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Suppose you are inside a perfectly spherical mirror. You shoot one beam of light and it reflects on the walls of the mirror. Considering the intensity is constant will the beam of light hit you again?

I can shoot it from anywhere inside the sphere and I am a point.

One answer mentions that the beam will always come arbitrarily close. How to prove this?

I thought of this question while sleeping. But even after a day of thinking about this problem I couldn't solve it.

Note

The only answer is brief and does not explain or prove what is says. Looking for detailed answers preferably using illustration and valid mathematical proofs for the statements they make. Would like to read about possible generalizations and variations of this question. Please don;t restrict yourself to this specific situation and do not hesitate to write long answers which give a wide overview of these type problems. Bounty-awarded answer is likely to be long, detailed and self-sufficient mathematically in terms of proving the facts it states.

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    $\begingroup$ Are 'you' a point? $\endgroup$ – Shuri2060 Jul 23 '17 at 13:51
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    $\begingroup$ Did you shoot the beam of light from the center of the sphere? $\endgroup$ – Joel Reyes Noche Jul 23 '17 at 13:52
  • $\begingroup$ I can shoot it from anywhere inside a sphere and I am a point $\endgroup$ – Agile_Eagle Jul 23 '17 at 13:56
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    $\begingroup$ Maybe you consider the plane defined by the first light beam, and going thru the center of the sphere. This way it's obvious that you can consider your problem in the circle defined by the sphere and the plane. I hope you can conclude that the answer to your question is "no". $\endgroup$ – ama Jul 23 '17 at 14:25
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    $\begingroup$ Related: "The case of Captain America's shield: a variation of Alhazen's Billard problem". This question asks, in the current context, how to aim the beam so that it returns to the source in "$n$" bounces. (The only answer ---mine--- considers only the case of $n=2$.) $\endgroup$ – Blue Jul 31 '17 at 4:22
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By the laws of optics, the reflections will occur in the plane formed by the initial ray and the center of the sphere, and this is a planar problem.

Now consider the circle where you are standing and the successive intersection points of the reflected rays with it.

enter image description here

By symmetry, these points will be located at angles that follow an arithmetic progression, and the angular delta from the initial point is $n\alpha\bmod1=\{n\alpha\}$, where the angles are expressed in turns. Now it is clear that the ray will come back iff $\alpha$ is a rational number. If it is irrational it will never come back exactly but you can always find an $n>0$ such that $\{n\alpha\}<\epsilon$ for any $\epsilon$.

The situation is similar for the second intersection points with the ray, at angles $n\alpha+\beta$, where $\beta$ is independent of $\alpha$*. But then, it may turn out that $n\alpha+\beta$ is a integer by coincidence so that even with irrational $\alpha$ there can be an exact return (only one because $n'\alpha+\beta$ cannot be another integer).

To summarize:

The return angles are $\{n\alpha\}$ and $\{n\alpha+\beta\}$, corresponding to

  • irrational $\alpha\to$ zero or one exact return plus infinitely many close ones, or

  • rational $\alpha\to$ infinitely many exact periodic returns but no close ones.


*As said somewhere else, the figure has two degrees of freedom: the (relative) distance to the center and the (relative) starting ray direction.

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Each time the light reflects it then traverses a chord, and all such chords will be the same length because of Snell's Law of reflection. If this length is $2r\sin(\theta /2)$ for $\theta=$ some rational multiple of $\pi$, the beam will trace out a convex or stellated polygon and hit you again. Else the beam misses you, but it fills in a ring whose outer boundary is on the sphere and whose inner boundary is inside the sphere (corresponding to how closely a chord of fixed length comes close to the center). This causes the beam to come arbitrarily close if given enough reflections.

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Here's a partial answer, with diagrams and further elaborations expected to follow soon.

In general, the answer should depend on the radius of the sphere, your position inside the sphere, and the direction in which you fire the laser.

But by symmetry, I believe we can reduce the number of parameters in this problem to two:

  1. First, decide how far away from the center of the sphere you are. We can measure the distance as a fraction of the radius, so if you are $r$ units away from the center of a sphere of radius $R$, call the distance $\rho = r/R$.

  2. Next, decide where to direct your light. This amounts to choosing a single angle $\theta \in [0,\pi]$.

For an explanation as to why you can completely determine the answer from just knowing $\rho$ and $\theta$, see the note [*] below.

