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Taylor & Wheeler on Thomas rotation

  1. Aug 7, 2010 #1
    I have a couple of questions about exercise **103 (yes, a two-star problem!) in Taylor & Wheeler's Spacetime Physics.

    In part (a), it says "For an atom [itex]\beta_r \leq Z/137[/itex] (Ex. 101), and for small Z, [itex]\beta_r \ll 1[/itex]. Therefore [itex]\tan(d\phi) \approx d\phi \approx -\beta_r^2 \sin(\alpha)[/itex]." But it's also said that in Newtonian mechanics there is no Thomas precession. And Newtonian mechanics is supposed to be the mechanics of negligibly small beta. Are Taylor and Wheeler talking implicitly about two degrees of negligible smallness (one negligibly small compared to the other), that is: one in which beta is so small that [itex]d\phi = 0[/itex], and another, the subject of this exercise, in which beta is small enough to ignore some aspects of the problem (if so, which exactly?) but not small enough to ignore all precession.

    [tex]\beta_a \ll \ll 1 \Rightarrow d\phi = 0[/tex]

    [tex]\beta_b \ll 1 \Rightarrow \tan(d\phi) \approx d\phi \approx -b_r^2 \sin(\alpha)[/itex]

    where [itex]\beta_a \ll \ll 1[/itex] and [itex]\beta_b \ll 1[/itex] means that [itex]\beta_a \ll \ll \beta_b[/itex].

    *

    Parts (a), (b) and (c) mention two inertial frames of reference: a "laboratory frame" (unprimed), and a "rocket frame" (primed). Borrowing Kevin Brown's notation, I'll call them K and K', and use K'' for the rest frame of ball B. The rocket frame, K', is equated with the rest frame of ball A which is moving right along the x axis of K, according to figures 130 and 131. In the scenario shown in Fig. 131, is the x' axis parallel to the x axis according to K? Is the x' axis parallel to the x axis according to K'? Is the x' axis parallel to the projection of the spin of B onto the xy-plane? Does this last question have a frame-independent answer; if not, how would it be answered wrt K and K'?

    *

    Which frame is Fig. 132 drawn according to? Since the angle alpha is defined in the laboratory frame, K, I guessed it would be K. In support of this is the fact that no obvious distortion of the polygon is shown, no squashing in the direction of motion of A. On the other hand, part (c) seems to be contrasting something being parallel (in part (a) and Fig. 131) with something being perpendicular (in part (b) and Fig. 132); and it's only in K' that the spin projection of B is parallel to the direction of movement, which suggests that Fig. 132 is according to K', perhaps with alpha representing a different angle. I'm presuming the angle alpha is changed by a boost from K to K'.

    *

    I don't understand how the beginning of part (c)--"As the electron moves around its orbit, the projection of its spin axis onto the xy plane of the orbit will occasionally be parallel to its direction of motion (a), and occasionally perpendicular to its direction of motion (b)"--relates to the actual examples in (a) and (b).

    In part (a), the spin projection of electron B is not parallel to its direction of motion, or to that of A, whose direction of motion is said to be effectively the same as that of B when alpha is "small". This is the electron whose spin projection is analysed in part (a) and illustrated in Fig. 131. This is the electron whose spin projection changes direction when we boost from K to K', as part (a) shows. If it had been parallel to the direction of motion of A and K', then surely a boost in this direction would not rotate its spin projection, so [itex]d\phi[/itex] would be zero, not [itex]-\beta_r \sin(\theta)[/itex]. Isn't this exemplified by electron A, whose spin projectin is parallel to the direction of the boost, and indeed not rotated by the boost?

    In part (b), and Fig. 132, both electrons, and in partricular B, have spin projections perpendicular to the boost from K to K' (i.e. to the direction of motion of A), and so their direction is unchanged by that boost, so it might be argued that it doesn't matter. But if both electrons, and in particular B, had been parallel to the boost in K in Fig. 131, then that wouldn't have made a difference to their direction either, and they'd have been parallel in K' too. So how do parts (a) and (b) actually contrast parallel and perpendicular?

    So far, the discussion has involved only a boost from K to K'. There hasn't been any analysis of the scenario wrt K'', so presumably the fact that the spin projection of B is not perpendicular or parallel to the direction of motion of B itself is not directly significant to these questions, I guess because their approximation is going to approximate towards the motion of A rather than the motion of B.
     
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  3. Aug 7, 2010 #2

    bcrowell

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    What edition of the book do you have? As far as I can tell from Amazon, there have been two editions:
    https://www.amazon.com/Spacetime-Physics-Edwin-F-Taylor/dp/0716723271/ref=sr_1_1

    The second edition is the one I own, and it has figures and problems numbered using a notation like x-y, where x is a chapter number and y is an identifying number. It also doesn't seem to have stars on problems. Thomas precession isn't in the index, and there is no problem like the one you describe in chapter L, on the Lorentz transformation, which is where I'd assume this topic would occur.
     
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  4. Aug 7, 2010 #3
    I think mine must be the first edition. Copywrite 1963 and 1966, W.H. Freeman & Co. It's divided into three chapters, and Ex. 103 is the second to last exercise of Chapter 2, "Momentum and Energy", I guess because it refers back to Ex. 101 which deals with the angular momentum of an electron.
     
  5. Aug 7, 2010 #4

    bcrowell

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    Hmm...the second edition has a Chapter 7, "Momenergy," but there's no such problem in it.

