Bonded interactions

\[\begin{split}\begin{split} V_{bonded} = &\sum_{(i,j)\in bonds}\frac{k^s_{ij}}{2}(r_{ij}-r_{ij}^0)^2 + \\ &\sum_{(i,j,k)\in angles}\frac{k^{\theta}_{ijk}}{2}(\theta_{ijk}-\theta_{ijk}^0)^2 + \\ &\sum_{(i,j,k,l)\in dihedrals}k^{\phi}_{ijkl}(1-\cos(n\phi_{ijkl} - \phi_{ijkl}^0)) + \\ &\sum_{(i,j,k,l)\in impropers}k^{\psi}_{ijkl}(\psi_{ijkl}-\psi_{ijkl}^0)^2 \end{split}\end{split}\]

Harmonic bond potential

Harmonic bond potential usually used to describe covalent bonds. Its mathematical formula is:

\[V_{bond}=\sum_{(i,j)\in bonds}\frac{k^s_{ij}}{2}(r_{ij}-r_{ij}^0)^2=\sum_{(i,j)\in bonds}V_{bond}^{ij}\]

Here, the sum goes over all bonded pairs \((i,j)\). The potential parameters \(k^s_{ij}\) and \(r_{ij}^0\) are normally based on atom types for atoms \(i\) and \(j\): for each possible combination of types of bonded atoms, there is a set of the two parameters in the forcefield. The current distance between atoms \(r_{ij}\) depends on the coordinates of the two connected atoms. Let us compute the interatomic force for two atoms, connected by a harmonic bond. The potential function term that describes this binary interaction is:

\[V_{bond}^{ij}=\frac{k^s_{ij}}{2}(r_{ij}-r_{ij}^0)^2\]

Hence, for each bonded interaction, each atom will be experiencing the force:

\[\mathbf{f}_{bond_{ij}}^i=\nabla_iV_{bond}^{ij}=\frac{\partial V_{bond}}{\partial r_{ij}}\nabla r_{ij} = -k^s_{ij}(r_{ij}-r_{ij}^0)\frac{\mathbf{r}_{ij}}{r_{ij}}\]

One can show that \(\mathbf{f}_{bond_{ij}}^j = -\mathbf{f}_{bond_{ij}}^i\), which corresponds to the third Newtons law.

Harmonic angle potential

Exercise

Derive the formula for forces that act on each of three particles, connected by harmonic angle potential.