As a result, though, this problem seems to be a 2D problem, fundamentally the same as firing a laser from within a 2D unit circle, as opposed to an arbitrarily-sized sphere.

(To see that the problem is 2D, consider two rays: the ray from the center of the circle to your current position, and the ray from your laser in the direction the light travels. Those two rays define a plane, and the bouncing light will always remain within that plane. The intersection of the plane with the sphere is, of course, a circle.)

The 2D problem has a nice solution explained here: Reflected rays /lines bouncing in a circle?

Here is an explanation of the diagram in that picture, to relate it to this problem:

  • $C$ is your position.
  • The points $A$ and $B$ are on the surface of the circle chosen so that (1) ray $\vec{AB}$ is parallel to the direction your laser is fired, and (2) ray $\vec{AB}$ passes through your position $C$.
  • The radius which perpendicularly bisects segment $\vec{AB}$ defines a coordinate system.
  • Angle $\alpha$ is the angle between the radius and the point where the light meets the circle for the first time.
  • Angle $\beta$ is the (signed!) angle between your position $C$ and the mirror image point $C^\prime$ on the other side of the radius.

See diagram for more details.


[*] Here is a short explanation as to why two numbers $\rho$ and $\theta$ completely determine the problem. Of course, from the description of the problem, we need to know six numbers in order to decide whether the light will hit or miss you:

  1. The radius of the sphere (1 number).
  2. Your position in the sphere (3 numbers; e.g. cartesian coordinates)
  3. The direction your ray is pointing (2 numbers; e.g. a pair of angles)

However, we can cut down on these numbers because of symmetry: the fact that observers in different positions and orientations will agree about whether the light will hit or miss you.

  • Nominally, we need three numbers to determine your position in the sphere. These could be $\langle x,y,z\rangle$ cartesian coordinates, or $\langle r, \vartheta, \phi\rangle$ spherical coordinates. However, we can reduce this to a single number $r$ because the answer to whether the light hits you or not does not depend on the angles $\vartheta$ and $\phi$ at all.

    We have symmetry of your angular position. Suppose you fire a beam of light from a particular configuration in a sphere. In your coordinate system, your position in the sphere makes an angle $\theta$ in the $xy$ plane and an angle $\phi$ in the $xz$ plane. But we could easily rotate the coordinate system (or you could imagine rotating all of 3D space). The light will still hit or miss you the same way it did before. The fact that the answer to this problem does not depend on your angular position means that it only depends on your distance $r$ from the center of the sphere.

  • Nominally, the problem depends on the size of the sphere $R$ and your distance $r$ from the center. But we can show that in fact, the problem only depends on their ratio $\rho \equiv r/R$.

    We have symmetry of length scale: Suppose you fire a beam of light from a particular configuration in a sphere. In your coordinate system, you are in a sphere of radius $R$ at a distance $r$ from the center. But an observer who is far away (or who has a differently-scaled coordinate system) would measure those distances as being scaled by some factor $\gamma$, measuring the sphere of radius $\gamma R$ and your distance from the origin as as $\gamma r$. But you will both agree as to whether the light will hit or miss you. And you will both agree on the value of $\rho \equiv r/R$. Hence the single parameter $\rho$, rather than both $r$ and $R$, will be enough to determine whether the light hits or misses you.

  • Nominally, we can direct our light ray toward any point in the sphere. To specify a point on a sphere requires two numbers (like a pair of angles $\langle \vartheta, \phi\rangle$) or three numbers and one constraint (like choosing an $\langle x,y,z\rangle$ coordinate on the surface of the sphere $x^2+y^2+z^2 = R^2.$)

    But we have symmetry of the azimuthal angle. Suppose you fire a beam of light from a particular configuration. Consider the ray $\vec{A}$ from the center of the sphere to your current position within the sphere. If you decide to spin in place around the axis $\vec{A}$, pointing your laser in a new direction, the light will still hit or miss you in exactly the same way. Hence that spin angle $\phi$ doesn't affect your answer— only the other angle $\theta$ does. (Only the "height" at which you aim your laser matters; your spin angle $\phi$ doesn't.)

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Let's look at the simplest cases: you shine your beam at the point on the wall closest to you, or at the point on the wall farthest from you, or you're at the center of the sphere. In any of these cases, the beam will bounce right back and hit you. There.

Now let's assume that you don't do any of that.

A good first step is to simplify the problem.

