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

[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)

• I think OP is asking for a concrete worked example, but this SE post appears to do the general case using the commutation relations for the generators of the Lorentz algebra. So very relevant, but possibly not a duplicate. – jacob1729 Nov 23 '19 at 14:57
• @jacob1729 ,while I really appreciate your reply as well as providing me the topic-related link, I - unfortunately - don't have the required mathematical (and special relativity) competency to understand what's going on in that post. I was looking more along the lines of a matrix example but I haven't gotten any luck finding one yet ... Regardless, thanks for the help! – Athenian Nov 23 '19 at 15:36
• I don't have time to write out a full answer but as an exercise for the OP: consider a rod aligned with the $x$ axis and moving in the $+y$ direction. Boost to a new frame $S'$ moving along $x$ at speed $\beta$ in $S$. You should see that the rod is not parallel to the $x'$ axis. – jacob1729 Nov 23 '19 at 16:06
• @jacob1729 , thank you for taking the time to help me! I sincerely appreciate it. I now have a slightly better idea how two perpendicular Lorentz boosts equal to a rotation after a boost. I'll be sure to think about your comment and see how should I provide my example for the question. Otherwise, to reinforce my understanding of what you have stated here, I'll still try my best to find an example of the OP here. – Athenian Nov 23 '19 at 16:24
• Related : General matrix Lorentz transformation. – Frobenius Nov 23 '19 at 17:24

1 Answer

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.

• @Athenian I fixed a typo in the formula for the general boost matrix. One of the subscripts on an $n$ was wrong. If that caused you to have problems verifying the example, I apologize. – G. Smith Nov 24 '19 at 18:35
• Thank you for notifying me of the changes! Your explanation was incredibly clear and I sincerely appreciate the effort you went through to provide me such a stellar example! Thanks to this, I was able to learn a lot regarding Lorentz boosts and transformations. – Athenian Nov 24 '19 at 18:55