# How to obtain Thomas Precession from lie algebra of the Lorentz group?

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it seems to be possible that you can get the Thomas Precession just through the commutation relations of the Lorentz group. With Thomas Precession i mean, that in general the product of two boosts is a boost with a rotation. The exercise 15b) in this book Lie Groups, Lie Algebras, and Some of Their Applications formulates my Problem pretty good.

I get into some details. Let $\mathsf{O}(n;k)$ be the general orthogonal/ pseudo orthogonal group with Lie algebra $\mathsf{so}(n;k)$. I already esatblished a decompositon:

$\mathsf{so}(n;k) = \mathsf{so}(n) \oplus \mathsf{so}(k) \oplus \mathsf{b}(n;k)$ with $\mathsf{b}(n;k)$ beeing the symmetric elements of the lie Algebra, thus the matrices of the form: $\begin{pmatrix} 0 & B \\ B^{tr} & 0 \end{pmatrix} \ \text{with} \ B\in \mathbb{R}^{n \times k}.$ I also showed $[\mathsf{so}(n),\mathsf{so}(k)] = 0$, $[\mathsf{so}(n),\mathsf{b}(n;k)] \subseteq \mathsf{b}(n;k)$, $[\mathsf{so}(k),\mathsf{b}(n;k)] \subseteq \mathsf{b}(n;k)$, $[\mathsf{b}(n;k),\mathsf{b}(n;k)] \subseteq \mathsf{so}(n) \oplus \mathsf{so}(k)$. I also esablished the fact that the exponential map is bijective from $\mathsf{b}(n;k)$ into the sets of boosts(symmetric, positive elements of $\mathsf{O}(n;k)$).

I want to show with that knowledge that the product of two Boosts is a Boost followed by a rotation. My first try was to write the boosts as exponentials of elements in $\mathsf{b}(n;k)$ and then use BCH formula like:

$e^{A}e^{B} = e^{A + B + \frac{1}{2}[A,B] ... }$, but i can't see how the commutator relations from above provide the desired result.

This post imported from StackExchange Physics at 2015-11-16 14:26 (UTC), posted by SE-user Ursus
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