example, which is not a dumb question at all, you could, indeed, set things up so you have two inertial paths = two geodesic paths between start and stop events, that have different 'proper time' measure along them. There is no way to correctly treat this example from a pure special relativity point of view - it is strictly a general relativity phenomenon as described. In special relativity it is simply impossible for two inertial paths in spacetime to have two intersections; it GR it is possible. When it happens, generally the paths will have different proper time. In your case, assuming only one main gravitating mass is involved, the close flyby path will have shorter proper time.

Further, to correctly compute this, you could not use SR formula. You would find the two geodesics and integrate the proper time expression (metric contracted with path tangent vector twice), based on a given geometry (e.g. Schwarzschild geometry).

[Edit: to respond to an earlier comment, A could easily be inertial the whole time. Let B be on a free floating space station, A accelerates before reaching B, and is inertial as of when A and B synch their clocks; both remain inertial the whole time until A meets B again].

5. Feb 7, 2012

### Karl Coryat

Thank you for the thoughtful responses. I thought I was missing something obvious.

So if I understand it correctly, even though the ship is in free-fall, is following a geodesic, and no acceleration is felt locally, the onboard clocks do "feel" the gravitation in the sense that the curvature of space inexorably alters the proper time that's ticked off by the clock. Is that right?

6. Feb 7, 2012

### PAllen

Pretty much. To a certain approximation, in simple cases in weak gravity, as noted in an earlier post, you could separate this into effect of relative speed and effect of gravitational time dilation. However, gravity is responsible for both effects being able to contribute to a twin paradox (in this set up). It contributes gravitational time dilation plus it curves the geodesic so a meet up is possible, and only because of this can you have the relative sped contribute to a twin paradox.

7. Feb 7, 2012

### Staff: Mentor

As PAllen has explained, this is a GR problem, not an SR problem - you bought that along with the gravitational slingshot. But you may be interested to know that there is a variant of the twin paradox problem that has no acceleration in it but can be solved using only special relativity (hence, is mathematically waaaaaay simpler).

We have *two* spaceships carrying observers, one far off to the left and moving right, and the other even farther off to the right and moving left. The earth is somewhere in between. The right-moving observer passes the earth, and as he does he and the earthbound observer zero their clocks.

At some later time, the right-moving and the left-moving observers will pass each other (this is the equivalent of the turnaround in the standard twin paradox) and the left-moving observer sets his clock to match the right-mover's clock.

When the left-mover passes the earth, he and the earth-bound observer compare times.

There's no acceleration involved, but the asymmetry between the two paths is obvious. You'll get a very satisfying resolution if you just use the Lorentz transforms to calculate the x and t coordinates of the three interesting points (right-mover passes earth, right-mover meets left-mover, left-mover passes earth) in all three(!) reference frames. The only subtlety is that you have to be very careful about where (0,0) is in the left-mover's frame; right-mover and earth-bound have the same (0,0) point.

8. Feb 7, 2012

### ghwellsjr

But if you analyze the entire scenario in any of these three frames, instead of insisting that the "moving clock" jumps between frames, then it won't matter where the origins are. In fact, you can use any other frame, such as one moving at twice the speed of the movers in any direction using any origin.

9. Feb 7, 2012

### PAllen

I really don't accept Nugatory's variant as a real differential aging problem. You don't actually have two people, or two lumps of radioactive matter, or whatever, that come back together demonstrating physical difference in time passed (one 30 years older than the other; one almost all decayed, the other hardly at all). It is just a clock synch trick. It reproduces the math but not the physics.

In SR with flat spacetime (including exclusion of non-trivial topologies - cylindrical or toroidal universes), there can be no differential aging of material bodies that separate and meet without one or both experiencing proper acceleration.

10. Feb 7, 2012

### ghwellsjr

But it does make clear that the aging isn't caused by the acceleration and I accept it as a legitimate variant of the twin paradox.
I'm glad you mentioned both bodies experiencing acceleration because we can have another variant of the Twin Paradox in which both of them accelerate exactly the same except that one returns home immediately while the other one continues far away from home before matching the acceleration of his twin and returning home a lot later. This clearly shows that it's not the acceleration that causes the differential aging but rather time spent at the relatively higher speed that causes the differential aging.

11. Feb 7, 2012

### PAllen

Acceleration is a necessary condition in flat spacetime for differential aging to occur. I believe we agree on this. As to cause, I am very familiar with variants like you refer to, and I also agree acceleration doesn't 'cause' the differential aging, despite it being necessary. However, I also don't think it is meaningful to talk about 'time spent at a higher speed'. Whose time? Higher speed relative to what? As has been discussed on other threads, localizing where on a spacetime path the 'missing time' is, is just as meaningless as drawing two different length paths on paper between two points, and declaring which part of the length is extra. You could have no 'near inertial' parts at all on either path, and still get differential aging.

12. Feb 7, 2012

### ghwellsjr

I'm sorry, I guess I should have repeated my previous post to supply the answer to your concerns:
Does that answer all your questions?

13. Feb 7, 2012

### PAllen

Nope. And my answers would just ditto what I said before. So this will have to be another of those areas where we agree to disagree.

