Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

There is no twin paradox - mathematical proof

  1. Dec 20, 2007 #1
    For the twin paradox to be considered a true paradox the framing of the scenario must be stringent, that is to say we cannot permit assumptions to be ignored. Therefore I must start with a short description of the twin paradox followed by identification of the inherent assumptions.

    From Einstein Light (http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm): [Broken]

    (The author goes on to explain that asymmetry resolves the paradox, this is an not entirely satisfactory explanation.)

    There are a few assumptions, which are perhaps only obvious when one takes time to search for them.

    "(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.

    By placing Joe on Earth we hide the other assumption, which is that we also share the same frame as Joe.

    The at-a-distance observations of each other's clock potentially hides an assumption of instantaneous transmission of information. I doubt the author intended that, but it must be remembered that information is not transmitted instantaneously.

    "(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.

    Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

    Let me provide an analogous scenario.

    Joe floats in a space suit with two clocks.

    Jane sits at one end of an extremely long structure with another two clocks. At the other end of the structure is a beacon. According to Jane, the structure has a length of L. Joe knows this.

    Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).

    Four events are noteworthy:

    1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

    2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.

    3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.

    4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.

    Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Joe and Jane were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to the third observer (the reader).

    There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part that asymmetry lies.

    Jane's calculations:

    Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.

    Joe's calculations:

    Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this is the case.

    Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.

    The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):

    t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
    =γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
    =γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
    =γ^2.(c^2.L/v + Lc)/(c^2-v^2)

    but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,

    t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
    =(c^2.L/v + Lc)/c^2
    =L/v + L/c

    The same process can be used to calculate that the second period elapsed is L/v-L/c. The total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the same as calculated by Jane.

    There is no disagreement and there is no paradox, merely a poorly frame scenario.



    (If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc and http://www.geocities.com/neopolitonian/sr.doc)
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 20, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I haven't read the whole post, so you will have to discuss the details of your proof with someone else. I would just like to add that no proof is needed. The special theory of relativity is just the claim that space-time can be mathematically represented by Minkowski space. This claim, or any other physical theory for that matter, can't be proved or disproved mathematically. We need experiments to tell us if a theory is an accurate model of reality or not.

    The only time when this is not true, is when the "theory" contradicts itself, but in that case, it would be completely wrong to call it a theory. "Nonsense" would be a more appropriate word.

    Is there any way that SR could be nonsense? Many people don't realize that since SR is the simple claim I mentioned above, the only way a logical inconsistency could be present in the theory is if it's already present in the mathematical definition of Minkowski space. But Minkowski space is just the set [itex]\mathbb{R}^4[/itex] along with a few functions. If you know what those functions are, you also know that they can't introduce contradictions into the theory. So if there are contradictions in SR, they would have to be present in the definition of real numbers. But real numbers can be constructed from rational numbers, rational numbers can be constructed from integers, and integers can be constructed from the axioms of set theory. So if there really was a twin paradox, it would be the downfall of pretty much all of mathematics.

    I might as well take this opportunity to post a space-time diagram that I made for a discussion in another forum.

    http://web.comhem.se/~u87325397/Twins.PNG [Broken] (Why doesn't the "img" tag work?)

    This is the Matlab code that generated the graphs, in case someone wants to play around with it:

    Code (Text):

    clf; axis equal; axis([0 40 0 40]); ylabel('t (years)'); xlabel('x (light-years)'); hold on;

    plot(x,x,'k:'); plot(x,-x+36,'k:');
    plot(x,1.25*x,'k'); plot(x,-1.25*x+40,'k');

    A=6; plot(x,sqrt(x.^2+A^2),'g');
    A=12; plot(x,sqrt(x.^2+A^2),'g');

    A=6; plot(x,20+sqrt(z.^2+A^2),'g');
    A=12; plot(x,20+sqrt(z.^2+A^2),'g');


    plot(x,6,'c'); plot(x,12,'c'); plot(x,20,'c'); plot(x,26,'c'); plot(x,32,'c');

    s=20:0.01:40; plot(16,s,'k');
    Last edited by a moderator: May 3, 2017
  4. Dec 20, 2007 #3
    Goddag yxskaft, Fredrik!

