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Inversion Angle Potentials

 

The inversion angle and associated vectors

The inversion angle potentials describe the interaction arising from a particular geometry of three atoms around a central atom. The best known example of this is the arrangement of hydrogen atoms around nitrogen in ammonia to form a trigonal pyramid. The hydrogens can `flip' like an inverting umbrella to an alternative structure, which in this case is identical, but in principle causes a change in chirality. The force restraining the ammonia to one structure can be described as an inversion potential (though it is usually augmented by valence angle potentials also). The inversion angle is defined in the figure above - note that the inversion angle potential is a sum of the three possible inversion angle terms. It resembles a dihedral potential in that it requires the specification of four atomic positions.

The potential functions available in DL_POLY_2 are as follows.

1. Harmonic: (harm)
   

   $$U(\phi_{ijkn})= {1\over 2} k (\phi_{ijkn} - \phi_0)^2$$ (2.51)

   
   
2. Harmonic cosine: (hcos)
   

   $$U(\phi_{ijkn})={k\over 2}(cos(\phi_{ijkn}) -cos(\phi_{0}))^{2}$$ (2.52)

   
   
3. Planar potential: (plan)
   

   $$U(\phi_{ijkn})= A \left [ 1 - \cos (\phi_{ijkn})\right]$$ (2.53)

   
   

In these formulae $\phi_{ijkn}$ is the inversion angle defined by

$$\phi_{ijkn}=\cos^{-1}\left \{\frac{\mbox{$\underline{r}$}_{ij}\cdot\mbox{$\underline{w}$}_{kn}}{r_{ij}w_{kn}}\right \},$$ (2.54)

with

$$\mbox{$\underline{w}$}_{kn}=(\mbox{$\underline{r}$}_{ij}\cdo... ...{$\underline{\hat{v}}$}_{kn})\mbox{$\underline{\hat{v}}$}_{kn}$$ (2.55)

and the unit vectors

$$\mbox{$\underline{\hat{u}}$}_{kn} = (\mbox{$\underline{\hat{r}}$}_{ik}+\mbox{$\underline{\hat{r}}$}_{... .../ \vert\mbox{$\underline{\hat{r}}$}_{ik}+\mbox{$\underline{\hat{r}}$}_{in}\vert$$

$$\mbox{$\underline{\hat{v}}$}_{kn} = (\mbox{$\underline{\hat{r}}$}_{ik}-\mbox{$\underline{\hat{r}}$}_{... ... \vert\mbox{$\underline{\hat{r}}$}_{ik}-\mbox{$\underline{\hat{r}}$}_{in}\vert.$$ (2.56)

As usual, $\mbox{$\underline{r}$}_{ij}=\mbox{$\underline{r}$}_{j}-\mbox{$\underline{r}$}_{i}$ etc. and the hat $\mbox{$\underline{\hat{r}}$}$ indicates a unit vector in the direction of $\mbox{$\underline{r}$}$. The total inversion potential requires the calculation of three such angles, the formula being derived from the above using the cyclic permutation of the indices $j\rightarrow k \rightarrow n \rightarrow j$ etc.

Equivalently, the angle $\phi_{ijkn}$ may be written as

$$\phi_{ijkn}=\cos^{-1} \left \{ \frac{ [(\mbox{$\underline{r}... ...mbox{$\underline{\hat{v}}$}_{kn})^{2}]^{1/2}}{r_{ij}}\right \}$$ (2.57)

Formally, the force on an atom arising from the inversion potential is given by

$$f_{\ell}^{\alpha}=-\frac{\partial}{\partial r_{\ell}^{\alpha}}U(\phi_{ijkn}),$$ (2.58)

with $\ell$ being one of $i,j,k,n$ and $\alpha$ one of $x,y,z$. This may be expanded into

$$-\frac{\partial}{\partial r_{\ell}^{\alpha}}U(\phi_{ijkn}) = \left\{\frac{1}{\sin(\phi_{ijkn})}\right\} \frac{\partial}{\partial \phi_{ijkn}}U(\phi_{ijkn})\times$$

$$\frac{\partial}{\partial r_{\ell}^{\alpha}} \left\{\frac{[(\mbox{... ...r}$}_{ij}\cdot\mbox{$\underline{\hat{v}}$}_{kn})^{2}]^{1/2}} {r_{ij}}\right \}.$$ (2.59)

Following through the (extremely tedious!) differentiation gives the result:

$$f_{\ell}^{\alpha} = \left\{\frac{1}{\sin(\phi_{ijkn})}\right\} \frac{\partial}{\partial \phi_{ijkn}}U(\phi_{ijkn})\times$$ (2.60)

$$\left\{-(\delta_{\ell j}-\delta_{\ell i})\frac{cos(\phi_{ijkn})} ... ...{\alpha}\} \phantom{\left\{\frac{a_{a}^{a}}{a_{a}^{a}}\right\}} \right. \right.$$

$$+ (\delta_{\ell k}-\delta_{\ell i}) \frac{\mbox{$\underline{r}$}_... ...\mbox{$\underline{\hat{u}}$}_{kn}))\frac{r_{ik}^{\alpha}}{r_{ik}^{2}} \right \}$$

$$+ (\delta_{\ell k}-\delta_{\ell i}) \frac{\mbox{$\underline{r}$}_... ...\mbox{$\underline{\hat{v}}$}_{kn}))\frac{r_{ik}^{\alpha}}{r_{ik}^{2}} \right \}$$

$$+ (\delta_{\ell n}-\delta_{\ell i}) \frac{\mbox{$\underline{r}$}_... ...\mbox{$\underline{\hat{u}}$}_{kn}))\frac{r_{in}^{\alpha}}{r_{in}^{2}} \right \}$$

$$\left . \left .+ (\delta_{\ell n}-\delta_{\ell i}) \frac{\mbox{$\... ...hat{v}}$}_{kn}))\frac{r_{in}^{\alpha}}{r_{in}^{2}} \right \} \right ] \right \}$$

This general formula applies to all atoms $\ell=i,j,k,n$. It must be remembered however, that these formulae apply to just one of the three contributing terms (i.e. one angle $\phi$) of the full inversion potential: specifically the inversion angle pertaining to the out-of-plane vector $\mbox{$\underline{r}$}_{ij}$. The contributions arising from the other vectors $\mbox{$\underline{r}$}_{ik}$ and $\mbox{$\underline{r}$}_{in}$ are obtained by the cyclic permutation of the indices in the manner described above. All these force contributions must be added to the final atomic forces.

Formally, the contribution to be added to the atomic virial is given by

$${\cal W}=-\sum_{i=1}^{4}\mbox{$\underline{r}$}_{i}\cdot\mbox{$\underline{f}$}_{i}$$ (2.61)

However it is possible to show by thermodynamic arguments (cf [26],) or simply from the fact that the sum of forces on atoms j,k and n is equal and opposite to the force on atom i, that the inversion potential makes no contribution to the atomic virial.

If the force components $f_{\ell}^{\alpha}$ for atoms $\ell=i,j,k,n$ are calculated using the above formulae, it is easily seen that the contribution to be added to the atomic stress tensor is given by

$$\sigma^{\alpha \beta}=r_{ij}^{\alpha}f_{j}^{\beta}+ r_{ik}^{\alpha}f_{k}^{\beta}+r_{in}^{\alpha}f_{n}^{\beta}$$ (2.62)

The sum of the diagonal elements of the stress tensor is zero (since the virial is zero) and the matrix is symmetric.

In DL_POLY_2 inversion forces are handled by the routine INVFRC.

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