How does the duration of acceleration affect the Twin Paradox?

[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.

 

[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.

 

[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.

 

[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].

 

[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?

 

[PAllen]

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?

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.

 

[Nugatory]

T
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?

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.

 

[ghwellsjr]

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.

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.

 

[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.

 

[ghwellsjr]

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.

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.

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.

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.

 

[PAllen]

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.

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.

 

[ghwellsjr]

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.

I'm sorry, I guess I should have repeated my previous post to supply the answer to your concerns:

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.

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.

Does that answer all your questions?

 

[PAllen]

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?

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.

 

[Nugatory]

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).

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.

 

[Nugatory]

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.

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.

 

[ghwellsjr]

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.

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.

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.

Do you disagree with any of this? If you do, then I am not agreeing to disagree with you.

 

[Nugatory]

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.

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.

 

[Nugatory]

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.

 

[ghwellsjr]

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.

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.

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.

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?

 

[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.

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.

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.

 

[Karl Coryat]

Thank you Nugatory, I regret the title. I was expecting to be really embarrassed when I read the answer, but I'm glad to see it has sparked discussion. I think that this effect gravity has on proper time is more interesting than the usual "rubber sheet" GR model you see everywhere.

 

[harrylin]

[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.

That is the original version of the problem, and at the time it wasn't considered a paradox but an illustration. However, the reasoning is entirely different from the one you here advance. You can read the elaborate discussion here, starting from p.47:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
In a nutshell, any change of velocity has an absolute sense.

Note that that discussion ignores the effect from gravitation which was not known at the time; however as long as the velocity is very great that effect may be neglected, as the spaceship is most of the trajectory away from strong gravitational fields and also the stay-at home isn't in a strong field.

 

[PAllen]

That is the original version of the problem, and at the time it wasn't considered a paradox but an illustration. However, the reasoning is entirely different from the one you here advance. You can read the elaborate discussion here, starting from p.47:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
In a nutshell, any change of velocity has an absolute sense.

Note that that discussion ignores the effect from gravitation which was not known at the time; however as long as the velocity is very great that effect may be neglected, as the spaceship is most of the trajectory away from strong gravitational fields and also the stay-at home isn't in a strong field.

Except the key feature of the OP was the idea that both paths were inertial, so that locally, in closed lab, change of velocity could not be detected. Even if you try to bring charges into it, the infinitesimal radiation from an orbiting charge is not locally detectable - it requires global detection (classically, at least; it is this point that helps reconcile results that free fall charges radiate; and other results that they don't. The best answer is they do, but the radiation, classically, needs a non-local measurement to detect.). It is uniquely gravity and the principle of equivalence that create this case where you have differential aging without locally detectable change of velocity. Thus, much of Langevin's discussion is off point.

Thus, while the main effect could be approximated using SR formulas, it is only because of gravity that the OP formulation could exist at all - two paths with no proper acceleration meeting at two events. This is a very interesting formulation that helps clear up many misunderstandings about applying simple SR rules (an inertial path maximizes proper time) to GR where they are not always true.

 

[PAllen]

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

This is the part I disagree with. Consider a twin variant where neither twin is ever moving inertially, but they separate and come back together with different ages. Now pick some arbitrary inertial frame for the analysis. Let us try to apply this phrase (bolded above). What 'time' do we use? Not proper time, because you could find that the twin that aged less spent less proper time traveling fast (in this frame). This would typically be the case. Do we use coordinate time? Then the statement is wholly ambiguous and unhelpful, in general. Consider that, while neither twin is ever inertial (due to continuous changes in direction), twin A is always moving at speed .4c in this chosen inertial frame. Suppose twin B is moving .1c for 80% of the coordinate time between separate and meet up, and at .99999c for 20% of the coordinate time. The twin A has been moving faster 'most of the time', and the average coordinate speed for twin A is also greater. Yet twin B would age less. (And this scenario is definitely realizable by introducing smooth switchbacks in the A twin path).

So, just as with lines on a paper (where I think there is nothing more meaningful to say than that the longer line is longer), in relativity you cannot really say anything more than that when you have two different world lines intersecting twice, one may have less proper time than the other. In SR, you can also say that if one is inertial, it will definitely be the one with more proper time. In GR, you cannot even say this (except locally).

 

[harrylin]

Except the key feature of the OP was the idea that both paths were inertial, so that locally, in closed lab, change of velocity could not be detected. [...]