    You might want to read posts before replying, since I make no claim to prove or disprove your statement: "The special theory of relativity is just the claim that space-time can be mathematically represented by Minkowski space."

    SR is entirely safe, I am just trying make people aware that there is no twin paradox and that you can demonstrate that fact mathematically.


  5. Dec 20, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I read enough to understand that. My point is that you don't have to use a thought experiment to refute the twin "paradox". A thought experiment is always carried out within a theoretical framework, and in this case it's sufficient to only think about what the framework is to complete the proof.
  6. Dec 20, 2007 #5
    I am a little unclear on whether you agree that the twin paradox is poorly stated, and hence irrelevant, or do you think that the travelling twin in the "paradox" can explain why she ages less than the nominally stationary twin. Remember the paradox revolves around the idea that both of the twins can consider themselves to be stationary and the other to be in motion. If you state that the travelling twin knows herself to be travelling, then you destroy the grounds of the paradox entirely.

    Your image indicates that B is aware that she is in motion ("The rocket turned around a minute ago"). The image also seems to be fundamentally wrong. A ages 20 years on each leg and B ages 12 years on each leg. There is no 2 minute discontinuity in which A ages 25.6 years. I would appreciate other people's input on that, but there is no requirement that I am aware of for B to return to A in order for relativistic effects to manifest.

    If merely changing directions made you age so significantly then pilots, ferry captains, postmen and bus and train drivers would all be dead and waving your hand would cause it to age significantly faster than the rest of your body. (Procreation would also represent a danger to parts of the male party.)

    It seems that you might benefit from reading my post in entirety since I believe it addresses the twin paradox issue far more safely than the introduction such deadly discontinuities.


  7. Dec 20, 2007 #6


    User Avatar
    Science Advisor
    Gold Member

    'stationery' and 'moving' in special relativity are frame-dependent qualities of the motion of an object/observer, but 'inertial' and 'non-inertial' are absolute qualities of the motion of an object/observer.

    As you've noted which twin is stationery and which is not is pretty much irrelevant, however the twin paradox arises because one twin is inertial and the other is non-inertial.

    When, as viewed by an inertial frame, 'the rocket turns around' it means that the rocket is non-inertial. Whether the rocket is 'in motion' is a moot point, that the rocket is non-inertial is not. (btw the twin that 'turns round' will age less than the other twin).

    When the ageing occurs again is a moot point it's enitrly dependent on how we compare the clocks when they are at a distance to each other and as we know in relativity simulatenity fails with disnatce.

    Simulatneity is still absolute though when thereis no distance involved so as when the twins start and stop the clocks they are conincidnetal we can say absolutely that the clocks have measured different lengths of time between the same events

    The relatvistic corrections for the motion of pilots, ferry captains, etc are small as to be insignicficant (plus as I said the twin who changes directions actually ages less)

    Your post doesn't really address the twin paradox.

    The introduction of discontinous velocites is in the twin paardox is unphysical yes, but it's simply done for ease of calculation as it doesn't qualitively effect the result. The twin paardox can be demsontsrated with finite accelartions, it's just slightly more difficult to model and the basic result is the same.
  8. Dec 20, 2007 #7
    Try reading my first post, ie post #1, rather than my replies to Fredrik.

    By the way, I think your definition of inertial is open to question. I deliberately tried to avoid using the term "inertial" because the "stationary" observer is equally as inertial as the "travelling" observer. Without an absolute frame of refence, you can't talk about absolute qualities.

    But I repeat, try reading my first post, please respond to that rather than responding to my responses to the responses of others.


    Last edited: Dec 20, 2007
  9. Dec 20, 2007 #8


    User Avatar
    Science Advisor
    Gold Member

    I have already read your first post.

    Why delibrately avoid using the term inertial? We're talking about special relativity and both the postulates of special relativity specifically reference 'inertial observers'. If you need me to expand on this I will do.

    Some quantities/qualitities are frame dependent some are not, so it is not wrong per se to talk about absolutes.

