Why does Wien's displacement law not hold for frequency?

[Jezuz]

Wien's displacement law states that the wavelength of highest intensity in the radiation from a blackbody is something like:

\lambda_{max} = \frac{2.898*10^{-3}}{T}

in meters, where T is the temperature given in kelvins.

If you try to transform this law into frequency one would expect that we should have:

f_{max} = \frac{c}{\lambda_{max}}

but apparently this is not the case! Why is it like that?
I mean, if you have a blackbody radiation field it will have a maximum of intensity at some frequency, but shouldn't that frequency coincide with the wavelenght for which it has the maximum intensity?

Please help!

 

[George Jones]

Wein's law results from finding a maximum in the Planck distribution u \left( \lambda \right). Given u \left( \lambda \right) and \lambda \left( f \right) = c/f, a new function \tilde{u} \left( f \right) = u \left( \lambda \left( f \right) \right) of frequency can be defined, but \tilde{u} \left( f \right) is not the Planck distribution in the frequency domain. If it were, then f_{max} and \lambda_{max} would correspond.

\int_{0}^{\infty} u \left( \lambda \right) d\lambda = \int_{\infty}^{0} \tilde{u} \left( f \right) \frac{d \lambda}{df} df = - \int_{0}^{\infty} \tilde{u} \left( f \right) \frac{d \lambda}{df} df

Consequently,

- \tilde{u} \left( f \right) \frac{d \lambda}{df}

needs to be maximized to find f_{max}.

Regards,
George

 

[Jezuz]

Of course!
You need to have the two integrals you wrote equall so that the energy (intensity) that is radiated in a certain frequency range is the same as the intensity of the corresponding wavelenght, right?

 

[George Jones]

Of course!
You need to have the two integrals you wrote equall so that the energy (intensity) that is radiated in a certain frequency range is the same as the intensity of the corresponding wavelenght, right?

Yes.

Note that I corrected a silly mistake in my previous post.

Regards,
George