# How to deal with the force of the form $\delta (\mathbf{r}) u(\mathbf{r})$, $u(\mathbf{r})$is the interaction potential

Now I have such an expression for potential energy: $$U_j(\mathbf{r}_j) = \int_{\mathbf{r}_j-\frac{\mathbf{L}^b}{2}}^{\mathbf{r}_j+\frac{\mathbf{L}^b}{2}} d\mathbf{r} \sum_{\mathbf{n}} \sum_{i\ne j}^{N} \delta(\mathbf{r}-\mathbf{r}_i + \mathbf{n} \mathbf{L}^B) u_{j}^b(\mathbf{r}-\mathbf{r}_j)$$ $$\mathbf{r}_j(x_j,y_j,z_j), \mathbf{L}^b(L_x^b,L_y^b,L_z^b), \mathbf{n}(n_x,n_y,n_z), \mathbf{L}^B(L_x^B,L_y^B,L_z^B)$$ are all three dimensional vectors. $$n_x,n_y,n_z \in \mathbb{Z}$$. Also, $$\mathbf{n L}^b = (n_x L_x^b,n_y L_y^b,n_z L_z^b)$$. Note $$\mathbf{L}^b \ne \mathbf{L}^B$$, this is the heart of the problem. Find the expression for the force on this potential energy： \begin{align} -F_j(\mathbf{r}_j) =& \dfrac{d U_j(\mathbf{r}_j)}{d\mathbf{r}_j} \notag \\ =&\sum_{\mathbf{n}}\sum_{i\ne j} \delta(\mathbf{r}_j + \dfrac{\mathbf{L}^b}{2} - \mathbf{r}_i + \mathbf{n}\mathbf{L}^B ) u_j^b(\dfrac{\mathbf{L}^b}{2}) \notag \\ & - \sum_{\mathbf{n}}\sum_{i\ne j} \delta(\mathbf{r}_j - \dfrac{\mathbf{L}^b}{2} - \mathbf{r}_i + \mathbf{n} \mathbf{L}^B ) u_j^b(-\dfrac{\mathbf{L}^b}{2}) \notag \\ +&\int_{\mathbf{r}_j-\frac{\mathbf{L}^b}{2}}^{\mathbf{r}_j+\frac{\mathbf{L}^b}{2}} d\mathbf{r} \sum_{\mathbf{n}} \sum_{i\ne j }^{N} \delta(\mathbf{r}-\mathbf{r}_i + \mathbf{n} \mathbf{L}^B) \dfrac{d u_{j}^b(\mathbf{r}-\mathbf{r}_j)}{d\mathbf{r}_j} \notag \end{align} where $$u_j^b(\mathbf{r})$$is a two-body interaction potential and its mathematical form is known. But I don't know how to deal with The first term on the right-hand side of the equation. How to convert the term containing $$\delta(\mathbf{r})$$ into a form that can be directly calculated

• Your notation doesn't make sense. You cannot directly differentiate the potential $U(\mathbf{r})$ w.r.t. a vector like d$\mathbf{r}$. Jun 20 at 1:24
• @Yejus While there are problems in the notations, I don't think this is one of them. $\partial U/\partial\vec{r}$ is a common notation for the gradient. Jun 20 at 1:32
• @Yejus This is another way of writing the gradient of potential energy without any problems. Jun 20 at 2:01

There are a number of issues with the notation being used here. An integral of the form $$\int_{0}^a dx~\delta(x) f(x)$$ isn't well-defined since the lower integration limit exactly coincides with the place the delta function blows up (see this and this answer for a discussion of this). Also, I'm not sure what you mean by $$\int_0^\mathbf{r+b} d\mathbf{a}~(\cdot)$$. It's written similar to a 1D integral but the notation seems to hint that all quantities are vectors.

With these two issues in mind, my best guess is that this is what you have in mind. Let $$\mathbf{r},\mathbf{a},\mathbf{b}\in\mathbb{R}^d$$. Then \begin{align} U(\mathbf{r}) &= \int_{0^-}^{r_1+b_d}da_1 ... \int_{0^-}^{r_d+b_d}da_d~\delta(\mathbf{a})u(\mathbf{a}-\mathbf{r}), \end{align} where the notation $$\int_{0^-}^a dx~(\cdot)$$ stands for $$\lim_{\epsilon\nearrow 0}\int_{\epsilon}^a dx~(\cdot)$$. In that case, you can first simplify $$U(\mathbf{r})$$: \begin{align} U(\mathbf{r}) &= u(\mathbf{r}+\mathbf{b}-\mathbf{r}) = u(\mathbf{b}), \end{align} meaning that the force $$\mathbf{F}(\mathbf{r}) = -\boldsymbol{\nabla}U(\mathbf{r}) = 0$$ if $$\mathbf{b}$$ is independent of $$\mathbf{r}$$.

Response to New Edits

Again, first evaluate the integral in $$U_j(\mathbf{r}_j)$$ before trying to calculate the gradient. \begin{align} U_j(\mathbf{r}_j) &= \sum_{\mathbf{n}} \sum_{i\neq j}\int_{x_j - L_x^b/2}^{x_j + L_x^b/2} dx \int_{y_j - L_y^b/2}^{y_j + L_y^b/2} dy\int_{z_j - L_z^b/2}^{z_j + L_z^b/2} dz ~ \delta(\mathbf{r}-\mathbf{r}_i+\mathbf{n}\mathbf{L}^B) u_j^b(\mathbf{r}-\mathbf{r}_j)\\[5pt] &= \sum_{\mathbf{n}} \sum_{i\neq j} u_j^b(\mathbf{r}_i-\mathbf{r}_j-\mathbf{n}\mathbf{L}^B)\times \mathcal{I}(\mathbf{r}_i,\mathbf{r}_j,\mathbf{n}), \end{align} where I've defined the function $$\mathcal{I}(\mathbf{r}_i,\mathbf{r}_j,\mathbf{n}) = \begin{cases} 1 & \text{if }\mathbf{r}_i - \mathbf{r}_j \in\text{cuboid centered at }\mathbf{n}\mathbf{L}^B\text{ with side lengths }\mathbf{L}^b\\[5pt] 0 & \text{otherwise} \end{cases}$$ You can now proceed with calculating the gradient. I'll leave that calculation to you. As a hint, handling the $$\mathcal{I}$$ function is straightforward since it's zero in one region and one in the other. So just consider each case separately.

• Thanks, I just changed the lower bound to 0 to simplify the problem. The actual problem is this, the upper and lower bounds of the integral have variables. How should this be handled please Jun 20 at 1:48
• @ZhaoDazhuang Please first clarify what you mean by the notation $\int_\mathbf{a}^\mathbf{b} d\mathbf{r}~(\cdot)$. Is it a volume integration like I assumed or is it some kind of contour integration? Jun 20 at 2:06
• This integral is a volume integral with $\mathbf{r}_j$ as the body center of the cuboid and $\mathbf{L}^b$ as the side length Jun 20 at 2:12
• Thanks a lot for your help Jun 20 at 2:13
• $$\int_{\mathbf{a}}^{\mathbf{b}} d \amthbf{r} (\codt) = \int_{a_x}^{b_x) dx \int_{a_y}^{b_y} dy \int_{a_z}^{b_z} dz (\cdot)$$ Jun 20 at 2:31