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

Physics Asked on October 29, 2021

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 $$B_y(theta_2)B_x(theta_1)=begin{pmatrix}
coshtheta_2 & 0 & -sinhtheta_2\
0 & 1 & 0\
-sinhtheta_2 & 0 & coshtheta_2
end{pmatrix}begin{pmatrix}
coshtheta_1 & -sinhtheta_1 & 0\
-sinhtheta_1 & coshtheta_1 & 0\
0 & 0 & 1
end{pmatrix}hspace{1.89cm}\=begin{pmatrix}
coshtheta_2coshtheta_1 & -coshtheta_2sinhtheta_1 & -sinhtheta_2\
-sinhtheta_1 & coshtheta_1 & 0\
-sinhtheta_2coshtheta_1 & sinhtheta_2sinhtheta_1 & coshtheta_2
end{pmatrix}.$$
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 $$B_{hat n}(theta)=R_z^{-1}(phi)B_y(theta_2)B_x(theta_1)\=left(
begin{array}{ccc}
1 & 0 & 0 \
0 & cosphi & sinphi\
0 & -sinphi & cosphi\
end{array}
right).left(
begin{array}{ccc}
coshtheta _2 & 0 & -sinhtheta _2\
0 & 1 & 0 \
-sinhtheta _2 & 0 & coshtheta _2\
end{array}
right).left(
begin{array}{ccc}
coshtheta _1 & -sinhtheta _1 & 0 \
-sinhtheta _1 & coshtheta _1 & 0 \
0 & 0 & 1 \
end{array}
right)\
={small left(
begin{array}{ccc}
coshtheta_1coshtheta_2 & -coshtheta_2 sinhtheta_1 & -sinhtheta_2\
-cosphisinhtheta_1-coshtheta_1 sinphisinhtheta_2 & cosphicoshtheta_1+sinphisinhtheta _1sinhtheta_2 & coshtheta _2sinphi\
sinphisinhtheta _1-cosphicoshtheta _1sinhtheta _2 & cosphisinhtheta _1sinh theta _2-coshtheta _1 sin phi & cosphicoshtheta_2\
end{array}
right)}$$

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.

2 Answers

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} = arctanleft(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.

Answered by Lucas Baldo on October 29, 2021

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.

Answered by G. Smith on October 29, 2021

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