Yes, and I put it to the OP's attention that the original formulation is exactly the case that he/she described. Perhaps you did not read it?
Here you are:

it is sufficient that our traveler consents to be locked in a projectile that would be launched from Earth with a velocity sufficiently close to that of light but lower, which is physically possible, while arranging an encounter with, for example, a star that happens after one year of the traveler's life, and which sends him back to Earth with the same velocity.

 

[PAllen]

Yes, and I put it to the OP's attention that the original formulation is exactly the case that he/she described. Perhaps you did not read it?
Here you are:

I missed that because it is on page 50, not 47. Page 47 is all about the importance of detectable change of velocity. Further (p.50), (not surprisingly for 1911) it fails to emphasize the local non-detectability of the acceleration, having previously emphasized the absolute character of acceleration. Thus the description here would fail to satisfy the conundrum raised in the OP (which is inherently GR in posing: inertial (=free fall) are supposed to be equivalent, so what gives here?).

 

[harrylin]

I missed that because it is on page 50, not 47.

Sorry, I should perhaps not have written "from p.47" but "from p.47".

Page 47 is all about the importance of detectable change of velocity. Further (p.50), (not surprisingly for 1911) it fails to emphasize the local non-detectability of the acceleration, having previously emphasized the absolute character of acceleration. Thus the description here would fail to satisfy the conundrum raised in the OP (which is inherently GR in posing: inertial (=free fall) are supposed to be equivalent, so what gives here?).

It's not clear if the OP was looking for an SR-based answer or a GR-based answer. I referred him to an SR-based answer according to which local detectability is irrelevant.

 

[ghwellsjr]

This is the part I disagree with. Consider a twin variant where neither twin is ever moving inertially, but they separate and come back together with different ages. Now pick some arbitrary inertial frame for the analysis. Let us try to apply this phrase (bolded above). What 'time' do we use? Not proper time, because you could find that the twin that aged less spent less proper time traveling fast (in this frame). This would typically be the case. Do we use coordinate time? Then the statement is wholly ambiguous and unhelpful, in general. Consider that, while neither twin is ever inertial (due to continuous changes in direction), twin A is always moving at speed .4c in this chosen inertial frame. Suppose twin B is moving .1c for 80% of the coordinate time between separate and meet up, and at .99999c for 20% of the coordinate time. The twin A has been moving faster 'most of the time', and the average coordinate speed for twin A is also greater. Yet twin B would age less. (And this scenario is definitely realizable by introducing smooth switchbacks in the A twin path).

So, just as with lines on a paper (where I think there is nothing more meaningful to say than that the longer line is longer), in relativity you cannot really say anything more than that when you have two different world lines intersecting twice, one may have less proper time than the other. In SR, you can also say that if one is inertial, it will definitely be the one with more proper time. In GR, you cannot even say this (except locally).

Why did you chop off my quote in mid sentence? I gave the answer in the rest of the sentence. Here's my full quote:

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.

I didn't say you could use an average speed. I said you have to take each interval of coordinate time where each twin is traveling at different speeds and calculate the proper time accumulated for that interval and then add all the accumulated times. That's what I meant by "time spent at a higher speed". It was only to counter the argument that the differential aging is attributed to a brief period of acceleration as opposed to the total time spent at a different speed.

If you want to calculate the age difference for one twin traveling in a circle, then it is going to be very difficult to transform that into a different reference frame because you cannot use the simplified x-only version and you'll have to approximate the circle with a large number of events, but it can be done.

I don't understand why you think this is a problem or why you are disagreeing with me. Have you actually done one of these scenarios as I described?

 

[PAllen]

Why did you chop off my quote in mid sentence? I gave the answer in the rest of the sentence. Here's my full quote:

I didn't say you could use an average speed. I said you have to take each interval of coordinate time where each twin is traveling at different speeds and calculate the proper time accumulated for that interval and then add all the accumulated times. That's what I meant by "time spent at a higher speed". It was only to counter the argument that the differential aging is attributed to a brief period of acceleration as opposed to the total time spent at a different speed.

If you want to calculate the age difference for one twin traveling in a circle, then it is going to be very difficult to transform that into a different reference frame because you cannot use the simplified x-only version and you'll have to approximate the circle with a large number of events, but it can be done.

I don't understand why you think this is a problem or why you are disagreeing with me. Have you actually done one of these scenarios as I described?

It is the 'capsule summary' phrase I don't like - it leads to false expectations like the scenario I gave . The longer form is effectively what I've been saying in contrast to the capsule summary. I prefer to word it as "accumulate proper time along the paths" (in any frame you want). By the time you clarify that 'time spent at higher speed' means add up coordinate time increment by 1/gamma for that increment (over the whole path), you just have a long winded way of saying 'accumulate proper time over the path'.

