Building up the free energy: repulsion

Let's start by looking at the ideal gas free energy: \begin{align} F_{\text{ideal}} &= -NkT\left(\ln\left(\frac{n_QV}{N}\right) + 1\right) \end{align} This free energy depends on both volume and temperature (also \(N\), but let's keep that fixed). The temperature dependence is out front and in \(n_Q\). The volume dependence is entirely in the logarithm. When we add the repulsive interaction, we can wave our hands a bit and argue that the effect of repulsion is to keep atoms from sitting too close to one another, and that results in each atom having less volume it could be placed in. The volume available for a given atom will be the total volume \(V\), minus the volume occupied by all the other atoms, which we can call \(Nb\) where \(N\) is the number of atoms, and \(b\) is the excluded volume per atom. You might argue (correctly) that the excluded volume should be \((N-1)b\), but we will be working in the limit of \(N\gg1\) and can ignore that fine distinction. Making this substitution gives us \begin{align} F_{\text{with repulsion}} &= -NkT\left(\ln\left(\frac{n_Q(V-Nb)}{N}\right) + 1\right) \end{align} This free energy is going to be higher than the ideal gas free energy, because we are making the logarithm lesser, but there is a minus sign out front. That is good, because we would hardly expect including repulsion to lower the free energy.

In your homework you will (incidentally) show that this free energy gives an internal energy that is identical to the ideal gas free energy, which bears out what I said earlier about repulsion affecting the entropy rather than the energy.

Adding attraction: mean field theory

When we want to add in attraction to the free energy, the approach we will use is called mean field theory. I prefer to talk about it as first-order thermodynamic perturbation theory. (Actually, mean field theory is often more accurately described as a poor approximation to first-order perturbation theory, as it is common in mean-field theory to ignore any correlations in the reference fluid.) You know perturbation theory from quantum mechanics, but the fundamental ideas can be applied to any theory, including statistical mechanics.

The fundamental idea of perturbation theory is to break your Hamiltonian into two terms: one that you are able to solve, and a second term that is small. In this case, in order to derive (or motivate?) the van der Waals equation, our reference would be the system with repulsion only, and the perturbation would be the attraction between our atoms. We want to solve this purely classically, since we don't know how to solve the energy eigenvalue equation with interactions between particles included.

