# Thermal radiation and emission spectra

by klawlor419
 P: 111 I have some questions regarding thermal radiation spectra of different objects. Is thermal radiation of objects due to electromagnetic radiation alone? Does my desk which is at room temperature emit as a blackbody with the proper factor for emissivity? Where is the "cavity" that the radiation is trapped inside for my desk? Is it the intermolecular spacing of the atoms in the desk? I have been thinking about this and getting confused or missing concepts. I think the theory has kind of clouded my understanding of the way that objects radiate, or just shown how little intuition I have of the way in which things thermally emit. Also, this is probably a really dumb question but why is it that we consider a blackbody to a perfect absorber of electromagnetic radiation? Why should black be any different than for example red? It is technically a color in the "visible" spectrum Thanks ahead of time for any help with my confusion
 P: 20 The exact thermal spectrum of an object is determined by molecular properties, like which functional groups it has. In the infrared spectrum specifically, many molecules have a distinct emission spectrum. Absorption and emission are essentially the same. "Absorptivity" means the fraction of the radiation falling on an object is absorbed vs. reflected. "Emissivity" means the fraction of radiation that is emitted vs. that which goes to heating up the object. A black body has an absorptivity and emissivity of 1. Radiation will fall on to some theoretical black body at all wavelengths, but will be emitted as a smooth spectrum (poisson distribution probably) centered on some frequency much lower than the light falling on it, which is probably white light with equal intensities at all frequencies. The "cavity" is just a thought experiment to help you understand what a black body is- it's just so that the radiation and the temperature of the black body can reach equilibrium before the radiation is re-emitted. Your desk might appear red because it absorbs all frequencies of light but is not capable of emitting very much blue or green light at its normal temperature. If you heated your desk to a few thousand degrees, it would probably emit light in blue and green because more of the light coming out of it is coming out as a black body spectrum (not including fire of course, it would be on fire too by the way). Black bodies appear black at normal ambient temperatures but are really putting out radiation at lower frequencies (infrared) and would become bright if you heated them up.
P: 111
 Quote by colliflour The exact thermal spectrum of an object is determined by molecular properties, like which functional groups it has. In the infrared spectrum specifically, many molecules have a distinct emission spectrum. Absorption and emission are essentially the same. "Absorptivity" means the fraction of the radiation falling on an object is absorbed vs. reflected. "Emissivity" means the fraction of radiation that is emitted vs. that which goes to heating up the object. A black body has an absorptivity and emissivity of 1. Radiation will fall on to some theoretical black body at all wavelengths, but will be emitted as a smooth spectrum (poisson distribution probably) centered on some frequency much lower than the light falling on it, which is probably white light with equal intensities at all frequencies. The "cavity" is just a thought experiment to help you understand what a black body is- it's just so that the radiation and the temperature of the black body can reach equilibrium before the radiation is re-emitted. Your desk might appear red because it absorbs all frequencies of light but is not capable of emitting very much blue or green light at its normal temperature. If you heated your desk to a few thousand degrees, it would probably emit light in blue and green because more of the light coming out of it is coming out as a black body spectrum (not including fire of course, it would be on fire too by the way). Black bodies appear black at normal ambient temperatures but are really putting out radiation at lower frequencies (infrared) and would become bright if you heated them up.
So are you saying that an objects emissions spectra are the collective effect of the way in which its microscopic constituents emit as well as the macroscopic emissivity or absorptivity of the material. Is that correct? Do individual molecules have emissivity or absorptivity? Or is the emissivity and absorptivity a property that is seen only in macroscopic matter, i.e. chunks of metal or my desk?

Lets say that we have two spherical pieces of metal that are exactly same size and composition. One is coated in red pigment and the other is coated in black pigment. Lets say that they are heated by a torch and brought up to a very high temperature well below their melting point. What are the differences in the way these would emit? Is the emissivity a function of frequency for pigments like this? Would there be a certain range of frequencies missing in the $$u_w=\text{spectral radiancy function}$$?

I can see that effecting the way in which the two spheres emit because it would decrease the total energy emitted by the red pigmented object. Still though it seems really non-intuitive to me that those spheres would emit differently. Maybe they don't? Thanks for the help

P: 20

## Thermal radiation and emission spectra

The emission spectra is measured as a collective effect, but each molecule has a distinct spectra and has its own microscopic sort of absorptivity and emissivity. If you take a sample of this molecule and turn it into a gas and measure the spectrum, you might actually get different values than if it was looked at when it was a liquid or gas. Odd scattering effects can happen with thin layers of a substance, like oil on water, or with a microscopic texture on a substance that makes it reflective vs a flat looking color. These are all optics phenomena and happen because the temperature is low compared to the black body temperature required to produce visible light, the object's absorption and emission not being in equilibrium. I hadn't really thought about this before, but the microscopic geometry can affect emissivity and absorptivity, apparently pyramidal divots make an object have increased emissivity...

As to your thought experiment comparing spheres heated by a torch, i see what you're getting at, but a better example would be two different filament wires in an incandescent light bulb. Once the metals are heated to glow in the visible spectrum, yes, each type of metal will have a distinct and different emission spectrum and have different overall emissivity. Now suppose you took two of the same incandescent bulb with an ideal black body for a filament. The light coming from the filaments and going to the surface of the bulbs would be a smooth spectrum. If the coatings were thick enough, the red coating would be reflecting light from the room, not emitting visible light from the filament inside. The black coated bulb would be reflecting no visible light, but both would be radiating infrared because of the radiation absorbed from the filaments inside. Assuming you could put enough wattage into both bulbs, the red and black bulb would both glow red hot (or white hot) depending on how hot they got exactly. The peaks and troughs in the emission spectra, the function you talked about, would be missing some frequencies, but to the unaided human eye, they would look like all the frequencies were coming through. You'd need to look through a spectroscope to see dark bands in the frequency spectrum. Once you looked through the spectroscope, you'd see that the red and black pigments do in fact have different spectra. Which means, of course, emissivity as a function of frequency. To my knowledge, emissivity is always a function of frequency, but when someone says they have determined an emissivity for an object in general, it's an average over the part of the electromagnetic spectrum that's of interest. You would certainly have a different emissivity and absorptivity for radio waves vs visible light.

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