# How to estimate the mobility constant and the surface energy parameter in the Cahn-Hilliard equation?

As a mathematician, I want to simulate phase separations with the Cahn-Hilliard equation $$\frac{\partial c}{\partial t} = M \Delta \bigg(\frac{\partial \mu}{\partial c} - \kappa \Delta c \bigg),$$ where $$c$$ is the molar fraction, $$\mu$$ is the chemical potential (given in the unit $$\text{energy}/\text{length^2}$$), $$M$$ is the mobility constant, $$\kappa$$ is the surface energy parameter and $$\Delta$$ is the Laplacian. (Since my simulations are in 2D, the units are correspondingly chosen with the unit length of $$1\mu\text{m}.$$)

Question 1: How do we get reasonable (good) estimates for $$M$$ and $$\kappa,$$ let's say for a certain mixture such as $$\text{Na}_2\text{O-SiO}_2$$? Are there standard references?

Literature and Problem: In the paper of Kim and Sanders (2020) the surface energy parameter is estimated by $$\kappa = \frac{\mu - \text{common tangent}}{(\partial c / \partial x)^2}$$ while the mobility is expressed as $$M = \frac{D \cdot N_A}{\partial^2 \mu / \partial c^2}$$ where $$D$$ is the diffusion coefficient (given in $$\text{length}^2/\text{time}$$) and $$N_A$$ is the number of particles per unit area (i.e. per $$\text{length}^2$$). According to these definitions, the corresponding unit for the mobility is $$\frac{\text{length}^2}{\text{energy} \cdot\text{time}}.$$ (length is e.g. $$\mu\text{m},$$ energy is e.g. $$\text{kJ},$$ time is e.g. $$\text{min}$$). For the unit of the surface energy parameter we have just
$$\text{energy}.$$ The surface energy parameter estimate seems to have the correct unit due to $$\bigg(\frac{\text{energy}}{\text{length}^2} - \text{energy}\frac{1}{\text{length}^2}\bigg).$$ However, the estimate of Kim and Sanders for the mobility constant seems to be incorrect, because $$\frac{1}{\text{time}} \not = \underbrace{\frac{\text{length}^2}{\text{energy} \cdot\text{time}}}_{\widehat = M} \cdot \underbrace{\frac{1}{\text{length}^2}}_{\widehat = \Delta} \cdot \underbrace{\frac{\text{energy}}{\text{length}^2}}_{\widehat = (\frac{\partial \mu}{\partial c} - \kappa \Delta c)}.$$

This former question was already concerned with the units of the Cahn-Hilliard equation where the mobility $$M$$ was determined to have the unit $$\frac{\text{length}^4}{\text{energy} \cdot \text{time}},$$ which in fact would lead to the correct unit.

Question 2: Do I misunderstand something?

• You multiplied by an extra $\frac{1}{length^2}$ Commented Nov 11, 2022 at 23:24
• Yes, but I have to do that. The ''extra'' $\frac{1}{\text{length}^2}$ comes from the Laplacian in front of the bracket in the Cahn-Hilliard equation. Commented Nov 12, 2022 at 2:54
• I edited the corresponding displaymath-line to make the origin of the ''extra'' $\frac{1}{\text{length}^2}$ clear. Commented Nov 12, 2022 at 3:04
• Do you know for a fact that the units of chemical potential given are $\frac{energy}{length^2}$? It usually has $\frac{J}{mol}$ units Commented Nov 12, 2022 at 4:17
• The chemical potential is usually normed by the molar area $A_m$ (unit $\text{length}^2/\text{mol}$), i.e., we consider $\frac{1}{A_m} \mu$ and hence $\frac{\text{mol}}{\text{length}^2} \cdot \frac{\text{energy}}{\text{mol}}$ Commented Nov 12, 2022 at 5:35