lj potential added to basics chapter

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@@ -1,2 +1,39 @@
 \chapter{Basics}
 
+\section{Molecular dynamics simulations}
+
+
+\subsection{Potentials}
+
+\subsubsection{The Lennard-Jones potential}
+
+The L-J potential is a realistic two body pair potential and is of the form
+\begin{equation}
+U^{LJ}(r) = 4 \epsilon \Big[ \Big( \frac{\sigma}{r} \Big)^{12} - \Big( \frac{\sigma}{r} \Big)^6 \Big] \, \textrm{,}
+\label{eq:lj-p}
+\end{equation}
+where $r$ denotes the disatnce between the two atoms.
+
+The attractive tail for large separations $(\sim r^{-6})$ is essentially due to correlations between electron clouds surrounding the atoms. The attractive part is also known as {\em van der Waals} or {\em London} interaction.
+It can be derived classically by considering how two charged spheres induce dipol-dipol interactions into each other, or by considering the interaction between two oscillators in a quantum mechanical way.
+
+The repulsive term $(\sim r^{-12})$ captures the non-bonded overlap of the electron clouds.
+It does not have a true physical motivation, other than the exponent being larger than $6$ to get a steep rising repulsive potential wall at short distances.
+Chosing $12$ as the exponent of the repulsive term it is just the square of the attractive term which makes the potential evaluable in a very efficient way.
+
+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 $\vec{r}$)
+\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 half the negative well depth.
+The $i$th component of the force $F^j$ on particle $j$ is obtained by
+\begin{equation}
+F_i^j = - \frac{\partial}{\partial x_i} U^{LJ}(r) \, \textrm{.}
+\label{eq:lj-f}
+\end{equation}
+