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Your preferred way of explaining this point about the twin paradox?

  1. Mar 1, 2014 #1

    bcrowell

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    A student asked me a question in class last week about the twin paradox, and I found that although I knew a valid answer, it wasn't an answer that my class had the background to understand, and I wasn't immediately able to come up with one that would work for them.

    The standard difficulty in understanding the twin paradox is that it seems like either twin could be considered the one at rest, and therefore there is symmetry between them, and no way to determine which twin should end up younger when they're reunited. The standard resolution of this is that the symmetry is broken by the fact that the traveling twin's motion is noninertial. This all works fine, and it works for students who know what the students in my class know.

    But my student raised a question that I'll state in the following way. Let E be the twin on earth and T the traveling twin. If there were symmetry, then the difference in age at reunion, [itex]\delta t=t_E-t_T[/itex], would have to be zero. But the lack of symmetry doesn't immediately prove that δt is nonzero or that it has a particular sign. How, then, do we know that it's nonzero and positive?

    If you have the concept of the metric, then the answer is straightforward. But for the level at which I'm teaching this class (algebra-based freshman physics for life science majors), I don't teach them about the metric. They do know about the Lorentz transformation in graphical form.

    Without the metric, I guess one natural way to explain the result is to say that the Lorentz transformation involves t and x coordinates, and we need to go back through the steps needed in order to establish t and x coordinates by techniques such as Einstein synchronization. These techniques involve signaling. We don't normally think of signaling as an idea that is an intrinsic part of a discussion of the twin paradox, but maybe it's necessary in this approach. It seems likely to be complicated, although maybe once you got the argument to work it could be simplified.

    A nice idea that I got from Rindler's SR book is to explain this by appealing to length contraction. T sees the distance from the Earth to the turnaround point as being length-contracted, but E doesn't. Therefore T correctly expects [itex]t_T[/itex] to be shortened relativistically compared to [itex]t_E[/itex]. This argument has the virtue of being very simple, but it seems to add some complication. I view length contraction is being a statement about the world-sheet of a ruler, which means that it's inherently more complicated and less fundamental than time dilation, which is a statement about the world-line of a pointlike clock.
     
    Last edited: Mar 1, 2014
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  3. Mar 1, 2014 #2
    what I would do is use an analogy with euclidean space where the shortest distance between two points is a straight line with the very important caveat that in Minkowsky space the longest distance between two points is the straight line (as measured by the proper time). The advantage of such answer is that it bypasses the need to explain the concept of metric by using the intuitive distance in euclidean space with the additional plus of sparking their curiosity about Minkowsky space. The disadvantage is that such a handwaved answer really isn't an answer at all.
     
    Last edited: Mar 1, 2014
  4. Mar 1, 2014 #3

    stevendaryl

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    E and T? I have trouble following thought experiments if the characters' names aren't "Alice" and "Bob".

    It seems to me that for E to calculate t_T just involves time dilation:
    [itex]t_T = t_E \sqrt{1-\frac{v^2}{c^2}}[/itex]

    Is the question just: why can't T reason the same way, to get
    [itex]t_E = t_T \sqrt{1-\frac{v^2}{c^2}}[/itex]

    It's not enough to just say: the time dilation formula only applies if the times and velocities are measured with respect to an inertial coordinate system?
     
  5. Mar 1, 2014 #4
    Maybe you can use the explanation with spacetime diagrams here (ctrl-f for "twin paradox"): http://www.damtp.cam.ac.uk/user/tong/relativity/seven.pdf

    It explains how as the traveling twin turns around, his surface of simultaneity shifts such that the stationary twin appears to age very rapidly.
     
  6. Mar 1, 2014 #5
    But a proof isn't really required since presumably one was already given one when demonstrating time dilation. The point of the paradox really is simply to show that there actually is no paradox due to the asymmetry between the paths taken by Travis (the twin that travels) and Earl (the one that stays on Earth)
     
  7. Mar 1, 2014 #6

    bcrowell

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    I think this is a perfectly reasonable approach, but I think it's an approach in which the metric is considered fundamental.
     
  8. Mar 1, 2014 #7

    WannabeNewton

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    http://arxiv.org/pdf/gr-qc/0104077v2.pdf

    So it's not that cut and dry unfortunately and while the subtleties associated with that approach can be addressed through Ben's suggestion of using light signals in post #1, fleshing it out would require complicated machinery as you can see in the paper.

    I personally like Steven's suggestion.
     
  9. Mar 1, 2014 #8

    bcrowell

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    Maybe that does it. Nice and simple!
     
