Non-Linear optics vs The 2nd Law of Thermodynamics

[Andrew Mason]

hmmm27:

You seem to be intent upon finding an argument around the second law of thermodynamics. The second law does not state that a high energy photon cannot spontaneously be emitted from a cold body and spontaneously absorbed by a hot body. It is a statistical law about macroscopic states. It simply says that at a macroscopic level, heat flow will not occur spontaneously from a cold body to a hot body. This is the case whether you are dealing heat transfer by contact or by radiation. You are not going to succeed in finding a flaw in the second law. But you might benefit greatly from trying to understand what Dale and others are saying.

AM

 

[hmmm27]

Emission is EMISSION.

yes thanks, now I know it's possible to both underline and bold the same text passage. Would it help if I said "the fluorescing mechanism is compromised by the presence of overwhelming thermal energy, and emissions cannot be produced in that manner" ?, instead ?

And, that your previous post was directed at me, not giving some other noob comeuppance for unhelpful answers...
-----
Okay so you mean his 1st law of spectroscopy, yes ?

"A solid, liquid, or dense gas excited to emit light will radiate at all wavelengths and thus produce a continuous spectrum."

And... what ? A spike or dip in a graph can be part of a continuity.

If you mean BB radiation, the things I read on the subject state quite specifically that things like the Sun and an incandescent light are _close_ to being BB bodies, which would lead one to believe they aren't... which could lead one to suspect that the regular emission spectra is superimposed on top of it in some manner.
-----
I have no clue what you mean by

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

or how it relates. No attempt is being made to state that the number of photons doing anything is the same as the number of photons doing something else.
-----
"NO" (in response to "will it work ?")

Okay that I understand. I actually understood that coming into this thread. It's the "why" I'm curious about.

hmmm27:

You seem to be intent upon finding an argument around the second law of thermodynamics [...] It is a statistical law about macroscopic states [...] You are not going to succeed in finding a flaw in the second law. But you might benefit greatly from trying to understand what Dale and others are saying.

I'm not intent on it; I have in fact broached what I think is a possible solution to (my) problem, several times.

Did you understand the explanation I gave you, of the device/process, and understand how one might think it would work ?

It doesn't seem that way, but I _am_ trying to understand, from a viewpoint other than a tautology (I don't mean the 2LT is a tautology, I mean saying something won't work because it's a given that it won't work is a tautology).

I've pretty much exhausted their patience, however.

 

[Bystander]

Can you explain "thermal equilibrium?"

 

[hmmm27]

Can you explain "thermal equilibrium?"

Pretty sure I can do that much: I understand this bit. It means that the flow of energy between two objects connected by an energy conduit of some kind is the same in both directions. To quote Wikipedia...

"Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat."

which could probably be better stated using "net heat flow" instead of "heat flow". Photons are constantly flying back and forth: (radiative) equilibrium is reached when the power flow from A to B is the same amount as the power flowing from B to A.

As far as the definition is concerned, it could be a bunch of green photons going one way and (a larger amount of) red photons going the other if you can pull that off; the important bit is that the power is the same in both directions.

So far, so good ?

(or do you mean "how is thermal equilibrium reached in the idea" ? which is a slightly longer explanation)

 

[Bystander]

if no heat flows between them when they are connected by a path permeable to heat."
... which is incorrectmpletely stated: it should say "no net heat" as regards radiative energy,

"Radiant/radiative" energy is/equals "heat."

So far, so good ?

If we can agree on the changes?

 

[hmmm27]

heheh, ok. typos'r'us.

 

[Bystander]

Next: are you familiar with Boltzmann?

 

[hmmm27]

I noticed the name while auditing some articles on black bodies.

 

[Bystander]

Boltzmann distribution?

 

[Dale]

yes, and by that standard of measurement, at some frequencies it could be greater than one.

I think this is the key. It is never greater than 1. A perfect absorber is 1 and you can't do better than perfect. And the coefficient is the same for emission and absorption.

The maximum peak of an emitter can only bring the emission coefficient up to 1, not beyond. Meaning that (without doing work on it) a system at a given temperature cannot emit more energy at any frequency than a blackbody at that same temperature.

noting that bb emissions have nothing to do with bb absorptions

They have everything to do with each other. That is the whole point of the hyperphysics page. The absorption coefficient is always exactly equal to the emission coefficient. They are the same thing. That is necessary for the 2nd law and follows from QM, as described in the hyperphysics page.

I don't see why that would hold true as a basic premise, though agreed at non-fluorescent-specific emission frequencies

The reason this holds true as a basic premise is explained in the hyperphysics page. It holds for all frequencies, including the fluorescent-specific emission frequency.

