Lennard-Jones on a Line

Imagine a particle (the ``probe'') fixed at a distance $a$ from a line of populated by particles . The probe and particles on the line interact via a pairwise potential:

$$u(r) = 4\left[r^{-12}-r^{-6}\right]$$ (360)

This is the Lennard-Jones potential in reduced units. The particles on the line do not interact with each other, but they remain fixed on the line.

Notation. A probe particle lies at a distance $a$ from a line (the $x$-axis) populated by Lennard-Jones particles. The line is divided into ``cells'' indexed by $j$, and the interaction of the probe with a particle in cell $j$ depends on the distance $r_j$.

Let us now write the total potential. We do not know exactly where the fixed particles are on the line, so we do the following. Divide the $y$-axis into an infinite number of finitely-sized divisions, or ``cells,'' indexed by a counter $j$. Let the quantity $n_j$ be the occupation number of division $j$, which is 0 if the division is unoccupied, and 1 if it is occupied. We must make the assumption that the cells are so small that there can either be one or no particles in any cell. We can write the total potential by summing over all cells:

$$\mathscr{U} = \sum_j u\left(r_j\right) n_j$$ (361)

where $r_j$ is the distance from the probe to cell $j$. Now we can pass into the continuum limit by considering that $n_j \rightarrow n\left(x\right)dx$, where $n\left(x\right)$ is the density of particles on the line, and thus $n(x)dx$ is the number of particles on the line between $x$ and $x + dx$. So, in a continuum representation:

$$\mathscr{U}\left(a\right) = \int_{-\infty}^{\infty} u\left[r\left(x\right)\right] n\left(x\right) dx$$ (362)

$$= \int_{-\infty}^{\infty} u\left[\left(a^2+x^2\right)^{1/2}\right] n\left(x\right) dx$$ (363)

where we have used the Pythagorean theorem: $r^2 = a^2 + x^2$. We seek a representation $\mathscr{U}(a)$, that is, the potential of the system as a function of the distance of the probe to the line.

Now, let us assume that the density of particles on the line is uniform and equal to 1 particle per unit length:

$$n\left(x\right) = 1$$

We can express Eq. 363 as

$$\mathscr{U}\left(a\right) = 4\underbrace{\int_{-\infty}^{\in... ..._{-\infty}^{\infty} \frac{dx}{\left(a^2+x^2\right)^{6}}}_{I_2}$$ (364)

At this point we turn to integral tables. Dwight [28] reports on p. 213 in integral # 856.21

$$\int_0^\infty \frac{dx}{\left(a^2+x^2\right)^n} = \frac{1\cd... ...2\cdot 4\cdot 6\cdots \left(2n-2\right)} \frac{\pi}{2a^{2n-1}}$$ (365)

Both integrals $I_1$ and $I_2$ in Eq. 365 are even. Applying Eq. 366, we find

$$I_1 = \frac{63\pi}{256}a^{-11} \mbox{and}$$ (366)

$$I_2 = \frac{3\pi}{8}a^{-5}.$$ (367)

This leaves us with

$$\mathscr{U}\left(a\right) = \frac{3\pi}{2}\left(\frac{21}{32}a^{-11} - a^{-5}\right).$$ (368)

Note that the potential now displays an ``11-5'' distance dependence. The figure below shows $\mathscr{U}(a)$ together with the raw Lennard-Jones 12-6 potential (Eq. 361).

The 11-5 potential $\mathscr{U}\left(a\right)$ as a function of distance to the line, $a$, together with the 12-6 Lennard-Jones potential used to derive it.