The first simplification I will make is this: there exists a plane containing the entirety of the path of your light beam; that plane also contains you and the center of the sphere.

The light will hit the side of your sphere and reflect off of it based on the angle of incidence of the beam on the plane tangent to the sphere at that point. Since that tangent plane is normal to a line drawn from the center of the sphere to the point of incidence, you end up with the first reflection of the beam remaining in the plane with you, the point of incidence and the center of the sphere. Now if you pick a point on your beam's path between its first and second points of reflection and think of it as a new source, you know that the second reflection of the beam will end up on a plane which includes your first beam reflection and the center of the sphere, for the same reason that your first beam reflection is in the same plane as your original beam and the center of the sphere. Since these two planes share your first reflected beam and the center of the sphere, you know that they must be the same plane. By induction, you'll find that every segment of your beam exists on one plane, which also includes you and the center of the sphere.

So the simpler problem, which will have the same answer, is this: If you shine a beam of light at the wall of a mirrored circular enclosure from any point inside the enclosure, will the light hit you?

Well, let's again look at the simplest case remaining: your first angle of incidence is a rational multiple of 2π.

In this case, your beam will hit you. This is because its path loops, which it will not do otherwise.

In this and the remaining cases, you will find that every time your beam reflects, the angle of incidence is the same. You can check this by drawing any chord on a circle and drawing line segments from the endpoints to the circle's center. Since the resulting triangle is an isosceles, its two base angles are the same. When you shine your beam in the circular enclosure, every beam segment (after the first) is a chord between two points on your enclosure's wall, and the two base angles on its isosceles are the angle of incidence on one reflection and the angle of reflection on another; since angle of incidence equals angle of reflection, every angle of incidence must be equal. Think of an arrow that points in the direction of motion of your light beam. The arrow would rotate by twice your angle of incidence each time your beam reflected; it would either always go clockwise or always go counterclockwise. After a full cycle is completed on a looped path, the arrow will be pointing in the same direction it was when it started. Thus, for the beam's path to loop, its total rotation—which would be an integer times twice your angle of incidence—would have to be an integer multiple of 2π; thus, the angle of incidence would have to be a rational multiple of 2π. So if your angle of incidence is not a rational multiple of 2π, the path will not loop.

But if your angle of incidence is a rational multiple of 2π, the path loops. A little more geometry shows that every beam segment (again barring the first) is the same length, no matter what that initial angle is. The length of a chord on a circle can be determined entirely by the radius of the circle and the angle between the chord and a line segment drawn from the center of the circle to an endpoint on the chord, so the constant angle of incidence directly implies that every beam segment has the same length. Given a circle and the fact that a chord of a given length is drawn in a given direction, there are only two possible chords for those criteria, one of which has the center of the circle to its left (by the drawing direction) and the other of which has the center of the circle on its right. The center of the circle will always be on the same side of your beam (from the beam's perspective). Since your beam will have multiple segments that go in the same direction due specifically to your angle of incidence, it follows that your beam will traverse the same chords repeatedly in the same direction—which means that your path will be looped. So in this case, yes, your beam will hit you.

So in the remaining cases, your initial angle of incidence is not a rational multiple of 2π, and so your path will not loop.

Let's start with a case in which your beam will certainly not hit you: your starting beam is perpendicular to the line containing you and the center of the circle.

In this case, the distance between you and the center of the circle is as close as any of your beams will pass to the center of the circle. Each beam segment will pass this close to the center of the circle exactly once, and that point will be its midpoint; thus, since a chord can be defined entirely by the location of its midpoint in the circle, only the chord on which your initial beam segment lies will have you on it. Since your path does not loop, your beam will not traverse this chord again; thus, your beam will not hit you.

The final cases are those in which your angle of incidence is not a rational multiple of 2π and your beam is not perpendicular to the line containing you and the center of your circle. Your beam's path will not loop.

Given a point in a circle and a length, if that length is greater than the length of the shortest chord that goes through that point and shorter than the diameter of the circle. then there are exactly two chords of that length that go through that point. So given the length that your beam segments take and your location in the circle, if your beam hits you, it hits you on the chord of that length that goes through you, but does not contain your initial beam segment, because your path will not be retraced. If there exist nonnegative m, n such that the angle between these two chords (the one that the line from the circle's center to you goes through) is equal to (m*2*the angle of incidence) minus 2πn, then your beam will include that chord and hit you; otherwise, it never will. This is because after m reflections, your beam's direction will have rotated by (m*2*the angle of incidence), which will be 2πn plus the difference in angle between your initial and final beam segments. This will select the final beam segment as the chord that will hit you. Otherwise, your beam will never traverse that chord.

