general form of a Lorentz transformation in $\mathbb{R}^2$

Question

Is it true that every Lorentz transformation acting on $\mathbb{R}^2$ is of the form \begin{pmatrix} \cosh(s) & \sinh(s) \\ \sinh(s) & \cosh(s) \end{pmatrix} for some $s \in \mathbb{R}$ ? If yes then why is that the case ? I understand there is an analogue, in that every orthogonal transformation on $\mathbb{R}^2$ can be written as \begin{pmatrix} \cos(s) & -\sin(s) \\ \sin(s) & \cos(s) \end{pmatrix} for some $s \in [0,2\pi)$.

Answer 1

So, a Lorentz transformation on $\mathbb R^{1+n}$ is by definition a linear map $L\colon \mathbb R^{1+n}\to \mathbb R^{1+n}$ that preserves the quadratic form $$ \eta(t, x)=t^2-\lvert x\rvert^2.$$ In your question you are interested in $n=1$, in which case $\eta$ reduces to $t^2-x^2$. If you now let $$\tag{1} \tilde{t}=\cosh(s) t + \sinh(s)x, \qquad \tilde{x}=\sinh(s)t+\cosh(s)x, $$ you see that $$\eta(\tilde t, \tilde x)=(\cosh^2(s) - \sinh^2(s))\eta(t, x)=\eta(t,x),$$ because $\cosh^2(s) - \sinh^2(s)=1$, which is the analogous of the familiar relation $\cos^2(s)+\sin^2(s)=1$.

So indeed that matrix you wrote is a Lorentz transformation. Are there any other? Yes there are: the following $$ \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ is a Lorentz transformation and it is not in the family you wrote. However, this last one has determinant $-1$. The transformations (1) all have determinant $+1$.

Actually, the transformations (1) can be characterized as the only positive orthochronous Lorentz transformations. That means that they are the only transformations $L$ with $\det L>0$ (hence $\det L=1$) and such that $L(t, x)=(\tilde t, \tilde x)$ satisfies $\tilde t>0$, provided $t>0$ and $\eta(t, x)>0$; these vector are called "positive time-like" as you noted in comments.