# Notation of $\nabla_{ij}V_{ij}$ Referring to a Potential

There's a brief section in Goldstein's Classical Mechanics book in chapter 1 that derives some useful basic mechanics things. In talking about the total internal energy of the system, there's a passage which I've had explained to me multiple times but which always confuses me when I want to refer to it, on the internal potential energy of a system. (ignore the first sentence fragment at the beginning of the passage!)

I'm very confused about what exactly "with respect to" means. Is the notation just imprecise? I'm almost sure I've used this notation before in a problem, but in everything else I do I make sure to write gradients as $\nabla V$. What's this $\nabla_i V$? Are we considering $V_{ij}$ as a function of six variables (supposing $\vec{r}_i$ and $\vec{r}_j$ are 3D vectors), so that $\nabla_i$ refers to the partial derivatives/gradient of (say) the first three arguments, and $\nabla_j$ refers to the second three?

[come to think of it, I'm almost sure that's the answer, but since I've written the question can anyone reaffirm it?]

$V_{ij}$ is indeed a function of six real variables, $\vec r_i$ and $\vec r_j$. $\nabla_i$ is the gradient with respect to $\vec r_i$.
For translationally invariant problems, $V_{ij}$ must be a function of $\vec r_j-\vec r_i$. $\nabla_{ij}$ is the gradient with respect to this vector. To make this a bit clearer, consider just the $x$ coordinate of that gradient, $\frac\partial{\partial x_{ij}}$. This is easier to calculate considering the components of $\nabla_i$, which need a second variable, say $\vec R=\vec r_i+\vec r_j$, to work with. You have $$\frac{\partial}{\partial x_{i}} = \frac{\partial x_{ij}}{\partial x_{i}}\frac{\partial}{\partial x_{ij}}+\frac{\partial X}{\partial x_{i}}\frac{\partial}{\partial X}=-\frac{\partial}{\partial x_{ij}}$$ since $\frac{\partial}{\partial X}\equiv0$ for allowable potentials.