Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Wien's displacement law.

  1. Mar 10, 2006 #1
    Wien's displacement law states that the wavelength of highest intensity in the radiation from a blackbody is something like:

    [tex] \lambda_{max} = \frac{2.898*10^{-3}}{T} [/tex]

    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:

    [tex] f_{max} = \frac{c}{\lambda_{max}} [/tex]

    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!
     
  2. jcsd
  3. Mar 11, 2006 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    [tex]\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[/tex]

    Consequently,

    [tex]- \tilde{u} \left( f \right) \frac{d \lambda}{df}[/tex]

    needs to be maximized to find [itex]f_{max}[/itex].

    Regards,
    George
     
    Last edited: Mar 11, 2006
  4. Mar 11, 2006 #3
    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?
     
  5. Mar 11, 2006 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes.

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

    Regards,
    George
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook