Regarding the classical Rayleigh-Jeans Law for blackbody radiation, why is the transverse component of the E field parallel to the walls of an assumed cubical cavity taken as zero? For non-zero conductivity σ of the walls,they are opaque to the incident radiation at least partly.However that does not mean an infinite attenuation such that E becomes zero instantaneously.The only way to reconcile this is to assume infinite conductivity for the walls.That means then that they are perfect reflectors. But for a blackbody produced in this particular fashion(heated body with a hole to the outside) it was my understanding that the walls should be at least partly absorbing(i.e. finite conductivity) to the incident radiation so that it is eventually absorbed by the walls to fulfill the blackbody's prime condition of perfect absorptivity. Where am i going wrong?Any help is appreciated. Edit:Later in the proof I also realise that the average energy of the standing wave can not be kT unless the walls are partly absorbing as the standing waves need to be in equilibrium.Although I think this bit of info is immaterial because it was eventually proved wrong.