Can you ever obtain a pure rotation from composing Lorentz transformations?

Question

An exercise asks one to show that given $v, u$ speeds much smaller than $c$ and oriented orthagonally, the composition of the lorentz boosts $B(\mathbf{v})B(\mathbf{u})B(\mathbf{-v})B(\mathbf{-u})$ is a rotation, and find the angle and direction of rotation.

Let $B(\mathbf{v})B(\mathbf{u})$ be written as $R(\mathbf{\epsilon})B(c\mathbf{a}/\gamma)$, and $B(\mathbf{-v})B(\mathbf{-u})$ as $B(-c\mathbf{b}/\gamma)R(\epsilon)$. Then the full product is $R(\mathbf{\epsilon})B(c\mathbf{a}/\gamma)B(-c\mathbf{b}/\gamma)R(\epsilon)$. If this is a pure rotation, then $B(c\mathbf{a}/\gamma)B(-c\mathbf{b}/\gamma)$ has to be a pure rotation, I don't see why this is.

Is the rotation aspect here the result of an approximation? In general, can you produce pure rotation through a product of boosts?(edit: yes from naturallyInconsistent's answer)

Doing it via. bruteforcing the matrix, I took the frame where $\mathbf{v}$ and $\mathbf{u}$ are aligned with the x and y axis respectively and got the following matrix for the spatial part of the transformation. Indeed $\mathbf{e_z}$ is an eigenvector and seems to be the axis of rotation, but I couldn't seem proceed to simplify it as $\gamma_v^2(1-\gamma_u^2v)$ goes to $1+v$ whilst $\gamma_u^2(1-\gamma_v^2u)$ goes to $1+u$ for small $u$, $v$. \begin{bmatrix} \gamma_v^2(1-\gamma_u^2v) & \gamma_v\gamma_uvu(\gamma_v+\gamma_u-\gamma_v\gamma_u) & 0 \\ -\gamma_v\gamma_uvu & \gamma_u^2(1-\gamma_v^2u) & 0 \\ 0 & 0 & 1 \end{bmatrix}

I am trying to find a $\theta$ s.t. this is equal to \begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0\\ \sin(\theta) & \cos(\theta) & 0\\ 0 & 0 & 1\\ \end{bmatrix}

Is this what one would usually look for?

Answer 1

Ever? Always, of course, for infinitesimal boosts.

Review your Wigner rotations but consider the evidently superior 2$\times$2 matrix representation of the spinor map, which protects you from the busy 4×4 matrix confusions you appear to be laboring under. Utilize rapidities, $$ \vec u = c \hat x \tanh \zeta, \qquad \vec v = c \hat y \tanh \xi, $$ so, for all boost speeds/rapidities, you just have the group commutator $$ B(\vec v)B(\vec u)B(-\vec v)B(-\vec u)= e^{\xi \sigma_2/2} e^{\zeta \sigma_1/2}e^{-\xi \sigma_2/2}e^{-\zeta \sigma_1/2}. $$

The equivalent of this is worked out explicitly in this answer, but for small rapidities this collapses to a triviality, necessarily the algebra commutator augmenting the identity; to lowest (quadratic) non vanishing order, it is but $$ (1+\xi \sigma_2/2 +...) (1+\zeta \sigma_1/2 +...)(1-\xi \sigma_2/2 +...)(1-\zeta \sigma_1/2 +...)\\ = 1+ \zeta \xi ~i\sigma_3/2+... $$ a small rotation around the $\hat z$ axis by an angle $\zeta \xi$.

Now, utilizing the linked answer, you can always arrange that $$ B(-c \tanh\! f ~~(\hat{x} \cos\phi +\hat{y} \sin\phi))~B(\vec v)B(\vec u) = R(\theta \hat z), $$for finite rapidities, without any approximations. $\phi$, θ and f are the functions of v and u specified there.