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Next: Geometric Interpretation and Phase Up: Phase Equilibria Previous: Phase Equilibria

Subsections


Free Energy Minima and Phases

Phase Coexistence

Homogeneous state:
All part of my system are alike
Inhomogeneous state:
Some parts are different:
\begin{figure}
 \InputIfFileExists{2phases.pstex_t}{}{}
 \end{figure}
Definition:
Macroscopic states, coexisting at different conditions, are called phases. Each phase exists in the thermodynamic limit $N\to\infty$
Question:
Is microphase separation in co-polymers a true phase equilibrium?
\begin{figure}
 \InputIfFileExists{microphase.pstex_t}{}{}
 \end{figure}
Answer:
No, we do not have separate independent macroscopic phases! This is just one microscopically inhomogeneous macroscopic phase.

Free Energy Minimum

Problem

Suppose our system is at constant temperature. Total free energy is

 
A(N1,N2,V1,V2,T) = A1(N1,V1,T) + A2(N2,V2,T)

(1)

with  
 \begin{displaymath}
 N_1+N_2=N,\qquad V_1 + V_2 = V \end{displaymath} (2)
We want to minimize (1) under conditions (2).

Method of Lagrange Multipliers

We want to minimize f(x,y) under conditions g(x,y)=0. Construct a new function

\begin{displaymath}
\tilde f(x,y,\Lambda) = f(x,y) - \Lambda g(x,y)\end{displaymath}

and minimize this as function of independent variables x, y, $\Lambda$. Minimizing by $\Lambda$ we obtain g(x,y)=0, i.e.

\begin{displaymath}
\min_{\Lambda} \tilde f (x,y,\Lambda) = f(x,y)\end{displaymath}

Example:
I want to fence a rectangular lot of the given area A. How can I save money on the fence?
\begin{figure}
 \InputIfFileExists{fence.pstex_t}{}{}
 \end{figure}

Solution:
For an $x\times y$ lot the area is A=xy, the fence is f=2x+2y long. We want to minimize

\begin{displaymath}
f(x,y)=2x+2y\to\min,\quad xy-A=0
 \end{displaymath}

We minimize:

\begin{displaymath}
\tilde f(x,y,\Lambda) = 2x+2y - \Lambda(xy-A)\to\min
 \end{displaymath}

and obtain

\begin{displaymath}
\left\{
 \begin{aligned}
 2 - \Lambda y &= 0\\  2 - \Lambda x &= 0\\  xy - A &= 0
 \end{aligned} \right.
 \end{displaymath}

The result is $x=y=\sqrt{A}$--I'd better make a square lot.

Minimization

To minimize free energy (1) under conditions (2), we minimize a new function  
 \begin{displaymath}
 \begin{gathered}
 \tilde A = A_1(N_1,V_1,T) + A_2(N_2,V_2,T)\\  - \Lambda_N(N_1+N_2-N) -
 \Lambda_V(V_1+V_2-V)\end{gathered}\end{displaymath} (3)

Minimization with respect to V1 and V2:

\begin{displaymath}
\left\{
 \begin{aligned}
 \left(\frac{\partial A_1}{\partial...
 ...ial V_2}\right)_{N_2,T}
 -\Lambda_V &= 0
 \end{aligned} \right.\end{displaymath}

The derivatives here are just (minus) pressures! We obtained:
1.
In equilibrium both phases have equal pressures (mechanic equilibrium) P1=P2=P
2.
Lagrange multiplier $\Lambda_V$ is minus pressure of the system: $\Lambda_V = -P$

Minimization with respect to N1 and N2:

1.
Both phases have equal chemical potentials $\mu_1=\mu_2=\mu$
2.
Lagrange multiplier $\Lambda_N$ is the chemical potential of the system: $\Lambda_N=\mu$

Conclusion

In coexisting phase P, T, $\mu$ are equal (as well as other thermodynamic fields).

Slightly Different View:

Let's forget about phase 2. The condition for N1 for phase 1 can be obtained from minimization a function

\begin{displaymath}
\tilde A_1(N_1,V_1) - \mu_0 N_1\end{displaymath}

with external pressure and chemical potential P0 and $\mu_0$.In equilibrium $\tilde A$ is $\Omega$-potential (we already derived this in previous lectures ).

Stability Condition:

The extremum should be a minimum $\Rightarrow$ second derivatives are positive!

General Situation

Suppose we have an extensive variable xi. It has conjugated field

\begin{displaymath}
X_i = - \left(\frac{\partial S}{\partial x_i}\right)_{V,N,\d...
 ...= 
 \left(\frac{\partial A}{\partial x_i}\right)_{V,N,T,\dots} \end{displaymath}

In equilibrium Xi is constant in all phases, and the matrix $\left\lVert \beta_{ik} \right\rVert$ with

\begin{displaymath}
\beta_{ik} = - \frac{\partial^2 S}{\partial x_i \partial x_k} =
 \frac{\partial^2 A }{\partial x_i \partial x_k}\end{displaymath}

is positive (in equilibrium S has maximum, and A minimum!). We can obtain this by minimizing

\begin{displaymath}
\tilde A = A - X_i^{(0)}x_i\end{displaymath}

with external value Xi(0)
next up previous
Next: Geometric Interpretation and Phase Up: Phase Equilibria Previous: Phase Equilibria

© 1997 Boris Veytsman and Michael Kotelyanskii
Thu Oct 2 21:02:12 EDT 1997