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Reverse Twin paradox ?

  1. Nov 15, 2008 #1
    HI My question is what would the outcome be if the twins started out in 2 ships traveling at uniform .6c relative to earth and one of them then took her ship and traveled to earth and back???
    Which twin would be older???
     
  2. jcsd
  3. Nov 15, 2008 #2

    CompuChip

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    Does it matter?

    I mean, if they started out from some frame stationary with respect to the earth, then your problem would be completely equivalent to the original paradox (with roles of earth and distant target interchanged).
    Now replace the stationary frame by an equivalent one (like, ohh, say, one moving at 0.6c relative to the earth).
     
  4. Nov 15, 2008 #3

    Dale

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    The twin that moves inertially the whole time will always be the oldest since the longest timelike interval between two events is a straight line.
     
  5. Nov 15, 2008 #4
    This is the second time this week you ask this question. Asking the same question multiple times in different topics is only going to clutter this forum.

    https://www.physicsforums.com/showthread.php?t=271847
     
  6. Nov 16, 2008 #5
    Hi----Apologies to all.
    I was very tired when I posted this query , both physically and mentally from trying to get a handle on how acceleration relates to time dilation, if it does not itself cause or change it. I realized later how silly it was and that it did not change the basic question at all.
    I will try and remember this and give more consideration before posting in the future.
     
  7. Nov 16, 2008 #6
    Hang on a second - unless I'm misunderstanding something, haven't DaleSpam in this thread and vizart in the other thread given the opposite answer of what MeJennifer gave in the other thread?

    Isn't it, as DaleSpam states here, that the one who remains in the original inertial frame is always older and the one who accelerates out of the original frame and then accelerates back into it is always younger?
     
  8. Nov 16, 2008 #7

    JesseM

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    It doesn't really matter what frame they were at rest in before they separated from one another, all that matters is which one moves inertially between the time they separate and the time they reunite. For example, suppose you have ship A and B moving inertially next to each other, then A accelerates instantaneously to a new velocity, after which they move apart inertially. After they have moved apart for some time, B accelerates back in the direction of A, and eventually they reunite. In this scenario A will have aged more because A moved inertially between the moment they separated and the moment they reunited, it's completely irrelevant that A was the one to accelerate initially and cause them to first move apart.

    But checking the other thread it does look like MeJennifer got it wrong, the one who accelerated in that scenario would be younger when they reunited, not older.
     
  9. Nov 16, 2008 #8
    But of course, my mistake, I meant to say younger not older.
     
  10. Nov 16, 2008 #9
    I am sorry ,,,I reposted because I have not seen the first post and still do not register it as being posted. I assumed that for some reason it had not actually gotten through.
    In any case I am sorry I posted either one.
     
  11. Nov 16, 2008 #10
    In this case both will be of same age. In one case A accelerated fast (instantaneously) and for B, he accelerated slowly later to catch up with A. The time gain(?) should be related to the acceleration rate. So before A accelerates to catch up, B is younger. Bu after B catches up with his own accelerations, both are of same age
     
  12. Nov 16, 2008 #11

    JesseM

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    No, the time gain has nothing to do with the acceleration rate, only with the geometry of the two paths in spacetime. As an analogy, if two cars are driving side by side, then car A swerves and then moves straight in a different direction while B continues to move in the same direction (so they are now moving apart), then later car B swerves again so it crosses paths with A at a later point, it will always be true that B travelled a greater distance (as measured by his odometer) between the moment A separated from B and the moment they reunited, because A travelled in a straight line from the position where they began to move apart (immediately after the swerve) to the position where they reunited, and a straight line is always the shortest distance between two points on a 2D plane. The time along worldlines in 4D spacetime is pretty similar, except that here a "straight" (constant speed and direction) worldline between two points in spacetime always has a larger time as measured by a clock moving along that wordline than any non-straight wordline.
     
  13. Nov 16, 2008 #12
    I still have a problem with this.
    I can understand if A does not accelerate, but only B swerves away and later swerves back to meet A. In which case A will be older.
    But as mentioned in your example, aren't both A and B accelerate (swerve) some time or other. shouldn't both have same aging?
    Could you suggest some reference I could look into and have a better understanding.
     
  14. Nov 16, 2008 #13

    JesseM

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    I wasn't talking about who was older, I was using an analogy that was just about the physical distance travelled by 2 cars on a 2D plane--the analogy doesn't involve relativity at all. Do you agree that if we look at two points on the plane--the point where A and B separated, and the point where A and B reunited--and we see that A travelled in a straight line from the first point to the second, while B travelled on a path with a bend in it (so that B's path would look like two sides of a triangle while A's would look like the third side), then because "the shortest distance between two points in a plane is a straight line", we can be sure that A's path will have a shorter distance? Do you agree that this is true regardless of whether it was A or B that initially swerved to cause them to begin to separate, assuming that prior to that they had been travelling next to each other along the same path? Remember, no relativity here, this is just a question about plane geometry.
     
