
Mechanically, a potential
A(r) was the difference
A(r_{2}) –
A(r_{1}) and equal to the work needed to go
from r_{1} to
r_{2}. 


The decisive point was that A(r)
did not depend on the particular way chosen
to go from r_{1} to
r_{2} as long as the acting forces F
(r) were given as the derivative of the potential
A(r) with respect to the coordinates. We have 


F(r) 
= 
– grad [A(r)] = – 
[A(r)] = –

æ
è 
¶A
¶x 
, 
¶A
¶y 
, 
¶A
¶z 
ö
ø 



These are the well known relations for mechanic
(and electrostatic) potentials. If one knows the potential and the momentum of
a massive (and charged) particle, one knows everything one can know and needs
to know. 


The history, i.e. how the particle came to its present
position r with the potential
A(r) and momentum, is totally irrelevant. 


Potential and momentum
together then define the state of the
particle. 

We may first generalize the idea of a potential by
allowing generalized coordinates,
i.e. any variables (and not just space coordinates) that describe the state of
a system. 


This allows to treat thermodynamic systems consisting of
many particles, where individual
coordinates loose significance and average values describing the system take
precedence. 

Lets consider the
free energy as a first example. It is a
thermodynamic potential and at the same
time a state function, i.e. it describes
completely the state of systems with the
generalized coordinates temperature T, volume V and
particle numbers N_{i}. The mechanical potential by
itself, in contrast, is not a state
function (we would need the momentum, too, to describe the state of a
mechanical system). 


A change of state thus necessarily demands a change in at
least one of the three generalized coordinates from the above example. 


In formulas we write 


F 
= 
thermodynamic
potential 
= 
F(V, T, N_{1}, ...
N_{i}) 




The N_{i} denote the number of particles
of kind i. 

Is it
that easy? Can we elevate all kinds of functions to the state of thermodynamic
potentials and state functions? 

The answer, of course, is
No! The statement that the function
P(x_{i}) is a potential with respect to the
generalized coordinates x_{i} is only true if a number of conditions are met.
For the case of equilibrium (which requires
that nothing changes and therefore that all ¶P/¶x_{i} = 0) those requirements
are: 


The values of the generalized coordinates describe the state
of the system completely (this means you
have the right number and the right kinds
of coordinates). 


The value of P(x_{i}) for a set
of values of the generalized coordinates is independent from the path chosen to arrive at the
particular state and thus from the history of the system. In formulas: 


DP 
= 
change in P
between a state 1
and a state 2 
= 
2
ó
õ
1 
dP 
= 
2
ó
õ
1 
¶P
¶x_{1} 
· dx_{1} + 
2
ó
õ
1 
¶P
¶x_{2} 
· dx_{2} + ... 
:= P_{2} –
P_{1} 




i.e. the difference depends only on the starting and end state. 

This can only be true if
dP is a total
differential of P, i.e. 


dP 
= 
¶P
¶x_{1} 
· dx_{1} + 
¶P
¶x_{2} 
· dx_{2} + 
.... 




i.e. the changes in the generalized coordinates describe
unambiguously the total change in P. 

The numerical values of the partial derivatives of
the thermodynamic potentials describe the "effort" it takes to change
the potential by fiddling with the particular coordinate considered. This can
be understood as a generalized force. If
the derivative happens to be taken with respect to a space coordinate of the system, the generalized force is a real mechanical force  it describes the change of
energy with space as it should. 


If the derivative happens to be taken with respect to a
particle number, however, the resulting
quantity is not called, e.g., "particle number
changing force" (which would have been a perfectly good name),
but chemical
potential, which might be a bit confusing at times. 

How do we know if a given function is a
thermodynamic potential and a state function? For a given function (in
differential form), it is not necessarily obvious if it is a total
differential. You have to resort to physical or mathematical reasoning to find out. Lets first look
at an example of physical reasoning: 

The first
law of thermodynamics was defined as follows: 




Are these three differential
quantities total differentials and thus state function and thermodynamic
potentials, or are they not? 


Physical reasoning tells
us that dU must be a total
differential because U must be a thermodynamic state function 
otherwise we can construct a perpetuum
mobile! Lets see why: 


If we start from some value U_{1},
characterized by as many variables x_{i} (= coordinates)
as you like (e.g. pressure p_{1}, volume
V_{1}, temperature T_{1}) and than
move to a second value U_{2} by adding, for example, a
heat quantity Q; we have described a path 1 to move from
U_{1} to U_{2}. 


If we return to the same values of the generalized coordinates
by a different path; e.g. by now extracting some mechanical work, but have a
value U_{1}' that is not identical to
U_{1}, we are now in a position to construct a cycle
between the states 1 and 2 as characterized by the generalized
coordinates that allows us to extract work in every cycle  we have a perpetuum mobile. 


So dU must be a total differential  and this
requires that dQ and dW are not total differentials! Because it is entirely
possible to go from one state of U to another one by adding
different amounts of Q and W  the path described
by these circumstances must not matter! 


DQ =
Q_{2} – Q_{1} or DW = W_{2} –
W_{1} thus are not independent of the path, therefore they
are not state functions and their
differentials cannot be total differentials. This is essentially tied to the
fact that the system entropy may change in these cases. 

Now lets turn to mathematical reasoning. Obviously, if a function
P(x_{i}) of several variables
x_{i} is given, it is always possible to calculate the
total differential dP(x_{i}). But the reverse is not necessarily true: 


If P is given in differential form, it always
can be written as 


dP(x,y) 
= 
g(x,y) · dx + f(x,y) ·
dy 




We used only two variables x and y
for simplicities sake. The functions g(x,y) and
f(x,y) are arbitrary functions  what ever you like is
allowed. 

If
dP is a total differential, we have necessarily 


g(x,y) 
= 
¶P
¶x 



f(x,y) 
= 
¶P
¶y 




On the other hand for all
functions P(x,y) the following equalities must obtain 


¶
¶y 
æ
ç
è 
¶P
¶x 
ö
÷
ø 
= 
¶
¶x 
æ
ç
è 
¶P
¶y 
ö
÷
ø 




This requires that 




Only if the above relation is fulfilled, is
dP(x,y) = g(x,y)dx + f(x,y)dy a total differential.
So we have a simple checking procedure for a given differential function (a
mathematical rule) to find out if it is indeed a total differential. 

