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

Question

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.

Answer 1

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 $B_{32}$:

\begin{equation} \sin(\phi) \cosh(\theta_2) = \cos(\phi)\sinh(\theta_1) \sinh(\theta_2) - \sin(\phi) \cosh(\theta_1), \end{equation} which leads to

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

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

Answer 2

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.