-The new silicon dumbbell partner is the one located at $\frac{a}{4}\hkl<1 1 -1>$ compared to the initial one.
-The carbon atom resides in the \hkl(1 1 0) plane along the path.
+During this migration the carbon atom is changing its silicon dumbbell partner.
+The new partner is the one located at $\frac{a}{4}\hkl<1 1 -1>$ relative to the initial one.
+Two of the three bonds to the next neighboured silicon atoms are preserved while the breaking of the other bond and the formation of a new bond is observed.
+The carbon atom resides in the \hkl(1 1 0) plane.
+This transition involves the bond-centerd configuration.
+Due to symmetry it is enough to consider the transition from the bond-centered to the \hkl<1 0 0> configuration.
+In the second path, the carbon atom is changing its silicon partner atom as in path one.
+In addition, the dumbbell configuration is changing
+