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Twin paradox, new version

  1. Sep 18, 2011 #1
    We think of two very similar scenarios:

    1. Two twins are hoovering with their spacecrafts at an earthlike distance A from the Sun and with no velocity with respect to the Sun. Then one of the twins suddenly accelerates to 100 km per second in some direction. He is now the accelerated twin. After a while when he reaches a Plutolike distance from the Sun B the accelerated twin stops so the two twins are at rest with respect to each other. Then both takes off at 100 km/s and meets in the middle and stops and compare their clocks. They agree that less time have elapsed for the twin that has been travelling at an accelerated speed for a longer period of time.

    2. Two twins are flying their spacecrafts in parallel just next to each other at 100 km/s with respect to the Sun. Suddenly, at an earthlike distance A from the Sun, one of the twins accelerates by 100 km/s so that he now is at rest in relation to the Sun . He is now the accelerated twin. The other twin happily flies on at constant velocity in his inertial reference frame until he reaches a Plutolike distance from the sun B where he suddenly accelerates by 100 km/s in the same direction that his twin accelerated. The two twins are now at rest with respect to each other and the Sun. Then both takes off at 100 km/s and meets in the middle and stops and compare their clocks. Now what will their clocks show? Will the clock of the twin that did not accelerate until he reached B show more or less elapsed time than his twin that started off by accelerating?

    We ignore the fact that thet have spent time at different gravitational potential and care only of the effect of velocity on time.

    (By mistake I put this in the wrong forum at first, sorry for the repeated posting)
  2. jcsd
  3. Sep 18, 2011 #2


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    The velocity profiles are the same from separation to reunion in the two scenarios. Therefore the difference in ages will be the same. Neither twin is inertial.
  4. Sep 18, 2011 #3


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    Your two scenarios are exactly the same as far as Special Relativity goes and so the one that you are calling the accelerated twin is the one that has less elapsed time.

    From your other post on the thread that got locked, you addressed two issues:

    1) You quoted a website that stated that in the Twin Paradox, if one twin remained inertial and the other twin accelerated, then the accelerated twin will experience less time. You objected to this and produced these two scenarios to illustrate your point. But in these two scenarios, both twins are accelerating, and so that simple explanation from the website doesn't apply. It is true that acceleration does not cause one twin to experience less time, it is being at a higher speed for a longer time that causes less elapsed time when they reunite, but this can only happen for the accelerated twin if the other one never accelerates.

    2) You stated that gravitational fields are important in analyzing Special Relativity time dilation and that a third body with which to describe the speeds of the two twins is necessary. Your two sceanios are an attempt to illustrate this.

    However, since you have stated that you want to ignore any gravitational effects, you don't need the sun in your scenario. Let me boil your two scenarios down into one equivalent scenario:

    Two twins are at rest with respect to each other. One of them takes off at 100 km/s with respect to the other for some period of time and then stops with repect to the other one. Then they both take off toward each other at 100 km/s and stop at the midpoint where they compare clocks. Of course the one that accelerated first will have less elapsed time.

    This might be easier to see and analyze if you just change the order in which events happen:
    Two twins are at rest with each other and both take off together at 100 km/s in some direction. (Their clocks remain in sync although both time dilated according to their initial rest state.) Then one of them stops with respect to their initial rest state. (His clock is no longer time dilated according to their initial rest state.) The other one (whose clock continues to be time dilated) continues away for some period of time and then stops and returns to the other twin at the same speed. When they compare clocks the one that continued on has less elapsed time.
    Last edited: Sep 18, 2011
  5. Sep 18, 2011 #4
    You misread the crucial part, in the first scenario it is the accelerated twin that travels from A to B but in the second scenario the accelerated twin is the one that stays at A.

    The question I am trying to adress is whether the crucial part is who has undergone acceleration from the original inertial reference frame or who is travelling with the highest velocity with respect to the sun...

    As you say "it is being at a higher speed for a longer time that causes less elapsed time". Yes but higher speed in relation to what? True, in the classical twin test it does not matter. It both twins take off in the same direction at speed k*c with respect to the sun, this frame is the inertial frame. Now one twin takes off at a speed of (b-k)*c in the opposite direction, all speeds measured in relation to the sun. Then he stops and chases after the first twin at a speed of (b+k)*c until he reaches him. Now it happens that it does not really matter what the numerical value ok "k" is. One twin spends all the time at velocity k*c and the other twin spends half the time at velocity (k+b)*c and the other half at velocity (k-b)*c. As long as (k+b) < 1 (choosing otherwise we would have superluminal speeds) the twin that has accelerated will always experience more elapsed time, no matter how we choose k and b.

