Cahn and Hilliard define the energy of an interface:

the difference per unit area of interface between the actual free energy of the system and that which it would have if the properties of the phases were continuous throughout

(Free Energy of a Nonuniform System. I. Interfacial Free Energy, John W. Cahn, and John E. Hilliard, The Journal of Chemical Physics 28, 258 (1958); doi: 10.1063/1.1744102)

And it is possible to determine exactly the form and energy of the interface in 1D for the potential
$$P=-\phi^2+\phi^4 + \epsilon ^2 (\nabla \phi)^2$$ which is following a $\tanh$ function with conditions at infinity of the form $\phi=\pm \phi_0$ (see for instance Chaikin and Lubensky : Principles of condensed matter physics p.596).

My question is: if $\phi$ is a concentration of an element and we have an interface between a phase with $\phi=\phi_0$ and a phase where that component is absent. So it is not an interface between 2 minima of the free-energy function! How do we define the energy of that interface?

  • 1
    $\begingroup$ If it's not an interface between 2 minima of the free energy then we are not in equilibrium and we shouldn't use equilibrium concepts to describe the system. I think that you should think of the "other" phase as a very diluted gas phase rather than a "vacuum" $\endgroup$
    – lr1985
    Commented Oct 15, 2018 at 10:18
  • $\begingroup$ @lr1985 I'm not sure I understand. I do not see how thinking of the other phase as a very diluted gas changes anything. When you look at the function $-\phi^2+\phi^4$ where $\phi$ has to be positive, being very diluted or being empty is the same. It looks like in that case, there are 2 ways to get to a minimum : either by going to the negative \phi, which is not physical, or to create some matter to get to the \phi positive different than 0, which is not possible in many cases... $\endgroup$
    – J.A
    Commented Oct 15, 2018 at 10:26

1 Answer 1


In this kind of simplified model, $\phi(x)$ is defined as a suitable order parameter, which does not necessarily have to be interpreted as a density or concentration. In a case like a ferromagnet, it can be the local spin density, and you can have two phases with (legitimately) positive and negative values $\pm\phi_0$. But for a system like liquid + gas, with densities $\rho_{\ell}$ and $\rho_g$, you just define a mean density $\overline{\rho}=\frac{1}{2}(\rho_{\ell}+\rho_g)$ and set $\phi(x)\propto\rho(x)-\overline{\rho}$, or more precisely $$ \phi(x) = \frac{2\phi_0}{\rho_{\ell}-\rho_g} \bigl(\rho(x)-\overline{\rho}\bigr) $$ so that $\rho(x)=\rho_{\ell}$ or $\rho_g$ corresponds to $\phi(x)=\pm\phi_0$. In this way the mathematics of solving the simplified problem becomes equivalent to solving one with more realistic densities.

Typically in density functional theory, which is based on a free energy functional, if one omits the squared-gradient term (or the weighted nonlocal density term) describing the interface, and sets $\rho(x)=\rho$, one is left with a free energy function (of the density, which no longer depends on position) describing the homogeneous one-phase system. At a state point where both phases can coexist you expect this free energy function to have two minima, one describing each phase. Both phases will have a positive $\rho$, unless something is amiss.

Chaikin and Lubensky are using a toy model, intended to illustrate the approach, so $\phi(x)$ need not be positive. As I mentioned, you can subtract a constant from $\rho(x)$ and re-write the functional in terms of $\delta\rho(x)=\rho(x)-\overline{\rho}$ which can be identified with $\phi(x)$. For a realistic system, you are unlikely to get a nice simple quartic expression in $\phi$, but the rewritten bulk term should look qualitatively similar (two minima separated by a maximum).


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