I might have been flaky on a few details, but I hope this answers your question.

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As has been mentioned, one should consider the plane formed by the source of the light, the first contact point, and the center of the sphere. The light beam will be confined to this plane, because the vector orthogonal to this plane will be tangent to the sphere at every point of contact. Now the problem is reduced to a similar argument, but considering a circle intead of a sphere.

Using Snell's Law (as Oscar Lanzi has pointed out), the length of each chord formed by the light beam will be the same. Now consider the angle 'a' around the circle formed by one of the chords. We will define the initial chord along which the beam was fired to be the 0'th chord, the next one to be the 1st chord, etc.

There are now two possibilities:

1) If 'a' is some rational multiple of 2*pi, say (p/q)*2pi (assume gcd(p,q) = 1), then we are guaranteed that the (n+q)'th chord will simply be the n'th chord rotated around the center by an angle of p*2pi. Since p is an integer, the (n+q)'th chord will therefore be identical to the n'th chord, meaning the 0'th chord will be identical to the q'th chord. Therefore, intersection will occur every q*b reflections for every non-negative integer b.

2) If 'a' is an irrational multiple of 2pi, let m[0] be the least positive integer such that:

m[0]*a mod 2pi < a.

We are guaranteed that such an m[0] exists, indeed m[0] will simply be the least positive integer such that m[0]*a > 2pi. Notice this implies that the angle between the m[0]'th chord and the 0'th chord is less than 'a'.

We can simply repeat this process by defining m[n+1] to be the least positive integer multiple of m[n] such that:

m[n+1]*a mod 2pi < m[n]*a mod 2pi

The geometrical interpretation of this is that the angle between the m[n+1]'th chord and the 0'th chord is less than the angle between the m[n]'th chord and the 0'th chord (for every n > 0). We can prove the existence of such an m[n+1] by letting j be the least positive integer such that:

j*(m[n]*a mod 2pi) > 2pi;

m[n+1] is simply j*m[n].

Define k[n] to be the angle between the m[n]'th chord and the 0'th chord, where k[0] is let to be 'a'. It is simple to see that the sequence k[n] is positive and decreasing. This means the sequence is convergent; in other words:

As n -> infinity , k[n+1]-k[n] -> 0.

So, for every real number e > 0 , there exists an n such that k[n+1]-k[n] < e.

Since the m[n+1]'th chord is at an angle k[n+1] from the 0'th chord, and the m[n]'th chord is at an angle k[n] from the 0'th chord, this implies that as n goes to infinity, the angle between chord m[n+1] and chord m[n] approaches 0. Notice that for any positive integers b,c, and an integer d (not greater than min(b,c)) the angle between chord b and chord c is identical to the angle between chord (b-d) and chord (c-d). This implies that the angle between chord m[n+1] and chord m[n] is identical to the angle between chord (m[n+1]-m[n]) and chord 0.

As stated, for any real e>0, there exists an n such that the angle between the m[n+1]'th chord and the m[n]'th chord is less than e, implying the angle betweeen chord (m[n+1]-m[n]) and chord 0 is less than e. This guarantees arbitrary precision.

It could also be the case that one chooses wisely the initial firing point in a way that 'a' is an irrational multiple of 2pi, but the initial firing point is met anyway; consider, for instance, the case where one fires the beam of light from the intersection of chord 0 and chord m[0]. In this case, the beam of light will clearly pass through the firing point, but will only do so once, since a given point on a given chord is shared only with exactly one other chord (of the same length), and for 'a' being an irrational multiple of 2pi, we are guaranteed that no chord is ever repeated.

Sorry for the lack of fancy formatting, I'm new here and haven't had a chance to learn how to do that properly.

I realize this is a bit of a tome, so feel free to ask me about any uncertainties that might arise!

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  • $\begingroup$ Not sure why this was downvoted; I mean, it might not be the prettiest comment, but it's the only one thus far that explains how and why the light beam will get arbitrarily close to the source. This site confuses me :( $\endgroup$ – WhiteboardFunk Jul 31 '17 at 23:48
  • $\begingroup$ It is badly formatted $\endgroup$ – Agile_Eagle Aug 11 '17 at 3:08

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