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Impure twin's paradox

  1. May 11, 2008 #1
    Quite often the explanation to the twins paradox is stated as acceleration breaks the symmetry or one twin feels acceleration while the other does not. I would like to add a new twist that analyses this explanation.

    Consider a push-me-pull-you rocket with two engines, one at either end so that it can accelerate and decelerate in a straight line without having to turn around. One twin (Peter) accelerates off into space in his fancy rocket for 10 years (by his clock) at a constant proper acceleration of 9.8m/s/s. He then turns off the rear engine and switches on the reverse engine mounted on the front and de-accelerates for a further 20 years at a constant proper acceleration of 9.8m/s/s before finally swapping engines again for the last 10 year segment of his journey to arrive back at Earth. For his entire 40 year journey Peter has experienced a proper acceleration of 9.8m/s/s while his twin Paul ,that remained on Earth ,also experienced proper acceleration of 9.8m/s/s for the entire period Peter was away due to the gravity of the Earth. Ignoring trivial split seconds when Peter was switching engines over, will both the twins have aged by the same amount (40 years) because they have both experienced the same acceleration?
  2. jcsd
  3. May 11, 2008 #2
    I like the out-of-the-box thinking, but thats not quite right.
    Acceleration causes gen. rel. and other complex effects... different relative velocities (w/out acceleration) cause standard special rel. time dilation.

    The person on earth, isn't experiencing 9.8m/s^2 acceleration all the time, unless they're jumping up and down alot. For the most part, they're just chillin, and being accelerated around the sun.

    The key to the twin paradox, is that if ever the twins are to be reunited - you can distinguish between which twin was accelerated - and accurately predict who will age how much.
  4. May 11, 2008 #3
    Can you see why he won't get very far?
  5. May 11, 2008 #4


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    There's some flaws and mistaken assumptions in your experiment.

    1] A person standing on Earth under 9.8m/s^2 does NOT experience the same level of time dilation as a person in a rocket under 9.8m/s^2. If they did, anyone that left the Earth's gravity well would observe dramatic relativistic effects, wouldn't they? For example, astronauts on the Moon would have come back significantly younger than their Earthbound counterparts.

    In fact, living at the bottom of a gravity well results in a dilatory factor so small that it can only be detected with the finest of instruments.

    2] A smaller error: having an engine on each end is no different from turning the rocket around. This is a misunderstanding of the causes and effects of acceleration on relativistic time dilation. Consider: you can do the same relativistic experiments without the rocket turning around or even stopping its engines; the rocket can do a giant loop, accelerating all the time.
    Last edited: May 11, 2008
  6. May 11, 2008 #5
    I totaly disagree with you on that Dave.
  7. May 11, 2008 #6
    What little i know of gen. rel. says that a gravitational field is equivalent to an accelerating reference frame... i think they would have the same effect - as for general relativistic effects. But again, not the same effects for special relativity.
  8. May 12, 2008 #7
    Yes, I can see that in order to take off he would have to exceed the acceleration due to Earth gravity. Can we assume for the sake of argument, that he exceeds it by a miniscule amount initially, to get away from the Earth and then as he gets further away from Earth and the gravity acceleration drops off, that he can start applying more acceleration to maintain a constant proper acceleration equal to that on the surface of the Earth?
    Last edited: May 12, 2008
  9. May 12, 2008 #8


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    Well! That's told him!

    Since Dave gave a detailed explanation of his position, perhaps you would like to explain why you disagree.
  10. May 12, 2008 #9
    1] I was not looking for dramatic gravitational relativistic effects. I was just trying to make the acceleration aspect as near as equal as possible for the two observers to "isolate the velocity effect.

    2] I only put the engine on each end to avoid adding complication due to rotation which is another form of acceleration. For the same reason a giant loop would introduce rotatational acceleration which complicates things. For example using the Sagnac effect which involves motion in a loop it is possible for an observer to determine his absolute (rotational) motion whereas he cannot determine his absolute linear motion in a straight line.

    If we can minimise the effects purely due to acceleration the main effect should be entirely due to linear velocity which would be greater than 0.9c for the rocket twin after 10 years (of proper time) of constant Earth like acceleration.
    Last edited: May 12, 2008
  11. May 12, 2008 #10


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    No. The time dilation does not rely on just the magnitude of the acceleration they experience.
    For Paul's case, the time difference due to gravitational time dilation is related to the difference in gravitational potential between Peter and himself. Since Peter is always at a higher potential than Paul, Peter's clock always runs fast a little due to this effect. Peter also measures this same difference in time due to this effect becuase he is always aware of his relative gravitational potential with respect to Paul.