How does the "thermal flux" being "large" have anything to do with it ?

In any radiative heat interaction there is energy going both ways. There is energy going from the cold object to the hot object. But there is more energy from the hot object to the cold object. So it isn't enough to point out that there is energy flux from cold to hot, you also have to consider the flux backwards and compare how large they are.

My rough estimation shows the thermal flux to be about 20 times greater than the fluorescent flux.

 

[hmmm27]

Boltzmann distribution?

Mmmm... a formula that shows the particle distribution of energy levels at a given temperature. Interesting, you'd think that some levels wouldn't be a completely linear sequence.

I think this is the key. It is never greater than 1. A perfect absorber is 1 and you can't do better than perfect. And the coefficient is the same for emission and absorption.

I understand the concept of AC referring to a specific atom or molecule: simply a measurement of how many photons are absorbed compared to the total that impinge in some manner. If they're all absorbed, then "1". If half are reflected or transparented, then 0.5 . Easy. (though there may be some question for a future time about the "transparented" bit, which doesn't have any bearing, currently)

(oh, I wrote "atom or molecule" because I had ac/ec confused with "gray body" which involves gross surface structure)

EC seems a bit more complex : a measurement of photons emitted... in comparison to what ? The number of the appropriate level drops that occur without a photon being produced ? (ie: resulting in energy of some kind being passed somewhere else). How could that be measured ? or is it inferred from the AC. As far as AC=EC, while the interesting actions are symmetrical: photon absorbed<>photon produced... the other ones: reflection or energy transfer, don't seem to be (to me, at this point), though it certainly is an attractive proposition.

[a later point and it does make sense: pardon the crudity of explanation but the whatever that stops a photon from being absorbed is the same thing that the energy bounces off of and decided to go somewhere else, instead. okay, it doesn't make complete sense, barely partial sense, but it's a start]

What is interesting and could be relevant later on is that a photon can hit a molecule, be absorbed, then either be re-emitted (different photon, same frequency) or changed into vibration. EC determines that probability, also determines it for a relaxation of either kind (emit photon or warm the surroundings) achieved from a level, caused by vibration instead of a photon. So the probability of an impinging photon directly resulting in an emitting photon is EC²

That was quite a bit of work. I hope I'll need it.

The maximum peak of an emitter can only bring it up to blackbody, not beyond.

Which is where I'm stuck up to my arse, obviously. Why does the combination of a pre-existing spectral line and thermal radiation not cause a bump on top of the bb curve.

noting that bb emissions have nothing to do with bb absorptions

They have everything to do with each other.

How can I, or why can't I, cut/paste nested quotes ?

I meant that absorption distribution has nothing to do with emission distribution, except in terms of power equalization.

At the fluorescence wavelength the material is a good emitter, and hence also a good absorber.

So the fluorescing mechanism is non-sequitur to the act of absorption and emission if taken as separate processes. It simply notes that a bumping up results only in a partial relaxation due to some internal mechanism or relationship.

Okay, that's not "simple" in any respect. Is there something special about the half-level ?

According to some earlier ramblings of mine , "fluorescence" therefore doesn't actually need an incoming photon: it could be caused by an internal energy transfer to reach the higher energy state, then fall to the half-level.

In any radiative heat interaction there is energy going both ways. There is energy going from the cold object to the hot object. But there is more energy from the hot object to the cold object. So it isn't enough to point out that there is energy flux from cold to hot, you also have to consider the flux backwards and compare how large they are.

My rough estimation shows the thermal flux to be about 20 times greater than the fluorescent flux.

So, let's be on the same page and note that the first paragraph has nothing to do with the second.

Re: the first: with the premise I walked in with (and haven't fully let go of) - that a fluorescent spike and dip would be added to a thermal curve in some manner, rather than be incorporated into it - it's trivial to design a theoretical contraption using bandpass filter(s), that sequesters (more) heat at one end. Using that premise, before equilibrium is reached, the temperature of C as measured from F (through the filter) is hotter than the temperature of F as measured from C(ttf). At equilibrium - when the power flows are the same, F as measured at F is hotter than C measured at C. Rinse and repeat for the relationship between F and H. (Cold side, Fluorescer, Hot side)

Re: the second: good to know. Without reference it's meaningless of course (it was Rhodamine101 fluorescing from/to 576/600nm, 300K I think. There was an interesting sidenote concerning "Karstens, who merely showed that QY is independent of temperature", without specifying the temperature range tested at, nor stating if QY reading was normalled by excluding predicted bb radiation at those temperatures. I don't have my 1871 J.Phys.Chem handy to check the original).At which point it looks like I'm in violation of forum rules, if massively misunderstanding something counts as a "personal theory".

