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Thomas Rotation

  1. Jan 11, 2010 #1
    Can anyone explain "Thomas Rotation" to me? If one applies two Lorentz transforms in sequence (even in the same direction!), the result is not the same as the sum of the two in one transform. For example, one transform for Vx=5 km/s, followed by another for Vx=5 km/s, is not the same as one transform for Vx=10 km/s.
    Thanks.
     
  2. jcsd
  3. Jan 11, 2010 #2

    Hans de Vries

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    Thomas precession is the result of non-simultaneity:

    If you accelerate (boost) a moving object from its restframe in a direction
    orthogonal to it's motion, then the boost is not simultaneous in the our frame.

    The boost occurs at different times at different places and the object gets
    skewed, multiple skews becomes rotation. See the image below from my book:
     

    Attached Files:

  4. Jan 11, 2010 #3
    Thanks, but the question concerned two successive transforms in the same direction. (No accelerations, and no orthogonal motion, etc.)
     
  5. Jan 11, 2010 #4

    bcrowell

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    There is no Thomas rotation in that case.

    Hans de Vries, I like the diagram as an intuitive aid, but it has some features that don't really match up correctly with SR. After the first boost, the upper and lower edges are still horizontal, so this is a Galilean boost (which preserves simultaneity) rather than a Lorentz boost. Also, the whole diagram is in a plane, but you need at least three axes for Thomas rotation (x, y, and t). The result should be that two boosts along non-parallel axes are equivalent to a boost plus a rotation, whereas your diagram would seem to imply that two boosts are equivalent to just a rotation.
     
  6. Jan 11, 2010 #5

    Hans de Vries

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    The drawing is correct as far as I can tell. The arrows indicate velocities
    (not accelerations or positions). This means that the accelerations are not
    orthogonal to the velocities but it makes the images simpler to understand.

    The first image is moving in the y-direction. The second image is moving in
    the x-direction and the final Image is moving in the minus y-direction.

    The (instantaneous) accelerations happen simultaneously in the rest frame
    of the image but at different times at different places in our frame. This
    is what is causing the skews.

    You could check this against formula's 11.114 and 11.115 in Jackson were
    the first has a skew (A12=A21) and the second has a rotation (A12=-A21).


    Regards, Hans
     
  7. Jan 12, 2010 #6

    bcrowell

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    Hi, Hans -- Ah, I see. I was interpreting the photos as graphs in an (x,t) plane, but I guess they're really graphs in an (x,y) plane. I suppose that should have been obvious to me, since a photo does, after all, represent a spatial "snapshot" of simultaneity.

    So if I'm understanding correctly, the first acceleration is in the direction of the vector (1,-1), and the second acceleration is in the (-1,-1) direction.

    But this still doesn't quite make sense to me. It seems to me that there should be Lorentz contractions, but there are none that I can see. Say you have spatial coordinates u and v, and a time coordinate t. Then a boost along the u axis preserves area in the (u,t) plane, but doesn't preserve area in the (u,v) plane. All three of your photos have the same area, but it seems to me that they shouldn't.

    Or am I still misinterpreting? Does each skew in your diagram represent the action of a single, specific Lorentz transformation acting on a certain polygonal set of points (projected into the x-y plane), or does it represent something else?

    Is there some text or a caption that goes with the figure?
     
  8. Jan 14, 2010 #7

    Hans de Vries

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    Hi, bcrowell

    The Lorentz contraction is indeed left out as not to complicate the basic mechanism.
    It's valid to do so at lower speeds (v/c) since non-simultaneity is linear in (v/c) while
    the Lorentz contraction in quadratic in (v/c) at lower speeds.

    Regards, Hans
     
  9. Jan 14, 2010 #8

    bcrowell

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    I see. But then I suppose the large angle of rotation isn't really consistent with the approximation that v is small.

    -Ben
     
  10. Jan 14, 2010 #9

    Hans de Vries

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    That's correct. I could make it smaller, The section isn't final yet.

    Regards, Hans
     
  11. Jan 14, 2010 #10

    bcrowell

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    I think making v smaller would have made it harder for me to appreciate the direct intuitive and visual appeal of the diagram. Why not just use the big v, but depict the Lorentz contractions accurately?
     
  12. Jan 14, 2010 #11

    tiny-tim

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    Hi exmarine! :smile:

    (this thread seems to have got a little bit side-tracked :rolleyes: …)

    I think you may be confusing "Thomas precession" with the (nameless, so far as I know) special relativity formula for adding velocities in one dimension.

    Yes, the transformation for Vx = p followed by another for Vx = q in the same direction is not Vx = p + q (unless p = -q, of course :wink:).

    I don't fancy trying to explain it myself, so here's a link … http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity" :smile:
     
    Last edited by a moderator: Apr 24, 2017
  13. Jan 17, 2010 #12
    Thanks tiny-tim. That's correct, I mis-labeled my query I guess. And I had forgotten the relativistic velocity addition stuff. But the side-tracked discussion is also interesting!
     
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