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About the Non-paradoxicality of the Twin Paradox

  1. Oct 6, 2007 #1
    When I wrote my paper on Special Relativity in my final year of high school, I explained the Twin Paradox as follows:

    Twin A stays on earth, twin B takes off to some star S. (I'm making this a very short version because I'm assuming we all know the structure of the paradox)

    Twin B measures less time passes on the way S and on the way back. But, using the Lorentz transformations, we find that even less time is supposed to have passed on earth (due to the equality of the reference frames). However, this is fixed because during the turnaround, suddenly a lot of time has passed on earth (this is as it was explained to me here). This all works out perfectly if you calculate things by putting B in a new frame of reference after the turnaround.

    My SR teacher explained it like this last week:

    Since B doesn't have a constant velocity, he doesn't have a valid view in Special Relativity, so we toss his observations out.

    He used a third observer to show that twin A was right.

    Now my question is, who has a better point of view, I or my teacher. Of course, I fully agree that he is right in saying B doesn't have a true inertial frame of reference, but then I wonder why it works out so perfectly if you do what's not allowed.
    Also, though Special Relativity is only valid for constant velocities, it would appear from these results that it is able to say something about differences of relativistic effects between different constant velocities, i.e. if the velocity changes, the effects of the Lorentz transformations change accordingly.

    To illustrate, my teacher said that if the velocity changed from v to -v, that earth's location would suddenly be shifted (where I reasoned that earth's time would suddenly jump forward, I'm not really sure why he arrived at position and I at time), which he claimed indicates the invalidity of B's point of view. However, doesn't it make perfect sense for the time to have shifted so? (I'm not really sure about the position, I don't really know how he got to that, it's puzzling me a little, but I'm assuming it's rather the same thing as the shift in time)

    For example, if B slows to a halt, then what he formerly perceived as so many lightyears will now seem like a bigger stretch of space, because the effect of length contraction is gone, so the spatial axes of A and B are synchronized (suddenly moving Earth back, now I see. The position part is easier to understand physically than the time part, though I don't see how one should use it to explain the paradox). It would seem most logical to me (in fact, I don't really see any other possibility) if the effect of length contraction changes continually when the velocity changes continually, thus there isn't really the problem of "it only works if you assume the acceleration in instantaneous".

    Alright I'm getting more confused as I type all this, so for the sake of keeping my post as clear and understandable as possible, I'll stop here and see what you all want to say about this. I'm afraid that my post isn't as clearly set up as I intended it to be, because my thoughts kept wandering around the subject as I was typing, but I hope you will be able to understand what I'm saying.
  2. jcsd
  3. Oct 6, 2007 #2


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    I suspect the two of you are just talking past each other -- the two of you think you're discussing slightly different things, and so you're misinterpreting each other.

    (e.g. I think everything he says is right... if he's interpreting "B's point of view" to mean the flawed argument that leads to the paradox)

    But if your description is accurate, then you seem to have the better point of view. There are several ways to use differential calculus to make sense of the notion of "B's frame of reference"; one of the obvious ones behaves exactly as you describe: as B slows, Earth rapidly moves away (and its clocks run very rapidly), then as B accelerates again, Earth rapidly moves closer (and its clocks run very rapidly). Finally, B settles on a constant return velocity, and things continue to behave like an inertial frame of reference.

    It's interesting to consider something on the far side of the star too; suppose the star was at the midpoint between Earth and planet Zelda. Then as B turns around he sees:
    Planet Zelda smoothly reverses direction and then moves rapidly away, turns around and moves rapidly closer, then reverses direction and starts moving away normally. During this time, the clocks on planet Zelda slow down and eventually run backwards at a rapid pace, then revert to running forwards again.
  4. Oct 6, 2007 #3


    Staff: Mentor

    Special Relativity does just fine with accelerating objects including accelerating clocks and rods. What Special Relativity cannot handle easily is non-inertial coordinate systems. In any inertial coordinate system twin A has a longer spacetime interval between the departure and reunion events than does twin B. Even if both twins accelerate in some complicated manner SR's spacetime interval can be used in any inertial coordinate system to determine what each twin's clock will read at the reunion.
  5. Oct 6, 2007 #4
    I'm not quite sure what you mean. What do you think the difference between what we are discussing?

    Perhaps I will try to draw the distinction between his arguments and my own a little more clearly, without adding all my side-thoughts.

    His argument: The problem of the paradox is solved because B's velocity is not constant, thus SR does not apply. You don't have to worry about B's point of view, or in other words, the reference frames of A and B are not symmetrical.

    My argument: I agree that he is right in what he says, but if you do consider B's point of view, it still works out.

    I don't have time to write out how I did it, because I'm not fluent in Latex and I would have to translate it to English, but if I remember correctly, what I did was mostly based on this post by jtbell:


    Ah, so I'm not crazy. But, even though this notion popped into my head, I don't fully comprehend it and I'm hoping you'd be so kind as to clarify some things to me.

