Special Relativity - Perpendicular Boosts Equaling to a Rotation after a Boost

Question

[Question] I recently read that two perpendicular Lorentz boosts equal to a rotation after a boost. Can anyone here show me an example of this happening? Thank you for your time and assistance!

Source: None - (not a homework question)

Answer 1

Here is an explicit example. The matrix rows and columns are in the usual order $t,x,y,z$.

$$ \left( \begin{array}{cccc} \frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\ 0 & 1 & 0 & 0 \\ -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{cccc} \frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \frac{4 \sqrt{3}}{7} & -\frac{1}{7} & 0 \\ 0 & \frac{1}{7} & \frac{4 \sqrt{3}}{7} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{cccc} \frac{4}{3} & -\frac{2}{3} & -\frac{1}{\sqrt{3}} & 0 \\ -\frac{2}{3} & \frac{25}{21} & \frac{2}{7 \sqrt{3}} & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{7 \sqrt{3}} & \frac{8}{7} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$

I will let you confirm the equality, and that the left-hand-side represents a boost by $c/2$ along the $x$-direction followed by a boost by $c/2$ along the $y$-direction, and that the right-hand-side represents a boost by $\sqrt{7}c/4$ in the direction $(2/\sqrt{7},\sqrt{3/7},0)$ followed by rotation around the $z$-axis by $\cos^{-1}(4\sqrt{3}/7)$ or $8.21$ degrees.

It helps to have the formula for a general boost matrix, which is

$$ \left( \begin{array}{cccc} \gamma & -\gamma \beta n_x & -\gamma \beta n_y & -\gamma \beta n_z \\ -\gamma \beta n_x & 1+(\gamma-1)n_x^2 & (\gamma-1)n_xn_y & (\gamma-1)n_xn_z \\ -\gamma \beta n_y & (\gamma-1)n_yn_x & 1+(\gamma-1)n_y^2 & (\gamma-1)n_yn_z \\ -\gamma \beta n_z & (\gamma-1)n_zn_x & (\gamma-1)n_zn_y & 1+(\gamma-1)n_z^2 \\ \end{array} \right). $$

To get a Wigner rotation, the two boosts don't have to be perpendicular; they just have to be non-colinear. Their composition can also be expressed as a rotation followed by a boost, rather than a boost followed by a rotation. If you express the composition of the boosts as a rotation followed by boost, the resulting rotation will be the same as before, but the resulting boost will be different. For example,

$$ \left( \begin{array}{cccc} \frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\ 0 & 1 & 0 & 0 \\ -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{cccc} \frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{cccc} \frac{4}{3} & -\frac{1}{\sqrt{3}} & -\frac{2}{3} & 0 \\ -\frac{1}{\sqrt{3}} & \frac{8}{7} & \frac{2}{7 \sqrt{3}} & 0 \\ -\frac{2}{3} & \frac{2}{7 \sqrt{3}} & \frac{25}{21} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \frac{4 \sqrt{3}}{7} & -\frac{1}{7} & 0 \\ 0 & \frac{1}{7} & \frac{4 \sqrt{3}}{7} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$

Now the boost direction is $(\sqrt{3/7},2/\sqrt{7},0)$.

If you do the two original boosts in the opposite order, you'll get different results since they don't commute.

ADDENDUM: Wondering how to decompose the product of a general Lorentz transformation into a boost and a rotation? See this related question.