# Lorentz group in 1+1 dimension

Consider the Minkowski 2D metric $$\eta = \text{diag}(-1, 1)$$. The Lorentz group is the set of matrices such that, for a transformation $$\Lambda$$, we get

$$\eta = \Lambda^T \eta \Lambda$$

This means

$$\begin{eqnarray} \Lambda^\dagger \eta \Lambda &=& \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\\ &=& \begin{pmatrix} c^2 - a^2 & cd - ab \\ cd - ab & d^2 - b^2 \end{pmatrix} \end{eqnarray}$$

So that we get the equalities

$$\begin{eqnarray} cd &=& ab\\ c^2 &=& a^2 - 1\\ d^2 &=& 1 + b^2 \end{eqnarray}$$

We can replace $$c$$ with, let's say at random, $$\sinh(u)$$, as it is a bijection on $$\mathbb{R}$$, so that

$$\begin{eqnarray} \sinh^2(u) + 1 &=& a^2 \end{eqnarray}$$

meaning that $$a = \pm \cosh(u)$$. Similarly with $$b = \sinh(v)$$, we get

$$d^2 = 1 + \sinh^2(v)$$

so $$d = \pm \cosh(v)$$. Now there are four possible cases to treat $$cd = ab$$, depending on the signs of $$a$$ and $$d$$ : $$++$$, $$+-$$, $$-+$$ and $$--$$.

For $$++$$ :

$$\begin{eqnarray} cd - ab &=& \sinh(u) \cosh(v) - \cosh(u) \sinh(v)\\ &=& \sinh(u - v) \end{eqnarray}$$

Meaning $$u = v$$, this is just the orthochronous oriented Lorentz transform,

$$\begin{eqnarray} \Lambda = \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}$$

For $$+-$$ :

$$\begin{eqnarray} cd - ab &=& -(\sinh(u) \cosh(v) + \cosh(u) \sinh(v))\\ &=& -\sinh(u + v) \end{eqnarray}$$

Meaning that $$u = -v$$. From the parity properties of hyperbolic functions, that is

$$\begin{eqnarray} \Lambda = \begin{pmatrix} \cosh(u) & -\sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}$$

For $$-+$$ :

$$\begin{eqnarray} cd - ab &=& \sinh(u) \cosh(v) + \cosh(u) \sinh(v)\\ &=& \sinh(u + v) \end{eqnarray}$$

Meaning that $$u = -v$$.

$$\begin{eqnarray} \Lambda = \begin{pmatrix} -\cosh(u) & -\sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}$$

with similar proofs, the rest will be

$$\begin{eqnarray} \Lambda = \begin{pmatrix} -\cosh(u) & \sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}$$

Now the first matrix is entirely fine as it is the $$SO^\uparrow(2)$$ matrix, and the second one is a Lorentz transformation with parity reversal. But the rest doesn't seem to correspond to the usual matrices, which are the Lorentz transform composed with parity and time reversal :

$$\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -\cosh(u) & \sinh(u) \\ -\sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}$$

$$\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \cosh(u) & -\sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}$$

$$\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -\cosh(u) & -\sinh(u) \\ -\sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}$$

where is the error here?

• Maybe because the sign of $u$ is not determined, so that your third Lorentz matrix is in fact the same as the first of the last three matrices, with $u\rightarrow-u$. And similarly your fourth Lorentz matrix is just the last of the last three matrices. – chaostang Aug 9 at 11:17
• True, flipping the signs of the last two would give back all the relevant matrices – Slereah Aug 9 at 11:18