1
$\begingroup$

I was reading Fundamentals of Atmospheric Radiation by Craig F. Bohren and Eugene E. Clothiaux and in page 20-23 they discuss the idea that the maximum of the Planck's distribution varies depending on whether you plot it vs wavelength or vs frequency. They conclude saying "There is, in general, no invariant maximum for a distribution function. This may be unpalatable but it is a fact of life, in the nature of distribution functions."

I still don't see how that is possible. I mean if the spectral density is larger at a determined wavelength why does that change if the distribution is plotted against frequency?

$\endgroup$
2
  • $\begingroup$ Frequency and wavelength are inversely proportional to each other, $f=c/\lambda$, so the distribution definitely looks different. You see however to misunderstand the conclusion or where it comes from. Perhaps you could re-edit your question by adding more details, starting with the distribution intself. $\endgroup$ – Roger Vadim Apr 12 at 15:58
  • $\begingroup$ What I mean is that if you obtain the maximum of the Planck's distribution plotted vs wavelength you get λmT = 2902 μmK. If you do the same for the Planck's distribution plotted vs frequency and then convert the maximum frequency to wavelength you get λmT = 5107 μmK. So basically the "maximum wavelength" is different depending on what the distribution is plotted against. $\endgroup$ – propagator Apr 12 at 16:03
4
$\begingroup$

The values being plotted are "intensity per unit wavelength" or "intensity per unit frequency", which are different things. So it's not just a transformation on the x-axis; the actual quantity being plotted is different.

It's scaled by the derivative of wavelength over frequency (which is not a constant and therefore affects the maximum). If $E$ is the energy up to some frequency/wavelength, the first quantity is $dE/d\lambda$ while the second is $dE/d\omega$.

$\endgroup$
1
$\begingroup$

There are different variants of the Planck's distribution function:

  • spectral density of radiance of blackbody over frequencies:

$$B_\nu(\nu,T)=\frac{2h\nu^3}{c^2}\frac1{e^{\frac{h\nu}{kT}}-1},\tag1$$

  • the same density, but over wavelengths:

$$B_\lambda(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac1{e^{\frac{hc}{\lambda kT}}-1}.\tag2$$

These are different densities that are supposed to be integrated over different quantities to get total radiance in a given spectral range: $B_\lambda$ is to be integrated over a range of wavelengths, while $B_\nu$ over a range of frequencies. This is also reflected in the dimensions of these densities.

If you have e.g. $B_\nu$ but need to integrate over wavelengths, you'll have to use an appropriate change of variables of integration. E.g. to get a total radiance in the range between 400 nm and 500 nm, you can integrate as

$$ R=\int\limits_{400\,\mathrm{nm}}^{500\,\mathrm{nm}} B_\nu(\nu,T)\,d\nu =\int\limits_{600\,\mathrm{THz}}^{749\,\mathrm{THz}} -\frac{d\nu}{d\lambda}\;B_\nu\left(\frac c\lambda,T\right)\,d\lambda \equiv\int\limits_{600\,\mathrm{THz}}^{749\,\mathrm{THz}} B_\lambda(\lambda,T)\,d\lambda. \tag3 $$

In the middle expression above the minus sign is just for convenience, so that the integration goes from smaller frequency to larger (and this minus also gets into $B_\lambda$ on the RHS and into $(2)$).


Why do we need different functions for different scales? It's because at different points in wavelength a unit interval of wavelengths will have different interval in frequencies. E.g. consider the range 400 nm to 401 nm, its width being 1 nm. After you find corresponding frequencies, you'll get respectively 749.5 THz and 747.6 THz, with a difference being about 1.9 THz. Now consider the range of 700 nm to 701 nm, also of width 1 nm. Corresponding frequencies are 428.3 THz and 427.7 THz, with a difference of about 0.6 THz.


Now, you can indeed plot e.g. $B_\nu$ on a wavelength plot as $B_\nu(c/\lambda,T)$. This wouldn't be wrong, but it would be misleading. The readers would expect to see the density plotted correspond to the scale on the axes, while in fact it doesn't.

$\endgroup$
1
  • $\begingroup$ @ChiralAnomaly fixed, thanks. $\endgroup$ – Ruslan Apr 13 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.