In said section, Jackson gives us this equation that relates any physical vector

**G**in a rotating vs. non-rotating reference frame:

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

where

[itex]\boldsymbol{\omega}_T = \frac{\gamma^2}{\gamma+1}\frac{\mathbf{a}\times\mathbf{v}}{c^2}[/itex]

"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

[itex]\left(\frac{d\mathbf{x}}{dt}\right)_{nonrot} = \mathbf{v}[/itex]

and

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

So far so good. Now, this implies that [itex]\boldsymbol{\omega}_T \times \mathbf{x} = \mathbf{v}[/itex]. So if we can get this result from the definition of [itex]\boldsymbol{\omega}_T[/itex], we're golden. However, if you use the fact that [itex]|a| = \frac{v^2}{|x|}[/itex] 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

[itex]\boldsymbol{\omega}_T \times \mathbf{G} = (1-\gamma)\mathbf{v}[/itex]

So this is definitely a contradiction. Because it implies that [itex]\mathbf{v} = (1-\gamma)\mathbf{v}[/itex]. 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.