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

I Twin Paradox Help

  1. Sep 4, 2016 #1
    Hello. Thank you to everyone who helped me with my previous post where I had general questions on special relativity. Several users suggested that I start by trying to understand the relativity of simultaneity which I did. Although I haven't mastered it, I understand it better now. Also, someone on another post recommended the book "Relativity Visualized" by Lewis Epstein which was also very helpful and uses R.O.S. as the foundation.

    Regarding this post, I have been stuck on trying to understand the twin paradox. I've been trying to work up to an understanding of it. However, I've watched and read probably over ten different online lessons explaining the twin paradox, but I still don't understand it. Here are a few things that I'm trying to clarify:

    1. In Epstein's book and in other lessons that I've seen, he uses a space time diagram to try and illustrate why one twin ages less than the other when they meet again. I'm not sure how to draw a space time diagram in here (I'd appreciate help on that as well), but the diagram is very simple so I'll just describe it: Earth twin has his X and Y coordinate system. He see's space twin get off into a rocket. On earth twin's diagram, he will see space twin moving at some angle upwards and also to the right. That will be represented by a diagonal line moving in the positive X and Y directions. Then, on the return trip, the earth twin will see the space twin moving in a diagonal line upwards and to the left. When the space twin returns to earth, the earth twin will have aged more because he went straight through time up the Y axis, whereas space twin went in a bent line to the right and to the left back to the Y axis.

    That all sounds logical, but the question is how would space twin's diagram of earth twin's "journey" look like (since earth twin is the one moving from space twin's perspective). It seems to me that it would be a mirror image of earth twin's diagram. Space twin would see earth twin first moving up to the left, and then, on the return trip, up and to the right. If that's the case they should be the same age when the meet again. If we drew side by side a space time diagram for each twin, how would those diagrams look compared to one another?

    2. Can this all be explained with special relativity? If so, are there explanations that don't involve acceleration and the Doppler effect?

    3. As a related question to question 1, I watched the minute physics video on the twin paradox:
    Firstly, he says, as an answer to the asymmetry problem, that time rotates. I am very confused by what it means for time to rotate?? When he draws the diagram from the space twin's perspective, the lines are at a different angle then the lines from the earth twin's perspective. Why is that? Also, he draws the diagram from the perspective of how the earth twin perceives the time of the space twin, and then compares that to how the space twin views his own journey's time (proper time). Shouldn't we be comparing the earth twin's perspective of space twin's time with space twin's perspective of earth twin's time, and not with space twin's perspective of his own time?

  2. jcsd
  3. Sep 4, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First, I'd forget about that video and all like it. It's impossible to learn anything subtle from a video in the style of a corn-flake advert, with everything gabbled at break-neck speed.

    Second, you can only really understand the twin paradox once you have nailed the basic concepts of SR. It's not the other way round: you can't (IMO) use the twin paradox to help learn the basic concepts of SR.

    Third, if you can't understand the basic concepts of SR (spacetime diagrams, ROS, Doppler effect) from a formal text, then it's hard for anyone on here. Your question seems to me to reveal (I'm sorry to say) that you haven't really grasped the basics.

    The key point you're missing is that the twin who accelerates is consantly changing their (inertial) reference frame (during the acceleration and deceleration phases), so they cannot use the same inertial reference frame/spacetime diagram throughout. The twin who stays at home can, of course, use the same inertial frame throughout.

    One simple approach to the paradox is to trust the stay-at-home twin's calculations (you know you can trust them, because you've nailed the concepts of inertial reference frames and time dilation). And, you can simply say that you don't know how to do the calculations for the moving twin because of the periods of accelerating and decelerating reference frames.

    It's nice, of course, to be able to analyse the problem from the travelling twin's perspective. But, logically, that isn't necessary. If you can prove it from the stay-at-home twin's perspective, you don't have to repeat the proof another way. One proof is enough.

    However, ...

    In my view, one simple way to understand things from the travelling twin's perspective is to analyse what they actually see (this is essentially just the Doppler effect calculated by hand, as it were). For example, if B moves at ##\frac{4}{5}c## for a distance of ##4## light years, hence ##5## years (in A's frame), then B sees only ##1## year pass on A's clock on the outward journey and ##9## years pass on A's clock on the return journey. And, because of length contraction, the journey time out and back for B is only 6 years in their own frame.

    If your algebra and arithmetic are up to it, you can work all that out on a bit of paper for yourself. Or, you can trust the Doppler equations, which are effectively what you are verifying.

    In any case, this verifies what you know from the analysis in A's frame: that 10 years have passed on A's clock and only 6 years on B's during the out and back journey that B made.
  4. Sep 4, 2016 #3
    The space twin cannot draw one Minkowski diagram of Earth twin's trip because the space twin is not inertial for the whole trip. He has to draw two separate diagrams, one covers the outbound trip until right before the turnaround, the second will cover the inbound trip right after the turnaround. But to figure out how to put the two together you'd need some math. Do you know how to use the Lorentz transformations? Try a numeric example.
  5. Sep 4, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    PS perhaps the most fundamental asymmetry between A and B lies in the distance between the starting point and the turnaround point. A will measure the distance between these as ##4## light years throughout. Whereas, B will initially measure this distance as ##4## light years, but after his/her acceleration to ##\frac{4}{5}c## will measure the distance as only ##2.4## light years. When he/she stops at the turnaround point, the distance is once more ##4## light years and so on. Hence, there is no doubt that B is changing his/her IRF and A is not.

    And, perhaps, that alone shows why the situation between A and B is not symmetrical and there is actually no paradox.
  6. Sep 4, 2016 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  7. Sep 4, 2016 #6


    User Avatar
    Science Advisor
    Gold Member

    I think this is the point that the OP is alluding to that is unsatisfying within SR. Yes, the traveling twin accelerates because it's motion is referenced to the rest of the universe. But how does measuring velocities wrt the rest of the universe make the "stationary" twin's inertial reference frame better? Yes, the traveling twin can feel his acceleration and deceleration because his velocity is changing wrt the rest of the universe. But why does that happen? Einstein wrote about these type of questions as motivation for GR. I have seen a thread somewhere in this forum that discussed this in depth.
  8. Sep 4, 2016 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No need to look outside at the rest of the universe.
    Suppose each ship has within it a ball sitting on a level frictionless table.
    While the stay-at-home twin's ball will remain in place, the traveling twin's ball will not when that twin turns to return home.

    In the most general version of this twin paradox, the point is that
    elapsed wristwatch time between separation and reunion events depends on the worldline path between those events.
    The inertial observer will log the longest elapsed time.
    (The Euclidean analog of this is that the arc-length along a path joining two points is shortest for the straight-line path.)
  9. Sep 4, 2016 #8
    When I worked out this example, at 4/5c, γ would be 1.667. So, if 5 years went by for A, 3 would go by for B. On the way back, I think, another 5 years would go by for A, and another 3 for B. That's all from A's perspective.

    From B's perspective, if 3 years went by for him, then 1.8 years will have gone by for A during the outward journey if I did it right. On the way back, another 3 will have gone by for B, and he'll see another 1.8 pass for A. So, when they come back together (and I'm sure I'm wrong on this based on how it sounds!), A will think he's aged 10 years and that B has aged 6, and B will think that he's aged 6 years and that A has aged 2.6 years! When A looks in the mirror, 10 years have gone by, but when B looks at him, he's only aged 2.6 years. That doesn't make sense to me but when I do the calculations it's what I get.

    How could B see 9 years pass on A's clock on the return journey when he only saw three years of his own time passing? This is the opposite effect of time dilation. Should he be seeing less time pass for A since A is moving relative to him? If the answer involves the fact that B is accelerating, I thought the twin paradox could be understood without taking the acceleration factor into account (e.g. if we assume the period of acceleration was very small). If I'm wrong, and we do need to understand the effects of acceleration on the problem (which would involve changing gamma factors during the trip), please let me know. Wouldn't that be getting into general relativity?

    But then how can the earth twin draw one diagram? The space twin is accelerating away from him just like the earth twin is accelerating away from the space twin?

    I just took a look. I think at my level it will take me time to understand. It uses hyperbolas which is complicated for me. Even the two dimensional graphs present me some problems! For example, in a stationary frame's coordinate system, the X and Y axis are at a 90 degree angles from one another like we usually see. If another frame is moving relative to the first coordinate system, its X and Y axis will be bent inward. Wont the person moving, however, see their X and Y axis the same way the person in the first frame saw his axis (i.e. at 90 degree angles from one another)? I'm assuming its the person in the stationary frame that sees the person in the moving frame as having tilted axis?

    Epstein, for the first few chapters, helped a lot. I stopped understanding when he got into the twin paradox.
  10. Sep 4, 2016 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The hyperbolas are there, just like circles are there in Euclidean geometry.
    However, the insights I referred to don't make explicit use of the hyperbolas.
    In my case, I have developed an alternate structure to encode the hyperbolas.
    The result is an accurate, physically-motivated visualization of the time elapsed along a piecewise-inertial worldline
    without explicitly using the Lorentz Transformation formulas.

