Mathematical trick in Statistical Physics

The other day I was looking into some Statistical Physics and ended up doing an exercise that was kinda cute. The exercise in question is on Mandl's Statistical Physics, Second Edition page 66.

The first time I did the exercise I did it the normal way, but then I noticed something that ended up simplifying my calculations and here I am posting it.

A system has three energy levels $ {E_1= \epsilon } $, $ {E_2=2 \epsilon } $ and $ {E_3=3 \epsilon } $ with degeneracies $ {g(E_1)=g(E_3)=1} $, $ {g(E_2)=2} $. Find the heat capacity of the system.

Don't worry if you don't understand all the terms in here since the gist of this post is not Statistical Physics but the mathematical trick I ended up using.

To solve this problem we need to calculate the partition function $ {Z} $.

$ \displaystyle \begin{array}{rcl} Z &=& \displaystyle \sum_{E_r}g(E_r)e^{-\beta E_r} \\ &=& 1+2e^{-\beta \epsilon}+e^{-\beta 2\epsilon} \\ &=& e^{\beta \epsilon}+2+2e^{-\beta \epsilon} \\ &=& 2(1+ \cosh (\beta \epsilon)) \end{array} $

After having calculated the partition function we have to calculate the average energy, $ {\bar{E}} $, of this system. By definition it is:

$ \displaystyle \begin{array}{rcl} \bar{E} &=& -\dfrac{\partial}{\partial \beta} \log Z \\ &=& -\dfrac{\partial}{\partial \beta} \log 2(1+ \cosh (\beta \epsilon)) \\ &=& -\dfrac{2 \epsilon \sinh (\beta \epsilon)}{2(1+\cosh (\beta \epsilon))} \\ &=& -\dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)} \\ \end{array} $

Keeping in mind that in Statistical Physics it is $ {\beta = 1/(kT)} $ the heat capacity, $ {c} $, is (this is where I started using the trick):

$ \displaystyle \begin{array}{rcl} c &=& \dfrac{\partial \bar{E}}{\partial T} \\ &=& k \dfrac{\partial \bar{E}}{\partial (kT)} \\ &=& k \dfrac{\partial \bar{E}}{\partial (1/\beta)} \\ &=& -\beta ^2 k \dfrac{\partial \bar{E}}{\partial \beta}\\ &=& \beta ^2 k \dfrac{\partial}{\partial \beta}\left( \dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \end{array} $

Notice that up until now we haven't calculated a thing at all. All that we have done is just to change variables in order to ease the difficulty in the derivative we'll have to calculate.

Taking the derivative with respect to $ {T} $ in the last expression isn't that hard, but it sure is boring and if one isn't careful errors are expected to creep in.

On the other hand the last expression is much easier to differentiate, but since we are caught up in the moment we'll just do one more change of variable.

$ \displaystyle \begin{array}{rcl} c &=& \beta ^2 k \dfrac{\partial}{\partial \beta}\left( \dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \\ &=& \beta ^2 \epsilon ^2 k \dfrac{\partial}{\partial (\beta \epsilon)}\left( \dfrac{ \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \\ &=& x ^2 k \dfrac{\partial}{\partial x}\left( \dfrac{ \sinh x}{1+\cosh x}\right) \\ &=& x ^2 k \dfrac{\cosh x + \cosh ^2 x - \sinh ^2 x}{(1+ \cosh x)^2} \\ &=& x ^2 k \dfrac{\cosh x + 1}{(1+ \cosh x)^2} \\ &=& \dfrac{x ^2 k}{1+ \cosh x} \end{array} $

Yes, in this case the simplification wasn't that great but I think that one shouldn't lose sight of the fact that this type of reasoning can greatly simplify some other physical derivations.

As an afterthought just let me point out the fact that we cold get a much simpler life if we just remembered a very easy algebraic identity:

$ \displaystyle \begin{array}{rcl} Z &=& 1+2e^{-\beta \epsilon}+e^{-\beta 2\epsilon} \\ &=& (1+e^{-\beta \epsilon})^2 \\ \end{array} $

With this expression the calculation of $ {\bar{E}} $ and $ {c} $ is a lot easier and the reader is urged to try it out.