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\line{Physics 352 \hfill {\bf Midterm} \hfill \today}
\medskip \hrule \medskip
{\narrower\noindent You may consult your own notes, my handouts, and 
Chandler's text.  To be turned in at the end of class.  If something is 
unclear, please ask. 
\par}
\noindent Total scores: 
29, 27, 25, 22, 21, 21, 20, 19, 13, 11, 8

\problem{classical pendulum} 
A mass $m$ is attached to a support point by a cord of length $\ell$ in a 
gravitational acceleration $g$.
The mass is free to oscillate in the $x$ or $y$ direction, and it is in 
equilibrium with an energy reservoir at temperature $T$.  The temperature is 
small, so that the oscillator is only weakly excited.
\item{a)} If the oscillator is now put in contact with a slightly hotter 
reservoir at temperature $T + \Delta T$, an average amount of heat energy 
$\Delta Q$ flows into it.  Find the heat capacity $C\definedas
\Delta Q/\Delta T$.  Here numerical factors are important.
\item{b)} If the temperature is too large, the oscillator is not
weakly excited, and your result found in a) becomes inaccurate.  For 
what range of $T$ is the formula accurate---\ie $T\muchlessthan what$?  Here numerical factors are not 
important.
\par\noindent Scores: 10, 10, 8, 7, 7, 5, 3, 2, 2, 1

\noindent{\bf solution} 
\item{a)} For this classical system the configurations are specified by 
giving the generalizaed co-ordinates of the system and their conjugate 
momenta.  We may specify the position of the pendulum by the angles 
$\theta_x$ and $\theta_y$.  The energy is then given by 
$$ 
E = \half m g \ell (\theta_x^2 + \theta_y^2) + \half m \ell^2 (\dot 
\theta_x^2 + \dot \theta_y^2)
.$$
The conjugate momentum of this system are $p_x = \partial E/\partial \dot 
\theta_x = m \ell^2 \dot \theta_x$ and similarly for $p_y$. The energy is 
quadratic in all the system variables.  Thus the equipartition theorem 
applies, and each of the 4 degrees of freedom has an average energy of 
$\half T$.  
$\expectation{E} = 4 \half T = 2T$.  If the temperature increases, the heat 
$\dslash Q$ can be written $d\expectation{E}+ \dslash W$.  But since there is no work 
done,  $\dslash Q = d\expectation{E}$, and $C \definedas dQ/dT = 
d\expectation{E}/dT$.  Since $\expectation{E}= 2T$, so $C=2$.
\item{b)} In order for the quadratic energy above to be valid, the angles 
$\theta$ must be small.  Thus $\expectation{E}$ must be $\muchlessthan 
mg\ell$.  Thus the temperature must satisfy $T\muchlessthan mg\ell$.

\problem{odd rubber band} The Helmholtz free energy $A(L)$ of a rubber band 
which is constrained to be {\it uniformly} stretched to length $L$ has the 
form
\def \LL {\tilde L}
$$
A(L) = A_0(- \LL^2 + \LL^4 + 1)
,$$
where $A_0$ is a constant and $\LL \definedas L - L_0$ is the length 
relative to some given $L_0$.  Thus $\LL$ can be negative.  For a certain 
range of $L$, this rubber band cannot be uniformly stretched in stable equilibrium.  
\item{a)}Find this range in terms of the parameters given.
\item{b)}If $\LL = 0$, find the Helmholtz free energy $A$ for the 
nonuniform, equilibrium system.
\par\noindent Scores: 10, 8, 4, 4, 4, 4, 3, 3, 2, 1, 1

