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, I often noticed that the mobility is given in $\frac{\text{m}^2}{\text{V} \, \text{s}}$ which I cannot reformulate in the required unit. In addition, I noticed that this Wikipedia entry considers the Cahn-Hilliard equation above (with a specific $\hat f$ and) with a diffusion coefficient which is given in $\text{length}^2 / \text{time}.$ Do I misunderstand something?

  • $\begingroup$ I think your mol% should be mole fraction and thus dimensionless? $\endgroup$ May 26 at 21:17
  • $\begingroup$ Actually, that does not change anything (of the problem), does it? $\endgroup$
    – Henning
    May 26 at 21:25

1 Answer 1


This question is addressed here. For the units, $e$=energy (eg J), $t$=time, $l$=length.

The Cahn-Hilliard equation after the variational derivative takes the form$$\dfrac{\partial c}{\partial t}=V_{m}\nabla\cdot M_{i} \nabla\left(\dfrac{\partial f_{loc}}{\partial c_i}+\dfrac{\partial E_d}{\partial c_i}-\kappa_i \nabla^2c_i\right)$$where $V_m$ is the molar volume of the reference state of the material with units of $l^3/mol$.

The units of this equation are$$\dfrac{1}{t}=\dfrac{l^3}{mol}\dfrac{1}{l}\dfrac{l^2\cdot mol}{te}\dfrac{1}{l}\left(\dfrac{e}{l^3}+\dfrac{e}{l^3}-\dfrac{e}{l}\dfrac{1}{l^2}\right)$$ where the units of $M_i$​ are $(l^2 mol)/(et)$. Note that some models include the $V_m$ in the mobility term, such that it has units of $l^5/(te)$.

The key bit is that last part about combining $V_m$ into $M_i$, which changes the units from what you'd expect for $M_i$. In your version of the equation, there is no separate $V_m$ term, so you need to account for it in your units for $M_i$.


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