Wigner Rotation

I'm trying to show that the composition of two Lorentz boosts produces a boost and a rotation using the generators from the Lorentz Group. If $\vec{K}$ denotes the Lorentz Boost generators and $\vec{S}$ denotes the rotation generators then two successive Lorentz Boosts in the $x$- and then $y$-directions is given by $$e^{-\xi_yK_y}e^{-\xi_xK_x}$$ How do I go on from here to prove the result is another boost and a rotation? I know I have to make use of the commutator $$[K_i,K_j]=-\epsilon_{ijk}S_k$$ but I'm not sure how to proceed.

• – Frobenius Apr 25 '18 at 21:37
• – Cosmas Zachos Apr 26 '18 at 9:35

Well, you are meant to compose the two exponentials of the noncommuting boost generators, possibly by the CBH expansion identity, or else by simply expanding each exponential to a few low orders in the ξ rapidity parameters, as the WP article suggests, and utilize the commutators. In your case, you need $[K_y,K_x]=S_z$, $[K_y,S_z]=-K_x$, and $[K_x,S_z]=K_y$, which happen to close to an SU(2)!
That is, to say, $$e^{-\xi_yK_y}e^{-\xi_xK_x}\\ = \Large e^{-\xi_yK_y-\xi_xK_x +\frac{\xi_y\xi_x}{2} [K_y,K_x]- \frac{\xi_y\xi_x}{12}(\xi_y [K_y,[K_y,K_x]] + \xi_x [K_x,[K_x,K_y]] )+... }\\= \Large e^{-\xi_yK_y-\xi_xK_x +\frac{\xi_y\xi_x}{2} S_z+ \frac{\xi_y\xi_x}{12}(\xi_y K_x + \xi_x K_y )+... } ~~.$$ In point of fact, the $O(\xi^4)$ term in the exponent vanishes in this particular case, and the leading non-vanishing one is $O(\xi^5)$, which you could easily compute.
There is a closed form for such an SU(2) expression, (or, better, the Gibbs composition formula), so a linear combination of the three generators with coefficients to all orders in ξ. The complete expression is a product of a boost and a Wigner rotation, $\exp(\theta S_z)\exp(a K_y +b K_x)$, with $\tan \theta= (\sinh \xi_x \sinh \xi_y )/(\cosh \xi_x +\cosh \xi_y)$, and $a(\xi_y, \xi_x)$ and $b(\xi_x, \xi_y)$ messy Gibbs parameters, and amounts to an SU(2) group identity. (Observe this ordering elements, when combined as above, yields the composite single element found above, to this order in the ξs, $\exp(\xi_x\xi_y S_Z/2+...) =\exp (\xi_y K_y/3 -\xi_x K_x/3+...)$, where the lagging new composite boost is asymmetric in x and y, as it should be!)
To sum up, combining two boosts along different axes x and y produces a Lorentz transformation which is not just a pure boost, the $O(\xi)$ sum of boosts, but also an $O(\xi^2)$ rotation, the Wigner rotation underlying the Thomas precession.
You may also observe that, if you undid these two boosts in the same order, you would get a doubled Wigner rotation as the leading, $O(\xi^2)$, remnant, $$e^{-\xi_yK_y}e^{-\xi_xK_x} e^{\xi_yK_y}e^{\xi_xK_x} \\ = e^{ \xi_y\xi_x [K_y,K_x] + O(\xi^3) +... }= e^{ \xi_y\xi_x S_z + O(\xi^3) +... } ~~.$$ This is called the group commutator, whose logarithm is the corrected Lie algebra commutator, a double Wigner rotation.