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Rayleigh jeans formula

  1. Feb 13, 2006 #1
    I am to derive the incorrect Rayleigh-Jeans formula from the correct Planck formula to show why plank's constant does not appear in the Rayleigh-Jeans formula. I should also recall the Stefen-Boltmann Law

    here's what I have but I'm stuck...

    Rayleigh-Jeans formula: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}[/tex]

    Planks formula: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}[/tex]

    so I am thinking I am somehow supposed to get: [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = 1[/tex] but I dont know how to even begin. any ideas?
     
  2. jcsd
  3. Feb 13, 2006 #2

    Physics Monkey

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    First, the Planck formula is [tex] u(\lambda) = \frac{8 \pi}{\lambda^4} \frac{ h c / \lambda}{e^{hc/\lambda k T} - 1} [/tex]. You have an extra factor of [tex] k T [/tex] in your Planck formula which is the source of the confusion. It should now be a simple matter to obtain the Rayleigh-Jeans formula by taking the appropriate limit.
     
  4. Feb 14, 2006 #3
    so that should lead me to prove that...

    [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

    I am sorry but it is still not apparent how I should go about solving this... as T nears infinity, i get infinity on both sides, so does that prove that [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]?

    and how does the Stefan-Boltzmann Law come into play?
     
  5. Feb 14, 2006 #4

    George Jones

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    For Rayleigh-Jeans, you need to look at what happens the [itex]\lambda[/itex] becomes large.

    For Stephan-Boltzmann, you need to look at the total contribution from all wavelengths, i.e., you need to look at

    [tex]\int_{0}^{\infty} u \left( \lambda \right) d \lambda.[/tex]

    Make a change of integration variable so that [itex]T[/itex] does not appear explicitly in the integrand. Evaluating the resulting integral requires some specialized knowledge of special function. If you only need to show proportionality to [itex]T^4[/itex], then the integral need not be evaluated. If you need the proportionality constant, use software (e.g., Maple) or tables to evaluate the integral

    Regards,
    George
     
    Last edited: Feb 14, 2006
  6. Feb 14, 2006 #5
    [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

    when lambda --> infinity, the left hand side becomes 0/0 so I applied hospital's rule to get:

    [tex]\frac{hc(-\frac{1}{ \lambda^2})}{\frac{hc}{kT} e^{hc/ \lambda K T}} = kT[/tex]

    [tex]\lambda \rightarrow \inf[/tex]

    [tex]\frac{hc}{\frac{hc}{kT}}=kT[/tex]

    [tex]kT=kT[/tex]

    I did not use the stefen-boltzmann law, did I do this correctly?
     
    Last edited: Feb 14, 2006
  7. Feb 14, 2006 #6

    George Jones

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    I guess I mistunderstood - I thought you wanted to derive Rayleigh-Jeans and Stefan-Boltzmann from Planck.

    To derive Rayleigh-Jeans, expand as a series the exponential in Physics Monkey's expression, and find what happens when [itex]\lambda[/itex] becomes large, but not infinite.

    Regards,
    George
     
  8. Feb 14, 2006 #7
    the Rayleigh-Jeans formula is: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}[/tex]

    while Planck's formula consists of Rayleigh-Jeans but includes [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}[/tex] instead of [tex]kT[/tex]

    so what I did was set [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex] and solved for when the wavelength was really big, which shows that indeed [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

    plugging this into Planck's equation, I will get the Rayleigh-Jeans formula:

    [tex]u(\lambda)=\frac{8 \pi }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}[/tex]

    [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

    Rayleigh-Jeans formula: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}[/tex]

    did it do this right? was I supposed to include the stefen-boltzmann law somewhere in there?
     
  9. Feb 14, 2006 #8

    George Jones

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    I don't think so - it appears that you've just gone round in a circle.

    You need to start with

    [tex]u \left( \lambda \right) = \frac{8 \pi hc}{\lambda^{5} \left( e^{hc/ \lambda K T} - 1 \right)},[/tex]

    do what I suggested in my previous post, and arrive at

    [tex]u \left( \lambda \right) = \frac{8 \pi k T }{\lambda^{4}}.[/tex]

    Regards,
    George
     
  10. Feb 16, 2006 #9

    dextercioby

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    If T is large, then the exponent in [itex] e^{hc/ \lambda kT}[/itex] is small and you can use Bernoulli's formula

    [tex] e^{x}=1+x [/tex] valid for "x" very small.

    Daniel.
     
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