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Cross posted on Math SE

I am trying to understand virtual sites in MD simulations, and I came across this configuration:

Virtual sites diagram

Here, coordinate $\mathbf{s}$ represents the virtual site, which is formed by three other atoms $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$. The distance between atom $\mathbf{i}$ and the virtual site $\mathbf{s}$ is $|\mathbf{d}|$. The position of atoms $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are $\mathbf{r}_i$, $\mathbf{r}_j$, and $\mathbf{r}_k$, respectively.

In this case, the virtual site ($\mathbf{r}_s$) is in the plane of the other three particles at a distance of $|\mathbf{d}|$ from $\mathbf{i}$ at an angle of $\theta$ from $\mathbf{r}_{ij}$. Atom $\mathbf{k}$ defines the plane and direction of the angle.

How should I get the position of $\mathbf{r}_s$ using the other three atom positions and an angle?

I know the equation for $\mathbf{r}_s$, but I couldn't understand how it is derived. Any help is appreciated.
$$ \mathbf{r}_s = \mathbf{r}_i + d \cos\theta\, \frac{\mathbf{r}_{ij}}{|\mathbf{r}_{ij}|} + d \sin\theta\, \frac{\mathbf{r}_{\perp}}{|\mathbf{r}_{\perp}|}\tag{1}, $$ in which $$ \mathbf{r}_{\perp} = \mathbf{r}_{jk} - \frac{\mathbf{r}_{ij} \cdot \mathbf{r}_{jk}}{\mathbf{r}_{ij} \cdot \mathbf{r}_{ij}}\, \mathbf{r}_{ij}. \tag{2} $$

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  • $\begingroup$ I think it's already mentioned in gromacs manual and the subsequent paper. $\endgroup$ Oct 5, 2023 at 20:22
  • $\begingroup$ @RoshanShrestha why not show where in the manual and where in the paper it's answered? Some people might not even know which paper to seek! $\endgroup$ Oct 5, 2023 at 21:38
  • $\begingroup$ Wouldn't it just be $\mathbf{r_i} + \frac{d \left(\mathbf{r_j}- \mathbf{r_k}\right)}{|\mathbf{r_j}- \mathbf{r_k}|}$? If you already have $\mathbf{r_i}$, then why would an angle $\theta$ that has nothing to do with the length $d$ matter? $\endgroup$ Oct 6, 2023 at 0:09
  • $\begingroup$ My bad, thanks @NikeDattani So, in the book Molecular Liquids Dynamics and Interactions, there is a chapter written by H.J.C. Berendsen and W.F. van Gunsteren from Page 475-500 on Molecular dynamics simulations: techniques and approaches. $\endgroup$ Oct 6, 2023 at 6:10
  • $\begingroup$ @RoshanShrestha, Yes, you are correct. I went through that book thoroughly. I did manage to derive for simple cases (as given in the book), but for this configuration, I couldn't. $\endgroup$
    – Vasista
    Oct 6, 2023 at 6:57

2 Answers 2

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The formula (1) in the question:

$$ \mathbf{r}_s = \underset{\text{term A}}{\mathbf{r}_i} + \underset{\text{term B}}{d \cos\theta\, \frac{\mathbf{r}_{ij}}{|\mathbf{r}_{ij}|}} + \underset{\text{term C}}{d \sin\theta\, \frac{\mathbf{r}_{\perp}}{|\mathbf{r}_{\perp}|}}\tag{1} $$

can be understood as going to the position of $s$ by starting at the position of $i$ (term A) and adding the two legs (terms B and C) of the right-angled triangle with $\mathbf{r}_{is}$ as its hypotenuse (equivalently, stating $\mathbf{r}_{is}$ in the orthonormal basis vectors of terms B and C).

Term B is the unit vector $\hat{\mathbf{r}}_{ij}$, and term C is the unit vector coplanar and perpendicular to $\hat{\mathbf{r}}_{ij}$. The (not yet unit) vector $\mathbf{r}_\perp$ in term C is constructed to lie in the desired plane but be perpendicular to $\mathbf{r}_{ij}$ by subtracting from $\mathbf{r}_{jk}$ its projection into $\mathbf{r}_{ij}$ (a Gram-Schmidt orthogonalization). This is the formula (2) in the question.

The below diagram may help visually:

Geometry of atoms k, j, i, and virtual site s, showing the vector is as the sum of basis vectors parallel and perpendicular to in.

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  • $\begingroup$ Useful relations to help understand this answer: $\sin(\theta - \pi)=-\sin(\theta),\cos(\theta-\pi)=-\cos(\theta).$ $\endgroup$ Oct 13, 2023 at 15:34
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I have posted the same question in Mathematics StackExchange and here is the reply.

Suppose that $d=|\mathbf{d}|$ and $\mathbf{r}_{ij}\ne \mathbf{0}$. Then in the plane of the other three particles with the coordinates origin at $\mathbf{r}_i$ and the basis of consisting of the unit vector $\frac{\mathbf{r}_{ij}}{|\mathbf{r}_{ij}|}$ and some orthogonal to it unit vector $\mathbf{r'}$, the virtual site has coordinates $(d\cos\theta,d\sin\theta)$. Thus $$\mathbf{r}_s = \mathbf{r}_i + d \cos\theta\, \frac{\mathbf{r}_{ij}}{|\mathbf{r}_{ij}|} + d \sin\theta\, \mathbf{r'}.$$ The vector $\mathbf{r'}$ belongs to the plane spanned by $\mathbf{r}_{ij}$ and $\mathbf{r}_{jk}$, so $\mathbf{r'}=\lambda_i\mathbf{r}_{ij}+\lambda_k\mathbf{r}_{jk}$ for some real $\lambda_i$ and $\lambda_k$. Moreover, $0=\mathbf{r'}\cdot \mathbf{r}_{ij}=\lambda_i\mathbf{r}_{ij}\cdot \mathbf{r}_{ij}+\lambda_k\mathbf{r}_{jk}\cdot \mathbf{r}_{ij}$. So $\lambda_i=-\lambda_k\frac{\mathbf{r}_{jk}\cdot \mathbf{r}_{ij}}{\mathbf{r}_{ij}\cdot \mathbf{r}_{ij}}$. Thus $$\mathbf{r'}=\lambda_k\left(-\frac{\mathbf{r}_{jk}\cdot \mathbf{r}_{ij}}{\mathbf{r}_{ij}\cdot \mathbf{r}_{ij}}\mathbf{r}_{ij} +\mathbf{r}_{jk}\right)=-\lambda_k\mathbf{r}_{\perp}.$$ Since $|\mathbf{r'}|=1$, we have $|\lambda_k|=\frac{1}{|\mathbf{r}_{\perp}|}$. It remains to determine the sign of $\lambda_k$. It corresponds to the direction of the angle $\theta$ between $\mathbf{r}_{ij}$ and $\mathbf{r}_{s}-\mathbf{r}_{i}$. In particular, when $\theta=\frac{\pi}{2}$, we have $\mathbf{r'}=\mathbf{r}_s-\mathbf{r}_i$. But this direction is not very clear to me, maybe, because there is no coordinate system in the supplementary picture.

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