Thermodynamics¶
This post discusses how phase equilibrium is established. In particular, we discuss multi-component saturating systems which spontaneously form droplets at zero temperature. This was specifically motivated by the discussion of the conditions for droplet formation in Bose-Fermi mixtures 1801.00346 and 1804.03278. Specifically, the following conditions in 1804.03278:
\begin{gather} \mathcal{E} < 0, \qquad \mathcal{P} = 0; \tag{i}\\ \mu_b\pdiff{P}{n_f} = \mu_f\pdiff{P}{n_b}; \tag{ii}\\ \pdiff{\mu_b}{n_b} > 0 , \qquad \pdiff{\mu_f}{n_f} > 0, \qquad \pdiff{\mu_b}{n_b}\pdiff{\mu_f}{n_f} > \left(\pdiff{\mu_b}{n_f}\right)^2. \tag{iii} \end{gather}1801.00346: https://arxiv.org/abs/1801.00346 1804.03278: https://arxiv.org/abs/1804.03278
import mmf_setup;mmf_setup.nbinit()
Brief explanation:
We shall explain $\mathcal{E} <0$ later, but it says that the energy density of the droplet should be less than the energy density of the vacuum. $\mathcal{P} = 0$ states that the droplet should have the same pressure as the vacuum. (Here we neglect surface tension). Restricting $\mathcal{P} = 0$ defines a curve in the $n_b$, $n_f$ plane. Minimizing the energy density along this curve can be accomplished with a Lagrange multiplier to hold $P$ fixed:
$$ \pdiff{}{n_j}\left(\mathcal{E}(n) - \lambda P\right) = 0, \qquad \pdiff{\mathcal{E}(n)}{n_j} \mu_j = \lambda \pdiff{P}{n_j}, \qquad \frac{1}{\mu_j}\pdiff{P}{n_j} = \lambda. $$The last equation is simply a rearrangement of (ii). Finally, for the state to be stable, the energy density must be positive-definite:
$$ [\mat{H}]_{ij} = \frac{\partial\mathcal{E}}{\partial n_i \partial n_j}. $$In general, this means all eigenvalues must be positive, which, for 2x2 matrices this is the same as requiring the trace and determinant to be positive.
For now we consider $T=0$. Let $n_i \in \{n_b, n_f\}$ be the densities of the various species. The micro-canonical ensemble can be described by the energy density $\mathcal{E}(n) = \mathcal{E}(n_f, n_b)$. Other thermodynamic variables can be obtained by differentiating, including the chemical potentials $\mu_i$ and applying the Legendre transform to obtain the pressure $P$:
$$ \mu_i = \pdiff{\mathcal{E}(n)}{n_i}, \qquad P(\mu_i) = \sum_i \mu_i n_i[\mu] - \mathcal{E}(n[\mu]). $$To go backwards, note:
$$ \pdiff{P(\mu)}{\mu_j} = n_j[\mu] + \sum_i \mu_i \pdiff{n_i[\mu]}{\mu_j} - \sum_{i}\overbrace{\pdiff{\mathcal{E}}{n_i}}^{\mu_i}\pdiff{n_i[\mu]}{\mu_j} = n_j, \qquad \mathcal{E}(n) = \sum_i \mu_i[n] n_i - P(\mu[n]). $$Typically, one does some sort of calculation of microscopic states, ending up with an energy density $\mathcal{E}(n)$ as a function of the various densities $n$. This is not the end of the story due the possibility of forming mixed phases. These occur if the function $\mathcal{E}(n)$ is not convex. In this case, a mixed phase consisting of a fraction $x$ of a volume being in state $\mathcal{E}(n_1)$ with volume fraction $1-x$ being in state $\mathcal{E}(n_2)$. The thermodynamic volume would thus have total energy density
$$ \mathcal{E}\bigl(xn_1 + (1-x)n_2\bigr) = x\mathcal{E}(n_1) + (1-x)\mathcal{E}(n_2). $$This is a form of Maxwell Construction. These phase combinations must also be considered in conjunction with the pure phases, and the minimum energy density phase must be chosen. This will result in an energy density $\mathcal{E}$ that is the convex hull of the previous microscopic state, with mixed phases appearing long sections of zero curvature denoting first-order phase transitions.
Single Component Example¶
Here we demonstrate this explicitly with a single component. We use an equation of state mocking up nuclear matter:
$$ \mathcal{E}(n) = n\left(\epsilon_0 + \frac{K_0}{2}\delta^2 + \frac{Q_0}{3!}\delta^3\right), \qquad \delta = \frac{n-n_0}{3n_0},\\ \epsilon_0 \approx -16\mathrm{MeV}, \qquad n_0 \approx 0.16\mathrm{fm}^{-3}, \qquad K_0 \approx 230\mathrm{MeV}, \qquad Q_0 = 9(18\epsilon_0 + K_0) \approx -522\mathrm{MeV}. $$Note: The value of $Q_0$ was simply chosen so that $\mathcal{E}(0) = 0$: more accurate functional forms will have a different values. The other parameters have physical meaning: the saturation density $n_0$, saturation energy-per-particle $\epsilon_0$ and nuclear incompressibility $K_0$.
From this, we find the thermodynamic relations:
$$ \mu = \mathcal{E}'(n) = \epsilon_0 + \frac{K_0}{2}\delta^2 + \frac{Q_0}{3!}\delta^4 + \frac{n}{3n_0}\left(K_0\delta + \frac{Q_0}{2}\delta^2\right)\\ P = \mu n - \mathcal{E} $$%pylab inline --no-import-all
e0 = -16.0
n0 = 0.16
K0 = 230.0
Q0 = 9*(18*e0 + K0)
print Q0
n = np.linspace(0,2*n0,100)
d = (n-n0)/3./n0
E = n*(e0+K0/2.*d**2 + Q0/6*d**3)
plt.plot(n, E/n)
plt.axis([0, 0.2, -20, 0])