    The acceleration due to Peter's acceleration however is different in many ways.

    1. It changes direction. While Peter is accelerating away from Paul, Paul is ''below' Peter and Peter measures Paul's clock as running slow. However, when Peter starts to accelerate towards Paul, Paul is "above" him, and he measures Paul's clock as running faster. And since he is furhter from Paul during this period he sees Paul's clock gain more time than it lost during the accelerating away phase. You can think of Peter as seeing his acceleration as a Uniform gravity field that extends through out the universe.

    2. Only Peter is aware of this "acceleration field" that causes this time dilation between himself and Paul, unlike the gravity field of Earth that both are aware of.

    3. The Earth's gravity falls off by the square of the distance. The "acceleration field" felt by Peter is uniform and doesn't fall off with distance at all.

    All these combine to cause a difference in the elapsed time experienced by Peter and Paul.
  12. May 12, 2008 #11


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    If you read the OP, you'll see that it seems he considers 9.8m/s^2 on Earth to produce the same effect as in a rocket under 9.8m/s^2 acceleration. The rocket, after a month will be doing near .99 c - enough to produce dramatic dilation effects when observing the universe around them. but a person standing on Earth for a month does NOT observe these same effects when observing universe around him.

    All I am doing is dispelling what seems to be a misunderstanding of the OP.

    The OP seems to think that both observers are undergoing the same effects, and that the only difference between them is when the rocket stops and turns over and that it is this change in acceleration alone that causes the relativistic discrepancy.
    Last edited: May 12, 2008
  13. May 12, 2008 #12


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    Would it hurt your scenario any to put Paul in a 1g centrifuge in space for the duration? (e.g. on a space station) That way you wouldn't have to bring in any GR effects and could do everything using SR.
  14. May 13, 2008 #13


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    First, let's look back the original twin paradox. After we've calculated that the travelling twin will age less than the earthbound one, we are faced with an objection, which can be stated as follows: There is a symmetry between the two twins, and so it is impossible to assign one twin a different elapsed time than the other, because there's no way to answer the question "which one gets the shorter time?".

    But, in fact, one of the twins experiences acceleration at some point (ie, he is a non-inertial observer), and this breaks the symmetry and thus resolves the paradox. Note that acceleration does not "cause" the time dilation (in fact, I don't see how anything can be said to cause time dilation, any more than something can cause one distance to be shorter than another). It simply provides a means to dismiss the naive argument outlined above.

    In your modification, acceleration cannot be used to break the symmetry, but there are clearly other things that can, such as the background spacetime geometry (eg, one observer experiences a constant curvature while the other does not). Thus, again, there is no paradox in the result that the travelling twin ages less (which can be seen by noting the time dilation (w.r.t. a distant stationary observer) on the surface of the earth is clearly negligible).

    Another modification of the twin paradox that really does remove the asymmetry is to have the twins orbiting a star at the same radius but in opposite directions. Then each is in free fall, and each sees the other in motion, so that your first thought might be that each twin would predict the other to have picked up less proper time than himself after they meet up again. Of course, a more careful analysis with GR shows this does not happen. For example, this can be reconciled with the doppler effect analysis shown http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler", if you take into account the bending of light as it passes near the star.
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  15. May 13, 2008 #14
    The best way to dismiss a naïve argument is to explain the actual mechanism, not wave your hands vaguely in the direction of acceleration. I applaud your clear statement that "acceleration does not "cause" the time dilation", but it could have been a little clearer since the question is about the lack of symmetry not about time dilation per se and you have implied that the acceleration does cause a symmetry break. It doesn't.

    What causes the symmetry break is that the terming of the twin paradox includes an assumption that is rarely made explicit.

    The "travelling" twin travels to a spot which is at rest relative to the "stationary" twin. This gives the "stationary" twin precedence, since the "travelling" twin is moving not only relative to the stationary twin but also relative to the destination.

    I doubt you can remove this assumption, but you can think about it a slightly different way.

    Put one twin at one end of a long rod (length L) and make the other end of the rod the destination. The other twin "travels" along the rod at velocity v. According to the twin on the rod, the travelling twin will take L/v to travel along the rod. According to the other twin, the rod which is relativistically contracted, passes by in a period of

    L/v . sqrt (1 - v^2/c^2)

    If the rod is sufficiently long, and the velocity sufficiently low, so that L/v is a sufficiently high value, we can ignore the time taken to decelerate/accelerate so that the travelling twin "travels" along the rod at a velocity of -v. (Note that I do not say that the travelling twin decelerates/accelerates, it could be that the rod does so.)

    When the twins reunite, the rod bound one will have a "dilated clock" showing more time elapsed.

    It doesn't matter whether it was the "travelling" twin who was "really" moving (ie moving relative to a third observer) or the rod with the other twin on it, or even both. The result will be the same.

    As I have pointed out elsewhere, acceleration is an indication that there is a symmetry break, but it is not the culprit behind the twin's paradox. Since it could be either the rod or the "travelling" twin who decelerates/accelerates, this cannot be the cause of one being subjected to time dilation more than the other.


  16. May 13, 2008 #15
    I can not see how this example, without any acceleration, simplifies the situation? Let's try to re-interpret this example as follows:
    Two observers are holding a long road (i.e. they are stationary w.r.t. each other), and another observer is on the rod, so that he can travel through the rod, from one observer to another, and at the start is alongside first observer at the first end of the rod.
    Now, both stationary (w.r.t one another) observers starts moving in a direction so that, the "on rod" observer feels that he is moving away from first observer, and at the end of the two observer's journey, ends up alongside second observer at the other end of the rod. It is obvious for him to think that, not the two observers (stationary w.r.t. each other) moved, but he moved through the rod, and reached from one observer to another observer (or from one end to the other end of the rod).
    Now in this situation, how do two stationary (w.r.t. each other) observers get any precedence?

    Edit: "he" refers to the "on rod" observer throughout the explanation.
    Last edited: May 13, 2008
  17. May 13, 2008 #16


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    Why do you think this is the case? The spot he travels to is just a point in spacetime, with no preferred notion of velocity. It could just as well be another ship travelling toward the earth, which he happens to pass at the moment he is stationary w.r.t. the earth.

    This is not true. If it is the observer with the rod that is accelerating (call him R), he will be the one who ages less. The analysis taking account of the rod is a little more complicated, but basically, after R starts turning around, it will take a while for the other end of the rod to get this information (which can travel no faster than the speed of light). So the other end of the rod will not always be at rest w.r.t. R, and the rod will not have a constant length, so that the argument used when the rod is truly at rest does not work.
  18. May 15, 2008 #17
    Relative to the frame in which the "travelling" twin is at rest, there is no separation between the spot (actually an "event") where (and when) he was collocated with the "stationary" twin and the point (ie event) he "reaches" before the relative velocity goes negative. The spatial separation only applies to other frames than that in which the "travelling" twin is at rest. Yes, the event could be collocation with another ship but, even so, that event has both temporal and spatial components. When I only discuss "the point to which the 'travelling' twin travels", it is only spatial component that I am referring to.

    I did mention this in another thread. I should have updated my post.

    People are so welded to the idea that acceleration somehow causes the symmetry break that they are very resistant to my attempts to eliminate the acceleration and show that the symmetry break remains.

    The fact that it takes a while for both ends of the rod to achieve the same velocity and be at rest with respect to each other is why I have the clocks stop and allow the rod to change velocity properly (if indeed it is the rod which changes direction and not the "travelling" twin) before restarting the clocks on the return trip.

    You don't have to have a real rod either. Think of two sentient beacons which are stationary with respect to each other with a rest separation of L and a "traveller" who moves between them at velocity v. When the traveller passes the first beacon, a message is sent to the second beacon. After a period of L/c, that message is received and the beacon's clock starts timing (from an offset of L/c).

    When the traveller passes the second beacon, the second beacon stops its clock, sends a message to the first beacon and waits for a period of L/c before accelerating to a velocity of v relative to the traveller (so that the relative velocity of the traveller with respect to the beacons is now -v). The first beacon accelerates to a velocity of v with respect to the traveller as soon as the signal is received in such a way that the two beacons remain at rest with respect to each other (ie according to a preplanned acceleration schema).

