9-3 Lennard-Jones potential

[EN CONSTRUCCION]

Functional form

The 9-3 Lennard-Jones potential is related to the standard Lennard-Jones potential.

It takes the form:

$V(r)={\frac {3{\sqrt {3}}}{2}}\epsilon \left[\left({\frac {\sigma }{r}}\right)^{9}-\left({\frac {\sigma }{r}}\right)^{3}\right].$

The minimum value of $V(r)$ is obtained at $r=r_{min}$, with

• $V\left(r_{min}\right)=-\epsilon$,

• ${\frac {r_{min}}{\sigma }}=3^{1/6}$

Applications

It is commonly used to model the interaction between the particles of a fluid with a flat structureless solid wall.

Interaction between a solid and a fluid molecule

Let us consider the space divided in two regions:

• $x<0$: this region is occupied by a diffuse solid with density $\rho _{s}$ composed of 12-6 Lennard-Jones atoms

with paremeters $\sigma _{s}$ and $\epsilon _{a}$

Our aim is to compute the total interaction between this solid and a molecule located at a position $x_{f}>0$. Such an interaction can be computed using cylindrical coordinates ( I GUESS SO, at least).

The interaction will be:

$V_{W}\left(x\right)=4\epsilon _{sf}\rho _{s}\int _{0}^{2\pi }d\phi \int _{-\infty }^{-x}dz\int _{0}^{\infty }{\textrm {dr}}\left[\sigma ^{12}{\frac {r}{(r^{2}+z^{2})^{6}}}-\sigma ^{6}{\frac {r}{(r^{2}+z^{2})^{3}}}\right].$

$V_{W}\left(x\right)=8\pi \epsilon _{sf}\rho _{s}\int _{-\infty }^{-x}{{\textrm {d}}z}\left[{\frac {\sigma ^{12}}{10(r^{2}+z^{2})^{5}}}-{\frac {\sigma ^{6}}{4(r^{2}+z^{2})^{2}}}\right]_{r=\infty }^{r=0}.$

$V_{W}\left(x\right)=8\pi \epsilon _{sf}\rho _{s}\int _{-\infty }^{-x}{{\textrm {d}}z}\left[{\frac {\sigma ^{12}}{10z^{10}}}-{\frac {\sigma ^{6}}{4z^{4}}}\right];$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V_{W}\left(x\right)=8\pi \epsilon _{sf}\rho _{s}\left[-{\frac {\sigma ^{12}}{90z^{9}}}+{\frac {\sigma ^{6}}{12z^{3}}}\right]_{z=-\infty }^{z=-x};}

$V_{W}\left(x\right)={\frac {4\pi \epsilon _{sf}\rho _{s}\sigma ^{3}}{3}}\left[{\frac {\sigma ^{9}}{15x^{9}}}-{\frac {\sigma ^{3}}{2x^{3}}}\right]$

[TO BE CONTINUED]