Non-Linear optics vs The 2nd Law of Thermodynamics

[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...

---

Let's start with definitions, simplified for clarity.

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.

----

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.

----

And conduct the experiment...

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

Place the ball within the sphere.

That's it.

----

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).)

----

Result

The ball gets colder.

----

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.

 

[Vanadium 50]

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".

 

[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.

 

[Drakkith]

Which breaks the 2nd Law of Thermodynamics.

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?

 

[hmmm27]

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?

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.

 

[Drakkith]

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.

 

[Vanadium 50]

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.

 

[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.

 

[Bystander]

The OP is asking why Kirchoff's other law, radiative equilibrium, is still obeyed at equilibrium; https://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation

 

[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 statement. 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.

 

[Dale]

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?

 

[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)

 

[Dale]

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.

 

[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)

 

[Bystander]

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

 

[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).

 

[Dale]

The mechanism doesn't rely on a quantum yield of 1.0

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

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)

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

 

[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)

 

[Bystander]

if it works at all

No.

 

[Dale]

Okay, let me ramble a bit...

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.

 

[Andrew Mason]

The concept of a one-way mirror is causing me difficulty. In order for the proposed system to work, outer body must allow some photons from the blackbody to pass through ( i.e. photons at the "fluorescent" frequency) but not absorb any incoming radiation at all, including radiation at that fluorescent frequency. I expect that such a mirror is not possible.

This reminds me somewhat of the Feynman lecture on entropy using the ratchet and pawl attached to a vane. The idea was to have molecules tighten the spring by having only molecules striking the vane in one direction move the vane. He showed that it does not work.

AM

 

[hmmm27]

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"..

So, for the given example - unexcited except by ambient thermal radiation - 5% of the emission at that frequency will be from fluorescence ? (I realize that for those specific parameters, emission and absorption is negligible)

The concept of a one-way mirror is causing me difficulty. In order for the proposed system to work, outer body must allow some photons from the blackbody to pass through ( i.e. photons at the "fluorescent" frequency) but not absorb any incoming radiation at all, including radiation at that fluorescent frequency. I expect that such a mirror is not possible.

It's not a "one way mirror" at the component level - more like a greenhouse effect using non-linear optics. In the original post, the inside of the sphere gets colder (though, if you change the bandpass to the absorption frequency of the fluorescer, it gets hotter inside, which is more greenhouse-y).

Dale simplified the admittedly confusing original contraption into...

Cold End <> Fl. Absorption-Frequency Pass Filter <> Fluorescer <> Fl. Emission-Frequency Pass Filter <> Hot End.

...which makes for a much clearer net flow, linearly from left to right - rather than being folded back on itself as the original post.

We can see that AF (flourescent absorption frequency) photons pass trom the cold end to the fluorescer which changes them into EF (fluorescent emission frequency) photons which can't go back, and pass into the hot end.

There is backwash - thermally produced photons from the middle travel back through the filter to the cold end; likewise there's travel from the hot end to the middle, but the flows don't equalize until the hot end is hot and the cold end is cold.

 

[Dale]

the flows don't equalize until the hot end is hot and the cold end is cold.

No, I don't think this is correct. The thermal flux is so large that it keeps them even.

Do you have any references that support your idea, because my references indicate that it is not large enough even for the most highly fluorescent material I could find.

 

[hmmm27]

Well, it seems to work at a pop-sci level of understanding, so I thought I'd try to see how far it could get before a(n unknown to me, as a newb) bit kicked in and said "no" ... without last principles (ie: a "Law") being used as a substitute for reasoning.

No, I don't think this is correct. The thermal flux is so large that it keeps them even.

I'm not unhappy if "hot" and "cold" means a temperature difference of 1x10^-27K or something, or a similarly infinitesimal energy potential between the two ends. (and non sequitur if quantum variations mean that sometimes the cold end might be hotter, just that the centerpoint has moved)

Is that what you mean ?

Do you have any references that support your idea, because my references indicate that it is not large enough even for the most highly fluorescent material I could find.

If thermal and fluorescent emissions aren't additive (ie: whichever one would be the highest wins : 2 + 1 = 2), then on the short wavelength side of the peak there isn't enough fluorescent absorption for the emission to top already existing thermal emission. On the long wavelength side... ?

But, I think I'm starting to get it...
So... a fluorescing mechanism that operates by moving from an arbitrary energy level 3 to level 2, is prevented from operating because the level is at 11 thanks to thermal energy, and will never dip down far enough for the mechanism to operate: all absorbed photons go straight to thermal usage.

 

[Dale]

without last principles (ie: a "Law") being used as a substitute for reasoning.

Huh? Physical laws and experiment are the sole basis for reasoning in science. I don't understand either "last principles" or "substitute" in this context.

I'm not unhappy if "hot" and "cold" means a temperature difference of 1x10-27K or something, or a similarly infinitesimal energy potential between the two ends. (and non sequitur if quantum variations mean that sometimes the cold end might be hotter, just that the centerpoint has moved)

At those levels there is always some departure from classical thermodynamics. You would have to use the fluctuation theorem from statistical mechanics ( https://en.m.wikipedia.org/wiki/Fluctuation_theorem ).

Note, there is nothing special about a fluorescent material in this regard, just that level is so small that a non classical treatment is required for any material.

 

[hmmm27]

No, I don't think this is correct. The thermal flux is so large that it keeps them even.

I don't actually understand what you mean by that; is what I tried to say in the previous post (fluorescence is impossible within the thermal envelope because the energy level of the molecule will never ever get low enough for the mechanism to happen) relevant ? or more like "So what ? the bias is only going to be 0.00000000000001%" . . . or something else.

The idea involves the cold side having less energy than the hot side, thus "cold" and "hot". The (popular understanding of the) mechanism of fluorescence, combined with appropriate filters supports the idea.

Do you have any references that support your idea, because my references indicate that it is not large enough even for the most highly fluorescent material I could find.

What idea ? the original post ? no, of course not.

Huh? Physical laws and experiment are the sole basis for reasoning in science. I don't understand either "last principles" or "substitute" in this context.

Trying to be funny, or wry at least ("last principles" is etymological, not scientific in nature : if you've finally got around to carving a principle for something in stone as a law, then it's the "last principle"... <waits for polite snicker>...) : I'm still worried somebody's going to pop up and say "you can't because the second law of thermodynamics says you can't" and that will be it. A premise is not a reason.

Note, there is nothing special about a fluorescent material in this regard, just that level is so small that a non classical treatment is required for any material.

Which was probably regarding the 10 to the minus 27 figure which I pulled right out of the nearest orifice, not an actual figure. But also not relevant.

Sorry if I'm being confusing or seem petulant - it's not deliberate.

aside - is this subforum where this post belongs ?

 

[Dale]

Sorry if I'm being confusing or seem petulant - it's not deliberate

It would really help if you would focus. It seems like you spend more effort on being "wry" than on either trying to understand the reasoning presented to you or to explain where you don't understand it. It is hard to help if you don't tell me where I am going off-track in my explanation.

I don't actually understand what you mean by that; is what I tried to say in the previous post (fluorescence is impossible within the thermal envelope because the energy level of the molecule will never ever get low enough for the mechanism to happen) relevant ? or more like "So what ? the bias is only going to be 0.00000000000001%" . . . or something else.

The idea involves the cold side having less energy than the hot side, thus "cold" and "hot". The (popular understanding of the) mechanism of fluorescence, combined with appropriate filters supports the idea.

So I found a good hyperphysics page that might help. It is not specifically on fluorescence, but it made me realize that fluorescence is not a fundamentally different phenomenon from normal thermal absorption and re-radiation.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/absrad.html

So a fluorescent material takes energy radiated at a higher wavelength, absorbs it, excites some internal states, and later re-radiates it at a lower wavelength. The same thing happens when a blackbody absorbs visible light and then radiates it in the infrared, just not as obviously. The difference is that a fluorescent material is a "graybody" rather than a blackbody, meaning that at some frequencies the emissivity coefficient is less than one. At the fluorescence wavelength the material is a good emitter, and hence also a good absorber. That makes it come to a thermal equilibrium by exchange at that frequency, but doesn't shift the equilibrium.

 

[Bystander]

Kirchoff's other law

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

No.

Is anybody home?

 

[hmmm27]

It would really help if you would focus. It seems like you spend more effort on being "wry" than on either trying to understand the reasoning presented to you or to explain where you don't understand it. It is hard to help if you don't tell me where I am going off-track in my explanation.

In general terms, the bits leading up to the last sentence of a paragraph you write makes sense, then the conclusion doesn't. Or I ask for a confirmation of what my interpretation of what you've said is correct, and... I get nothing. Like...

So, for the given example - unexcited except by ambient thermal radiation - 5% of the emission at that frequency will be from fluorescence ? (I realize that for those specific parameters, emission and absorption is negligible)

to which you replied

Nothing.

or, to be more exact...

hmmm27 said: ↑

the flows don't equalize until the hot end is hot and the cold end is cold.​

​

No, I don't think this is correct. The thermal flux is so large that it keeps them even.

How does the "thermal flux" being "large" have anything to do with it ? Also, since it's the entire post, please explain how you can pretend not to understand it.

Do you have any references that support your idea, because my references indicate that it is not large enough even for the most highly fluorescent material I could find.

To which I asked "what idea ?" and you declined to reply.

etc.

So I found a good hyperphysics page that might help.

I think I'm familiar with what it says... probably.

So a fluorescent material takes energy radiated at a higher wavelength, absorbs it, excites some internal states, and later re-radiates it at a lower wavelength

check, it's the basis for the thread.

The same thing happens when a blackbody absorbs visible light and then radiates it in the infrared, just not as obviously.

check, again, noting that bb emissions have nothing to do with bb absorptions except in terms of overall power equality.

The difference is that a fluorescent material is a "graybody" rather than a blackbody, meaning that at some frequencies the emissivity coefficient is less than one.

yes, and by that standard of measurement, at some frequencies it could be greater than one. Thus "fluorescence". check. (or I don't understand the concept of a "gray body", fair enough)

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

I don't see why that would hold true as a basic premise, though agreed at non-fluorescent-specific emission frequencies. Nor do I see the relevance, either way.

That makes it come to a thermal equilibrium by exchange at that frequency, but doesn't shift the equilibrium.

And I'm completely back to "What ?". What is "it", and what is it in "thermal equilibrium" with ? Since you rejigged my original post to one that makes the flow easier to follow, you don't mean the contraption which is the subject.

At this point, any problem I may be having seems to be an understanding of how the emission of thermal energy precludes fluorescent emission, if indeed that's the case. To which end I've typed the same possible answer twice, with no reply from you.
I do realize that, this being the Internet and all, you're under no financial or moral obligation to be anything but a jerk.

I do apologize if it's me being one.

 

[Bystander]

emission of thermal energy precludes fluorescent emission.

Emission is EMISSION.

 

[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.