 

[Dale]

I understand the concept of AC

Yes, your description sounds good to me.

EC seems a bit more complex : a measurement of photons emitted... in comparison to what ?

In comparison to a blackbody. The point is that at a given temperature and wavelength (without work) a material cannot absorb half of what a blackbody does without also emitting half of what a blackbody does.

So the probability of an impinging photon directly resulting in an emitting photon is EC2

At thermal equilibrium and without work, I think that is right.

Why does the combination of a pre-existing spectral line and thermal radiation not cause a bump on top of the bb curve.

The bump isn't a wavelength that has a higher emissivity than a blackbody surrounded by wavelengths with equal emissivity. It is a wavelength with emissivity close to a blackbody surrounded by wavelengths with much lower emissivity. The material has an available energy transition at the fluorescence wavelength, so it emits and absorbs well at that wavelength, but not at nearby wavelengths.

I meant that absorption distribution has nothing to do with emission distribution, except in terms of power equalization

OK, I misunderstood. I was talking about absorption and emission coefficients and you were talking about absorption and emission spectra.

Is there something special about the half-level ?

Not that I am aware of, except that it is a relatively big jump with no available intermediate transitions. The lack of intermediate levels is what makes the material have reduced absorption and emission at nearby wavelengths.

 

[hmmm27]

in comparison to a black body

Well... that's wrong, innit : it should be "compared to the total number of relaxations", no ?

The contraption description is now a forum-friendly

An enclosed system, consisting of a cold end with a thermal mass, and a hot end with a fluorescer, separated by a filter, bandpass at the fluorescer's absorption wavelength.

 

[Dale]

Well... that's wrong, innit : it should be "compared to the total number of relaxations", no ?

No, it is correct. Compared to a blackbody, which is a perfect absorber and the best thermal emitter possible.

The contraption description is now a forum-friendly "An enclosed system, consisting of a cold end with a thermal mass, and a hot end with a fluorescer, separated by a filter, bandpass at the fluorescer's absorption wavelength".

If the fluorescent relaxation state was more persistent, then upshifts could happen as well as downshifts.

No. Since the fluorescent material absorbs well at the absorption wavelength it also emits well at that wavelength. So the fluorescent material heats up the cold thermal mass.

 

[hmmm27]

No. Since the fluorescent material absorbs well at the absorption wavelength it also emits well at that wavelength. So the fluorescent material heats up the cold thermal mass.

Short answer: That's specifically in reference to the last paragraph, as applied to the first, right ? I edited that out seeing your response, after you quoted it, but before you edited your response to include it. Or something like that. The two paragraphs weren't actually related.

I deleted it, pending further rumination on energy states. What I jotted down looks like a very elegant answer, but I've no basis to put it forward 'cuz I know squat about how energy states develop as temperatures rise. Could you recommend a web resource on the subject, that uses English as much as Greek ? (ie: formulas thoroughly explained).

If the short answer doesn't apply, then the long answer is where I try to explain that if the fluorescence is the only absorber at a certain wavelength then (given the normal lack of upshifts in the mechanism), the emission coefficient of the absorption wavelength should be the inverse of the quantum yield of the fluorescence.

Again, thank you for taking the time to converse.

 

[Dale]

I edited that out seeing your response, after you quoted it, but before you edited your response to include it.

Sorry about that. For multi-part replies sometimes I add the parts by edit, but usually I try to do that soon enough that it doesn't cause exactly that problem.

given the normal lack of upshifts in the mechanism

I just realized that in your previous comment (the one I said "No, ..." to earlier today) you may have meant upshift in energy and I was thinking of upshift in temperature. So I am not sure what you mean by that.

Could you recommend a web resource on the subject,n

I am partial to Susskinds lectures. Here is the first in the series

 

[hmmm27]

So I am not sure what you mean by that.

This explanation isn't really based on anything except elegance, but should be mildly entertaining.

For normal fluorescing, let's call the energy states a,b,c for base, excited and relaxed.

Normal fluorescence goes a>b>c, then a dissipation back to a. Absorption is at the a,b gap (difference in energy levels), emission at the b,c gap.

If the c state were more persistent - didn't leak as much; stuck around awhile, whatever - then there could be b>c>a fluorescence where absorption is at the b,c gap and emission at the a,c gap. Which is a symmetric reversal of normal fluorescence.

If the number of emissions of each type fit into the bb spectrum, there would be no (net) biasing of the distribution spectrum.

So, if an increase in temperature causes the aforementioned to happen, then everybody's happy (except maybe the guy who built the contraption that just lost its ambient greenhouse-effect capability).

Elegant. Also convenient... too convenient.