ps6 problem 2 done.

env/.aspell.en.pws

@@ -1,4 +1,4 @@
-personal_ws-1.1 en 468 
+personal_ws-1.1 en 469 
 CDF
 Google
 tuples
@@ -8,8 +8,8 @@ ParametericPlot
 Indistinguishability
 Exponentiating
 extremum
-Fick's
 Ficks
+Fick's
 cgs
 phasor
 linestyle
@@ -88,8 +88,8 @@ Eikonal
 Goldstein's
 GPS
 colinear
-Hmm
 hoc
+Hmm
 Peeter's
 Benard
 ijk
@@ -118,6 +118,7 @@ ket
 online
 LDE
 ipu
+Boson
 ipx
 telecentric
 Thywissen
@@ -161,8 +162,8 @@ Hestenes's
 parametrizations
 imaginaries
 reparametrize
-nlm
 n'l'm
+nlm
 PDE
 Lut
 quantized
@@ -189,8 +190,8 @@ arctan
 entropic
 invertible
 pion
-OuterMorphism
 outermorphism
+OuterMorphism
 QFT
 rescaling
 spinors
@@ -312,8 +313,8 @@ df
 LIGO
 spacetime
 dH
-dj
 BT
+dj
 Routhian
 dk
 dL
@@ -341,9 +342,9 @@ Prandtl
 ia
 isync
 inferometer
+eV
 ib
 iB
-eV
 orthonormalization
 ic
 ie
@@ -358,16 +359,16 @@ im
 Dekker's
 ji
 unnormalized
-KE
 jj
+KE
 elastostatics
 amino
-jk
 ip
+jk
 indices
 kj
-kk
 iu
+kk
 mc
 mE
 iz
@@ -380,8 +381,8 @@ Strang's
 xyz
 mk
 resistive
-kx
 mn
+kx
 Eulerian
 anticommutes
 n'l

notes/METADATA

@@ -4758,6 +4758,11 @@ Generate figures for continuum mechanics problem set II figure 1.  Using Show an
    path => 'phy456/mathematica/24.4.3.newAttempt.nb',
    WHAT => qq(A new attempt at Desai 24.4.3 from scratch.  This one has an error, as did the original.  The original is now fixed.),
 },
+{
+   DATE => 'March 30, 2013',
+   path => 'phy456/mathematica/basicStatMechProblemSet6Problem2.nb',
+   WHAT => qq(Some rough calculations and plots that were discarded for ps6 p2.  Mathematica functions of interest: Map, Evaluate, Flatten, pure functions, Assumptions),
+},
 # not all of these are committed to the repo.  Some are, but are not described here.
 #blogit/imageProcessingExperimentation.cdf
 #blogit/streamSlowCurveFit.cdf

notes/blogit/basicStatMechProblemSet6Problem2.tex

@@ -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}
 }

notes/peeter_prologue_print2.tex

@@ -83,8 +83,8 @@ \section*{}
 \colorSectionsForArticle
 
 % (a), (b), (c), ... numbering in ex (exercise pulled in by peeters_layout)
-%\renewcommand{\QuestionNB}{\alph{Question}.\ }
-%\renewcommand{\theQuestion}{\alph{Question}}
+\renewcommand{\QuestionNB}{\alph{Question}.\ }
+\renewcommand{\theQuestion}{\alph{Question}}
 
 % have \cref fixup in book_layout.sty.  Use this macro instead:
 \newcommand{\eqnref}[1]{eq. (\ref{#1})}