$\renewcommand{\Im}{\operatorname{Im}}\renewcommand{\Re}{\operatorname{Re}}$In a short paper by H. Renggli, On holomorphic mappings of annuli into annuli, 1976, I found the following passage:
Let $A$ denote the annulus $\{ z \in \mathbb C: 0 < r < |z| < R \}$. Using a conformal mapping of the universal covering surface of $A$ onto the unit disk one defines in $A$ a non-Euclidean metric (with constant Gaussian curvature $-1$).
The universal covering surface of $A$ is just the strip $$S = \{\, z \in \mathbb C : -\log R < \Im z < -\log r \,\},$$ and the covering map is given by $z \mapsto e^{iz}$. I'm not entirely sure how Renggli wants to proceed here. Of course by the Riemann mapping theorem, there exists a biholomorphic map $S \to \mathbb D = \{z \in \mathbb C : |z| < 1\}$, and we can use this to pull-back the hyperbolic metric from $\mathbb D$ to $S$. However, it is not clear at all if this metric descends to $A$.
Question 1. Is it possible to write down a biholomorphic map $S \to \mathbb D$ explicitly?
My idea to obtain a hyperbolic metric on $A$ is the following: The shift $S + i\log R$ is contained in the upper half plane $\mathbb H = \{z \in \mathbb C : \Im z > 0\}$, so that we may restrict the hyperbolic metric on $\mathbb H$, $$ds^2 = \frac{dx^2 + dy^2}{y^2}$$ to $S + i\log R$. Since this metric is invariant under the group of deck transformations $\operatorname{Deck} = \{\, z\mapsto z + 2\pi k : k \in \mathbb Z\}$, it descends indeed to $A$. In that setting however, $S$ does not map onto the unit disk.
Anyway, in the following Renggli claims that the circle $$z(t) = \sqrt{Rr} e^{it}, \quad 0 \leq t \leq 2\pi$$ in $A$ has length $$\label{eq:1}L = \frac{2\pi^2}{\log \frac R r}.\tag{1}$$
Question 2. How can I see this?
Suppose I have any circle $t \mapsto \rho e^{it}$ with $r < \rho < R$. This lifts to the curve $t \mapsto t + i (\log \frac R \rho)$ in $S + i \log R$. Hence the length of that curve should be $$L = \frac{2\pi}{(\log \frac R \rho)^2} = \frac{2\pi}{\big(\log \frac {\sqrt R}{\sqrt r}\big)^2} = \frac{8 \pi}{\big(\log \frac{R}{r}\big)^2},$$ where I have substituted $\rho = \sqrt{Rr}$. This seems to be at odds with \eqref{eq:1}, though. Is my calculation correct?
You need to rescale your $S+ i \log R$ by $\frac{\pi}{\log \frac{R}r}$ to get to the strip $$ S_0 = \{z\,|\,\operatorname{Im} z\in (0, \pi)\}, $$ which can be mapped to the upper half-plane ${\mathbb H}$ by the exponential function.
The final section map is $$ f: A\to {\mathbb H};\ z \mapsto e^{i\pi\frac{\log\frac{z}{r}}{\log\frac{R}{r}}}. $$ Note that $\log\frac{z}{r}$ is well defined up to adding a multiple of $2\pi i$, and that would multiply the image by $$ e^{i\pi \frac{i2\pi}{\log\frac{R}{r}}} = e^{-\frac{2\pi^2}{ \log\frac{R}{r}}}. $$
Of course, it is the inverse of this that is the covering map ${\mathbb H}\to A$. So to summarize, $$ A = {\mathbb H}/(w \sim \lambda w),\ \lambda = e^{\frac{2\pi^2}{ \log\frac{R}{r}}}>1, $$ as discussed in many posts, for example, Riemann Surface of an Annulus.
Note that this $\lambda$ is apparently related to the arc-length (1) of the middle circle in the post.
To be concrete, let me compute the length of the middle circle $$ z(t)=\sqrt{Rr}e^{it},\quad 0\leq t\leq 2\pi. $$ The image is the curve $$ w(t)=e^{i\pi\frac{\log\Big(\sqrt{\frac{R}{r}} e^{it}\Big)}{\log\frac{R}{r}}} =e^{i\frac{\pi}{2}-\frac{\pi t}{\log\frac{R}{r}}}=ie^{-\frac{\pi t}{\log\frac{R}{r}}}, \quad 0\leq t\leq 2\pi. $$ So the image curve lies on the $y$-axis in $\mathbb H$ between $ie^{-\frac{2\pi^2}{\log\frac{R}{r}}}$ and $i$. Note the hyperbolic length on the $y$-axis from $ia$ to $ib$ is $$ \int_a^b \frac{dy}y = \log \frac{b}{a},\quad 0<a\leq b. $$ So the length we are after is $$ \frac{2\pi^2}{\log\frac{R}{r}} $$ as desired.
To be complete, I also write down the induced hyperbolic metric from $$ \frac{|dw|}{\operatorname{Im} w} \quad\text{on}\quad {\mathbb H} $$ to the annulus $A=\{z\,|\, r<|z|<R\}$ as $$ \frac{\frac{\pi}{\log\frac{R}{r}}}{\sin\Big(\frac{\pi}{\log\frac{R}{r}}\log\frac{|z|}{r}\Big)}\frac{|dz|}{|z|}. $$ Note that if $R=1$ and we let $r\to 0^+$, this approaches the standard hyperbolic metric on the unit punctured disk as $$ \frac{|dz|}{\big|\log|z|\big||z|}, $$ by $$ \sin\Big(\pi\frac{\log|z|-\log r}{-\log r}\Big) = \sin\Big(\pi\frac{\log|z|}{\log r}\Big) \sim \pi\frac{\log|z|}{\log r}$$ as $r\to 0^+$, since $\sin(\theta + \pi)=-\sin(\theta)$.