\noindent {\bf solution}
\item{a)} The system has an additive variable $L$.  The total $L$ must 
partition itself over the rubber band so as to minimize the thermodynamic 
potential $A$.  Thus if the rubber band is divided into two pieces 
containing a fraction $\alpha_1$ and $\alpha_2 = 1-\alpha$ of the rubber, 
the lengths $\alpha_1 L_1$ and $\alpha_2 L_2$ of the two parts must 
distribute themselves so as to minimize the total free energy.
The free energy of piece $1$ must be $A_1 = \alpha_1 A(L_1)$.  Since $A$ is 
extensive, taking a fraction $\alpha_1$ of the rubber and an equal fraction 
$\alpha_1$ of the length must result in a fraction $\alpha_1$ of the free 
energy.  Adding the free energies of the two pieces
$$
A = \alpha_1 A(L_1) + \alpha_2 A(L_2)
.$$
If the stretching is to be uniform then the length of the left side 
$\alpha_1 L_1$ must be a proportionate fraction of the total length: \viz 
$\alpha_1 L$.  Thus uniform stretching means $L_1 = L_2 = L$.  If the stable 
state is the uniform state its free energy must be lowest:
$$
A(L) \lessthanorequal \alpha_1 A(L_1) + \alpha_2 A(L_2)
,$$
where $L$ is the sum of the two sublengths: $L=\alpha_1 L_1 + \alpha_2 L_2$.  
This is the weighted average of $L_1$ and $L_2$: $\expectation{L}$.  In other words, the 
condition for uniform stretching is the convexity condition
$$
A(\expectation{L}) < \expectation{A}
.$$
As we have seen from a handout, this condition is true if and only if a 
straight line between two points of the $A(L)$ curve lies entirely above 
that curve.  In the given function $A(L)$ this property does not hold for 
points between the two minima.  For $L$ values in this range one may find a 
straight line connecting points on the $A(L)$ curve that lies below the 
point in question.  Thus the free energy may be lowered by forming two 
unequally stretched subsystems of corresponding lengths $L_1$ and $L_2$.  
(These lengths may be found by the Maxwell construction described in Chapter 
2.  The resulting $A(L)$ for the (possibly nonuniform) system must be 
convex.  It is the convex envelope of the initial $A(L)$ function that was 
constrained to be uniform.)
\item{b)}  $\tilde L = 0$ lies in the unstable range.  Here the free energy 
may be lowered by having half of the system at one state of minimum $A$ and 
half at the other.  The total free energy is $A_{min} = A(\tilde L = 
1/\sqrt{2}) = \frac 3 4 A_0 $.  This is also the free energy obtained from 
the constrained one by the Maxwell construction of forming its convex 
envelope.

\problem{freely-jointed chain} A certain molecule can be represented as a 
freely-jointed chain of a large number of segments $N$.  Each segment has a 
fixed length $a$, but may rotate in any direction in three-dimensional 
space. The chain is in thermal equilibrium with a reservoir at temperature 
$T$. A tension $F$ in the $z$ direction is applied to the two ends of the 
chain.  
\item{a)} Find an appropriate thermodynamic potential for one segment of the 
chain.
\item{b)} Find average extension $\expectation{L}$ of the chain in the $z$ 
direction, assuming $F\muchlessthan kT/a$.  
\item{c)} Find the mean-squared fluctuation $\expectation{(L - 
\expectation{L})^2}$ assuming $F\muchlessthan kT/a$.  
\par\noindent Scores: 6, 5, 5, 5, 5, 5, 2, 2, 0, 0, 0