I absolutely agree there is no location for the missing time, and argue against conceptions like the time is lost at a turnaround.

 

[ghwellsjr]

It is the 'capsule summary' phrase I don't like - it leads to false expectations like the scenario I gave . The longer form is effectively what I've been saying in contrast to the capsule summary. I prefer to word it as "accumulate proper time along the paths" (in any frame you want). By the time you clarify that 'time spent at higher speed' means add up coordinate time increment by 1/gamma for that increment (over the whole path), you just have a long winded way of saying 'accumulate proper time over the path'.

I absolutely agree there is no location for the missing time, and argue against conceptions like the time is lost at a turnaround.

Then we both agree that the 'capsule summary" phrase saying that the time differential occurs during the acceleration is really bad, correct?

Personally, saying "accumulate proper time along the paths" is no different than saying "the time on the clocks as they travel" and doesn't off any additional explanation unless you say how you calculate the time on the clocks based on the speed in a frame.

 

[PAllen]

Then we both agree that the 'capsule summary" phrase saying that the time differential occurs during the acceleration is really bad, correct?

Personally, saying "accumulate proper time along the paths" is no different than saying "the time on the clocks as they travel" and doesn't off any additional explanation unless you say how you calculate the time on the clocks based on the speed in a frame.

Agree. In my mind, "accumulate proper time" is associated with a notion of line element, and this one notion applies with full generality to SR or GR; whether the line element is:

d tau^2 = d t^2 - (dx^2 + dy^2 + d z^2)/ c^2 [which trivially gives (1/gamma) dt by rearrangement]

or the more general GR metrics, you have one method, one concept.

 

[ghwellsjr]

Agree. In my mind, "accumulate proper time" is associated with a notion of line element, and this one notion applies with full generality to SR or GR; whether the line element is:

d tau^2 = d t^2 - (dx^2 + dy^2 + d z^2)/ c^2 [which trivially gives (1/gamma) dt by rearrangement]

or the more general GR metrics, you have one method, one concept.

But since proper time is the time displayed on any clock, saying "accumulate proper time" is no different than "explaining" the twin paradox by saying "look at the clocks of each twin at the start of the scenario and at the end of the scenario and that will tell you how much each one aged". It doesn't matter what you associate "accumulate proper time" with because it's the problem we're trying to explain, not the explanation of a different problem. I'm trying to at least associate the (dilated) times on the clocks with their speeds relative to a single frame of reference, not to each other.

 

[PAllen]

But since proper time is the time displayed on any clock, saying "accumulate proper time" is no different than "explaining" the twin paradox by saying "look at the clocks of each twin at the start of the scenario and at the end of the scenario and that will tell you how much each one aged". It doesn't matter what you associate "accumulate proper time" with because it's the problem we're trying to explain, not the explanation of a different problem. I'm trying to at least associate the (dilated) times on the clocks with their speeds relative to a single frame of reference, not to each other.

This is really good example of the inherent pitfalls (into which I've frequently slipped) of trying to intuitively express a technical concept. With all positive intent, we start with:

attempt 1) Twin that spends more (coordinate) time at higher speed in some frame ages less. Problem: example in post #24 shows this is wrong.

attempt 2) Twin with higher average speed in some frame ages less. Problem: false by same example

attempt 3) Twin that spends more time at higher gamma ages less. Problem: also false by same example. Twin A had higher gamma 80% of the time but ages more not less.

attempt 4) Twin with higher average gamma ages less. This can also be false. Consider twin A goes at speed with gamma of 3 the whole time. Consider twin B goes at speed with gamma of 1.5 half the time and gamma of 6 half the time. Twin B has higher average gamma (3.75) , but twin B ages more rather than less (assuming coordinate, e.g. t=6 over the paths, A ages 2, B ages 2.5).

attempt 5) Twin with lower average 1/gamma ages less. This is trivially true, but also just says "twin with lower average aging per coordinate time ages less", rather un-profound. We arrive, as the only true statement we can make, one twin ages less and here is how you calculate it.

This underscores my belief that all attempts at a simple rule (other than the formula itself) for which twin ages less run into trouble, even in SR.

 

[R Power]

If acceleration is necessary for differential aging, how can twin's paradox be solved without involving general relativity? i.e solely on the basis of STR.

 

[PAllen]

If acceleration is necessary for differential aging, how can twin's paradox be solved without involving general relativity? i.e solely on the basis of STR.

Erase from your memory the false statement than SR cannot deal with acceleration.