I was taught today that the Electromagnetic wave Theory is unable to explain black body radiation. The example that was given to me: When a metal is heated, it emits different frequencies of light as it gets hotter. If electromagnetic wave theory was correct, it would not be so, the frequency (color) of light would remain the same, but only the intensity will change. I don't understand why this is so.

My logic: Electromagnetic waves occur when a charged body oscillates in a electric and magnetic field. If the metal is provided with more energy (in the form of heat) won't the charged body vibrate faster, thus changing the frequency of the light emitted?

• It's true that classical wave theory can't explain black body radiation because it predicts an ultraviolet catastrophe. You need to bring in quantum theory to get the correct curve. Is that what you mean? Jul 3 '14 at 14:47
• @JohnRennie I am fine with the fact that it doesn't explain it. I'll take a look at the link you have provided me. However. that is not my question. My question is why should a heated metal, according to the electromagnetic wave theory, emit only a single frequency of light regardless of how hot it is. Jul 3 '14 at 14:56
• I don't know any argument for why a heated rod should only emit a single frequency of light. If you can provide a reference to that argument I'll have a look. Jul 3 '14 at 15:00
• That seems to be the problem, it is what my teacher taught me, and I there was no reference to any source. I will try to look for one in the textbook though. Give me a few seconds. Jul 3 '14 at 15:01
• This question appears to be off-topic because it is appears to be based on a misunderstanding Jul 3 '14 at 15:28

Suppose you have an object which is a perfect absorber/emitter of electromagnetic radiation, a.k.a. a "black body". Suppose we try to compute the electromagnetic power radiated by this object using statistical mechanics and classical electromagnetic theory. We would find that the object emits electromagnetic radiation at all wave lengths, and the wave length dependence of the emitted power would go like

$$P_{\text{classical}} (\lambda)\,d\lambda \propto \lambda^n$$

where $\lambda$ is wave length and $n<0$. The problem here is that integrating at $\lambda \rightarrow 0$ diverges and you get infinite power, which is obviously wrong.

If you redo the calculation assuming that you still have waves, but that each mode of the electromagnetic field can only have discrete quantities of energy in it, you get a $P(\lambda)$ which contains finite power, and more importantly, is reproduced in experiment! This "quantum" theory still has waves, but the energy in each wave comes in discrete chunks.