Day: April 14, 2017

Circumference to radius ratio

Posted on by arkajad

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.

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:

(1)   \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

(2)   \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

(3)   \begin{equation*}C=2\pi \sinh r,\end{equation*}

and

(4)   \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.