Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If you happen to have the Third Edition of Classical Electrodynamics by John David Jackson, turn to section 11.8, as that's where I'm getting all this from. If not, you should still be able to follow along.

In said section, Jackson gives us this equation that relates any physical vector G in a rotating vs. non-rotating reference frame:

$\left(\frac{d\mathbf{G}}{dt}\right)_{nonrot} = \left(\frac{d\mathbf{G}}{dt}\right)_{rest frame} + \boldsymbol{\omega}_T \times \mathbf{G}$

where

$\boldsymbol{\omega}_T = \frac{\gamma^2}{\gamma+1}\frac{\mathbf{a}\times\mathbf{v}}{c^2}$

"where a is the acceleration in the laboratory frame," according to the textbook. Also, gamma is defined using v, the velocity of the particle as measured in the lab frame.

Ok. So I decided to check this by setting G = x, the position vector, for a particle that is undergoing circular motion in the laboratory frame. So we have

$\left(\frac{d\mathbf{x}}{dt}\right)_{nonrot} = \mathbf{v}$

and

$\left(\frac{d\mathbf{x}}{dt}\right)_{rest frame} = 0$ because the particle doesn't have any velocity in its own frame (right?).

So far so good (I think). Now, this implies that $\boldsymbol{\omega}_T \times \mathbf{x} = \mathbf{v}$. So if we can verify this using the definition of $\boldsymbol{\omega}_T$, we're golden. However, if you use the fact that $|a| = \frac{v^2}{|x|}$ for circular motion as well as the fact that a is perpendicular to v, and that a is (anti)parallel to x, and carefully apply the right hand rule, you'll find, after the algebraic dust settles, that

$\boldsymbol{\omega}_T \times \mathbf{G} = (1-\gamma)\mathbf{v}$

So this is definitely a contradiction. Because it implies that $\mathbf{v} = (1-\gamma)\mathbf{v}$. Can anyone tell me where this went horribly horribly wrong? I worked on this with my professor for two hours yesterday and we couldn't figure it out.

share|improve this question

3 Answers 3

up vote 3 down vote accepted
+100

This is a great question because it brings up some real subtleties that Jackson glosses over -- or at least doesn't make as clear as he could. Different objects have different transformation laws, and the position vector you describe simply does not transform according to the Thomas precession law.

Ultimately, Thomas precession is only useful for vectors that should be measured relative to a "parallel propagated" frame -- that is, a frame carried along with the particle around the circle without any extraneous rotation. (More precisely, the technical name for this carrying is Fermi-Walker transport.) In the original application, the relevant vector was the spin of the electron as it moved around the nucleus. (A more familiar model that's frequently referred to for Thomas precession is a gyroscope, where its spin axis should stay pointing in the "same" direction as it is moved around.) Note that an electron carries its spin along with it, in some sense; it doesn't need to refer to any particular point outside the electron. But the position vector is simply not in this class; it should not be transported with the particle, and you should not use Thomas precession to relate its components in different frames.

Now, you might ask what the transformation law should be, if not the Thomas precession law. Well, you're looking at the time derivative of the position vector, so you're just talking about a velocity as measured in two different frames. You know from basic special relativity that this is described by the velocity-addition formula (with a pretty trivial application in this situation). If observer $n$ (nonrot) measures something moving at velocity $u$, and observer $r$ (rest) is moving with respect to $n$ at velocity $v$, then $r$ measures that thing moving at velocity \begin{equation} u' = \frac{u-v}{1-uv/c^2}~. \end{equation} Here, if the thing being measured is the particle, then $u=v$, so $u' = 0$ is the velocity of the particle in its own rest frame. So, in this case, the appropriate transformation law for the position "vector" $\mathbf{x}$ in your notation is just \begin{equation} \left( \frac{d\mathbf{x}}{dt} \right)_{nonrot} = \left( \frac{d\mathbf{x}}{dt} \right)_{rest} + \mathbf{v}~, \end{equation} which you already knew.

