 \magnification\magstep1 \nopagenumbers%\parskip=12 pt plus 4pt minus 4pt  
\def \chapno{2} %roman for chapters; arabic for problem sets; 
\newcount\notenumber \notenumber=1
\def\problem#1{\par\medskip\noindent \chapno .\the\notenumber \global\advance
\notenumber by 1 ~{\sl #1}\quad}
\def\?#1{ } 
\line{Physics 352 \hfill Spring, 1995 \hfill Problem Set \chapno \hfill Due
date: 
Th. April 13}\smallskip\hrule\bigskip

\problem{Additivity of the statistical entropy}  A system consists of two
independent subsystems, with configurations labeled by $c$ and $\tilde c$.   For
example, the untilde subsystem may be a container of gas, while the tilde
subsystem may be a magnet in the next room.  Since the two systems are 
independent, the probability of a combined configuration $(c,\tilde c)$ of the
combined system is simply the product of two independent probabilities: $p(c)
q(\tilde c)$.  Show that the statistical entropy $S\definedas -\sum_c p(c) \log
p(c)$ of the combined system is the sum of those of the two subsystems.

\problem{Positivity of statistical entropy in a discrete system} Show that the
entropy
$S\definedas -\sum_c p(c) \log p(c)$ is greater than zero except when $p(c)$ has
no randomness---\ie it is 1 for one configuration and zero for all the others.
You may use the convexity properties of the entropy discussed in class.

\problem{Smearing and entropy increase}  
{\bf revised 4/12.  For information. 
You need not do this  revised version.}
One way to view an ergodically allowed process is to consider the time
development of any particular state $c$ when the system is allowed to approach
equilibrium.  Even if the system is initially prepared to be with certainty in
state $c$, it need not be in that state with certainty at a later time, if it is
approaching equilibrium.  Instead, it may be in configuration $c'$ with
probability $q(c,c')$.  The ``conditional probabilities" $q$ have the
normalization $\sum_{c'} q(c,c') = 1$, for any $c$.  
Thus if the
initial system is in the configuration $c$ with probability $p(c)$, the final
probabilities $\tilde p(c')$ are given by $\sum_c p(c) q(c,c')$.  Suppose that the
$q$'s obey ``detailed
balance": $q(c, c') = q(c', c)$.  
\item{a)} Show that 
the final set of probabilities obtained
in this way must have a greater entropy $S$ than the initial set unless all
the probabilities remain unchanged: $\tilde p(c) = p(c)$ for all $c$.  This
means that such processes would be adiabatically allowed in the sense defined in class.
%\omit{
\item{\bf solution}  
Because the conditional probabilities $q$ are symmetric, they are normalized 
with respect to their first index as well as their second: $\sum_{c} q(c,c') =
1$.  This means that the final probabilities $\tilde p(c')$ may be viewed as a 
{\it weighted average} of the $p(c)$'s: $$\tilde p(c') = \sum_c p(c) q(c,c')$$
with $\sum_{c} q(c,c') =1$ for all $c'$.  That is, $\tilde p$ has the form
$\expectation{p}_{q}$, where the averaging is performed using the weights $q$.
But because the entropy $S$ is convex, it must increase under averaging so long
as the $\tilde p$'s aren't the same as the $p$'s: $S([\expectation{p}_q)] >
S([p])$.  

