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In old quantum theory max Planck derived the energy density of black body radiation , for that purpose he calculated the no. Of modes in the cavity and those are actually standing waves ,different standing waves are different modes enter image description here

And if length is L and wavelength is λ then L=nλ/2

And from intuition I can say these no. Of modes will be finite and so in integer number (If I say that number of modes is 5 , means there could be 5 different possible of modes or orientation of standing wave but ,There is no meaning of no. Of modes being 2.5 or 1.3 like these fractions ,they are only in inegers)

But Planck calculated no. Of modes is 8πν^2/c^3

enter image description here Now here I think" 8πν^2/c^3" is not an integer (as π is not integer) Then how can no. Of modes be fraction??

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    $\begingroup$ You are ignoring the "per unit frequency" and "per unit volume" aspects. You have a continuum of possible frequencies, and there is no necessity that the number of modes per frequency interval be integral. $\endgroup$ – Raziman T V Dec 29 '16 at 9:33
  • $\begingroup$ Then what is the definition of mode? What is the meaning of being the no. Of mode fraction like 2.5 ? What the heck does it even mean? $\endgroup$ – user101134 Dec 29 '16 at 15:30
  • $\begingroup$ Consider your example of waves on a string. The modes correspond to $\lambda_n = 2L/n$. Suppose you now ask how many modes exist between $\lambda$ and $\lambda + d\lambda$. The answer is clearly $2L/\lambda - 2L/(\lambda + d\lambda) \approx 2L d\lambda / \lambda^2$ and thus the number of modes per unit wavelength is $2L / \lambda^2$. Now let $L= \pi m$ and $\lambda = 1 \,\rm{micron}$, don't you have the same irrationality problem? $\endgroup$ – Raziman T V Dec 29 '16 at 16:03
  • $\begingroup$ Yes here is the problem here also no. Of modes is fraction ,what's the matter ?how can no. Of mode be fraction? $\endgroup$ – user101134 Dec 29 '16 at 17:15
  • $\begingroup$ Because we are not counting modes but number of modes per unit wavelength interval. This happens because you are converting a discrete problem into a continuous space. The number of modes per allowed wavelength is still 1, but the density of allowed modes per unit length varies. $\endgroup$ – Raziman T V Dec 29 '16 at 17:33

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