\noindent{\bf solution:}
\item{a)} To find the thermodynamic potential we take the log of the 
partition function $Q \definedas \sum_c f_c$, where $f_c = e^{-\beta E_c}$.  
The energy $E_c = F L$, where $L$ is the length projected in the direction 
of the force.  If a given link makes an angle $\theta_i$ with the $F$ axis, 
then $E_c = \sum_i F a \cos \theta_i$.  Since the energy is a sum of link 
energies, we may take each link to be an independent subsystem.  Its 
partition function is the sum of $f_i$ over all orientations of the link.
$$ Q_i = \integral d\Omega e^{-\beta F a \cos \theta}
= \integral 2\pi d (\cos \theta) e^{-\beta F a (\cos \theta)}
= 2\pi (\beta F a)^{-1} (e^{\beta F a} - e^{-\beta F a})
.$$
Finally the thermodynamic potential for a link $A_i = -T \log Q_i$.  The 
thermodynamic potential for the whole chain $A = N A_i$.
\item{b)} $\expectation{L} = \sum_c L e^{-\beta F L} / Q = \partial A 
/\partial (\beta F)$.  For small $\beta$ $Q_i$ may be expanded:
$$
 Q_i \goesto {2\pi(2 \beta F a + \frac 2 6 (\beta F a)^3) \over \beta F a}
= 4\pi(1 + \frac 1 3 (\beta F a)^2) 
.$$
Thus 
$$
\expectation{L} = -N\partial \log Q_i/\partial (\beta F) 
= \frac 2 3 (\beta F a) a
.$$
\item{c)} From Chapter 3, 
$$
\expectation{(L - \expectation{L})^2} = \partial \expectation{L}/\partial 
(\beta F) = \frac 2 3  N a^2
,$$
independent of $F$.  (As expected for an extensive system, 
$\expectation{(L - \expectation{L})^2}/\expectation{L}^2 \goesas 1/N$ .)   

\problem{density matrix}  A sample of independent spin-one atoms is 
prepared in a state described by the density matrix
$$\hat \rho_a \definedas \left (\matrix{
.1& 0& 0   \cr
0& .1& 0  \cr
0& 0& .8   \cr
}\right )
\quad {\rm ~or~}\quad
\hat \rho_b \definedas \frac 1 {20} \left (\matrix{
10& 0& 0 \cr
0& 9& 3 \cr
0& 3& 1 \cr
}\right )$$ 
\item{a)} Find the entropy (per atom) of the system described by $\hat 
\rho_a$ 
\item{b)} Find the entropy (per atom) of the system described by $\hat \rho_b$.

\item{c)} Which system---a or b---could not evolve into the other spontaneously 
in a closed system, by releasing constraints.  
\par\noindent Scores: 10, 10, 10, 10, 10, 9, 8, 7, 6, 0, 0

\noindent {\bf solution}
\item{a)} If $\hat \rho = \sum_c \ket{c} p_c \bra{c}$, the $S = -\sum_c 
p_c \log(p_c)$.  For $\hat \rho_a$ the matrix is diagonal, so the 
eigenvalues $p_c = (.1, .1, .8)$.  Thus $S = -(.1 \log .1 + .1 \log .1 + .8 
\log .8) = .639$.
\item{b)} The eigenvalue equation $\det(\hat \rho_b - p_c \hat 1) = 0$ can 
be
written 
$$
\det \left ( \matrix{10 - \lambda& 0& 0\cr
0& 9 - \lambda& 3\cr
0&  3& 1 - \lambda}
\right ) = 0
,$$
where $\lambda = p_c/20$.
Expanding,
$$ 
0 = (10 - \lambda)[(9 - \lambda)(1-\lambda) - 9]
.$$
Evidently one solution is $\lambda=10$ (already obvious since $\hat \rho_b$ 
was block-diagonal).  For the other solutions, we expand the $[...]$:
$$0 = [...] = 
9 - \lambda - 9 \lambda + \lambda^2 - 9
= \lambda(\lambda -10)
.$$
Thus $\lambda = (10, 10, 0)$ and $p_c = \lambda/20 = (.5, .5, 0)$.
$S = -.5 \log(.5) - .5 \log .5 = \log 2 = .693$.
\item{c)} An isolated system cannot spontaneously evolve into one of lower 
entropy.  Since $S_b > S_a$, system b cannot spontaneously evolve into state 
a.  (No one guarantees us that system a can evolve spontaneously into system 
b.)

\end
                                        notes
ok      question on adiabatically allowed change
        question on equipartition
ok      question on composite probability
        question on monte carlo
        question on legendre transform
        question on convexity
ok      question on fluctuations
ok      question on density matrix