# I Non-Linear optics vs The 2nd Law of Thermodynamics

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1. Sep 3, 2016

### hmmm27

Hi, rank newbie here, with my first post.

This one is something I figure every first year student comes up with at some point, but I don't know enough keywords to Search for an answer. (I'm not a student except in the category "of life": this isn't assigned homework)

I figger, using a bit of non-linear optics, the 2nd law of Thermodynamics can be circumvented. Obviously this isn't true, so let's take a looksee...

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Temperature - overall power emission per areal unit.

Black Body - absorbs all radiation, emits a predefined spectrum, related directly to power, with some non-sequitur to the experiment formula.

Fluorescer - transforms a certain input range of frequencies into a certain output range. Basically, when photons are absorbed, a bit of internal heat is produced and lower frequency photons emitted.

Bandpass mirror - passes only a certain frequency range, reflects the rest.

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Next, the components of the device, also simplified, also for clarity...

- A black-body ball, partially painted with a fluorescer (polkadots or stripes, your pick).

- A bandpass mirror sphere, larger than the ball.

Required is that the fluorescence output be the same frequency as the mirror's bandpass.

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And conduct the experiment...

Within an environment of a non-specified BB background radiation temperature...

Place the ball within the sphere.

That's it.

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What happens...

First let's backtrack a bit and compare the spectra of environment and ball before the experiment starts. Both integrate the same - the power outputs/temperatures are identical - but the spectra are different. While the environment spectrum is strictly BBR, the ball's spectrum should look like a spike was added to a lower-temperature BBR.

(To bridge narrative somewhat seamlessly into the next bit, we can see that at the spike frequency the ball will be more energetic than the environment)

The sphere only allow equilibration at its bandpass, which is at the spike. So, when the ball is placed within the sphere, there is more radiation exiting than entering, to start with.

Eventually, power equilibrium (in vs out) at the bandpass/spike is reached, as the spike diminishes in size.

(To clarify, the mechanism for the spike's decrease, but not cessation is the BB bit of the ball continuing to emit in the fluorescing absorption range (which reflects back onto the ball from the inside of the mirror).)

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Result

The ball gets colder.

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So....

a) within the given framework of simplifed definitions and magically perfect components, will it work ?
b) what property stops it from working in a real world ? with real, non-theoretical substances/components.

Note that there's no "what will reduce the efficiency" question: the experiment was kept deliberately simplistic. In the grand scheme of things it shouldn't matter that bandpass mirrors have different properties at different angles, or that perfect black bodies don't exist. The only requirement is that the bandpass include some of the fluorescent emission and exclude some of the fluorescent absorption ranges.

Last edited: Sep 3, 2016
2. Sep 3, 2016

Staff Emeritus
In all this complication your question seems to be "I have a system initially in equilibrium with a thermal bath of photons. Can I decrease the entropy of the system by increasing the entropy of the photons?" The answer is "yes".

3. Sep 3, 2016

### hmmm27

Thanks for answering. Sorry for the "compIexity"; I set up the problem to try and prevent non-sequitur answers.

Okay, let me make it simple...

Three components:
- a bandpass mirror sphere
- a black-body ball
- a fluorescing ball.

The fluorescing absorption freqs are outside the bandpass range,
The fluorescing emission freqs are inside the bandpass range.
All ranges are within the ambient envelope.

Place the balls inside the sphere.
They get colder.

Which breaks the 2nd Law of Thermodynamics.

Last edited: Sep 3, 2016
4. Sep 3, 2016

### Drakkith

Staff Emeritus
How so? What exactly is your system and is it isolated? What happens to the entropy of the system compared to the entropy of the environment?

5. Sep 3, 2016

### hmmm27

I don't mean to be rude - I don't understand all the terms that you take for granted - but what do you mean by "system" that isn't covered by the first post, or the one you responded to ? It doesn't matter if it's isolated (the contraption within a large mirror-sphere) or not (floating in a space of ambient background radiation).

It breaks 2LT by moving heat from a cold spot to a warm spot. Put your beer inside and it gets cold. Drill a hole and put a Peltier in between a ball and the outside and you've got "free energy", constantly replenished.

6. Sep 3, 2016

### Drakkith

Staff Emeritus
My apologies. I was trying to bounce off of V50's post, but I think I misunderstood something when I posted. I'll step out of this and let others more knowledgeable in this area help you.

7. Sep 3, 2016

Staff Emeritus
Drakkith is right - if you want to discuss the 2nd law, you have to follow the entropy. It's going from the inner sphere to the photons streaming out of the outer sphere. As for why it's cooling down, for the same reason a warm sphere in empty space cools down. It's radiating away energy.

8. Sep 3, 2016

### hmmm27

Yes, it is radiating away energy. That's what it does. That's its raison d'etre.

But it isn't a "warm sphere in empty space cooling down"

The inside is getting colder than the outside. More energy is leaving the sphere than entering it, thus the inside is getting colder.

