X-Git-Url: https://hackdaworld.org/gitweb/?a=blobdiff_plain;f=posic%2Fthesis%2Fbasics.tex;h=6addfab29058d243a7890eb7c5e1369c2e5c2079;hb=633db250e42a5724c851597ee938cfa93df0665b;hp=d3c91aa6ec22a6ea8d3649f7bb0be189b4c8a27d;hpb=08f9c1435f017ec1f9202e0632cfb8989613e308;p=lectures%2Flatex.git

diff --git a/posic/thesis/basics.tex b/posic/thesis/basics.tex
index d3c91aa..6addfab 100644
--- a/posic/thesis/basics.tex
+++ b/posic/thesis/basics.tex
@@ -94,16 +94,16 @@ Chosing $12$ as the exponent of the repulsive term it is just the square of the
 The constants $\epsilon$ and $\sigma$ are usually determined by fitting to experimental data.
 $\epsilon$ accounts to the depth of the potential well, where $\sigma$ is regarded as the radius of the particle, also known as the van der Waals radius.
 
-Writing down the derivation of the Lennard-Jones potential in respect to $x_i$ (the $i$th component of the distance vector ${\bf r}$)
+Writing down the derivative of the Lennard-Jones potential in respect to $x_i$ (the $i$th component of the distance vector ${\bf r}$)
 \begin{equation}
 \frac{\partial}{\partial x_i} U^{LJ}(r) = 4 \epsilon x_i \Big( -12 \frac{\sigma^{12}}{r^{14}} + 6 \frac{\sigma^6}{r^8} \Big)
 \label{eq:lj-d}
 \end{equation}
 one can easily identify $\sigma$ by the equilibrium distance of the atoms $r_e=\sqrt[6]{2} \sigma$.
 Applying the equilibrium distance into \eqref{eq:lj-p} $\epsilon$ turns out to be the negative well depth.
-The $i$th component of the force $F^j$ on particle $j$ is obtained by
+The $i$th component of the force is given by
 \begin{equation}
-F_i^j = - \frac{\partial}{\partial x_i} U^{LJ}(r) \, \textrm{.}
+F_i = - \frac{\partial}{\partial x_i} U^{LJ}(r) \, \textrm{.}
 \label{eq:lj-f}
 \end{equation}
 
@@ -172,50 +172,27 @@ It is of the form:
 \begin{eqnarray}
 b_{ij} & = & \chi_{ij} (1 + \beta_i^{n_i} \zeta^{n_i}_{ij})^{-1/2n_i} \\
 \zeta_{ij} & = & \sum_{k \ne i,j} f_C (r_{ik}) \omega_{ik} g(\theta_{ijk}) \\
-g(\theta_{ijk}) & = & 1 + c_i^2/d_i^2 - c_i^2/[d_i^2 + (h_i - \cos \theta_{ijk})^2]
+g(\theta_{ijk}) & = & 1 + c_i^2/d_i^2 - c_i^2/[d_i^2 + (h_i - \cos \theta_{ijk})^2] \\
 \end{eqnarray}
 where $\theta_{ijk}$ is the bond angle between bonds $ij$ and $ik$.
 This is illustrated in Figure \ref{img:tersoff_angle}.
 
 \printimg{!h}{width=8cm}{tersoff_angle.eps}{Angle between bonds of atoms $i,j$ and $i,k$.}{img:tersoff_angle}
 