    Your first post is erroneous and does not address the twin paradox, precisley because you have not recgnised that both observers are not 'equally as inertial'.

    I am sorry I do not want to go through your first post in fine comb detail, it is rather long. But the 'error' is clear - you haven't recognised you cannot escape the differemce between inertial and non-inertial observers. The equations you use apply to inertial frames only, whereas (at least) one of the twins must be non-inertial.
  10. Dec 20, 2007 #9
    So we have chosen the frame so that the earth is in origo and in rest, and we ignore the orbiting around sun, so that this is an inertial frame.

    I guess you mean to point out, that the distant location [itex]x\in\mathbb{R}^3[/itex] in the spatial space is chosen so that it doesn't depend on time, in this frame. Or if we consider its world line [itex]x(t)[/itex] in the four dimensional space time, it is a straight line upwards.

    In other frames the point [itex]x(t)\in\mathbb{R}^3[/itex] would change as function of time, and in your language "it would not share the same frame".

    Okey, I understood you correctly?

    And then we note, that the Joe is in rest in this chosen frame too.

    We talk about the speed in the chosen frame. Since the Joe and the distant location are rest in this frame, talking about the speed in this frame, or about the speed relative to the Joe and the distance location, are obviously equivalent. This is so trivial claim, that I have difficulty seeing what should be considered as assumptions and what as consequences.

    So far everything you have said seems to be fine, although it was a bit laborous to translate your explanations into more understandable form. I must tell you, that you are underestimating physicists. Your remarks contain nothing new to a person who knows relativity.

    I seriously doubt this! It matters very much who is accelerating. Are you sure you are aware of the fact, that the Jane will be younger than Joe after the trip?

    I think I'll post this now, and see rest of your post next. To avoid too long post.
    Last edited: Dec 20, 2007
  11. Dec 20, 2007 #10
    No, I never said the structure was attached to anything other than Jane and the beacon. I did later say they would move relative to each other in such a way that it is not possible for either to know which was moving. That pretty much eliminates the possibility of rotational movement on Jane's part.

    Given this fundamental misunderstanding. I will not respond to anything else in your post.

    The setup is such that it is analogous to moving between two predetermined, fixed points in space (as Earth and "some distant location" are assumed to be in standard formation of the twin paradox). You can assume that the universe is empty with the exception of Joe, Jane and the structure plus a mischievous god which gives them the relative motion described, if that makes it easier.


  12. Dec 20, 2007 #11
    Check the scenario and the maths which follows, note that I completely eliminated the question of acceleration. Note further that despite eliminating acceleration such that we can never know who accelerated, Jane ends up younger than Joe.


  13. Dec 20, 2007 #12
    I deleted the post now, so that it is not causing confusion anymore.
  14. Dec 20, 2007 #13
    I avoid the term inertial because some people misunderstand it to mean being in motion. It doesn't. A stationary frame also has inertia - inertia is what makes it necessary to expend energy to set the frame in motion. So, the term "inertial observers" basically means that the observers won't change velocity. That velocity can be zero.

    You don't need to expand on this, you need to check your terms.


  15. Dec 20, 2007 #14
    "Joe floats around in a space suit, Jane is located on a long structure, there is a beacon in the other end, and then Joe and Jane pass each others twice with relativistic velocities v and -v."

    I think nobody can understand what is happening there. Explain it with coordinates
  16. Dec 20, 2007 #15
    btw. I took today two difficult exams, and I feel like I've deserved a free evening that I can waste doing nothing useful.

    Wwwhoooo....... okey. :rolleyes: One should always check links in the original post, it seems. Let's see if I manage post this before a lock hits the thread.

    Page 6:

    The use of prime is fully consistent, since the equations 8 and 9 are in fact equivalent, as you can verify by exercising your algebra skills by solving [itex]\Delta x[/itex] and [itex]\Delta t[/itex] out of the equations 8, or by solving [itex]\Delta x'[/itex] and [itex]\Delta t'[/itex] out of the equations 9.