9. Sep 3, 2016

### Bystander

10. Sep 3, 2016

### hmmm27

Actually, it's the "Clausius Statment" referred to in the Wk article on the 2nd law of thermodynamics.

"Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."

Technically, there needs be yet another ball which doesn't actually do anything odd, outside the bandpass sphere and inside the reflective enclosing sphere, to meet the linguistic requirements of that statment. The outside ball gets warmer, the inside, colder. So now there's.... 5 objects: a wholly enclosing mirror sphere, enclosing a ball and a bandpass sphere, the latter enclosing a fluorescing ball and a black body ball. This isn't getting any easier, is it.

In terms of entropy, if I understand the usage, the inside is growing entropic while the outside grows enthalpic.

11. Sep 3, 2016

### Staff: Mentor

This is an interesting question. I think that the spheres and mirrors and balls and environment details are a little confusing, so let me see if I can simplify it a bit.

There is no conduction or convection, only radiative heat transfer. There is a hot reservoir and a cold reservoir each radiatively connected to a fluorescent material via a bandpass filter. The filter to the hot reservoir passes the emission frequency and the filter to the cold reservoir passes the absorption frequency.

Is this a simplification which still gets at the essence? And your concern is that the fluorescent material should absorb energy from the cold reservoir and then emit it to the hot reservoir?

12. Sep 3, 2016

### hmmm27

Cold end <> fl.absorption pass filter <> fluorescer <> fl.emission pass filter <> hot end

yup, same thing... you also managed to skip the black-body bit, which is a relief. Much clearer. Thankyou.

Starting off from all temperatures the same, one end will get hot, the other cold. A photonic Maxwell's Demon device.

(It's probably worth mentioning that fluorescers aren't symmetrical - if you shoot the emission frequency at it, you don't get the absorption frequency back)

Last edited: Sep 3, 2016
13. Sep 3, 2016

### Staff: Mentor

So a fluorescent substance has a certain quantum yield, i.e. a certain ratio of emission photons produced per excitation photon absorbed. A black body spectrum at a given temperature similarly has a certain ratio of photons at two specified frequencies. You would only get energy transfer if the fluorescent quantum yield ratio were higher than the black body frequency ratio. I don't know for sure, but I would expect that is impossible from first principles.

14. Sep 3, 2016

### hmmm27

Way I understand it, an excitation freq photon hits a fluorescer molecule, the molecule spits out an emission freq photon and keeps the change (in the form of vibrational energy)

15. Sep 3, 2016

### Bystander

Emissivities are not unity. "Photon" numbers are NOT conserved.

16. Sep 3, 2016

### hmmm27

That's over my head... were you talking to Dale or myself ? The mechanism doesn't rely on a quantum yield of 1.0 (there are many things that can lower efficiency).

17. Sep 3, 2016

### Staff: Mentor

No, but it does rely on a yield greater than the ratio of black body photons at the two different frequencies.

In QM everything is probabilistic. There is a chance of that happening, but there are many other possibilities also, which reduces the yield.

18. Sep 4, 2016

### hmmm27

Okay, let me ramble a bit....

Emissive radiation is caused by relaxation of a molecule from a higher state to a lower one... hmm but relaxation isn't necessarily fully to a discrete level down if it's thermal radiation, because of all the bouncing around (lol, sorry).

Regardless, if conditions were right for a fluorescer to release a fl.absorption-frequency photon (by thermal cause), would there not be a chance it would get stuck and release a fl.emission-frequency photon, instead ? ie: the path is there, already, like a detente. Therefore, wouldn't a fluorescer's "BB" spectrum have a bit of a dip at the absorption frequency, and spike at the emission frequency.

(or not, and you've already told me that a couple times, whatever you meant about ratios)

Also, by definition a BB cutoff isn't actually a "cutoff"; frequencies still exist beyond the peak, so it's not necessary for both absorption and emission freqs to be behind the curve, so to speak, if getting swamped by thermal radiation is the issue.

(I'm not concerned that this thing works well, I'm just curious if it works at all, by stacking the deck, even ever so slightly)

19. Sep 4, 2016

### Bystander

No.

20. Sep 4, 2016

### Staff: Mentor

Rather than ramble, let's walk through a specific example. For a given fluorescent compound with a given excitation wavelength and a given emission wavelength there is a specific quantum yield. Usually it is less than 1, but there are some compounds with a quantum yield of 1. For example Rhodamine 101 at 576 nm excitation and 600 nm emission (see p 2219 at http://iupac.org/publications/pac/pdf/2011/pdf/8312x2213.pdf).

Now at 300 K there are more than 20 blackbody photons at 600 nm for every single blackbody photon at 576 nm. So even with a 1:1 excitation to fluorescence conversion, there are more than 20:1 thermal photons at the emission wavelength compared to the excitation wavelength meaning that the thermal flux is about 20 times greater than the fluorescent flux. So any energy that is momentarily moved "up" a temperature gradient would immediately be moved back "down".

Now, maybe I should focus on energy flux instead of photon density, but I don't think that the situation would change.