Intractable math / physics puzzle: Light in a perfectly mirrorred sphere.

May 15, 2006 2:49 PM

You are suspended in a large, hollow sphere. The inside of the sphere is perfectly mirrored. There is nothing else in the sphere, and all you carry is a flashlight. You turn the flashlight on. What do you see?

I first read this question in one of Spider Robinson's "Callahan's Crosstime Saloon" novels in high school, and it has dogged me ever since. (I don't have a copy on me anymore, so feel free to correct me on any inaccuracies.) The character who poses it states that no one has ever been able to give him (and thus Robinson himself, we have to assume) an answer with any confidence. I've had the same experience -- no one has any idea what answer -- and there are many initially plausible ones -- might be correct. I know that it's possible to work out, but my high school AP Physics class didn't provide me with the tools I need to solve it.

If you can provide a convincing answer verbally, go for it, but if you can prove the result via math and / or modelling, I'd be elated -- I just want to get the question settled once and for all, so I can sleep at night.

Extra qualifications: Let's say that you're suspended such that your vantage point (that is, your eyes) are exactly in the center of the sphere, and that the sphere is perfectly reflective. On the chance that the answer will significantly change depending on variables such as the size of the sphere, the position of the flashlight, etc., choose whichever make the question easiest for you to model and / or solve. (I suspect that the essence of the result will stay the same regardless, however.)

Thanks!

posted by tweebiscuit to Science & Nature (39 answers total)

Response by poster: Don't worry too much about the practicalities -- if you want to posit that your vantage point is the same as the light's point of origin, don't worry that the flashlight would be blocking your view or anything. Boiled down, all that matters is (as far as I can tell):

1) The sphere is perfectly mirrored.
2) There is a light source inside the sphere.
3) There is a vantage point inside the sphere.
4) There is at least one solid, light-absorbing object inside the sphere.
posted by tweebiscuit at 2:51 PM on May 15, 2006

You see your reflection.

Are you asking us to verbally describe the way in which the image would appear distorted?
posted by JekPorkins at 3:00 PM on May 15, 2006

I don't understand why this is so mysterious? Isn't the inside hemisphere facing you going to reflect you back distorted and upside down, as does the bowl of a spoon, to the extent you are illuminated by the light reflected back from the flashlight. Bigger sphere, less light?
posted by A189Nut at 3:01 PM on May 15, 2006

If you we inside the sphere I imagine you'ld eventually aborb all the light bouncing around.
posted by blue_beetle at 3:04 PM on May 15, 2006

TB wants to know if the point of light from the flashlight will be reflected, and how many times.

Seems to me that the reflected light will appear all over the inner surface of the sphere.
posted by yclipse at 3:06 PM on May 15, 2006

Ha! Not that I am any help, but I have also wondered about this.
posted by skryche at 3:06 PM on May 15, 2006

Response by poster: Sure. Or an approximation thereof -- the problem is that so many answers are plausible. For example, answers I've heard:

1) The entire sphere would be filled with light. (Probably the least likely, but makes sense to some.)
2) You'd just see a big mishmash (or wash of color, or something likewise indistinct), because a massive number of rays would be reflected off each point in the sphere.
3) It would just be like looking at yourself in funhouse mirror.

As a layman's thought experiment, it's pretty impossible to decide between these options, which is why I'm looking for a harder proof.
posted by tweebiscuit at 3:07 PM on May 15, 2006

Response by poster: (Sorry, the "sure" was a response to JekPorkins.)

The point of light from the flashlight will definitely be reflected, and will definitely be reflected a massive number of times (remember that the surface is perfectly reflective) in all directions.
posted by tweebiscuit at 3:08 PM on May 15, 2006

A perfectly reflective sphere would not allow any energy through, cleaving the entire Universe into two.

It is, then, a closed system.

It seems (at first blush) that energy released from the flashlight would accumulate, diffusing in the air and increasingly generally illuminating the interior until the light ran out, which would then bounce around inside forever.

No physical explanation is going to make much sense, though, since what happens when a photon (or any other particle, including the atoms of the air or your hand) hits the perfectly reflective wall is not entirely clear.

