Subsections

• General Condition
• Stability and Metastability
• Concave and Convex Functions
• Once again Legendre Transforms
• Fluctuations of Volume. Gauss Distribution

Thermodynamic Stability

General Condition

The system is stable if the total entropy $S_{\text{tot}}$ has maximum. Equivalently, $R_{\min}$ is positive for any deviation.

It means that:

• At $T=\mathit{const}$, $V=\mathit{const}$ A should be minimal
• At $T=\mathit{const}$, $P=\mathit{const}$ G should be minimal
• At $P=\mathit{const}$, $S=\mathit{const}$ H should be minimal

Stability and Metastability

There are two kinds of minima--local and global

• If the system is neither in a global, nor in a local minimum, it is unstable
• If the system is in the global minimum, it is stable
• If the system is in the local minimum, it is metastable: it can spend a lot of time there, but eventually it will leave for the global minimum. This is called partial equilibrium.

Glasses--many metastable minima:

Concave and Convex Functions

A body at constant pressure and temperature. Equilibrium condition: $G\mapsto\min$

What is the equilibrium volume V for the given pressure?

$$A(V)+P_0V \mapsto \min$$

It means that:

1.

First derivative should be zero:

$$P = -\left(\frac{\partial A}{\partial V}\right)_T = P_0$$

This is mechanical equilibrium condition

2.

Second derivative should be positive:

$$\left(\frac{\partial^2 A}{\partial V^2}\right)_T \gt$$

This is convexity condition.

Definition:

A function f(x) is convex if f''(x)>0 and concave if f''(x)<0:

Mnemonics:

Concave = ``no coffee''

We proved, that A is a convex function of volume.

If $(\partial^2 A/\partial V^2)_T\gt$, we can make this point V a minimum by changing P. If $(\partial^2 A/\partial V^2)_T<0$, we cannot help! Since

$$\left(\frac{\partial A}{\partial V}\right)_T = -P$$

we proved that

$$\left(\frac{\partial P}{\partial V}\right)_T <0$$

General Statement:

Thermodynamic potentials are convex functions of their extensive arguments.

Once again Legendre Transforms

We now can prove that Legendre Transform of thermodynamic potentials is always possible!

$$p=f'(x), \quad p=p(x) \leftrightarrow x = x(p)$$

To invert p(x), we need $p'(x)\ne0$. But this follows from the convexity condition p'(x)=f''(x)>0!

Theorem:

Thermodynamic potentials are concave functions of their intensive variables.

Proof:

If g(p) is the Legendre transformation of f(x), then g'(p)=-x and

$$g''(p)= - \frac{dx}{dp} = -\left(\frac{dp}{dx}\right)^{-1} <0$$

Fluctuations of Volume. Gauss Distribution

We know the average volume of a given body V₀. What is the probability to obtain V?

We start from the formula:

$$\Prob(V)\sim\exp(-\Delta G/kT)$$

We obtain:
$$\begin{multline*} \Delta G = \Delta A + P\, \Delta V =\ \left(\frac{\partial ... ...^2 A}{\partial V^2}\right)_{T}\,(\Delta V)^2 + P\,\Delta V+\dots\end{multline*}$$

Since

$$\left(\frac{\partial A}{\partial V}\right)_T = -P$$

we are left with

$$\Prob(V)\sim\exp\left[ - \frac{1}{2kT}\left(\frac{\partial^2 A}{\partial V^2}\right)_{T}\,(\Delta V)^2\right]$$

Isothermic compressibility:

We define

$$\alpha_T = - \frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$$

This is an intensive variable (why?).

Then

$$\frac12\left(\frac{\partial^2 A}{\partial V^2}\right)_{T} =... ...(\frac{\partial P}{\partial V}\right)_T = \frac{1}{V\alpha_T}$$

and

$$\Prob(V)=\frac{1}{\sqrt{2\pi kTV \alpha_T}}\exp\left[-\frac{(\Delta V)^2}{2kT V\alpha_T} \right]\,dV$$

This is called Gauss Distribution

Average values:

$$\left\langle \Delta V\right\rangle = 0,\quad \left\langle (\Delta V)^2\right\rangle = kTV\alpha_T$$

and

$$\frac{\sqrt{\left\langle (\Delta V)^2\right\rangle}}{V} = \frac{\sqrt{kT\alpha_T}}{\sqrt V}$$

Once again,

$$\alpha_T\gt,\quad \left(\frac{\partial P}{\partial V}\right)_T <0$$

Once again maximal term method: at $V\to\infty$ the width becomes zero!