So I was up last night in bed reading. The material isn’t germane to the question, suffice it to say it concerned black bodies and wasn’t Entertainment Weekly.
I understand the reason that black bodies are called that: Since they absorb all wavelengths of radiation equally, they are thus capable of emitting all wavelengths and thus generate a smooth, emission-line free spectrum. But do black bodies really exist. I was under the impression that light from a hot body (no jokes, now) was created by excited electrons transitioning to lower energy states. Wouldn’t the spectrum from such a process generate a spectrum of emission lines, albeit very densely packed from the variety of states possible?
The question also brings in cavity radiation. Isn’t the initial radation being bouced around in the cavity from the electrons changing energy states? Or is there some other process involved in radiation generated by heat that I’m not aware of, or have perhaps forgotten about.
No need to dumb down (much) your replies as I have a pretty good math background. I just feel like I’m missing something, here.
No, an ideal black body is one which reflects no radiation whatsoever. Of course, such an entity does not exist in the real world, but it is a useful concept, nevertheless.
True, just as the pressure of a gas in a balloon is due to a large number of densely packed billiard-ball-like collisions imparting momentum to the wall. The point of statistical mechanics (the field of physics that the blackbody specturm is derived from) is that you can ignore the microscopic properties of the system and make arguments based on the ensemble of allowed states, without needing to know how the system can get between them. (Of course, if they can’t, then you have to be careful.)
My solid state physics is a bit rusty, but I think that in a metal, for example, there actually exists a continuum of bound states for the electrons. In general, light is caused by the acceleration of charged particles; moreover, light is essentially a special kind of electromagnetic field, so charged particles get accelerated by light. It’s all very circular.
If you’d like a more precise treatment of how statistical mechanics works, I highly recommend the standard text in the field. “Fundamentals of Statistical and Thermal Physics”, by F. Reif. It’s pretty damn good, although it has the disadvantage that it was written in 1965 and reads as such. If you’ve got a bit of physics background (a course or two in college, ideally), you shouldn’t have much trouble. Blackbody radiation is discussed in Chapter 9.
I was under the impression, however misguided, that a black body only had to be capable of absorbing radiation of any wavelength. Whether it does or not in a particular instance is not essential for its blackness, if you will.
Belatedly, thank you John Mace and Gaudere for the help. Always nice to see you.
Isn’t this where classical statistical mechanics broke down when trying to explain black body radiation? I’m thinking of the runaway UV effect, here.
I guess I’m really trying to get at the source of a pure continuum of radiated wavelengths. We’re talking about thermally generated radiation. Is there some other mechanism at work other than electron state transitions, or is it some aspect of QM operating? The latter I could believe, if that was in fact the source of the correct explanation for cavity and black body radiation.
Yes. It’s collisions (interactions) between the matter and the radiation.
Consider a star. Forget photospheric absorption for a minute; the star is a good blackbody. But its energy comes from the proton-proton fusion chain. This reaction doesn’t produce visible light; it produces MeV gamma-rays at a few characteristic energies. So why don’t we see an emission spectrum with (highly Doppler broadened) lines at these frequencies? Because in a stellar interior, the mean free path of a photon is on the order of 1cm, and so in order to escape the star, a photon would undergo on the order of a jillion interactions which change its energy. This process is called thermalization of the radiation field, because it will tend to bring the radiation to a blackbody spectrum at the temperature of the matter it interacts with.
There are four processes which are involved in thermalization: electron scattering, free-free, bound-free, and bound-bound. Because the vast majority of the matter in a star is ionized, electron scattering and free-free dominate; there are very few bound atoms to interact with. The process you mention, electron state transitions (bound-bound), is the only one that occurs at energies defined by the atom’s bound energy levels. Photoionization (bound-free) has a minimum energy, but an electron can be ionized with extra energy. Electron scattering and free-free are only constrained by the electrons’ Fermi-Dirac statistics.
I don’t think any of the replies get at the answer the OP is looking for.
Blackbodies don’t radiate because of electrons changing states, they radiate because the atoms and hence the electrons near their surfaces are vibrating (and not colliding - they are already always in “contact” in the same sense atoms can be said to “collide”).
For the purpose of thermal radiation a body is a blackbody if all its little charge centers are vibrating with equal ability at all frequencies and so coupling their motion into electromagnetic radiation. By “equal ability” I mean each frequency that could be present given the motivating temperature is, in fact, present in full swing.
I hope this is accurate, even within the context of quantum mechanics, which would just add the idea that the vibrations (phonons) and the electromagnetic radiation are packaged or quantized. The packages have a broad continuum of allowable sizes and do not break the blackbody emission spectrum into discrete bins.
So is the “thermal jiggling” that Napier describes an example of bound-bound radiation emission? I’m thinking of my fireplace late at night when there can be a big pile of coals surrounding a gap between the grate and bed. With a really good fire the light is just about white in color. Am I getting decent cavity radiation to read by?
Spectral lines are never infinitely sharp.
The fundamental tenets of quantum mechanics account for the minimum width (natural broadening) of a spectral feature, while various physical processes further broaden the line profile:
•Natural broadening [symbol]D[/symbol]E = h/2[symbol]pD[/symbol]t. Where [symbol]D[/symbol]t is the lifetime of the excited state.
•Thermal Doppler broadening. When we say that a gas is at a certain temperature T we mean that the gas particles move about with random velocities characterized by (avg)mv[sup]2[/sup]/2 = 3kT/2. The resultant doppler shift [symbol]Dl/l[/symbol] is ~v/c (neutral hydrogen at T = 6000K moves at a mean speed of ~12 km/sec so 12/ 300,000 = 4e-5; hence the thermal doppler width of the Balmer [symbol]a[/symbol]line (6356 angstrom) is approximately 1/4 angstrom (6356X 4e-5).
•Collisional broadening.
The energy levels of an atom are perturbed by neighboring particles, especially charged particles like ions and electrons (the Stark effect). In a gas these effect are random, and they result in a broadening of spectral lines. [-My text is too wimpy to have the equation here]
•The Zeeman effect
When an atom is placed in a magnetic filed the atomic enery levels each separate into three or more sublevels -this is the Zeeman effect. Where before we had a single atomic transition and a single spectral feature, we now have three or more closely spaced lines. [Again, the text is too wimpy too give equations, and a good thing too or I’ve flunked the course.]
•Turbulence broadening.
Resulting from the macroscopic flow of gases. This adds onto the thermal doppler effects already mentioned.
(excerpted from Smith and Jacobs; Introductory Astronomy and Astrophysics, 1973)
When you add all these effects together in a gas, much less a solid, you end up getting the smooth curve called continuum radiation.