Jastro correlation factor

In this section we explicity compute the gradient and Laplacian of the the corellation factor. We begin by calculating the gradient terms. Note that R is a 3N dimensional vector, so that the gradient with respect to R will have 3N components. For clarity, the 3Ndimensional vector will be represented by N 3-dimensional vectors, $\{{\mathbf r}_i\}$. The $i^{\text{th}}$ component of the gradient is simply

$$\nabla_i = \frac{\partial}{\partial x_i} \hat{x_i} + \frac{\p... ...tial y_i} \hat{y_i} + \frac{\partial}{\partial z_i} \hat{z_i}$$ (11)

We will explicity calculate the x_i component and symmetry considerations will give us the remainder. We begin by expanding the notation in our expression for U_ij.

$$U_{ij}({\mathbf R}) = \sum_{i<j} \frac{a_{ij} \left[(x_i - x_... ... - x_j)^2 + (y_i - y_j)^2+ (z_i - z_j)^2\right]^{\frac{1}{2}}}$$ (12)

Since the summation is given for i<j, to calculate the $i^{\text{th}}$component, we will need to sum over the remaining j's.

 

$$(\nabla_i)_x \sum_{i<j} U_{ij} \sum_{i<j}\partial_{x_i} \left[\frac{a_{ij} \left[(x_i - x_j)^2 +... ...left[(x_i - x_j)^2 + (y_i - y_j)^2+ (z_i - z_j)^2\right]^{\frac{1}{2}}} \right]$$ =

$$\sum_{i<j} \left[ \frac{a_{ij} r_{ij}^{-1}(x_i - x_j)}{1 + b_{ij}r_{ij}} - \frac{a_{ij} b_{ij}(x_i - x_j)}{(1 + b_{ij} r_{ij})^2} \right]$$ =

$$\sum_{i<j} \left[ \frac{a_{ij}}{1 + b_{ij}r_{ij}} \left(r_{ij}^{-1} - \frac{b_{ij}}{1 + b_{ij}r_{ij}} \right)(x_i - x_j) \right]$$ =

$$\sum_{i<j} \left[ \frac{a_{ij}}{1 + b_{ij}r_{ij}} \left(\frac{r_{ij}^{-1} + b_{ij} - b_{ij}}{1 + b_{ij}r_{ij}}\right)(x_i - x_j) \right]$$ =

$$\sum_{i<j} \left[\frac{a_{ij}}{r_{ij}(1 + b_{ij}r_{ij})^2} (x_i - x_j) \right]$$ = (13)

With the form for the x component, we can generalize the calculation to the $i^{\text{th}}$ component of the gradient.

$$\nabla_i \sum_{i<j} U_{ij} = \sum_{i<j} \frac{a_{ij}}{r_{ij}(1+b_{ij}r_{ij})^2} ({\mathbf r}_i - {\mathbf r}_j)$$ (14)

For each gradient term $\nabla_{{\mathbf r}_i}$ we need to to sum over all $j \neq i$.

Now we move on to the Laplacian. We begin by expanding out (13).

$$(\nabla_i)_x \sum_{i<j} U_{ij} = \sum_{i<j} \frac{a_{ij}(x_i ... ...}[(x_i-x_j)^2 + y_{ij}^2 + z_{ij}^2]^{\frac{1}{2}} \right\}^2}$$ (15)

This calculation is quite involved, so we begin by taking the derivative of the denominator.

$$\partial_{x_i} \ \mathrm{denom}$$ = r_ij^-1(1+b_ijr_ij)²(x_i - x_j) + 2 b_ij(1 + b_ijr_ij)(x_i - x_j)

= (1 + b_ij)(r_ij^-1 + 3 b_ij)(x_i - x_j) (16)

With this derivative, we can compute the full second derivative.

$$\partial^2_{x_i} U_{ij} \frac{a_{ij}}{r_{ij}(1+b_{ij}r_{ij})^2} - \frac{(1 + b_{ij}r_{ij})(r_{ij}^{-1} + 3b_{ij})(x_i - x_j)^2 a_{ij}}{r_{ij}^2(1 + b_{ij}r_{ij})^4}$$ =

$$\frac{a_{ij}}{r_{ij}(1 + b_{ij}{r_ij})^2} \left[ 1 - \frac{(r_{ij}^{-1} + 3b_{ij})(x_i - x_j)^2}{r_{ij}(1 + b_{ij}r_{ij})} \right]$$ = (17)

Now, we may sum over components to generate the Laplacian with respect to r_i. This summation changes the 1 in brackets to a 3 and the (x_i - x_j)² to r_ij².

$$\nabla_i^2 U_{ij} \frac{a_{ij}}{r_{ij}(1+b_{ij}r_{ij})^2} \left[ 3 - \frac{(r_{ij}^{-1} + 3b_{ij})r_{ij}^2}{r_{ij}(1 + b_{ij}r_{ij})}\right]$$ =

$$\frac{a_{ij}}{r_{ij}(1+b_{ij}r_{ij})^2} \left[\frac{3 + 3 b_{ij}r_{ij} - 1 - 3b_{ij}r_{ij}}{1 + b_{ij}r_{ij}} \right]$$ =

$$\frac{2 a_{ij}}{r_{ij}(1 + b_{ij}r_{ij})^3}$$ = (18)

We remember that we must sum over all $i \neq j$ to calculate the full Laplacian.