I mean, the repulsive electromagnetic force between objects in close proximity is what keeps solids from falling through each other. Actual objects are only there stochastically, this wall would have to exist from eternity. The boundary between it and the space it encloses is inconceivable.
posted by sonofsamiam at 3:13 PM on May 15, 2006

Response by poster: Well, sonofsamiam -- I did say, after all, not to worry about the practicalities all too much. ;)

You bring up a good point, though, that it is indeed a closed system, so the light should grow -- notwithstanding the solid, light-absorbing object in the sphere (you), which you forgot about.
posted by tweebiscuit at 3:24 PM on May 15, 2006

If you look though just one eye which you keep at the centre, You will see the your eye - no matter which direction you look, all surfaces are reflecting exactly back to your eye. So you would see a giant, largely dark/red surface (since your pupil is dark, but depending on how bright the flashlight is, you will likely see a very low level of "red-eye"), unless you looked down at yourself, wherein you would see your body by torchlight as normal.

You will be able to see your eye because the flashlight is not at the centre, thus the light from the flashlight will bounce off the mirror at odd angles and off your clothes at all angles (and so on), producing a very even ambient light under which you are able to see your eye.

If you are wearing nothing, or wearing all clothes of a single colour, this dominant colour will tint the colour of the ambient light, which in turn will affect the colour your eye appears to be on the inside of the sphere. (I would assuming that the margin of error are low enough that parts of your eye other than your pupil will be visible in the mirror)

If you have both eyes open, with the centre of the sphere between them, you will see a mix that is mostly huge eyes and enormous nose and surrounding parts of the face.

To get an idea of what the inside of the sphere would look like, you can generate something vaguely similar using a small parabolic mirror from a flashlight instead of a giant sphere. In the sphere, the viewpoint and the focal point are the same, and your eye is thus at both. With the parabola, they are different, so put an object at the focal point, hold it at a distance and look squarely into it, and you will see the entire surface reflecting only the thing that is at it's centre. If you put something that looks like an eye at that point, you'll see what an eye looks like when spread over the entire surface of a mirror.

Have you ever looked at your own retina in a mirror? That is another thing you could try to get a sense of what you would see.
posted by -harlequin- at 3:35 PM on May 15, 2006

I would imagine when the light hits a non-perfectly-reflective surface (you), some of the light will be absorbed as heat. So over time, you will absorb (as heat) all the chemical energy in the flashlight batteries.

Also of note, if we assume the sphere is reflective to all radiation, the heat put out by your metabolic processes will build up inside too. If this is dangerous to you probably depends on the size of the sphere.

As to what you will see, it would be you in a funhouse mirror, but weirdly enough, I don't think it matters how big the sphere is, once it's slightly bigger than you.
posted by Capn at 3:39 PM on May 15, 2006

I think sonofsamiam makes it too difficult. There is at least one solid light-absorbing object inside the sphere, which means that the whole thing will be a blackbody after the light runs out. If we can assume, as would be natural, that every ray hits the light-absorber after a finite number of reflections, "infinite reflectance" doesn't differ qualitatively from "very high reflectance". It is a ray-tracing problem, and I think that's what it's meant to be. What you see probably depends a lot on where your eyes are and where you hold the flashlight.
posted by springload at 3:40 PM on May 15, 2006

If the light source is located at the center then the beam will hit the mirror orthogonally and be reflected directly back to the center. Depending on the exact level of "practicality" involved here, that either means it hits you on its way back, or passes to the diametrically opposite point. In the first case you see one illuminated patch on the sphere, which probably looks like nothing more than a bright light (you can't see your reflection in it - like shining a light directly at a mirror in a dark room). If you imagine yourself as a disembodied observer then you can see two illuminated patches diametrically opposite one another.

If you move the flashlight off-center then a variety of paths for the light-beam are possible, from closed paths that leave a finite number of patches illuminated, to paths that illuminate an entire circle of the sphere (remaining entirely in one plane), to path that become dense on the surface, illuminating the whole sphere. It would not be hard to cook up a little 2-d illustration of this with a light source inside a circle that would still illustrate the possibilities in 3-d just fine.

