In the blackbody radiation experiment, the inside of the body is metal so its considered a conductor inside, shouldn't the electric field be zero, so we would be left with only magnetic field instead of E.M

  • $\begingroup$ What about a mirror or a microwave? Just because its metal doesnt mean it somehow kills electromagnetic radiation (i.e. light or heat). What should be zero is the tangent component of an electrostatic field (as the charges can move freely) $\endgroup$
    – Bort
    Dec 18 '15 at 15:22
  • $\begingroup$ Just as Bort said, the tangent component should be zero and that means that you have standing waves inside your metal box. I recommend you check out Eisberg "Quantum mechanics", he have a partially detailed answer in this topic, however it's a little long and you require some waves and E&M, thermodynamics solid Background. $\endgroup$
    – Keith
    Dec 19 '15 at 14:25
  • 1
    $\begingroup$ Please define "the blackbody radiation experiment". We cannot read your mind :) $\endgroup$
    – DanielSank
    Jan 15 '16 at 6:19

For the electromagnetic waves, the electric field parallel to the wall must go to zero if the wall is a perfect conductor. An Electromagnetic wave , which is a self propagating transverse waves of oscillating electric and magnetic fields. In case of a metal wall,

the metal has electrons free to move through the entire solid. This is why metals can conduct electricity. It is also why they are shiny. These unattached electrons oscillate together with large amplitude in response to the electrical field of an incoming light wave. They themselves then radiate electromagnetically, just like a current in an antenna. This radiation from the oscillating electrons is the reflected light. In this situation, little of the incoming radiant energy is absorbed, it is just reradiated, that is, reflected. We in this case, get standing waves.

Ref: http://galileo.phys.virginia.edu/classes/252.old/black_body_radiation.html


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