# On reflection properties of convex regions

It is well known that any ray of light passing thru a focus of an ellipse will pass thru the other focus after a single reflection from the ellipse boundary. If $$A$$ and $$B$$ are the foci of an ellipse, this property of rays holds both ways (those passing thru $$A$$ meet at $$B$$ and vice versa).

1. Is there a closed convex region $$C$$ with the property: there exists a pair of points $$A$$ and $$B$$ within $$C$$ such that all rays thru $$A$$ will reflect once on $$C$$ and pass thru $$B$$ but not all rays thru $$B$$ will pass thru $$A$$ after one reflection from $$C$$?

2. Is there a closed convex region $$C$$ such that: there is a pair of points $$A$$ and $$B$$ in the interior such that all rays thru $$A$$ pass thru $$B$$ after exactly 2 reflections from $$C$$? It is sufficient for the ray going directly from $$A$$ to $$B$$ to get reflected exactly twice from somewhere on $$C$$ and then pass thru $$B$$.

Note 1: Question 2 can have 'one-way' (convergence only of rays thru $$A$$ at $$B$$) and 'both-ways' variants.

Note 2: If not explicit constructions, even existence/non-existence arguments could be sought as answers to these questions. One can also ask if relaxing convexity has any implications.

• you need $C$ to be smooth, right?
– erz
Jun 25 '20 at 22:19
• The guess is that a candidate C needs to be smooth for it to have a chance of achieving the above properties. Jun 26 '20 at 6:46
• but we cannot even define reflection at the non-smooth points
– erz
Jun 26 '20 at 7:18
• I understand. Yes. The ellipse is smooth. So, one is looking for other curves in the same spirit. Jun 26 '20 at 12:38

I will assume your regions are closed and have differentiable boundary, otherwsie reflections are not defined for certain directions.

## Question 1: a no-go result.

The answer to Question 1 is no: the reflection property you are interested in is symmetric in $$A$$ and $$B$$.

For every ray $$v$$ originating at $$A$$, write $$f_A(v)$$ for the (unique, by convexity) point where $$v$$ hits the boundary of the convex region. Because of convexity, the map $$f$$ is a bijection between the unit circle and the boundary of the region. If $$v$$ is now a ray originating at $$B$$, similarly define $$f_B(v)$$ to be the unique point of the boundary hit by a ray from $$B$$ in direction $$v$$. If $$x$$ is a point on the boundary, write $$n_x$$ for the normal to the boundary at that point.

By hypothesis, every ray through $$A$$ reflects through $$B$$ after hitting the boundary once. This is the same as saying that for every $$x$$ on the boundary the directions $$f_A^{-1}(x)$$ and $$f_B^{-1}(x)$$ form the same angle to $$n_x$$. The statement is symmetric in $$A$$ and $$B$$, so your condition holds for all reflections from $$A$$ to $$B$$ if and only if it holds for all reflections from $$B$$ to $$A$$.

## Question 2: a no-go result.

The answer to Question 2 is also no: at some point the two reflections must collapse to one.

Indeed, consider any convex region and let $$f_A(v)$$, $$f_B(v)$$ and $$n_x$$ as before. Because the boundary is closed and differentiable and the region is convex, there must exist an $$x$$ on the boundary such that $$n_x = -f_A^{-1}(x)$$, i.e. a ray from $$A$$ towards $$x$$ reflects back through $$A$$. Either $$B$$ is on the way of the reflection, in which case we're done, or the ray must pass through $$A$$, hit the boundary again at another point $$y$$ and from there reflect through $$B$$. But then the ray from $$A$$ in direction $$f_A^{-1}(y)$$ reflects at $$y$$ and then goes through $$B$$ after a single reflection.

## Question 2: relaxing assumptions.

The answer to Question 2 can become yes if some assumptions are relaxed. In the example(s) below, we must allow a finite number of directions in which the second focus is hit only after a single reflection (because of the the no-go result above).

Cofocal parabolas

For this first example we have to allow the points A and B to coincide (must as the circle is a degenerate case of an ellipse where the foci coincide). Then the answer is yes: joining two parabolic arcs with the same focus and axis of symmetry (but opposite vertices wrt the focus) always yields a convex shape with the desired property. Indeed, every ray from the focus (point A) will reflect exactly twice before reaching the focus again (point B), with the exception of two rays parallel to the axis of symmetry of the parabolas. Note that the two rays orthogonal to the axis of symmetry of the parabolas hit the boundary in a non-differentiable point, but the reflection is nevertheless well-defined by continuation.

Half-ellipse and two half-circles

For this second example we have to allow non-convex shapes and we have to relax the requirement of two reflections: let's allow the ray from A to travel through B after the first reflection, as long as it always goes through B after the second reflection.