-In order to calculate the forces the derivation of the potential with respect to $x^i_n$ (the $n$th component of the position vector of atom $i$ $\equiv$ ${\bf r}_i$) has to be known.
-This is gradually done in the following.
-The $n$th component of the force acting on atom $i$ is
-\begin{eqnarray}
-F_n^i & = & - \frac{\partial}{\partial x_n} \sum_{j \neq i} V_{ij} \nonumber\\
- & = & \sum_{j \neq i} \Big( \partial_{x_n^i} f_C(r_{ij}) \big[ f_R(r_{ij}) + b_{ij} f_A(r_{ij}) \big] + \nonumber\\
-& & + f_C(r_{ij}) \big[ \partial_{x_n^i} f_R(r_{ij}) + b_{ij} \partial_{x_n^i} f_A(r_{ij}) + f_A(r_{ij}) \partial_{x_n^i} b_{ij} \big] \Big)
-\end{eqnarray}
-The cutoff function $f_C$ derivated with repect to $x^i_n$ is
+Here comes an explanation, energy per bond monotonically decreasing with the amount of bonds and so on and so on \ldots
+
+The force acting on atom $i$ is given by the derivative of the potential energy.
+For a three body potential ($V_{ij} \neq V{ji}$) the derivation is of the form
 \begin{equation}
-\partial_{x^i_n} f_C(r_{ij}) =
-  - \frac{1}{2} \sin \Big( \pi (r_{ij} - R_{ij}) / (S_{ij} - R_{ij}) \Big) \frac{\pi x^i_n}{(S_{ij} - R_{ij}) r_{ij}}
-\label{eq:d_cutoff}
+\nabla_{{\bf r}_i} E = \frac{1}{2} \big[ \sum_j ( \nabla_{{\bf r}_i} V_{ij} + \nabla_{{\bf r}_i} V_{ji} ) + \sum_k \sum_j \nabla_{{\bf r}_i} V_{jk} \big] \textrm{ .}
 \end{equation}
-for $R_{ij} < r_{ij} < S_{ij}$ and otherwise zero.
-The derivations of the repulsive and attractive part are:
-\begin{eqnarray}
-\partial_{x_n^i} f_R(r_{ij}) & = & - \lambda_{ij} \frac{x_n^i - x_n^j}{r_{ij}} A_{ij} \exp (-\lambda_{ij} r_{ij})\\
-\partial_{x_n^i} f_A(r_{ij}) & = & \mu_{ij} \frac{x_n^i - x_n^j}{r_{ij}} B_{ij} \exp (-\mu_{ij} r_{ij}) \textrm{ .}
-\end{eqnarray}
-The angle $\theta_{ijk}$ can be expressed by the atom distances with the law of cosines:
-\begin{eqnarray}
-\theta_{ijk} & = & \arccos \Big( (r_{ij}^2 + r_{ik}^2 - r_{jk}^2)/(2 r_{ij} r_{ik}) \Big) \\
-\partial_{x^i_n} \theta_{ijk} & = & 
-\frac{-1}{\sqrt{1 - ((r_{ik}^2+r_{ij}^2-r_{jk}^2)/2r_{ik}r_{ij})^2}} \times \nonumber\\
- & & \times \Big( \frac{4 r_{ik}r_{ij} (2 x^i_n - x^k_n - x^j_n) + 2(x^j_n - x^i_n)\frac{r_{ik}}{r_{ij}} + 2(x^k_n - x^i_n)\frac{r_{ij}}{r_{ik}} }{4 r^2_{ik} r^2_{ij}}\Big) \label{eq:d_theta}
-\end{eqnarray}
-Using the expressions \eqref{eq:d_cutoff} and \eqref{eq:d_theta} the derivation of $b_{ij}$ with respect to $x^i_n$ can be written as:
-\begin{eqnarray}
-\partial_{x^i_n} b_{ij} & = &
-- \frac{1}{2n_i} \chi_{ij} \Bigg( 1 + \beta_i^{n_i} \Bigg[ \sum_{k \ne i,j} \bigg( f_C(r_{ij}) \omega_{ik} \Big( 1 + \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2} \Big) \bigg)^{n_i} \Bigg] \Bigg)^{-\frac{1}{2n_i} - 1} \times \nonumber\\
-&& \times n_i \beta_i^{n_i} \Bigg[ \sum_{k \ne i,j} f_C(r_{ik}) \omega_{ik} \Big( 1 \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2}  \Big) \Bigg]^{n_i -1} \times \nonumber\\
-&& \times \sum_{k \ne i,j} \Bigg( \omega_{ik} \Big( 1 + \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2} \Big) \partial_{x^i_n} f_C(r_{ik}) + \nonumber\\
-&& + f_C(r_{ik}) \omega_{ik} (-1) \frac{c_i^2}{(d_i^2 + (h_i - \cos \theta_{ijk})^2)^2} \times \nonumber\\
-&& \times 2 \Big( h_i - \cos \theta_{ijk} \Big) \sin \theta_{ijk} \partial_{x^i_n} \theta_{ijk} \Bigg)
-\end{eqnarray}
+The force is then given by
+\begin{equation}
+F^i = - \nabla_{{\bf r}_i} E \textrm{ .}
+\end{equation}
+The details of the Tersoff potential derivative can be seen in appendix \ref{app:d_tersoff}.
 
+And here comes why we use it. Advantages and disadvantages compared to other interaction potentials, maybe this is best at the very end of all potentials \ldots
 
 \subsubsection{The Brenner potential}