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SR Paradox

  1. Oct 6, 2004 #1
    Hi all,
    I know that this isn't a "flaw" in SR. So, I'm not here claiming to have found an amazing flaw in SR. lol.
    I actually read about this flaw I think, and thus, probably read the solution, however, it was some time ago.
    So, can anyone enlighten me here please? :)

    It's related to time dilation and thus Frame of Reference:

    It's said that the time between to objects, which are moving relative to each other changes.
    e.g. Person A flies away from person B at 0.9c. For # years. He comes back, and is 10yrs younger than person B.
    Okay, fine.. time dilation, sure. :)
    However; person B is moving relative to person A, thus, person Bs time should slow down relative to person A, right? (or speed up? (i can't remember which one lol :P)).

    So, who's time slows down?

    Thanks in advance, :biggrin:
  2. jcsd
  3. Oct 6, 2004 #2


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    http://mentock.home.mindspring.com/twins.htm [Broken]
    Last edited by a moderator: May 1, 2017
  4. Oct 6, 2004 #3
    Hi Nate,

    I checked out the website you posted. At first glance it seems like maybe a valid explanation and the numbers crunch correctly.

    However, the trajectories don't match.

    In the first Bob goes out and comes back relative to the initial rest frame of Ann.

    In the second, Ann keeps going in the same direction relative to the initial rest frame of Bob. Sending Bob after her at higher velocity doesn't change this fact.

    The second problem is not the same as the first even though the numbers track.

    Try this explanation.


    Last edited: Oct 6, 2004
  5. Oct 6, 2004 #4
    Person A ages less. It's not a symmetrical situation, because A has to accelerate to take off, and decelerate to land
    However, person B is always in an inertial reference frame
    Last edited: Oct 6, 2004
  6. Oct 6, 2004 #5
    Its not so simple a situation. Acceleration just changes the level of the symmetry, but I don't see how it breaks the symmetry. The symmetry is still there at the new velocity. Just on the basis of acceleration alone, you cannot determine whose viewpoint is correct.

  7. Oct 6, 2004 #6
    The thing is: person B is ALWAYS in a inertial reference frame
    person A is NOT ALWAYS in an inertial reference frame
    So there's not symmetry
  8. Oct 7, 2004 #7
    Yet, you get the same dilation effects if person A is traveling over a specified space interval without acceleration. It is a matter of understanding that by specifying the space interval you break the symmetry.

  9. Oct 8, 2004 #8
    resolving the twin paradox

    Good question, Doc. To understand how the apparent paradox is resolved, it is of great help to realize that SR dictates not just two distortions, length contraction and time dilation, but a third distortion as well, one that can be termed "time dissynchronicity". Without that third, it is difficult to envision a resolution. I do not recommend complicating the picture with accelerations, especially as it is entirely unnecessary.

    Here is the full story: http://www.sysmatrix.net/~kavs/kjs/addend4.html
    Last edited by a moderator: Apr 21, 2017
  10. Oct 8, 2004 #9
    Hello Ostren,

    In the example on the website you gave, I don't see how the readings on the buoy and earth clock can be different.
    They are in the same reference frame.

    The two ends of the interval are at different times only when measured on the clock of the moving observer.

    The clock starting points are at different times in the moving frame, but they tick off the same interval.

    Last edited: Oct 8, 2004
  11. Oct 8, 2004 #10
    Yes, but it is from someone else's reference frame they are being perceived. It's fundamental Lorentz transform: separated clocks of a moving elongated frame read different times. I dunno why, but the fools have adopted the verbiage, "The Relativity of Simultaneity", to describe this... no wonder it winds up being obfuscated.

    Not really. That observer can have an entire cosmos filled with comoving clocks belonging to her native frame, and of course they will all read the same time. So it's not *precisely* as you state. The relatively moving observer reckons the Earth-to-buoy clocks to be out of sync. Note that I use the term "reckons" very frequently. In order for lightspeed to be on par from everyone's perspective, this reckoning must be assessed. And in fact, it is truly coming across that way. The spread-out clocks of an alien frame are not in sync, and that's perfectly real. It's hard to detect, because there are so many other dynamics at play at relativistic speeds.. the corrections for image transmission delays are nontrivial. That's why I choose to use the term "reckons". After adjusting for other dynamics, when you open your eyes and LOOK, those alien clocks will be out of accord.

    Right.. I said that in my "Background" section: those clocks advance at the same time-dilated rate.

