A hyperbolic rotation is what we get when we slide all the points on the hyperbola along by some angle: https://en.wikipedia.org/wiki/Hyperbolic_angle. I want to do the same thing for a different function.
Consider the following transformation matrix that slides points on the curve along the curve $\ln(x)\ln(y)=1,$ with $s\in \Bbb R,$ and $x,y \ne 0,1.$
$$ T_s = \begin{bmatrix} e^{e^s} & 0 \\\ 0 & e^{e^{-s}} \end{bmatrix}.$$
Here $s$ equals the area of the sector bounded by two rays from the origin to the curve $\ln(x)\ln(y)=1.$ I calculated the area of one such sector in the following way:
$$ s=\int_0^{e} x-\frac{x}{e} ~dx ~+ \int_{e}^{e^{\phi}} \exp\bigg(\frac{1}{\ln(x)}\bigg)-\frac{x}{e}~dx. $$ where $\phi$ is the golden ratio and the rays are $y=x$ and $y=\frac{x}{e}.$
We can relate $T_s$ to another transformation matrix $H_s$ where $H_s$ is a standard hyperbolic rotation matrix that slides points on the hyperbola along the hyperbola $xy=1.$
$$H_s=\begin{bmatrix} e^s&0\\ 0&e^{-s} \end{bmatrix}.$$
So $T_s$ and $H_s$ are related by $T_s=\exp(H_s).$ This means, via the exponential map, $T_s$ is an image of hyperbolic rotation. And $\ln(x)\ln(y)=1,$ when plotted on a $\log-\log$ plot becomes a hyperbola. We can see this by letting $u=\ln(x)$ and $v=\ln(y)$ resulting in the hyperbola $uv=1.$
Q: While discussing this with two others, we came to the conclusion that in essence, $T_s$ and $H_s$ both describe the phenomena of hyperbolic rotation. Is this an accurate conclusion? How can I improve my understanding of the matter?