    Page 7:

    You have replaced the usual spacetime interval with a new concept of magnitude of journey, which seems to be the Pythagorean distance in four dimensional space, and then assume that this distance is invariant in boosts.
  17. Dec 20, 2007 #16
    Whose coordinates?

    According to Jane, she sits at (0,0,0,t) and at a distance L away from her at the other end of the structure is a beacon. To make it easier on herself, she nominates the length of the structure as her x-axis so that the beacon sits at (L,0,0,t).

    According to Joe, he sits at (0,0,0,t).

    Event 1 is at (0,0,0,t1).

    Event 4 is at (0,0,0,t4).

    Events 2 and 3 are at either
    • (L,0,0,t2 according to Jane) and (L,0,0,t3 according to Jane) ), or
    • (0,0,0,t2 according to Joe) and (0,0,0,t3 according to Joe),
    depending on whether you listen to Jane or Joe. Note that L is the length of the structure at rest (which it can only be in Jane's frame).

    The issue is that either
    • Joe moves past Jane, along the structure and past the beacon with a speed of v and then later moves back past the beacon, along the structure and past Jane again, also with a speed of v, or
    • Jane and her structure move past Joe in one direction with a speed of v, and then back again in the other direction also with a speed of v.
    Neither of them can say with certainty which of them does the moving. Despite that, the time spent moving past is each other is such that less time elapses for Joe, irrespective of whether he moves or not.

    People will argue about simultaneity, but the mathematics doesn't support that being a problem.

    Try to work through the mathematics, don't trust that I got it right but don't just assume that I got it wrong either. Work it through and come to your own conclusion.

    I hope this clarifies things a little.

  18. Dec 20, 2007 #17
    The use of the prime is consistent between equations 8 and 9, because as you say they are equivalent but only in so much as they are the same equation from the perspectives of two different frames of reference. Please check the use to which the equations are then used, taking these two perspectives from two different frames of reference and then using them in one single frame of reference. That is the inconsistency. I don't claim there is an inconsistency before that usage.

    Not really, I start at the beginning and work forwards, identifying assumptions as I go and eliminating the ones that we all know not to be valid. What I don't do, and I can understand that some people get upset about it, is use the ramifications of other derivations as the basis for my derivation. That is to say, I am not so much assuming that "Pythagorean distance in four space" is invariant in boosts, I merely not assuming that it isn't invariant.

    It was a while since I last worked on that paper, but as I recall I was discussing the motion of one photon and the different interpretations of the photons movement in two different frames of reference. A single photon can only make one journey, irrespective of how it is perceived by different observers. If you want to say that makes Pythagorean distance in four space invariant, then I cannot argue against it.

    Without taking refuge in other derivations, can you explain why Pythagorean distance in four space must not (or should not, or perhaps cannot) be invariant? The mathematics works for my derivation without that requirement.


  19. Dec 20, 2007 #18
    neopolitan if i understand you correctly joe moves past jane with a velocity u (in janes frame) and somehow returns later though the mechanism for such reversal is left unkown.

    the problem with this is that the only way for joe to return would be if there was a gravitational field to bring him back (the only way to keep him inertial), however in this instance joe would be effected by gravitational time dilation effects which would be unkown) and if joe ever underwent acceleration than the equations of special rleativity take on a different form, and his time would run slower.
  20. Dec 20, 2007 #19
    Hi Luke,

    Did you check the first post, in that I make clear that I am only measuring the time elapsed between two pairs of events. Anything that happens outside of the time which elapses between each of the pairs of events is immaterial. Yes, Joe or Jane may be affected by any gravitation used to turn them around, but their clocks will be stopped during that time.

    The twin paradox is usually framed in such a way that you are only talking about two inertial journeys, one there and one back. I just make that blatantly obvious and show that you expect one of the twins to experience more time elapsing than the other one.

    The mathematics, if you work it through, shows that the twins will not expect anything else.


  21. Dec 20, 2007 #20


    User Avatar
    Science Advisor
    Gold Member

    'inertial observers' are referenced in the postulates of relativity so the term needn't be avoided. Inertial observers do not experince acceleration so do not change direction (as viewed by another inertial observer).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook