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

I Simultaneity and the Twin Paradox

  1. Jul 30, 2018 #1
    Here is a diagram from the Wikipedia. I'm sure you understand what this represents. Those who don't, can go to the Wikipedia page (https://en.wikipedia.org/wiki/Twin_paradox) and see all the details.
    485px-Twin_Paradox_Minkowski_Diagram.svg.png
    After my prior post, I have a better appreciation for simultaneity. Originally, when I looked at this diagram, I looked at it using the common meaning of "simultaneous". I thought it was telling me that where the red and blue lines cross the space time lines of the stationary and traveling twins, the twins exist in the same moment from both twin's perspectives.

    My insight is that the correspondence is just for the traveling twin. The stationary twin will draw different lines. The word (on the chart itself!) "simultaneity" really tripped me up. I'll be honest; even when the diagrams included the stationary twin's idea of simultaneous, I didn't really understand what I was looking at. There's seeing and there is understanding.

    I'm sure this is so ingrained with experienced physicists that they may not understand why throwing these diagrams at beginners doesn't make things crystal clear. I mean, you even tell the poor beginner what the lines means and he still doesn't get it. Diagrams like this that omit the other twin's simultaneity lines tend to support the wrong idea.

    Yes, there is a question here. Two, actually.

    The diagram above assumes the traveling twin can change directions instantaneously. I want to check my understanding. The stationary twin can look at a giant clock attached to the traveling twin. He records the clock and then corrects the recording for light delay. I believe he sees the traveling twin's clock running slow by the same factor on each leg of the trip.

    The stationary twin also has a clock that the traveling twin can record. The traveling twin corrects his recording for light delay. He also sees his stationary twin's clock running slower and by the same amount on each leg. However, at turn-around, he sees his twin's clock jump forward in time. Do I have it right or am I still missing something?

    The re-alignment of the spacetime axes of the traveling twin resolves the twin paradox. But I realize I still have a hole in my understanding. If any frame can be considered the stationary frame, is there some perspective in which the traveling twin could assume he was motionless for the whole trip and that the "stationary" twin went out and back? (By the way, I have seen a video that showed just that: the traveling twin stationary while the Earth moves back and forth and claiming that, since everything is relative, either viewpoint is equivalent.)

    Could we redraw the Minkowski diagram from the traveling twin's point of view?

    My own attempt at an answer is that, if you change your frame, you can't treat it as though you are motionless and everyone else changed frames. Even if you had an inertialess drive and could switch directions instantly, you could still tell that you changed direction and not the rest of the universe. And if you wanted to draw a Minkowsky diagram for the traveling twin, you would need two, one for each direction.

    I can say those words, but I have to admit to not having an intuitive understanding of why it's true. My best guess is that maybe inertialess drives are impossible because it would create this very confusion. Someone who didn't see the inertialess drive activated wouldn't be able to tell if they just experienced a change in direction or the rest of the universe did, and the twin paradox would turn into a real paradox. You can't resolve it if everything is symmetrical. Everyone has to be able to agree that the traveling twin was the one who changed direction.

    Corrections and clarifications welcome.
     
  2. jcsd
  3. Jul 30, 2018 #2

    Nugatory

    User Avatar

    Staff: Mentor

    There is such a frame, but it is not an inertial frame so none of the straightforward equations like the time dilation formula can be used to compare times registered by a clock at rest in that frame with times registered by the earth clock.

    The best way to understand what each twin sees is by working through the Doppler Effect explanation in the Twin Paradox FAQ: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html. Just imagine that each flash of light is actually an image reflected from the clock at the source (or equivalently, a radio message reporting the time on that clock when the message is sent). For simplicity, assume that the turnaround is instantaneous and happens between flashes (any other assumption complicates the picture but introduces no new physical insight).
     
    Last edited: Jul 30, 2018
  4. Jul 30, 2018 #3

    Nugatory

    User Avatar

    Staff: Mentor

    No, you've been misled by people putting too much emphasis on the acceleration.

    The acceleration is actually a red herring; the important point is that the two twins took different paths through spacetime and these paths had different lengths. The acceleration only comes into the picture because
    1) We can't send the twins on two different paths between the separation and reunion events without accelerating at least one of them, so there's going to be some acceleration somewhere as we describe the thought experiment.
    2) We can point to the acceleration and say "look, it's not symmetrical so there's nothing inherently paradoxical about them having different experiences" (but we're still left explaining why they do have different experiences, and the acceleration doesn't help with that).

    "The traveller was subjected to an inertialess drive" works just as well as "the traveller was subjected to acceleration" for both 1 and 2.

    As an aside, it is possible to do an acceleration-free version of the twin paradox. When gravity is involved, it is possible to send the twins on different paths through spacetime between the separation and reunion events without accelerating either twin. We usually don't introduce the problem that way because it requires some additional background on the relationship between gravity and acceleration, and je problem is already confusing enough.
     
  5. Jul 30, 2018 #4

    Bandersnatch

    User Avatar
    Science Advisor
    Gold Member

    You can be very insistent on doing that, and just put the entire path of the travelling twin on a straight line. But, as mentioned by Nugatory above, since the travelling is necessarily changing inertial frames, doing so introduces a discontinuity as shown on the second of the three graphs in the link below:
    http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html
    The discontinuity is there because the outbound and the inbound trips of the travelling twin are objectively different inertial frames, with non-zero relative velocity, so the graph showing them both on one straight line is just two different graphs glued together. Compare with the third graph in the link to see the situation from the inbound trip perspective. You should see the half that makes up the inbound part of the second graph.
     
  6. Jul 30, 2018 #5
    Corrrection: My favorite version of the twin paradox involves no acceleration. I've read articles and watched videos that have reinforced that the fact that one twin experiences acceleration is not relevant. So, the people I've been paying attention to have not been misleading me. I have to take credit for that all by myself. :-)

    If there were such a thing as an inertialess drive, could a traveler detect that he has changed directions (or re-aligned his spacetime axes) and not the rest of the universe (assuming the traveler had no idea if the drive had engaged or even that his ship carried an inertialess drive)? If so, how? Or am I asking some silly question?

    What if a cosmic prankster put an inertialess drive on the Earth and pulled away from the twin on the ship and then turned around and came back. Would the ship twin be fooled or would he know that his Earth twin is the one who will be younger when they reunite?

    Edit: I meant a version with three observers. not the more complicated one with gravity, which I have not seen.
     
  7. Jul 30, 2018 #6

    PeterDonis

    Staff: Mentor

    You're assuming that without the inertialess drive, the traveling can detect that he has "changed directions" and not the rest of the universe, presumably by feeling his own proper acceleration. But feeling proper acceleration does not, in itself, mean that it's you who have "changed directions" and not the rest of the universe. You only know that when you meet up with your twin and see that he has aged more than you. The same would be true if you used an inertialess drive.
     
  8. Jul 30, 2018 #7

    Dale

    Staff: Mentor

    Yes, this is correct. By the way, the fact that you explicitly mention correcting for the light delay is an indication that you do have a grasp on some of the important issues that arise in SR.

    This question is not as straightforward as the previous one. The reason is as follows:

    In the case of the stationary twin the correction for light delay is simple, the first postulate says that the speed of light in vacuum is equal to c in any inertial reference frame. However, the travelling twin's frame is not inertial so it is not clear how to correct the observations for the light delay. The simultaneity convention used above is what I call the naïve simultaneity convention, and it simply uses the momentarily co-moving inertial frame's definition of simultaneity. In the "gap" and in the "overlap" regions the speed of light is no longer c, in fact all speeds of all objects in those regions are undefined.

    One alternative is to use radar coordinates, which avoids those problems and results in a non-inertial reference frame where the speed of light is still c at all events. However, the drawings are a little different and can be surprising:

    https://arxiv.org/abs/gr-qc/0104077

    Note, this is not the only alternative to the naïve simultaneity convention, but it is one that I personally like.

    Yes, the above paper shows things from the traveling twin's frame. However, regardless of who is considered "motionless" the traveling twin is always the non-inertial one and the stationary twin is inertial.
     
  9. Jul 30, 2018 #8
    I know I have two questions. Let me try to clarify the first.

    I like the video transmission idea, with corrections for light delay. That means that one can't really review the video until the trip is complete. The trip takes 10 years for the Earth twin, 6 for the traveling twin. The traveling twin always sees his Earth twin's clock running slow at a constant rate. Therefore, the traveler has a video recording 6 years long that shows less than 6 years passing on Earth. But the recording shows Earth 0 and the start and Earth 10 at the end. So, to my simplistic mind, the video should show a discontinuity.

    If instead of a constant velocity and an inertialess turn-around, we use more typical acceleration, we can steal this diagram from Wikipedia:

    [​IMG]

    This one show simultaneity from both viewpoints. Between traveling twin's years 4 and 5, it looks like time would appear to be running faster for the Earth twin than the traveling twin.

    Dale just posted something that might affect my interpretation (about adjusting for light delay). The paper he refers to is rather advanced for me, but perhaps it addresses the question--I'm not sure.
     
  10. Jul 30, 2018 #9
    Let's say that we use this magic inertialess drive that people say it's OK to use. The traveling twin never accelerates or rather, the only acceleration is a change in direction at turn-around. Otherwise, he moves at a constant velocity. Can we now say that the traveling twin's frame is inertialess and so we can correct for light delay just as on Earth? Or do I not understand what you mean when you say the traveling twin's frame is not inertial?

    Or we could just use two travelers, one outbound, one inbound, traveling at constant speed relative to Earth, who meet at the destination point. Outbound traveler passes inbound traveler the video he's recorded so far. Inbound traveler continues the recording. In this scenario, the gap is a lot less disturbing. Looking at it this way, it does seem that if we use the magic inertialess drive, there will be a gap and if we use conventional acceleration, the traveler will see the Earth clock run faster as he turns around.

    But, hey, I've been wrong before. :-)
     
  11. Jul 30, 2018 #10

    Ibix

    User Avatar
    Science Advisor

    The problem is that it's non-trivial to correct for light travel time near the turnaround. You can't use the distance travelled by light from the outbound frame because that's wrong for the inbound frame due to length contraction, and vice versa. And because the two frames don't agree on what "at the same time as the turnaround" means there's no unique way to work out for what events you should use which frame.

    One neat solution is Dolby and Gull's radar coordinates, which Dale linked above.
     
  12. Jul 30, 2018 #11

    Dale

    Staff: Mentor

    I don't agree that it is OK to use in SR. You could use gravity to do it, but then you would need to use GR.
     
  13. Jul 30, 2018 #12

    PeterDonis

    Staff: Mentor

    This. In the presence of an "inertialess" drive, you have to be more careful in defining what an "inertial" frame is, but you can still do it based on properties like the behavior of light signals. For example, at the "turnaround", the traveling twin sees the Doppler shift of light signals coming from the stay-at-home twin change abruptly (instantaneously if the turnaround is like the diagram in post #1; over a finite interval of time if the turnaround is like the diagram in post #8) right when the turnaround happens. But the stay-at-home twin only sees the Doppler shift of light signals coming from the traveling twin change much later. (The article @Nugatory linked to in post #2 goes into this.) Going into this in more detail would require defining "inertial frame" in terms of the underlying flat spacetime geometry, which might be a bit too complex for this discussion. But it can be done.

    You don't need an "inertialess drive" or conventional acceleration to "turn around" in this scenario. The outbound traveler could send the inbound traveler his video using light signals; the inbound traveler just loads the received data into his own video camera and goes on from there. The video will show the Doppler shift of light signals arriving from Earth changing abruptly (as described above) either way.

    Also, the "gap" you speak of is not something that's directly observed at the turnaround. It's something that has to be constructed from the data, and that construction requires assuming a simultaneity convention. In other words, the "simultaneity planes" drawn in the diagrams in posts #1 and #8 are not something anyone directly observes; they're constructions from the data, and the way they're drawn in those posts is not the only way to construct them.

    One big advantage of the Doppler shift approach (described in the article @Nugatory linked to, and which I've briefly described here) is that it doesn't require any such constructions. It doesn't even require a simultaneity convention. It just focuses on the differences between what the two twins actually see (as in, observe directly with their eyes and instruments) to explain why their clocks read different elapsed times when they meet up again. And this way of doing the analysis doesn't even care how, physically, the turnaround is accomplished, or about "frames" or anything else. You could work things out the same way using the Doppler shift approach if the traveling twin used a distant planet as a gravitational "slingshot" to reverse course (meaning he would feel zero acceleration even without having to use an inertialess drive).
     
  14. Jul 30, 2018 #13

    PeterDonis

    Staff: Mentor

    It's unphysical, yes, but I think it's fair to emphasize the point that how, physically, the turnaround is accomplished, including whether the traveling twin feels acceleration during it, is irrelevant to the calculation of differential aging.
     
  15. Jul 30, 2018 #14

    Ibix

    User Avatar
    Science Advisor

    You can make the acceleration phase negligible, however, just by increasing the distance travelled to tens of thousands of light years. You can accelerate to relativistic speeds in a year or so at relatively sedate accelerations.

    That said, I agree with you. The point is that the travelling twin is not at rest in a single inertial frame. "Inertialess" in the E.E. "Doc" Smith sense of being able to stop and maneuver instantaneously is just confusing in this context, to my mind. Not to mention probably unphysical (what do I mean by "stop" for example?)
     
  16. Jul 30, 2018 #15

    Dale

    Staff: Mentor

    I don't like it, because to me more fundamental than all of SR is being able to identify an inertial frame from a non-inertial frame. Allowing non-physical inertialess drives destroys that. Since that is more basic than SR, then I am not willing to entertain it as anything other than sci-fi entertainment.
     
  17. Jul 30, 2018 #16

    PeterDonis

    Staff: Mentor

    This is a fair point. I think such a definition can still be formulated based on spacetime geometry, as I mentioned before, but I agree that allowing an inertialess drive basically removes the physical connection between the motion of objects and spacetime geometry, so it puts the latter on a much shakier foundation.
     
