Three-body potential

Fig. 11 Schematic representation of the angle potential (left panel) and the shape of the potential function as a function of angle \(\theta\) (right panel). The minimum of the harmonic potential corresponds to the angle \(\theta_0\).

Introducing a triplet of atoms \((i,j,k)\), forming a particular angle Fig. 11 and a set of angles in a system, \(A\), the angle potential can be written as:

(12)\[V_{angles} = \sum_{(i,j,k) \in A}\frac{k_{ijk}^{\theta}}{2}(\theta_{ijk}-\theta^0_{ijk})^2\]

The force, acting on a particular atom \(l\) due to the angle potential is then

(13)\[\begin{split}\begin{split} \mathbf{f}_{l}=-\nabla_{l}V_{angles} &= -\nabla_{l}\left(\sum_{(i,j,k) \in A}\frac{k_{ijk}^{\theta}}{2}(\theta_{ijk}-\theta^0_{ijk})^2\right)\\ &= -\sum_{(i,j,k) \in A}k_{ijk}^{\theta}(\theta_{ijk}-\theta^0_{ijk})\nabla_{l}\theta_{ijk} \end{split}\end{split}\]

Since, from the atomic coordinates, it is easier to compute \(\cos\theta_{ijk}\) and \(\sin\theta_{ijk}\), let us switch to a \(\cos\) and \(\sin\) representation as follows:

(14)\[\theta_{ijk}=\arccos\left(\cos\theta_{ijk}\right),\]

and

(15)\[\nabla_{l}\theta_{ijk} = \nabla_{l}\left(\arccos\left(\cos\theta_{ijk}\right)\right) = -\frac{1}{\sqrt{1-\cos^{2}\theta_{ijk}}}\nabla_{l}\cos\theta_{ijk} = -\frac{\nabla_{l}\cos\theta_{ijk}}{\sin\theta_{ijk}}.\]

Let us denote vectors that connect particles \(i\), \(j\) and \(k\) in a triplet as \(\mathbf{r}_{ji}=\mathbf{r}_{i}-\mathbf{r}_{j}\) and \(\mathbf{r}_{jk}=\mathbf{r}_{k}-\mathbf{r}_{j}\); and the respective distances as \(r_{ji}=|\mathbf{r}_{ji}|\) and \(r_{jk}=|\mathbf{r}_{jk}|\) Then

(16)\[\cos\theta_{ijk}=\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{|\mathbf{r}_{ji}||\mathbf{r}_{jk}|}=\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}}\]

Here, \(\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}\) is a dot product of vectors \(\mathbf{r}_{ji}\) and \(\mathbf{r}_{jk}\). It is clear, that \(\nabla_{l}\cos\theta_{ijk}\) will vanish if \(l\) does not belong to the triplet \((i,j,k)\). Considering all three cases (\(l=i\), \(l=j\) and \(l=k\)) separately and using Eq. Equation 16 we obtain:

(17)\[\begin{split}\begin{split} \nabla_{l}\cos\theta_{ijk}\bigg|_{l=i}&=\nabla_{l}\left(\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}}\right)\bigg|_{l=i}=\left(\frac{1}{r_{ji}r_{jk}}\nabla_{l}\left(\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}\right)-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}^{2}r_{jk}}\nabla_{l}r_{ji}\right)\bigg|_{l=i}=\\ &=\frac{1}{r_{ji}r_{jk}}\mathbf{r}_{jk}-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}^{2}r_{jk}}\frac{\mathbf{r}_{ji}}{r_{ji}}=\frac{1}{r_{ji}}\left[\frac{\mathbf{r}_{jk}}{r_{jk}}-\cos\theta_{ijk}\frac{\mathbf{r}_{ji}}{r_{ji}}\right]\text{,}\\ \end{split}\end{split}\]

