# The quantum-drives-classical loop. Is this perpetual?

Discussion in 'Physics & Math' started by al onestone, Jul 15, 2013.

1. ### al onestoneRegistered Senior Member

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I'm constructing this thought experiment with two basic "rules of thumb" as concerns quantum mechanics;

1) QM systems tend to exhibit a zero-valued activity that always occurs on average. For example, the famous "zero point energy" that exists in a solid in the form of thermal heat (phononic vibration) even at absolute zero temperature. There is always a residual average activity occurring in QM systems.

2) The propagation of light (quantized EM fields) in the absence of any loss. The EM wave requires no acceleration, and it doesn't experience loss due to anything like "friction". Photons are a transfer of energy from point A to point B with negligible loss (only radiation pressure, etc.)

This is an attached illustration of the thought experiment (very simple):

View attachment 6385

There are two layers/regions, one is metallic and one is an optical cavity (vaccuum). The upper region is the optical cavity and it is sealed off with 100% reflective mirrors. The optical cavity has an optical diode (half way) which only allows photons to pass from the right side of the cavity to the left. The lower layer is a metal with adiabatic walls around it everywhere accept the surface which is adjacent to the optical cavity. So photons may enter the optical cavity from the metal across their common surface. The metal can emit and absorb photons from the cavity, otherwise no energy is exchanged between the two regions and no energy is able to escape the two regions.

I assume no photo-electric effect because this would require the photons to be of an energy which is equal to a sum of the bandgap (which would be sufficient to delocalize the electron and allow it to conduct) added to "the work function" which is the minimum energy required to remove a delocalised electron from the surface of the metal. The emitted photons are, at maximum, an energy equal to the bandgap of the metal. Therefore no photoelectric effect.

The photons enter the cavity after emission from the metal (which happens naturally at an average rate). More photons wind up in the left side of the cavity due to the optical diode in the middle of the cavity. So more photons are now absorbed on the left side of the metal than on the right side.

So what happens to this energy? Does it remain on the left side of the metal? Remember, the metal must "return to thermodynamic equillibrium" and in so it must use a portion of its energy to do so.

My guess is that the metal channel propagates something like an electric current (left to right) to return the energy to the right side until equilibrium is established. This would be a net movement of holes to the right and electrons to the left. However, it could also manifest as a net movement of "higher energy current carrying electrons" from left to right and "lower energy non-current carrying electrons" from right to left, with no net current.

Because the metal always emitts photons into the cavity then the cycle is continuous. Ergo perpetual motion?

For a more concise explanation, see my webpage on this thought experiment:

http://violationprotocol.webs.com/propagation-protocol

3. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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I just so happened to look into something like this recently, out of a renewed interest in thermodynamics and statistical mechanics. I'm pretty sure this scenario is ruled out because the optical diode itself is subject to thermal fluctuations, so you can't actually construct a perfect one-way device in practice. There's no established method in theory or practice to ratchet energy out of the quantum vacuum and lower the entropy.

5. ### Q-reeusValued Senior Member

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3,383
The imo glaring weakness in your scheme is the unstated assumption here of an optical diode as 'reactive isolator' - i.e. where essentially lossless reflection occurs at diode junction for reverse propagating photons. That holy grail would allow energy gain in a very straightforward way using just ordinary waveguide + 'optical diode' (or microwave frequency counterpart). Such possibility was theoretically and experimentally investigated back in the 1950's at the dawn of microwave nonreciprocal ferrite device technology - with definitively negative conclusions.

Actual isolators work via nonreciprocal absorption. And given your photons are thermal anyway, I see no prospect of any optical diode behavior even for the usual absorptive isolator. Have a good read here re 'reactive isolator':
www.lucent.com/bstj/vol38-1959/articles/bstj38-6-1427.pdf
Just to spice this up a bit: A few years ago a new type of unidirectional ferrite mode was discovered, and it sure appears to allow 'reactive isolation'. If you want further details, PM me.

7. ### al onestoneRegistered Senior Member

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223

The diode doesn't have to be 100% operational. Even if it is slightly operational in this situation then the photons (energy) must propagate in the loop.

8. ### al onestoneRegistered Senior Member

Messages:
223

I looked at the paper you've linked, and there they use a partial ferrite waveguide with a magnetic field. They analyse all the modes for each direction of propagation and find that under certain conditions there is a "diode like" performance of the waveguide. However, they find that the "diode" performance which should not allow propagation in one direction ends up absorbing for that direction. But does this prevent my scheme from working?

