# Understanding 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, 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?

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

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 $$M_i$$.