I am seeking clarification of the following terms:

  • Black body radiation
  • Thermal radiation
  • Thermal light source

At the first glance, the first two are the same thing, and they are the radiation emitted by the third. However, once we define them more rigorously, they do not seem identical anymore:

  • Black body radiation is an equilibrium state of a photon gas, from where Planck's law readily follows.
  • Thermal radiation is the radiation emitted by thermal light sources.
  • Thermal light sources are light sources in a state of thermal equilibrium.

My point is that, although black body radiation is an excellent first approximation for the light emitted by thermal light sources, it is only an approximation. Indeed, the radiation emitted by an incandescent lamp, heated object or a star is not in equilibrium - otherwise, the source would be absorbing as much radiation as it emits, i.e., it would be not a light source but a black body. Furthermore, the term thermal light source arguably refers not to the properties of the radiation, but to the mechanism of the emission: the material of the source is in the state of thermal (quasi)equilibrium, but the source is not in equilibrium with the surrounding radiation. In fact, this is manifestly the case for most of the light sources that are conventionally referred to as thermal: a heated object cools down, incandescent lamp does not give light, if a current doesn't continuously heat it, and the stars keep radiating only because of the constant supply of energy due to nuclear reactions.

To summarize: there seems to be much ambiguity about what is called thermal in this context. I will appreciate clear definitions, preferably supported by references.

Remarks On the question of whether BBR is an equilibrium state and what is a black body: Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, emitted by a black body (an idealized opaque, non-reflective body).

An example of a light source in thermal (quasi)equilibrium, but not in thermal equilibrium with the radiation is a gas lamp, which emits radiation corresponding to a specific atomic transition.

A related question

  • 1
    $\begingroup$ Note that sources (like the Sun) can be in thermal equilibrium without absorbing as much radiation as they emit. But that requires an alternate source of power. $\endgroup$
    – ProfRob
    Jun 9 at 6:03
  • $\begingroup$ @ProfRob If energy is being transferred from a source to radiation, I would call it a steady state rather than equilibrium. $\endgroup$ Jun 9 at 6:20
  • $\begingroup$ Thermodynamic equilibrium just means that thermodynamic state variables (like temperature) aren't changing on relevant timescales. $\endgroup$
    – ProfRob
    Jun 9 at 6:26
  • 1
    $\begingroup$ @ProfRob This is a steady state, not necessarily an equilibrium one. I will think about formulating it as a separate question. $\endgroup$ Jun 9 at 8:41
  • $\begingroup$ @ProfRob Thermodynamic equilibrium is much more static than just thermal equilibrium. The latter requires just constant temperature distribution; the former requires that energy flux vanishes everywhere. $\endgroup$ Jun 9 at 13:10

Black body radiation

Originally, radiation coming out of a black body - an hypothetical body that absorbs all radiation that comes its way, reflects nothing, and radiates its own radiation. When this radiation is radiated into open space, while it has spectrum of thermodynamic equilibrium radiation, it is not itself radiation in state of thermodynamic equilibrium, because its intensity is highest at the blackbody surface and gets weaker with distance, so there is obvious source and obvious sink of the radiation. However, if such blackbody was enclosed in a cavity inside material medium (like inside Earth or ceramic oven), radiation in the space between the two bodies would come close to equilibrium radiation, with same density and spectrum everywhere.

Thermal radiation

Every body emits EM radiation due its charged constituents accelerating/changing state all the time. When the amount of radiation in some range of frequencies can be explained just by thermal motion of the particles forming the body, we say radiation in that spectral range is thermal, for example, red glow of steel when heated by a gas fire to a high temperature, or visible light of incandescent bulb. This is so when intensity of radiation is lower or equal to intensity of blackbody radiation at the same temperature. The body does not have to be in thermodynamic equilibrium for us to say its radiation in some spectral range is thermal radiation; it is enough that it has well defined temperature.

When the radiation is stronger than blackbody at same temperature would produce (light from a discharge tube such as neon lights or fluorescent bulb, or light from a light-emitting diode, or light of a fluorescent dye), we say this is not purely thermal radiation, although it may have some thermal component.

Thermal light source

This is a source of light (visible EM radiation) whose light spectral properties (in visible spectrum range) can be explained just by thermal emission, i.e. any body that radiates less or at most as much as black body at the same temperature.

  • $\begingroup$ Thanks. If black body radiation is not an equilibrium state, then it is only approximately described by Planck's law? $\endgroup$ Jun 9 at 5:09
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    $\begingroup$ Planck's law is accurate model for spectrum of blackbody radiation at the surface of the blackbody, but not for its directional properties or spectral properties far from the blackbody. $\endgroup$ Jun 9 at 11:52
  • $\begingroup$ True blackbody radiation is isotropic. That is: $J_\nu = B_\nu$. It is in complete thermodynamic equilibrium with the matter in its vicinity. $\endgroup$
    – ProfRob
    Jun 9 at 13:45
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    $\begingroup$ You cannot view true blackbody radiation from outside a blackbody, only rough approximations to it like the solar photosphere or a tungsten filament lamp. The cavity with a hole approximation only works if the hole is so small that it doesn't disturb the isotropy of the radiation in the cavity. We have to be inside the blackbody to get isotropic blackbody radiation - like the interior of the Sun or the cosmic microwave background. $\endgroup$
    – ProfRob
    Jun 9 at 14:06
  • 1
    $\begingroup$ @ProfRob en.wikipedia.org/wiki/No_true_Scotsman You are overgeneralizing radiation of a black body into radiation in thermodynamic equilibrium. $\endgroup$ Jun 9 at 14:20

Blackbody radiation $\neq$ thermal radiation.

Blackbody radiation is emitted by objects that are in complete thermodynamic equilibrium and is isotropic. The population of energy states of the matter constituents and the energy distribution of photons is stationary and described by a single temperature. The matter and radiation are in equilibrium.

Thermal radiation is characterised by local thermodynamic equilibrium. Local patches of matter can be characterised by a temperature and the radiation emitted by that matter, which would be called thermal radiation, could also be characterised by that temperature. However, the matter and radiation need not be in equilibrium. Thus whilst all blackbody radiation is thermal radiation, the opposite is not true.

We could use the Sun as an example. The interior radiation field of the Sun is extremely close to a blackbody. The mean free path of a photon is of order $10^{-3}$ m, which is tiny compared with the solar radius. Photons are absorbed and remitted within a region that is very close to isothermal so the radiation field is (almost) isotropic and characterised by the temperature of the matter which is emitting it. The structure is stable on timescales of billions of years, which is orders of magnitude longer than the cooling timescale, and thus complete thermodynamic equilibrium of the radiation and matter is an excellent approximation.

The corona of the Sun emits thermal radiation. The plasma has temperatures in excess of $10^6$ K and emits photons via free-free (bremsstrahlung), free-bound (recombination) and bound-bound processes in the UV and X-ray regions. Whilst this radiation is thermal, and is characterized by the (local) temperature of the plasma, the photons have very long free paths and would usually escape. This means the radiation field and matter at any point in space are not in equilibrium. The radiation spectrum has the lines and recombination edges generated by the emission processes and not the smooth Planck blackbody distribution. These spectral features are diagnostic of the plasma temperature - hence the radiation is thermal, but not blackbody.

The photosphere of the Sun is an interesting intermediate example. It is often approximated as a blackbody, but it is not one. The reason for this is that, by definition, the photosphere is where the mean free path of photons becomes long enough to allow them to escape. Whilst local thermodynamic equilibrium is usually a reasonable assumption, the radiation field is no longer strongly coupled to the matter and is anisotropic. In practice that means when we look at the Sun, we see to different depths depending on the wavelength and so see plasma emitting at a variety of temperatures and the spectrum is no longer given by the Planck function.

Thermal light sources will not be blackbodies in general. They will only be approximately blackbodies if they are in thermodynamic equilibrium at a single temperature and they are opaque to all radiation.

EDIT: There seems to be some dispute over whether blackbody radiation must be isotropic and in thermal equilibrium with the surrounding matter. It is, and I refer the interested reader to Chapter 1 of Radiative Processes in Astrophysics by Rybicki & Lightman.

  • $\begingroup$ local equilibrium is a really good way to describe it! Thank you also for discussing the Sun - one of the questions that bother me is how an equilibrium is established between the matter and the radiation, before the radiation escapes. $\endgroup$ Jun 9 at 6:08
  • $\begingroup$ @RogerVadim the radiation doesn't escape. The mfp is 1 mm. $\endgroup$
    – ProfRob
    Jun 9 at 6:23
  • $\begingroup$ > Blackbody radiation is emitted by objects that are in complete thermodynamic equilibrium. -- What you mean is correct, but this is an unfortunate formulation. Blackbody radiation is radiation of an hypothetical blackbody, a body with temperature that does not reflect any radiation coming its way but absorbs everything and emits its own radiation. This body does not have to be in any kind of equilibrium with radiation that it is absorbing. $\endgroup$ Jun 9 at 12:58
  • $\begingroup$ What you mean there is rather radiation in state of thermodynamic equilibrium, which is a different thing; we need some enclosed space or dense medium to allow for the radiation to come to thermodynamic equilibrium with matter inside. $\endgroup$ Jun 9 at 12:59
  • $\begingroup$ @JánLalinský actually it does. That is why the surface of the Sun is not a blackbody but the interior is. Complete Thermodynamic equilibrium is certainly a condition for blackbody radiation and the temperature of the radiation has to be the same as the temperature of the matter. $\endgroup$
    – ProfRob
    Jun 9 at 13:00

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