Wednesday, February 26, 2014

quantum statistics

can you write the Fermi-Dirac-distribution as a superposition of Bose-Einstein-distributions at different temperatures? is the inverse possible as well?

1 comment:

  1. this funny result can be explained by using the addition theorem of the hyperbolic tangent and cotangent,
    \begin{equation}
    \tanh(x) + \coth(x) = 2\coth(2x)
    \end{equation}
    which follows directly from the addition theorem,
    \begin{equation}
    \tanh(x+y) = \frac{\tanh(x)+\tanh(y)}{1+\tanh(x)\tanh(y)}
    \end{equation}
    by setting $x=y$. by writing the hyperbolic functions in terms of exponentials,
    \begin{equation}
    \tanh(x) = \frac{\exp(2x)-1}{\exp(2x)+1}
    \end{equation}
    you can show that
    \begin{equation}
    \frac{1}{\exp(x)-1} - \frac{2}{\exp(2x)-1} = \frac{1}{\exp(x)+1}
    \end{equation}
    such that the Fermi-Dirac-distribution is the superposition of two Bose-Einstein-distributions at two temperatures separated by a factor of two. because the addition theorem of the cotangent is equivalent to that of the tangent, you can't do the inverse.

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