1. Jul 15, 2016

### Bennie Grezlik

Forgive my naivete, but I've struggled with the twin paradox for a long time.

If there is no privileged frame of reference, why does time dilation apply only to the travelling twin? I have been told (possibly erroneously) that the stay-at-home represents the entire universe, but this seems to give the stay-at home a privileged reference. What if half the universe moved away from the other half at near C. Who's the traveler and who's the stay-at-home?

2. Jul 15, 2016

### Ibix

There are many threads on this. The simplest explanation is that your watch measures the distance (the "interval") you travel through spacetime, like the odometer on your car measures distance through space. Since the twins take different routes through spacetime the distances can be different, so their watches show different elapsed times.

You are correct that there are no special frames, and references to things like the outside universe are irrelevant to resolving the paradox. The stay at home twin is special within the context of the scenario, however. Because she did not accelerate she was always moving inertially. That is guaranteed to produce a longer elapsed time than any other person who is not co-moving with her.

3. Jul 15, 2016

### Staff: Mentor

You will want to read and review it carefully, then try some of the many other threads here on twin paradox, see if that helps clear things up.

The only thing that I would add is:
The differential aging that shows up in the twin paradox isn't the same thing as the time dilation you see in special relativity. If you're just considering time dilation, the travelling twin considers the stay-at-home twin's clock to be running slow and the stay-at-home twin considers the traveller's clock to be running slow, for exactly the reason that you gave: There is no preferred frame, so they both entitles to consider themselves at rest while the other one is moving.

Nonetheless, the traveller ends up younger. This is because both traveller and stay-at-home are moving between the same two points in spacetime: the event at which they separate and the event at which they rejoin. However, they take different paths through spacetime; the length of the paths is different, and because they're paths in spacetime not space a different length means a different amount of time elapsed on the path. Reference frames don't need to come into it at - all observers in all frames agree about the length of the various paths through spacetime.

4. Jul 15, 2016

### mathman

Te short answer: the traveler goes from the stay at home from to a moving frame, then stops (frame change), then gets moving back home (change frame), and finally stops (frame change)

5. Jul 15, 2016

### Orodruin

Staff Emeritus
While technically it could be argued that this is an explanation, it only explains why there can be a difference. It also gives the false impression that it is the act of acceleration (frame change) itself that is responsible for the time dilation (it is not, it is the geometry of space-time).

6. Jul 16, 2016

### mathman

I do not claim the frame change is responsible for anything. My point is that there is a difference between the stay at home and traveler and the difference in time can be calculated from the frame changes.

7. Jul 22, 2016

### Jeronimus

I agree with this answer if one considers the frame changes at the turn around phase. In the initial acceleration phase, the travelling twin also changes frames (infinitely many until he reaches his final velocity), but their clocks remain synced locally when considering an almost instantaneous acceleration where the change in the x and t positions are negligible small

If you watch this video (the video contains links to the software, if you want to run this yourself)

You can see how in the left diagram, representing the say at home twin, the white clock, representing a clock that travels with the travelling twin, is running slower in respect to the cyan coloured clock, representing a clock that resides with the stay at home twin.
The white clock ALWAYS runs slower in all stages (stage 1 to 6) in the case of the left diagram compared to the cyan clock.

In the right diagram, you can see the blue clock, representing the stay at home twin's local clock, measured from the perspective of the travelling twin, running slower up until stage 2. Up until stage 2 the scenario is completely symmetric. Basically both could consider themselves the "stay at home twin". Both measure each others' clocks to run slower.

It is stage 3 and 4, the stage in which the travelling twin accelerates back towards the stay at home twin, in which the travelling twin measures the blue clock to speed up and overtake his own (white clock).
So yes, it is the frame changes at stage 3 and 4(the acceleration phase back towards earth) which result in the travelling twin's clock to display a lesser clock count, and it is only stage 3 and 4 in which the travelling twin measures the stay at home's clock to run _faster_. At all other stages both measure/observe each other's clocks to run slower by the same factor.

To create those two diagrams, only the Lorentz transformation formulas were used, mapping events happening at x,t on the left diagram to x',t' on the right diagram and vice versa.
The Lorentz transformation formulas as well as the time dilation and length contraction formulas can all be derived mathematically from just the two basic postulates of SR.
So what you see in this video is a result of those two postulates.

