Qualifier

Elastic Interaction

The lattice model that we begin to study in this paper is a generalization of the model introduced by us in [18]. We consider a discrete system, where we don't allow any type of superposition of particles. When a deposition attempt is made on a grid location with all the nearest neighbors unoccupied, the particle diffuses to a local minimum of the potential. In this chapter we concern ourselves only we the calculation of the form of this potential due to the deformation of the substrate.

A surface distribution of forces is generated that, in turn, induces an interaction between particles mediated by the lattice deformation. We can establish the form of those interactions from theoretical considerations in the elasticity of solids. We derive the form of the potential in accordance with Politi et al [49] and Marchenko [36]. We then proceed to calculate the total energy using a form of the potential to be given by Eq. 3.1. This is the most generic form, since we can intuitively see that the interaction should also be proportional to some intrinsic property of the particles, such as mass or charge. As we will see, the interaction constant depends on the elastic properties of the substrate's material [48], the lattice constant, and the lattice mismatch.

$\displaystyle E_{ij}=\frac{Gm_im_j}{r^3}$

(3.1)

Elasticity Arguments

**Figure 3.1:** Elastic deformation of a substrate. The circles represent the original positions of the particles and the diamonds, the final positions due to the elastic deformation. From[52]
$\resizebox{5cm}{!}{ \includegraphics{elastic}}$

If we consider a small elastic deformation of an initially flat surface given, at each point, by a vector of components , we have that the change is surface energy, $\Delta E_S$ , is given by:

$\displaystyle \Delta E_S=-\int f_\mu u_\mu\mathrm{d}S_0, \mu=1,2$

(3.2)

where the integral ranges over the original flat surface and the $f_\mu$ terms are the tangential forces that can be written as the divergence of the surface stress tensor, $\beta_{\mu\nu}$ .

$\displaystyle f_\mu= \frac{\partial \beta_{\mu\nu}}{\partial x_\nu}$

(3.3)

Assuming a point distribution of forces, we obtain:

$\displaystyle f_\mu=A_{\mu\nu} \frac{\partial \delta(\mathbf{\rho})}{\partial x_\nu}$

(3.4)

where we assume that the defect is located at $\mathbf{\rho}=0$ . Following [30,36] we can determine than isotropic defects ( $A_{\mu\nu}=m\delta_{\mu\nu}$ ) repel in accordance with:

$\displaystyle E(\rho)=\frac{1-\sigma^2}{\pi E}\frac{m^2}{\rho^3}$

(3.5)

where is the Young Modulus and $\sigma$ is the Poisson ratio.

Energy Calculation

We can conceive of $E_{ij}$ , the interaction energy between particles

and

, as being a symmetric matrix with a null main diagonal:

$\displaystyle E_{ij}=\left(\begin{array}{llll} 0 & E_{12} & \ldots & E_{1n} \\... ...vdots & & \ddots & \vdots \\ E_{n1} & E_{n2} & \ldots & 0 \end{array}\right)$

(3.6)

Hence, the total energy possessed by all particles must be:

$\displaystyle E_{T}=\sum_{i<j}E_{ij}$

(3.7)

where

denotes that the sum is taken over all pairs of particles, which corresponds to the sum over all elements bellow the main diagonal of our matrix. We must, however, exercise some care when applying the previous relations since we can only simulate a finite system of size, say,

. This implies that we must use periodic boundary conditions that correspond mathematically to considering that we have, instead of an isolated system, an infinite number of contiguous replicas of our system.

To perform the calculation of the effective potential in two dimensions we will separate the total value of the energy in its two contributions, one due to the interaction of a particle with its replicas and another one due to the interaction between two particles and the replicas of one of them. We will follow the procedure outlined in [8] and we will use the following definition:

$\displaystyle r^{-\frac{n}{2}}=\frac{1}{\Gamma (\frac{n}{2})}\int_0^\infty e^{-ru}u^{\frac{n}{2}-1}~\mathrm{d}u$

(3.8)

Interaction of a particle with its replicas

**Figure 3.2:** A particle and its replicas in 2D
$\resizebox{5cm}{!}{ \includegraphics{poten2d-1}}$

To calculate this contribution to the energy we need to evaluate a sum of the form:

$\displaystyle E_i=\sum_{k_1,k_2}\frac{Gm_i^2}{[(k_1 L)^2+(k_2 L)^2]^\frac{3}{2}}$

(3.9)

where we define $G\equiv \frac{1-\sigma^2}{\pi E}$ and

and

cannot be simultaneously null. Using Eq. 3.8 and simplifying, we obtain:

$\displaystyle E_i=\frac{Gm_i^2}{\Gamma(\frac{3}{2})L^3}\int_0^\infty \sum_{k1,k2} e^{-(k_1^2+k_2^2)u} \sqrt{u}~\mathrm{d}u$

(3.10)

