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Approximation problem

  1. Aug 7, 2008 #1
    I'm having problems understanding how

    \frac{e^{-\hbar \omega / 2k_BT}}{1-e^{-\hbar \omega / k_BT}}

    approximates to

    k_BT/ \hbar\omega


    T >> \hbar\omega/k_B

    Seems like it should be simple but don't quite see how to arrive at this result.


    I have tried using taylor expansions of [tex]exp(-x)[/tex] and [tex]1-exp(-x)[/tex] and just using the first expansion term since if [tex]T>>\hbar\omega/k_B[/tex] then [tex]\hbar\omega/k_BT[/tex] should be small. This seems to give the right answer but i'd be interested in knowing if indeed my method is ok and if there are alternate methods.
    Last edited: Aug 7, 2008
  2. jcsd
  3. Aug 7, 2008 #2
    If you call
    [tex]x = - {{\hbar \omega } \over {2 k_B T}}[/tex]
    (and [tex]x \to 0[/tex] when [tex]T >> \hbar\omega/k_B[/tex])

    then your expression is equivalent to
    [tex]{{e^x } \over {1 - e^{2x} }}[/tex]

    Utilizing the known limit
    [tex] \mathop {\lim }\limits_{x \to 0} {{e^x -1} \over x} = 1[/tex]

    you can write
    \mathop {\lim }\limits_{x \to 0} {{e^x } \over {1 - e^{2x} }} = \mathop {\lim }\limits_{x \to 0} {{e^x } \over {\left( {1 - e^x } \right)\left( {1 + e^x } \right)}}\left( {{{e^x - 1 } \over x}} \right) \to -{1 \over {2x}}[/tex]

    So the expression near zero goes like
    [tex]-{1 \over {2x}}[/tex]
    that means that the original expression goes like
    [tex]{{k_B T} \over {\hbar \omega }}[/tex]
  4. Aug 8, 2008 #3
    Thanks, the way i did it was equivalent it seems, but yours was a lot more clearer..

    thanks again.
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