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Black body radiation, question arising from a book

  1. Sep 14, 2010 #1

    fluidistic

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    In the book "Introduction to the structure of matter" by Brehm and Mullin, page 78. They say "It should be empathized that there is no reason for the peak positions [tex]\mu _m[/tex] and [tex]\lambda _m[/tex] in the respective distributions to be connected by the relation [tex]c=\mu \lambda[/tex]. They are talking about the blackbody frequency and wavelength spectra. On a graph the independent variable would be either [tex]\mu[/tex] or [tex]\lambda[/tex] and the dependent variable, [tex]M_ \mu (T)[/tex] or [tex]M _\lambda (T)[/tex].

    I do not understand at all how, for a given temperature, a body would emit a peak of frequency [tex]\mu _m[/tex] that does not correspond to a peak of wavelength given by the formula [tex]\lambda _m =\frac{c}{\mu _m}[/tex].
    Can you explain to me what's going on?
     
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  3. Sep 15, 2010 #2

    Born2bwire

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    I know I am going to explain this poorly but...

    It has to do with the fact that we look at the energy over a bandwidth of frequency or wavelength. These peaks are the peaks in the spectral energy densities, which I will call the function \rho. By itself, \rho is meaningless because it is a density. If you wish to find the actual energy you need to integrate it across the frequency or wavelength. Thus, the actual infinitesimal energy is
    [tex] \rho(\nu)d\nu [/tex]
    [tex] \rho(\lambda)d\lambda[/tex]
    This is kind of like the same thing that happens when we talk about wavefunctions in that the magnitude is the probability density but the meaningful probability is |\Psi(x)|^2dx.

    Ok, so the peak energies should be the same, thus,

    [tex] \rho(\nu)d\nu = \rho(\lambda)d\lambda [/tex]
    [tex] \rho(\nu)\left| \frac{d\nu}{d\lambda} \right| = \rho(\lambda) [/tex]
    [tex] \rho(\nu) \frac{c}{\lambda^2} = \rho(\lambda) [/tex]

    Now using this relation you can properly account for the difference in the peak energy densities.
     
  4. Nov 30, 2011 #3

    fluidistic

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    Sorry for being so late on this, but thank you.
    However can you explain me why [itex]\big | \frac{d \nu }{d \lambda} \big |=c/\lambda ^2[/itex]?
    Because there's a problem in Brehm's book that asks me to calculate the product [itex]\nu _m \lambda _m[/itex] and to realize that it's not worth c. Though I must "consider the derivations of the dependence of [itex]\nu _m[/itex] and [itex]\lambda _m[/itex] on the temperature".
     
  5. Nov 30, 2011 #4

    Born2bwire

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    Because the relationship between frequency and wavelength is [itex]c = \lambda \nu [/itex]. Thus,
    [tex] \frac{d\nu}{d\lambda} = \frac{d}{d\lambda} \frac{c}{\lambda} = -\frac{c}{\lambda^2} [/tex]
     
  6. Dec 1, 2011 #5

    fluidistic

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    Ah ok thank you.
    So does this mean that the peak in the frequency [itex]\nu[/itex] is shifted by a factor [itex]\frac{c}{\lambda ^2 }[/itex] compared to the peak in the wavelength [itex]\lambda[/itex]?
     
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