Classically, we would begin by writing down energy, and then we would work out the partition function by summing over all possible microstates in the canonical ensemble. A logarithm would then tell us the free energy. The energy will be \begin{align} E &= \sum_i^{\text{all atoms}} \frac{p_i^2}{2m} + \frac12\sum_{ij}^{\text{atom pairs}}U(|\vec r_i-\vec r_j|) \end{align} where \(U(r)\) is an attractive pair potential, which is to say, a potential energy of interaction between each pair of atoms. The first term is the kinetic energy (and is the same for the ideal gas), while the second term is a potential energy (and is zero for the ideal gas). The partition function is then \begin{align} Z &= \frac1{N!} \int\!\!d^3r_1\int\!\!d^3r_2\cdots\int\!\!d^3r_N \notag\\&\quad \int\!\!d^3p_1\int\!\!d^3p_2\cdots\int\!\!d^3p_N e^{-\beta\left(\sum \frac{p_i^2}{2m} + \frac12\sum U(|\vec r_i-\vec r_j|)\right)} \\ &= \frac1{N!} \int\!\!d^3p_1\int\!\!d^3p_2\cdots\int\!\!d^3p_N e^{-\beta\left(\sum \frac{p_i^2}{2m}\right)} \notag\\&\quad \int\!\!d^3r_1\int\!\!d^3r_2\cdots\int\!\!d^3r_N e^{-\beta\left(\frac12\sum U(|\vec r_i-\vec r_j|)\right)} \end{align} At this point I will go ahead and split this partition function into two factors, an ideal gas partition function plus a correction factor that depends on the potential energy of interaction. \begin{align} Z &= \frac{V^N}{N!} \int\!\!d^3p_1\int\!\!d^3p_2\cdots\int\!\!d^3p_N e^{-\beta\left(\sum \frac{p_i^2}{2m}\right)} \notag\\&\quad \frac1{V^N}\int\!\!d^3r_1\int\!\!d^3r_2\cdots\int\!\!d^3r_N e^{-\beta\left(\frac12\sum U(|\vec r_i-\vec r_j|)\right)} \\ &= Z_{\text{ideal}} \frac1{V^N}\int\!\!d^3r_1\cdots\int\!\!d^3r_N e^{-\beta\left(\frac12\sum U(|\vec r_i-\vec r_j|)\right)} \\ &= Z_{\text{ideal}}Z_{\text{configurational}} \end{align} Now we can express the free energy! \begin{align} F &= -kT\ln Z \\ &=-kT\ln(Z_{\text{ideal}}Z_{\text{configurational}}) \\ &= F_{\text{ideal}} -kT\ln Z_{\text{configurational}} \end{align} So we just need to approximate this excess free energy (beyond the ideal gas free energy). Let's get to the approximation bit. \begin{align} Z_{\text{config}} &= \int\!\!\frac{d^3r_1}{V}\cdots\int\!\!\frac{d^3r_N}{V} e^{-\beta\left(\frac12\sum U(|\vec r_i-\vec r_j|)\right)} \\ &\approx \int\!\!\frac{d^3r_1}{V}\cdots\int\!\!\frac{d^3r_N}{V} \left(1-\sum_{ij}\frac{\beta}2 U(r_{ij})\right) \end{align} At this point I have used a power series approximation on the exponentials, under the assumption that our attraction is sufficiently small. Now we can write this sum in a simpler manner, taking account of the symmetry between the different particles. \begin{align} Z_{\text{config}} &= 1 - \frac{\beta}2\sum_{ij}\int\!\!\frac{d^3r_1}{V}\cdots\int\!\!\frac{d^3r_N}{V} U(r_{ij}) \\ &= 1 - \frac{\beta N^2}2\int\!\!\frac{d^3r_1}{V}\int\!\!\frac{d^3r_2}{V} U(r_{12}) \\ &= 1 - \frac{\beta N^2}2\int\!\!\frac{d^3r}{V}U(r) \end{align} At this stage, I've gotten things way simpler. Note also, that I did something wrong. I assumed that the potential was always small, but the repulsive part of the potential is not small. But we'll ignore that for now. Including it properly would be doing this right, but instead we'll use the approach that leads to the van der Waals equation of state. To continue... \begin{align} F_{\text{excess}} &= -kT\ln Z_{\text{conf}} \\ &= -kT\ln\left(1 - \frac{\beta N^2}2\int\!\!\frac{d^3r}{V}U(r)\right) \\ &\approx kT\frac{\beta N^2}2\int\!\!\frac{d^3r}{V}U(r) \\ &= \frac{N^2}2\int\!\!\frac{d^3r}{V}U(r) \\ &\equiv -\frac{N^2}{V} a \end{align} where I've defined \(a \equiv -\frac12\int\!\!d^3r U(r)\). The minus sign here is to make \(a\) a positive quantity, given that \(U(r)<0\). Putting this together with the ideal gas free energy modified to include a simple repulsion term, we have the van der Waals free energy: \begin{align} F_{\text{vdW}} &= -NkT\left(\ln\left(\frac{n_Q(V-Nb)}{N}\right) + 1\right) - \frac{N^2}{V}a \end{align}

van der Waals equation of state

Small groups

Solve for the van der Waals pressure, as a function of \(N\), \(V\), and \(T\) (and of course, also \(a\) and \(b\)).

Answer

\begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_{T,N} \\ &= \frac{NkT}{V-Nb} - \frac{N^2}{V^2}a \end{align}

This equation is the van der Waals equation of state, which is often rewritten to look like: \begin{align} \left(p + \frac{N^2}{V^2}a\right)(V-Nb) = NkT \end{align} as you can see it is only slightly modified from the ideal gas law, provided \(a\ll p\) and \(Nb\ll V\).

