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Homework Help: Thick as a planck and making a spectra of myself

  1. Mar 6, 2008 #1
    for hv>>kT how does exp(hv/kT) compare to 1?
    I understand hv >>KT leads to an exponential fall in brightness but why did Planck introduce 1 in his equation.
    and only for values hv<<kT can this exponential be expanded!
    Thanks for any help!
     
  2. jcsd
  3. Mar 6, 2008 #2
    I have no clue how to help you, however that might be one of the catchiest titles I've ever seen XD
     
  4. Mar 6, 2008 #3

    CompuChip

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    I'm not sure about the question, but you can always expand the exponential in
    [tex]\exp(\hbar\nu / k T) = 1 + \hbar\nu / kT + \mathcal O((\hbar\nu / k T)^2)[/tex]
    where the last term indicates the error you would make if you would just terminate after the first two terms.
    Now what happens if [itex]\hbar\nu \gg k T[/itex] and if [itex]\hbar\nu \ll k T[/itex]?

    BTW I agree, the title is not very descriptive but definitely caught my eye.
     
  5. Mar 6, 2008 #4
    Thanks for looking; I have been looking at Rayleigh Jeans Formula and I need to understand what happens for hv>>kT
     
  6. Mar 6, 2008 #5
    Catastrophe?

    So, Rayleigh-Jeans says that the energy density is [tex] U(\nu) = \frac{8 \pi k T \nu^4}{(c^4)} [/tex] and Planck says [tex]U(\nu)d\nu = \frac{4 \hbar \nu^3}{c^3 (\exp(\hbar\nu / k T)-1)}[/tex].

    When [tex] \hbar \nu >> kT, exp(\hbar \nu / kT) >> 1[/tex], so you can treat it as
    [tex]U(\nu)d\nu = \frac{4 \hbar \nu^3}{c^3 \exp(\hbar\nu / k T)}[/tex]

    Now you can do a limit as [tex] \nu[/tex] goes to infinity for the Rayleigh-Jeans and the Planck equations and compare what happens at really high frequencies.
     
  7. Mar 9, 2008 #6
    Thanks 82
    I actually thought that the result would be of little consequence;
    I'll look closely at the differentiation!
    Obviously now that I see it, with h=constant and K=constant, any overall increase over 1 must mean the value of v increases against T which I assume doesn't change either because I should be looking at Dv against T. (D is supposed to be delta).
    Is the 1 there because of the nature of exp(1)?
    Will any of the differentiation has to employ the exponential rules or product rule?
    Thanks Red
     
  8. Mar 9, 2008 #7

    olgranpappy

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    it's much larger than 1
    to account for the fact that photons are bosons.
    in a taylor expansion about zero... true. But for hv>>kT the planck formula itself can be expanded about zero in the variable [itex]x=e^{-hv/kT}[/itex] which is small and the first term of the expansion is given in klile82's post.
     
  9. Mar 9, 2008 #8

    olgranpappy

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    what differentiation? do you mean to say "derivation"?
     
  10. Mar 9, 2008 #9
    Thanks to all you good guys!

    Is the Planck function mentioned above, valid for high frequency range where v>>kt/h?
    What would a graph look like; v against kt/h?...an exponential rising upwards and rapidly from 0. How would the range on the y axis appear 10^-1 to 10^-10 for example or the otherway about?

    Red
     
  11. Mar 9, 2008 #10

    olgranpappy

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    what the heck? have you read any of the previous posts? what was planck's whole point?
     
  12. Mar 9, 2008 #11
    Why the 1?

    The 1 isn't a hack, it's there because of an identity used in the derivation. Basically, as you go through the derivation you wind up with a term that looks like the sum from n=0 to infinity of [tex] exp(-nh\nu / kT) [/tex]. There's an identity that says that the sum from n=0 to inifity of x^n is \frac{1}{1-x} [/tex] as long as x< 1. That's where the 1 in the denominator comes from.

    It's really helpful, if you're comparing the Rayleigh-Jeans equation to the Planck equation, to graph them as a function of v on the same graph. You should be able to see the difference right away. Just choose a value for T and use the standard values for c, k and h. You can do this easily in something like Mathematica. You can do this fairly easily in excel, too.
     
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