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Twin paradox question

  1. Nov 22, 2011 #1
    Lets say we have triplets on earth A B and C.
    B and C goes from earth together in the same direction at some relativistic speed lets say lorentz factor one milion.
    Now after one year passes from the point of A, brother B decides to stop.
    Now after one more year from the point of A passes brother C stops too.
    Then both B and C sets course to meet each other at midway at some nonrelativistic slow speed.
    the question is: When B and C meets which one of them is older?

    if your answer is "B is older than C" then please explain why, because from viewpoint of C the B was one who was accelerating away from him. So im thinking this solution is violating the postulate about physical laws being same independently of uniform motion (that violation being someone accelerating away from you is actually aging faster than you).

    if your answer is "C is older than B" then please explain how is this possible from viewpoint of A. From my understanding of the formulas the longer distance you go in relativistic speed the more time you avoid (from the viewpoint of A). So im thinking this solution is violating the same postulate.

    Thanks for answers.
  2. jcsd
  3. Nov 22, 2011 #2


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    First, you left out a small piece of the problem specification: what is B doing during the one year (from A's viewpoint) after he stops and before C stops?

    Second, have you tried to actually calculate the answer? Try it in A's frame first.
  4. Nov 22, 2011 #3
    In mentioned time B just sits there keeping the same distance from A.

    No i haven't but im starting to feel i will have to.
  5. Nov 22, 2011 #4


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    That's what I thought but I wanted to make sure.

    Well, given the clarification just above the answer should be obvious from A's frame: B is only moving for one year from A's viewpoint and then is at rest for one year, while C is moving for the full two years; after that B and C both age at the same rate.

    Now try calculating the answer (or at least outlining the calculation) from C's frame. (For bonus points, you can do it from B's too, but doing it from C's should clear up the issue in your OP.)
  6. Nov 22, 2011 #5


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    On re-reading the OP I should clarify this point. I'm not sure which "postulate" you are referring to, since there are several possibilities:

    (1) There is a postulate that says that physical laws are the same for all observers, regardless of their state of motion. But this postulate does *not* translate into any simple statement like "people who are accelerating away from you always age slower". You have to look at the actual details of each case--in particular, since you can only directly compare ages with someone when you're at the same location, you have to consider the full motion of the other person, not just the part where they're accelerating away. (Also, as you've stated the problem, B and C spend almost all their time in uniform motion, not accelerated; the only accelerations are the brief periods where they start and stop, relative to A. So you can really arrive at an answer without considering acceleration.)

    (2) There is a postulate that says that uniform motion can't be distinguished, locally, from rest; but that only applies to observers that aren't accelerating (i.e., that don't feel acceleration).

    (3) There is a postulate that says that acceleration can't be distinguished, locally, from a gravitational field; but that is only relevant where there is gravity involved, and my understanding is that you don't intend there to be any gravity in this problem.
  7. Nov 22, 2011 #6
    One important thing to point out is that in STR the acceleration is not relative.

    Einstein's principle of relativity states that 'physics are same' in two inertial systems.
    In principle it is possible to distinguish whether you are accelerating or not.

    Edited: Actually, the more 'important' acceleration in this 'paradox' is when the C is slowing down. The quote was not meant to be specific.
  8. Nov 22, 2011 #7
    Well sorry but thats how i get it, in fact thats exactly how i get it. I cant see how age difference comparable purely between B and C depends on their speed to some A.. If it was true then nothing would be relative. Ergo the principle of relativity wouldnt work because physical laws would be dependent on their state of motion.
    Also to solve their location problem i said after the experiment they move slowly to each other... So after 2 years into experiment from point of A their age difference must be constant no?
  9. Nov 22, 2011 #8
    So it is possible for B to tell the difference between
    acceleration with respect to C and deceleration with respect to universe?
  10. Nov 22, 2011 #9
    It does not.
    How it looks from the instant when the B stopped?
    The C is moving away from him at great speed. The C is 'aging' slower than him. The 'C' does see it same. (The B is moving away from him and is aging slower.) But the one who will 'slow down' (one who will change his speed) is the C. From viewpoint of B nothing special happens, he is in the inertial frame the whole time. But from C's viewpoint his decceleretion appreciable.
    One thing, especially important for this gedanken experiment is that notion of simultaneity (of presence) is changing. In the process of decelaretion, from C's viewpoint, the B has aged considerably.
  11. Nov 22, 2011 #10
    I am convinced it is. For example the Newton's first law does not hold in non-inertial frames.
    (Although using it is not the best procedure to determine inertial frame, since it is hard to determine whether the body is free.)

    The wording 'with respect to universe' is not the best.
    In STR it is postulated that there exist special, so called inertial frames of reference. They are equivalent in some sense. They are 'moving' with constant speed with respect to each other. But there is not any special one, the frame of universe.
    And in principle, you can tell difference between 'being in' inertial and non-inertial frame of reference.
    Last edited: Nov 22, 2011
  12. Nov 22, 2011 #11
    Yes nice, sounds reasonable, but in order to stop it is B who must turn his engines ( or brakes if they could work in vacuum) on and can feel the acceleration . So then again if they were to meet which would be older? I thought it has to be the one twin who stays home( doesnt experience acceleration) But that clearly doesnt make sense from viewpoint of A for me :[
  13. Nov 22, 2011 #12
    Yes, B stops, and during this deceleration he also 'sees' this change in what he perceives as the present. But since, during the process of slowing down, B and C are at (nearly) the same position, it does not have an appreciable effect with respect to 'relation' between B and C.
    (It does, however, matter, once we are interested how A looks from B's point of view.)

