Fundamental idea
Think about thermodynamics
The observable properties of systems containing very large numbers of particles depend only on specifying a few macroscopic properties - internal energy, temperature, volume, pressure, density, etc.
Most experiments measure averaged properties - over many particles and over the time of measurement.
That is why it is hopeful to be able to calculate what we need to know, for macroscopic systems.
Principle of equal a priori probabilities
A system with fixed , and (isolated system) is equally likely to be found in any of its energy states.
Hence, to find the probability of a system in a particular macrostate, we calculate the degeneracy. For macroscopic systems, the degeneracies of energy levels are superastronomically large; the most probable states have overwhelmingly large degeneracies compared to other states.
Bridging with thermodynamics
Energy transfer between two systems - temperature, entropy
Note that when the total energy is , the total degeneracy becomes . It turns out more convenient to consider the logarithm of , which is extensive.
An extensive variable scales linearly with the system size. e.g. , , , ; an intensive variable is independent of system sizes. e.g. , , .
That and entropy are both maximized at equilibrium, and that they are both extensive, mean that they are proportional. Experimental measurements gave the constant of proportionality .
When we fix the total energy between the two systems, depends on either or . For a maximum,
Defind , then the above equality says that thermal equilibrium implies .
We have known from thermodynamics that thermal equilibrium also mean temperatures are the same.
Using a fundamental equation, , we get
System at constant temperature - Canonical ensemble
System in thermal equilibrium with a large heat bath. Total energy is fixed. The probability that the system is in state is
By Taylor expanding ,
and the definition of ,
The denominator is the canonical partition function
Average energy, Helmholtz energy and equilibrium
Now the way to compute the average energy is
Note the similarity of the numerator to , so in fact,
To obtain the expectation of any linear term in Hamiltonian, differentiate the relevant ln partition function w.r.t the quantity it couples to ( in this case).
Write that in terms of : .
Compare with the thermodynamic relation , Helmholtz analogue of Gibbs-Helmholtz equation.
We arrive at the bridge relation:
Again from the Taylor expansion on ,
to maximize , we need to minimize
Now we recover that, under constant and , is minimized at equilibrium.
Classical statistical mechanics
In the classical limit, it is possible to approximate sum with integrals. Hence the expectation values can become
and the partition function becomes
Evaluate the kinetic energy part by factorising into separate components, and use Gaussian standard integral to find
This is as important as the bridge relation.
Fluctuations
In most cases, any fluctuation about the expectation value is ignored. How valid is this?
It is possible to either work out the probability distribution of energy is a Gaussian, with variance
Or, the same result can be achieved using the fact that a second derivative of ln partition function gives the negative variance. Then we can relate heat capacity through the thermodynamic definition
Therefore, in the thermodynamic limit, i.e. , the ratio
Other ensembles
| Ensemble | Fix |
|---|---|
| Microcanonical | N,V,E |
| Canonical | N,V,T |
| Isothermal-isobaric | N,P,T |
| Isoenthalpic-isobaric | N,P,H |
| Grand | µ,V,T |
Using different ensembles will always arrive at the same result. Ensembles are equivalent in the thermodynamic limit. Choices are made based on how much it facilitates the derivation and calculation of a specific problem.
Pressure
To investigate pressure, we need to allow exchange of volume between system and bath. At equilibrium, maximum of entropy is achieved.
From thermodynamics, .
Then, . Combining with the fact that temperatures are equal, we conclude pressures must be equal at equilibrium.
With the fundamental relation for Helmholtz energy ,
Pressure of interacting particles
From partition function
Consider a cubic box with edge length . Using scaled coordinates , and use the equation above
Working through the change in coordinates, and define a force , we can arrive at
Pairwise additive forces
If the intermolecular forces are pairwise-additive, then
By permuting the indices and calling Newton’s 3rd law,
where . This is important because we may know the form of intermolecular forces!
Write in a dimensionless form, we get the compressibility factor
The sum for is omitted as in a homogeneous system, the average contribution of any one particle is the same.
Law of corresponding states
Generally, if the interactions of several types of atoms or molecules are of the same funcional form , then the reduced potentials, are all equal. Reduced quantities are expressed in the units of and .
Similarly, the compressibility factor of different substances can be made to lie on the same curve by rescaling the temperature and density, if their potentials are of the same functional form.
