# Why the cavity radiation has the form of standing waves?

I'm reading about the cavity radiation in the context of blackbody theory. I'm asking myself: WHY do we describe this radiation by the use of standing waves? Why can't they be not-standing, maybe reflecting along some strange paths inside the cavity itself?

• Why the downvote? This seems to me to be a reasonable question. Oct 18, 2013 at 15:48
• For completeness have a look at this article hyperphysics.phy-astr.gsu.edu/hbase/mod6.html Nov 17, 2013 at 18:56

The situation you describe can be expressed as a linear combination of all the possible waves inside the cavity that are zero at the boundary (perhaps an idealization at very high frequencies). Now, an arbitrary wave bouncing around is then a sum of standing waves each one with a different phase factor and amplitude (Fourier series). In black body theory, we are concerned with all possible waves, and it is just convenient to do the calculation in a basis of standing waves.

More about boundary conditions being necessary for thermal equilbrium:

We only require that the walls have nonzero conductivity because walls of zero conductivity would be transparent; they would generate no currents in response to the electric fields of incoming radiation and the radiation would pass through unaffected. For walls having nonzero conductivity, the transverse electric field strength at the walls is $E=0$ because modes with $E\neq 0$ at the walls are lossy. Only those standing waves with $E=0$ at the walls will persist after some time $t \gg a/c$.

In other words, this idea of "absorbing and remitting" is, in the classical model, is how reflection occurs. Finally, remember that it isn't the inside walls of the cavity that we consider a perfect absorber (black body), but the opening itself (that is much smaller than the size of the cavity). The opening is the black body. The cavity is one way to model a black body. Remember that the walls won't radiate isotropically (another requirement), but the opening will.

• Your answer is well written and sufficiently detailed. But I'm wondering what you mean when you say that the opening is the black body and not the cavity. I'd like to be given a precise answer since this whole topic is obsessing me at times. May 3, 2017 at 21:01
• If the area of the opening is very small compared to the cavity, then the chance of an incident photon reflecting off of an inside wall and coming right out again is very small, which means the photon's energy stays inside the cavity long enough to come to some sort of equilibrium with the walls. In that way the reflectivity of the hole is zero, even if that of the walls isn't. Which is what makes it a perfect absorber. May 3, 2017 at 22:38

Because, as in a guitar string, the wave has to be zero at the boundaries. At the boundaries, the radiation is absorbed and re-emitted. This process creates the standing waves inside the cavity.

• I'm not sure that's a compelling argument. The requirement for a node at the boundary applies if the wave is reflected, but in a BB cavity the wave is not reflected. It is absorbed and a wave with an unrelated phase may be emitted some unrelated time later. Oct 18, 2013 at 15:50
• Well I left a comment here about cavity radiation, as described by Max Planck; and somehow, within the hour it magically disappeared. Being of sound mind, I presume that if I reposted it, the result would be exactly the same. The censorship rules here are quite baffling; and anonymous too.
– user26165
Oct 18, 2013 at 21:31