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!)

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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?]


1 Answer 1


$V_{ij}$ 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.

I hope this makes it clearer.


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