By defining[66]:

$\displaystyle \theta_3(q)=\theta_3(0,q)=\sum_{k=-\infty}^{\infty} q^{k^2}$

(3.11)

where $\theta_3(p,q)$ is the well known Jacoby elliptic function of the third kind and simplifying, we obtain:

$\displaystyle E_i=\frac{Gm_i^2 \pi^\frac{3}{2}}{\Gamma(\frac{3}{2})L^3}\left\{ ... ...3^2(e^{-\pi\alpha})-1\right] \sqrt{\alpha}~\mathrm{d}\alpha-\frac{2}{3}\right\}$

(3.12)

Defining the function $\phi_m(x)$ such that,

$\displaystyle \phi_m(x)=\int_1^\infty\beta^me^{-\beta x}\mathrm{d}\beta$

(3.13)

and expanding $\theta_3^2$ as a Taylor series and summing over all particles, we obtain the final result for the self-interaction energy.

$\displaystyle E_1^{(2)}=\sum_i\frac{Gm_i^2}{\Gamma(\frac{3}{2})L^3}\left\{\sum_... ...\infty\left[a_n\phi_{1/2}(n\pi)+a_n\phi_{-3/2}(n\pi)\right]+\frac{1}{3}\right\}$

(3.14)

Interaction between two particles

**Figure 3.3:** Coordinate relations between a deposited particle, a previously existing one, and its replicas in 2D
$\resizebox{5cm}{!}{ \includegraphics{poten2d-2}}$

In this case, the sum we must perform is (see Fig. 3.3):

$\displaystyle E_{ij}=\sum_{k1,k2}\frac{Gm_im_j}{[(a+k_1L)^2+(b+k_2L)^2]^\frac{3}{2}}$

(3.15)

Simplifying:

$\displaystyle E_{ij}=\frac{Gm_im_j}{L^3}\sum_{k1,k2}\frac{Gm_im_j}{[(x+k_1)^2+(y+k_2)^2]^\frac{3}{2}}$

(3.16)

Where

and

. Using Eq. 3.8 we find that:

$\displaystyle E_{ij}=\frac{Gm_im_j}{\Gamma(\frac{3}{2})L^3}\int_0^\infty\sum_{k1,k2} e^{-[(x+k_1)^2+(y+k_2)^2]u} \sqrt{u}~\mathrm{d}u$

(3.17)

Considering just the summations:

$\displaystyle \sum_ke^{-(k+a)^2u}=\sum_ke^{-(k^2+a^2+2ak)u}=e^{-a^2u}\theta_3(-\mathrm{i}a u,e^{-u})$

(3.18)

and using our previous expression, we obtain:

**Figure 3.4:** The approximations used in the integration of $\theta _3(q)$ . A curve labeled as Integral represents the result of Eq. 3.20 and the curve labeled Series represents the series expansion of Eq. 3.21
$\resizebox{10cm}{!}{ \includegraphics{theta3}}$

$\displaystyle E_{ij}=\frac{Gm_im_j}{\Gamma(\frac{3}{2})L^3}\int_0^\infty e^{-(x... ...a_3(-\mathrm{i} x u,e^{-u})\theta_3(-\mathrm{i} y u,e^{-u})\sqrt{u}~\mathrm{d}u$

(3.19)

According to [6,64,63],

$\displaystyle \sum_{k=-\infty}^\infty e^{-u(k^2-2ak)}=\int_{-\infty}^\infty e^{-u(x^2+2 a x)}~\mathrm{d}x+R_a=e^{a^2 u}\sqrt{\frac{\pi}{u}}+R_u$

(3.20)

Where the error,

, varies with

but is always inferior to

. The value of

tends to 0 when

tends to infinity. Due to this behavior of the error, this approximation is only useful within a certain interval, $u\in~]0,-\log 0.3]$ , where the error is always inferior to

. In the remaining domain of integration, the function can be well approximated using the first few terms of the series given by:

$\displaystyle \sum_{k=-\infty}^\infty e^{-(k^2+2ak)u}\approx\sum_{k=-3}^3 q^{k^2+2ak}$

(3.21)

with $q\equiv e^{-u}\in\left[0,0.3\right]$ . Where the error is also always inferior to 0.001. Using

$\displaystyle \begin{equation} \pi\int_{0.3}^1\frac{q^{r^2-1}}{\sqrt{-\log q}}... ...c{1}{2}}}{4}\frac{1-Erf\left[1.09726\sqrt{1+a}\right]}{(1+a)^2} \end{equation}$

We finally obtain that the total energy is given by:

$\begin{multline} E_E=\sum_i\frac{Gm_i^2}{\Gamma(\frac{3}{2})L^3}\left\{\sum_{n=... ...1+2yk_2}\right]}{(r^2+k_1^2+k_2^2+2xk_1+2yk_2)^2}\right]\right\} \end{multline}$