  15. Nov 16, 2008 #14
    You wrote
    "For example, suppose you have ship A and B moving inertially next to each other, then A accelerates instantaneously to a new velocity, after which they move apart inertially. After they have moved apart for some time, B accelerates back in the direction of A, and eventually they reunite."
    First A accelerates away (or A swerves away). Later B accelerates (or sverves) to catch up with A. In this case in plain 2D - A's path is straight line than say 30deg angle to the right. B's path is a longer straight line (since he swerves later) and a steeper angle (say 45 deg) to the right so that he meets A's path.

    This case is different from 'A keeps going straight and B swerves out and then later swerves in to meet A again'
     
  16. Nov 16, 2008 #15

    JesseM

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    I'm talking about B's entire path from the point in the plane where A and B depart from one another (because A swerves) to the point in the plane where A and B reunite. B's path is not a single straight line between these points, it's two line segments at different angles. For example, suppose A and B were both initially moving vertically (0 degrees), then A swerves so he's moving at 30 degrees to the right of the vertical, while B continues to move along a vertical path at 0 degrees. Later, B swerves so he's moving at 45 degrees to the right of the vertical, so he's able to meet with A again. In this case, A's path from the point where they depart to the point where they reunite is a single 30-degree line segment, but B's path from the point where they depart to the point where they reunite is a combination of a 0-degree line segment and a 45-degree line segment. That's why I said B's path looked like two sides of a triangle while A's path looked like the third side. Since A follows a straight line between these two points, while B follows a path that has a bend in it, we know A's path between these points will have a shorter distance that B's path between the same two points since in 2D plane geometry a straight line is always the shortest distance between two points.

    Note that it would make absolutely no difference to the problem if A and B had initially been moving at 30 degrees to the right of vertical, then they departed from one another because B swerved to move at 0 degrees from the vertical, then later B swerved again to move at 45 degrees--the question of who swerves initially doesn't change the fact that A has a single 30-degree line segment between the point where they depart and the point where they reunite, while B's path between the same two points is a combination of a 0-degree segment and a 45-degree segment. We are only interested in the length of the two paths between the point where they depart and the point where they reunite, the direction they were moving prior to departing one another is irrelevant.
     
  17. Nov 16, 2008 #16

    Dale

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    Hi mviswanathan

    I highly recommend actually drawing the spacetime diagram. It will make the geometry quite obvious. Just remember, for Euclidean distance "the shortest distance between two points is a straight line", but for timelike Minkowski intervals "the longest interval between two events is a straight line".
     
  18. Nov 16, 2008 #17

    Dale

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    Usually for introductory problems we only deal with instantaneous accelerations. The reason is simple, instantaneous accelerations make polygonal spacetime diagrams where the intervals can be summed, but gentle accelerations make smooth curves where you need to evaluate an integral to calculate the interval. But conceptually there is no difference, you are still just measuring the interval along the worldline.
     
  19. Nov 17, 2008 #18
    I tried to draw the diagram in Space time (with space of only x axis, considering y and z do not change). The result is interesting, since different from what I assumed and stated on 16th.
    I am atteching a JPG file of the diagram.
    with "proper time, T" based on the relation
    dT2 = dt2 - dx2 - dy2 - dz2
    In case -1, with A and B stationary with respect to the reference frame:
    For A
    T (a to b) = sqrt(100^2 - 0) = 100
    T (b to d) = sqrt(200^2 - 75^2) = 185.4
    Hence total time as measured(aged) by A = 285.4
    For B
    T (a to c) = sqrt(200^2 - 0) = 200
    T (c to d) = sqrt(100^2 - 75^2) = 66.1
    Hence total time as measured(aged) by A = 266.1
    Hence B is younger when they meet again

    Working similarly for Case -2, with A and B intially moving equal uniform velocity with respect to the reference frame:
    For A
    T (a to b) = sqrt(100^2 - 25^2) = 96.8
    T (b to d) = sqrt(200^2 - 100^2) = 173.2
    Hence total time as measured(aged) by A = 270
    For B
    T (a to c) = sqrt(200^2 - 50^2) = 193.6
    T (c to d) = sqrt(100^2 - 75^2) = 66.1
    Hence total time as measured(aged) by A = 259.7
    Here again B is younger when they meet again

    Based on http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

    Thanks
     

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  20. Nov 17, 2008 #19

    CompuChip

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    Now try Lorentz transforming the reference frame of case -2 in which they are moving at equal velocities to the reference frame of case -1 (i.e. "bend" back the "narrowed" axes of your second diagram). Aren't the diagrams the same?
     
  21. Nov 20, 2008 #20
    The differences between Case-1 and Case-2 was bothering. I took the advice:
    I updated the diagram with Lorentz transformation. With Case -1 as the base, I corrected the coordinated for 1, b, c and d in Case-2. The case-1 is with respect to v/c as 0.25.
    With r = 1 / sqrt(1- 0.25^2)
    x2 = (x1 + 0.25t1) * r
    t2 = (t1 + 0.25x1) * r
    The result is interesting
    With "proper time, T" based on the relation
    dT2 = dt2 - dx2 - dy2 - dz2
    For BOTH Case -1 and Case -2, I get SAME results:

    Time measured by A from a to d through b = 285.4
    Time measured by B from a to d through c = 266.1

    For Lorentz Transformation, I referred to
    http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html relativity/Lorentz transformation
     

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