From what I understood, a Blackbody is a perfect absorber, and hence also a perfect emitter since it is in thermal equilibrium. One example of a blackbody is a box with interior walls painted in black. What I don't understand is why the wave must have zero amplitude at the walls of the box. If it is nodes at the walls, then it never gets absorbed, right? But shouldn't a black body absorb all radiation?


2 Answers 2


The box with interior walls painted in black is not the blackbody. The realization of the surface of a black body is the surface of a small hole (small with respect to the size of the box) on the wall of such a box.

As clearly stated in wikipedia page:

Any light entering the hole is reflected or absorbed at the internal surfaces of the body and is unlikely to re-emerge, making the hole a nearly perfect absorber. When the radiation confined in such an enclosure is in thermal equilibrium, the radiation emitted from the hole will be as great as from any body at that equilibrium temperature.

When modeling the set of electromagnetic (em) waves in the cavity, there are a couple of points which allow significant simplifications:

  1. provided the cavity is large enough, details of the em radiation inside become less important. In a more technical words, the density of states at thermodynamic limit becomes independent on the shape of the cavity and on the boundary conditions. Therefore, one is enabled to chose the simplest case, without loss of generality.
  2. When interested in equilibrium properties, the precise mechanism at the basis of thermal equilibrium becomes irrelevant, thus one has not to take into account explicitly the real absorption/emission by the atoms on the wall. The final effect of thermal equilibrium (whatever is the underlying mechanism) is taken into account by the equilibrium probability function for the em modes. Interestingly, a similar approach is used when dealing with the equilibrium properties of the perfect gas. Strictly speaking the perfect gas does not have a mechanism for thermalization (no collision between molecules). What is implicitly done in the usual statistical mechanics treatment is to take for granted that an unspecified mechanism exists allowing the system to reach thermal equilibrium, even if it is not explicitly present in the Hamiltonian. It is enough to make the hypothesis that the mechanism exists and is effective in establishing thermal equilibrium. At that point it can be switched off.
  • $\begingroup$ yes, thank you. But why does the wave have to have nodes at the walls of the box? $\endgroup$ Jul 13, 2019 at 17:40
  • $\begingroup$ @KoutaDagnino This is something which has to do with the set up of a workable model for the radiation inside the box. The simplest model is a set of stationary waves in the box. Which require the presence of odes at the walls. $\endgroup$ Jul 13, 2019 at 17:49
  • $\begingroup$ Thank yous o much. So when deriving the Planck equation the simplest model is considering stationary waves inside a box, if I understood correctly. If so, then they have zero amplitude at the walls, right? How does the wave get absorbed then? $\endgroup$ Jul 13, 2019 at 18:13
  • $\begingroup$ @KoutaDagnino I'll add something on this issue to my answer to clarify this point. $\endgroup$ Jul 14, 2019 at 8:20
  • $\begingroup$ Thank you, clarified a lot. $\endgroup$ Jul 14, 2019 at 14:00

For thermal radiation inside a cavity the waves do not need to have, and usually do not have, nodes at the walls of the box.

There is a widely-used picture in simple presentations of the theory which suggests the waves have nodes at the walls. In fact this picture is just offering a way to count how many linearly independent modes of the field there can be. Standing-wave boundary conditions is one way to do it; periodic boundary conditions is another way. In the latter method the waves do not have nodes at the walls. But in either case what one gets from this part of the analysis is the density of states. The actual waves in the box do not need to have nodes at the walls, but you can imagine forming them by first using the waves you have counted (by whichever argument) and then adjusting the phases to move the nodes along. When one adjusts the phase of the waves like this, the number of states, and therefore the density of states, does not change.


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