1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Approximation problem

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

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

    approximates to

    [tex]
    k_BT/ \hbar\omega
    [/tex]



    when

    [tex]
    T >> \hbar\omega/k_B
    [/tex]

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

    *update*

    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
    [tex]
    \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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Approximation problem
  1. Integral Approximation (Replies: 3)

Loading...