# Thomas Precession, Angular Momentum, and Rotating Reference Frames

1. May 23, 2013

### Izzhov

If any of you 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 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.

So far so good. Now, this implies that $\boldsymbol{\omega}_T \times \mathbf{x} = \mathbf{v}$. So if we can get this result from 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 parallel (really antiparallel) 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 today and we couldn't figure it out.

2. May 23, 2013

### Mentz114

Could it be that you dropped some $\gamma$ factors in $|a| = \frac{v^2}{|x|}$ ?

3. May 23, 2013

### Izzhov

I don't believe I did, because if the book is to be believed both a and v are being measured in the lab frame, which would mean there's no Lorentz transformations going on there.

4. May 23, 2013

### pervect

Staff Emeritus
Wasn't v supposed to be the velocity in the lab frame?

5. May 23, 2013

### Mentz114

V is a relative velocity, it doesn't matter what frame it's measured in.

Another point is that $\omega$,$v$ and $x$ are mutually orthogonal if the origin is the centre of rotation. $v$ and $x$ are in the plane of rotation and $\omega$ points away from it.

Last edited: May 23, 2013
6. May 23, 2013

### Bill_K

Izzhov, You misunderstand the meaning of the Thomas precession. It is not simply the usual difference between a rotating frame vs the lab frame. It is a relativistic effect ON TOP OF the usual rotation. In the nonrelativistic limit it goes to zero. Hence the (γ -1) factor.

7. May 23, 2013

### Izzhov

Now I'm confused. Are you telling me that $\left(\frac{d\mathbf{x}}{dt}\right)_{nonrot}$ is not, in fact, the velocity as measured in the lab frame? If not, what is it?