# Is it possible to make a closed mirror system?

1. Jan 21, 2010

### Edward Solomo

Hello, my name is Edward Solomon, after much experimentation and calculation I have failed to make a system that can reflect light in a closed system.

Now I am not naive enough to believe I can make a true closed system. There is an absorption and conversion to heat each time light strikes a mirror. Also, as my mirrors are not smoothest mirrors available, not all of the light is reflecting exactly where it should be.

My actual goal was to create a system that could concentrate a high density of light by having it reflect in a circular pattern. Let our Light Source be Z, let mirror one be A, let mirror two be B and let mirror 3 be C.

My goal is for the light to travel as follows.

Z to A to B to C to A to B to C to A. . . with the light striking the same spot of its respective mirror that it had hit before. You would get something that appears like this.
Z-----A
xxxxxx/\
xxxxx / \
xxxxC---B

This would cause a concentration of light whose luminosity that would be equal to L times the original luminosity, where L = limit of the series (A/B)^n where is A/B is the statistical mean percentage of the light reflected when absorption and mirror roughness are factored in.

The most obvious problem that arose was that the light source itself, Z, obstructs the light. Thus this system only seems to be achievable if the light source itself is a mirror (and in that case we would only need two mirrors).

So Z to A to Z to A to Z to A to Z to A. . .

Z---A

Let us assume that between the absorption and the imperfect surface of the mirror that we get a statistical mean of 90% reflection of the incoming light per reflection. Thus after one reflection (assuming the light source remains on the entire time) we would have 190% of our original luminosity. After the second reflection (of the source mirror) we would have 271% of the original luminosity. As the series of (9/10)^n converges on 9, after exactly 65 reflections we would have 999.04% of our original luminosity, or nearly ten times as much luminosity of the original light source.

Of course this experiment is fictional and not at all possible as the source cannot be a reflective surface, as far as I know.

So my question is:
Is it possible by using a set mirrors of different shapes (planar/parabolic/hyperbolic/etc) to create a closed system, one that can avoid including the light source to the first mirror, Z to A, in the path of a closed system.

2. Jan 21, 2010

### Phrak

One example is a ring laser gyroscope. Another is simply a laser with neither mirror designed to leak, where the beam travels back on itself. Neither is perfect, of course.

3. Jan 21, 2010

### bp_psy

How are you going to make the light not go from Z to A to B to C to A to Z ?

Also what do you mean by obstructs the light?

4. Jan 21, 2010

### Redbelly98

Staff Emeritus
No, the source will always be part of the light path. People who make such configurations use a source that amplifies instead of absorbs the light as it passes through; this is a laser.

5. Jan 22, 2010

### Phrak

Taking this question at face value, but consider that he's asking about a circulating light path that does not include the light source. Suppose we direct a laser through an aperture in a perfect mirror. It cycles once, bouncing off more purfict mirrors, then bounces off the mirror with a hole in it. The beam is shaped so that more of the amplitude2 misses the hole and cycles around for another pass.

Can the beam continue to circulate such that energy leakage back out of the aperture is less the energy put through the aperture?

I'm fairly sure the answer must be "No, the energy leaking out of a perfectly reflective box over time, no matter its shape, will equal the energy input", but I can't prove it.

I'm equally sure there's a clever proof of this, but I don't know what it is.

What do you think?

Last edited: Jan 22, 2010
6. Jan 22, 2010

### Edward Solomo

This is indeed what I am asking.

7. Jan 22, 2010

### Phrak

Wonderful Edward! I'd like to know too. This is a question I've had off-and-on for a very few decades.

8. Jan 23, 2010

### Redbelly98

Staff Emeritus
If we think in terms of geometric rays, consider the following.

The beam enters through an aperture in the mirror. After some number of round trips, the beam path makes a closed cycle that misses the aperture. But in a closed cycle, the beam path should be reversible ... meaning the reverse path should also be a closed cycle that misses the aperture. But this contradicts the fact that the reverse path must eventually exit through the aperture.

In reality there is another complication, since diffraction will cause a beam to spread out if you are using flat mirrors.

9. Jan 23, 2010

### Edward Solomo

If we could bend the path of the light source with medium (such as water) I wonder if we can still entertain the idea of a closed cycle.

Let the light source, Z, be aimed at a transparent container of water in such a way that it bends and strikes Mirror A. The light then reflects to mirror B, which travels to Mirror C, however we must reflect the light from C in such a way that it passes through the container of water and exits the container in manner that allows the beam to join with the original beam exiting from Z. This could be accomplished by shaping the sides on the container and possibly adding another lens inside the container of water itself. I'm going to see if I can make it work on paper.

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Last edited: Jan 23, 2010
10. Jan 23, 2010

### Redbelly98

Staff Emeritus
The ray path across a change in medium is still a reversible path, so my argument from two posts ago still applies. And we are ignoring the complications of diffraction that happens in a real beam.

11. Jan 23, 2010

### diazona

I'd expect that the amount of energy in the box would tend toward some equilibrium, in which the power input by the laser equals the power output from thermal radiation. The energy in the box can't continue to increase forever, since eventually it would explode or collapse into a black hole or something, and it can't decrease forever because if it ever hits the ground state of the EM field, presumably the energy can't go any lower. The only possibility left is equilibrium.

12. Jan 23, 2010

### Edward Solomo

Right, lets pretend that we did have a perect closed system, where only heat absorption was a factor. Let us say that after each reflection, 10% of the light absorbed, or 90% of the light remaining. So If we use a two mirror system, it must be reflected two times to complete a cycle through the system. Thus we would have (.9)^2 left of the original light beam, which would be 81% of the original beam.

However since the light source is continuous, it would be followed by a beam at 90%, and anew beam, at 100%, thus we would have 100% + 90% + 81% of our original luminosity, which is 271%.

The equilibrium of this closed system would be equal to: 1 + the limit of the series (9/10)^n

Which would be 1 + 9 = 10, thus any laser that we create cannot exceed 10 times the strength of the light source. Since this is relatively weak, we don't have to worry about our apparatus melting or exploding.

Of course no system has yet been devised that could achieve such a circular pattern to being with.