Subsections

• Phase Coexistence
• Free Energy Minimum
  • Problem
  • Method of Lagrange Multipliers
  • Minimization
    • Minimization with respect to V₁ and V₂:
    • Minimization with respect to N₁ and N₂:
    • Conclusion
    • Slightly Different View:
    • Stability Condition:
  • General Situation

Free Energy Minima and Phases

Phase Coexistence

Homogeneous state:

All part of my system are alike

Inhomogeneous state:

Some parts are different:

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?

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)

with  

$$N_1+N_2=N,\qquad V_1 + V_2 = V$$ (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

$$\tilde f(x,y,\Lambda) = f(x,y) - \Lambda g(x,y)$$

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

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

Example:

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

Solution:

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

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

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

Minimization

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

$$\begin{gathered} \tilde A = A_1(N_1,V_1,T) + A_2(N_2,V_2,T)\\ - \Lambda_N(N_1+N_2-N) - \Lambda_V(V_1+V_2-V)\end{gathered}$$ (3)

Minimization with respect to V₁ and V₂:

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

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

$$\tilde A_1(N_1,V_1) - \mu_0 N_1$$

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 x_i. It has conjugated field

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

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

$$\beta_{ik} = - \frac{\partial^2 S}{\partial x_i \partial x_k} = \frac{\partial^2 A }{\partial x_i \partial x_k}$$

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

$$\tilde A = A - X_i^{(0)}x_i$$

with external value X_i⁽⁰⁾