Getting the emissivity from scratch

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fluidistic
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I have checked around 10 Solid State Physics and Condensed Matter textbooks, including the classic "Ashcroft and Mermin" and I've noticed that the "emissivity" of a solid is a totally neglected subject.
This leaves me entirely knowledgless about how to compute the emissivity from scratch. I do not even know whether it is a purely electronic phenomenon for metals, and/or an atomic/ionic phenomenon for semiconductors and insulators.
I have found the paper https://www.researchgate.net/profil...d_Disorder/links/561d25b508aecade1acb3409.pdf which suggests that the emissivity can be found by calculating the dielectric constant, which makes entirely sense to me because it is related to the EM absorption of the material. But then it is said that such a quantity can be calculated using an ab initio approach. I do not know whether this is valid for any material, specially non-metallic ones. What about materials without free electrons? How would one approach the problem?
 

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DrClaude
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This leaves me entirely knowledgless about how to compute the emissivity from scratch.
Could this even be done? Emissivity is a surface property, so it will depend a lot on how the surface is prepared. For instance, a metallic surface will have a different absorptivity depending on how much it is polished (or if there is an oxide layer at the surface).

https://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html
 
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fluidistic
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Could this even be done? Emissivity is a surface property, so it will depend a lot on how the surface is prepared. For instance, a metallic surface will have a different absorptivity depending on how much it is polished (or if there is an oxide layer at the surface).

https://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html
Sure it can be done. You can consider a perfectly flat surface for a simpler model, then add a flat oxidized slab of a certain thickness to complicate the model and eventually simulate surface roughness if you want. My goal is not to compute the most realisitc case. It's to know where to even start, with the most simplistic model.
 
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f95toli
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I believe the bulk dielectric constant or some materials can -at least in principle- be calculated using time-dependent DFT. That said, I'm not sure if this is also true for a surface.

Note that the dielectric properties of real materials (as opposed to idealized crystals) are very, very complicated and frequently dominated by defects and impurities. Hence, whereas I am sure it is possible to in principle calculate some of these properties; I wouldn't be too surprised if the calculated values were way off those obtained from experiments.
Example very clean sapphire (Al2O3, a material I am very familiar with) has a bulk loss tangent of 10^-6. The theoretical value is about 10^-11 (at T close to 0 k); i.e. even in the best crystals you will find that the the dielectric properties are completely dominated by impurities and defects.
 
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fluidistic
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I believe the bulk dielectric constant or some materials can -at least in principle- be calculated using time-dependent DFT. That said, I'm not sure if this is also true for a surface.

Note that the dielectric properties of real materials (as opposed to idealized crystals) are very, very complicated and frequently dominated by defects and impurities. Hence, whereas I am sure it is possible to in principle calculate some of these properties; I wouldn't be too surprised if the calculated values were way off those obtained from experiments.
Example very clean sapphire (Al2O3, a material I am very familiar with) has a bulk loss tangent of 10^-6. The theoretical value is about 10^-11 (at T close to 0 k); i.e. even in the best crystals you will find that the the dielectric properties are completely dominated by impurities and defects.
I see, thanks for the information. Usually for metals the effects of impurities near room temperature and above are virtually invisible. I do not think the emissivity would behave differently for this particular case. Also, I have noticed that the emissivity of most metals increases with temperature, somewhat linearly at least up to 1500 K. I am sure there is a simple explanation involving the electrons/phonons/scattering, etc. I wish someone with the knowledge would explain that behavior.
 
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Dr_Nate
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A material's interaction with light can be characterized as a function of frequency by its reflectivity ##(R)##, transmissivity ##(T)##, and it absorptivity ##(A): T + R + A = 1##. When the object is in thermal equilibrium with its surroundings Kirchoff's law of thermal radiation (which is a special case of conservation of energy) tells us that absorptivity equals emissivity. (https://www.optotherm.com/emiss-physics.htm)

If you want to calculate the emissivity for a real material you'll need accurate band structure calculations. However, if say you were happy to know the emissivity of an opaque material from measured reflectivity data then the emissivity ##(E)## is: ##E = 1 - R##.

Using the Drude-Lorentz model, you could also invent a series of oscillator strengths, resonance frequencies, scattering frequencies to give you a dielectric function. From the dielectric function it rather straight forward to turn that into ##R##, ##T##, or ##A/E##.
 
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Dr_Nate
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Also, I have noticed that the emissivity of most metals increases with temperature, somewhat linearly at least up to 1500 K. I am sure there is a simple explanation involving the electrons/phonons/scattering, etc. I wish someone with the knowledge would explain that behavior.
You are probably looking at engineering values for emissivity, which often give a single value for the emissivity. It actually changes as a function of frequency.

As temperature goes up, more phonons will be created thus there will be more electron-phonon scattering. This will broaden the thing we call the plasma edge (which only occurs in metals). Due to the high density of electrons at the Fermi surface, the plasma edge is usually in the ultraviolet. Below the plasma edge the metal is usually highly reflective. Above the plasma edge the metal has a low reflectivity. In the visible with increased scattering the reflectivity will go down and because ##\varepsilon(f) = 1 - R(f)## the emissivity will go up.

I doubt the change in emissivity will be exactly linear with temperature because it will depend on many details.
 

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