How Does This Quantum Mechanics Approximation Problem Work?

[Narcol2000]

I'm having problems understanding how

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

approximates to

<br /> k_BT/ \hbar\omega<br />
when

<br /> T &gt;&gt; \hbar\omega/k_B<br />

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 exp(-x) and 1-exp(-x) and just using the first expansion term since if T&gt;&gt;\hbar\omega/k_B then \hbar\omega/k_BT 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.

 

[lemma28]

If you call
x = - {{\hbar \omega } \over {2 k_B T}}
(and x \to 0 when T &gt;&gt; \hbar\omega/k_B)

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

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

you can write
<br /> \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}}

So the expression near zero goes like
-{1 \over {2x}}
that means that the original expression goes like
{{k_B T} \over {\hbar \omega }}

 

[Narcol2000]

Thanks, the way i did it was equivalent it seems, but yours was a lot more clearer..

thanks again.