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

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.

So I don't know what to do to put this off hold so I'm putting more content here. Let's get even more specific. Consider the Lorentz-Group $\mathsf{O}(1;3)$. Then let $K_{1},K_{2},K_{3}$ be the generators of the boosts and $L_{1},L_{2},L_{3}$ be the generators of the rotations. The commutation relations are: $[K_{i},K_{j}] = \epsilon_{ijk}L_{k}$, $[L_{i},L_{j}] = - \epsilon_{ijk}K_{k}$, $[L_{j},K_{i}] = \epsilon_{ijk}K_{k}$. Now consider :

$e^{K_{1}}e^{K_{2}} = \exp( K_{1} + K_{2} + \frac{1}{2}L_{3} + \frac{1}{12}(-K_{2} -K_{1}) - \frac{1}{24} \cdot 0 - \frac{1}{720} ( -K_{1} + K_{2} ) ...$

The pattern I see here is that there are only $K_{1},K_{2},L_{3}$ in the exponential. There could be a pattern in the coefficients, but i dont know where the coefficients of the BCH formula come from. I don't know if that's even the right approach to that question.

## closed as off-topic by Danu, ACuriousMind♦, user36790, Sebastian Riese, Kyle KanosNov 14 '15 at 11:51

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• Well, just write the boosts as exponentials of their corresponding algebra elements and use the usual formulae for the Lie algebra exponential. – ACuriousMind Nov 13 '15 at 18:40
• What do you mean with " the usual formulae for the Lie algebra exponential"? If I use BCH-Formula + Commutation Realtions its not clear for me to get the desired form. – Ursus Nov 13 '15 at 19:17
• – Cosmas Zachos Apr 26 '18 at 22:59
• – Cosmas Zachos Apr 28 '18 at 14:50