Difference in thermal radiation between condensed matter and gases

Question

In Quantum physics of atoms, molecules, solids, nuclei, and particles by Robert M. Eisberg and Robert Resnick, the authors write:

Matter in a condensed state (i.e., solid or liquid) emits a continuous spectrum of radiation.

What happens in a gas?

Answer 1

If the gas is atomic or molecular, then there is very little continuum opacity, since absorption of a photon would usually take place at the relatively distinct wavelengths associated with atomic and molecular transitions. Rayleigh scattering has a continuous cross-section, but is an elastic process.

Since there is a directly proportional relationship between the effectiveness of emission and absorption at each wavelength, then in such cases the gas is likely to be transparent, or optically thin at most wavelengths. Almost all its emission will be in the form of discrete lines, with perhaps a continuum of scattered light from whatever illuminates it. A blackbody, as a minimum requirement, must absorb all radiation incident upon it, so such gases cannot emit as blackbodies.

If there was really a lot of such gas, then things like line broadening mechanisms come into play. There is a tiny amount of opacity at a broad range of wavelengths, even for discrete transitions, because of doppler broadening, pressure broadening, the Stark effect and others. This small opacity, combined perhaps with the increased optical path length caused by scattering, could make the gas optically thick at a broad range of wavelengths. In which case it might approximate a blackbody. I can't think of any examples.

If you heat and ionise/partially ionise the gas, then the options for continuum opacity are considerably increased. Free-free absorption, inverse bremsstrahlung, Compton scattering, photoelectric absorption, pair production are examples of mechanisms that produce a continuum opacity. It requires far less quantity of such gases (or plasmas if you insist) to become optically thick and approximate blackbodies. Examples include the visible surface of stars, or even better, the interiors of stars - which we can't see at any wavelength, because all wavelengths are absorbed!

The basic rule of thumb is - if radiation can pass through an object then it will not emit blackbody radiation

EDIT: Perhaps the best example of the differing properties of ionised and atomic gases is in the early universe. When the universe was hotter than $\sim 3000$ K, there were sufficient free electrons and free-free opacity to keep the universe opaque at all wavelengths - leading to the emission of blackbody radiation. The universe cooled, the electrons and ions combined to form atoms, and the universe became transparent to radiation. That is why the cosmic microwave background looks like blackbody radiation at $\sim 3000$ K, redshifted into the microwave region by the expansion of the universe.

Answer 2

From Basics of Radio Astronomy, published by NASA's Jet Propulsion Laboratory in April 1998, Chapter 3 - The Mechanisms of Electromagnetic Emissions (emphasis added):

In solids, the molecules and atoms are vibrating continuously. In a gas, the molecules are really zooming around, continuously bumping into each other. Whatever the amount of molecular motion occurring in matter, the speed is related to the temperature. The hotter the material, the faster its molecules are vibrating or moving.

Electromagnetic radiation is produced whenever electric charges accelerate—that is, when they change either the speed or direction of their movement. In a hot object, the molecules are continuously vibrating (if a solid) or bumping into each other (if a liquid or gas), sending each other off in different directions and at different speeds. Each of these collisions produces electromagnetic radiation at frequencies all across the electromagnetic spectrum. However, the amount of radiation emitted at each frequency (or frequency band) depends on the temperature of the material producing the radiation.

[...]

Any matter that is heated above absolute zero generates electromagnetic energy. The intensity of the emission and the distribution of frequencies on the electromagnetic spectrum depend upon the temperature of the emitting matter. [...]

The author provides a table on page 3 of the PDF (page 21 in the book):

Thus, according to the JPL, gases do indeed emit radiation in all wavelengths (according to their temperature), just like liquids and solids do.

Answer 3

In a solid or liquid the electrons in the valence and conduction bands are roughly uniformly distributed so we have a continuous distribution of electrons throughout the whole material. By contrast atoms or molecules in a gas are widely separated by vacuum (where there are no electrons).

The reason this matters is that the radiation mentioned in the quote, black body radiation, is mainly produced by oscillations in the electron density. Thermal motion makes the atoms oscillate and this produces changes in the electron density. This in turn produces oscillating electric dipoles, and those dipoles emit the black body radiation.

In a gas at everyday temperatures and pressures this can't happen because the gas atoms or molecules are too widely spaced. There is no continuous distribution of electrons to oscillate, and as a result gases do not emit black body radiation. The radiation emitted by gases generally consists of sharp lines related to rotational or vibrational transitions.

If you compress a gas to very high density then it can emit black body radiation much as a liquid does. However this normally requires extreme pressures not found outside physics labs or astronomical bodies.