-with $N_{\text{Si}}$ and $n_{\text{Si}}$ being the number of Si atoms and the Si density respectively of the corresponding material.
-Due to a slightly lower Si density of 3C-SiC compared to c-Si an increase of x \% of the total volume would be expected for precipitate with a radius of 3 nm embedded in
-
-Calc expected increase due to Si density mismatch ...
-Obviously the surrounding matrix is chosen big enough to exclude size effects ...
-Nice, since obviously matrix is big enough to exclude size effects in the system in which pbc are applied, we can consider it single precipitate in a infinite Si matrix.
-A new peak for the silicon pairs arises at 0.307 nm.
-It is identical to the peak of the C-C distribution around that value.
-It corresponds to second next neighbours in 3C-SiC, which applies for Si as well as C pairs.
-The bumps of the Si-Si distribution at higher distances, which are marked by green arrows and do not exist in plain c-Si, can be explained in the same manner.
-They correspond to the fourth and sixth next neighbour in 3C-SiC.
-Again, these peaks apply to Si and C pairs and indeed it is easily identifiale how the C-C peaks at contribute to the bumps observed in the Si-Si distribution.
-
-4.34 \AA{} compared to 4.36 \AA{}.
+with the notation used in table \ref{table:md:sic_prec}.
+The lattice constant of plain c-Si at $20\,^{\circ}\mathrm{C}$ can be determined more accurately by the side lengthes of the simulation box of an equlibrated structure instead of using the radial distribution data.
+By this a value of $a_{\text{plain c-Si}}=5.439\text{ \AA}$ is obtained.
+The same lattice constant is assumed for the c-Si surrounding in the precipitate configuration $a_{\text{c-Si of precipitate configuration}}$ since peaks in the radial distribution match the ones of plain c-Si.
+Using $a_{\text{3C-SiC of precipitate configuration}}=4.34\text{ \AA}$ as observed from the radial distribution finally results in an increase of the initial volume by 0.12 \%.
+However, each side length and the total volume of the simulation box is increased by 0.20 \% and 0.61 \% respectively compared to plain c-Si at $20\,^{\circ}\mathrm{C}$.
+Since the c-Si surrounding resides in an uncompressed state the excess increase must be attributed to relaxation of strain with the strain resulting from either the compressed precipitate or the 3C-SiC/c-Si interface region.
+This also explains the possibly identified slight increase of the c-Si lattice constant in the surrounding as mentioned earlier.
+As the pressure is set to zero the free energy is minimized with respect to the volume enabled by the Berendsen barostat algorithm.
+Apparently the minimized structure with respect to the volume is a configuration of a small compressively stressed precipitate and a large amount of slightly stretched c-Si in the surrounding.
+
+In the following the 3C-SiC/c-Si interface is described in further detail.
+One important size analyzing the interface is the interfacial energy.
+It is determined exactly in the same way than the formation energy as described in equation \eqref{eq:defects:ef2}.
+Using the notation of table \ref{table:md:sic_prec} and assuming that the system is composed out of $N^{\text{3C-SiC}}_{\text{C}}$ C atoms forming the SiC compound plus the remaining Si atoms, the energy is given by
+\begin{equation}
+ E_{\text{f}}=E-
+ N^{\text{3C-SiC}}_{\text{C}} \mu_{\text{SiC}}-
+ \left(N^{\text{total}}_{\text{Si}}-N^{\text{3C-SiC}}_{\text{C}}\right)
+ \mu_{\text{Si}} \text{ ,}
+\label{eq:md:ife}
+\end{equation}
+with $E$ being the free energy of the precipitate configuration at zero temperature.
+An interfacial energy of 2267.28 eV is obtained.
+The amount of C atoms together with the observed lattice constant of the precipitate leads to a precipitate radius of 29.93 \AA.
+Thus, the interface tension, given by the energy of the interface devided by the surface area of the precipitate is $20.15\,\frac{\text{eV}}{\text{nm}^2}$ or $3.23\times 10^{-4}\,\frac{\text{J}}{\text{cm}^2}$.
+This is located inside the eperimentally estimated range of $2-8\times 10^{-4}\,\frac{\text{J}}{\text{cm}^2}$ \cite{taylor93}.