# Understanding the Cahn-Hilliard equation in terms of units

The Cahn-Hilliard equation may be formulated as $$\frac{\partial c}{\partial t} = M \nabla^2 \left(\frac{\partial \hat f}{\partial c}\right),$$ where $$c : \Omega \to [0,100]$$ describes the concentration (mol-%) of an interesting substance, $$M$$ is the mobility coefficient (for simplicity $$M$$ is assumed to be constant) and $$f$$ is the generalised free energy per unit volume, i.e., $$\hat f$$ depends on the concentration $$c$$ and higher derivatives of $$c$$ (see e.g. Novick-Cohen & Segel (1984), p. 278 -- 282).

Problem: If I take a look at the units of this equation, I am confused. According to the equation, we have on the LHS $$\frac{\text{mol-%}}{\text{s}}.$$ For the unit of $$\nabla^2 \left(\frac{\partial \hat f}{\partial c}\right)$$ on the RHS, I get $$\frac{\text{J}}{\text{mol-%}\,\text{m}^2}.$$ Hence, the mobility constant $$M$$ should be given in $$\frac{\text{m}^2 \, \text{mol-%}^2}{\text{J} \, \text{s}}$$ to end up with $$\frac{\text{mol-%}}{\text{s}}$$ on the RHS. However, the mobility is given in $$\frac{\text{m}^2}{\text{V} \, \text{s}}$$ which I cannot reformulate in the required unit.

When you say "the mobility is given in $$\frac{\text{m}^2}{\text{V} \, \text{s}}$$," I'm guessing this refers to a mobility found from another source and in another context than the Novick–Cohen and Segel paper.
• Thanks! In fact, I thought that ''mobility'' is an universal notion (different from the so-called ''diffusion coefficient''). So your answer seems helps to make things clearer. However, since in this Wikipedia entry the CH equation is definitely given with a diffusion coefficient with units of $\text{length}^2 / \text{time}$ I think that I misunderstand something. May 26, 2022 at 20:50
• Well, that's a different formulation in which $c$ is a nondimensional parameter. For that matter, Eq. (3) here formulates the problem with no coefficient at all. May 26, 2022 at 21:55