    In my opinion, 2D graphs are essential. (Doing geometry with just words is difficult.)

    In fact, you can play with a GeoGebra file I created for a talk:
  11. Sep 4, 2016 #10


    User Avatar

    Staff: Mentor

    It can be, and indeed that is by far the easiest way to understand it. Once you understand it under the assumption that the period of acceleration is very small (essentially, B's path in the space-time diagram is two straight diagonal lines forming a sharp angle at the turnaround point), you understand it. Including the time spent accelerating (so that B's path is a hairpin curve around the turnaround point) complicates the calculation without adding any new physical insight.

    If you haven't seen http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html, take a look at it now. Pay particular attention to the "time gap" and "spacetime diagram" sections.

    And here is a question/hint for you.... B sees three years pass on his clock before he reaches the turnaround, so if we assume near-instantaneous acceleration, one nanosecond before the turnaround B's clock reads three years minus one nanosecond. One nanosecond after the turnaround, B's clock reads three years plus one nanosecond. Using the frame in which B is at rest, what does A's clock read at the same time that B's clock reads three years minus one nanosecond? Using the frame in which B is at rest, what does A's clock read at the same time that B's clock reads three years plus one nanosecond? Note that these are different frames because B's velocity relative to A is different in the two cases.

    (And note also the appearance of the phrase "at the same time", which suggests that relativity of simultaneity is involved. It usually is).
    You are not wrong, and no general relativity is involved.
  12. Sep 5, 2016 #11


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I really don't know what it will take to make you understand that the situation is not symmetrical and only one twin is accelerating. Acceleration is absolute, not relative. The Earth twin remains in a single IRF throughout; the space twin does not.

    Until you understand that, you will just keep repeating the same fallacy that each twin can measure the other's motion in the same way. And then patently there is no way out of the paradox.

    How you measure things from the space twin's perspective is problematic. But you really must stop insisting that the space twin remains in a single IRF throughout.
  13. Sep 5, 2016 #12


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Whether yet another explanation will help, I'm not sure, but here's my take on how ROS (leading-clocks lag) allows you to analyse the motion from the space twin's perspective.

    First, you have A & B at rest on Earth and C at rest a distance of 4 light years away at the turnaround point. Initially, A, B & C all share the same IRF.

    But, after B has accelerated to ##\frac{4}{5}c##, B has changed his IRF and in B's reference frame things are no longer the same. A and B are still (approximately) colocated at the start and still at time ##t = 0## (approximately) in each of their reference frames. This is "when" B starts the journey, and in B's new IRF.

    1) C is now only 2.4 light years away.

    2) The time at C in the A-C IRF is not ##t=0##. From B's perspective, when he starts his journey, the time in the A-C frame at C is ##t_C = \frac{16}{5} = 3.2## years.

    In other words, from B's perspective, the clocks at A and C are no longer synchronised. And the time at C (in the A-C IRF) has already advanced by 3.2 years. If you don't understand this, you need to go back and master ROS.

    Now, B's journey takes 3 years in his IRF and the clocks at A and C advance only by 1.8 years according to B (normal time dilation). So, when he arrives at C (and before he slows down), the clocks at A and C read 1.8 and 5 years respectively.

    If he slows down quickly, then his clock reads 3 years when he stops, C's clock reads 5 years. And now, because once again B has changed IRF's, he finds that A's clock also reads 5 years, as A, B and C once again share the same IRF.

    Exactly the same thing happens on the way back, except the roles of A and C are reversed. So, when B arrives back at A and stops, he finds 6 years have elapsed by his clock and 10 years on the clocks at A and C.

    You can also see from this that you don't need the return journey to resolve the paradox. By changing IRF's twice, B has experienced less time than A & C during the outward journey.