Free energy and phase behaviour
Equilibrium
At equilibrium, the entropy reaches a maximum, . From the fundamental relation of ,
as , and can vary independently, the coefficient for each must be equal between the two phases. That is , and must be equal.
Then, consider a second order transfer of energy, .
Therefore it can be deduced that
Heat capacity is never negative.
Similarly, by considering the Helmholtz energy at minimum, and give a second order transfer in volume, it can be deduced that compressibility is never negative.
Coexistence
If we have a analytical form of a Helmholtz energy , the coexistence can be determined graphically.At constant , pressure and chemical potential need to equate between the two phases.
- Using , same gradient guanrantees equal pressure;
- Using , same intercept gives equal chemical potential.
A common tangent can be used to find the coexisting phases.
Note that as compressibility is positve, must be convex - any concave part represents unstable systems (these exist usually because the form of is approximate). Therefore, for the region between the common tangent construction,
- Convex part is metastable - there is an energy barrier to separate;
- Concave part is unstable - spinodal.
Phase diagrams
For one component systems, are the variables to determine the state. Hence the phase diagram is actually a surface in the space.
Critical point is the point beyond which liquid-vapour transition is continuous.
Thermodynamic integrations
There are two classes of TD variables
- ‘Thermal’ variable - depend on accessible volumes in phase space: , , etc.
- ‘Mechanical’ variables - depend on positions and momenta: , , etc.
We can measure the averages of mechanical quantities, but nothing about thermal quantities. Fortunately, the derivatives of thermal quantities can be mechanical quantities, so the values of thermal variables can be worked out by integrations.
Note that choosing to ‘measure’ thermal variables by e.g. ‘observing’ phase transitions is not reliable, due to factors including metastability.
Thermodynamic perturation theory
Gibbs-Bogoliubov inequality
Use a linear parameterisation of the potential energy function
where is a reference, and is the system of interest. Then we can compute
First look at the second derivative. It is a negative of a variance, hence is negative. Therefore, inside this region, the function is concave.
Then, integrate the first derivative from 0 to 1,
Combining with the concaveness, this expression gives an upper bound to , i.e.
As with all perturbation problems, the closer is chosen to represent , the more accurate the bound is; but usually the harder to compute.
Ising model and mean-field theory
The Ising model is a classic in describing magnetic systems. Its Hamiltonian is
where the index runs through neighbours of . This involves interactions between spins, which may not be easy to deal with, so we turn to a mean-field approximation of the Hamiltonian
Here it is easy to work out the partition function and expectation of spin
To use G-B inequality, we also need to calculate the ‘gradient at ‘, . Since in the reference system, spins are not correlated, this is very easy.
where is the number of nearest neighbours.
By minimizing the expression, we find the best value for the parameter by differentiation.
With graphical method or otherwise, we can see that
- At high temperature, there is only one solution;
- Below a certain temperature, Curie temperature, there are two solutions and the mean magnetization is non-zero;
- Once in a branch, it is unlikely to enter the other branch by fluctuation - spontaneous symmetry breaking.
- The mean field Curie temperature is higher than the actual value due to ignoring the correlation between spins.
Van der Waals equation
An example where the 2nd derivative in G-B vanishes - so the ‘=’ sign is taken in G-B.
Hard-sphere gas
The potential is
Approximate that each sphere excludes a volume of , where as its own volume is . In the dilute limit, it is assumed (drastically) that no two excluded volumes overlap.
After some clever algebra, the partition function, and hence pressure is given by
Note
- This diverges as ;
- This is due to the crude no exclusion approximation, hence only valid at low density.
Mean-field attraction
To model gas better, we need an attraction. This is introduced by a mean-field attraction potential, which is very weak, but very long-ranged (Poisson-like).
- On average, particles within interaction range;
- Each contributes ;
- As , the fluctuation goes to 0, hence the first order G-B becomes exact.
Therefore, the free energy becomes
where is a Van der Waals parameter, along with .
Therefore, the pressure is
By rearranging, we have the Van der Waals equation of state
Note that in reduced units, this can also be written as
Maxwell equal-area construction
The Van der Waals state P-V phase diagram shows that
- Above critical temperature, the curve is monotonic, so there is only one fluid phase;
- Below critical temperature, a single horizontal line, representing a common pressure, can cross the curve three times - called the Van der Waals loop.
At coexistence,
Now use thermodynamic integration to express the free energy , we get the condition
Hence, the equilibrium phases at the given temperature make the area enclosed above and below the pressure line to be the same.