I read that thermal emission/radiation are the ones whose spectra is similar to that of a Black body radiation. Also thermal radiation depends solely on the temperature of the object. How can we consider Free-Free emission and spectral line emission as thermal emission processes? kindly explain


1 Answer 1


Thermal radiation $\neq$ blackbody radiation.

Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature.

That does not necessarily imply that the radiation field is in equilibrium with the matter at the same temperature. It is this latter equilibrium that is required in order for the radiation field to approximate to a blackbody.

So the following statements are true:

Blackbody radiation is thermal radiation. Thermal radiation is not necessarily blackbody radiation.

An example of thermal, but non-blackbody, radiation would be thermal bremsstrahlung. The particle velocities in the gas follow a Maxwellian distribution at a characteristic temperature and this determines the spectrum of the emitted free-free radiation. However, the bremsstrahlung radiation is able to escape from the gas without further interaction - in other words the gas is "optically thin". The result is a thermal bremsstrahlung spectrum that is not at all like a blackbody spectrum, especially at frequencies $<kT/h$.

On the other hand, the bremsstrahlung process is purely caused by the interaction of free electrons and ions. If you allow there to be atoms and partially ionised atoms in the gas, then it is feasible for those to be in equilibrium at the same temperature. The population of the various energy states in the atoms and ions is determined by this temperature, and transitions between these levels produces spectral emission lines in a completely predictable, temperature-dependent, way.

  • $\begingroup$ plus $-i^2$ for exemplary, concise yet complete explation! $\endgroup$
    – uhoh
    May 5, 2016 at 5:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .