Half a croissant, on a plate, with a sign in front of it saying '50c'
h a l f b a k e r y
Ambivalent? Are you sure?

idea: add, search, annotate, link, view, overview, recent, by name, random

meta: news, help, about, links, report a problem

account: browse anonymously, or get an account and write.



Maxwell's Demon's Demon

Done with photons, instead of molecules. Also, not a real demon.
  (+3, -1)
(+3, -1)
  [vote for,

(For the purposes of illustration, the established scientific principles and physical properties are simplified somewhat, but the relevant bits are retained. It's a thought experiment, at this stage)

Start with the concepts of...
Black Bodies:
A black body absorbs all radiation (radio, infrared, light, x-rays, etc.) thrown at it, regardless of frequency/wavelength and power. As a whole it emits a spectrum directly related to the power it's emitting.

A fluorescer absorbs radiation at a specific frequency range, and emits radiation in a lower range - when a molecule is hit by a photon it translates some of the energy into vibration, and spits the rest out as a lower-energy photon.

Radiation bounces off a reflective surface, a perfect one reflects everything.

Bandpass Filter
Most everything is reflected, except a specified range of frequencies, which is let through (the "band" that's being "passed").

The radiative power, proportional to the area it's being emitted from.


The equipment:

A perfect mirror shell
A bandpass mirror shell
A black body bowling ball
A black body bowling ball with a stripe of fluorescent coating on it.

A bowling ball fits snugly within a sphere without touching the inner surface.

The environment is an infinite open space which radiates a black body spectrum of an unspecified temperature.


Some preliminary baselines.

We point our thermal sensor and spectral analyser at the four items:

The perfect mirror just reflects all the radiation tossed at it. It's the same temperature and spectrum as the environment.

The bandpass mirror reflects everything except at the bandpass range. But, since that's being passed through from the other side, it's currently the same result as above.

The black body absorbs all the radiation thrown at it, emitting a black body spectrum. Since the environment is a black body spectrum, both temperature and spectrum read as the same as the environment.

The fluorescent black body is the odd one out. Even though it's the same temperature, the spectrum is a bit skewed from that of the environment, looking like that of a slightly lower temperature black body, with a spike added at the fluorescent emission frequency, to compensate.


Right, let's mix and match shells and bowling balls and see what happens.

Mirror Shell + BB Bowling Ball

A bit boring. The shell is still the same temperature and spectrum as before, and so is the bowling ball. The only interesting thing is that you can change the outside environment, making it much much hotter or much much colder, and the bowling ball is still the same temperature as it was before, because the radiation it's emitting is bouncing off the inside the shell and back into it, and the outside radiation isn't coming in.

Mirror Shell + Fluorescent BB Bowling Ball

Also a bit boring, sorry. The shell still shows the same temperature and spectrum as before, and so does the bowling ball (a black body spectrum with a spike).

Bandpass Shell + BB Bowling Ball.

Well at least something happens with this match. If the outside temperature changes, so does that of the bowling ball, even though the only radiation going in and out is just that at the bandpass frequency, so it takes awhile.

Slightly interestingly, though predictable, is that you can crank the outside radiation temperature through the roof: raging cascades of photons of every frequency but, if there's no radiation at the bandpass frequency then not only will the bowling ball not get any hotter, it will actually get colder as its emitted bandpass frequency leaks out with nothing to replace it. Cool.

Likewise you can take a very low-powered laser (at the bandpass frequency) and heat the bowling ball up to a ridiculous temperature. The temperature will only stop rising when the amount of bandpass frequency going out matches that of the amount coming in. (This is vaguely the "greenhouse effect".)

Okay, go get a coffee, take a leak, come back.

Bandpass Shell + BB Fluorescent Bowling Ball.

As soon as the fluorescent bowling ball is placed in the shell, something strange happens. The temperature of the Shell apparently increases ! Why ? you ask with bated breath. Recall the baseline preliminary: the bandpass shell has the same temperature and spectrum as the environment because it's directly reflecting most of the radiation from the environment, except the bandpass frequency which is coming through from the other side.

But the Fluorescent Bowling Ball has a spike at that frequency*. It's radiating more energy at the bandpass frequency than the environment does. So the temperature, as measured from outside the shell, has increased.

But only for awhile. Eventually, because less energy is getting into the shell than is coming out, the bowling ball loses some thermal energy, until the transfer amounts equilibriate.

