I was reading this post and I think I am still confused. A blackbody $radiator$ is not in equilibrium, by definition, since it is radiating. If so, why does the blackbody spectrum accurately describe the radiation, since, after all, it is derived assuming the body is in equilibrium with the electromagnetic field?

And if I'm wrong, and the "blackbody spectrum" does not accurately describe the radiation, then what is the correct spectrum from a blackbody radiating in vacuum?

Edit: Just as a guess, I would think that the equilibrium assumption should be correct if one can show that the rate of energy exchange between the system and the EM field (perhaps within the volume of the system) is much larger than the rate at which the radiation escapes. The rates probably depend on the volume and surface area of the system, and so the equilibrium assumption will be correct in the limit that the volume of the system is large. But I'm not sure what are the exact quantities that need to be compared.

Edit: You probably need to use the Thomson cross section $\sigma \sim \frac{\alpha^2}{m^2}$ and multiply it by the energy exchange $T^2/m$ and by the number density $n$ of charged particles and their velocity $\sqrt{T/m}$ and by the number of photons $T^3 V$ in the volume $V$, and then compare this to the energy radiated from the Stefan Boltzmann law $T^4 A$ where $A$ is the surface area of the system. But unfortunately I find that you need a system with a size of order $10^{12}$ meters.

  • $\begingroup$ A related question: How does radiation become black-body radiation? $\endgroup$
    – Roger V.
    Commented Apr 4, 2022 at 13:35
  • $\begingroup$ Yes, it's actually the same question, sorry I didn't see it. Should I close my question? Or move my second edit to there as a comment or an answer? $\endgroup$ Commented Apr 4, 2022 at 14:01
  • $\begingroup$ I don't think it is exactly the same - there are more nuances here, so I wrote an answer... also I am not quite happy with its clarity, perhaps I will review it later. $\endgroup$
    – Roger V.
    Commented Apr 4, 2022 at 14:02
  • $\begingroup$ Tip: Let's not have posts look like revision histories. $\endgroup$
    – Qmechanic
    Commented Apr 5, 2022 at 12:24

1 Answer 1


Black body and black body radiation
The black body is defined as an object that absorbs all the radiation incident on it - i.e., none of this radiation is reflected. This however does not mean that it does not emit any radiation - it does emit and this radiation has black body spectrum. The emitted and the absorbed energy are then equal, so that object itself remains in thermal equilibrium.

So defined black body is actually a historical artefact from pre-quantum times, which led to deriving the Planck's law. In fact, a more robust definition of black body radiation is that of a radiation in thermal equilibrium (i.e., photon gas in thermal equilibrium - aka, Boltzmann distribution). Deriving the Planck's law then becomes a simple exercise... provided that one is at ease with quantum mechanics and the occupation number representation.

What kind of radiation is actually emitted by bodies
Black body spectrum used to describe objects, such as stars, human body or fire in an oven is actually an approximation. The approximation assumes that the body is (a) in thermal equilibrium, (b) that it can emit at any wave-lengths, and (c) that emitted energy is too small to significantly change the thermal state of the body. Indeed, if the body were not in thermal equilibrium, it could emit at some frequencies more than at others - e.g., a neon lamp. The same neon lamp can serve as an example of an object that emits only certain wave lengths. Finally, if the body loses a lot of energy via radiation, it would cool down and its radiation spectrum would shift in time towards lower frequencies. However, the approximation works admirably well in many cases, whereas deviations from this approximation allow, e.g., to determine the chemical composition of stars.

Equilibrium state vs. steady state
Actually, that object receives the same amount of energy as it emits is not sufficient for it to be in thermal equilibrium - i.e., its state may not be describable by Boltzmann statistics, and it may not be in equilibrium with its environment.

E.g., the Earth is heated by the Sun, which can be considered a black body at temperature of about 6000K. The Earth then emits some heat to vacuum, which can be considered as an environment at temperature 0K. The Earth is neither in thermal equilibrium with the Sun, nor with the vacuum, but the relaxation processes on Earth can be considered rather fast, and its state can be considered as an equilibrium state at some temperature intermediate between 0K and 6000K (which is gradually rising). So Earth is in a quasiequilibrium steady state.

The same reasoning can be applied to a human body, see, e.g., here and here.

See also:
Does fire emit black-body radiation?
How does radiation become black-body radiation?
Black body vs. Thermal radiation
How do photons reach thermal equilibrium with the walls of the blackbody cavity?

  • $\begingroup$ Why do you say a Neon lamp is not in thermal equilibrium? Is it getting hotter or cooler? The reason a Neon lamp emits mostly at discrete freaquencies is because it is optically thin. It would not absorb all radiation incident upon it, and that is why it isn't a blackbody. $\endgroup$
    – ProfRob
    Commented Apr 5, 2022 at 16:22
  • 1
    $\begingroup$ There's electric current flowing, and there's a directed flow of energy from electrons to atoms to radiation to environment outside. It is a steady state, but not thermodynamic equilibrium. $\endgroup$
    – Roger V.
    Commented Apr 5, 2022 at 18:45

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