1. Oct 10, 2012

### Rishavutkarsh

ok it's about the twin paradox the guy on spaceship goes at 90% of C (say) but since both the twins see each other's clocks equally slow and if one sees other at 90%C then the other one also sees his twin at 90%C in opposite direction. then how on returning the twin on spaceship is younger?
is there something I am missing?

2. Oct 10, 2012

### HallsofIvy

That's the whole point of the "twin paradox" and why it is called a "paradox"- the symmetry.

But the "returning" part means there cannot be symmetry. In order for the twin who goes out an comes back to be able to return, he must accelerate. While velocity is relative, acceleration is not.

3. Oct 10, 2012

### harrylin

Any question remains?

4. Oct 10, 2012

### TrickyDicky

I read a few twin paradox threads, including the one suggested in this thread, and I am not sure a consensus has been reached about it's solution. It seems half the people claims the key is in the acceleration of the travelling twin, and the other half says acceleration has nothing to do, (not to mention those that want to solve it introducing GR).
So what is the official PF position?:-/

5. Oct 10, 2012

### ghwellsjr

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his earth twin will age 2.2942 times as much. Simple, isn't it?

6. Oct 10, 2012

### Rishavutkarsh

well it surely is but isn't it said that a person sees the clock fastest which is at rest compared to him so can he really see the other clock running 4.359 (or whatever)
also is there droppler's effect also involved in this in any way?

7. Oct 10, 2012

### Rishavutkarsh

that's what is the point of dispute some say it's due to acceleration while others say that it has nothing to do

8. Oct 10, 2012

### harrylin

The official PF position?

In the multitude of answers some subtle points may have gone unnoticed, and while there was no disagreement about this issue as far as I am aware, I can understand the need for clarification.

1. As remarked in the first full discussion of the "twin" scenario in the literature (from both points of view, and in which the stay-at-home is approximately at constant inertial motion), the symmetry is broken by a change in velocity of the traveler.

2. The acceleration that is required for this change in velocity has itself no effect on clock rate; clock rate is calculated as function of the clock's speed.

If not well understood, from point 1 one may think that acceleration itself causes a difference in clock rate, but that is wrong; or from point 2 one may think that the traveler may be considered the whole time at rest in an inertial frame, but that is wrong too.

Depending on which reference frame you choose, speeds appear differently; and if one chooses two different reference frames for the traveler (typically the ones in which the traveler is in rest during different periods), then one has to do a Lorentz transformation when one makes the switch. And different frames attribute different times to distant events.

9. Oct 10, 2012

### Jeronimus

I am not sure i agree fully with this.

Taking the situation, earth with the twins at the same place with synced clocks, i can see there being an observer which is also at the same place shortly before the acceleration AND who is within an inertial reference frame at rest which after the acceleration will see both rockets moving away of each other at the same velocity seen from his point of view. He would see both twins accelerating symmetrically.

If such a thing is possible, then how is acceleration not relative? It is not symmetric for all inertial reference systems. Does breaking the symmetry in some inertial reference frames imply that something is absolute rather than relative? As i understand it, it does not.

10. Oct 10, 2012

### ghwellsjr

No, you are wrong. The third inertial observer that you describe will see both twins prior to acceleration traveling in the same direction at the same speed with respect to him. Then he will see just one twin accelerate so that he ends up traveling in the opposite direction at the same speed as before. The other twin remains inertial.

So before acceleration, he sees both twins traveling towards him at a constant speed. Then when they get to him, one of the suddenly reverses direction while the other one continues on just as before and then he sees both of the traveling away from him at the same speed.

But what has this to do with the topic of this thread?

11. Oct 10, 2012

### ghwellsjr

Your problem is that you misrepresented what the twins will see. They will not both see the other ones clock going slower than their own as I pointed out in post #5. You need to understand that every one of the other explanations will agree on what the twins see of the other ones clock. There is no dispute about what the explanations do, all the explanations work and agree on what the twins see which is what you asked about. The only dispute is about how some people insist that their personally favorite explanation is better than all the others but anyone who understands relativity will agree that all these explanations are equally valid.

You need to quit misrepresenting the Twin Paradox and the different explanations of the Twin Paradox.

12. Oct 10, 2012

### Jeronimus

removed for edit

13. Oct 10, 2012

### TrickyDicky

Ok, this is helpful. My own understanding is along these lines.
To me the asymmetry comes simply from the fact that in SR inertial frames are preferred over non-inertial ones, and the inertial one is the one that always experiments the longest proper time.
What is intriguing IMO is that by introducing the two different kind of frames and thus the asymmetry that dissolves the
paradox, and in the scheme that considers accelerations for the travelling twin as instantaneous, it introduces a way to see velocities as absolute, something that is mathematically impossible when using only inertial frames in SR.

14. Oct 10, 2012

### Jeronimus

I was imagining an observer which is at free fall i guess, for whom both rockets would look like they are accelerating. Not sure if this can be considered an inertial frame of reference. Anyway, i will not further this as it gets too complicated for me.

Last edited: Oct 10, 2012
15. Oct 10, 2012

### harrylin

More or less so: as determined in an inertial frame, a clock that is at rest accumulates more proper time than one that is moving, and that was already found in Einstein's 1905 paper. And as a matter of fact, the first "twin" illustration with observations from both sides was part of an article that explained these things:
- https://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
Actually, as you can see in the above link, it was the very purpose of that example to illustrate that acceleration is "absolute" in an observational sense (not clear what you mean with "a way to see velocities as absolute", except if you basically mean the same as him). And it became known as a "paradox" because Einstein tried to make accelerations also "relative" in the development of GR.