Of course, the reason the system is in equilibrium is because the inside of the shell is now colder than the outside.

Sir Isaac Newton, you owe me a beer.


* I suppose I should have mentioned earlier that the bandpass on the shell is set to the fluorescent emission frequency. Oh well.

FlyingToaster, Nov 06 2015

The scary story of the man who did not believe in deamons https://storybird.c...ble-from-rebbi-nah/
and would not even write them so. (on Storybird) [pashute, Nov 09 2015]


       Let me guess, the energy gained from one cycle of opening the shell, inserting the ball, waiting for the new equilibrium, opening the shell and extracting the ball, is a wee bit less than the effort expended?
pocmloc, Nov 06 2015

       ^ the easier way would be to include the concept of thermoelectriciy in the build, sticking it between the ball and the outside through a hole in the shell. Further annoyance could be achieved by using the electricity to run a heater, outside.
FlyingToaster, Nov 06 2015

       No demon = no bun. Firm but fair.
not_morrison_rm, Nov 06 2015

       It seems unable to separate hot things from cold things.
travbm, Nov 06 2015

       OK, let me re-guess, that the energy expended in inventing the idea is more than the energy harvested from the device.
pocmloc, Nov 06 2015

       <sigh> looks like it: I burned through at least 5-6 buns of calories typing it out. Woulda been more like a dozen if I hadn't already taken a few stabs at it, previously.   

       The maths, at least as general principles, line up, and reality mostly cooperates as well. Most of the idealized physical properties in the post are just for ease of visualization: for instance...   

       There's no such thing as a perfect black body, but any substance which has identical'ish absorption and emission spectra (ie: almost everything) would work.   

       Bandpass mirrors, in reality, aren't as cut and dried as is made out in the post but, again, the only real necessity is that the mirror reflect some wavelength(s) present that aren't in the fluoro-emission spectrum. Easy peasy.   

       Which brings us to the the hook that makes it all work. It would fall over if fluorescence was symmetrical; if its emission spectrum was the same as its absorption spectrum, but it isn't: pumping emission wavelengths back into it will not produce absorption wavelengths out of it.   

       So that seems to be kosher for the experiment, as well.   


// It seems unable to separate hot things from cold things//

       Since that's exactly what it purports to do (the ball inside the shell gets cold), your reading or my writing isn't up to par. What part is vague ?
FlyingToaster, Nov 06 2015

       Thermodynamics wasn't Newton.   

       Other than that, this definitely doesn't work, now I just have to figure out why. I think it's something to do with a fluorescent object not operating as a black body otherwise, but I can't swear to it.   

       There's also the issue that a bandpass filter absorbs some of the out of band energy hitting it rather than reflecting it (This is inherent in the nature of a bandpass filter, no "perfect mirror except this wavelength" can exist). Also relevant in this regard is that the pass band of a filter varies with angle. I'm not sure how it's relevant, however.
MechE, Nov 08 2015

       Ah... well, somebody owes me a beer.   

       The only (theoretical) requirement for the mirror is that it reflect at least one existing wavelength that isn't fluoroemissive and pass at least one that is. the "bandpass" thing is just sleight of hand to focus people on the energy in vs energy out thing. As is the "black body" requirement.   

       (as an aside, a Bell jar + a chunk of carbon black on an insulating pedestal + a toy laser might make an interesting demonstration of the greenhouse effect).
FlyingToaster, Nov 08 2015

       Complete lack of chanting, skulls, chicken blood, and orgies in the moonlight. [-]
Voice, Nov 08 2015

       Actually, the fluorescent ball isn't able to emit more energy than it absorbs at a given frequency unless it has some other energy to work with. Your uphill waterfall isn't going to work.
WcW, Nov 09 2015

       Okay, let's do the shortened version :). All the objects are perfectified to focus on the mechanism.   

       Ambient is a black body radiation.
The sphere is a bandpass at freq y, reflecting everything else.
The ball is a black body
The stripe of paint on the ball is a fluorophor that absorbs a photon at frequency x (which is within the black body envelope of ambient) and emits a photon at freq y, absorbing the difference as heat.

       Note that the bandpass frequency is the same as the fluoroemission frequency, and the fluoro absorption frequency is within the ambient thermal spectrum (or it won't work).
part 2 in a minute...
FlyingToaster, Nov 09 2015

       Now measure the ball (with stripe) and the sphere, separately against ambient.   