    Loosely speaking, the effects of time dilation from B's perspective were outweighed by the effects of ROS. So, although A & C's clocks ticked slower throughout from B's perspective, there was the ROS over the distance between A and C to take into account, and when you account for this, you find that more time passed on A and C's clocks than B's during the journey. (This is all from B's perspective.)
  14. Sep 5, 2016 #13
    This is actually starting to make sense! I'm going to need to think about it more over the next few days because it's still so strange. At the beginning, A, B, and C all saw their clocks at t = 0. Then, once B starts moving at .8c, C's clock all of a sudden jumps from t=0 to t=3.2 years! Can you please also post the general equation for calculating the lag between two clocks which are positioned along the direction of motion?
  15. Sep 5, 2016 #14


    User Avatar

    Staff: Mentor

    C's clock doesn't really jump. All that is happening is that when B starts moving B's definition of "at the same time" changes, and along with it his notion of what C's clock reads "at the same time" that B's clock reads some given value..
  16. Sep 5, 2016 #15


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The "leading clocks lag" rule means that (using this case as an example) as B travels from A to C at speed ##v##, clocks that are synchronised in the A-C frame are not synchronised in B's frame and:

    ##\Delta t = \frac{vD}{c^2}##

    Where ##\Delta t## is the time difference and ##D## is the proper distance (i.e. the distance in the A-C frame) between the "synchronised" clocks.

    In particular, if ##t=0## in the A-C frame when B leaves A, and ##t_b = 0## (in B's IRF) when B leaves A, then ##t = +\Delta t## at C when ##t_b = 0## (at C).

    What this means is:

    If B could carry out an experiment to determine what C's clock read "when" (according to B's IRF) he left A, he would find ##t_C = +3.2## years.

    Whereas, A would conclude that when (according to A's IRF) B left him ##t_C = 0##

    This is a key aspect of ROS and, as has been mentioned before, this perhaps the most difficult aspect of SR to grasp.

    I'm sure a good text book would explain this much better than I have in a few lines here!
  17. Sep 5, 2016 #16

    Mister T

    User Avatar
    Science Advisor

    Time doesn't rotate. The time axis rotates as shown in the diagrams. The reason has to do with the way you draw lines of constant time, they take into account relativity of simultaneity. Two events that are simultaneous are connected by a line of constant time. But if you are moving relative to me, my line of constant time and your line of constant time are rotated relative to each other because events that are simultaneous to me are not simultaneous to you.
  18. Sep 5, 2016 #17
    Thank you.

    Introducing C into the picture was very helpful.

    Well the example got me thinking: Say A and C are two 30 year olds in their IFR, and they are standing the same distance apart like A and C in PeroK's example. When B is at rest relative to A and C, he will also see them as both being 30. Now, if B starts going at at a fast enough speed, and the time it takes him to reach that speed is near instantaneous, then it's possible that he'll see C as a 90 year old and A as still being 30. Also, if I understand it right, C himself would see himself as a 30 year old when B would see him as 90! Does that mean that if someone flew by me fast enough they'd see me as a 90 year old??

    Thank you for clarifying this. I am going to need to learn the rules for drawing the diagrams (e.g. what the relationship is between the speed of the moving frame and the angle that the axis shifts.
  19. Sep 6, 2016 #18


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're still making two fundamental mistakes. The first is that these things are not what you "see", but the results of measurements. The second is that the lack of synchronicity is due to the distance between A and C: B cannot be in both places at the same time. So, B can never directly observe A (locally) and C (locally) at the same time.

    The way I would think about this is to have a fourth person, D, at rest with respect to B. So that B and D share an IRF and D passes C "when" (in the B-D IRF) B passes A. Let's move away from the twin paradox for a moment and assume that B and D have established that they are at rest with respect to each other and (from their perspective) A and C are 2.4 light years apart and moving towards them at ##\frac{4}{5}c##.

    Now, B records the time on A's clock when he passes A; and D records the time on C's clock when he passes C. Then, they communicate the results to each other.

    B communicates: "I passed A at ##t_B = T## (say) and A's clock read ##t_A = 0##.

    D communicates "I passed C at ##t_D = T## (this establishes that B was at A when D was at B - in the B-D frame) and C's clock read ##t_C = 3.2 years##

    This constitutes a "measurement" in the B-D frame of the lack of sychronicity of the A-C clocks. Neither B nor D can do it easily by themselves, but between them they can carry out this measurement.

    Meanwhile, A and C measure their clocks as synchronised in their frame and, in their IRF, B passing A and D passing B were events that were 3.2 years apart. In other words: simultaneity is relative.

    I suggest you go back to your text book and master ROS. I don't think one can "half" understand ROS: you either understand it completely or you don't!
  20. Sep 6, 2016 #19


    User Avatar

    Staff: Mentor

    Go back and read Einstein's train example showing the relativity of simultaneity again. Instead of the lightning strikes, we have passengers A and C celebrating their 30th birthdays at each end of the train, and B is the observer on the train, in the middle. B accelerates by jumping off the train and ending up at rest on the platform.