    But, given a certain b, will the difference in elapsed time always be the same, no matter how we choose k? (maybe this is actually the case, then I have to rethink things...)

    I did not state such a thing. I stated we will ignore the fact that the two twins will spend different amount of time at different gravitational potential, which makes things more complicated, and only consider the implication of their different velocities on measured time.

    You are missing the point. In scenario two the two spacecrafts moves together at 100 km/s with respect to the sun. We define this to be the rest state, since the fact that they are moving with respect to the local gravity field should not matter according to some. Now one of the twins accelerates with 100 km/s towards the sun (which means he stops with respect to the sun). The other stays at rest in the original reference frame until he reaches B.

    The whole point of my "paradox" is that you can choose which twin is the accelerated and who is the inertial...
  6. Sep 18, 2011 #5


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    In both scenarios both twins are non inertial.

    No, you cannot. Neither twin is inertial in either of your scenarios.
  7. Sep 18, 2011 #6


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    The problem stated can be analyzed using general relativity, and Ii expect that one will find that the twins have aged differently if one does so, though I haven't crunched the numbers.

    To take a simpler example where no calculations are necessary, a "stationary" twin, accelerating to hold themselves in place (and hence not inertial) at the orbital distance of pluto from the sun will have a faster-running clock than the same twin hovering at the orbital distance of the Earth, due to gravitational time dilation.

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html has an approximate analysis (used to analyze a different experiment, but one which has been acatually performed, the well-known Haeffle-Keating experiment) which could be modified to give an APPROXIMATE answer. The approximations used wouldn't matter for this particular brand of the thought experiment, but might be misleading if they are taken as exact - a better discussion to get the exact answers would require a modest background in GR.

    Someone else might be motivated to provide a more detailed discussion, at this point I'd need two things to be motivated to do more - one is to get a better sense of the OP's background, to see at what level the discussion should occur at. The second is some indication on the OP's part that they're interested learning about SR and GR, rather than making mistakes based on misunderstandings (which is OK, everyone has to learn), and then insisting that they're right after the correct answer is pointed out (which is not OK, and is a general problem that comes up with people giving "false" paradoxes, they become emotionally attached to them which prevents them from learning).
  8. Sep 18, 2011 #7
    To be specific, from the time that the twins separates at A and until one of the twins reaches B you can choose which twin has been inertial so far. Agreed?

    I make the case that the clock onboard the spaceship going to B during this time will tick slower because it has higher velocity with respect to the sun. According to SR nobody knows which clock tics faster because of something?

    Now the twin that travelled to B suddenly stops. That will not effect his clock in any way besides that the two twins clocks now start running at the same speed, agreed? ( The twin at B can of course not be inertial any more). The two twins may hold up their clocks with its digital display for the other to see. Since it takes time for light to travel both will see the clock of the other having less elapsed time then their own clock. However if it is acutally the case that the clock of the twin that have travelled to B have ran slow, the two twins will notice it because the difference between their own time and the time they see that the other twin has is not the same for both twins...
  9. Sep 18, 2011 #8


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    No. At any point you can choose which twin is stationary, by choice of coordinate system. Velocity is relative. However, acceleration is absolute, so you cannot choose whether or not a twin is inertial. They are either inertial or not, and all frames will agree. No choice of coordinate system will change an inertial twin into a non inertial twin nor vice versa.
  10. Sep 18, 2011 #9
    First scenario. We let both twins stay stationary at A. This is the inertial frame. One twin suddenly accelerates to a speed of 100 km/s and travels at that velocity until he reaches B.

    Second scenario. We let both twins be at A both moving towards B at a velocity relative to the sun of 100 km/s. This is the inertial frame. One twin suddenly stops as compared to the sun, but accelerates with 100 km/s relative to the restframe. The other twin stays with his inertial frame until he reaches B.

    To me these situations should be physcally equivalent as far as the tickingrates of the twins clocks go, even though the intertial twin and the accelerated twin swithces place in the two scenarios. But you think otherwise?
  11. Sep 18, 2011 #10


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    I have drawn spacetime diagrams of the two scenarios you gave in post #1. As you can see they are almost identical apart from the very beginning, at the bottom.

    In either diagram the red line is noticeably longer than the blue line. In spacetime diagrams it's a general rule-of-thumb that the longest line represents the shortest time. So in both cases the red observer takes less time than the blue observer. The difference between the two scenarios makes very little difference.

    Note that when DaleSpam says "neither twin is inertial" what he means is that neither twin is inertial throughout the whole journey. Each twin is inertial at those times when he is not accelerating, but there is no inertial frame in which either twin is at rest for the whole journey.