Common tangent

Most approaches require us to work with the Helmholtz free energy rather than the pressure equation. If we plot the Helmholtz free energy versus volume (with number fixed) the pressure is the negative slope. We also need to ensure that the chemical potential (or Gibbs free energy) is identical at the two points.

The common tangent construction is very commonly used when looking at multiple crystal structures, when you don't even know which ones are stable in the first place.

Note

The common tangent construction also works when we plot \(F\) versus \(n\) or \(N\).

Gibbs free energy

Another approach to solve for coexistence points is to plot the Gibbs free energy versus pressure, each of which can be computed easily from the Helmholtz free energy. When we plot the Gibbs free energy versus pressure, we find that there is a crossing and a little loop. This loop corresponds to metastable and unstable states, and the crossing point is where the two phases (liquid and vapor, in our case) coexist.

As we increase the temperature, we will find that this little loop becomes smaller, as the liquid and vapor densities get closer and closer. The critical point is where it disappears.

Why does \(G\) look like this?

We had a good question about what the “points” represent in the Gibbs free energy curve. We can understand this a bit better by thinking a bit about the differential of \(G\): \begin{align} dG &= -SdT + Vdp \end{align} This tells us that the slope of the \(G\) versus \(p\) curve (at fixed temperature) is just the volume of the system. Since the volume can vary continuously (at least in the Helmholtz free energy we constructed), this slope must continuously change as we follow the path. That explains why we have pointy points, since the slope must be the same on both sides of the curve. The points thus represent the states where the pressure has an extremum, as we change the volume. In between those two extrema is the range where increasing volume causes the pressure to increase. These states are mechanically unstable, and thus cannot be observed.

A continuous phase transition

To make things concrete, let us assume an order parameter with inversion symmetry, such as magnetization or electrical polarization. This means that \(F_L\) must be an even function of \(\xi\), so we can write that \begin{align} F_L(\xi,T) &= g_0(T) + \frac12g_2(T)\xi^2 + \frac14 g_4(T)\xi^4 + \cdots \end{align} The entire temperature dependence is now hidden in the coefficients of the power series. A simple example where we could have a phase transition, would be if the sign of \(g_2\) changed at some temperature \(T_0\). In this case, we could do a power series expansion of our coefficients around \(T_0\), and we would have something like: \begin{align} F_L(\xi,T) &= g_0(T) + \frac12\alpha(T-T_0)\xi^2 + \frac14g_4(T_0)\xi^4 \end{align} where I am ignoring the temperature dependence of \(g_4\), under the assumption that it doesn't do anything too fancy near \(T_0\). I'm leaving \(g_0(T)\) alone, because it causes no trouble, and will be useful later. I'm also going to assume that \(\alpha\) and \(g_4(T_0)\) are positive. Now we can solve for the order parameter that minimizes the free energy by setting its derivative to zero. \begin{align} \left(\frac{\partial F_L}{\partial \xi}\right)_T &= 0\\ &= \alpha(T-T_0)\xi + g_4(T_0)\xi^3 \end{align} This has two solutions: \begin{align} \xi &= 0 & \xi^2 &= (T_0-T)\frac{\alpha}{g_4(T_0)} \end{align} If \(T>T_0\) there is only one (real) solution, which is that the order parameter is zero. Thus when \(T>T_0\), we can see that \begin{align} F(T) &= g_0(T) \end{align} exactly, since \(\xi=0\) causes all the other terms in the Landau free energy to vanish.

In contrast, when \(T<T_0\), there are two solutions that are minima (\(\pm \sqrt{(T_0-T)\alpha/g_4(T_0)}\)), and one maximum at \(\xi=0\). In this case the order parameter continuously (but not smoothly) goes to zero. This tells us that the free energy at low temperatures will be given by \begin{align} F(T) &= g_0(T) - \frac{\alpha^2}{4g_4(T_0)}(T-T_0)^2 \end{align}

Small groups

Solve for the entropy of this system when the temperature is near \(T_0\).

