Black body radiation, question arising from a book

[fluidistic]

In the book "Introduction to the structure of matter" by Brehm and Mullin, page 78. They say "It should be empathized that there is no reason for the peak positions $$\mu _m$$ and $$\lambda _m$$ in the respective distributions to be connected by the relation $$c=\mu \lambda$$. They are talking about the blackbody frequency and wavelength spectra. On a graph the independent variable would be either $$\mu$$ or $$\lambda$$ and the dependent variable, $$M_ \mu (T)$$ or $$M _\lambda (T)$$.

I do not understand at all how, for a given temperature, a body would emit a peak of frequency $$\mu _m$$ that does not correspond to a peak of wavelength given by the formula $$\lambda _m =\frac{c}{\mu _m}$$.
Can you explain to me what's going on?

 

[Born2bwire]

I know I am going to explain this poorly but...

It has to do with the fact that we look at the energy over a bandwidth of frequency or wavelength. These peaks are the peaks in the spectral energy densities, which I will call the function \rho. By itself, \rho is meaningless because it is a density. If you wish to find the actual energy you need to integrate it across the frequency or wavelength. Thus, the actual infinitesimal energy is
$$\rho(\nu)d\nu$$
$$\rho(\lambda)d\lambda$$
This is kind of like the same thing that happens when we talk about wavefunctions in that the magnitude is the probability density but the meaningful probability is |\Psi(x)|^2dx.

Ok, so the peak energies should be the same, thus,

$$\rho(\nu)d\nu = \rho(\lambda)d\lambda$$
$$\rho(\nu)\left| \frac{d\nu}{d\lambda} \right| = \rho(\lambda)$$
$$\rho(\nu) \frac{c}{\lambda^2} = \rho(\lambda)$$

Now using this relation you can properly account for the difference in the peak energy densities.

 

[fluidistic]

Sorry for being so late on this, but thank you.
However can you explain me why $\big | \frac{d \nu }{d \lambda} \big |=c/\lambda ^2$?
Because there's a problem in Brehm's book that asks me to calculate the product $\nu _m \lambda _m$ and to realize that it's not worth c. Though I must "consider the derivations of the dependence of $\nu _m$ and $\lambda _m$ on the temperature".

 

[Born2bwire]

Because the relationship between frequency and wavelength is $c = \lambda \nu$. Thus,
$$\frac{d\nu}{d\lambda} = \frac{d}{d\lambda} \frac{c}{\lambda} = -\frac{c}{\lambda^2}$$

 

[fluidistic]

Ah ok thank you.
So does this mean that the peak in the frequency $\nu$ is shifted by a factor $\frac{c}{\lambda ^2 }$ compared to the peak in the wavelength $\lambda$?