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If I wanted to surround myself with mirrors how would I need to arrange them so that if I stood in the middle of them I couldn't see myself in any reflection?

For simplicity let's just say we only do the sides, not the ceiling and the floor.

How many panels would I need and in what configuration?

It seems to me that only a star formation could work, where all the mirrors are angled slightly outwards from me, but I'm not sure that at some point I would be reflected.

Is this even possible? Any ideas?

UPDATE:

Based on all the responses below I believe the problem is that I cannot have any angle facing me that is at 180 degrees or less.

For example, this is at about a 30 degree angle and doesn't work (thanks @mmesser314):

enter image description here

And this one is exactly 180 degrees and obviously will not work:

enter image description here

But from 181 degrees onward light is reflected away, like this:

enter image description here

So if we could figure out a configuration where I cannot see any angle that is less than 180 degrees I should never be able to see my reflection. This is what I came up with. Can anyone tell me if this (or some variation of it) seems valid?

enter image description here

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    $\begingroup$ What about placing those mirrors with their backsides to you? $\endgroup$
    – Georg
    Jan 27, 2015 at 11:39
  • $\begingroup$ funny, but not what I meant :-) $\endgroup$
    – Andy
    Jan 27, 2015 at 12:28
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    $\begingroup$ I think it's an interesting question concerning angles of reflectivity. Besides, how cool would it be to stand in the middle of an infinite space without seeing yourself? $\endgroup$
    – Andy
    Jan 27, 2015 at 12:59
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    $\begingroup$ One approach is to make the arms a fractal where light takes an infinitely long path out, and so never returns. I don't know if that is possible in a finite box. $\endgroup$
    – mmesser314
    Jan 28, 2015 at 17:18
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    $\begingroup$ This is getting farther and farther from physics. It is more and more a pure math puzzle. $\endgroup$
    – mmesser314
    Jan 28, 2015 at 17:19

3 Answers 3

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This would work, but you might think it is cheating.

enter image description here

If there can be no gaps, I would guess it is not possible. Your star idea won't work. Light that leaves you will bounce up the V and back out. The ray shown misses you. But if you look at a fan of rays, some will go to the left of you and some to the right. There is always one that will hit you.

enter image description here

If you use a smooth curve instead of planar mirrors, you cannot avoid seeing at least a point. This shows a ray that goes to the left and another that goes to the right. Some ray in between will hit you.

enter image description here

You might try to avoid it by putting mirror in the way to block that ray. But that piece has the same problem.

enter image description here

You might try to make a mirror where all the rays are deflected to the left. But that is a spiral, not a closed curve.

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  • $\begingroup$ Wow, a lot of effort in this answer, and many interesting things to consider. Thanks. What I meant by star formation is like this: http://etc.usf.edu/clipart/37700/37716/04-star_37716_lg.gif. Don't you think this would work? $\endgroup$
    – Andy
    Jan 27, 2015 at 15:26
  • $\begingroup$ ooops, I guess your second point answers my question. But what If we could eliminate that convergence point by adding a smaller star pattern behind each of the four convergence points? I don't have time now but I'll add an image later this evening illustrating what I mean. $\endgroup$
    – Andy
    Jan 27, 2015 at 15:36
  • $\begingroup$ Reread the second point. There is no 'convergence point', the light rays do not tend towards the corners of the star, so it is not a problem resulting from there being a corner. Most of the returning light rays come back from hitting the walls of the star's arms. There must be light rays going in all directions from the initial point, at least one ray will always hit you. $\endgroup$ Jan 27, 2015 at 16:28
  • $\begingroup$ Hi Michal, thanks for all your input. Your answers helped me come up with a possible solution. Please take a look at my edits above and tell me if it might work. $\endgroup$
    – Andy
    Jan 27, 2015 at 23:37
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Here I'll show that in 3d configuration, it is impossible. However the argument could be adapted to 2d also, if we used 2d black-body radiation instead.