  18. Jul 30, 2018 #17

    Dale

    Staff: Mentor

    @Freixas as @PeterDonis indicates, an inertialess drive is unphysical. I would not pursue it too much. If you already know what you are doing, then you can carefully formulate the twins scenario correctly (see post 12). However, if you do not already know what you are doing, then you are likely to get things wrong. I most emphatically recommend that you not pursue it further until you have mastered special relativity and at least begun general relativity. You should (IMO) not begin this until you can understand the first two chapters of Sean Carroll's "Lecture Notes on General Relativity".

    I will not answer any questions based on a non-physical inertialess drive.
     
  19. Jul 30, 2018 #18

    Dale

    Staff: Mentor

    Yes, or you can just use "future technology" to allow arbitrarily high accelerations. A sharp turnaround is OK to me, but I just want to make sure that accelerometers still detect the acceleration.
     
  20. Jul 30, 2018 #19
    The radar approach doesn't seem to forbid the speed-up. I have to confess that most of the paper makes my eyes glaze over, so I can't be sure what Figure 5 represents. In Region I, which appears to be the region I am most interested in, it sure looks like the lines of simultaneity (from Barbara's view) make it so that Alex's clock runs faster than Barbara's.

    The thing that attracted my attention to this is that the first concept one hears is that each observer views the other's time as running slower., so it's only at the turn-around that I'm seeing cases where that's not so. Or appears not to be so.

    As I am more an engineer than a mathematician, it seemed like recording the whole thing on video should make it all clear. Ibix said "The problem is that it's non-trivial to correct for light travel time near the turnaround." and Dale doesn't like inertialess drives. How about the following?
    • B passes A at some constant velocity relative to A. A and B set their clocks to 0 as they pass. A begins to transmit video of his clock to B who records it for later.
    • Some time later, B passes C, who is traveling towards A also at velocity v. B passes his video recording and clock time to C, who immediately begins his own recording of A's signal.
    • C passes by A and passes his recording and clock time to A and continues on. Let's say this conjunction happens at time T. We expect that the time that C passes to A will be T', where T' < T.
    • As both A and C were traveling at constant speed, A can calculate the light delay for every frame and adjust as needed. To make this easier, every frame embeds A's time and the recording medium records the local time at which it was received.
    Well, maybe I'm wrong about how easy it is to calculate, but there seem to be a lot of articles which claim that it can be done (the Wikipedia page, for one). Nobody is accelerating and nobody changes their frame of reference at any point.

    I sit down at a theater to watch this fascinating video. What do I see? I mean, I will see something, right? I know the recording starts with a clock showing 0 and ends with a clock showing T. The length of the recording can't exceed T'. T seconds over a period of T' (where T' < T) would suggest that B and C perceived A's clock as running fast at some point (or the tape had the equivalent of an "intermission").

    Note that I am not drawing any simultaneity planes (at least, I don't think I am). I am just watching a video. The only adjustments made to the video were for light delay.

    There could be a lot of ways to answer this that avoid talking about what the video shows. The video creation assumes that you can adjust for light delay. Everyone in this scenario is moving at constant speed, so that seems to me to be reasonable, but if it's not, let me know. Other assumption about passing clocks and videos across at conjunction don't seem completely out of line, at least not for these kinds of thought experiments, but if that's flawed, go ahead and point it out. However, assuming that the problem is set up properly, the answer I'd interested in is about what the video would show.

    As it doesn't appear there would be any discontinuity in the video (C picks up right where B left off). I suspect the problem is that C is picking up video that really was intended for B, so that the correction for light delay gets processed in the wrong reference frame and that somehow skews the result. I still wouldn't mind knowing what the video would show. I'd love to calculate it myself, but I'm more likely to write a software simulation and then see what happened.
     
  21. Jul 30, 2018 #20

    Ibix

    User Avatar
    Science Advisor

    Depends what they do with the video.

    The problem with this set up is that video cameras produce 25 images per second in their rest frame. Each video image will therefore show the clock one twenty fifth of a second advanced from its previous state, because the clock being videoed and the clock controlling the camera shutter are at rest with respect to one another.

    Due to the increasing distance between A and B, B will wait a long time between receiving successive images. Due to the decreasing distance between A and C, C receives images at very short intervals.

    What you see when you watch it then depends on how you interpret those incoming images. Do you show them at the spacing you received them? In that case you get the same effect as an old hand-cranked projector that is turning very slowly for half the journey and then super quickly for the second half. Or do you show them every 25th of a second? In that case you just see an ordinary video of a clock - the only odd thing is that (by B and C's combined wristwatches) they were away for a year but have two years' of footage, because A says the trip took two years and it's A's clock that dictates how many images were sent.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook