calculate the mean friction force denoted by $F^{(1)}$ and $F^{(2)}$. To calculate $F^{(1)}$ we determine the Hellmann-Feynman forces acting on the individual atoms in our *ab-initio* calculation [@feynman:39]. We perform an arithmetic average of all force components of the upper slab parallel to the sliding direction following the path described above, e.g., as indicated in Fig. \[fig:2\] by the dotted line. The contributions of the unrelaxed and the relaxed forces to $F^{(1)}$ are marked by the superscripts $U$ and $R$ while $N_{\mathrm{U}}$ and $N_{\mathrm{R}}$ are the total numbers of sampling points on the unrelaxed and relaxed energy curves yielding $$F^{(1)}= \frac{1}{N_\mathrm{U}} \sum_{i=1}^{N_\mathrm{U}} F^{\mathrm{U}}_{i} + \frac{1}{N_\mathrm{R}} \sum_{j=1}^{N_\mathrm{R}} F^{\mathrm{R}}_{j} \quad .
\label{eq:F1}$$ In contrast, $F^{(2)}$ is defined via the sum over the gains and losses of energy along the sliding path divided by the path length $l$, $$\begin{aligned}
F^{(2)} & = & \frac{1}{l}\Bigg[ \sum_{i^{\prime}=1}^{N_{\mathrm{max}}} \left(E^{\mathrm{U}}_{i^{\prime}} - E^{\mathrm{R}}_{i^{\prime}} \right) - \sum_{j^{\prime}=2}^{N_{\mathrm{min}}} \left(E^{\mathrm{U}}_{j^{\prime}} - E^{\mathrm{R}}_{j^{\prime}}\right) - \nonumber \\
 & & - \sum_{i^{\prime}=1}^{N_{\mathrm{max}}} \sum_{j^{\prime}=2}^{N_{\mathrm{max}}} \left(E^{\mathrm{R}}_{i^{\prime}}-E^{\mathrm{R}}_{j^{\prime}}\right) \Bigg] \quad .
\label{eq:F2}\end{aligned}$$ The index $i^{\prime}$ iterates over all maxima in the unrelaxed energy curve and $j^{\prime}$ covers the minima while the energies are denoted by $E^{\mathrm{U}}$ (unrelaxed) and $E^{\mathrm{R}}$ (relaxed). $N_{\mathrm{min}}$ ($N_{\mathrm{max}}$) is the number of minima (maxima) 