I'm struggling to answer one of the problems in the book Notes on Geometry by Elmer Rees. The question is as follows:

Consider a disc $D^2$ made of a material such that the speed of light at a point $p$ is proportional to the Euclidean distance of $p$ from the boundary. Prove that the light rays are the hyperbolic lines of the Poincaré model.

After trying to solve this using purely mathematical ideas (the book is a classic mathematical geometry graduate text), I found the only way to proceed was by using a Snell's Law (a law of physics and so not in the book). This makes me nervous that I'm perhaps missing the point!

However this does allow me to relate $r$ (the general distance of $p$ from the centre of the disc O, $k$ (the distance of $p$ from O when the light is travelling parallel to the boundary) and the general angle $\theta$ between Op and the direction of motion. The relationship I have derived is: $$1-r = (1-k)\sin\theta$$ Which looks really promising since then $\theta=\pi/2$ when $r=k$ and $\theta=0$ or $\pi$ when $r=1$, as required for a circular path (i.e. hyperbolic lines of Poincaré model). However I still can't show the path of the light is actually circular.

So I have a few questions:

  1. Am I using the right approach, or is there a better way that avoids Snell's Law?
  2. If my approach is correct, is my equation relating $r$ and $\theta$ correct?
  3. If this is right, how can I show the motion is along a circular path?

Many thanks in advance!

  • 1
    $\begingroup$ I think Snell’s Law follows from either of the following: light takes the fastest path between two points; or the path is orthogonal to the family of level curves of the function that gives the minimum time to each point. $\endgroup$ – David K Jul 20 '19 at 11:55

I think I have found the right solution: I thus deleted my previous answer to avoid any possible confusion.

To compute the light path, we can take advantage of Fermat's principle: among all paths, light chooses the one that takes the minimum amount of time. In polar coordinates $(r,\theta)$ the line elements is $ ds=\sqrt{dr^2+r^2d\theta^2}$ and the speed of light inside the unit disk is $v=\alpha(1-r)$, where $\alpha$ is some positive constant. If $k$ is the minimum distance of the light ray from the center, the path must be symmetric with respect to the point nearest to the center and the time light takes to travel from that point to the boundary of the disc is: $$ T=\int_{path}{ds\over v}=\int_k^1{\sqrt{1+r^2\dot\theta^2}\over\alpha(1-r)}dr, $$ where I set $\dot\theta=d\theta/dr$. Unfortunately, this integral diverges: this might be a signal that the model proposed in that book is not realistic, or that there was some misprint.

We can however circumvent this problem by integrating up to $1-\epsilon$ (with $\epsilon>0$), finding the solution and letting then $\epsilon\to0$. To find the function $\theta(r)$ giving minimum time we must solve Euler-Lagrange equation, which doesn't depend on $\epsilon$ and can be written as: $$ {d\over dr}{\partial\over\partial \dot\theta} \left({\sqrt{1+r^2\dot\theta^2}\over\alpha(1-r)}\right)=0, $$ that is: $$ {r^2\over 1-r}{\dot\theta\over\sqrt{1+r^2\dot\theta^2}}=a, $$ where $a$ is a constant which depends on boundary conditions. We can safely choose that half of the path where $\dot\theta\ge0$ (which entails $a\ge0$), square the above equation and solve for $\dot\theta$: $$ \dot\theta={a\over r}{1-r\over \sqrt{r^2-a^2(1-r)^2}}. $$ We can find $a$ by noticing that at $r=k$ we have $dr/d\theta=1/\dot\theta=0$. That entails $k^2-a^2(1-k)^2=0$, that is: $a=k/(1-k)$. Substituting that into the above equation we finally get: $$ \tag{*} \dot\theta={k\over r}{1-r\over\sqrt{(r-k)(r+k-2kr)}}. $$ It is convenient to choose the reference frame so that $\theta(k)=0$. We can then readily integrate $(*)$ to find $\theta(r)$: $$ \tag{**} \theta(r)=\int_k^r{k\over x}{1-x\over\sqrt{(x-k)(x+k-2kx)}}\,dx. $$ The result can be written in closed form: I computed it with Mathematica but haven't been able to cast it yet in a real simple form.

What matters, however, is that this result is not "a hyperbolic line of the Poincaré model", i.e. a circle perpendicular to the boundary of the unit disk. For a point on such a circle on can easily derive that its polar coordinates obey: $$ \theta_{circ}=\arccos\left({k\over r}\cdot{1+r^2\over1+k^2}\right), $$ which gives: $$ \dot\theta_{circ}={k\over r}{1-r^2\over\sqrt{(r^2-k^2)(1-k^2r^2)}}. $$ This is different from equation $(*)$ found above and one can check that the integrand appearing in the expression for $T$ is indeed smaller using $(*)$ than using the expression above.

One can see below some plots of the light paths given by solution $(**)$, for different values of $k$.

enter image description here


As noted by David K in a comment below, taking $v=\alpha(1-r^2)$ and repeating the same steps as above, from the Euler-Lagrange equation one gets an expression for $\dot\theta(r)$ which is the same as that for $\dot\theta_{circ}$ above. In that case, then, one actually finds as solution an arc of a circle, perpendicular to the boundary of the unit disk. This is then probably the form which was intended in the book.

