# Black body radiation, question arising from a book

1. Sep 14, 2010

### 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?

2. Sep 15, 2010

### 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.

3. Nov 30, 2011

### 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".

4. Nov 30, 2011

### Born2bwire

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

5. Dec 1, 2011

### 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 $\lambda$?