%\input twocolumn
\line{Physics 352 \hfill \today \hfill Handout A}
\beginsection Convexity property of thermodynamic potentials 

The extremal properties we have noted for the entropy can be expressed in a more
general and powerful way. The entropy is a so-called ``convex" function of its
variables. We will have reason to exploit the convexity of the entropy later. In
any case, it is good to know about this property. In addition, it applies
generally to thermodynamic potentials.  We first discuss the meaning of
convexity in the simplest case of its effect on $S(E)$. Then we generalize to
arbitrarily many variables and to other thermodynamic potentials.

In class we considered constraining variables $y$ of a system divided into
subsystems. In an isolated overall system, for example, one may consider two
systems 1 and 2 which initially are isolated from each other. We suppose that
each subystem is a sample of the same substance, which has a well-defined $S(E,
{\bf X})$. Then the two subsystems have their own energies $E_1$ and $E_2$. If
the constraint of isolating the two systems is released, then the entropy must
increase:
$$
S(E_1 + E_2, 2{\bf X}) > S(E_1, {\bf X}) + S(E_2, {\bf X})
. \eqno(1)$$ 
The same is true if three
subsystems are involved, as we see by using the above result twice:
$$S(E_1 + E_2 + E_3) > S(E_1 + E_2) + S(E_3) > S(E_1) + S(E_2) + S(E_3) .$$ 
Evidently this reasoning can be extended to larger number $k$ of subsystems are
involved $S(\sum_{i=1}^k E_i, k{\bf X}) > \sum_{i=1}^k S(E_i, {\bf X})$. By dividing
both sides by $k$ and using the extensivity of $S$ we may convert the sums
into averages:
$$S(\frac 1 k \sum_{i=1}^k E_i, {\bf X}) > \frac 1 k \sum_{i=1}^k S(E_i, {\bf X})
{\rm~or~} S(\expectation{E}, {\bf X}) > \expectation{S(E, {\bf X})}.$$ 
Finally, we could imagine a continuous sampling of energies where the fraction of
subsystems between energy
$E$ and $E + dE$ is
$p(E) dE$. Now the limiting form of $\frac 1 k \sum_i$ becomes
$\integral dE~ p(E)$.  Then the average $E$ or
$\expectation{E}$ is
$\integral dE~ p(E) E$. When $S(\expectation{E}, {\bf X}) > \expectation{S(E, {\bf X})}$
holds for any distribution $p(E)$, we say that the function is a {\it convex}
function of its argument.  

The convexity property extends to all the system co-ordinates that
$S$ depends on---\viz ${\bf X}$. Like $E$, the ${\bf X}$ variables may be divided among
subsystems and then allowed to pass between these subsystems. Thus we may say
generally that $S$ is a convex function of all its variables:
$$S(\expectation{E}, \expectation{{\bf X}}) > \expectation{S(E, {\bf X})}.$$

Convexity implies negativity of second derivitives, as noted in Chandler.  To
see this, we choose $E_1$ and $E_2$ very close together.  In this neighborhood
$S(E)$ is well approximated by its Taylor expansion:
$$S(E_1) - S(\expectation{E}) \goesto {dS \over dE} (E_1 - \expectation{E}) 
+ \half {d^2 S \over dE^2} (E_1 - \expectation{E})^2.$$

Thus 
$$S(E_1) + S(E_2) - 2 S(\expectation{E}) \goesto
{dS \over dE} (E_1 + E_2 - 2\expectation{E}) +
\half {d^2 S \over dE^2} [(E_1 - \expectation{E})^2 + (E_2 - \expectation{E})^2]
$$

The $dS/dE$ term vanishes, since $E_1 + E_2 = 2\expectation{E}$.  The
left side is $2 \expectation{S} - 2 S(\expectation{E})$; because
$S$ is convex, this left side must be negative.  But the squared energies in
$[...]$ must be positive.  Thus the second derivitive is negative.  The same
reasoning  would have worked if the two subsystems had differed in one of the
${\bf X}$ variables instead of in energy.  Thus all second derivitives of $S$
$\partial^2 S / \partial x_i^2$ must be negative.  More generally, we could
have defined mixed variables $\tilde {\bf X}$ that were linear combinations of $E$
and ${\bf X}$.  Then changing \eg $\tilde x_1$ slightly would amount to changing $E$
and the ${\bf X}$'s simultaneouly.  The convexity of $S$ again leads to the
conclusion that $\partial^2 S/\partial \tilde x_i^2 < 0$.  The negativity of
second derivitives holds for an arbitrary choice of system co-ordinates.

In deriving this second derivitive information, we used our convexity property
only for nearly identical states.  But remarkably, this second-derivitive
information is just as powerful as the full convexity property.  We may derive
the full convexity from the negativity of second derivitives.  To see this, we
consider states with $E_1$ and $E_2$ at an arbitrary distance from each other. 
The convexity property can be represented graphically by Figure 1.
\midinsert
$$
{\scaledpicture 6.82in by 5.88in (convex scaled 600)}
$$
\endinsert
Because the second derivitive is negative everywhere between $E_1$ and $E_2$,
the straight line must lie entirely below the curved line representing $S(E)$. 
Thus the midpoint of the straight line, whose height is $\expectation{S}$ must
lie below the corresponding point on the curve, whose height is
$S(\expectation{E})$.  From this fact we may again apply the reasoning after
Eq. 1 to conclude that any function
$S(E)$ must be convex in
$E$ if its second derivitive is negative everywhere.  Now, the second derivitive
of our
$S$ is negative with respect to any variable $\tilde x_i$.  Thus we could have
drawn the corresponding graph of $S$ \vs $\tilde x_i$ and concluded that $S$
was convex in $\tilde x_i$.  This means that negativity of the second
derivitives everywhere is enough by itself to give global convexity.  Since the
negative-second-derivitive property is equivalent to convexity, it is often
{\it called} convexity.

The convexity property is preserved for all thermodynamic potentials---$E(S,
{\bf X}), A(T, {\bf X})$, etc.  But since these potentials tend to a minimum
instead of a maximum, their second derivitives are positive rather than
negative---they are ``convex up" instead of ``convex down".  Chandler shows that
the energy second-derivitives are positive.  From the above reasoning, this
means that
$\expectation{E(S, {\bf X})} < E(\expectation{S},
\expectation{{\bf X}})$.  The same is true for the other potentials when they are
stated in terms of their proper variables. 

\end