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hmmm27
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heheh, ok. typos'r'us.
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.hmmm27 said:yes, and by that standard of measurement, at some frequencies it could be greater than one.
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.hmmm27 said:noting that bb emissions have nothing to do with bb absorptions
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.hmmm27 said:I don't see why that would hold true as a basic premise, though agreed at non-fluorescent-specific emission frequencies
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.hmmm27 said:How does the "thermal flux" being "large" have anything to do with it ?
Bystander said:Boltzmann distribution?
Dale said: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.
Dale said:The maximum peak of an emitter can only bring it up to blackbody, not beyond.
How can I, or why can't I, cut/paste nested quotes ?Dale said:They have everything to do with each other.noting that bb emissions have nothing to do with bb absorptions
At the fluorescence wavelength the material is a good emitter, and hence also a good absorber.
Dale said: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.
Yes, your description sounds good to me.hmmm27 said:I understand the concept of AC
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.hmmm27 said:EC seems a bit more complex : a measurement of photons emitted... in comparison to what ?
At thermal equilibrium and without work, I think that is right.hmmm27 said:So the probability of an impinging photon directly resulting in an emitting photon is EC2
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.hmmm27 said:Why does the combination of a pre-existing spectral line and thermal radiation not cause a bump on top of the bb curve.
OK, I misunderstood. I was talking about absorption and emission coefficients and you were talking about absorption and emission spectra.hmmm27 said:I meant that absorption distribution has nothing to do with emission distribution, except in terms of power equalization
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 said:Is there something special about the half-level ?
in comparison to a black body
No, it is correct. Compared to a blackbody, which is a perfect absorber and the best thermal emitter possible.hmmm27 said:Well... that's wrong, innit : it should be "compared to the total number of relaxations", no ?
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 said: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.
Dale said: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.
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.hmmm27 said:I edited that out seeing your response, after you quoted it, but before you edited your response to include it.
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.hmmm27 said:given the normal lack of upshifts in the mechanism
I am partial to Susskinds lectures. Here is the first in the serieshmmm27 said:Could you recommend a web resource on the subject,n
Dale said:So I am not sure what you mean by that.