# Poincaré invariance of linearly polarised plane wave

I am reading a book that just quotes the Lie group generators and the discrete subgroups that leave a linearly polarised plane wave unchanged.
And I have no idea how to derive them.

## Context

The electromagnetic field tensor $$F_{\mu\nu} = \partial A_{\mu}/\partial x_{\nu} - \partial A_{\nu}/\partial x_\mu$$ is:

$$\left ( \begin{array}{ccc} 0& E_1 & E_2 & E_3 \\ -E_1 & 0 & H_3 & -H_2 \\ -E_2 & -H_3 & 0 & H_1 \\ -E_3 & H_2 & -H_1 & 0 \end{array} \right ),$$

and a Poincaré transformation $$g = (L|v)$$ leaves the field invariant because $$L.F.L^{T}(g^{-1}x) = F(x).$$

However the gauge potential $$A$$ is not left invariant, but gauge transformations come to the rescue: $$L.A(g^{-1}x) = A(x) + \partial \chi.$$

## Problem at hand

Now the book makes an example: a linearly polarised plane wave with wave vector $$q$$ along the $$x_2$$ axis, electric field along $$x_3$$ axis and magnetic field in the $$x_1$$ direction: $$E_3 = H_1 = A\cos(2\pi q.x) \quad \text{with} \quad q^{\mu} = (\omega, 0, \omega, 0) .$$

Then it just states:

This field has as symmetry group a Poincaré transformation with translations $$a$$ such that $$q^\mu a_\mu = 0 \,\,\text{mod}\,\,1$$, and a point group which consists of a Lie group with generators:

$$L_1(t) = \left ( \begin{array}{ccc} 1+\frac{1}{2}t^2 & \sigma & -\frac{1}{2}t^2 & 0 \\ t & 1 & -t & 0 \\ \frac{1}{2}t^2 & t & 1-\frac{1}{2}t^2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right ), \qquad L_2(t) = \left ( \begin{array}{ccc} 1+\frac{1}{2}t^2 & 0 & -\frac{1}{2}t^2 & t \\ 0 & 1 & 0 & 0 \\ \frac{1}{2}t^2 & 0 & 1-\frac{1}{2}t^2 & \rho \\ t & 0 & -t & 1 \end{array} \right ),$$ and a discrete subgroup generated by the transformations $$2_y = \left ( \begin{array}{ccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right ), \quad 2_x'= \left ( \begin{array}{ccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right ), \quad m_x = \left ( \begin{array}{ccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right )$$

I have $$\color{red}{\text{no idea}}$$ how the above matrices are derived.
Does anyone have an intuition on how this was done?

• Are you sure you don't have some typos? What are $\rho,\sigma$? Should they be $t$? Even if so you might have some errors in $L_2$. – Tal Sheaffer Feb 1 at 19:17
• There was a tiny typo in $L_2$, first row, you are right. Anyway I don't know what $\sigma$ and $\rho$ are supposed to be. The book just quotes them, I was hoping they were known things in the field of Lie algebras and Poincaré groups... – SuperCiocia Feb 2 at 0:25
• This is on page 45, Aperiodic Crystals by Ted Janssen – SuperCiocia Feb 2 at 0:27
• Honestly, I don't get it. I think the author may have an error. For one thing, $L F L^{\rm tr} = F$ implies $\det (L) = \pm 1$ (as is generally the case for Lorenz transformations) but from a quick calculation one sees this is only true for $\sigma = \rho = t$. However, even then the metric isn't preserved, e.g. $L_1 g L_{1}^{\rm tr} \neq g$. And neither is the chosen $F$. – Tal Sheaffer Feb 3 at 15:18