There is much more to this however, especially when it comes to understanding where the "now" is for each of the twins in respect to the other. When one twin at x,t, being his "now"(the present), asks "what is the other twin doing right _now_"

By analysis of the diagrams, it certainly does not seem to be located in what we call simultaneous or the simultaneity axis.
Hence, the blue clock, the travelling twin interprets as the stay at home clock he calculates/measures(observes) is not really the "now" of the stay at home twin.

Anyway, i already took this too far considering the scope of the question, so i will leave it at that.

Last edited: Jul 22, 2016
8. Jul 22, 2016

### Matter_Matters

Very short answer - the twin that has departed is in a non inertial frame - hence he/she can tell that is was them that was in motion. It's non inertial because to return home they would need a change in velocity - no constant motion and hence non inertial.

9. Jul 22, 2016

### Jeronimus

No, that is not really an answer. In the initial acceleration phase, locally the clocks remain in sync if we consider a near instantaneous acceleration in which the t and x position changes are negligible small.

Up until the turn around phase, the situation remains completely symmetric. Both could consider themselves the "stay at home twins". In fact, if instead of what we defined as the travelling twin, the stay at home twin would decide to accelerate towards the other twin, it would be the stay at home twin (not staying at home any more) which would end up younger.

There is something relative about acceleration but the whole matter is so complex, only a few will be able to properly wrap their mind around it.

I believe that it is possible to accelerate someone, without him noticing he is accelerating. And it would not be a gravity field which can establish this, because a gravity field gets stronger towards the acceleration direction but what would be required is a field that gets weaker towards the acceleration direction. Not sure what to call it. Maybe an anti gravity field.

10. Jul 23, 2016

### Staff: Mentor

Please bear in mind the PF rules about personal theories. Unless you can give an acceptable reference, you should not be discussing speculations like this.

11. Jul 23, 2016

### Orodruin

Staff Emeritus
You can set up exactly the corresponding situation in GR, where two freefall observers separate and then reunite. The issue is not the acceleration itself, it is the geometry of the space-time. The acceleration does imply a difference in the history of the observers though.

Compare to asking if a curved path is longer because you are turning or because you can place more meter sticks along it. The latter is the more fundamental.

12. Jul 23, 2016

### pixel

Isn't it the acceleration that causes the curved path in spacetime for the traveling twin?

13. Jul 23, 2016

### stevendaryl

Staff Emeritus
Well, we can turn this from speculation to a technical physics question, which is surely appropriate for this forum:

Suppose you have an observer in orbit near a massive planet. You use huge rockets to accelerate the planet non-inertially. The observer will then be dragged along by gravity, but won't ever feel any acceleration forces. (There are probably some constraints along the lines that the acceleration of the planet cannot be too large, or the observer will be left behind.)

This problem is General Relativity, rather than Special Relativity, but my intuition is that if you did a twin-paradox with the planet--Accelerate it to near the speed of light, let it travel for 10 years, then bring it back to its original location--then the time dilation experienced by the observer will be roughly the same as in the SR twin paradox, even though he never "feels" any acceleration.

14. Jul 23, 2016

### Staff: Mentor

What you are proposing is not what I understood Jeronimus to be proposing. He talked about "anti-gravity", and about a behavior of the "gravity field" that, AFAIK, is not possible for any source that I'm aware of.

The latter is not something you calculate; it's something you stipulate in your statement of the problem. In other words, you are hypothesizing an observer who never feels any proper acceleration--they don't have a rocket attached to them, they don't get pushed by a huge EM field, etc. But given that hypothesis, you then have to calculate the observer's actual motion--you certainly can't just assume that he will stay in the same orbit about the planet (even with the caveat that we keep the planet's proper acceleration "small enough"). By adding whatever it is that gives the planet proper acceleration, you are changing the spacetime geometry, and that will change the observer's motion as well. Whatever it is that is pushing the planet has to have an energy comparable to the planet's, and therefore produces spacetime curvature comparable to that produced by the planet. So the observer's "orbit" will now be affected as much by the "pushing" energy as by the planet's.

Of course there is also the more fundamental point that describing the object as "accelerating, but not feeling any acceleration" equivocates between two different meanings of "acceleration". That's true no matter what the object's motion turns out to be.

15. Jul 23, 2016

### stevendaryl

Staff Emeritus
I don't think that those considerations actually change anything. Regardless of how it is achieved, it is certainly possible (in principle) to accelerate something the size of a planet (maybe by throwing asteroids at it from far away). So I'm pretty sure that there is nothing in principle impossible about a planet accelerating in a straight line to nearly the speed of light, drifting at that speed for 10 years, and then being accelerated in the opposite direction so that it returns in 20 years. Given that there is a planet that travels in such a way, that implies some time-dependent spacetime curvature. Then the question becomes: what are the geodesics of this spacetime, and are there any geodesics that remain close to the planet the full time.