In practice, since the lens of the flashlight has some extent, not all the light rays it sends out actually orginate at the center of the sphere. For a given orientation of the flashlight, I believe the set of points that give rise to a dense path is actually dense itself, so - with the degree of 'idealization' I think you're intending - some one of those rays will have a dense path and entirely illuminate the sphere.
posted by Wolfdog at 3:48 PM on May 15, 2006

So you would see a giant, largely dark/red surface

This was a confusing choice of words, as you would not see a giant surface in the sense of knowing where the mirror was, I meant that it will fill your field of view, and the distance away from you that it appeared to be would be deceptive. (Since we're not used to seeing our eyes, there is no visual cue as to distance, so with one eye open, you would have only the focus of your eye to judge. With both eyes open, you have an additional cue from the binocular vision, which would suggest that the image is behind the mirror, not on it, since the distance to the mirror is half the distance from eyes to mirror to nose).
posted by -harlequin- at 3:48 PM on May 15, 2006

On the chance that the answer will significantly change depending on variables ... choose whichever make the question easiest for you to model and / or solve.

If the flashlight is sufficiently powerful, wouldn't you be instantly burnt to a crisp?
posted by dersins at 3:56 PM on May 15, 2006

Yeah, that bit from Robinson's Callahan's always bothered me.

I think that someone tries to explain it here with a crude POVray dealy.

Apparently, the size of the sphere (sub-lightsecond or sufficiently big such that it'd take time for the reflection to reach the observer) makes a difference.
posted by porpoise at 4:07 PM on May 15, 2006

I have no answer; except to point out that the retina must necessarily absorb photons in order to "see." So at minimum, that part of the system is not perfectly reflective.

Presuambly, the rest of the person is also not perferctly reflective -- this means that photos absorbed by the clothing or skin gets converted to heat, i.e. wavelengths too long for us to see.

As a parting shot: there are no such things as perfect mirrors. Check the cavity ringdown literature.
posted by NucleophilicAttack at 4:10 PM on May 15, 2006

Here's another attempt.

From reading about integrating spheres, I'd imagine that since your body is inside sucking up photons (since you aren't perfectly reflective), I'd imagine it's end up being a muddy hazy blur.
posted by porpoise at 4:20 PM on May 15, 2006

I just did some playing around in a 3d program(maya software render).
this is using only 10 reflections. I have a point light in the centre and a single plank running through my sphere as an object

Using plain old raytracing in these programs wont really work with the energy buildup aspect of the scenario. 3D programs generally dont trace all the light in the scene, they trace back from the camera to save cpu power.

Actually know that I think about it my light is probably blooming a bit(i set it very very low), which wouldnt happen with an absolutly perfectly reflective sphere.

Im sure this isnt accurate but its pretty and might be a bit of an approximation.
posted by phyle at 5:21 PM on May 15, 2006

sorry about that, let me try again-

posted by phyle at 5:22 PM on May 15, 2006

When all your bodies infrared energy was reflected back onto you, you would see Angels as you died of hyperthermia.
posted by Megafly at 6:48 PM on May 15, 2006

I used to wonder about this when I was a kid -- and never could wrap my head around it. In my hypothetical, though, there was a perfectly (internally) mirrored sphere with only some kind of light source inside. For example, if a single flash bulb went off in the sphere, and you broke the sphere open a year later, would you see the flash then? How would that differ if instead of a single flash, you had a continuous light emitter inside?

Anyway...
posted by GregW at 6:57 PM on May 15, 2006

Asuming absolute reflectivity, I'd go for the flash coming out once the sphere is breached. As for a continuous light emitter, there'd be more and more photons bouncing back and forth.

Asuming absolute reflectivity...

phyle - that's really cool. If you set up increasing reflections, do you think it would "wrap around" (each subsequent reflection of the plank seems to move a little bit) and overlap itself?

So, given a lot more reflections, it's basically look like a jumbled blur?
posted by porpoise at 7:28 PM on May 15, 2006

There's really only one solution to this problem.
posted by Slarty Bartfast at 8:00 PM on May 15, 2006

Since your bubble wouldn't cause any dissipation of heat, and the light that struck you would turn to heat, wouldn't you eventually catch fire? (Which causes more light...)
posted by unreason at 8:01 PM on May 15, 2006

Since your bubble wouldn't cause any dissipation of heat, and the light that struck you would turn to heat, wouldn't you eventually catch fire? (Which causes more light...)

