Source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html

We arrive at this result:

$$N(\nu)=\frac{4}{3}\frac{\pi L^3\nu^3}{c^3}.$$

So now we have a function representing the number of modes for a frequency $\nu$, that means I can plug any frequency $\nu$ and know the number of modes $N(\nu)$ corresponding to that frequency, is that right?

But in the proof, they differentiate $N(\nu)$ with respect to $\nu$


Why did we have to differentiate? We already got the function $$N(\nu)=\frac{4}{3}\frac{\pi L^3\nu^3}{c^3}$$ that can give us a clue of how intense the radiation is for a frequency $\nu$, can someone explain why taking the derivative is important?

  • $\begingroup$ I could swear there was a question yesterday on the same topic. $\endgroup$
    – DanielC
    Mar 1, 2019 at 18:09
  • $\begingroup$ Yes, I think it was me, but there's a difference between both questions. $\endgroup$
    – khaled014z
    Mar 1, 2019 at 18:56

2 Answers 2


$N(\nu)$ is the total number of modes for all frequencies, and the derivative is the number of modes within $\nu\pm \frac 1 2 d\nu$.

It is very similar to saying the number voxels (if you will) in a sphere of size $r$ goes like:

$$N(r) = \alpha \frac 4 3 \pi r^3$$

while the number of voxels in a shell of radius $r$ and thickness $dr$ is:

$$ n(r) = \frac{dN}{dr} = \alpha(4\pi r^2) $$

($\alpha$ is just some constant).

  • $\begingroup$ I understand this, but what I'm saying is, we already have the function that tells us how many modes there are for every frequency (this essentially tells us that intensity increases with increasing frequency if $N(\nu)$ increases) , then what is a derivative good for, unless we're looking for a relation between some parameter that I think is called 'spectral intensity' vs the frequency, is that right? $\endgroup$
    – khaled014z
    Mar 1, 2019 at 18:55

You want to sum the mean energy density of each mode over all modes in order to find the total energy density in the cavity: $$u =\int \langle u\rangle dN.$$ If the mean energy depends on the frequency you may switch the integration variable to $\nu$, and then the derivative appears: $$u = \int \langle u\rangle (dN/d\nu) d\nu.$$ The failure of classical physics is giving us a $\nu$-independent mean energy per mode, which clearly makes the integral divergent. Planck's formula introduces a cutoff for high frequencies, guaranteeing the integral to be finite.


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