16. Jul 23, 2016

### Staff: Mentor

Sure, but whatever is the source of the acceleration must also be causing comparable spacetime curvature to the planet itself. Even if you use a bunch of asteroids, each individually small, they would, collectively, have to have a total energy similar to the planet's (counting whatever energy source is being used to accelerate the asteroids themselves).

My intuitive guess is that there won't be; objects that started out in orbit around the planet won't be able to keep up, and objects that end up in orbit about the planet when it is moving at nearly the speed of light (relative to its original starting point) will have been captured by the planet recently.

To actually solve this mathematically is well beyond what I am prepared to take on, since it would need to be done numerically (AFAIK there are no exact solutions even close to the scenario we are discussing).

17. Jul 23, 2016

### PAllen

But, per GR definition, any geodesic paths are non-accelerated, inertial paths. Then, this all boils down to the utterly trivial notion in GR that two inertial paths between two events will normally span different proper times. At least for a test body in GR, the statement "accelerated motion that you don't feel" is a mathematical self contradictions. Don't feel -> geodesic; accelerated is defined as having nonzero magnitude of 4-acceleration, -> non-geodesic.

Last edited: Jul 23, 2016
18. Jul 23, 2016

### stevendaryl

Staff Emeritus
Yes. This would just be an example of that more general statement. But the point is, for people talking about the twin paradox, that whether or not you "feel" any acceleration is not enough to figure out whether you age more or less than someone else who takes a different path.

19. Jul 23, 2016

### Orodruin

Staff Emeritus
Yes, but it is the velocity that matters for the computation of proper time. This is what I alluded to with the length example. You can turn the wheel a lot when driving, but to compute the distance travelled you would still use the number of times the wheels turned.

20. Jul 23, 2016

### Staff: Mentor

It is, but that's a bit of a red herring. The important point is that the twins are on different paths through spacetime. The acceleration only comes into the picture because there's no practical way of sending two twins initially at rest relative to one another on different paths through spacetime without accelerating one or both of them.

There are variants of the twin paradox in which we accelerate both twins in the exact same way but at different times: for example they both start out in the same direction but one of them turns around and returns to earth sooner than the other. Now the accelerations are identical in magnitude and duration, but the paths through spacetime have different lengths and the aging will still be different.

Last edited: Jul 23, 2016
21. Jul 24, 2016

### m4r35n357

Is it not fair to say it is a major red herring; after all, the requirement for both twins to be initially at rest is most commonly dispensed with before the analysis to keep things to three frames of reference? So all we really need is for two particles in free-fall to coincide at two events.
Then we can look at gravitational examples like intersecting orbits, where there is no proper acceleration in either case but they age unequally. At least that is how I interpreted @stevendaryl's comment #18.

22. Jul 24, 2016

### PAllen

I interpret Nugatory's comment to refer to SR. It helps in these discussions to always be explicit about whether gravity is allowed to 'exist' for purposes of discussion.

23. Jul 24, 2016

### pixel

Can you provide a reference to that example?

24. Jul 24, 2016

### robphy

...then there is the "Cosmological Twin Paradox (PF, 2004)" or "Twin Paradox in a Closed Universe (PF, various)" for a flat but not simply-connected spacetime that is locally R^4 (e.g. a cylindrical spacetime). There, no proper acceleration is needed... but there are unequal proper times along geodesics between the same pair of events.

25. Jul 24, 2016

### Staff: Mentor

I don't know if I've ever seen it in print, although I wouldn't be surprised to find it as an exercise in some textbook somewhere - it's pretty straightforward.

Using coordinates in which the home planet is at rest, twins Alice and Bob both leave at the same time and accelerate to the same speed, .5c. Bob travels outbound for one year at .5c, turns around and returns also at .5c, and once home waits for Alice. Alice travels outbound for two years, turns around, and is back home four years after their common departure. All of their speed changes are done with the same acceleration.

Alice travels two segments, from (t=0,x=0) to (2, 1) and then from (2,1) to (4,0). Elapsed proper time is $2\sqrt{3}$ years.

Bob travels three segments, from (0,0) to (1,.5), then from (1,.5) to (2,0), and then from (2,0) to (4,0). Elapsed proper time is $2+\sqrt{3}$ years.