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Brown's derivation of Thomas rotation

  1. Aug 1, 2010 #1

    I'm about half way through this derivation of Thomas precession for simple circular motion, and have got stuck at the equation

    [tex]\Delta \theta=-2 \pi \frac{v^2}{1-v^2}\left \langle \cos^2 \theta \right \rangle_{mean}[/tex]

    I think this represents the total change in angle of a line segment that's moved once around the whole circumference an n-sided polygon at constant speed.

    [itex]\theta[/itex] is defined as the "limiting value of both [itex]\theta_1[/itex] and [itex]\theta_2[/itex] as [itex]\alpha[/itex] goes to zero". I think this represents the change in angle of the line segment as it changes direction at the first vertex, or maybe at each vertex, I'm not sure.

    The angles [itex]\theta_1[/itex], [itex]\theta_2[/itex] and [itex]\alpha[/itex] are defined in the second diagram on the page.

    I don't understand the expression [itex]\left \langle \cos^2 \theta \right \rangle_{mean}[/itex].

    [itex]\theta[/itex] was defined as a constant, but this seems to be treating it as a variable. Is theta being used in this equation as a variable for the limiting values of each pair of adjacent angles? But then if [itex]\theta_1[/itex] and [itex]\theta_2[/itex] approach the same limit as each other, and [itex]\theta_2[/itex] and [itex]\theta_3[/itex] also approach the same limit as each other, won't all of these limiting values be the same? Or is it allowing the possibility that alpha varies, i.e. that this is not a regular polygon, and would it be necessary to treat theta as a variable in that case, even if the equation is only being used to derive the limit as n goes to infinity and [itex]\alpha[/itex] to 1?

    (I'm still hazy about the principles according to which some distinctions can be "neglected" while others must be observed in such proofs involving limits.)
  2. jcsd
  3. Aug 1, 2010 #2
    [tex]\left \langle \cos^2 \theta \right \rangle_{mean}=\frac{cos^2(\theta)}{n}[/itex].
  4. Aug 1, 2010 #3
    [tex]\theta_2-\theta_1=-\cos(\theta)^2 \frac{v^2}{1-v^2} \alpha[/tex]

    "The total precession resulting from one complete revolution around the n-sided polygon is n times the mean value of [itex]\theta_2-\theta_1[/itex] for each of the n vertices of the polygon."

    By analogy with this anglular-brackets-subscript-mean notation, perhaps "the mean value of [itex]\theta_2-\theta_1[/itex]" (also a constant for a given alpha?) signifies

    [tex]\left \langle \theta_2-\theta_1 \right \rangle_{mean}=\frac{\theta_2-\theta_1}{n}[/tex]

    but why wouldn't the total precession simply be [itex]n(\theta_2-\theta_1)[/itex]?

    After all, if "mean" is defined as "divide by n",

    [tex]\Delta \theta = \frac{n(\theta_2-\theta_1)}{n} = \theta_2-\theta_1[/tex]

    [tex]=-\cos^2(\theta) \frac{v^2}{1-v^2} \alpha = -2\pi \frac{v^2}{1-v^2} \left \langle \cos^2(\theta) \right \rangle_{mean}[/tex]

    since [itex]n = 2 \pi / \alpha[/itex]. But then in the limit as [itex]\alpha[/itex] goes to 0, [itex]\theta_2 = \theta_1[/itex], and there'd be no total precession, which would contradict the whole argument of the page.
  5. Aug 1, 2010 #4
    Theta is not defined as a constant, it is defined (for any given incremental segment around the circle) as the limiting value of theta1 and theta2, i.e., the endpoints of that segment. The expression you mentioned is just what it says it is, the mean of cos(theta)2 over the entire circle from theta = 0 to 2pi. This is the quantity inside the square brackets of the very next equation on that page.

    Yes, that's how theta is defined on that page.

    No. Think of a circle with N equidistant points placed uniformly around the circumference. As N increases, the distance betwen any two neighboring points goes to zero, but this does not imply that all the points converge on a single point as N increases. The distance between neighboring points goes to zero, but the number of points goes to infinity, and the total range is the product, which is constant.

    Yes, to understand that page, you need to have at least some grasp of basic calculus concepts, or at least some intuitive sense for such things.
  6. Aug 1, 2010 #5
    Which circle is this? The circle that the polygon is approximating, or some other more abstract circle? I'm puzzled as to how we know that theta will necessarily take on each of these values. It seems to suggest that if we move that little line ab around a circle of any size at any speed, b would always rotate exactly one full circle, pivoting around a. Obviously that's not the case, so I must be completely misunderstanding what theta represents.

    So does the limit that theta represents consist of infinitely many values (one for each corner of the polygon, or point on the circle) as n goes to infinity? Could you describe this idea in the conventional notation for defining a function: "theta : A --> B, such that..." (where A and B are sets). I'm still very confused about what's going on here.

    I was thinking: theta1 = theta2, and theta2 = theta3, would normally imply that theta1 = theta3. But I maybe I should be thinking of it as function theta : RxR-->R, theta(theta1, theta2) = x, theta(theta2,theta3) = y, etc. where x and y are real numbers between 0 and 2pi, not necessarily equal to each other.

