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What is the physical explanation for boosts not commuting?

  1. Jan 15, 2010 #1
    I don't understand why an x boost followed by a y boost is different from a y boost followed by an x boost.
    Surely only the resultant velocity matters, and isn't time reversal a symmetry of SR?
    Similarly, why is the matrix for three simultaneous boosts any different from the product of boosts in the x,y, and z directions?
    Could someone clarify?
    Thanks a lot!
     
  2. jcsd
  3. Jan 15, 2010 #2

    bcrowell

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    Well, time reversal is also a symmetry of quantum mechanics, but qm has non-commuting operators.

    One way to explain mathematically the existence of the Thomas rotation is that a boost in the x direction mixes up the x and t coordinates, and a boost in the y direction mixes y and t, so it's not surprising that the final result has sime mixing of x and y.

    Another way to get at it is to imagine manipulating a rod, and considering the effect of the fact that if the accelerations are applied simultaneously everywhere on the rod in one frame, that isn't seen as simultaneous in another frame: http://www.mathpages.com/rr/s2-11/2-11.htm

    I think you can see this via the rod idea. If you do a boost along x, and then along y, the implication is that when you do the y boost, it's being done simultaneously at all points on the rod. Since the rod has already been boosted, the rod's own idea of simultaneity doesn't match up with yours. This disagreement about simultaneity doesn't occur if there's only one boost.
     
  4. Jan 15, 2010 #3

    Dale

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    Hi Rearden, welcome to PF.

    I think the basic reason is mathematical rather than physical. Boosts do not form a group.
     
  5. Jan 16, 2010 #4
    Thanks for the help :) makes things much clearer
     
  6. Jan 16, 2010 #5

    D H

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    The same thing occurs with rotations in three space. Consider a book. Denote the x axis as being in the direction the print goes on a page, the y axis as up and down page (positive up), and the z axis as boring a hole through the pages of the book. Hold the book horizontally. Now make a positive (counterclockwise) 90 degree rotation about x followed by a +90 degree rotation about y. The book will be oriented vertically, ready to insert in a bookshelf. If you reverse the order of the rotations (y followed by x) you will get a different orientation. The binding will be oriented horizontally rather than vertically.

    The reason for this asymmetry is the same as for boosts: Rotations do not form a group.
     
  7. Jan 16, 2010 #6
    This is not correct. Rotations do form a group, but this group is non-commutative.

    Eugene.
     
  8. Jan 16, 2010 #7

    D H

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    Sorry, my bad. Rotations do form a group -- a non-commutative one. The same goes for boosts. Boosts+rotations similarly form a non-commutative group (the Lorentz group), adding translations yields the Poincoire group. It is the non-commutativity that creates the situation described in the OP.
     
  9. Jan 16, 2010 #8

    bcrowell

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    Well, if you take what he asked in the first place, and just translate it into more mathematical language, he's asking two things: (1) why don't boosts form a group, and (2) why do you need to add rotations to the set in order to make a group. It's not really answering his question to say that boosts don't form a group because boosts don't form a group.

    There has to be a physical reason because this is physics. If there was a mathematical reason and no physical interpretation, it would indicate that the math was being applied incorrectly. Seems to me that the physical reason here is the lack of agreement on simultaneity, as discussed in detail at the mathpages link. If you think there's something wrong with the physical analysis at the mathpages link, I'd be interested to hear it.
     
  10. Jan 16, 2010 #9
    It is this mixing of space and time that is most interesting to me. I still do not feel I have any intuitive understanding of this. This mixing seems to be the source of heated exchanges between people about time being constant or not constant. It is spacetime that is constant neither time nor space are constant. Again I still have not developed an intuitive feel for this.
     
  11. Jan 16, 2010 #10

    D H

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    I disagree.

    The answer is not that boosts don't form a group. That is irrelevant. The answer is that
    • Boosts don't commute; this is a mathematical consequence of how boosts are defined and how they combine and
    • That "only the resultant velocity matters" is a false expectation. The OP is bringing too much baggage from our slow-going everyday world into the realm of special relativity.
     
  12. Jan 16, 2010 #11

    bcrowell

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    The way I would put it is this. In Galilean relativity, a boost mixes up time and space coordinates to make a new space coordinate,
    [tex]
    x'=x+vt \qquad ,
    [/tex]
    and there's nothing spooky and mystical about it at all. Different observers in Galilean relativity disagree about whether two events happened in the same place. In SR, the same thing happens with time as well: a boost mixes up time and space coordinate to make a new time coordinate. Therefore different observers disagree on whether two events happened at the same time.

    If you apply this to the Thomas rotation, it makes it clear why there is no such effect in Newtonian physics. In Galilean relativity, an x boost mixes x and t to make the new x. But if you then do a y boost, it just mixes y and t, neither of which was affected by the previous x boost.

    And the way boosts are defined and the way they combine have physical interpretations. Therefore there is also a physical interpretation for the Thomas rotation.
     
  13. Jan 17, 2010 #12
    Boosts are basically rotations in spacetime. Since rotations in ordinary 3D space don't commute, I don't know why one would expect rotations in 4D spacetime to commute. You could ask why rotations in space don't commute, but its something you can just observe for yourself by rotating a book or a tennis ball with a pencil jammed into it. Boosts not commuting might even be a direct consequence of space rotations not commuting, since different observers don't always agree on which direction is space and which is time.
     
  14. Jan 18, 2010 #13

    Physics Monkey

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    One might ask the same question in a simpler example, like rotations, as MaxwellsDemon suggested. Or we could even ask why translations commute. I do like the observation that you can just go and look and see that rotations don't commute while translations do. Still, I think we can ask for additional physical insight into the problem.

    The simplest case has got to be translations. Experimentally, translations commute (you can check by walking in squares in a parking lot, though you might get you some strange looks). However, they don't have to commute. If you try walking in big squares on a hilly lawn, you'll find that translations don't commute. I think it's reasonable to say that this is physically because the ground is now curved. Another way to say it is that in some sense the action of translation depends non-trivially on where you start (when there is curvature).

    The same kind of reasoning works for boosts. In Galilean relativity boosts are just like translations. In particular, the space of velocities looks just like [tex] R^3 [/tex] and there is no curvature. What happens in special relativity? Now velocities are limited to be less than or equal to one in magnitude (in units with [tex] c = 1 [/tex]). The space of velocities looks like a ball in [tex] R^3 [/tex]. But this is no ordinary ball, because the boundary of the ball is infinitely far away in a precise sense. For example, it takes an infinite number of small boosts to reach the speed of light. This idea can be translated into a precise notion of a metric on the space of velocities. This metric makes the space of velocities into a curved hyperbolic space. Thus from the physical insight that there is a maximum speed, we should not be surprised that boosts don't commute.

    You can also make precise the idea that boosting depends on where you start in a non-trivial way, but I don't have time to elaborate right now.
     
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