In Euclidean geometry the curvature of a line or curve in the reciprocal of the radius of the kissing circle

In hyperbolic geometry we cannot use this as measure of curvature (what is the curvature of an hypercycle?)

What measure can we use instead as measure of curvature?

The natural notion to use is Geodesic curvature which makes sense for curves on any Riemannian manifolds. The name comes from the fact that geodesics have zero curvature.

For example, on the hyperbolic plane with Gaussian curvature $-1$, horocycles have geodesic curvature $1$. Indeed, let's consider Poincaré half-plane model with metric $(dx^2+dy^2)/y^2$. The line $y=1$ is a horocycle that is naturally parametrized by arclength, $\alpha(t) = (t, 1)$. The tangent vector is the unit vector that points to the right. This makes it appear as if the horocycle doesn't curve but we should use parallel transport to judge whether two vectors are parallel.

Take two points $A=(t, 1)$ and $B=(t+h, 1)$ on the horocycle and draw a geodesic between them: it's an arc of a circle with Euclidean radius $\sqrt{1+(h/2)^2}$. The tangent vector to $\alpha$ makes the angle $\sin^{-1}(h/2)$ with the geodesic at both $A$ and $B$, but it's in opposite directions. So, transporting the vector $\alpha'(t)$ from $A$ to $B$ along the geodesic, we see that at the point $B$ it makes the angle $2\sin^{-1}(h/2)$ with $\alpha'(t+h)$. Since the unit tangent rotates by $2\sin^{-1}(h/2)$ over the distance $h$, the geodesic curvature is $$ \lim_{h\to 0} \frac{2\sin^{-1}(h/2)}{h} = 1 $$

Disk model

On second thought, it's easier to use the disk model; the metric will be $4ds^2/(1-x^2-y^2)^2$ so the curvature is still $-1$. The diameter $(-1,1)\times \{0\}$ is a geodesic, and near the center $(0,0)$ its arclength parameterization moves approximately as $t\mapsto (t/2,0)$ when $t\approx 0$. So the parallel transport along this geodesic for small distances near center will be Euclidean, which implies that the geodesic curvature of any curve tangent to this geodesic at $(0,0)$ will be just $1/2$ of its Euclidean curvature at that point. (Here $1/2$ comes from the aforementioned speed of parameterization).

Summary: to compute geodesic curvature in the hyperbolic disk model, move the point of interest to the center by a Möbius transformation, and take $1/2$ of Euclidean curvature there. Examples:

  • Horocycles have geodesic curvature $1$, as shown by the horocycle $x^2+(y-1/2)^2=1/4$ that passes through $(0,0)$ and has Euclidean curvature $2$.
  • A hyperbolic circle of hyperbolic radius $R$ has geodesic curvature $1/\tanh R \in (1,\infty)$. Indeed, when such a circle passes through $(0,0)$, its point furthest from $(0,0)$ is at hyperbolic distance $2R$ from $(0,0)$. Solving $2\tanh^{-1} d = 2R$ yields $d=\tanh R$ for the Euclidean diameter of this circle, so its Euclidean curvature is $2/\tanh R$ and geodesic curvature is $1/\tanh R$
  • Thanks for your answer but I tried to do it the same way. with a ( hyperbolic ) circle with radius r and it doesn't seem to give a right answer ( I got 1/ sinh(r) and that doesn't seem to make sense because some circle will have the same curvature as an horocycle – Willemien Sep 16 '17 at 12:05
  • I added an easier way to compute these. – user357151 Sep 16 '17 at 13:37
  • Still don't see the error in my thinking, can you help? my idea was:curvature is (change in tangent angle) divided bij (arclength) - the circle has an circumference of 2 \pi \sihh(r) and so the curvature 1/sihh(r) – Willemien Sep 19 '17 at 14:09
  • The error is that you are measuring the change of tangent angle in Euclidean terms, instead of using parallel transport with respect to the hyperbolic metric. – user357151 Sep 19 '17 at 15:29
  • thanks, i noticed there are some errors in your answer (sin^{-1}(h/2) should be artan(h/2) or tan^{-1} (h/2) the metric of the poincare disk is ds^2 = (4 / (1-x^2-y^2))dx . also I found the same solution using hyperbolic trigonometry , is it okay if i add them to your answer? PS added links to this post at wikipedia > hyperbolic geometry > circles and hypercycles/ horocycles – Willemien Sep 24 '17 at 12:15

Using Hyperbolic geometry to calculate the geodesic curvature of a circle

If the circle has centre M and a radius $r$

Imagine 2 points on the circle $A$ and $B$

let $C$ be the midpoint of the segment (chord) $AB$

Then the triangle $\triangle ACM$ is a right angled triangle (right angle is $\angle ACM$) and $\angle M = \angle AMC = 1/2 \angle AMB $

For this triangle we have the following formula's

  • $ \tan(\angle CAM) = \cot(\angle M) / \cosh(r) $
  • $ \sinh(AC) = \sin(\angle M) \sinh(r) $
  • $ \tanh(MC) = \cos(\angle M) \tanh(r) $ (not needed)

The angle between the tangent at A and $AB$ is $$ \frac{\pi}{2} -\angle CAM = \frac{\pi}{2} -\arctan \left( \frac{\cot(\angle M)}{\cosh(r)} \right) = \arctan \left( \tan ( \angle M)\cosh(r) \right) $$

And the curvature is $ \lim_{\angle M \to 0} \frac{\arctan \left( \tan(\angle M)\cosh(r)\right)} {\sin(\angle M) \sinh(r)} $

Because $ \lim_{\angle M \to 0} \frac{ \arctan \left( (\tan(\angle M) C \right) }{\sin(\angle M)} = C $

The limit is $\frac{\cosh(r)}{\sinh(r)} =\frac{1}{\tanh(r)}$

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.