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 Kanos Nov 14 '15 at 11:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Danu, ACuriousMind, Community, Sebastian Riese, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Well, just write the boosts as exponentials of their corresponding algebra elements and use the usual formulae for the Lie algebra exponential. $\endgroup$ – ACuriousMind Nov 13 '15 at 18:40
  • $\begingroup$ 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. $\endgroup$ – Ursus Nov 13 '15 at 19:17
  • $\begingroup$ Linked . $\endgroup$ – Cosmas Zachos Apr 26 '18 at 22:59
  • $\begingroup$ Closely related. $\endgroup$ – Cosmas Zachos Apr 28 '18 at 14:50