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?

  • $\begingroup$ You multiplied by an extra $\frac{1}{length^2}$ $\endgroup$
    – Sam202
    Commented Nov 11, 2022 at 23:24
  • $\begingroup$ 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. $\endgroup$
    – Henning
    Commented Nov 12, 2022 at 2:54
  • $\begingroup$ I edited the corresponding displaymath-line to make the origin of the ''extra'' $\frac{1}{\text{length}^2}$ clear. $\endgroup$
    – Henning
    Commented Nov 12, 2022 at 3:04
  • 1
    $\begingroup$ 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 $\endgroup$
    – Sam202
    Commented Nov 12, 2022 at 4:17
  • 1
    $\begingroup$ 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 $\frac{\text{mol}}{\text{length}^2} \cdot \frac{\text{energy}}{\text{mol}}$ $\endgroup$
    – Henning
    Commented Nov 12, 2022 at 5:35


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.