Looking at the radiation from the sun (T=5800K) I got a little surprise which I do not understand.

I first calculated the energy density u and also the number of photons per unit volume ng. From this one gets the average energy per photon:

$$E_g = \frac{u}{n_g}$$

Converted to a wavelength one gets:

$$\tag{1} \lambda = 920\:\mathrm{nm}$$

The average frequency using

$$\nu_{\text{av}} = \frac{\int \nu \cdot u_{\nu} \cdot \mathrm{d}\nu}{\int u_{\nu} \cdot \mathrm{d}\nu}$$

Converted to a wave length I get $\lambda_{\text{av}} = 650\:\mathrm{nm}$.

Next I calculate the average wavelength using $$\lambda_{\text{av}} = \frac{\int \lambda\cdot u_\lambda \cdot \mathrm{d}\lambda }{ \int u_\lambda \cdot \mathrm{d}\lambda}$$ One gets $\lambda_{\text{av}2} = 920\:\mathrm{nm}$

Questions:

I was expecting some value which would be of course different from the value of the frequency distribution. This is actually the case. The same is of course true for the max of the distributions.

However to my surprise this is also equal to the the value of eq. (1). Why?

What is the physical meaning of the average frequency $\nu_{\text{av}}$?

If there is no particular meaning of $\nu_{\text{av}}$ it looks like the wavelength distribution has more physical meaning which I think is nonesense.

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I $\TeX$ed your formulas (please learn to do that yourself), but I couldn't make much sense of your calculations. Check them back, perhaps I misunderstood a few things... – leftaroundabout Oct 10 '13 at 23:48
Frank, what @leftaroundabout has done is write the math in the input language of the MathJax rendering engine which is a pretty good implementation of $\LaTeX$'s mathmode and which we have running on the site. You should examine the raw text to understand what he did (just click the edit button). – dmckee Oct 10 '13 at 23:49
You are working your statistics out from the Plank law, right? Off the top of my head, I should expect the mean energy to be $h$ times the mean frequency, since the expectation is a linear operator. – WetSavannaAnimal aka Rod Vance Oct 10 '13 at 23:57
What formula are you using to calculate the number of photons per unit volume? Don't confuse this with the density of states. – WetSavannaAnimal aka Rod Vance Oct 11 '13 at 1:43
Please revise your first formula. It does not make any sense. – mcodesmart Oct 11 '13 at 6:51