Last edited: Oct 10, 2012
16. Oct 10, 2012

### Staff: Mentor

First, a small quibble over terminology: it's not the "inertial frame" that experiences the longest proper time, it's the observer in *inertial motion*. (And strictly speaking, "longest proper time" means "longest proper time between two specific events.)

The more substantive issue I have with your way of putting it is that it does not generalize to curved spacetimes. Consider, for example, the following two observers: one on the International Space Station, and another "hovering" over the Earth at the same altitude as the ISS, but not orbiting the Earth (i.e., "hovering" motionless with respect to the distant stars). The first observer is inertial, the second is not; but if we pick two successive events where the two observers meet (between which the ISS completes one orbit), the second observer (the non-inertial one) will experience more elapsed proper time between them than the first (the inertial one).

So IMO a more general "solution" to the "paradox" is needed, and that is to simply understand that "elapsed time" is length along a curve, and that different curves between the same pair of points can have different lengths. The two twins in the original "paradox" travel on different curves between the same pair of points, so they experience different "lengths" (elapsed proper times). Once you're over that hurdle, figuring out which of the lengths is longer is just calculation; but the real hurdle is getting people to understand that "elapsed proper time" is just length along a curve. That concept generalizes easily to *any* timelike curve.

17. Oct 10, 2012

### TrickyDicky

I was thinking of your point 2. above. It introduces a way to assign velocities and rest preferentially to the twins based only on speed.

Last edited: Oct 10, 2012
18. Oct 10, 2012

### Staff: Mentor

I think the lack of consensus is not so much about the resolution of the paradox as it is about the best way to explain it to someone who doesn't understand relativity at the level of the Minkowski geometry. (If you do understand the Minkowski geometry there isn't any paradox to explain, it all just makes sense).

Last edited: Oct 10, 2012
19. Oct 10, 2012

### phyti

If each twin experiences an acceleration, they may accumulate the same or different amounts of time at reunion. This eliminates acceleration as a causative factor.

20. Oct 10, 2012

### TrickyDicky

Could be.
I'm not sure it is so straight forward, the introduction of non-inertial frames (curvilinear coordinates) in a flat space can lead to confusion( as the long countless threads on the paradox testify) even to people who is aware of the properties of Minkowski space.

21. Oct 10, 2012

### TrickyDicky

If both twins experience change of velocity (so both are noninertial), how do you decide wich one is older/younger at reunion?

22. Oct 10, 2012

### zonde

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

23. Oct 10, 2012

### Staff: Mentor

Ah... where's the acceleration in this example?

24. Oct 10, 2012

### TrickyDicky

We should keep in mind we are in the domain of SR, that is, flat Minkowski spacetime, regardless of the use of different type of reference frames/coordinates the physics should not change.

25. Oct 10, 2012

### Jeronimus

As i see it,

when the twins are at rest on earth, the acceleration which changes which frame the leaving twin is at rest in, is not the cause of him aging less.

One could imagine, that before the twin leaves earth, he places a long line of synced red colored clocks in front and behind him. Let's assume those clocks also have negative counters, with the acceleration taking place when the clock is at zero.

The frame earth and the twins are at rest in before the acceleration takes place is frame A.
Frame B is the frame the leaving twin will be at rest in after the acceleration.

Now assume that in frame B there is a guy named Bob, who also placed synced blue colored clocks in front and behind him. Just when Bob passes by the twins at vrel, the leaving twin accelerates instantaneous. Bob has set the clocks in such a way, that just when he passes by the twins, the clocks are at zero seen from within his rest frame.

Bob will accelerate instantaneous just when he passes by the twins, when all the blue clocks in his frame show zero.
The staying twin and Bob are now at the same place at rest in frame A, while the leaving twin is at rest in frame B moving at vrel = 0.9c relative to Bob/staying twin.

The leaving twin now at rest in frame B, will see the red clocks display higher counts the further away they are in front of him, and lower counts the further away they are at the back of his rocket.

Bob will see the equivalent, concerning the blue clocks he placed in his initial frame B BEFORE accelerating. At rest in frame A AFTER accelerating, Bob will see blue clocks in front of him are now showing higher counts, while behind him they show lower count. Higher/lower the further away.

Assume Bob and the twins are of the same age when the acceleration (instantaneous) events take place at t=0.

Bob is basically in the same position as the staying twin now, but went through the equivalent acceleration process the leaving twin went through.

Neither Bob nor the leaving twin seem to be any special in this regard. The situation to me looks symmetrical.
Therefore it is not the LOCAL acceleration at t=0 which is the cause of the age difference.

The acceleration/accelerations which occur non-local are the cause for the age difference once they meet up again.

In fact, instead of the leaving twin returning, the staying twin or Bob could change his mind, and decide to accelerate towards the leaving twin. In that case, Bob/the staying twin would have aged less.

The initial acceleration of the leaving twin made no difference in the aging. It was necessary however, to get a distance between the twins, allowing for the combination of acceleration and distance to cause the difference in aging.

So no, acceleration is not the cause of the difference in aging, but the distance to each other combined with acceleration which makes this happen.