    You say the clocks run very rapidly on earth during B's acceleration. I have, as I said, proven this mathematically, but I don't understand physically.

    You say Earth first moves away, then moves closer again. I'm assuming that both changes cancel each other out. I would expect so, because the effect of length contraction should be the same, no? So how does Earth's position change, as my teacher said? Of course, you could say it changed position if you considered B's axes to be flipped around, but that doesn't make much of a difference to human perception, I'd think. My teacher used the argument of the change of Earth's position to say "of course this is all nonsense". Am I missing something here?

    That sounds interesting. I'll have to take some time, though, to have a look at that. It doesn't strike me as very obvious at first glance.
  6. Oct 6, 2007 #5


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    You've clearly crossed the conceptual hurdle here; you know that B is not an inertial traveller, and are now wondering what his 'frame' really looks like. However, it is possible that your teacher may not have fully realized this!

    Geometrically, it's analogous to spinning around on a merry-go-round. If you're doing 1 revolution every 5 seconds, then your friend standing 10 meters away is whizzing by at 28 miles per hour. Your house that's a mile down the road? It's zipping around at over 4500 miles per hour! The moon sitting on the horizon? 61% faster than the speed of light!

    That's what happens in a frame rotating in a spatial plane. The effects we are seeing in our analysis of B's frame here are the manifestations of rotating in a space-time plane.

    And this hits upon something that's easy to forget: we don't see our entire frame: we only see the light reaching us from distant events. In order to talk about the time and place where distant events occurred, we have to invoke some sort of mathematical model to determine those numbers.

    Inertial frames are 'linear'; we can just define the big picture to act as a scaled up version of the little picture. But the same isn't true of a non-inertial frame.

    Incidentally, what is typically called a "frame" in SR is more properly called simply a coordinate chart. A frame really only makes sense near a single point; and in general relativity, a frame is defined in that way: it's merely a choice of coordinate axes in an "infinitessimal" neighborhood of a single point.
  7. Oct 6, 2007 #6
    That's not a very comforting thought. I am more inclined to think, anyway, that he may be more trying to keep things simple, since he is only teaching an introduction to SR.

    Though this makes sense to me in some degree, I'm not fully understanding what you're saying. You mean this happens because B spins around. I never really thought about it like that. What if B doesn't spin around but just travels backward?

    Now you've lost me. I understand what your words mean, but I'm not really sure what you mean to say by them, how they apply to what I said. By saying that it doesn't much mean anything to human perception, I meant that we can distinguish between back and front - if we turn around, an object will be at spatial co-ordinate -x instead of x, but we won't be baffled by that: we find it obvious that this should happen if we spin around.
  8. Oct 6, 2007 #7


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    Distance isn't something we see with our eyes; it's something we compute with our brains, or via other means. e.g. we count seconds between a lightning flash and a crack of thunder, or we shine a beam of light, and see how long it takes before we see the reflection.

    In terms of what person B literally sees -- if he trains his telescope on Earth, all he will see is that Earth clocks run slowly on the outbound trip, and run quickly on the inbound trip. He will also see that the light is redshifted on the outbound trip and blueshifted on the inbound trip. Both effects are due to the (relativistic) Doppler effect... though this would also be expected due to the classical Doppler effect. (But the classical version gets the amount of shift wrong)
    Last edited: Oct 6, 2007
  9. Oct 6, 2007 #8
    Quite right you are, Hurkyl, and perhaps in this lies my problem. I also explored the aspect of the relativistic Doppler effect in the Twin Paradox, and it in itself makes perfect sense, but I was never able to link the two solutions - using the Doppler effect or just the Lorentz transformations. There doesn't appear to be any "jumping" of anything through time or space. Is there a link between the two?
  10. Oct 7, 2007 #9
    One of the interesting aspects of SR and the Twin Problem is that there are several ways of arriving at the age difference - some authors claim it can only be analysed along the lines proposed by Einstein in his 1918 paper, others claim it can be completely treated by adding up the apparent time rates in the other twins frame; others claim it is always a simple application of the "invariance of the interval" applied to the outgoing journey and again to the incoming journey - you can also keep track of signals received an sent - The issue is: "what is the actual physics that results in two clocks having logged different times for a round trip journey by one twin" My own preference is to use the invariance of the interval and break the trip into two parts, calculate the time accumulated by the traveler's clock on the outbound journey and double the result for the round trip - when the clock of the stay at home twin is compared to the traveler's clock - the latter will have logged less time.
  11. Oct 8, 2007 #10


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    That is my preference too. Particularly since it (a) shows that SR can handle accelerating bodies and (b) can be generalized to arbitrary paths of the various twins.
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