Consider two distinct points A and B in the plane and let $$a > 0$$ be their distance. Without loss of generality, let the points be at $$(\pm a/2, 0)$$ in the Cartesian plane. Draw the positive y-coordinate half of an ellipse with foci A and B and having semimajor axis $$a$$ (so that the semiminor axis $$b$$ satisfies $$\frac{a}{2} = \sqrt{a^2-b^2}$$, i.e. $$b = \frac{\sqrt{3}a}{2}$$): $$y = \frac{\sqrt{3}}{2}\sqrt{a^2-x^2}$$ Draw the negative y-coordinate half of a circle around $$A$$ with diameter $$a$$: $$y = -\sqrt{\frac{a^2}{4}-\left(x+\frac{a}{2}\right)^2}$$ Draw the negative y-coordinate half of a circle around $$B$$ with diameter $$a$$: $$y = -\sqrt{\frac{a^2}{4}-\left(x-\frac{a}{2}\right)^2}$$

A ray from A with positive y direction will first reflect against the ellipse boundary, then reflect against the circle boundary below B and finally go through B (but it will have passed through B on its way from the ellipse boundary to the circle boundary). A ray from A with negative y direction will first reflect against the circle boundary below A, then reflect against the ellipse boundary and finally go through B.

## Bonus: additional "reflections" of a physical nature.

The answer to Question 1 is related to a property of classical geometric optics sometimes known as reciprocity: the path described by light is independent of the direction in which light travels through the path. Specifically, if a ray starting at $$A$$ in direction $$v$$ follows a path which ends at $$B$$ in direction $$w$$, then light starting at $$B$$ in direction $$-w$$ follows the same path (travelling in the opposite direction) and arrives at $$A$$ in direction $$-v$$.

It may interest you to know that this property fails in curved spacetimes: a ray starting at $$A$$ in direction $$v$$ may follow a path which ends at $$B$$ in direction $$w$$, but light starting at $$B$$ in direction $$-w$$ might not even reach $$A$$ at all. As a pictoresque (and extreme) example of this, consider light falling radially into a Schwarzschild black hole: rays starting a point $$A$$ outside directed radially towards a point $$B$$ inside will reach $$B$$, but rays starting from $$B$$ inside directed radially towards the point $$A$$ outside will never reach $$A$$.

• Hi Stefano, I like your argument for question 2) but the criteria does not strictly exclude the ray passing through the other focus after one reflection as long as it hits again after 2 reflections. I wonder, if we allow this, is there an argument that still works? As you can see I managed to find a curve that works for only 2 rays excluded for each focus. Jun 26 '20 at 15:08
• Indeed, as your (clever!) example shows a surface can be constructed with the required property as long as some singular directions are allowed where the two reflections collapse to one. Also, your remark made me notice that the question does not implicitly require all reflection directions to be defined, but only those from the two points, which makes one of my assumptions (differentiability) perhaps too strong. Jun 26 '20 at 15:15
• Thanks Stefano, note that I actually have to exclude some rays but an arbitrarily small percentage from each focus. I wonder therefore if there is an argument that shows that you have to exclude at least some measurable set of rays from any convex example? Jun 26 '20 at 16:10
• I see: if you could take the limit you'd get to a zero fraction, but for every finite shape you have a non-zero fraction of the rays that don't actually end up reflecting through B. Jun 26 '20 at 17:06
• I added a comment to your answer to this effect: it would be interesting to see if such a deformation is possible. After all, my proof only excludes finitely many points, so it doesn't rule out such a possibility. Jun 26 '20 at 17:18

The answer to question 1) is no:

Take any ray R passing through B. Since C is convex and B lies in the interior of C, R will intersect C at a point O. Now the ray $$\overrightarrow {AO}$$ lies in the interior of C and its reflected ray passes through B. But this means we have a path of light $$\overrightarrow {AO},\overrightarrow {OB}$$. Since C is smooth the tangent at O is unique. Hence we can reverse this to give a light path $$\overrightarrow {BO},\overrightarrow {OA}$$ and hence the ray R from B will reflect off C at O and pass through A.

Since R was arbitrary, any ray passing through B will reflect off C and pass through A.

For question 2) a good example is to take two parabolas facing each other with a common axis, where the parabolas intersect at points A and A':

Rays through the focus of either parabola, not passing through A or A' or hitting the other parabola first will be reflected twice and then pass through the other focus. If we move the foci apart we can get an arbitrarily large percentage of directions to work.

Thus the answer to 2) is yes if we allow an arbitrarily small percentage of rays from each focus to be excluded. This is the 2-way version.

Note that we can prove that such any example has to be 2-way in the same way as we did for question 1).

• I wonder whether there is a way to take the limit of this construction where the two parabolas are taken to infinity and then apply some transformation to the space which brings the shape back to a finite but preserving the horizontal directions followed by the rays between the two reflections. The resulting shape would still have two directions where a single, degenerate reflection happens, at the two points where the parabolas join, but those would be the only places where the two-reflection property fails. Jun 26 '20 at 17:11
• @StefanoGogioso I don't think this can be done unfortunately as any time you have such a parallel set of rays you have a parabolic curve and hence the curve can never become parallel to the parabola axis. This axis must join the two points A and B hence you could never achieve a complete curve without a point where the tangent is discontinuous. Jun 26 '20 at 21:35