    [If the original poster, DocWarezz has any difficulty following http://sysmatrix.net/~kavs/kjs/addend4.html, his questions are more than welcome.]
    Last edited by a moderator: Apr 21, 2017
  12. Oct 9, 2004 #11


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    The trajectories match right up until the point someone accelerates.

    The point is that either person can be the oldest when both twins next meet, depending on what happens next.

    Thus there is no paradox, in flat-space time it's unambiguously clear who ages more -the person who ages the most is the person who did not accelerate, whoever that was.
  13. Oct 9, 2004 #12
    Thanks all.
    I still don't get how you know who is accelerating.

    For example, A is accelerating relative to B as much as B is accelerating relative to A.
    How is the symmetry broken? Shouldn't As time dilate relative to B, equally, Bs time should dilate relative to A?

    I understand you've tried to explain this already, but I still don't understand it.. Could someone enlighten me some more please? :)

    Thank you,
  14. Oct 9, 2004 #13


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    Suppose someone draws a triangle on a flat sheet of paper. There is a mathematical result called the triangle inequality that says that the sum of two sides of a triangle will always be greater than the third side.

    This results basically arises from the fact that a straight line is the shortest distance between two points. When you make a detour, you can't possibly shorten the path.

    The situation in relativity and the twin "pardox" is very similar,with two important differences.

    The first difference is that we measure not distance, but the time it takes to traverse a path.

    The second difference is that the longest time interval (rather than the shortest) is a "straight line". Or to use the technical term, the longest time interval arises when a body travels along a "geodesic".

    So instead of saying the shortest distance between two points is a straight line, we say that the longest time between two points is a path known as a geodesic.

    It's very easy to tell when an object is travelling along a geodesic in flat space-time. The situation when gravity involved is only slightly more complex, but if you're having trouble understanding the idea, it's worthwhile understanding the flat space-time case first and saving the case where space-time is not flat (i.e. when there is gravity) for later.

    Moving along a geodesic is the natural motion of a body when no force is applied. When a force is applied to a body, the acceleration of the body is easily measured with an accelerometer, or even the "seat of one's pants". So there is no doubt about who is accelerating, and who is not accelerating. The body that has a force applied to it is the oen that's accelerating. Newton's laws work in an inertial frame that is not accelerating. Newton's laws do not work in an accelerating frame, one feels additional "inertial" forces. So if you are comfortable with the basic physics of Newton's laws and know what an inertial frame is, you shouldn't have any problem determining when a body is accelerating and when it's not accelerating. A body is an inertial frame is one that is not accelerating.
  15. Oct 9, 2004 #14
    As I said in my post #8 in this thread, accelerations only complicate the issue unnecessarily. Did you read my http://www.sysmatrix.net/~kavs/kjs/addend4.html ??

    Accelerations do NOT IN ANY WAY determine who is "actually" moving, as there is no such thing as unequivocal motion under Relativity. So just put that idea out of your mind, is my humble suggestion. Time dilation is NOT the entire story. If it were, then you would truly have a contradiction, because each twin's clock is slowed by time dilation relative to the other. Only by understanding the time dissynchronicity inherent in the Lorentz Transform can you come to appreciate that the paradox is resolved. So say I.

    What pervect answered you about triangles and trajectories -- I just don't know where all that's going except maybe he desires to confuse you further?
    Last edited by a moderator: Apr 21, 2017
  16. Oct 9, 2004 #15
    A and B have to cross a square of grass from one corner to the opposite one. A walks straight across the grass, while B thinks that walking on the grass is forbidden and walks along a path on two sides of the square. B walks faster so that A and B arrive at the opposite corner at the same time. (Thus A gets to the middle of the square just as B is turning the corner). A and B have pedometers and find that B has walked further than A.

    "That's because I walked directly, while your path had a bend in it" says A
    "But direction is relative, so how come it doesn't work the other way round" says B.

    Taking B's point of view, A is walking slower, and at a 45 degree angle to B, and so is only covers half the forward distance that B is covering. This is true on both halves of the walk. Also taking B's definition of forward, A turns through 90 degrees half way through the walk. Thus A should only have walked half as far as B. Paradox?

    In fact all aspects of the twin paradox are mirrored here, with accelerating becoming turning. One can have a "resolution without turns" if instead of turning B passes his pedometer to C who is already walking in the required direction.

    People don't think of this as being paradoxical, so why is there such a problem with the twin paradox?
  17. Oct 9, 2004 #16

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    Your analysis is fine, but realize that you still invoke acceleration: If you wish to have the traveling twin return and compare herself in age with her "stay at home" sister, she must change inertial frames. That's acceleration.