It's also worth mentioning a second subtlety that Jackson glosses over. He talks about the origin of the coordinates of the moving reference frame, but his derivation doesn't actually deal with the origin; it uses rotations and boosts, but no translations. Thus, the formula doesn't actually apply to vectors that specifically refer to the origin -- a fact that you hinted at by noting that the formula applies to "any physical vector $\mathbf{G}$" (emphasis mine). Strictly speaking, $\mathbf{x}$ is a mathematical device with no intrinsic physical meaning. You can use it to label points, so that you could talk about the electric field $\mathbf{E}(\mathbf{x})$ at that point, for example. In particular, its appearance in $\boldsymbol{\omega}_T \times \mathbf{x}$ really makes me nervous without even thinking about the content of the formula. On the other hand, when you differentiate it, you get rid of any dependence on the origin, so $d \mathbf{x} / dt$ is just fine.

To summarize, Thomas precession simply doesn't apply to the position vector. Instead, the relationship between velocity vectors in different frames is just given by the familiar velocity-addition law.

share|improve this answer
1  
The OP used a position "vector", not a displacement vector. –  Larry Harson May 29 '13 at 13:13
    
True. I've gone back and made my language more precise. –  Mike May 29 '13 at 13:23
    
Ok, so does this mean that if, for instance, I set $\mathbf{G} = \mathbf{v}$, the Thomas Precession formula should work? Or is velocity not parallel propagated either? –  Izzhov May 29 '13 at 14:14
    
I ask because I just tried doing that and the algebra still didn't work out. –  Izzhov May 29 '13 at 15:38
    
No, that again is not suited to "parallel propagation", because it refers to structure apart from the particle itself (the laboratory rest frame), much like the electric and magnetic fields. I think the problem is that you really need to use four-vectors to describe things properly in special relativity. This Thomas precession formula is specialized to the case where you have a spatial vector defined in two simple frames, so it's not really applicable to many things. In fact, spinning particles and gyroscopes are the only applications I can think of. –  Mike May 29 '13 at 15:53

I haven't fingured out how to completely solve the problem, but I have found two mistakes (of which I am absolutely certain of) in what you have done. Incorporating these two is still not giving the right answer, something else is wrong as well.

(1) You are using an incorrect expression for acceleration which is valid for classical cases only. That is, $\ a = \frac{\ v^2}{\ x}$ . The correct one is $\ a = \frac{\ v^2 \gamma}{\ x}$

These pages from Hartle's book Gravity will help. See example 5.6, equation 5.52 and whats written after it.

enter image description here

enter image description here

(2) $\ dt$ has to be corrected for. Use $\gamma\ {dt_{rest frame}}=d\tau_{rest frame}$ See the line following equation 11.119 in your textbook.

share|improve this answer
3  
This is still not giving the correct answer, something else is wrong too. –  Man May 27 '13 at 23:49
1  
For (1), does that mean that, in fact, $\mathbf{a} \neq \frac{d\mathbf{v}}{dt}$, but rather $\mathbf{a} = \gamma\frac{d\mathbf{v}}{dt}$? If so, point taken. However, I believe I've already accounted for (2), as the left hand side is in the lab frame, meaning you don't need to refer to proper time, and the derivative on the right hand side is zero, so multiplying it by a constant won't change it. Furthermore, all this only adds factors of $\gamma$, but the solution is off by a factor of $1 - \gamma$. –  Izzhov May 28 '13 at 14:45
    
(1) Yes, so that's settled. (2) I agree, it still doesn't give the answer. I am trying. –  Man May 28 '13 at 15:53
    
Have you had a course on relativity? Because that's what is messing up things. –  Man May 28 '13 at 15:54
    
Yeah, I've taken a seminar on special relativity, though we didn't get quite as advanced as learning Thomas Precession. I was trying to teach myself this more advanced material using Jackson's book when I came up with this problem. –  Izzhov May 28 '13 at 16:35

The thing is that the derivation of this formula implies that $\bf{\omega}_T$ describes additional precession due to relativistic effects:

Goldstein

Note that Jackson in the next equation adds $\bf{\omega}_T$ to $\frac{e\bf{B}}{mc}$ - precession due to magnetic field.

share|improve this answer
    
So are you saying the actual first equation in my question should be $\left(\frac{d\mathbf{G}}{dt}\right)_{nonrot} = \left(\frac{d\mathbf{G}}{dt}\right)_{rest frame} + (\boldsymbol{\omega}_T + \boldsymbol{\omega}) \times \mathbf{G}$, where $\boldsymbol{\omega}$ is the classical angular velocity? But if I go through the algebra with this new equation, I end up with $\mathbf{v} = (2-\gamma)\mathbf{v}$, so there still seems to be an error. –  Izzhov May 28 '13 at 15:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.