When I tried to derive the Wien's displacement law I used Planck's law for blackbody radiation:

$I_\nu = \frac{8 \pi \nu^2}{c^3} \frac{h \nu}{e^{h\nu/k_bT}-1}$

Asking for maximum:

$\frac{dI_\nu}{d \nu}=0:~0= \frac{\partial}{\partial\nu}(\frac{\nu^3}{e^{h \nu/k_bT}-1}) = \frac{3\nu^2(e^{h \nu/k_bT}-1) - \nu ^3h/k_bT \cdot e^{h\nu/k_bT}}{(e^{h \nu/k_bT}-1)^2}$

It follows that numerator has to be $0$ and looking for $\nu>0$:

$3(e^{h \nu/k_bT}-1) - h \nu/k_bT \cdot e^{h\nu/k_bT}=0$

Solving for $\gamma=h\nu/k_bT$:

$3 (e^\gamma-1) - \gamma e^\gamma=0 \rightarrow \gamma=2.824$

Now I look at the wavelength domain:

$\lambda = c/\nu:~ \lambda =\frac{h c}{\gamma k_b} \frac{1}{T}$

but from Wien's law $\lambda T = b$ I expect that $hc/\gamma k_b$ is equal to $b$ which is not:

$\frac{h c}{\gamma k_b}= 0.005099$, where $b = 0.002897$

Why the derivation from frequency domain does not correspond the maximum in wavelength domain?

I tried to justify it with chain rule:

$\frac{dI}{d\lambda} = \frac{dI}{d\nu} \frac{d \nu}{d \lambda} = \frac{c}{\nu^2} \frac{dI}{d \nu}$

where I see that $c/\nu^2$ does not influence where $dI_\lambda/d \lambda$ is zero.

  • $\begingroup$ that goes back into the definitions of frequency, wavelength and temperature; and thats how it works. Look at the last paragraphs of Wien's law on wikipedia where they make a compromise for frequency-wavelength combined correlation by selecting 4 from 3 and 5 as a middle value. And may be that analogous to near and far fields thing. [IMO + Wikipedia] $\endgroup$
    – user28737
    Commented Sep 29, 2013 at 7:54
  • 1
    $\begingroup$ After reading about how the Planck's law for wavelength is derived in [physics.stackexchange.com/q/13611 ] it became clear :) $\endgroup$ Commented Sep 30, 2013 at 18:29

1 Answer 1


Well Janis, it is not clear exactly what it is that you want to know.

Wien's displacement law, simply says that $\lambda_{\text{max}} \cdot T$ is a constant. That applies to the form of Planck's radiation formula, that gives the spectral radiant emittance in Watts, per square meter, per "Wavelength Interval"; where the wavelength interval could be meters of nanometers or microns, or any other useful wavelength increment; as a function of WAVELENGTH. Perhaps per micron is most common.

Now since $f_\lambda = c$, then one could simply say that $\frac{c T}{f_{\text{max}}}$ is a constant, where $f_{\text{max}}$ is the frequency corresponding to $\lambda_{\text{max}}$.

BUT that is not what is usually meant.

Some users who prefer to think of black body radiation in the frequency domain, want a plot as a function of frequency or wave number, and not a plot versus wavelength. This is reasonable since the photon energy is simply $h\nu$.

BUT, when spectral radiant emittance is plotted against frequency, the vertical axis units are different. The units are Watts, per square meter, per wave number increment, or per frequency increment, and NOT per wavelength increment.

As a result the spectral peak occurs in a different place depending on the way it is plotted.

For example, the spectral peak for black body radiation at the presumed mean surface temperature of the earth (288 K), is 10.1 microns, when the plot is on a wavelength scale. The corresponding curve plotted on a frequency basis, and emittance per wavenumber basis, peaks at an entirely different wavelength. I don't use the frequency or wave number form myself, so I never remember any of those numbers, but you can easily find such data on the web or in text books.

But the key is, one plot is radiant emittance per wavelength increment, and the other is radiant emittance per frequency increment.

And then of course it will be $\frac{f_{\text{max}}}{T}$ that is a constant for Wien's displacement law.


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