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I have some questions about blackbody radiation. In the classical calculation, we assume every mode has the same energy, then every mode should have the same probability, then why the curve of intensity is going up instead of being constant. Also, in the Plank formula, why the peak is not at lowest frequency (lowest frequency corresponds to lower energy, then in a fixed temperature, lower energy will have more modes so the intensity will be larger?) I understand the derivation of plank's formula but I really don't grasp the physics idea behind it.

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  • $\begingroup$ We don't assume that all modes have the same energy. Does that solve your problem? $\endgroup$ – CuriousOne Jul 6 '16 at 5:38
  • $\begingroup$ @CuriousOne Can you explain a little bit more? $\endgroup$ – lol Jul 6 '16 at 5:40
  • $\begingroup$ I am more asking what you mean. A system that has only one mode can't give you a spectrum. The only thing it can do is oscillate at one frequency. $\endgroup$ – CuriousOne Jul 6 '16 at 5:55
  • $\begingroup$ @CuriousOne I am talking about blackbody radiation, so of course the "system" does not only have one mode. $\endgroup$ – lol Jul 6 '16 at 5:58
  • $\begingroup$ So what do you mean by "in the classical calculation we assume every mode has the same frequency"? $\endgroup$ – CuriousOne Jul 6 '16 at 6:02
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As you correctly mentioned, using a classical approach leads to the wrong result, in which the distribution goes asymptotically to infinity as the frequency grows, this is called the ultraviolet catastrophe. It is true that classically each mode will have the same energy,but most of the energy in a natural vibrator will be in the smaller wavelengths and higher frequencies, where most of the modes are. So it is the density of modes that makes the curve non-flat. According to classical electromagnetism, the number of electromagnetic modes in a 3-dimensional cavity, per unit frequency, is proportional to the square of the frequency.

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  • $\begingroup$ Thank you for your reply. I figured this out and you are correct. $\endgroup$ – lol Jul 6 '16 at 21:21

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