14. Feb 8, 2012

### Staff: Mentor

This objection can be softened a bit by preparing three identical lumps of radioactive matter far in advance such that right-mover's and earther's chunks are equally decayed at their meeting, and right-mover's and left-mover's are equally decayed at their meeting. Now the difference in decay when left-mover and earth-bound compare is more interesting; left-mover's lump isn't the same right-mover's lump that earth-bound originally compared with, but last time we looked, left-mover's lump was identical to that one - and now it's not. I have A=B, B=C, but not A=C.
But your objection is still fair; now I'm just using the rate of decay as a clock and I still don't have two worldlines that intersect each other at two events.

Nonetheless, I find this version of the paradox to be helpful. It makes it clear (when you calculate the x-t coordinates of the start in left's frame and the end in right's frame) how both left-moving and right-moving can see the earth-bound clock running slow for the whole time - yet somehow the earth is where the most aging happened.

Your mileage may vary - if you don't like it as much as I do, my feelings won't be hurt.

15. Feb 8, 2012

### Staff: Mentor

You're right that the origins don't matter in the sense that you'll get the same answer no matter where they lie. But you will get a way cleaner and more intuitive picture if you choose the origins carefully:
1) You want them to lie on the world line of their observer. That happens automagically for earth-bound and right-mover because it's natural to choose their origin to be their initial meeting when they zeroed their clocks. But a poor choice of origin for left-mover will leave you schlepping around an annoying constant offset.
2) It's convenient to pick left-mover's origin so that his clock reads the same as right-mover's at the turnaround point.

You are also right that you can use any frame moving at any speed and you'll get a valid analysis. But much of the explanatory value of the exercise comes from seeing the coordinates of the three events in the reference frames of the three observers, so we might as well work with those frames.

16. Feb 8, 2012

### ghwellsjr

If you are talking about differential aging where the two bodies do not start out colocated and end up colocated, then I agree, it's meaningless. But this thread and all the discussion up to this point has been about the Twin Paradox where they start out together, separate, and come back together and I'm saying that if you agree to ignore gravity, then you can analyze the scenario in any single inertial Frame of Reference and the "time spent at a higher speed" is defined uniquely in that FoR and the "higher speed" is defined uniquely in that FoR and we're talking about the coordinate time of each body in that FoR and we apply Einstein's time dilation formula to convert coordinate time into proper time for each body and then we see how much proper time has accumulated for each body as it travels at different speeds according to the FoR for whatever coordinate time intervals from the time they separated until the time they reunite and we get the amount that each one aged and subtract them and we have the differential aging and no time disappeared or needs to be accounted for. You can make the paths as complicated as you want, they don't even have to go back to the original location and they don't even have to be at the same speed at the beginning and the end. This always works and it's real simple. And if you want, you can transform all the significant events into any other Frame of Reference and do the calculations all over again and you get the same accumulated ages for the two bodies and the same differential age between them.
Do you disagree with any of this? If you do, then I am not agreeing to disagree with you.

17. Feb 8, 2012

### Staff: Mentor

Hmmm.... Much as I hate to disagree with anyone who is supporting me ....

I'd rather say that this variant makes it clear that the twin paradox is due to the change in reference frame on the outbound and inbound legs. In flat space-time any such change must involve acceleration, so acceleration will always be associated with any SR twin paradox; this kinda takes the fun out of any discussion of whether acceleration causes the twin paradox or is associated with it.

I prefer to think of this variant as a way of not having to calculate the proper time through the acceleration (or, considering the post that started the thread, gravitational effect) at the turnaround point. Any day that I can avoid a line integral along a curved world line is a good day.

18. Feb 8, 2012

### Staff: Mentor

BTW.... In the unlikely event that Karl Coryat (original poster) is still with us.... You chose a really bad title for this thread. It is most assuredly NOT a dumb question.

19. Feb 8, 2012

### ghwellsjr

If you're going to analyze the entire scenario from a single reference frame, why would you consider two different origins? I agree, the initial meeting makes the most sense for the origin of the frame in which earth-bound is at rest but I'm saying you would track the motion of right-mover and then the motion of left-mover back to earth-bound all in this same frame. Then you can transform the three events into the frame where the right-mover is at rest and do it all over again and finally transform the same three events into the frame where the left-mover is at rest and do it a third time. If you want to change the origin to the contact event between right-mover and left-mover, you're going to have to put in an offset at some point rather than rely on the Lorentz Transform which kind of takes away the consistent process.
As long as an observer is at rest in a reference frame, it doesn't matter where the origin is, as he's not going to stay put at the origin beyond one moment in time so what does it matter if his location is not also all zeroes? I think there is more explanatory value in using just the Lorentz Transform to get the events from one frame to another. Besides, if you're going to move the origin, why should the event of right-mover contacting left-mover be considered more significant than the ending event where left-mover contacts earth-bound?

20. Feb 8, 2012

### ghwellsjr

Well, on this point I disagree. It is never necessary to use multiple reference frames to explain anything that is happening in a scenario. And doing so promotes the idea that each observer needs their own reference frame in which they are at rest. This is why the issue of the "missing time" that PAllen talked about comes up because jumping frames can cause jumps in time if done carelessly. Your idea of putting the origin at the meeting of right-mover and left-mover can eliminate this if done in a particular way. But stick with a single frame for the entire scenario and use the Lorentz Transform exclusively to get from one frame to the next and it's a non issue.

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