The precise formula of the Poincare-Bergman metric on the disc $\mathbb{D}$.

Question

Let $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. If $\gamma:[0,1]\rightarrow{\mathbb{D}}$ is a $C^1$ curve in $\mathbb{D}$, we define the Bergman length of $\gamma$ by $$l_B(\gamma)=\int_0^1\frac{|\gamma'(t)|}{1-|\gamma(t)|^2}dt.$$ For any $z,w\in{\mathbb{D}}$, we define the Bergman distance of $z,w$ to be $$\beta(z,w)=\inf\{l_B(\gamma):\gamma \text{ is a piecewise $C^1$ curve joining $z$ to $w$}\}.$$ My problem is showing that the Bergman distance between 0 and $r$ ($r\in(0,1)$ is real) is given by $\beta(0,r)=\frac{1}{2}\log\frac{1+r}{1-r}$.

I know that we need to show the curve $\gamma(t)=tr+i0,(0\leq{t}\leq{1})$ is the shortest in the Bergman metric among all curves of the form $\gamma(t)=v(t)+iw(t),(0\leq{t}\leq{1})$. But I failed to take infimum with so many symbols. Can someone show me the proof?

Answer 1

Recently, I thought about this problem again and I obtain a simpler proof. So I present the proof here.

For any parametric curve $\gamma$ joining $0$ and $r$, we have (since $1-|\gamma(t)|^2\leq{|1-\gamma(t)^2|}$) \begin{equation*} \begin{split} l_{B}(\gamma) &=\int_0^1\frac{|\gamma'(t)|}{1-|\gamma(t)|^2}\mathrm{d}t\geq\left|\int_0^1\frac{\gamma'(t)}{1-\gamma(t)^2}\mathrm{d}t\right|\\ &=\left|\int_0^r\frac{1}{1-z^2}\mathrm{d}z\right|=\left|\frac{1}{2}\log{\frac{1+r}{1-r}}\right|\\ &\geq{\frac{1}{2}\log{\frac{1+r}{1-r}}}. \end{split} \end{equation*} The last equality holds since $\frac{1}{1-z^2}$ is holomorphic in $\mathbb{D}$ and thus the integral is independent of the path. The curve given by $\widetilde{\gamma}:[0,1]\rightarrow\mathbb{D}$ via $\widetilde{\gamma}(t)=rt$ has Bergman length $$l_B(\widetilde{\gamma})=\frac{1}{2}\log{\frac{1+r}{1-r}}.$$ Therefore, $\beta(0,r)=\frac{1}{2}\log\frac{1+r}{1-r}$.

Answer 2

For a positive integer $n$, consider the annuli $$A_k = \{z : r(k-1)/n< |z| < rk/n\}, \quad k=1,\dots,n$$ Any curve from $0$ to $r$ contains disjoint subcurves connecting the boundary components of $A_k$, for $k=1,\dots,n$. The length of such subcurve is at least $$\frac{r}{n}\frac{1}{1-(r(k-1)/n)^2}$$ where the second factor is the infimum of the metric density on $A_k$. Therefore, the length of any curve from $0$ to $r$ is at least $$\frac{r}{n}\sum_{k=1}^n\frac{1}{1-(r(k-1)/n)^2}\tag{1}$$ and this is true for every $n$. As $n\to\infty$, $(1)$ converges to $$r \int_0^1 \frac{1}{1-(rt)^2}\,dt = \int_0^r \frac{1}{1-s^2}\,ds = \frac12\log\frac{1+r}{1-r}\tag{2}$$ giving a lower bound on the Poincaré distance $\beta(0,r)$. And since the radial segment attains this bound, the right hand side of $(2)$ is indeed $\beta(0,r)$.