\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<>>>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 $E$.