# Why does the blackbody radiate even at thermal equilibrium?

Everytime, I read about blackbody, I always get confused at the point where it is written

Under thermal equilibrium conditions , the blackbody radiation depends only on temperature. ..... . At equilibrium , electromagnetic waves bounce around with the walls of the cavity.

• If the incoming irradiation balances the outgoing, there is equilibrium – Steeven Nov 24 '14 at 20:07
• The cavity is an idealized model where only photons exist in equilibrium exchanging energy with walls of fixed temperature, . It is an analogue model for the observed black body radiation from matter. As @garyp says there is not equilibrium with the environment for the black body radiation that all matter emits at a given temperature. en.wikipedia.org/wiki/Black-body_radiation – anna v Nov 24 '14 at 20:16
• @Steeven: Sir, is the rate of radiation equal to rate of absorption or the amount of radiation equal to the amount of absorption?? Which is correct??? – user36790 Nov 25 '14 at 2:28
• Equilibrium means that nothing changes overall. As much heat that comes in every second must also leave every second. Else, there would be a heat "build up" so to say, and the temperature would change, which is not equilibrium. So you are talking about rates (that is amount per second), not amounts. – Steeven Nov 25 '14 at 8:21

The equilibrium mentioned in that quotation is between the radiation field and the walls of the container. The walls of the container are imagined to be held at some fixed temperature by some (unspecified) means. Under those conditions, the spectrum of the radiation in the cavity is that of a blackbody at the temperature of the walls. The radiation field and the walls exchange energy, but since they are in equilibrium the temperature of both remains the same. The walls radiate and absorb equally.

So far I haven't mentioned anything about radiation that leaves the cavity. In the ideal case of a closed cavity, the discussion is about the walls of the cavity and the field within. One could poke a small hole in the cavity, allowing some of the radiation to escape. In the limit that the hole is infinitesimally small, the disturbance caused by the hole is negligible, and the radiation that leaks out has very nearly a blackbody spectrum. This is a blackbody radiator. Usually the temperature of the blackbody is much higher than the surroundings into which the radiation radiates. Thus the blackbody radiator is not in thermal equilibrium with it's surroundings.

• +1. The rate at which the body radiates is equal to the rate at which it absorbs. The net flow is zero. It is in thermal equilibrium with the radiation field and not with the surroundings. But what would happen if there were no equilibrium? Why is it so important?? – user36790 Nov 25 '14 at 2:38
• You mean no equilibrium between the walls and the radiation? Then energy will be transferred between the walls and the field, while at the same time energy is redistributed within the walls and within the radiation until 1.) The walls are in a state of equilibrium (and thus has a well defined temperature) 2.) The radiation field is also in a state of equilibrium, and 3.) the walls are in equilibrium with the radiation. – garyp Nov 25 '14 at 16:17
• @MAFIA36790 I agree with everything garyp has said above. Just to add a bit: if the radiation field inside were not in equilibrium, and the external radiation would not be that of a blackbody (ie, thermal), instead would be whatever the (nonthermal) spectrum is of the current radiation field. – anon01 Oct 24 '16 at 15:11

Any physical body has many degrees of freedom, not only mechanical, but also the field degrees of freedom. The energy is distributed amongst these degrees of freedom, so the radiation (as field excitations) is always present in a body. The energy exchange is always on and in the equilibrium conditions what is "radiated" is "absorbed" with the same rate. It is a dynamical equilibrium, like vaporization and condensation in a closed volume containing a liquid. On average there is no energy "flow" in a certain direction, but it is due to equality of energy fluxes in the opposite directions.

• +1 . That means the rate at which it radiates is the same rate at which it absorbs. This is the dynamic equilibrium. But,sir, what about thermal equilibrium?? – user36790 Nov 25 '14 at 2:24
• Thermal equilibrium means time-independence of the average energy contained in each degree of freedom. In this condition the notion of temperature $T$ is well defined. That is why the black body radiation determined with $T$ solely. – Vladimir Kalitvianski Nov 25 '14 at 8:55

If a body has a positive temperature then it will emit radiation (over some very low limit perhaps).

Imagine a hollow sphere at a uniform temperature $T$. radiation will be emitted from the walls inside the sphere and absorbed by the same walls.

Consider a small part of this wall it loses energy by emission of radiation and gains energy by absorption of radiation from other parts of the wall. - Overall there is 'equilibrium' because all parts of the wall are at the same temperature and emit the same range of wavelengths with similar intensities.

What is this equilibrium all about?

as described above it is about exchange of energy between bodies of the same temperature.

Why is it important here??

Because if different parts of the system had different temperatures then they would emit different intensities/wavelengths of light - this would transfer energy from the hotter to the cooler body and mean that the spectrum of light was not 'blackbody radiation'

If there is thermal equilibrium, why does the body radiate??

All bodies will radiate photon unless they are really cold and very close to or at absolute zero temperature.

• +1 . Equilibrium means all the walls of the container are at same temperature. But when a part of the wall radiate energy, will not the temperature decrease there?? – user36790 Nov 24 '14 at 19:52
• yes and it absorbs radiation which will increase the temperature - the net effect is zero or .... equilibrium – tom Nov 24 '14 at 20:02
• So the rate of radiation is equal to the rate of absorption, right?? – user36790 Nov 25 '14 at 2:41
• @user36790 yes that is correct... – tom Nov 25 '14 at 9:40