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I read that it makes a difference whether you calculate $\frac{d \omega}{d \lambda}=0$ or $\frac{d \omega}{d \nu}=0$ in the sense that the maximum energy densit with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\nu_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation). Does anybody know how to explain this odd thing?

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Related: physics.stackexchange.com/q/13611/2451 and links therein. –  Qmechanic Sep 21 '14 at 12:39

2 Answers 2

Let's revisit what the term "spectral energy density" means.It means the amount of energy emitted in a infinitesimal $d \lambda$ or $d \nu$.

Now due the the relationship $\lambda=\frac{c}{\nu}$,we can find that $d \lambda=-\frac{c}{\nu^2}d \nu$.

The $\nu^2$ in the denominator leads to the given phenomenon.We can think of problem as finding the box($d \lambda$) which has the maximum height in the enter image description herePlanck curve.

Obviously the orange box is the required maximum.However if we were to represent the given graph in terms of the frequency,the size of boxes($d \lambda$ or $d \nu$) would change due to the $\nu^2$ in the denominator.As a result the box which corresponded to the maximum height in the wavelength curve may not correspond to the max. height in the frequency curve.This is exactly what happens and the graphs peak at two different points.

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There are many different choices for the variable $x$ describing the kind of plane harmonic EM wave (its frequency), and each may lead to different function $I_x(x)$ that is to express the original spectral distribution of the studied radiation. If new variable $x' = T(x)$ is introduced, this means that generally the equality

$$ I_x(x) = I_{x'}(x')~~~(1) $$

need not be satisfied. If it was, the maximum of $I_x$ would have $x$ corresponding to $x'$ that maximizes $I_{x'}$; but it often is not.

This is because the transformation to other spectral functions is done rather by requiring that $$ I_x(x) dx= I_{x'}(x')dx', $$ together with using prescription for the value of $$ \frac{dx'}{dx}, $$ which follows from the relation $T(x)$ defining $x'$ based on $x$.

Since these two functions $I_x, I_{x'}$ do not need to satisfy (1), their maxima may correspond to different kinds of EM wave, and then obviously the wave whose wavelength maximizes $I_{\lambda}(\lambda)$ need not be the same wave whose frequency maximizes $I_f(f)$.

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