Thomas precession Goldstein/Eisberg versus Taylor/Wheeler

  • #1
gjj
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Main Question or Discussion Point

I've looked at Taylor and Wheeler's Spacetime Physics Example 103 on the Thomas Precession and also the discussion of Thomas precession in Eisberg and Goldstein (3rd edition). Both treat the rotation angle gotten by the addition of 2 non-collinear velocities. The answers they get are different and I'm trying to figure out why. In doing the calculations I used the usual approximations [itex]{\beta _y} < < {\beta _x} < < 1[/itex], so for instance, [itex]\left( {\gamma - 1} \right)\frac{{{\beta _y}}}{{{\beta _x}}} \approx \left( {\frac{1}{{\sqrt {1 - \beta _x^2} }} - 1} \right)\frac{{{\beta _y}}}{{{\beta _x}}} \approx \left( {\frac{{\beta _x^2}}{2}} \right)\frac{{{\beta _y}}}{{{\beta _x}}} = \frac{{{\beta _x}{\beta _y}}}{2}[/itex].
In Taylor and Wheeler a meter stick (spin, gyroscope), aligned along the x-axis of S, is given a velocity [itex]{\beta _x}[/itex]relative to S. Its rest frame is then S'. It is then given a velocity [itex]\beta {'_y}[/itex] relative to the S'. After invoking the "simultaneity of relativity" (i.e. as viewed from S, the right side of the meter stick gets a y-velocity later than the left side) it turns out that the angle of rotation of the meter stick, as viewed from S, is a clockwise [itex]{\beta _x}{\beta _y}[/itex]. If the meter stick is taken around a circle it turns out that the clockwise rotation is between 0 and [itex]{\beta _x}{\beta _y}[/itex] depending on the orientation of the meter stick with respect to its motion. On average the rotation is [itex]\frac{{{\beta _x}{\beta _y}}}{2}[/itex].

Eisberg and Goldstein on the other hand do not look directly at the meter stick, but instead look at the rest frames through which the meter stick passes. The original rest frame is [itex]{S_1}[/itex], the next rest frame,
[itex]{S_2}[/itex], is found by giving a boost [itex]{\beta _x}[/itex] with respect to [itex]{S_1}[/itex] and finally
[itex]{S_3}[/itex] is found by giving a boost of [itex]\beta {'_y}[/itex] relative to [itex]{S_2}[/itex]. The rest frame
[itex]{S_3}[/itex] is found to rotate through an angle of [itex]\frac{{{\beta _x}{\beta _y}}}{2}[/itex] as seen by [itex]{S_1}[/itex] and this is then equated with the rotation of the meter stick. They don't have an oscillating term depending on the orientation of the meter stick with respect to the direction of motion.

So, does the meter stick rotate by [itex]{\beta _x}{\beta _y}[/itex] or by [itex]\frac{{{\beta _x}{\beta _y}}}{2}[/itex] when it aligned parallel to the motion? Does the Eisberg/Goldstein approach use the "relativity of simultaneity"? I can't see where they've used it and yet it would be strange if they get any rotation at all without its use? What happened to the oscillating term? Wouldn't the Eisberg/Goldstein approach yield the wrong answer if we did an astronomical calculation of a partial period?
 

Answers and Replies

  • #2
WannabeNewton
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So, does the meter stick rotate by [itex]{\beta _x}{\beta _y}[/itex] or by [itex]\frac{{{\beta _x}{\beta _y}}}{2}[/itex] when it aligned parallel to the motion?
The latter if one is in the non-relativistic limit. Without approximation the instantaneous precession angle is more complicated.

Does the Eisberg/Goldstein approach use the "relativity of simultaneity"?
Yes of course, just not explicitly.

I can't see where they've used it and yet it would be strange if they get any rotation at all without its use?
It's implicit in the Lorentz transformations used to go from one momentarily comoving inertial frame of the gyroscope to the next. It's no different from explicitly using the relativity of simultaneity as opposed to using the Lorentz transformations when solving any other relativity problem.

What happened to the oscillating term?
They did not calculate the precession angle as a function of the orbital parameter. They only calculated it for a sequence of two consecutive momentarily comoving inertial frames. In order to calculate the precession angle as a function of e.g. the time ##t## of the background global Lorentz frame one would have to do the calculation for an arbitrary continuous sequence of momentarily comoving inertial frames of the gyroscope. This can easily be done and in this the oscillating terms will show up. One can then average the result over an entire period or a partial period of the gyroscope orbit.