Answer

We can find the entropy from the free energy by considering its total differential \begin{align} dF &= -SdT -pdV \end{align} which tells us that \begin{align} -S &= \left(\frac{\partial F}{\partial T}\right)_{V} \end{align} Let's start by finding the entropy for \(T<T_0\): \begin{align} S_< &= -\frac{dg_0}{dT} -\frac{\alpha^2}{2g_4(T_0)}(T_0-T) \end{align} When the temperature is high, this is easier: \begin{align} S_< &= -\frac{dg_0}{dT} \end{align} This tells us that the low-temperature phase has an extra-low entropy relative to what it would have had without the phase transition. However, the entropy is continuous, which means that there is no latent heat associated with this phase transition, which is called a continuous phase transition. An older name for this kind of phase transition (used in the text) is a second order phase transition. Currently “continuous” is prefered for describing phase transitions with no latent heat, because they are not always actually second order as is this example.

Examples of continuous phase transitions include ferromagnets and superconductors.

An abrupt phase transition

To get an abrupt phase transition with a nonzero latent heat (as for melting or boiling), we need to consider a scenario where \(g_4<0\) and \(g_6>0\). This gives us two competing local minima at different values for the order parameter. (Note that an abrupt phase transition is also known as a first order phase transition.) \begin{align} F_L &= g_0(T) + \frac12\alpha(T-T_0)\xi^2 -\frac14|g_4(T)|\xi^4 + \frac16 g_6\xi^6+\cdots \end{align}

Small groups if we have time

Find the solutions for the order parameter, and in particular find a criterion for the phase transition to happen.

Answer

We want to find minima of our free energy... \begin{align} \frac{\partial F_L}{\partial \xi} &= 0 \\ &= \alpha(T-T_0)\xi - |g_4(T)|\xi^3 + g_6\xi^5 \end{align} One solution is \(\xi=0\). Otherwise, \begin{align} 0 &= \alpha(T-T_0) - |g_4(T)|\xi^2 + g_6\xi^4 \end{align} which is just a quadratic. It has solutions when \begin{align} \xi^2 &= \frac{|g_4(T)| \pm \sqrt{g_4(T)^2 -4g_6\alpha(T-T_0)}}{2g_6} \end{align} Note that this has four solutions. Two have \(\xi<0\), and show up because our free energy is even. One of the other solutions is a local maximum, and the final solution is a local minimum. For this to have a real solution, we would need for the thing in the square root to be positive, which means \begin{align} g_4(T)^2 \ge 4g_6\alpha(T-T_0) \end{align} It would be tempting to take this as an equality when we are at the phase transition. However, that is just the point at which there is a local minimum, but we are looking for a global minimum (other than \(\xi=0\)). This global minimum will require that \begin{align} F_L(\xi>0) < F_L(\xi=0) \end{align} which leads us to conclude that \begin{align} \frac12\alpha(T-T_0)\xi^2 -\frac14|g_4(T)|\xi^4 + \frac16 g_6\xi^6 &< 0 \end{align} We can plug in the criterion for an extremum in the free energy at nonzero \(\xi\) to find: \begin{align} \tfrac12\left(|g_4(T)|\xi^2-g_6\xi^4\right)\xi^2 -\tfrac14|g_4(T)|\xi^4 + \tfrac16 g_6\xi^6 &< 0 \\ \tfrac14|g_4(T)|\xi^4 - \tfrac13 g_6\xi^6 &< 0 \\ \frac14|g_4(T)| - \frac13 g_6\xi^2 &< 0 \end{align} At this point we would want to make use of the solution for \(\xi^2\) above that used the quadratic equation. We would then have eliminated \(\xi\) from the equation, and could solve for a relationship between \(|g_4(T)|\), \(g_6\), and \(\alpha(T-T_0)\).