  • $\begingroup$ Interesting. This graph of the light paths is at least qualitatively consistent with the physics (namely, the radius of curvature is always toward the slower part of the disk). I wonder how the exercise was actually intended to be done and whether there is a subtle error in it. The fact that it takes infinite time to reach the boundary may already have been recognized in that method; it merely claims that the paths are the hyperbolic lines, not that any path can be completed in finite time. So if the exercise is wrong it is probably some other error. $\endgroup$ – David K Aug 17 '19 at 15:46
  • 1
    $\begingroup$ Two things: first, I couldn't find any error in your calculations; second, I now have an independent hand-wavy intuitive reason to think the exercise's claim is wrong. I think the speed should have been given as $1-r^2$ rather than $1-r.$ $\endgroup$ – David K Aug 17 '19 at 16:49
  • $\begingroup$ You are right: I repeated the same reasoning with $v=\alpha(1-r^2)$ and one gets for $\dot\theta$ the same expression given above for $\dot\theta_{circ}$. I'll add that to the answer. $\endgroup$ – Intelligenti pauca Aug 17 '19 at 17:04

Snell's law as it is initially taught in high school physics applies to a ray of light passing through a single boundary between two uniform transmission media. There are only two speeds of light in this scenario, one on each side of the boundary.

Now consider two vectors normal (perpendicular) to the boundary between the media, one pointing into the medium with the higher speed and one into the medium with the slower speed. Notice that when the ray of light passes through a boundary between the media, it is always deflected toward the normal vector pointing into the slow medium, and away from the normal vector pointing into the fast medium.

If you have multiple transmission media layered so that the boundaries between media are a set of parallel planes, then you can easily extend Snell's law (by induction on the number of boundaries, if you like) to show that the sine of the incidence or refraction angle on one side of any boundary is proportional to the speed of light on that side of that boundary.

You can even extend this idea to a medium in which the speed of light varies continuously, provided that the places where the speed is $v_1$ all lie in a plane, the places where the speed is $v_2$ all lie in a plane, and those two planes are parallel for any speeds $v_1$ and $v_2.$

To apply the idea of the previous paragraph in two dimensions, you need parallel straight lines instead of parallel planes.

But that is not what you have. If the speed of light is $v_1$ at some point $p_1$ on the disk, the set of all points where the speed is $v_1$ forms a circle. Multiple speeds gives you a set of concentric circles.

The interaction of light with an infinite number of concentric circles is hard to visualize, so let's discretize the system for the time being by dividing the disk into concentric annular rings, each one having the same small width $\Delta r$. The figure below shows the two circular boundaries that separate three of those rings.

enter image description here

This actually shows just a small arc of each circle so that we can enlarge the figure enough to fit some labels between the two boundaries.

Suppose the radius of the inner circle in this figure (the arc through $A$) is $r$; then the speed of light in the ring inside $A$ is $(1 - r + \Delta r)c,$ the speed of light between $A$ and $B$ is $(1 - r)c,$ and the speed of light outside $B$ is $(1 - r - \Delta r)c,$ where $c$ is the speed of light at the center of the disk.

Now you can use Snell's Law to compare $\theta_1$ with $\theta_2$ or to compare $\theta_3$ with $\theta_4.$ But notice that $\theta_3 \neq \theta_2$; in fact, $\theta_3 = \theta_2 - \Delta \phi,$ where $\Delta \phi$ is the angle between the radii from the center of the disk to $A$ and to $B.$ Therefore you cannot rely on Snell's Law alone to compare $\theta_1$ with $\theta_4.$ If a ray of light travels along the path shown in red from the ring inside $A$ to the ring outside $B,$ the angles of incidence/refraction decrease not only because of Snell's Law, but also because the vectors normal to the boundaries that the ray crosses are not parallel.

The curvature of a differentiable curve is the rate at which the direction changes relative to the rate at which we travel along the curve. Note that "direction" is not the angle the curve makes as it crosses a radial line of some system of polar coordinates; the change in direction should be independent of any particular system of coordinates, but you can measure it by observing the change in angle between the tangent line and a fixed line (e.g. the $x$ axis of some coordinate system).

The path in the figure is not differentiable so we cannot speak of its curvature, but it has a corresponding property: starting at $A$ the path goes in one direction toward $B$; it travels a distance $\Delta s,$ and then it changes direction by the amount $\theta_4 - \theta_3$ (a negative number, indicating a rightward turn). If we reduce $\Delta r$ so that we have more rings, as $\Delta r$ goes to zero we should approach the shape of the path of the light across the disk when the speed varies continuously, and $(\Delta s)/\lvert \theta_4 - \theta_3\rvert$ should approach the radius of curvature at each point along the curve. The goal would be to show that on a single path, that limit is constant over the entire path.

I don't have a solution yet, but I think what I wrote at least gives some hope that the exercise you were given is correct. If instead we assumed (incorrectly) that $$ \frac{\sin\theta_4}{\sin\theta_2} = \frac{1 - r - \Delta r}{1 - r} $$ (notice how this compares an angle at $A$ with an angle at $B$), we would get results like the ones already found in another answer, which are not lines of the hyperbolic geometry of a Poincaré disk, thereby disproving the thing you were asked to prove.

Some additional, possibly random thoughts (since I still do not have a complete answer):

There is quite a lot written about the Poincaré disk on this site:

I think you have assumed your disk $D^2$ is a unit disk. If we take a point $P$ inside the unit disk and invert it through the unit circle, we get a point $P'$ on the same radial ray such that $\lvert OP\rvert\lvert OP'\rvert = 1.$ Then every circle through $P$ and $P'$ is a line of the Poincaré model. This set of circles is the elliptic pencil of a set of Apollonian circles, the unit circle belongs to the hyperbolic pencil of the same set of Apollonian circles.

Elsewhere in the various questions on the Poincaré disk I see that a circle around $P$ in the Poincaré model (a set of points all at an equal distance from $P$) is also a circle in the Euclidean metric, but with a center other than $P.$ Assuming your exercise is correctly put, I believe the the set of circles around $P$ (according to the Poincaré model) should be part of the hyperbolic pencil of the set of Apollonian circles described above, and they should correspond to the wave fronts of a disturbance that starts at $P$ and propagates at the speed of light (as defined in your exercise) in all directions.

If you can show that the hyperbolic pencil around $P$ (or at least the part inside the unit disk) is the set of Poincaré circles about $P$, then the elliptic pencil through $P$ (which is orthogonal to the hyperbolic pencil) would be the set of fastest paths from $P$ to any point, and therefore would be the paths of rays of light starting at $P.$

The speed of light defined by your exercise induces a distance metric defined by the time it takes light to travel from one point to another.

For an arbitrary point $P$ there is a Möbius transformation that preserves the unit circle but takes the origin to $P.$ If that transformation preserves the metric induced by the speed of light (does it?) then it takes the circles around the origin to the Poincaré circles about $P.$ Are those circles the hyperbolic pencil around $P$?

Update: It occurs to me that the Poincaré model of the unit disk implies that we can measure distances between points by integrating a distance metric along paths between the points. By several hand-wavy leaps I come to the idea that the metric is chosen so that its geodesics are the hyperbolic lines of the Poincaré model, and that if the speed of light is defined so that it is constant with respect to that metric, it will follow those geodesic paths. If the speed of light is not constant with respect to that metric then in some cases it will be refracted and will depart from a hyperbolic line. By a further leap (based on reading Distance in the Poincare Disk model of hyperbolic geometry and these course notes) I come to the idea that the metric should be proportional to $\frac1{1-r^2}$ where $r$ is the distance from the center, and therefore the speed of light should be proportional to $1-r^2$ if we want light to follow the hyperbolic lines.

Since the exercise says the speed of light is proportional to $1 - r$ rather than $1-r^2,$ it seems credible that the light will not follow hyperbolic lines in all cases. If the path of the light stays near the edge of the disk, however, its speed will be almost proportional to $1 - r^2$ (since the relative error of the missing factor $1 + r$ is small), producing a path that almost follows a hyperbolic line. So if we plot several paths as in Aretino's answer, the ones near the edge of the disk should be almost circular (which they appear to be).

Taking this together with the calculations in Aretino's answer, I'm inclined to agree that the exercise is incorrect.

  • $\begingroup$ You are right: the law to apply in this general case is mentioned here: physics.stackexchange.com/questions/43711/… (see also the answers). $\endgroup$ – Intelligenti pauca Jul 22 '19 at 13:15
  • $\begingroup$ @Aretino I’m glad we’re on the same page. I just wish I knew how the problem ultimately can be solved. I hope I do eventually. $\endgroup$ – David K Jul 22 '19 at 15:42
  • $\begingroup$ @David K, Aretino - thank you both for your help in trying to solve this problem! It follows a chapter on hyperbolic geometry in Rees' book, so surprising if we have to use formulas from physics that are not in the book. I wonder if there is a purely mathematical solution. But anyway, thank you both again, I really appreciate the help! $\endgroup$ – Martin W Jul 23 '19 at 6:48
  • $\begingroup$ Thanks to the help provided, I have an idea of how to solve the problem. However, there is a critical detail that I still need help with, and if unlucky it may mean my idea won’t work! As @DavidK pointed out, we can’t apply Snell’s Law where the lines of constant refractive index are not straight lines. So using a Möbius transformation (which preserves circles) from the disc to the upper half plane which flattens the concentric circles around O to horizontal lines in the upper half plane, we now can use Snell’s Law. And here I can show circular paths! But is there such a Möbius transformation? $\endgroup$ – Martin W Jul 27 '19 at 9:51
  • $\begingroup$ As I recall from my reading while gathering the "random thoughts", there is such a transformation. Of course you get higher speeds this way: the transit time between two horizontal lines has to be the same as the time between the concentric circles they were mapped from, which are generally much closer together. But you can apply Snell's Law then without needing to do any weird jiggery-pokery to account for turning of the normal vector. $\endgroup$ – David K Jul 27 '19 at 11:31

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