Gibbs measures

By NTZung, on April 9th, 2011

These are the notes that I’m taking for myself in order to learn some statistical mechanics.

The main reference is: Anton Bovier, Lecture notes on Gibbs measures and phase transitions (Bonn University)

In thermodynamics one has:

$E = E_{mech} + E_{chem} + E_{thermal}$

where $E_{mech}, E_{chem}, E_{thermal}$ are respectively the mechanical, chemical and thermal components of the energy.

$E_{thermal}$ is also denoted by $Q$ ,

$dE_{mech} = - p dV$

where $p$ is the presure and $V$ is the volume

$d E_{chem} = \mu dN$

where $N$ is the number of molecules, and $\mu$ is the average chemical potential per molecule

$dQ = T dS$

where $T$ is the temperature and $S$ is the volume

Thus we have, for a closed system:

$0 = dE = -pdV + \mu dN + TdS$

We will view the total energy $E$ as a function of three extensive variables $V, N, S$ : $E = E(V,N,S)$ . Then we have

$p = - \partial E / \partial V$ $\mu = \partial E / \partial N$ $T = \partial E / \partial S$

$p, \mu, T$ are called intensive variables. The last three equations are called equations of state.

Instead of viewing $E$ as a function of three variables $V,N,S$ one can also view it as a function of another set of three variables, e.g. $p,N,T$ , etc. We will assume that $E$ is a convex function, so that change is possible. (Legendre transformation). Problem: it can’t be convex ?! Howevern can have a nondegenerate Hessian ?!

Intorduce a function $G$ (of three variables $p,T,N$ ) such that

$dG = Vdp - SdT + \mu dN = d(Vp - ST + E)$

So in fact

$G = Vp - ST + E$ . This sunction is called the Gibbs free energy. Some other useful functions are:

Helmholtz free energy: $F = E - TS$

Enthalpy: $H = E + pV$

In some case $E$ is linear in $V$ : $p = \partial E/\partial V$ is a constant (then we can’t compute $V$ as a function of $p$ ). This situation is called a first order phase transition. (Example: condensation of vapor into water).

1-dimensional model:

$N$ particle, volume $V$ , pressure $f$ , each particle has moment $p_i$

Total energy function (conserved quantity): $H = fV + \sum p_i^2/2$ is a constant, $V \leq V_{max} = H/f$

Assume equi-distribution of all possible positions and momenta $x_i,p_i$ having a fixed $E$ , the level $V$ which has the highest probabality will be (assuming $N$ very large): $V = 2/3 . E/f$ . (This is the level that one will observe in reality).

The probability formula for $v \in dv$ is:

$P(v \in dv) = \frac{dv \exp(Ns)}{\int_0^{v_{max}} \exp(Ns)}$

where

$s = \ln v + 1/2 \ln (2m(u-fv))$ , $u = H/N, v = V/N, e = E/N, E = \sum p_i^2/2$ .

Equilibrium state:

$v = 2/3 u/f, T = 2e, {\partial e \over \partial s} = T.$

Everything works very nicely in this 1-dimensional model ! This model also indicates why the entropy is a log function, and what does it mean to be a micro-state.

ZetaMu

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