The amount of energy stored in typical flashlight batteries is not nearly enough to cause any significant damage to a person, even if absorbed all at once.
posted by Optimus Chyme at 8:44 PM on May 15, 2006

I think sonofsamiam gets close.

If the sphere is perfectly reflective, then for our purposes it's a closed system. So as the flashlight converts stored battery energy into light radiation, the light in the sphere just accumulates and it gets brighter and brighter (think microphone feedback).

Now, if "you" are in the room and you're a physically realistic object, like a human being with real skin and clothes and whatever, then you will be absorbing some of the light energy and converting it to heat. The RATE at which you do so will increase with the intensity of the light, so after some period of the flashlight dumping light energy into the room, the accumulated light intensity will reach an equilibrium; you will be absorbing just as much light as the flashlight is putting out. Now, mind you, that probably requires an extremely intense light level.

Mmmmkay, now what? You're absorbing light and converting it to heat. Assuming the sphere contains mass (say, you and the air and the flashlight), the temperature will start to rise, settling at the rate that the flashlight is dumping light energy into the room.

Then the flashlight runs out of battery power, the remaining light gets absorbed, converted to heat, and you end up with a sphere interior slightly warmer than what you started with. Specifically, the energy capacity of the battery (in joules) has been converted to heat.

Oh, and you'd see a glaring field of white in all directions, and be blind at the end of it all.

Thermodynamics rocks.

This is NOT what I planned to do this evening. Damn you, MeFi! Damn you, sanctimonious brain!
posted by intermod at 8:52 PM on May 15, 2006

porpoise- more reflections just makes the white glow get brighter and blow out. The reflections when you look at them are just like a carnival mirror thing, they tend to go out in trails to infinity, so adding more just makes the tiny trails longer, it doesnt really jumble up and get messy.
posted by phyle at 9:01 PM on May 15, 2006

"Oh, and you'd see a glaring field of white in all directions, and be blind at the end of it all."

Won't you see at most the same amount of light as if you pointed the flashlight directly into your eyes? Painful, yes, but a normal flashlight won't blind anybody. The reflections don't increase the power output of the flashlight; they just bounce it into your eyes (or body) where it is then absorbed.
posted by mbrubeck at 10:36 AM on May 16, 2006

The only light you could see is that that which your eye lens focuses on the retina. The only light that is entering your eye at an angle for this to happen is that which is perpendicular to the mirror surface, meaning ambient light bouncing off your retina. Ie you would see very little light at all if looking at the sphere from its centre.
posted by -harlequin- at 12:07 PM on May 16, 2006

You see a dim reflection.
( If it is a normal but perfect mirror, the light dissipates by continuous reflection within the large space). Now if its some supernatural atomically reflective mirror thats a different story, as s.o.s.i.a. said.
posted by uni verse at 5:11 PM on May 16, 2006

Response by poster: Wow, there are some fantastic answers here -- thanks, everybody! I think harlequin and Wolfdog got it right -- at the very least, I feel that I know the answer now.

The real problem with the question is that, as it was posed to me, it had too many variables for a meaningful answer. Harlequin's example pinpoints the most interesting one (in which we don't have to worry about "light feedback" and in which our eye is at the center) and gives an answer that feels right to me, since it takes into account the angles of reflection, the fact that our eye is at the focal point, and the affect of the other light on the essential image.

To summarize in my own words (and tell me if I've gotten this wrong) -- if your eye is at the center, it is in the focal point of the lens, so you see a clear image of your entire eye. The bouncing light reflecting off other parts of your body and other sides of the sphere willy nilly will diffuse and combine enough that it will only create an ambient glow, which will actually serve to further illuminate the image of your eye. The glow's color will change if the solid objects in the sphere (i.e., your body) are dominantly one color.

Wolfdog illuminated nicely why we don't want the light source to be at the center -- if the light source is at the focal point, all the light will simply bounce directly back to it! -- while clearly explaining how the angle of the light will affect how much of the sphere is illuminated. Neat!

Finally, all of this together has given me the impression that if both your eye and the light were off-center, you would simply see a diffuse glow on the sphere, since you would be at focal length for so little of it.