  10. Mar 1, 2014 #9
    What about Travis and Earl?
     
  11. Mar 1, 2014 #10
    Sure, constructing some sort of global coordinate system centered on the traveling twin is complicated, but I think figure 1 in that paper is not an unreasonable way to argue the limited point that the traveling twin will age less. My original link is actually a little bit careful and first considers the case where the outgoing traveler does not turn around but instead meets an incoming traveler; in that case there is surely no subtlety.
     
  12. Mar 1, 2014 #11

    pervect

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    The thing I've seen on PF is that students insist on trying to understand the viewpoint of the travelling twin, but can't handle the math for a constant acceleration turnaround and don't understand and/or don't believe the sudden "jump" in simultaneity that occurs with the turnaround.

    It might help to spend some time on Einstein's train and the relativity of simultaneity first, calculate the time gap first using the same arguments (you might call it the Andromeda paradox), then introduce the twin paradox last. Or not - I haven't really seen it tried.

    Someone once a long time ago mentioned a paper that studied how well students learned SR with various appraoches, but unfortunately I don't recall the author (and I'm not sure it addresses this specific issue).
     
  13. Mar 2, 2014 #12

    stevendaryl

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    Something that I find helpful in understanding relativity is to see exactly how much of the supposed paradoxes and weirdness of relativity have analogs in good old Euclidean geometry. Almost all of them do. For example, the twin paradox has an analog in Euclidean geometry.

    Suppose that you have a network of highways criss-crossing each other. Each highway has a system of distance markers--every 100 meters, there is a marker, with a number on it, and the numbers increase consecutively as you move down the highway.

    Now, if highway A crosses highway B, you can set up correspondences between those distance markers in the following way: You go to a marker on highway A. Call the number on it [itex]x_A[/itex]. Look perpendicularly to highway A toward highway B. Your line of sight will cross highway B at some point. Call the number on the closest marker on B [itex]x_B[/itex]. To simplify, let's assume that we renumber the markers so that [itex]x_A = x_B = 0[/itex] at the point where the roads intersect. Then the relationship between [itex]x_A[/itex] and [itex]x_B[/itex] will be:

    [itex]x_B = \sqrt{1+s^2}\ x_A[/itex]

    where [itex]s[/itex] is the slope of highway B, relative to highway A (slope = the tangent of the angle between them).

    This is just geometry, but here's the weird thing about it: It looks like it's saying that the distances along B increase at a greater rate than distances along A. But obviously, that effect is purely relative. You could just as well have chosen highway B as your reference, and for each distance marker along B, chosen the corresponding distance marker along A. You would have found that

    [itex]x_A = \sqrt{1+s^2}\ x_B[/itex]

    This seems like a contradiction. The first formula says that [itex]x_B[/itex] increases faster than [itex]x_A[/itex], while the second formula says that [itex]x_A[/itex] increases faster than [itex]x_B[/itex]. The resolution is that the correspondence between points along A and points along B make use of the notion of a perpendicular. [itex]x_B[/itex] is the point on highway B that is on the perpendicular to highway A at the point [itex]x_A[/itex]. But that line is only perpendicular to A, not to B. So they use different notions of perpendicular, and so the end up with different notions of which points correspond. So the "distance expansion" factor [itex]\sqrt{1+s^2}[/itex] is mutual and relative and it's all consistent.

    Now, suppose that highway B is not straight. Instead, it crosses A at one point at particular angle, then later makes a turn toward A, and crosses A again (for symmetry, assume that the angle is the negative of the original angle, so the slope during the second part is the negative of the slope during the first part). When the two roads cross a second time, how does the distance marker of highway B at that point compare with the distance marker of highway A?

    From the standpoint of highway A, the answer is simple: Since the value of [itex]s^2[/itex] is the same all along B, we can just use the "distance expansion" formula to get:

    [itex]x_B = \sqrt{1+s^2}\ x_A[/itex]

    So the distance along B is greater than the distance along A, even though they start and end at the same point.

    But the paradoxical part is this: why can't we compute things from the point of view of highway B? Then we would get the opposite answer:

    [itex]x_A = \sqrt{1+s^2}\ x_B[/itex]

    From the point of view of highway B, it seems that the distance along highway A should be longer. How do we resolve this paradox?

    The key to resolving it is to remember that the way that we set up the correspondence between points on A and points on B is by using perpendicular lines. When highway B makes a sudden turn, it's notion of what's perpendicular changes suddenly. So the notion of what point on A corresponds to what point on B changes suddenly. Imagine someone driving his car along highway B and looking out his side window to see the numbers on highway A go past. On the first leg of his journey, the numbers for A will increase steadily. But when he makes his turn at the bend in the road, he'll see the numbers on A rush back in the opposite direction. So after the turn, the number that he sees out his window will be smaller than the number that he saw before the turn. So when he crosses A for the second time, the distance marker that he sees on A will be smaller than the distance marker for B.