    And then back to your spaceships...
    We may say that the events "A celebrated his 30th birthday" and "C celebrated his 90th birthday" happened at the same time using the frame in which A and C are moving.

    We also may say that the events "A celebrated his 30th birthday" and "C celebrated his 30th birthday" happened at the same time using the frame in which they are at rest.

    Let's consider exactly what those statements mean. No matter where you are and how fast you are moving, you will eventually see all three birthday celebrations in your telescope. If your clock reads T when you see one of these celebrations and it happened at a distance Q light-years from you, you will say that the celebration happened at time T-Q; you're just correcting what you see for the light travel time.

    Even observers at rest relative to A and C may see the two 30th birthday celebrations in their telescopes at different times, because they may be different distances away. If A and C are one light-year apart, someone 1000 light years away will see one celebration happening when his clock reads 1001 years and the other happening when his clock reads 1000 years. That doesn't mean that the two events happened at different times, it means they happened at the same time and the light took longer to cover the greater distance between the more distant celebration and the telescope.

    However, no one who is moving relative to A and C will ever find that these two celebrations happened at the same time. As in the at-rest case, they will eventually see all three celebrations in their telescope, but when they allow for the light travel time as they calculate what their clock read at the same time that the celebration happened, they'll get different results for the two 30th birthdays.

    Furthermore, if we choose B's post-acceleration speed properly, observers who are at rest relative to B after the acceleration will find that A's 30th birthday celebration and and C's 90th happened at the same time after they allow for the light travel time. It's a good exercise to construct such a situation. (Hint: A and C must be at least 60 light-years apart in their rest frame).
  21. Sep 6, 2016 #20
    Since the two examples are similar, I'll combine them to see if I'm understanding R.O.S correctly.
    1. There are two frames: A-C and B-D. In the A-C frame, D's clock is the lagging clock, and in the B-D frame, A's clock is the lagging one.
    2. The proper length between B and D is less than the proper length between A and C. That must be true because in PeroK's example, according to the B-D frame, when the two frames past each other they were of equal length (A lined up with B and C lined up with D). Since B and D are measuring a contracted length of A-C and getting an = measure, A-C's proper length must be bigger.
    3. If, in the A-C frame, A and C celebrate birthdays on the same day, then in the B-D frame C must be older. When I say "on the same day" I am taking into account the time it takes for light to travel.
    4. In the B-D frame, there comes a point where the nose and tail end of the two frames line up (A with B and C with D). On a two by two checker board, A and B would be the bottom row and C and D would be in the top row. In the A-C frame, however, A and B will cross each other first, and then only later will C and D cross. That's because according to the A-C frame, B and D are compressed together, and it takes longer for D to get to C.
    5. All four of them have cameras. A and C, who are the exact same age in their frame, decide that exactly when they turn 30, they will take pictures of themselves blowing out the candles on their cake. We don't need to worry about light travel time because they took the pictures up close. It just so happens that as A is blowing out the candles, B is passing by and appears in the photo. When C takes his photo, D hasn't passed him yet and doesn't appear in the photo. A and B get together and compare photos. A appears in his photo with B (who is also taking a picture of A as described below), and C appears in his photo alone.
    6. B and D, whose clocks are synchronized in the B-D frame, agreed to take simultaneous photographs when B sees A blowing out the candles (they happen to know A's birthday is coming up and they know when that will be in their frame). They then get together and compare photos. B's photo will show pretty much the same thing (with differences only due to the angle the photo was shot from) that A's photo showed: a thirty year old A blowing out candles. D's photo will show C, but C will be older than 30. How much older depends on the velocity of the B-D frame in relation to the A-C frame.
    Is any of this right so far?
    7. Here is the most confusing part for me: Now say all for of them are in the same reference frame at rest with respect to each other. A and C are 29 and have their birthdays tomorrow. B and D, since they are both at rest with respect to A and C, measure A and C to be 29 also (after adjusting for light travel time). That same evening B and D suddenly start moving with respect to A and C at a speed close to the speed of light. The very next day, A and C will measure each other's age at thirty, but B and D will have measured A at 30, and C at 90. Is this what could happen if B and D went fast enough? If that's actually what could happen, then how can we reconcile that for B and D, C's clock, due to time dilation, is moving slower than their own, but yet, due to R.O.S, it's moving faster than their own?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Twin Paradox Help
  1. The twin paradox (Replies: 1)

  2. Help with twin paradox (Replies: 30)

  3. Twin paradox help (Replies: 52)