# A unusual derivation of thomas precession

I'm very confused about a strange derivation of the thomas precession. I hope someone can clarify me. We define 3 different frame : the laboratory frame, the rest frame of the particle and an inertial frame moving with the particle velocity at $$t = 0$$. After time $$dt$$ the velocity of the particle is changed and we can write it as $$\vec{v}_{0} + d\vec{v}$$. Now we know that in the laboratory frame the particle will rotate through an angle given by: $$$$d\alpha = \frac{\beta \times \dot \beta}{|\beta|^{2}} dt$$$$

We know the equation of spin precession in the inertial frame, so to find the change in the rest frame, because of the relative rotation between the C frame and the I frame (the C frame follows the particle), we have to delete the relative rotation.

$$$$d \textbf{s} = (d\textbf{s})_{I} - d\phi \times \textbf{s}$$$$

where s represent the spin vector and $$d\phi$$ the angle between I and C. Now the point it's that I can't understand how $$d\phi$$ is obtained; the demostration continues with:

at time $$dt$$ the old rest frame ($$t = 0$$) is oriented at an angle $$-d\alpha$$ relative to the new velocity $$\vec{v}_{0} + d\vec{v}$$. Second in the moving intertial frame, both these directions are rotating (same as the ‘new’ velocity) γ times faster than in the laboratory frame. Hence $$$$d\phi = \gamma d\alpha - d\alpha = (\gamma - 1)d\alpha$$$$

This passage leads to the correct form of the thomas precession, but I can't understand how can we subtract an angle relative to the laboratory ($$d\alpha$$) with an angle measured in the intertial frame I. And why this is the correct angle to put in the equation above. An image found in the book follows: