13.5: Conformal interpretation
Let us give another interpretation of the h-distance.
Consider the h-plane with the unit circle centered at \(O\) as the absolute. Fix a point \(P\) and let \(Q\) be another point in the h-plane. Set \(x = PQ\) and \(y = PQ_h\) . Then
\[\lim_{x\to 0} \dfrac{y}{x} = \dfrac{2}{1-OP^2}.\]
The above formula tells us that the h-distance from \(P\) to a nearby point \(Q\) is almost proportional to the Euclidean distance with the coefficient \(\dfrac{2}{1-OP^2}\) . The value \(\lambda(P)=\dfrac{2}{1-OP^2}\) is called the conformal factor of the h-metric.
The value \(\dfrac{1}{\lambda(P)}=\dfrac{1}{2} \cdot (1-OP^2)\) can be interpreted as the speed limit at the given point \(P\) . In this case the h-distance is the minimal time needed to travel from one point of the h-plane to another point.
- Proof
-
If \(P=O\) , then according to Lemma 12.3.2
\[\dfrac{y}{x}=\dfrac{\ln \dfrac{1+x}{1-x}}{x}\to 2\]
as \(x\to0\) .
If \(P\ne O\) , let \(Z\) denotes the inverse of \(P\) in the absolute. Suppose that \(\Gamma\) denotes the circle with the center \(Z\) perpendicular to the absolute.
According to the main observation ( Theorem 12.3.1 ) and Lemma 12.3.1 , the inversion in \(\Gamma\) is a motion of the h-plane which sends \(P\) to \(O\) . In particular, if \(Q'\) denotes the inverse of \(Q\) in \(\Gamma\) , then \(OQ'_h=PQ_h\) .
Set \(x'=OQ'\) . According to Lemma 10.1.1 ,
\(\dfrac{x'}{x}=\dfrac{OZ}{ZQ}.\)
Since \(Z\) is the inverse of \(P\) in the absolute, we have that \(PO\cdot OZ=1\) . Therefore,
as \(x \to 0\) .
According to 13.5.1, \(\dfrac{y}{x'} \to 2\) as \(x' \to 0\) . Therefore
as \(x \to 0\) .
Here is an application of the lemma above.
The circumference of an h-circle of the h-radius \(r\) is
\(2 \cdot \pi \cdot \sinh r,\)
where \(\sinh r\) denotes the hyperbolic sine of \(r\) ; that is,
\(\sinh r := \dfrac{e^r-e^{-r}}{2}.\)
Before we proceed with the proof, let us discuss the same problem in the Euclidean plane.
The circumference of a circle in the Euclidean plane can be defined as the limit of perimeters of regular \(n\) -gons inscribed in the circle as \(n \to \infty\) .
Namely, let us fix \(r>0\) . Given a positive integer \(n\) , consider \(\triangle AOB\) such that \(\measuredangle AOB=\dfrac{2\cdot\pi}{n}\) and \(OA=OB=r\) . Set \(x_n=AB\) . Note that \(x_n\) is the side of a regular \(n\) -gon inscribed in the circle of radius \(r\) . Therefore, the perimeter of the \(n\) -gon is \(n\cdot x_n\) .
The circumference of the circle with the radius \(r\) might be defined as the limit
\[\lim_{n\to\infty} n\cdot x_n=2\cdot\pi\cdot r.\]
(This limit can be taken as the definition of \(\pi\) .)
In the following proof, we repeat the same construction in the h-plane.
- Proof
-
Without loss of generality, we can assume that the center \(O\) of the circle is the center of the absolute.
By Lemma 12.3.2 , the h-circle with the h-radius \(r\) is the Euclidean circle with the center \(O\) and the radius
\(a=\dfrac{e^r-1}{e^r+1}.\)
Let \(x_n\) and \(y_n\) denote the side lengths of the regular \(n\) -gons inscribed in the circle in the Euclidean and hyperbolic plane respectively.
Note that \(x_n\to0\) as \(n\to\infty\) . By Lemma \(\PageIndecx{1}\),
\(\lim_{n\to\infty} \dfrac{y_n}{x_n} = \dfrac{2}{1-a^2}.\)
Applying 13.5.2, we get that the circumference of the h-circle can be found the following way:
\(\begin{array} {rcl} {\lim_{n \to \infty} n \cdot y_n} & = & {\dfrac{2}{1 - a^2} \cdot \lim_{n \to \infty} n \cdot x_n =} \\ {} & = & {\dfrac{4 \cdot \pi \cdot a}{1 - a^2} =} \\ {} & = & {\dfrac{4 \cdot \pi \cdot (\dfrac{e^r - 1}{e^r + 1})}{1 - (\dfrac{e^r - 1}{e^r + 1})^2} =} \\ {} & = & {2 \cdot \pi \cdot \dfrac{e^r - e^{-r}}{2} =} \\ {} & = & {2 \cdot \pi \cdot \sinh r.} \end{array}\)
Let \(\circum_h(r)\) denote the circumference of the h-circle of the h-radius \(r\) . Show that
\(\\text{circum}_h(r+1)>2\cdot \text{circum}_h(r)\])
for all \(r>0\) .
- Hint
-
Apply Proposition \(\PageIndex{1}\). Use that \(e > 2\) and in particular the function \(r \mapsto e^{-r}\) is decreasing.