You've also pointed this out,

You're right, my scheme will always require a diode which depends upon absorption occurring, and this absorbed energy will be liberated back to the right side of the cavity via photons and then they get back to the right side of the metal. I just noticed two links on wikipedia's section about optical isolators that discusses exactly the relationship between the Faraday isolator and thermodynamical law. (one is by Rayleigh from 1901, imagine that)

So now I don't know what to make of this comment,

All this being said, I still wonder if the absorption would occur in the diode. When the photons start gathering on the left side (with some being absorbed by the diode) why would the system force absorption at an increased rate in order to balance things thermodynamically when there is an obvious pathway for the energy through the metal which would be so much easier to traverse thermodynamically? It is less costly for the energy to return to the right side via absorption of the photons by the metal than by the diode (Faraday isolator). This would complete the loop.

In my scheme I am giving the photons(energy) another possible path which it can traverse to get back to the right side.

9. ### al onestoneRegistered Senior Member

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223
Another problem my scheme has is the polarization dependence of the diode. I have not taken this into account.

However, the problem related to absorption is not a problem. I noticed in the wikipedia link,

Here is the link if you're interested:

I am not officially convinced that my scheme does not work (obviously others are thinking something similar) but for now I will abandon this thought experiment.

10. ### Q-reeusValued Senior Member

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3,383
I believe so, and there was more I could and perhaps should have added before. Ferrite isolators work over a typically narrow bandwidth of say 2-5%, for a single coherent mode having a single polarization (I see you have addressed that in #6), all within a waveguide structure having transverse dimensions comparable to the wavelength. Compare to the situation of incoherent blackbody radiation - random polarizations, propagation directions, and huge frequency spectrum. Not much dice for even a tightly packed wall array of 'optimal size' micron dimensions optical isolators. There likely will be a truly meager level of nonreciprocal absorption going on. But as you in effect say, such absorption and random re-radiation should be thermal thus have the same spectra as incident flux, so one gains what for the effort?
If one accepts absorption re above process has no net effect on thermal distribution, we simply have a system in thermodynamic equilibrium, with no temp or chemical potential gradients, hence no net currents of any kind.
Thing is all possible paths exhibit reciprocity overall (owing to ambient thermal distribution) - so no net flows.
I threw it in for sake of completeness in a sense, but it diverges from topic really. Will discuss in PM.

11. ### Q-reeusValued Senior Member

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3,383
I covered that earlier, but to be fair, there are clever designs for polarization-independent isolators. However can't recall offhand if any net gain would result, as it may entail e.g. double the input area to achieve.
An interesting article which I recall having read quite a way back. The resolution described there by Lord Rayleigh may not be too clear, but amounts to saying that given there is no one-way absorption, there will be a double reflection occurring for the nominally 'reflected' ray - hence there is no NET one-way reflection and the final result is complete reciprocity of transmission.
Well here's something that may spark your interest. I came up with a planar variant of the following, but quite some years previous, so I claim a priority of sorts:
http://www.newscientist.com/article...-for-light-created-on-earth.html#.UeWWRG34Vfw
And some Googling will provide leads to free-to-read technical articles. More than one way to skin a cat - perhaps!

Last edited: Jul 17, 2013
12. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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Yes, Captain is correct. I've mentioned on these forums before that I command an intergalactic space trawler fishing for space salmon, and I'm damn proud of it

. As to the diode efficiency issue, basically the point I took out of my own reading into the subject was that there would be enough thermal fluctuations to prevent any net energy from being ratcheted out of the vacuum. It's basically a modern version of a thought experiment that goes back almost 150 years (have you ever heard of Maxwell's Demon?).

13. ### al onestoneRegistered Senior Member

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223
Maxwell's Demon? You mean the little guy that opens the door between the two thermo baths? Yeah I've heard of him. Sadly he is just a fictional character. I think the end result of this thought experiment is a learning experience for me. There is no perpetual motion scheme of the type I have mentioned that works.

However, I will be posting another scheme for the gravitation specialists soon. This new one might work.

14. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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The best kind of learning experiences are the ones where you pose challenges to the existing understanding (provided you allow yourself to be informed by the responses, of course). I usually learn the most in physics when I'm trying to understand something and the picture doesn't seem to add up, my calculations don't match with known results, etc., and then I gain valuable new insights by seeing where I went wrong.

Yeah it is sad the way physics puts so many restrictions on what we can do no matter how many resources we have at our disposal. A couple of weeks ago I overheard a couple of people on a plane talking about quantum teleportation, and one of them said he'd heard that an entire piece of steak had now been teleported- I had to butt in at that point and tell them the truth, but we all agreed the world would be a much cooler place if such things were indeed possible.

Well I wish you the best of luck, and hopefully you at least gain new insights even if it doesn't work out the way we'd all wish.