    This is almost completely analogous to the twin paradox for Special Relativity, except instead of highways, we have worldlines, and instead of measuring them by distance markers, we measure them using clocks, and instead of slopes we use velocities, and instead of

    [itex]x_B = \sqrt{1+s^2}\ x_A[/itex]

    we use

    [itex]t_B = \sqrt{1-(v/c)^2}\ t_A[/itex]

    and instead of using perpendicularity to determine the correspondence between points, we use simultaneity. Other than those differences, the situations are exactly the same :smile:
     
  14. Mar 2, 2014 #13
    I think the Doppler explanation is useful and intuitive.

    TwinParadox2.jpg

    It has the benefit that the students can count the birthdays of the twins directly off the chart, as well as graphically demonstrating the non symmetrical nature of what the twins experience. For example the twin that turns around sees an immediate change in the Doppler shift while there is a significant delay before the stay at home twin sees any change.It might be instructive to do a chart analysing the Newtonian Doppler case without time dilation, to contrast and compare.
     
  15. Mar 2, 2014 #14

    Jonathan Scott

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    I've been using an equivalent analogy of ants looking over their shoulder for a long time (mainly because I could draw it using ASCII characters). The point is that if you are going at the same speed as someone else but not in the same direction, then they are going slower in your direction than you are, and that applies both ways, but the solution to the paradox is obvious in that case.
     
  16. Mar 2, 2014 #15

    Meir Achuz

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    I don't think it helps students at any level without introducing the key notion of Minkowsky space.
    Just draw an obtuse triangle where the sum of the two short sides is longer that the long side.
    Then point out that in SR the proper time^2 is the distance^2-time^2, rather than all sums.
    No equations are necessary
     
  17. Mar 2, 2014 #16

    FactChecker

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    I think you are not giving them enough credit for abstracting the idea of "no preferred reference frame". They wonder why velocity is relative but acceleration is not. They can imagine that the "traveling" twin is actually stationary and the rest of the universe does all the motion, including the acceleration. The situation of the two twins are completely symmetric unless acceleration wrt the rest of the universe is used to say that the traveling twin does all the accelerating. The math of SR can assume that the traveling twin does all the acceleration and come up with an answer. But the answer to their real question requires more abstraction than just cranking the numbers through SR equations. This bothered Einstein as much as it bothers students. I think that GR is really required to answer it.
     
  18. Mar 2, 2014 #17

    WannabeNewton

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    But the point is you don't accelerate with respect to the universe. You don't accelerate with respect to anything. You just accelerate. Similarly you don't rotate with respect to anything. You just rotate. I don't need the distant stars to see if I'm accelerating or rotating. I just need an accelerometer and a compass of inertia. SR is just as non-Machian as Newtonian mechanics. But this gets into some age-old off-topic philosophical concepts.

    Well it isn't. GR is completely irrelevant both mathematically and (more importantly) conceptually with regards to the problem at hand.
     
  19. Mar 3, 2014 #18

    Dale

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    If you have already introduced the Lorentz transformation then you can sneak the metric in pretty easily. Simply draw a spacetime diagram (t,x) and look at all of the events with a constant t' for all possible v. This gives you a hyperbola. Talk about that hyperbola as a shape with constant "radius in time" just like a circle is a shape with a constant radius in space. Now, you can draw a hyperbola through the turnaround event and show that they have not aged as much as the stay at home twin.
     
  20. Mar 3, 2014 #19

    stevendaryl

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    I'm not sure what you mean. Using the Lorentz equations,

    [itex]t' = \gamma (t - vx/c^2)[/itex]

    the set of points with [itex]t' = K[/itex] is the graph

    [itex]x = \dfrac{c^2}{v} t - \dfrac{Kc^2}{\gamma v}[/itex]

    That's a straight line, not a hyperbola.
     
  21. Mar 3, 2014 #20

    Dale

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    I mean take the Lorentz transform for x'=0 and t'=K. Then you get:
    ##t=\frac{K}{\sqrt{1-v^2/c^2}}##
    ##x=\frac{-Kv}{c\sqrt{1-v^2/c^2}}##

    That is the parametric equation of a hyperbola parameterized by v.

    It is an easy of getting to the geometric content of the metric without actually introducing the metric.
     
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