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?
 A: 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.
A: 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.
