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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.

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

Note that this is neither true nor the real issue. The electromagnetic wave theory in classical physics predicts power at all wave lengths (all frequencies). The problem is that there's too much power at low wave lengths (high frequencies). This is not due to the wave nature.

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.

To recap, the thing that makes the classical electromagnetic theory fail is that it assumes that each mode of the electromagnetic field can have any level of energy in it. This leads to an infinite radiation power for a black body. In quantum theory, each mode's energy comes in discrete (not continuous) values, and this leads to a correct prediction for the wavelength-dependent radiated power.

The actual form of the radiated power predicted in quantum theory is the Planck law.

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### Gummy bears

Updated on July 04, 2020

• Gummy bears over 3 years

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?

• John Rennie over 9 years
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?
• Gummy bears over 9 years
@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.
• John Rennie over 9 years
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.
• Gummy bears over 9 years
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.
• Gummy bears over 9 years
@JohnRennie Here is the link to the PDF file: ncert.nic.in/NCERTS/textbook/textbook.htm?kech1=2-7 The concept I am talking about is on the end of page 36 into the beginning of page 37. Please take a look.
• John Rennie over 9 years
OK I've read section 2.3.2 pages 36-37, but nowhere does it say classical wave theory claims a heated body would emit the same wavelength regardless of temperature.
• Luboš Motl over 9 years
Dear @Gummybears, you are misreading. The text - and its pages 36-37 - are a flawless introduction to the Planck radiation. None of the sentences over there say that the black body radiation only occurs at a single frequency. Still, for each temperature, some range of frequencies - around a maximum-intensity frequency - dominates. All these things are calculable. I am afraid that every answer to your question will say exactly what your good textbook does, and you will misunderstand just like you misunderstand the textbook.
• Gummy bears over 9 years
@LubošMotl The fact is that my question is not from this textbook. My question is from what my teacher has told me. Upon reading the section that I said, I don't find the concept my teacher insisted upon in there. So my question remains, is it true that according to the wave theory, only the intensity of light emitted by a metal change as the temperature increases?
• Gummy bears over 9 years
@JohnRennie Yes, not reading it, I too am unable to find that concept. I think I should ignore what my teacher has told me.
• Gummy bears over 9 years
But then the question arises, why is the wave theory unable to explain the change of intensity? And how does the particle nature of light explain it?
• Luboš Motl over 9 years
Dear @Gummybears, the classical theory of waves or electromagnetism is predicting what the field is doing at temperature $T$, but the prediction is wrong. It is self-evidently wrong because it predicts that the total energy stored in the field at a finite temperature is infinite. It's called the "ultraviolet catastrophe". Quantum mechanics modifies the prediction at frequencies $f$ such that $hf\sim kT$ or $hf\gt kT$, adds an exponential decrease of the energy with frequency, and the resulting total energy in the field at temperature $T$ is then finite and dominated by a certain range of $f$.
• Luboš Motl over 9 years
The classical wave theory makes wrong predictions because it's a wrong theory. The prediction for the total thermal energy is infinite because the classical theory of waves infinitely overestimates the amount of "degrees of freedom" - dynamical variables that oscillate and that can store energy. Quantum mechanics modifies these predictions so that the results for measurable quantities become finite. The frequency at which the radiation is maximized is $f\sim kT/h$ but it's just an order-of-magnitude estimate - actually all frequencies are present in the black body radiation.
• Gummy bears over 9 years
• Luboš Motl over 9 years
The correct, quantum explanation is effectively a theory of particles because it implies that the energy carried by the field mode at frequency $f$ is always an integer multiple of the photon energy $E=hf$, i.e. $Nhf$. Because the energy is an integer multiple, the integer itself may be interpreted as the number of particles - photons - in the state.
• John Rennie over 9 years
This question appears to be off-topic because it is appears to be based on a misunderstanding
• DanielSank over 9 years
@Gummybears: No problem. Thanks for the positive response; it encourages good answers.