%      }

\problem{ Boltzmann distribution}  It was stated in class that the
probability that a system in thermal equilibrium is in a configuration $c$ of
energy $E(c)$ varies as $\exp(-\beta E(c))$.  This is true for any conserved
quantity in a small subsystem of a closed random system.  To illustrate this
point consider the following Museum-of-Science-and-Industry example.    Our
System is a $1000\times1000$ grid of $U=1,000,000$ cells.  We designate a
section of $V=10,000$ cells as our Subsystem.  Into the System a fixed total
of $N=1,000$ identical balls are distributed at random.  (The fixed number $N$
of balls corresponds to a fixed total energy. Just as different pieces of
energy in a system don't have an individual identity, neither do our balls. 
These cells are allowed to contain more than one ball.)  A specific $k$-ball
configuration $c_k$ of the Subsystem is then specified by stating the number
of balls in each of its $V$ cells.  We define $p(c_k)$ as the probability of a
configuration $c_k$ of the Subsystem with $k$ balls.  It is the fraction of
all System configurations in which the Subsystem has the configuration $c_k$. 
We define the configuration of the outer part of the system (which has $N-k$
balls) as $d_{N-k}$.  Then $p(c_k)=(constant) [\sum_{\{d_{N-k}\}} 1]$: the
probability of a given {\it single} Subsystem configuration is given by the
number of System configurations $d_{N-k}$ compatible with it. (Clearly any
configuration with $k$ balls has the same probability as any other.)  
For simplicity in what follows, you may assume that $k<<N$ (Subsystem is small)
and $N<<U$ (System is dilute).
\item{ a)}  
Given an $m$-ball configuration of the outer part of the system (the
``reservoir"), now many $m+1$ ball configurations can be made from it by adding
one ball?
\item{b)} Given an $m+1$ ball configuration of the reservoir, now many $m$ ball
configurations can be made by removing one ball?
\item{c)}
Show that the ratio $p(c_{k+1})/p(c_k)$ is independent of $k$.
(This can be done without calculating the 
(horrendous)
$[\sum_{d_{N-k}} 1]$ explicitly.)  What is this ratio?  {\bf d)} Show that
$p(c_k)$ has the form $p(k=0) \exp(-k/constant)$, and find the constant. 
%\omit{
%>>>>Needs to be fixed up.  replace S by some other symbol.  fix wording
\item{\bf Solution:}  a) The outer part of the system (the ``reservoir") has
$N-k\definedas m$ balls in
$U-V\definedas v$ sites.  The number of distinct configurations with $m$ balls
$\sum_{\{d_m\}} 1$ will be called the ``partition function" $z_m$.  We wish to
infer $z_{m+1}$ given $z_m$.  Any $m+1$-ball configuration may be made from 
some $m$-ball configuration by adding one ball.  Thus to count all $m+1$
configurations we may consider each $m$-ball configuration and ask how many
$m+1$-ball configurations can be made from it.  Whatever $m$-ball configuration
we start from, there are exactly $v$ $m+1$-ball configurations that can be made
from it, since every possible site to add the new ball results in a distinct
configuration.  Each of these $m+1$-ball configurations could have been made from
$m+1$ different $m$-ball configurations.  If we remove any of the $m+1$ balls, we
get an $m$-ball configuration.  If $v>>m$, the chance of two balls at a site is
negligible and so all these $m$-ball configurations are distinct. We may now
count the $m+1$ ball configurations as follows.  
\def \calS {{\cal S}}1) Form  a set $\calS=\{d_m\}$
containing all $m$-ball configurations.  2) pick an
$m$-ball configuration from $\calS$.  3) Form all the $m+1$-ball configurations
that can be made from it.  There are $v$ of these, as noted above.  3) for each
of these, produce all $m$-ball configurations that can be made by removing a
ball.  There are $m+1$ of these as noted above.  4) remove each of these
``redundant"
$m$-ball configurations from the set $\calS$.  5) continue with step 1--4 until
the set $\calS$ is exhausted.  In this way, all the $m+1$ ball configurations
are  produced.
Now, there were $z_m$ elements in $\calS$ to start with, and
$m+1$ are removed at every cycle.  Thus the counting process terminates in $z_m/(m+1)$
cycles.  Each cycle finds $v$ different $m+1$-ball configurations.  The total
found in the $z_m/(m+1)$ cycles is thus $z_m [v/(m+1)]\definedas z_{m+1}$.  We
recall that the desired probability $p(c_k)=(const) z_{N-k}$.  Thus we have found
that $p(c_{k+1})/p(c_k) = z_{N-k}/z_{N-k-1}=(N-k)/v$.  Since $N>>k$, this ratio
is practically independent of $k$, as we wished to show.
\item{}
If the reservoir is not dilute, the ratio $p(c_{k+1})/p(c_k)$ changes.  It
becomes a function of the volume fraction $N/U$ of the reservoir, because the
number of $m$-ball configurations obtained from an $m+1$-ball configuration is
no longer $m+1$ but $m+1$ times a function of $N/U$.  Still, the 
$p(c_{k+1})/p(c_k)$ remains independent of $k$.
%   }

\problem{Statistics of a Classical Dipole} A classical system has a dipole
moment $\vector \mu$.  For definiteness, we take it to be a magnetic dipole, so
that its energy of interaction with a magnetic field is $\mu\cdot B$.  The
configurations of this system are labeled by the directions in which $\mu$ may
point---\ie the unit sphere.  a) Find the partition function of such a dipole in
a magnetic field $B$ and at temperature $T$.  b) Find the average
$\expectation{\mu}$ and average energy, as a function of $B$ and $T$.  c) Find
the low-temperature $T$ dependence of the heat capacity $C_v= dE/dT$. 

\problem{Quantized dipole and magnetic cooling}  An certain electron in a solid
has a magnetic moment $\mu$, which may take on one of two equal and opposite
values---say, up and down.  (It cannot be oriented in any direction in space
like a classical dipole.)  a) Repeat part a) from the preceding problem for this dipole. 
b), c) Repeat Parts b and c.  d) If the external field $B$ is gradually
changed, this cannot change the probability that the dipole is up or down.  Find
the temperature change of the dipole upon reducing the external field from $B_1$
to $B_2$.

\problem{$f_c$ versus $f_E$:}  A system consists of two independent quantum
dipoles like those of the preceding problem.  Each dipole may have energy $\plusorminus
\epsilon$. The dipoles are in equilibrium with a heat reservoir; thus each
configuration $c$ occurs with a probability $const.~ e^{-\beta E_c}$.  Find the
probability that the the total energy is $E$.  Notice that this probability is
not exponential in $E$.  