(18)\[\begin{split}\begin{split} \nabla_{l}\cos\theta_{ijk}\bigg|_{l=k}&=\nabla_{l}\left(\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}}\right)\bigg|_{l=k}=\left(\frac{1}{r_{ji}r_{jk}}\nabla_{l}\left(\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}\right)-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}^{2}}\nabla_{l}r_{jk}\right)\bigg|_{l=k}=\\ &=\frac{1}{r_{ji}r_{jk}}\mathbf{r}_{ji}-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}^{2}}\frac{\mathbf{r}_{jk}}{r_{jk}}=\frac{1}{r_{jk}}\left[\frac{\mathbf{r}_{ji}}{r_{ji}}-\cos\theta_{ijk}\frac{\mathbf{r}_{jk}}{r_{jk}}\right]\text{, and}\\ \end{split}\end{split}\]

(19)\[\begin{split}\begin{split} \nabla_{l}\cos\theta_{ijk}\bigg|_{l=j}&=\nabla_{l}\left(\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}}\right)\bigg|_{l=j}=\\ &=\left(\frac{1}{r_{ji}r_{jk}}\nabla_{l}\left(\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}\right)-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}^{2}r_{jk}}\nabla_{l}r_{ji}-\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}^{2}}\nabla_{l}r_{jk}\right)\bigg|_{l=j}=\\ &=-\frac{1}{r_{ji}r_{jk}}\mathbf{r}_{jk}-\frac{1}{r_{ji}r_{jk}}\mathbf{r}_{ji}+\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}^{2}r_{jk}}\frac{\mathbf{r}_{ji}}{r_{ji}}+\frac{\mathbf{r}_{ji}\cdot\mathbf{r}_{jk}}{r_{ji}r_{jk}^{2}}\frac{\mathbf{r}_{jk}}{r_{jk}}=\\ &=\frac{1}{r_{ji}}\left[\cos\theta_{ijk}\frac{\mathbf{r}_{ji}}{r_{ji}}-\frac{\mathbf{r}_{jk}}{r_{jk}}\right] + \frac{1}{r_{jk}}\left[\cos\theta_{ijk}\frac{\mathbf{r}_{jk}}{r_{jk}}-\frac{\mathbf{r}_{ji}}{r_{ji}}\right]=\\ &=-\nabla_{l}\cos\theta_{ijk}\bigg|_{l=i}-\nabla_{l}\cos\theta_{ijk}\bigg|_{l=k} \end{split}\end{split}\]

Summirizing Eqs. Equation 13, Equation 15 and Equation 17 to Equation 19 we see that, one can compute three components of force \(\mathbf{f}_{i}\), \(\mathbf{f}_{j}\) and \(\mathbf{f}_{k}\), acting on each atom in the triplet \((i,j,k)\) due to the angle potential between atoms \(i\), \(j\) and \(k\) using the following relations:

(20)\[\begin{split}\begin{split} \mathbf{f}_{i}&=k_{ijk}^{\theta}(\theta_{ijk}-\theta^0_{ijk})\left(-\frac{1}{\sin\theta_{ijk}}\right)\frac{1}{r_{ji}}\left[\cos\theta_{ijk}\frac{\mathbf{r}_{ji}}{r_{ji}}-\frac{\mathbf{r}_{jk}}{r_{jk}}\right]\\ \mathbf{f}_{k}&=k_{ijk}^{\theta}(\theta_{ijk}-\theta^0_{ijk})\left(-\frac{1}{\sin\theta_{ijk}}\right)\frac{1}{r_{jk}}\left[\cos\theta_{ijk}\frac{\mathbf{r}_{jk}}{r_{jk}}-\frac{\mathbf{r}_{ji}}{r_{ji}}\right]\\ \mathbf{f}_{j}&=-\mathbf{f}_{i}-\mathbf{f}_{k} \end{split}\end{split}\]

Hence, the forces acting on three atoms connected by the angle potential are related. Consequently it is more efficient to use the modification of the potential pair based parallelization approach.