In Texas kids learn about the number “pi” in 6th grade. I do not know how it is in Kansas. But we are not in Kansas anyway, so we can learn it now.

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We are not in Texas, and we are not in Kansas. We are on the unit disk with hyperbolic geometry defined by the transitive action of SU(1,1) – as it was discussed in the recent series of posts. We know the line element:

\begin{equation}ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.\end{equation}

We can now ask this pressing question: what is the ratio of the circumference of a circle to its radius?

Any point of the disk is as good as any other, but in the standard coordinates $(x,y)$ that we are using the origin of the disk is most convenient. Let us introduce polar coordinates

\begin{eqnarray}

x&=&\rho\cos\phi,\\

y&=&\rho\sin\phi.

\end{eqnarray}

We need to express our line element in terms of $(\rho,\phi).$ We calculate

\[ dx =d\rho\cos \phi-\rho\sin\phi d\phi,\]

\[ dy =d\rho \sin\phi+\rho\cos\phi d\phi,\]

\[dx^2= d\rho^2\cos^2\phi-2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\sin^2\phi d\phi^2,\]

\[dy^2= d\rho^2\sin^2\phi+2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\cos^2\phi d\phi^2,\]

\[dx^2+dy^2=d\rho^2+\rho^2 d\phi^2,\]

\[ds^2=4\frac{d\rho^2+\rho^2d\phi^2}{(1-\rho^2)^2}.\]

Consider now a circle with $\rho=\rho_0$ and $\phi$ running from $0$ to $2\pi.$ What is the length $C$ of its circumference? We have to integrate $ds.$ Going around circumference $\rho$ is constant, so $d\rho=0.$ Therefore

\[C=\int_0^{2\pi}ds=\int_0^{2\pi}\frac{2\rho_0 d\phi}{1-\rho_0^2}=\frac{4\pi \rho_0}{1-\rho_0^2}.\]

What is the length $r$ of its radius? Along the radius $\phi$ is constant, so $d\phi=0$, so

\[r=\int_0^{\rho_0} ds=\int_0^{\rho_0}\frac{2d\rho}{1-\rho^2}=2\mathrm{arctanh}\rho_0.\]

From the last equation we get

\[\rho_0=\tanh\frac{r}{2}.\]

Substituting into the formula for $C$:

\[C=4\pi\frac{\sinh(r/2)/\cosh(r/2)}{1-\sinh^2(r/2)/\cosh^2(r/2)}=2\pi\frac{2\sinh(r/2)\cosh(r/2)}{\cosh^2(r/2)-\sinh^2(r/2)}.\]

Thus

\begin{equation}C=2\pi \sinh r,\end{equation}

and

\begin{equation}\frac{\mathrm{circumference}}{\mathrm{radius}}=\frac{C}{r}=2\pi\frac{\sinh r}{r}.\end{equation}

We can find this formula in Wikipedia article Hyperbolic geometry, in the section Circles and disks (putting there the curvature $R=-1$).

Looking at the graph of the function $\sinh r/r$

we see that the ratio circumference to radius on the disk is always greater than $2\pi.$ In fact the ratio is growing faster and faster with increasing $r$.