    Here's the google books page for the first edition: http://books.google.com/books?id=oB4xNQAACAAJ&dq=taylor+wheeler+%22spacetime+physics%22&hl=en&ei=0pJdTPqKKZG6sQO696WqCw&sa=X&oi=book_result&ct=result&resnum=6&ved=0CE4Q6AEwBTgU [Broken] There doesn't seem to be any way to access a scan of the pages.
     
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  6. Aug 9, 2010 #5
    Scans of the first three pages.

    The quality of the image has suffered a bit in the attachment process. In the left hand diagram of Fig. 131, the line segment between Q and xQ is labelled betartQ. The angle on the left is d phi. The other two angles are both called alpha.
     

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  7. Aug 9, 2010 #6
    ...and of the rest.

    The two rightmost labels in Fig. 133 are:

    [tex](L \cos \phi)(\beta_r^2 \sin \alpha)[/tex]

    and below that:

    [tex]\beta_r^2 \sin \alpha[/tex]
     

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  8. Aug 9, 2010 #7

    bcrowell

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    Yes, there are two degrees of smallness involved. They're asking you to compute the Thomas precession to lowest nonvanishing order in [itex]\beta[/itex]. One line below equation (130), they give a complicated equation relating [itex]d\phi[/itex] and [itex]\beta[/itex]. If you solve this equation for [itex]d\phi[/itex] and then expand it in a Taylor series in terms of [itex]\beta[/itex], the first nonvanishing term is of order [itex]\beta^2[/itex] (since only [itex]\beta^2[/itex] occurs in the equation). They're telling you to throw away the higher-order terms.
     
  9. Aug 9, 2010 #8

    bcrowell

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    Yes. You do get aberration effects in SR, i.e., directions and angles aren't invariant under Lorentz boosts. But by symmetry, a line can't be reoriented by a boost parallel to itself.

    Yes. This is basically equivalent to the previous question, except that the direction of the boost is reversed.

    No, the answer is frame-dependent.

    In K, it's not parallel (left-hand panel of figure 131, where the distinction between the directions of x and x' is irrelevant). In K' it is parallel (right-hand panel of figure 131).
     
  10. Aug 9, 2010 #9

    bcrowell

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    Yes, I think it's drawn according to K, for the reasons you gave. Although alpha would be frame-dependent, I don't think that matters here, because we're carrying out the analysis to lowest order in alpha; at the end, we're going to make the number of vertices on the polygon large, so that alpha becomes small. If you transformed fig 132 into K', I think B's spin would still be perpendicular to the segment of the polygon that A is sitting on top of. That's because K and K' only disagree on simultaneity of events that have different values of x, but when the spin axes are oriented along the y axis, they're lines of constant x. (I'm not totally certain I've got this one right. If you want to make sure, you could probably check by looking up equations for relativistic transformation of angles, and seeing what happens in this special case of a 90-degree angle w.r.t. the boost.)
     
  11. Aug 9, 2010 #10

    bcrowell

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    I don't quite follow you here. In (a), the spin is initially parallel to the direction of motion. This initial state is given by electron A.

    Electron A's spin *doesn't* change direction when you boost from K to K'. What changes direction is electron B's spin.
     
  12. Aug 9, 2010 #11
    If the lines denoted by x and x' axes were parallel according to an arbitrary reference frame, could they be not parallel according to another frame, related to the first by a boost in an arbitrary direction? Come to think of it, is my question ambiguous, even given the context? Would I need to say whether, by x and x', I meant lines in Euclidean space, or lines in Minkowski space? I'm guessing lines in Minkowski space which are parallel according to one inertial coordinate system would have to be parallel according to all, since Lorentz transformations preserve the metric tensor. And I'm guessing that if x and y and x' and y' are lines in space, then, say, x and x' can be made parallel (as defined by the Euclidean dot product) or otherwise by a boost from one coordinate system to another, just as the projection of the spin axis is made parallel to x and x' by the boost from K to K'.

    So what is the xy-plane? Is it a projection of spacetime (or one of its tangent spaces) such that events (or tangent vectors) which differ only by a t or z coordinate are treated as equivalent. Or does the definition change with a change of coordinates to "...which differ only by a t' or a z' coordinate"?

    If B's spin had been parallel to the x axis in fig 131, then B's spin would still be parallel to the segment of the polygon that A is sitting on top of after the transformation of coordinates to K'. That seems like the analogous thing to say about being parallel: as when perpendicular, no change. (c) says that the projection of the spin axis (of B) will sometimes be parallel to the direction of motion (of A and K', towards which the motion of B is to be approximated) and sometimes perpendicular. But it only says there will be no change when it's perpendicular. When parallel, I think (c) says the angle it makes with the direction of motion (of the boost from K to a frame travelling instantaneously with B) will change by [itex]-\beta_r^2 \sin \alpha[/itex]. But this equation was derived for the case shown in fig 131 where the spin projection of B is not parallel to the direction of motion along the x axis. So it doesn't seem to be comparing like with like.

    Alternatively, if it's talking about electron A, its spin is parallel to the direction of motion, and, as you said, doesn't change. And it's the fact of being parallel (or perpendicular) that ensures a spin projection doesn't change doesn't change, in which case, how can they state that for phi = 0 (parallel) there is a change of [itex]-\beta_r^2 \sin \alpha[/itex]?

    Hmm, I suppose the rule is "measure d phi after the boost". In (a), and fig 131, phi is rigged to be the same as d phi. What is the significance of that? It must be to make the different electrons A and B somehow as if they were the same electron. But that seems to suggest a single electron could have two different spin projections according to K which only coincide according to K'. Presumably the anomaly goes away when the approximation is completed--or more likely I'm still not getting the set-up.
     
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