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Getting Planck's Law in terms of frequency from wavelength

  1. Mar 21, 2015 #1
    Show that Planck's law expressed in terms of the frequency f is:
    u(f) = (8πf2/c3)(hf/(ehf/kT - 1))

    from the equation:

    u(λ) = (8πhcλ-5)/(ehc/λkT - 1)

    When I do this algebraically by simply plugging in λ = c/f, I get:

    u(f) = (8πhc-4)/(f-5(ehf/kT - 1)

    which clearly doesn't involve the correct powers of c and f.

    This is the same thing I get when going back and putting n(λ) = 8πλ^-4 in terms of the frequency and multiplying it by the average energy E bar = hf/(ehf/kT - 1) to get u(f).

    Going through all of these equations is confusing and I am having trouble putting it all together, but I figure that the problem is I need to go back and differentiate somewhere, since n(λ) is found from n(λ)dλ which involves the range between λ and λ+dλ, but it hasn't been converted to the range between f and f+df. I don't know how I would go about doing any of this...please help!
     
  2. jcsd
  3. Mar 21, 2015 #2

    DrClaude

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    Staff: Mentor

    Planck's law is a distribution function, so you can simply start by considering ##u(\lambda) d\lambda## (radiance per unit wavelength times unit wavelength), which you can easily convert to ##u(f) df##, for which you get ##u(f)##.
     
  4. Mar 21, 2015 #3
    You would have to put dλ in terms of df then right? If λ = c/f, then dλ = -c/f2 df?

    So I already found u(f) correctly (I think), so multiplying my equation for u(f) by df would get:

    u(f)df = (8πhc-4)/(f-5(ehf/kT - 1) (-c/f2 df)

    Dropping the df to get u(f) and simplifying:

    u(f) = (-8πhc-3)/(f-3(ehf/kT - 1) = (-8πhf3/c3)(ehf/kT - 1)

    Alright that is exactly the answer I need, except that my answer is negative because of the negative derivative. I'm not sure how to resolve that.
     
  5. Mar 21, 2015 #4

    DrClaude

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    Staff: Mentor

    If you integrate from wavelength ##\lambda_1## to ##\lambda_2##, with ##\lambda_2 > \lambda_1##, when happens when you convert that to frequency?
     
  6. Mar 21, 2015 #5
    You'll integrate from f1 to f2 with f1>f2?
     
  7. Mar 21, 2015 #6

    DrClaude

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    Staff: Mentor

    Yes, and considering that normally you want to integrate from low frequency to high frequency, what do you need to change?
     
  8. Mar 21, 2015 #7
    Ah, I understand. So could I indicate this by writing that:

    λ21 and f1>f2 thus when ∫u(λ)dλ → ∫u(f)df

    f2f1 u(f)df = -∫f1f2 u(f)df

    Thus u(f) = (-8πhf3/c3)(ehf/kT - 1) taken from f1 to f2 =
    u(f) = (8πhf3/c3)(ehf/kT - 1) taken from f2 to f1

    Or maybe a simpler way?
     
  9. Mar 22, 2015 #8

    DrClaude

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    I don't think there is a simpler way to demonstrate it.
     
  10. Mar 22, 2015 #9
    Thanks for your help!
     
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