Partial Derivative of a scalar (absolute distance) with respect to its position vector

Question

Imagine we want to take the partial derivative of a quantity, we will call it $\rho_i = f(F(r_{ij}))$ with respect to a particle's position vector, $\vec{r}_k$.

In mathematical terms, this would be expressed as:

$$\frac{\partial \rho_i}{\partial \vec{r}_k} = \frac{\partial F}{\partial r_{ij}}\frac{\partial r_{ij}}{\partial \vec{r}_k}$$

given the definition of $\rho_i$ and using the chain-rule. The $\frac{\partial F}{\partial r_{ij}}$ is a straight-forward calculation, but I run into trouble when trying to calculte $\frac{\partial r_{ij}}{\partial \vec{r}_k}$. Given that $r_{ij}$ is defined as $|\vec{r}_i-\vec{r}_j|$, how does one approach/conceptualize this?

The paper from which I am drawing this hypothetical scenario says that this derrivative is $(\delta_{ik}-\delta_{jk})\hat{r}_k$, where $\hat{r}_k$ is the unit vector, i.e. $\frac{\vec{r}_k}{|\vec{r}_k|}$.

I know and understand that the derrivative of a quantity with respect to a vector is the gradient of the quantity, as explained here, but I do not understand how and why the deltas appear given this notation setup. This probably has to do with my lack of intuition when it comes to Einstein Vector notation, and any help on the matter would be greatly appreciated.

By the way, here is the full image of the full derrivation which I am struggling to understand. The weird notation introduced above results from the need to introduce independent indicies from multiple summations. I have indicated the relevant parts of the derrivation in the image.

Answer 1

The problem boils down to calculating

\begin{eqnarray} \frac{\partial}{\partial \vec{r}_{k}} |\vec{r}_i - \vec{r}_j| &=& \frac{\partial}{\partial \vec{r}_{k}} \sqrt{(\vec{r}_i - \vec{r}_j) \cdot (\vec{r}_i - \vec{r}_j)} \\ &=& \frac{1}{2} \frac{1}{\sqrt{(\vec{r}_i - \vec{r}_j) \cdot (\vec{r}_i - \vec{r}_j)}} \frac{\partial}{\partial \vec{r}_k} [(\vec{r}_i - \vec{r}_j) \cdot (\vec{r}_i - \vec{r}_j)] \\ &=& \frac{1}{2 r_{ij}} \frac{\partial}{\partial \vec{r}_k} [(\vec{r}_i - \vec{r}_j) \cdot (\vec{r}_i - \vec{r}_j)] \\ &=& \frac{1}{2r_{ij}} \frac{\partial}{\partial \vec{r}_k}(\vec{r}_i \cdot \vec{r}_i - 2 \vec{r}_i \cdot \vec{r}_j + \vec{r}_j\cdot \vec{r}_j) \tag{1} \end{eqnarray}

We need to calculate the gradients of each term, which is actually not that difficult. You can use identities for the gradient, but because you mentioned you're struggling with Einstein's notation, let's do it with Einstein's notation (actually I'm abusing a bit of the notation in what follows)

\begin{eqnarray} \frac{\partial }{\partial \vec{r}_k} (\vec{r}_i\cdot \vec{r}_j) &=& \hat{e}_a \frac{\partial }{\partial x_k^a} (x_i^b x_j^b) \\ &=& \hat{e}_a \frac{\partial x_i^b}{\partial x_k^a} x_j^b + \hat{e}_a x_i^b\frac{\partial x_j^b}{\partial x_k^a} \\ &=& \hat{e}_a\delta_{ab}\delta_{ik}x_j^b + \hat{e}_a x_i^b \delta_{jk}\delta_{ab} \\ &=&(\hat{e}_a x_j^a)\delta_{ik} + (\hat{e}_a x_i^a)\delta_{jk} \\ &=& \vec{r}_j \delta_{ik} + \vec{r}_i\delta_{jk} \tag{2} \end{eqnarray}

Now we can go back to (1):

\begin{eqnarray} \frac{\partial}{\partial \vec{r}_{k}} |\vec{r}_i - \vec{r}_j| &=& \frac{1}{2r_{ij}} \frac{\partial}{\partial \vec{r}_k}(\vec{r}_i \cdot \vec{r}_i - 2 \vec{r}_i \cdot \vec{r}_j + \vec{r}_j\cdot \vec{r}_j) \\ &=& \frac{1}{2r_{ij}} (2\vec{r}_i\delta_{ik} - 2\vec{r}_i\delta_{jk} - 2\vec{r}_j\delta_{ik} + 2\vec{r}_j\delta_{jk}) \\ &=& \frac{1}{r_{ij}}[ \vec{r}_i (\delta_{ik} - \delta_{jk}) - \vec{r}_j (\delta_{ik} - \delta_{jk})] \\ &=& \frac{\vec{r}_i - \vec{r}_j}{r_{ij}} (\delta_{ik} - \delta_{jk}) \\ &=& \hat{r}_{ij} (\delta_{ik} - \delta_{jk}) \tag{3} \end{eqnarray}