    Attached Files:

  12. Sep 18, 2011 #11
    This was interesting and new to me. In general relativity the frame of someone "standing still" by using rocket engines or by resting his feet on the surface of some planet no longer constitutes an inertial frame?

    If you dig a cave in the centre of the earth you should be able to flow around stationary without using any force and midways between two stars they might pull you with equal but opposite forces so you do not have to use force to stay stationary.

    I do not do Riemann algebra. I did studie for four and a half years and have a degree in engineering physics. I am really just suggesting two things, and if someone will give me a counterexample I would be happy. If you send a vechicle out in a spherically symmetric gravitational field with an atomic clock, the assumption that will give correct time for that clock at all times is that the time varies with velocity and distance in relation to the centre of the spherically symmetric gravitational field. Setting the velocity of light constant (or radially varying depenging how you see it) with respect to the center of the same gravitational field will give correct result for for instance the gps-system.

    Now on Earth that would be the centre of the earth and out in the solar system, where the influence of the planets can be ignored (to the extent they could be ignored) that would be the gravity field of the sun.
  13. Sep 18, 2011 #12
    Hmm... nice diagrams... However, in the second scenario, we start off with the two spaceships moving with 100 km/s in relation to the sun as our inertial reference frame. In that frame, the twin that ends up at pluto does not move, instead he stays put and Pluto, for some reason, comes towards him. Similarily, the twin that seems to stay put at A actually chases the sun by 100 km/s, however the sun moves away at the same speed so he does not come any closer...

    That space-time diagram would look different.

    Well I understand that but I never intended to imply that any of the twin would be inertial throughout the journey.
  14. Sep 18, 2011 #13
    Why did you even put the Sun in the example if you are ignoring the gravitational potential? Remove the sun, put both out in open space away from all other bodies and restate the experiment.
  15. Sep 18, 2011 #14


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    There is no such thing as "the" inertial frame. There are an infinite number of inertial frames, all related to each other by the Lorentz transform.

    No, I don't think otherwise. They are equivalent. In each case there is one non-inertial twin with a delta-v of 100 km/s and one inertial twin.
  16. Sep 18, 2011 #15


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    As drawn from the initial inertial frame of the red observer, it would look something like this, below. (The diagrams aren't accurately to scale, in fact I've exaggerated the relativistic effects by choosing an even higher speed than 100 km/s.)

    It's still the case that the red line in the diagram is longer than the blue line, indicating a shorter time.

    Attached Files:

  17. Sep 18, 2011 #16


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    Something like the one I've attached.

    But it doesn't really matter which frame you examine things from; they will all agree to the same thing in terms of which clock shows less time when they meet up.

    That being the case, it makes sense to choose the easiest one. So let's choose the Sun's rest frame.

    Stage 1. both clocks are moving with respect to the frame at the same speed and show the same time dilation.
    Stage 2. Clock 1 comes to rest with respect to the Sun and shows no time dilation while clock 2 continues on at the same speed showing time dilation.
    Stage 3. Clock 2 come to rest with respect to the Sun. Both clocks show no time dilation.
    Stage 4. Both clocks move towards each other at the same speed. They show the same time dilation.

    For three of the stages: 1,3 and 4, the two clocks experience the same time dilation and thus accumulate the same time. Only in stage 2, do the clocks run at different rates, with clock 2 running slower, so during this time, Clock 2 accumulates less time. So at the end clock 2 is the one that will show less time when the clocks meet up again.
  18. Sep 18, 2011 #17
    Well good then, in both scenarios both twins begin in one rest frame of reference. Then one accelerates up to 100 km/s relative to that frame. After a while either, depending on the scenario, the twin that has not undergone any acceleration end up at Pluto or the twin that has undergone acceleration ends up at pluto.

    Now, up to the time when one of the twins end up at Pluto, whose clock have been ticking slower? The twin that has undergone acceleration? The twin that has not undergone acceleration? The twin that has been travelling with the highest velocity relative to the sun?

    Now if the twin that encounters pluto accelerates so that the twins have no relative velocity nothing special happens with the time elapsed on his clock, but the two clocks now tick at the same rate.

    Which clock will now show more elapsed time? If both twins hold their clock up in the air, so the twins can calculate (elapsed time on my clock minus elapsed time on my twins clock), which twin will calculate the largest difference?
  19. Sep 18, 2011 #18
    I am actually a bit worried... The question is, will the difference in time elapsed always be the same in both scenarios? I wonder if a dare do the math... It is not enough that one of the twins show more elapsed time in both scenarios, the amount of more elapsed time must be the same...
  20. Sep 18, 2011 #19


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    There is no frame invariant answer to that question. You need to specify the reference frame in which to make the comparison. In any given frame, it is the clock which is moving faster which has been ticking slower.