The proof is by contradiction, using thermodynamic arguments. Assume there exists some finite solid angle in which direction you wouldn't see yourself. Now this would imply that if you send light in this direction from your eye, it never reflects back to your body and never gets absorbed. Suppose now that your body (including your eye) is perfectly black, and it has a constant temperature $T$. Then some of the radiation emitted from your eye will never come back, implying it remains in the chamber. As more and more radiation will be compressed in the room, the radiation intensity at least at some place must diverge to infinity. This could be used to heat a hotter body with a colder one. Contradiction.

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    $\begingroup$ I like this argument. You don't even need to assume constant temperature for the object, as the result would be that the temperature of any object would asymptotically reach absolute zero as it loses energy to the non-reflecting region. Meanwhile, the non-reflecting region would approach some other non-zero temperature, resulting in a hot region gaining energy from a cold one. Such a setup would be a perfect refrigerator! $\endgroup$
    – Mark H
    Jan 27, 2015 at 18:05
  • $\begingroup$ Great answer, I like it. Except you said "Assume that there exists a finite solid angle in which direction you could send light and it never reflects back to you" so why do you then state "some of the radiation emitted will never come back"? Doesn't your first statement imply that ALL the radiation emitted will never come back? $\endgroup$
    – Andy
    Jan 27, 2015 at 23:49
  • $\begingroup$ @Andy: I've clarified my answer a bit. Well, the point was that if you put a black body in that place, some of the radiation would be emitted in that solid angle from which nothing comes back and some will be radiated in these directions from which the light returns. $\endgroup$
    – kristjan
    Jan 28, 2015 at 8:41
  • $\begingroup$ I disagree with this answer. If you continually add energy to a closed box, it will get hotter without limit. If there is a contradiction, it would be that at a certain point you could not add energy without losing more than you added. $\endgroup$
    – mmesser314
    Jan 28, 2015 at 17:04
  • $\begingroup$ Consider a circular mirror with you at the center. Somewhere off center there is a laser. We will make it perfectly transparent so that it only enters the problem as a light source. If the beam is not pointed directly at you or away from you, light will bounce around and around forever without ever hitting you. There may be thermodynamic problems, but not geometrical ones. $\endgroup$
    – mmesser314
    Jan 28, 2015 at 17:07
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Unless you use methods that you might consider cheating, it's almost certainly impossible. The reason why is that the Law of Reflection states that $\theta_i=\theta_r$. This means, in words that when a ray of light hits a reflective surface the angle that the ray of light makes with the surface before it hits (the incident angle, or $\theta_i$) is equal to the angle that the ray of light makes with the surface after it is reflected (the reflected angle, $\theta_r$). Out of convention, these angles are measured normal to the surface of the mirror. The diagram below shows reflection of a ray of light at some arbitrary angle.

Law of Reflection

Here's the kicker though: If the ray of light hits normal to the surface (perpendicular to the surface, then $\theta=0$, so the ray of light reflects straight back, retracing its path. In this thought experiment of yours, any ray of light that leaves your body and strikes a mirror normal to the surface like this will be seen my you. (More or less. I am treating you as a point particle, rather than a person with dimensions. This is ok because seeing your own eyes still counts as seeing you).

So now let's view the last paragraph in the negative. In order to not see yourself, there can not be a single ray of light that leaves your head and strikes a mirror perpendicularly. Try to imagnine being in a room that satisfies that criteria. A mirror sphere wouldn't work. A mirror cylinder wouldn't work. A regular polygon won't work. Even if you try to imagine some spiky, star shaped wall of mirrors, I believe the vertices still would likely act as corner reflectors. Unless I'm missing a shape really, really outside the box, the answer is no.

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    $\begingroup$ Don't forget that you're not a point particle, which means even if you can't see your head. There are feet, a torso, arms, the back of the head, etc. Light, uh, finds a way $\endgroup$
    – Jim
    Jan 27, 2015 at 16:47
  • $\begingroup$ How about my possible solution above? I think it goes under your "spiky, star shaped wall of mirrors" definition, but as long as there's no angle within my eyesight that is under 180 degrees, I should never be reflected... right? $\endgroup$
    – Andy
    Jan 27, 2015 at 23:40
  • $\begingroup$ In this problem, I was idealizing the person to a point. It makes the problem more interesting. $\endgroup$
    – mmesser314
    Jan 28, 2015 at 17:09

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