    When the second beacon passes the traveller (or the traveller passes the beacon), it restarts its clock. When the first beacon passes the traveller, it sends a signal to the second beacon. When that signal is received the second beacon stops the clock and winds it back an amount of L/v (as measured in the frame in which that beacon is at rest).

    The traveller merely starts a clock when first beacon passes, stops it when the second beacon passes, restarts it when the second beacon passes again and stops it for the last time when the first beacon passes for the second time.

    This eliminates the effect of the acceleration. You merely have measurements made during inertial motion.

    Despite this, the time elapsed for the "traveller" will be less than that elapsed for the beacons. Note that this will be the same is the traveller undergoes the acceleration, rather than the beacons.

    If it doesn't matter which undergoes acceleration, then it is not acceleration which is the culprit.

    Still, as said, people are so welded to the idea that acceleration plays a part that this will be argued against irrationally (by picking up on irrelevant details and ignoring the central issue). If you believe that acceleration is the culprit, then nothing is lost by making an attempt to properly analyse the scenario as presented and working out the consequences.

    If your belief that acceleration is to blame to the extent that it is an article of faith, then you are probably in the wrong place. Nothing is gained by blinding accepting received wisdom and defending points of view with rhetoric. Work out what the clocks read, then argue. Arguments along the lines of "you can't do that" or "4-vectors work for me" or "read a good essay" or "I know this stuff, you don't" just don't cut it.

    A few people have stated that they don't see the relevance of the scenario. The relevance is that the scenario is analogous to the twins' paradox with the acceleration removed (since time is only measured when there is no acceleration).


  19. May 15, 2008 #18


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    It's true, the event is at the same spatial location for the travelling observer as the departure event. But again, this says nothing about how the event is "moving", and in particular whether it is stationary with respect to the earthbound observer. This is simply not a question you can answer for events, only for worldlines.

    As for the rest of your post, it seems like your basic argument is something like "acceleration should not be presented as the resolution to the twin paradox because here's a variant without acceleration." But your variant is simply not the twin paradox, it's a completely new scenario.

    In the twin paradox, there is an apparent symmetry that makes the result seem impossible, but this symmetry is shown to be false by noting one of the observers is accelerating (ie, non-inertial). This is all that is being claimed with regard to acceleration - it is part of a resolution to a specific confusion in a specific example. On the other hand, in your setup there is a different, more obvious asymmetry, so the problem never arises.
  20. May 15, 2008 #19
    What I am saying is that that more obvious asymmetry exists in the twins' paradox.

    Can you explain why that asymmetry doesn't exist in the twins' paradox? If that is your contention, of course.



    PS - I note that you didn't work out what the clocks read, or at least you didn't mention that you had done so. Did you even bother thinking about it?
    Last edited: May 15, 2008
  21. May 15, 2008 #20
    Since it is entirely possible that no-one will be bothered to actually go through the calculations themselves, here is my working.

    Twins' paradox (stay at home twin Joe and travelling twin Jane):

    Joe stays on Earth while Jane travels to a distant planet, a distance of L away from Earth, with a constant velocity v. She reaches v by undergoing enormous acceleration. Once at the distant planet, Jane undergoes enormous acceleration to change her velocity to -v and travels back to Earth. When back she undergoes enormous deceleration to come to rest alongside Joe. The acceleration is enormous so that the time spent accelerating and deceleration is negligible when compared to the total time elapsed.

    According to Joe the time elapsed is 2L/v (+ negligible acceleration/deceleration time in Joe's frame). According to Jane the time elapsed is 2L/v . sqrt (1 - v^2/c^2) (+ negligible acceleration/deceleration time in Jane's frame).

    My "completely new scenario":

    See earlier post, https://www.physicsforums.com/showpost.php?p=1730961&postcount=17.

    Call the "traveller" Bob, the sentient beacons are Ann and Andy.

    According to Ann and Andy, they have a separation of L between them. Relative to them, Bob has a velocity of v when he passes between them the first time, and -v the second time. Therefore, the time elapsed on the second beacon's clock (Andy's clock) will be L/v - L/c + L/v + L/c = 2L/v

    According to Bob, the separation between Ann and Andy is contracted due to their relative motion, so that:

    L' = L . sqrt (1 - v^2/c^2)

    Since Ann and Andy move past Bob with a velocity of -v the first time and v the second time, the total time elapsed in this scenario, according to Bob, is 2L'/v = 2L/v . sqrt (1 - v^2/c^2)

    In short:

    Time elapsed according to Joe - 2L/v
    Time elapsed according to Andy - 2L/v

    Time elapsed according to Jane - 2L/v . sqrt (1 - v^2/c^2)
    Time elapsed according to Bob - 2L/v . sqrt (1 - v^2/c^2)

    Is my scenario really completely new? Note that Jane and Andy experienced acceleration, not Joe and Bob, but that Joe and Bob measure different times.

    A variation of my "compeltely new scenario":

    Rather than the beacons undergoing acceleration, the traveller does. Let's call the beacons Sue and Stu this time, and the traveller Tom.

    In the frame at which they are at rest, Sue and Stu have a separation of L. Tom passes by both of them twice, Sue first, with velocities v and -v. He undergoes acceleration between passes. Stu signals Sue when Tom passes by each time. Sue starts her clock when Tom passes, stops it when she receives Stu's signal, starts it again when she gets another signal from Stu and stops it when Tom passes again. Time elapsed -
    L/v - L/c + L/v + L/c = 2L/v

    According to Tom, in the frame in which he is at rest, the separation between Sue and Stu is contracted by their relative velocity (magnitude v) - L' = L . sqrt (1 - v^2/c^2)

    Tom starts his clock when Sue passes, stops it when Stu passes, restarts it when Stu passes in the opposite direction and stops it when Sue passes again. Total time elapsed on Tom's clock - 2L'/v = 2L/v . sqrt (1 - v^2/c^2)

    Summarising again:

    Time elapsed according to Joe - 2L/v
    Time elapsed according to Andy - 2L/v*
    Time elapsed according to Sue - 2L/v

    Time elapsed according to Jane - 2L/v . sqrt (1 - v^2/c^2)*
    Time elapsed according to Bob - 2L/v . sqrt (1 - v^2/c^2)
    Time elapsed according to Tom - 2L/v . sqrt (1 - v^2/c^2)*

    * indicates who underwent acceleration.

    Note that this variation results in precisely the same results as the twins' paradox (I acknowledge that I am neglecting the negligible acceleration/deceleration time, but I did that openly and with good cause.)

    Acceleration is not the culprit.


  22. May 15, 2008 #21


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    The apparent symmetry is that, from each twins point of view, the partner takes off, travels some distance L, turns around, and returns to the starting point. It would seem that if one twin would predict his partner to age less, then the other should make the same prediction. The problem with this logic is that one of the twins' frames is non-inertial (ie, there is some point where he is accelerating), so only one of them can use the usual formulas of special relativity to work out the dilation. The other will have to use slightly more complicated formulas, and if he does this correctly, he will indeed agree he will actually age less, even though it appears it is his partner who is travelling.

    No, I didn't see a need to. From the description I could tell enough to make my point that there is an obvious asymmetry, namely, one of the "twins" has a partner, and they undergo a synchronization procedure, while the other does not. In other words, I didn't doubt your calculations were correct, as the result was neither surprising nor bothering.

    If you can come up with an example with two pairs of observers which each do some kind of synchronization and you still get such a result, then I would check your calculation. But let me assure you that this will neither work (there will be no difference in time readings if there is a true symmetry, although if the situation is complicated enough we can waste a good amount of time trying to find the actual asymmetry/ calculation mistake if you did get a difference) or be necessary for this argument.

    Again, no one is claiming that "acceleration is the culprit". If you read what I've already written carefully, you'll see exactly what is being claimed, and how this has nothing to do with your example.
    Last edited: May 15, 2008
  23. May 16, 2008 #22
    I find this saddening. It could be a personal thing (ie directed at me, because you have made a judgement about me), but if as a homework advisor your standard response is "I didn't actually bother thinking about what you have said, but you are wrong" then you might be sending the wrong message. Even if it is a personal thing, you are still sending the wrong message.

    The question begged here is "why does undergoing a synchonisation procedure make a difference?"

    I would agree that this would be a complete departure from the twins' paradox. I also know as you do that you would run into simultaneity problems (two identical rods moving past each other will not have both ends aligned with both ends of the other rod simultaneously). I can't actually see how this would possibly come up with the same result since the scenario that I think you are hinting at is not analogous to the twins' paradox - I think you mean something like having both rods aligned at the start with one moving past the other ... but this is not possible, they can't be both aligned (with the ends of the rod collocated simultaneously) and in relative motion.

