2

In four dimensional space time an arbitrary element $\Lambda\in\text{SO}(3,1)^{\uparrow}$ of the Lorentz group may be decomposed in the following way: $$\Lambda=\tilde{R}\Lambda_x(\psi)R$$ where $\tilde{R},R$ are spatial rotations and $\Lambda_x(\psi)$ is a Lorentz boost in the x-t plane.

Is the analogous decomposition in three dimensional space-time $$\Lambda=\Lambda_x(\psi)R\text{ ?}$$ I dont think this can be true since this would imply that any 3D Lorentz transformation could be described by 2 continuous parameters but the Lie group $\text{SO}(2,1)^{\uparrow}$ has dimension 3. On the other hand, I have almost proved that the above identity holds from the polar decomposition of real 2x2 matrices $N=OP$ where $N\in\text{ SL}(2,\mathbb{R})$ is the double cover of $\text{SO}(2,1)^{\uparrow}$, $O$ is an orthogonal 2x2 matrix and $P$ is a positive definite symmetric 2x2 matrix.

The piece I am missing in this proof is that a Lorentz transformation is symmetric iff it is a pure Lorentz boost. I have read that this is true, but I am unsure of how to prove it.

EDIT

After reflecting on ZeroTheHero's answer, I realise that what I was proving using the polar decomposition theorem was the more general decomposition $$\Lambda=\text{R}(\alpha)\Lambda_{\hat{n}}(\psi)$$ given in his answer. So there is no contradiction here, this has three continuous parameters just as its Lie group does.

NormalsNotFar
  • 760

1 Answers1

3

The factorization you want is $$ \Lambda=R_z(\alpha)\cdot \Lambda_x(\psi)\cdot R_z(\gamma) $$ where $R_z(\alpha)$ is a rotation in the $xy$ plane by an angle $\alpha$, and $\Lambda_x$is a boost along $x$.

At the group level, the elements are products of rotations $R_z$ in the $xy$ plane and boosts $\Lambda_{\hat n}$ in the direction $\hat n$ in the $xy$ plane, so general element is trivially a product of the form $R_z(\tilde\alpha)\Lambda_{\hat n}(\psi)$.

If $\gamma$ is the rotation that takes $\hat x$ to $\hat n$, then $$ \Lambda_{\hat n}(\psi)=R_{z}^{-1}(\gamma)\Lambda_x(\psi)R_z(\gamma) $$ so that a general element is now factorized as $$ R_z(\tilde\alpha)R_z^{-1}(\gamma)\Lambda_x(\psi)R_z(\gamma)\, . $$ The first two factors are $\in SO(2)$ so can be combined into a single $R_z$ rotation to finally get $$ R_z(\alpha)\Lambda_x(\psi)R_z(\gamma)\, . $$

ZeroTheHero
  • 45,515