    One twin remains in the same inertial frame throughout the exercise; the other twin does not: she accelerates! That acceleration breaks the symmetry.
    He was explaining how someone traveling a curved worldline through spacetime ages less than someone who travels a straight worldline between the same spacetime points. In the case of the twins, the one that accelerates follows a curved worldline, while the one who stays at home remains on her straight (single inertial frame) worldline and thus ages more.
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  18. Oct 9, 2004 #17


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    What I'm trying to explain is abstract, but it's fundamentaly not very complicated - it's just geometry.

    I'll try some different words, perhaps they'll get through, (or perhaps it won't, we'll have to see).

    What is invariant in special relativity is not time, nor space, but the Lorentz interval. The Lorentz interval is the difference of the squares between the space interval, and the time interval multiplied by the speed of light.

    ds^2 = dx^2 - (c*dt)^2

    So the geometry of special relativity is Lorentzian geometry.

    Lorentz geometry is almost the same as Euclidian geometry, but not quite. There is that pesky minus sign in front of c*dt^2.

    If you remember your Euclidian geometry, you remember that distance is defined by

    ds^2 = dx^2 + dy^2 + dz^2

    (this is the Phythagorean theorem). There are no minus signs.

    This is what makes some of the results a little different for Lorentzian geometry than they are for Euclidean geometry, such as the fact that that a straight line is the longest time between two points. In Euclidean geometry, it's the shortest distance - it's that minus sign in front of the 't' that makes the difference.

    The key point is this. If one draws a diagram, called a space-time diagram, which is just a graph of position vs time, the twin paradox can be represented on this graph by a triangle.

    One observer moves along a "straight line" in the Lorentz geometry, called a geodesic. The other observer does not move along a straight line, he moves along two sides of a triangle.

    Two points determine a line in flat Lorentzian geometry, just as they do in Euclidian geometry. And the parallel postulate works too (as long as the geometry is flat) - so it is not possible for two different straight lines to join the same two points. Thus, when space-time is flat, there is exactly one observer between any two points who follows a geodesic. (Things get considerably more complicated when space-time isnt' flat, which is why I strongly suggest deferring questions about that case until the situation in flat space-time is understood).

    Nobody expects the sum of two sides of a triangle to be equal to the third side in Euclidean geometry. The only reason people expect that the sum of the time readings on two clocks not following a straight line will be equal to the time reading of a third clock that is following a straight line is because they (falsely) think that time is a fundamental invariant. It's not - the true invariant is the Lorentz interval.

    Note that the Lorentz interval measured along a path for any observer is proportional the elapsed time for that observer, because of the definition of the interval. In the body's own frame of reference, the distance it travels is zero. So we have

    ds^2 = -(c*dt)^2

    the Lorentz interval is -c^2 times the time interval.

    What's important in resolving the twin paradox is not actually acceleration. Acceleration shows up on a space-time diagram as the curved line. What's important is that one observer follows a straight line , and the other observer does not.

    This shows up in such things as the magnitude of the difference in the clock readings. The difference in the two clocks when they are re-united depends on the distance travelled, and the "angle" on the space-time diagram. The angle can be seen to be related directly to the velocity change with a little graphical work. The acceleration per se is not important except to the extent that it's product with time determines the total angle, the change in velocity.

    Acceleration may not be ithe ultimate "cause" of the time dilation, but it is a very convenient way of telling who is travelling along a straight line (a geodesic), and who is not travelling along a straight line. And it's not difficult to understand at all - acceleration can be "felt", the laws of physics are not and never have been independent of acceleration. This is why all Newtonian physics must be done in an inertial frame. Newtonian physics will not work in a non-inertial frame. So anyone who understands Newtonian physics should be able to tell, unambiguously, when a body is accelerating, and when it is not. An accelerating body will have a non-zero force on it in any inertial frame - and in a non-inertial frame, Newton's laws won't apply directly.
    Last edited: Oct 9, 2004
  19. Oct 9, 2004 #18
    And here we go again! Given two frames in inertial motion relative to each other, if one frame experiences an acceleration, this acceleration can be detected in the frame experiencing it and it can be said then that the accelerated frame's state of motion relative to the inertial frame's has changed.
  20. Oct 9, 2004 #19
    How's this? Relative velocity equally affects how each observer caclulates the expected time and space distortions ascribed to the other. When only one entity undergoes acceleration, that also affects those calculations but ONLY those made by the accelerating entity in assessing the distortions ascribed to the non-accelerating one. Nothing is concluded about states of motion, as they remain perfectly relative.
  21. Oct 9, 2004 #20


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