       The sphere is the same temperature and spectrum, though note that the bandpass photons aren't reflected from the environment, they're coming through from the other side.   

       The ball is the same temperature (energy emission) as ambient. The spectrum, because of the fluorophor, is a bit different - there's a spike at the bandpass frequency (and, since it's at equilibrium, the rest of the spectrum is a little subdued to compensate).
part 3 coming right up...
FlyingToaster, Nov 09 2015

       Before we put the ball in the sphere, let's remember that the ball's spectrum has a spike at the bandpass frequency, whereas ambient doesn't. At the bandpass frequency more radiation is coming from the ball than is coming from the environment. Also recall that the sphere only allows radiation in and out at the bandpass frequency.   

       With that in mind, what happens when you put the ball inside the sphere ?
and our startling conclusion...
FlyingToaster, Nov 09 2015

       Since the ball is, for the moment, emitting more energy at the bandpass frequency than the environment is, more energy is going out than coming in.   

       In fact, more energy will be going out than coming in until the spike matches the level of ambient at that frequency.   

       The only way that's going to happen is if the ball is (overall) colder than ambient.   

       (and a couple of odd notes: when the ball is first placed within the sphere, the sphere will measure as hotter than the environment because the measurement is the spike from the sphere and the rest of the spectrum reflected from ambient. Second, when things are at equilibrium, even though the ball is colder, the sphere will measure as the same temperature as the environment because now the spike from the ball is at the same level).
FlyingToaster, Nov 09 2015

       I understand that. I also know, based on the laws of thermodynamics that this effect cannot happen, as it would allow the extraction of energy from a uniform temperature (put a seebeck generator across the sphere). Therefore, there is something wrong with the concept.   

       And I think I just realized why. Now, the fluorescent object only receives radiation from the outside in it's emission wavelength (the pass band). And it only receives energy from the inside in the wavelengths it emits itself.   

       So the problem is that, a fraction of a millisecond after you close the shell (speed of light times the distance from the shell to the ball), it is no longer fluorescing. This is because a fluorescent object is not a true black body. It does not emit energy at it's excitation wavelength. Since it isn't emitting in it's excitation wavelength, nothing is reflected back to it in it's excitation wavelength, and it no longer fluoresces. Instead it continues to radiate as a black body with holes in it's spectrum at both the emission and excitation wavelengths.
MechE, Nov 09 2015

       Hmm, I thought I specified that the ball wasn't completely covered by the fluorophor, for exactly that reason. ahh...   

       I meant to write "some fluorescent coating", and wrote "some fluorescent coating", instead. Thankyou, English language.   

       Sorted, now reading "a stripe of fluorescing coating". The bare BB keeps the fluorophore fed: bandpass wavelengths come in and are absorbed by the BBBB, remitted as BBR, reflects off the shell and the absorption frequency of same absorbed by the fluorophore. The absorption spectrum of the fluorophore is within the ambient BB envelope, of course.
FlyingToaster, Nov 09 2015

       It still won't work, but give me a bit to figure out why in this case.   

       One possibility is that it can work, but that the fluorescent material has a limited lifetime (fluorescent materials do decay with fluorescence events). If that's the case, and the energy to reactivate the fluorescent material is equal to or greater than the energy of the temperature difference, then that would explain it. Once again, I can't be certain of the math in this case.
MechE, Nov 09 2015

       well.... I'm pretty sure that, unlike phosphorescence, fluorescence isn't entropy powered. Fluorescent light bulbs go, but I always thought that was because the gas envelope was compromised, not the coating.   

       I was going to say that the mechanical vibrations preclude absorbing photons in their frequency range, but that's how thermal transfer works. The only difference (I can imagine) between a normal molecule and a fluorescing one is the normal one absorbs the photon and distributes the energy with its compadres, while the fluorescing one absorbs a photon, emits a lower energy photon and distributes the difference.
FlyingToaster, Nov 09 2015

       I don't know about all of these bowling balls covered in flour, but I'm hungry for something dark. Maybe some blackbody chocolate.
RayfordSteele, Nov 09 2015

       Fluoresence degrades over time because not all of the fluorescing compound re-emit it's photon, some of them lock into the excited state. That being said, I'm not happy about that explanation for the 2nd law mechanism either. As I said, I'm still thinking about it.
MechE, Nov 09 2015


back: main index

business  computer  culture  fashion  food  halfbakery  home  other  product  public  science  sport  vehicle