# peeterjoot/physicsplay

ps6 problem 2 done.

 @@ -25,7 +25,7 @@ For Fermions, obtain the behavior of $f_\nu^\pm(z)$ for $z \rightarrow \infty$ again keeping the two leading terms. \makesubproblem{}{basicStatMech:problemSet6:2c} -For Bosons, we must have $z \le 1$ (why?), obtain the leading term of $f_\nu^\pm(z)$ for $z \rightarrow 1$. +For Bosons, we must have $z \le 1$ (why?), obtain the leading term of $f_\nu^-(z)$ for $z \rightarrow 1$. } % makeoproblem \makeanswer{basicStatMech:problemSet6:2}{ @@ -338,5 +338,106 @@ \makeSubAnswer{}{basicStatMech:problemSet6:2c} -TODO. +FIXME: why $z \le 1$. + +For the Boson case, we substitute $z = e^{-\alpha}$ to look at the $z \rightarrow 1$ case. + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:540} +G_\nu(e^{-\alpha}) += +\Gamma(\nu) +f_\nu^-(e^{-\alpha}) += +\int_0^\infty dx \frac{x^{\nu - 1}}{e^{x + \alpha} - 1}. +\end{dmath} + +For $\nu = 1$, this is integrable + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:560} +\frac{d}{dx} \ln\lr{ 1 - e^{-x - \alpha} } += +\frac{e^{-x - \alpha}} +{ 1 - e^{-x - \alpha} } += +\inv +{ e^{x + \alpha} - 1}, +\end{dmath} + +so that +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:580} +G_1(e^{-\alpha}) += +\int_0^\infty dx \frac{1}{e^{x + \alpha} - 1} += +\evalrange{ +\ln \lr{1 - e^{-x - \alpha} } +}{0}{\infty} += +\ln 1 +- \ln +\lr{1 - e^{- \alpha} } += +-\ln \lr{1 - e^{- \alpha} }. +\end{dmath} + +Taylor expanding $1 - e^{-\alpha}$ we have + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:600} +1 - e^{-\alpha} = 1 - \lr{ 1 - \alpha + \alpha^2/2 - \cdots}. +\end{dmath} + +Noting that $\Gamma(1) = 1$, we have for the limit + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:640} +\lim_{\alpha \rightarrow 0} G_1(e^{-\alpha}) +\rightarrow - \ln \alpha, +%= \ln (1/\alpha) +\end{dmath} + +or +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:620} +\lim_{z\rightarrow 1} f_\nu^-(z) += -\ln (-\ln z). +\end{dmath} + +For values of $\nu \ne 1$, the denominator is + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:660} +e^{\alpha + x} - 1 += (\alpha + x) + (\alpha + x)^2/2 + \cdots +\end{dmath} + +To first order this gives us + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:680} +f_\nu^-( e^{-\alpha} ) +\approx +\inv{\Gamma(\nu)} +\int_0^\infty dx \frac{1}{x + \alpha}. +\end{dmath} + +Of this integral Mathematica says it can be evaluated for $0 < \nu < 1$, and has the value + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:700} +\inv{\Gamma(\nu)} +\int_0^\infty dx \frac{1}{x + \alpha} += +\frac{\pi}{\sin(\pi\nu)} \frac{1}{\alpha^{1 - \nu} \Gamma (\nu )}. +\end{dmath} + +From \citep{abramowitz1964handbook} 6.1.17 we find + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:720} +\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}, +\end{dmath} + +with which we can write + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:740} +\myBoxed{ +f_\nu^-( e^{-\alpha} ) +\approx +\frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} }. +} +\end{dmath} }