Benedict Irwin edited untitled.tex  over 8 years ago

Commit id: d019c35c40c638dfc19204ea6f01b4ae0c26acef

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for any value of $n$! That is the entropy of each state is the same, independant of the level of excitation.  When adjusting for waves of any amplitude $\alpha$ we get \begin{equation}  \int_0^\pi \alpha^2sin^2(nx)\mathrm{ln} \alpha^2sin^2(nx) \; dx =\frac{\pi\alpha^2}{2}(1-\mathrm{ln}(4\alpha^2)) =\frac{-\pi\alpha^2}{2}(1-\mathrm{ln}(\frac{4}{\alpha^2}))  \end{equation}