# If boost matrices are always symmetric, why is it not so in my example?

Consider two consecutive boosts in $$2+1$$ dimensional spacetime, the first along the $$x$$-axis and the second along the $$y$$-axis. The net transformation is $$B_y(\theta_2)B_x(\theta_1)=\begin{pmatrix} \cosh\theta_2 & 0 & -\sinh\theta_2\\ 0 & 1 & 0\\ -\sinh\theta_2 & 0 & \cosh\theta_2 \end{pmatrix}\begin{pmatrix} \cosh\theta_1 & -\sinh\theta_1 & 0\\ -\sinh\theta_1 & \cosh\theta_1 & 0\\ 0 & 0 & 1 \end{pmatrix}\hspace{1.89cm}\\=\begin{pmatrix} \cosh\theta_2\cosh\theta_1 & -\cosh\theta_2\sinh\theta_1 & -\sinh\theta_2\\ -\sinh\theta_1 & \cosh\theta_1 & 0\\ -\sinh\theta_2\cosh\theta_1 & \sinh\theta_2\sinh\theta_1 & \cosh\theta_2 \end{pmatrix}.$$ Since the product boost $$B_y(\theta_2)B_x(\theta_1)$$ can always be written as the product of a rotation and a boost, I can write $$B_y(\theta_2)B_x(\theta_1)=R_z(\phi)B_{\hat n}(\theta)$$. Here, $$R_z(\phi)$$ is the rotation matirx in the $$xy$$ plane and $$B_{\hat n}(\theta)$$ is some boost matrix. By brute force calculation, I find that $$B_{\hat n}(\theta)=R_z^{-1}(\phi)B_y(\theta_2)B_x(\theta_1)\\=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos\phi & \sin\phi\\ 0 & -\sin\phi & \cos\phi\\ \end{array} \right).\left( \begin{array}{ccc} \cosh\theta _2 & 0 & -\sinh\theta _2\\ 0 & 1 & 0 \\ -\sinh\theta _2 & 0 & \cosh\theta _2\\ \end{array} \right).\left( \begin{array}{ccc} \cosh\theta _1 & -\sinh\theta _1 & 0 \\ -\sinh\theta _1 & \cosh\theta _1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\\ ={\small \left( \begin{array}{ccc} \cosh\theta_1\cosh\theta_2 & -\cosh\theta_2 \sinh\theta_1 & -\sinh\theta_2\\ -\cos\phi\sinh\theta_1-\cosh\theta_1 \sin\phi\sinh\theta_2 & \cos\phi\cosh\theta_1+\sin\phi\sinh\theta _1\sinh\theta_2 & \cosh\theta _2\sin\phi\\ \sin\phi\sinh\theta _1-\cos\phi\cosh\theta _1\sinh\theta _2 & \cos\phi\sinh\theta _1\sinh \theta _2-\cosh\theta _1 \sin \phi & \cos\phi\cosh\theta_2\\ \end{array} \right)}$$

If boost matrices are always symmetric (the general form can be found here), why is $$B_{\hat n}(\theta)$$ calculated above is not symmetric? A pointing out my mistake will also be much appreciated.

As stated in other answers, you did in fact get a symmetric matrix, but only for some given values of $$\phi$$. One way to look at this is noticing you started off with two degrees of freedom ($$\theta_1$$ and $$\theta_2$$) but were finishing with three. By then $$\textit{imposing}$$ that the matrix you obtained is symmetric (as you know it must be), you can eliminate this extra variable. The way I found easier to achieve this is by comparing the entries $$B_{23}$$ and $$B_{32}$$:

$$$$\sin(\phi) \cosh(\theta_2) = \cos(\phi)\sinh(\theta_1) \sinh(\theta_2) - \sin(\phi) \cosh(\theta_1),$$$$ which leads to

$$$$\phi_{boost} = \arctan\left(\frac{\sinh(\theta_1) \sinh(\theta_2)}{\cosh(\theta_1) + \cosh(\theta_2)}\right)$$$$

It's not that hard to also check that this value also makes the rest of the matrix symmetric.

• Since this is a homework-like question, it would have been better not to provide a complete answer. Jul 21, 2020 at 4:17
• Why did you put “boost” as a subscript on $\phi$? It’s a rotation angle, not a boost parameter. Jul 21, 2020 at 4:19
• Sorry. It wasn't tagged as homework, though. About the subscript, it's the angle of a rotation generated by boosts. I agree it may be misleading, I just couldn't think of something better. Jul 21, 2020 at 4:25
• @LucasBaldo Thanks. I appreciate your help. The mistake was that I was thinking that for any value of $\phi$, $B_n$ will be symmetric which is obviously wrong. Thank you very much. Jul 21, 2020 at 4:55

Just because a matrix isn’t obviously symmetric doesn’t mean it isn’t symmetric. Given $$\theta_1$$ and $$\theta_2$$, there is a $$\phi$$ which makes the final matrix above symmetric. For example, when $$\theta_1$$ and $$\theta_2$$ are rapidities corresponding to boosts to speed $$0.500c$$, $$\phi$$ is $$0.143$$. Put in the numbers and see.

The general algebraic solution for $$\phi$$ is a common homework problem for students learning about the Wigner rotation, so I am not going to provide it, in accordance with the site’s policies.