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\line{Physics 352 \hfill \today \hfill Handout B}
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\beginsection Solution to Problem 1.3

This amounts to demonstrating a mathematical property of Legendre 
transforms.  Thus we consider some function $f(Y, X)$ and its Legendre 
transform $g(Y, U)$ following p. 19 of Chandler.  Here $X$ and $Y$ are lists 
of variables and $U$ is the set of partial $X$ derivitives: $u_i = \partial 
f \partial x_i$.  We now suppose that if the variables $X$ are fixed, $f$ 
has a minimum for a certain $Y$.  We may say this is where $Y=0$.  We want 
to show there is a corresponding minimum in $g$.  We shall only consider 
local minimum properties here, since this is all that the book deals with 
and all that Callen deals with.  

Since $f$ is a local minimum, 
$$\Delta f \definedas f(Y, X) - f(0, X) > 0 
$$
for all $Y$'s sufficiently close to 0.
For simplicity, suppose that only one of the $Y$'s, denoted $y$, changes.  
Then we can write the change of $f$ as
$$\Delta f =\integral_0^y dy' \partial f/\partial y
.$$
Since the integral must be positive for all small-enough $y$'s, the 
integrand must be positive over at least part of that region.  Thus for 
small enough $Y$
$\partial f/\partial y > 0$.
Now consider the change of $g$:
$$g(Y, U) - g(0, U) = \integral_0^y dy' (\partial g/\partial y )_U
.$$
But the partial derivitive can be expressed in terms of the untransformed 
function $f$: a Legendre transform preserves partial derivitives with 
respect to untransformed variables.  Thus 
$$\Delta g \definedas g(Y, U) - g(0, U) = \integral_0^y dy' (\partial f/\partial
y )_{X(Y, U)} .$$
Here the subscript $X(Y, U)$ means that the derivitive is at constant $X$.  
But as $y$ changes during the integration, each of the partials must be 
taken at a {\it different} $X$ value: \viz, that which maintains the given 
value of $Y$ and the given $U$.  

Fortunately, there is no need to perform this integral.  Since $\Delta f$ is 
known to be positive, we have shown that  $(\partial f/\partial y )_X$ must 
be positive as well in a small region around $Y=0$.  Thus the integrand for
the $\Delta g$ integral is positive as well.  The integral itself must  be
positive, and so $\Delta g$ must be positive, as claimed.  

\beginsection Thermodynamic potentials and work

As with the potential  energy of a mechanical system, the value of a
thermodynamic potential is the {\it work}  required by an external agent to
change its variables.  This property is a  direct property of the differential
properties of Legendre transforms, as  discussed in Chandler p. 19.  Thus, \eg
for an  isolated system with an entropy $S$ and conserved system co-ordinates
$X$,  we might change the values of the $X$'s by applying some external force.  
For example if a system of volume $V$ were partitioned into two subsystems  of
volumes $V_1$ and $V - V_1$, we could change $V_1$ by moving a wall  between the
two subsystems.  But this would required; the amount needed  would be the change
in the energy $E(S, X)$---\ie the change of the  thermodynamic potential for the
system.  

Now if the system is coupled to a reservoir of \eg heat at temperature $T$ 
the appropriate thermodynamic potential is the Helmholtz free energy $A 
\definedas E - TS$.   If we change one of the $X$ variables by operating on 
our system, the work required is no longer the change in $E$, for now energy 
will be transferred to the reservoir as well as to the system.  
We consider the incremental work $\dslash W$ required for a small 
displacement one of the $X$ variables, denoted $x$.  In the original 
isolated system this $\dslash W$ was $f dx$, where the generalized force $f$ 
is simply $(\partial E/\partial x)_S$.  This force is a property of the 
system itself, irrespective of how $x$ might be changed.  It must remain the 
same if the change in $x$ is now made with a reservoir present so that $T$ 
rather than $S$ is fixed.  Thus in general $\dslash W = (\partial E/\partial 
x)_S dx$.  This work is readily related to the change of thermodynamic 
potential $A$.  The fundamental property of Legendre transforms that we 
noted in class was that they preserve derivitives with respect to 
untransformed variables like $x$.  Thus $(\partial A/\partial x)_T = (\partial 
E/\partial x)_S$.  Thus the differential work can be expressed as $\dslash W 
= (\partial A/\partial x)_T dx$.

We may now add these incremental works to obtain the total work for a finite 
change of $x$: $W = \integral_{x_0}^x dx' (\partial A/\partial x )_T 
$.  Since only the $x$ variable is changing within the integration, the 
right hand side is evidently $\Delta A \definedas A(T, X) - A(T, X_0)$.  The 
work done is equal the change of thermodynamic potential, as claimed.   It 
might appear that the same reasoning could be used to show that $W = 
 \integral dx' (\partial E/\partial x)_S  =_? \Delta E$.  But second equality 
is not valid.  Although the partial derivitives in the integral are each 
taken at constant $S$, the various $dx'$s all have different values of $S$
like the $X(Y, U)$ of the previous section.   Thus we cannot express this
integral as a change of
$E$.  
 
This work property justifies and generalizes the minimal property discussed in
Problem 1.3.  For any system with a constraint variable $Y$, it ought to require
positive work to change $Y$ from its equilibrium value 0 to a different value. 
This is true whether the change in $Y$ is small or large.  And it is true whether
the system is isolated or is attached to reservoirs.  Since $W$ is positive
for all changes of $Y$'s from equilibrium, the change of the thermodynamic
potential must be positive as well.