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.


1 Answer 1


Mobility is a very general term that refers to the speed obtained for a given driving force.

The mobility of a charge carrier driven to move by an electric field (in m/s per V/m) is completely distinct from the mobility of a particle driven to move by a gradient in the molar chemical potential (in m/s per J/mole-%-m).

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.

The two parameters are not equivalent, and numerical estimates and units of one cannot be equated to those of another. Does this answer your question?

  • $\begingroup$ 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. $\endgroup$
    – Henning
    May 26, 2022 at 20:50
  • $\begingroup$ 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. $\endgroup$ May 26, 2022 at 21:55
  • $\begingroup$ The unit of conecentration mol-% is dimnsionless. The number of moles is a pure number and so is "percent." $\endgroup$
    – mike stone
    Jun 2, 2022 at 13:53

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