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


Free Energy Minima and Phases

Phase Coexistence

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

Free Energy Minimum


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)


 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

\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.

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

I want to fence a rectangular lot of the given area A. How can I save money on the fence?

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

f(x,y)=2x+2y\to\min,\quad xy-A=0

We minimize:

\tilde f(x,y,\Lambda) = 2x+2y - \Lambda(xy-A)\to\min

and obtain

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

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


To minimize free energy (1) under conditions (2), we minimize a new function  
 \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:

 \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:
In equilibrium both phases have equal pressures (mechanic equilibrium) P1=P2=P
Lagrange multiplier $\Lambda_V$ is minus pressure of the system: $\Lambda_V = -P$

Minimization with respect to N1 and N2:

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


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

\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

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

\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

\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