    Actually, I don't really know what is your 'working knowledge' of STR. To understand what is going on you should have some qualitative idea how the line t'=0 (or t'=const) look like in unprimed coordinates for different speeds of primed frame. But, generally, if you are 'looking' from the inertial frame the whole time, you should get the right answer. (So you can compute it from A's point of view. And then think why/how it is consistent with what the B and C do see.)
  14. Nov 22, 2011 #13
    My knowledge is very limited. Thats why im here i guess :D I had a physics course in one term. Just to be clear only one thing i wanted to compare whas their age when they meet. Not what they see during experiment is happening to others. Anyway so whats the right answer? whos older? B or C ? :D
  15. Nov 22, 2011 #14
    What is going on during the process may be important to understand why you are getting the answer you got.

    But once again, in special theory of relativity, if you use the inertial frame for your calculations, you should get the right answer. So you can compute the thing from A' viewpoint to answer your question.

    In this case B should be older. But it is good to think it over by yourself, I may be wrong, and even if I am not, in this case process of learning why the answer is what it is is more important than answer itself. (I'm going to get some sleep now, so even if you are convinced that my answer is wrong, I won't be able to continue the discussion until tomorow. Bye.)
  16. Nov 22, 2011 #15

    No i dont think youre wrong i just can find where i am wrong :D Thinking about it im starting to think i underestimated the effect described by PeterDavis, that is what happenes when they are moving to meet each other to compare their ages.
    So im gonna outline my assesment from point A:
    at t=0 both ships start moving. (clock on A)
    At t1= 1 year. lorentz factor = 10^6 they move about one light year.
    then B stops at xb=1ly and C continues to go one more year to distance xc=2ly.
    In that point clock on ship B would tick 1 year / 10^6 + 1year times which is
    tB = 1year + 31.5s
    on ship C tC= 2 years / 10^6 = 31.5s*2= 63s
    so B would now appear about one year older than C;
    now they start to move to each other to match their ages ( We need them in one time frame again so we may compare their ages ). Their mutual distance is 1 ly. If they go to meet each other halfway and their speed is same then their time dilation must be same. So they meet and together they go back to earth and B is one year and 31.5s older than C.

    Now consider situation on ship C.
    They are stationary with B for some time, then they are running almost the same experiment as A did, only they are sending B away in the opposite direction.. yet when they meet with B, its B whos actually older.. Dont you see that as a problem ? Im going to have to speak with some physicist from school on that one
  17. Nov 22, 2011 #16


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    Did you try to do as I suggested earlier, and actually calculate (or at least outline the calculation of) what things look like from C's point of view? I'm guessing not, because your description above leaves out one highly significant fact, which should have become evident to you when you tried to actually work the problem: when C stops, relative to A, he *changes inertial frames*. (B does too, when he stops.) However, A remains in the *same* inertial frame the whole time. So the first calculation, from A's point of view, is straightforward: we run all the numbers in a single inertial frame, and we find, as you have, that C ages less than B, because he spends more time moving, in our single inertial frame, than B does.

    To work the problem from C's point of view, however, we have to face up to the fact that there is an event on C's worldline at which C switches inertial frames; so we simply can't run the straightforward analysis you describe from C's point of view, because that analysis assumes that we are in the same inertial frame throughout. (The same issue arises in the common "twin paradox" scenario, which has appeared quite a number of times on these forums.) Instead, to analyze things from C's point of view, we have to do it in two steps: first, the segment where C is moving away from A (and from B, after half the segment is done); and the segment where C and B are moving back together.

    The first segment is basically what you're describing above; C and B move together for half the segment, then B, from C's point of view, starts moving very, very fast in the opposite direction. However, from C's point of view, this segment doesn't last very long, because C is moving with a Lorentz gamma factor, relative to A, of about a million, so a year from A's point of view is only about 30 seconds from C's point of view. So the first segment of C's journey has C moving along with B for 30 seconds, then B moving away very fast in the opposite direction (Lorentz factor of a million) for 30 seconds, and then that segment is *done*.

    Now C turns around and starts moving back towards B. When that happens, C changes inertial frames; and that means that what B is doing "at the same time" changes drastically as well, because C's line of simultaneity changes. Right before he turned around, B was moving away very fast from C's point of view (in C's original frame); but right after C turns around, B is now stationary from C's point of view, and has been for about a year, in C's new inertial frame (which, since both B and C are moving very, very slowly from this point on, relative to A, is basically the same as A's inertial frame, in which B has been stationary for a year when C turns around). Another way of putting this is that the point on B's worldline which C considers to be "now" to him moves *forward* by a large amount (about a year) when C turns around. When C calculates what age he expects B to be when they meet up again, he has to take this into account; and of course that makes him predict that B will be older by about a year when they meet (because that about-a-year portion of B's worldline corresponded to just 30 seconds for C).
  18. Nov 22, 2011 #17


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    Here's the thing: When B or C accelerates( acceleration being any change of speed), it changes how they measure the universe around them. Clocks that in same direction from them as they are accelerating run faster and ones in the opposite direction run slower. The greater the distance between them and the distant clock, the greater the difference. This is due to the Relativity of Simultaneity.

    When A stops, it is accelerating (towards the Earth), but since B and C are right next to each other when this happens, B does not see this as effecting C.(assuming the acceleration is nearly instant.)

    However, when C stops, it is accelerating towards B, and it is separated from B by a great distance. This causes B to age very rapidly according to C while he is accelerating. So even though B aged more slowly than he did between the time of B stopping and and C stopping, after C stops, C will expect B to be older than himself, just like B expects C to have aged less. If you now bring B and together slowly, they both will agree that B is older.

    More Often than not, it is neglecting the effect of the Relativity of Simultaneity that ends up causing the problem when approaching these types of questions.
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