    Is this an analogy to explain limits, or a literal description of what's happening in this case? In this example, I would think of zero and infinity as limiting values. But these limits seem like constants. Which part of this example is the variable limit that corresponds to theta?
  7. Aug 1, 2010 #6

    Theta is defined as the angle which the little transported segment makes with the x axis of each of the successive frames aligned with the edges of the polygon. In other words, theta is the angle between the little segment and the local edge of the polygon. For small values of v, the segment's orientation changes by only a very small amount as it is transported around the polygon, but the angles of the edges of the polygon vary from 0 to 2pi, so the angle which the segment makes relative to those edges varies from 0 to 2pi. This is what the page means when it says "If the circumferential speed v is small compared with 1 ... the transported vector changes its absolute orientation only very slightly on one revolution. In this case it follows that theta varies essentially uniformly from 0 to 2pi as the vector is transported around the circle."

    Yes, you're misunderstanding. I think the thing you're missing is that, to analyse each segment of the polygon, the coordinate system is being re-aligned with the edge, and theta always represents the angle between the segment and the local edge.

    Yes, at each vertex, theta1 and theta2 are close together, and as the number of vertices increases they approach each other, and the common value they approach is called theta.

    Done, just copying what is said on that page.

    No, that's still not right. The angular distance between theta1 and theta3 also goes to zero as n goes to infinity. You need to stop thinking in terms of things being equal, and start thinking of the limit that the things approach as n increases, and you need to parameterize over some finite arc of the circle, so that the total angular range is held constant as the individual angles approaches zero and the number of those angles goes to infinity. This is just basic concepts of calculus.

    Both (assuming you are able to abstract away the rotation of the coordinate system).

    To each of the n points distributed uniformly around a circle, assign an angle theta, and note that the sum of the differences between the theta values of consecutive points equals 2pi. Now, the angle between any two consecutive points approaches zero as n increases, so we can assign a single value of theta to every pair of consecutive points (in the limit as n increases). In fact, for any fixed j, the difference between theta1 and thetaj approaches zero. But this does not mean that all n points approach the same angular position. It remains true that the points are uniformly distributed around the entire range from 0 to 2pi around the circle.

    Again, I think the key point you are missing in the discussion of Thomas precession is that there is a different coordinate system orientation for each edge of the polygon, so the variation in theta is mainly due to the rotation of the coordinate system.
  8. Aug 2, 2010 #7

    This is a very good summary. Being the one that I recommended the link in the first page, I feel a little guilty, it isn't the best , nor the clearest explanation of Thomas precession. "Taylor and Wheeler's "Spacetime Physics", problem 103 is a much better treatment. Same notation but a much better explanation.
  9. Aug 2, 2010 #8
    Ah, I see. Thanks! I didn't realise it was with respect to lots of differently oriented coordinate systems. The initial definition just seemed to be the angle it made with K'.

    Yeah, I'm finding it hard to hold in my mind the apparently contradictory images of the octagon, from diagram 2, and the continuous circle. I'm confused by the fact that the one boost it's described explicity, up to this point, is the one from K to K', with a velocity that makes a zero angle, exactly, to the x axis of K. In this respect, it's unique, since all of the other boosts are at some angle to the x axis of K. Brown and Taylor & Wheeler both seem to say that we can, in some sense, neglect the difference in angle between K' and K'', but obviously we can't neglect all aspects of this difference in angle, and treat is as completely equivalent to the boost from K to K', otherwise there'd be no total difference. As you say, it is a basic calculus thing: what can be neglected and when? I feel like I have a bunch of ad hoc rules-of-thumb for special cases, but frustratingly no general, secure, explicit knowledge of the principles. Maybe I need to do more examples and getting more experience and then it'll become obvious.

    Close, small, "small", nearly... These words worry me when I come to an unfamiliar topic because they usually signal that something is to be ignored, but possibly not in every aspect, or not at every stage in the proof, and I'm probably expected to decide by my own obviously unreliable intuition which bits and where.

    So does the example show a physical object rotating with respect to one coordinate system, K' or K? I guess this is the goal. And does it aim to do so by means of calculations involving infinitely many other coordinate systems, each moving in a different direction? I have a little bit of experience with the algebra of relating measurements in different coordinate systems moving parallel to a mutual axis, and have just worked through a few derivations of the formula for a boost in a general direction, but the idea of composing infinitely many of them is still a bit giddying.

    I think I see. Would it be right to say that expressions like "two consecutive points on a circle" and "two consecutive real numbers" are both meaningless, as such, unless some limiting procedure is specified.

    Yeah, I thought it was making some calculation involving a comparison between K and K', and maybe K'', and then generalising this one result by summing it somehow over the circle. I'm finding it hard to see what's changing and what's fixed; what is being compared to what. But thanks for all your help. Hopefully I'll get there eventually... And thanks starthaus for the reading suggestions.
  10. Aug 3, 2010 #9
    For example, what it the principle behind the choice of a first order approximation for sine and cosine in this example, as opposed to some other order of approximation?
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