Finally, there's the matter of energy absorption, the variables affecting which should really be written into the premise (i.e., "The light output of the flashlight is at equilibrium with the light absorption rate of your body"), lest it entirely ruin the optical problem that we're really interested in -- but it's still fun to think about.

Obviously, this is not a question with a discrete answer -- rather, it's a (very cool) physics problem that has a number of discrete outcomes depending on the variables in play. You guys have done a fantastic job of exploring those various outcomes and walking us through them -- thank you so much! Consider the question solved -- I'm going to sleep tonight.
posted by tweebiscuit at 3:01 PM on May 17, 2006

Response by poster: (Harlequin's final answer, I think, is the final piece of the puzzle, in that it explains why the image directly in front of your eye isn't totally overridden by all the other reflections bouncing the other way.)
posted by tweebiscuit at 3:04 PM on May 17, 2006

Response by poster: Hm. Actually, here's a counterargument from my sister, whose a physics student at Brown (and a brilliant one:)

ah! the focal length of a spherical mirror is 1/2 the radius, so the equation I mentioned, 1/f=1/do+1/di would suggest that the distance the image appears at is exactly where your eye is, but upside down. and the image won't be magnified. f=focal length, do=distance of the object(you), di=distance at which the image appears. I'm not sure about this point because I have not had much optics but I think that because your eye is going to be exactly where the image is reflected, you will see no distortion of the image like you see in a carnival mirror. So you'll just see yourself, at full size, upside down.

but that is for the case where light is just coming from the object, without any other source. Once you add the light source, I think it will overpower the reflection of yourself you see. Since the light emits way more light than you do (regardless of how bright the light is, since you are just emitting what you absorbed from the flashlight) and since the light is being bounced around continuously until it gets absorbed by you (unlike in the case of a mirror that is not entirely enclosed), you'll only see light, no image.

Also, if you were suspended at the focal point (which you're not) and had a powerful enough light source, you could burn yourself, just like you can burn things with a magnifying glass. But that wouldn't happen at the center of the sphere, like some people suggested. And I don't think ordinary flashlights could do it.

Woah! What a cool problem. I have been having fun thinking about it. Brown doesn't offer an optics course so I really don't know much about it. Maybe I can tell you more after grad school :P

Anyone care to respond? (harlequin, i'm especially interested in your response.)
posted by tweebiscuit at 6:00 PM on May 17, 2006

Response by poster: Yup -- turns out that harlequin can't be right -- the center of a spherical mirror is its image distance, not its focal distance. Parabolic mirrors aren't an apt comparison, because they have completely different focal lengths.
posted by tweebiscuit at 8:22 PM on May 17, 2006

Once you add the light source, I think it will overpower the reflection of yourself you see. Since the light emits way more light than you do [...] and since the light is being bounced around continuously until it gets absorbed by you

Draw the ray beams on paper with a circle representing a longitudinal slice of the sphere - it doesn't matter how much light is in the sphere, or where in the sphere the light bounces around, none of it can reach your eye if your eye is at the centre - there is simply no part of the mirror that could reflect light into the centre, unless that light came from the centre in the first place, the angles are wrong.

Light bouncing off your clothes reaches the centre, so if you looked down, you would see how brightly lit the sphere was by viewing the things within the sphere, but the mirror surface itself would appear quite dark, because very little light is emitted from the eye.

Try to draw a ray, any ray from any direction or location that hits the mirror and bounces into the centre of a perfect circle - it can't be done. Objects inside the sphere would be illuminated and visible, but the mirror surface would be dark. (With the exception of the part of the mirror occluded by the light source as viewed from the centre, but since it's occluded, that light isn't really visible either)
posted by -harlequin- at 9:50 PM on May 17, 2006

As to the image in the mirror, I imagine the light reaching your eye would be coherrent, but it's all a reflection of the same point, so whether that's an "image"...?
posted by -harlequin- at 9:56 PM on May 17, 2006

A buddy of Spider's (whose email address I am *not* giving out :-), to whom I pointed out this thread, notes that no one's mentioned whether the *air* (and, say, dust in it) is absorbing or reflecting any of the light energy...
posted by baylink at 12:42 PM on August 23, 2006