    Same as before, there is no frame invariant answer to that question.
    Last edited: Sep 18, 2011
  21. Sep 18, 2011 #20
    Well formally, you have to prove that the difference in elapsed time will be the same, no matter what frame one choses before you can really draw such a conclusion... After I get some sleep maybe I will get into that if no one does it for me.

    Okej, I write your version of scenario two, with respect to the rest frame of the twins at the beginning.

    Stage 1. Both clocks are at rest. Pluto is approaching by 100 km/s and the Sun is moving away by 100 km/s. The clock ticks at the same speed they are not time dilated.
    Stage 2. Clock 1 accelerates and chases after the Sun by 100 km/s. Clock 1 is now time dilated according to SR. Clock 2 is at rest.
    Stage 3. Pluto comes up next to Clock 2. Clock 2 chases after the Sun at 100 km/s. Both clocks now ticks at the same rate. Both clocks are at rest with respect to the sun. Both clocks are moving with 100 km/s relative to the rest frame.
    Stage 4. Clock 2 increases his velocity in the direction of the Sun by 100 km/s. Clock 1 increses his speed by 100 km/s towards him. Clock 2 now moves by 200 km/s relative to the inital rest fram and clock 1 is at rest. According to SR Clock 2 is now time dilated but not clock 1.

    Now as both scenarios must be equal.
    A. During stage 2. Which clock actually ticks at the highest rate? Is this a question that cannot be answered under special and general relativity? "It depends"?

    B. At stage 3 both twins can compare what their own clocks show with what they see on their distant twins clock. Which twin will calculate the lowest difference in time for the quantity (time as shown on my clock minus time as shown on my twins clock)?

    C. After the clocks come toghether again, the twin that made a close encounter with Pluto will show less elapsed time. Is this difference the same, according to both the rest frames we chose above?

    Does the prescense of the sun somehow break the symmetry? (Now I will give this a rest for a while.)
  22. Sep 18, 2011 #21


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    No, the presence of the sun is irrelevant since we are neglecting gravitation.
  23. Sep 18, 2011 #22


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    The conclusion is unavoidable unless you are willing to except physical contradiction.

    I'm an observer at rest with the Sun, I watch the clock and see that they started together and met up again with clock 2 having elapsed less time by a certain amount. We will assume that the clocks stop at this moment preserving their respective reading.

    You're an observer in the initial reference frame (moving at 100 km/sec with respect to the Sun). I stay in this reference frame until the two clocks reunite and stop running.

    A third observer travels with clock 1 and fourth with clock 2.

    Now, after the experiment is over we bring all the observers and clocks together. Since the clocks have frozen in their readings from the instant they reunited, bringing everyone together again will not change these reading.

    So unless everyone agrees as to what the clocks read when they all come together, there will be a contradiction. You can't have four observers sitting in the same room as the clocks all seeing different clock readings.

    actually, clock 2 now moves at 199.9999111... km/sec relative to the the initial rest frame. (addition of velocities theorem)
    yes it depends. From the initial frame clock 1 runs slow, and from the frame of clock 1, clock 2 runs slow. At this point there is no meaning to which one "actually" run slow.
    yes, as I explained above.
    Last edited: Sep 18, 2011
  24. Sep 19, 2011 #23


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  25. Sep 19, 2011 #24
    Forget the sun. Instead, put in inertial reference frame 3 traveling in which both observers are stationary at the beginning of the experiment.

    Experiment starts at Step one for both cases. Both clocks read the same.

    Step two. One observer accelerates in both experiments - his clock will slow down relative to frame 3.

    Case one, step three. The accelerating observer slows down. His clock runs slow relative to frame 3. His clock also ran slow when he initially accelerated, so his clock is definitely slower than the non-accelerating guy.

    Case two, step three. The other observer accelerates. Now his clock slows downs relative to frame 3. But all this time, the first guys clock is running slow (relative to frame 3) because he has a velocity already. So the second guy's clock eventually runs at the same rate after he finishes accelerating, but his clock was always running faster than the first guy. The first guys clock is still slower than the second guy.

    Step four, both cases. They both accelerate toward each other. This step is irrelevant. In either case, the two guys are stationary relative to each other at the start of step four. So, if they accelerate equally toward each other, neither clock will gain an advantage. The relative clock readings are given at the end of Step Three.
  26. Sep 20, 2011 #25
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