Subsections

• General Problem
• Energy for a Classical System
• Same or Different Particles?
• Factorization
• When This Does Not Work?
• Kinetic part. Maxwell Distribution
• Internal Part
• Epitome

Many-Particle Systems

General Problem

Suppose we have N identical particles 1, 2,.... We want to describe their thermodynamics--i. e. calculate partition function.

Mixtures:

N₁ particles of kind 1, N₂ particles of kind 2, ....

General formula:

Gibbs law  

$$Q = \sum_i \exp(-E_i/kT)$$ (1)

What is a state in this formula?

Quantum Answer:

I have a number of quantum states for each particle. The state is the set of occupation numbers n₁, n₂, ...

Consequence:

For quantum mechanics the grand canonical ensemble and $\Omega$ potential are more natural than canonical ensemble and free energy!

Classical Answer:

I need coordinate q_i and momentum p_i for each of my particles plus information about the internal degrees of freedom $\tau_i$ for each particle.

We will discuss only classical case.

Energy for a Classical System

• Kinetic energy. In the non-relativistic case it depends only on the momenta:

  $$K = \sum_{i=1}^N \frac{p_i^2}{2m}$$

• Potential energy. In the classical case it depends only on the coordinates:

  $$U = U(q_1,q_2,\dots)$$

• Internal energy. Depends on the internal degrees of freedom $\tau$ (often they are quantum--even if everything else is classical!):

  $$E_{\text{internal}} = \sum_{i=1}^N\epsilon(\tau_i),\quad \tau_i=1,2,\dots,k$$

Same or Different Particles?

Boltzmann principle:

If the particles are identical, exchanging them will not change the state. This is a consequence of Quantum Mechanics.

For distinguishable particles classical partition function is:

$$\begin{gathered} \frac{1}{(2\pi\hbar)^{3N}}\int d\tilde p\,... ...au} = \sum_{\tau_1=1}^k \dots \sum_{\tau_N=1}^k \end{gathered}$$

However, they are identical. States are overcounted--divide by the number of permutations:  

$$Q(N,V,T) = \frac{1}{N!}\frac{1}{(2\pi\hbar)^{3N}}\int d\tilde p\,d\tilde q\,\sum_{\tilde\tau} \exp\bigl(-E/kT\bigr)$$ (2)

This is called Boltzmann Statistics

Factorization

Since

$$E(\tilde p,\tilde q,\tilde\tau)= K(\tilde p) + U(\tilde q) + E_{\text{internal}}(\tilde \tau)$$

we can write:  

$$Q(N,V,T) = \frac{1}{N!}Q_{\text{kin}}Q_{\text{internal}}Z_N$$ (3)

Kinetic part

$$Q_{\text{kin}} = \frac{1}{(2\pi\hbar)^{3N}}\int\exp\bigl(-K(\tilde p)/kT)\,d\tilde p$$

Internal part

$$Q_{\text{internal}} = \sum_{\tilde\tau}\exp\bigl(-E_{\text{internal}}(\tilde\tau)/kT)$$

Configuration integral

$$Z_N = \int\exp\bigl(-U(\tilde q)/kT)\,d\tilde q$$

When This Does Not Work?

1.

Large quantum effects: we must include them honestly instead of as a kludge! But the kludge works for large temperatures (if the distance between quantum levels is $\ll kT$)

2.

Relativistic effects: at velocities about light speed $v\sim c$ or high gravitation fields we cannot separate kinetic and potential energy!

3.

Polarisability: sometimes we cannot separate U and $E_{\text{internal}}$.

4.

In non-Cartesian coordinates K might depend both on $\tilde p$ and $\tilde q$--we cannot separate it! In the systems with rigid bonds we must use non-Cartesian coordinates.

Kinetic part. Maxwell Distribution

Since

$$K = \sum_{i=1}^N \frac{p_i^2}{2m}$$

the kinetic part is fully factorizable:

$$Q_{\text{kin}} = \frac{1}{(2\pi\hbar)^{3N}}\int\exp\bigl(-K(\tilde p)/kT)\,d\tilde p = q_{\text{kin}}^N$$

with

$$q_{\text{kin}} = \frac{1}{(2\pi\hbar)^{3}}\int\exp(-\mathbf{p}^2/2mkT)\,d\mathbf{p}$$

Vector momentum $\mathbf{p}$:

$$d\mathbf{p}= 4\pi p^2\,dp$$

and

$$q_{\text{kin}} = \left(\frac{mkT}{2\pi\hbar^2}\right)^{3/2} = \frac{1}{\Lambda^3}$$

$\Lambda = \sqrt{2\pi\hbar/mkT}$ has the dimension of length. It is called thermal de Brogile wavelength.

Quantum interpretation:

particles have wavelengths $\lambda$depending on the momentum p. Thermal wavelength $\Lambda$ corresponds to $\sqrt{\left\langle \mathbf{p}^2\right\rangle}$.

Validity of Boltzmann Distribution:

Classical formulae work if

$$\Lambda^3 \ll \frac{V}{N}$$

Since partition function is factorized, the probability to have momentum $\mathbf{p}$

$$d\Prob(\mathbf{p})\sim\exp(-p^2/2mkT)\,d\mathbf{p}$$

For velocity $\mathbf{v}=\mathbf{p}/m$

$$d\Prob(\mathbf{v})=\left(\frac{m}{2\pi kT}\right)^{3/2}\exp(-mv^2/2kT)\,d\mathbf{v}$$

or, since $d\mathbf{v}= 4\pi v^2\,dv$

$$d\Prob(v)=4\pi\left(\frac{m}{2\pi kT}\right)^{3/2}\exp(-mv^2/2kT)v^2\,dv$$

This is called Maxwell distribution

Average velocity

$$\left\langle v^2\right\rangle = 3\frac{kT}{m}$$

Average kinetic energy per molecule:

$$\left\langle K\right\rangle = \frac{3kT}{2}$$

Another definition of temperature: in classical systems measures the average kinetic energy of thermal motion.

In classical systems kinetic part can be separated from other parts and calculated exactly!

Internal Part

The exact value depends on the molecular structure.

It is always factorizable:

$$Q_{\text{internal}} = q_{\text{internal}}^N$$

with

$$q_{\text{internal}} = \sum_{\tau=1}^k \exp\bigl(-\epsilon(\tau)/kT\bigr)$$

and $q_{\text{internal}}$ depends only on T!

Epitome

The kinetic part is calculated exactly--for all classical systems.

Calculation of the internal part is mostly a Quantum Mechanics problem.

Classical Statistical Thermodynamic is about calculation of the configuration integral

$$Z_N = \int\exp\bigl(-U(\tilde q)/kT)\,d\tilde q$$