Frequency and Wavelength peak for Wien's displaement law of a blackbody

Question

This is a question relating to Wien's displacement law for the Planck function. As we all know frequency and wavelength are related to the speed of light by:

$$\nu\lambda=c$$

However, why is it that:

$$\nu_{\mathrm{peak}}\lambda_{\mathrm{peak}}\neq{c}$$

Any explanations would be very much appreciated.

To all of the people wanting to know where this statement came from. It hasn't come from anywhere specific, is it a well known fact of the Planck function. $\lambda_{\mathrm{peak}}=0.290T^{-1}$ cm K and $\nu_{\mathrm{peak}}=5.88\times{10^{10}}T$ Hz K$^{-1}$.

Welcome to Physics SE. Look around, and please take the [tour]. For this particular question, a little more background would be useful, in particular the source for the statement you disagree with. — Jon Custer, Oct 06 '15 at 18:22
I don't disagree with the statement. I want to know why. This question has not been answered sufficiently in these forums. And the information given should be sufficient for someone who actually knows the answer. — DarthPlagueis, Oct 06 '15 at 18:23
where did you get that $\nu_{\mathrm{peak}}\lambda_{\mathrm{peak}}\neq{c}$ ? — , Oct 06 '15 at 18:26
It's in any physics textbook, but I can't find a sufficient explanation as to why this is. — DarthPlagueis, Oct 06 '15 at 18:28
@user3125347 - So, to repeat myself, what is your source for this statement? — Jon Custer, Oct 06 '15 at 18:28
Could you specifically point to a book and page? I do not think it is correct — , Oct 06 '15 at 18:31
possible duplicate of The strange thing about the maximum in Planck's law — Timaeus, Oct 06 '15 at 20:28

score 6 · Accepted Answer · edited Apr 13 '17 at 12:39

6

The maximum of the spectral flux per unit wavelength $$I(\lambda,T)$$ does not correspond by $\lambda\nu = c$ to the maximum of the spectral flux per unit frequency $$I(\nu,T)$$ since these two functions are related by $$ I(\lambda,T)\mathrm{d}\lambda = I(\nu,T)\mathrm{d}\nu$$ but are not the same function, so their maxima are not the same.

edited Apr 13 '17 at 12:39

Community

1

answered Oct 06 '15 at 18:34

ACuriousMind

124,833

our electrons went flying by each other... – Jon Custer Oct 06 '15 at 18:35
Can I assume that I(nu,T) is interchangeable with B_nu(T)? – DarthPlagueis Oct 06 '15 at 18:40
The only thing I'd add is that indeed the student is right to intuit that if $f(y)$ is maximized for some special $y^$ then $f(c/x)$ is maximized for $c/x^ = y^,$ so that indeed the problem is that these are not defined to be equal* densities (as in the same function) but equivalent densities (as in the same physics). – CR Drost Oct 06 '15 at 18:40
@user3125347: It's the spectral flux/radiance per unit frequency. If that's what your $B_\nu(T)$ is then yes, otherwise no. – ACuriousMind Oct 06 '15 at 18:42

score 3 · Answer 2 · answered Oct 06 '15 at 18:35

I believe that the issue is with the difference between evaluating the peak in Planck's law with respect to frequency vs with respect to wavelength. Since this is pointed out on the Wikipedia page, it seems a bit much to replicate the differentiation of the Planck distribution vs wavelength and frequency here. However, the point is that since wavelength and frequency are inversely related, the derivative of the energy density with respect to one or the other can, and will, result in a different answer.

Frequency and Wavelength peak for Wien's displaement law of a blackbody

2 Answers2