I suggest working through the following exercise (Rindler "Relativity: Special, General, and Cosmological" 2nd edition):
Rindler thomas precession.png


See also Bill_K's entry on calculating gyroscopic precession in flat space-time for an arbitrary sequence of momentarily comoving (with the gyroscope) inertial frames with axes fixed to the distant stars: https://www.physicsforums.com/threads/precession-in-special-and-general-relativity.768752/#post-4840026
 
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  • #3
gjj
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The latter if one is in the non-relativistic limit. Without approximation the instantaneous precession angle is more complicated.
Thanks for your help WannabeNewton. Why do you say that Taylor/Wheeler are not working in the non-relativistic limit? From what I can see both Goldstein/Eisberg and Taylor/Wheeler are working in the non-relativistic limit, so I'm not sure how that could account for the different answers.

Yes of course, just not explicitly.



It's implicit in the Lorentz transformations used to go from one momentarily comoving inertial frame of the gyroscope to the next. It's no different from explicitly using the relativity of simultaneity as opposed to using the Lorentz transformations when solving any other relativity problem.

If they are, in essence, doing the same calculation they should get the same answer. But there's that factor of 2 difference. As I say, I don't see that one is in the non-relativistic region and the other isn't.

They did not calculate the precession angle as a function of the orbital parameter. They only calculated it for a sequence of two consecutive momentarily comoving inertial frames. In order to calculate the precession angle as a function of e.g. the time ##t## of the background global Lorentz frame one would have to do the calculation for an arbitrary continuous sequence of momentarily comoving inertial frames of the gyroscope. This can easily be done and in this the oscillating terms will show up. One can then average the result over an entire period or a partial period of the gyroscope orbit.
Actually, I think that's what Rindler does, but his answer for the inclination doesn't seem to depend on the orientation of the spin, so there's no averaging to be done (see below).

I suggest working through the following exercise (Rindler "Relativity: Special, General, and Cosmological" 2nd edition):
View attachment 77048
What I notice in looking at this exercise is that what Rindler does is very similar to Eisberg and along the same lines as Goldstein. He gets the answer [itex]\delta \theta \approx vv'/2{c^2}[/itex] for the inclination [itex]\delta \theta [/itex] of S'' with respect to S caused by two non-collinear boosts. Notice that he doesn't define the orientation of the spin as being in any particular direction. In fact Rindler assumes that irrespective of the position of the spin along the circle (and therefore irrespective of its orientation with respect to the circle) the inclination, [itex]\delta \theta [/itex], of S'' with respect to S caused by two non-collinear boosts is the same. This allows him to multiply [itex]\delta \theta [/itex] by the period to find the total change in inclination for a complete circuit. He doesn't do any averaging to get his answer. On the other hand Taylor and Wheeler end up with the formula [itex]\tan (d\phi ) \approx d\phi \approx - {\beta _r}^2\sin \alpha [/itex] (see a few lines after equation 130). [itex]d\phi [/itex] is the same as Rindler's [itex]\delta \theta [/itex]. The magnitude of [itex]{\beta _r}[/itex] is essentially Rindler's v/c and [itex]{\beta _r}\sin \alpha [/itex] is Rindler's v'/c. So, in Rindler's terminology, Taylor and Wheeler find that when the spin is oriented along the direction of motion the change after the two non-collinear boosts is [itex]{\rm{vv'}}/{c^2}[/itex] Now, Taylor and Wheeler do the averaging of the spin orientation as the spin is taken around the circular path (because sometimes the spin is parallel to the direction of motion and sometimes it's perpendicular and sometimes it's in between and it matters in terms of the resulting inclination after the boosts). It is that averaging that results in the same final formula as Rindler gets for the total change of inclination [itex]{\frac{{\pi v}}{{{c^2}}}^2}[/itex] (see eq. 133). I don't see how you can explain away the factor of 2 difference between Taylor/Wheeler and Rindler as being due to one of them doing a relativistic calculation. It still seems to me that there's a disagreement of some sort and I'm not sure how to resolve it. I don't see any averaging in the Rindler calculation and I don't understand where and how you would put it into that calculation.
.
See also Bill_K's entry on calculating gyroscopic precession in flat space-time for an arbitrary sequence of momentarily comoving (with the gyroscope) inertial frames with axes fixed to the distant stars: https://www.physicsforums.com/threads/precession-in-special-and-general-relativity.768752/#post-4840026
Thanks. I'll take a look.
 
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