    Are you just being cute here?

    Hm, I can't be bothered thinking about your scenario, since it is wrong. But you have to carefully read what I write.



  24. May 16, 2008 #23
    There is a mistake here.

    "The apparent symmetry is that, from each twins point of view, the partner takes off, travels some distance L."

    This is wrong.

    According to one of the twins, the other travels a shorter distance (arbitrarily we can say the shorter distance is L.sqrt(1 - v^2/c^2), but the perceived distances could be L and L/sqrt(1 - v^2/c^2) such that L is the shorter of the distances).

    The error is, of course, that L is "some distance" relative to a specific twin. The symmetry that you claim is apparent simply does not exist, at all. The reason you think it does is because you assume that twins initially share the same rest frame, then one uses measurements made in what remains the other's rest frame and applies those measurements even when that other's rest frame has a relative velocity.


  25. May 16, 2008 #24


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    Let's say the stationary twin decides the other twin is a distance L away at the point he is turning around. At this point, the twin's are in the same frame, and so the second twin will also decide the first is a distance L away. The symmetry is still there. Although even if it weren't, there would be no problem with my argument, as all I'm saying is that acceleration can break the symmetry, not that it is the only thing that can (as in your example, where it doesn't).

    Let me try again. I read your scenario, and I understand it. There's no reason for me to question the calculation (one small part of what you've written), because I have no reason to think it's wrong, and it is not the problem I have with your argument. It would be a waste of my time. Here's an analogy if this is still bothering you. If you're a chemistry teacher, and someone hands you a biology test, should you still grade it, or is it enough to say, "this is the wrong test"? Don't take it personally.

    That being said, maybe I should have just done the calculation (or said I did) out of courtesy, as it would've taken me about a minute. I guess I just like sticking to my principle of not doing needless work, but if you were offended, I apologize.

    In any case, we don't seem to be getting anywhere. It's not even clear what you're argument is anymore, but it doesn't seem like you're misunderstanding anything important. In fact, it seems like what's going on is that you're misunderstanding what other people are saying and disagreeing with that, even though there's no one really saying it.
    Last edited: May 16, 2008
  26. May 20, 2008 #25
    How does the second twin determine the distance? I see two options, one is to come to rest relative to the first twin and do some sort of laser range finding, in which case, yes, L will be the distance for both. Another alternative is to use the relative velocity and time elapsed, in which case the distance will be less than L for the second twin.

    I think that fundamentally we are in agreement with a slight difference in focus. My argument was not originally with you, but with someone who stated that acceleration causes less time to elapse than for an observer in an unaccelerated frame. And what I am saying is that it is not the actual acceleration that is the cause.

    Here is my analogy:

    JFK did not complete his second term. Some people could say it was because he got shot.

    I argue that it was not because he got shot. Ronald Reagan got shot too, and he finished out his first term and also the second.

    Getting shot is like acceleration, since it can change you from one state (one frame) to another state (another frame).

    JFK did not complete his second term because he was dead. That is key. His state changed. What actually caused that state to change is really irrelevant other than historically. He would not have completed his second term if he died driving off a bridge, or flying a plane, invading the Bay of Pigs or even slipping on the soap in the shower. So long as he died somehow, he would no longer be suitable Presidential material. (By the way, you can look away when he dies too, to eliminate the cause of death: JFK alive one day, dead the next - cause unknown ... no longer able to be President.)

    Similarly, the acceleration we are discussing causes one twin to change state (being at rest in the frame in which the first twin and the destination are at rest to being at rest in a frame in which the first twin and the destination are not at rest). Similarly, you can look away when the change of state happens. Not matter how the change of state happens - acceleration, wormholes or grade-A magic - you get the same result.

    The acceleration per se does not cause the time elapsed to be less according to one twin. I don't think that you, StatusX, are arguing that it does. But many do (an example https://www.physicsforums.com/showpost.php?p=1728687&postcount=36"). The explanation as to why one twin experiences less time is associated does involve transition from one rest frame to another rest frame, back again, to a third rest frame and back to the original rest frame - and acceleration is the mechanism behind that - but it is the whole "not actually sharing the same rest frame the whole time" thing that is behind the "paradox".

    And despite what Doc Al says, it doesn't actually matter who accelerates - as long as one observer remains at rest with respect to the destination of the other, the "other" will experience less time.


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