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enter image description hereI am reading "Classical mechanics" by Goldstein, Poole and Safko, Third edition. Kindly please refer to page no 10, last paragraph.

They write

the subscript $i$ on the del operator indicates that derivatives are with respect to components of $\mathbf{r}_i$, ($\mathbf{r}_i$ is a position vector).

I only know one definition of the gradient operator and that is $$\mathbf{i}\frac{\partial}{\partial x}+\mathbf{j}\frac{\partial}{\partial y}+\mathbf{k}\frac{\partial}{\partial z}.$$ When applied to a scalar function, it calculates the slope of the scalar with respect to the $x$, $y$ and $z$ axes respectively these slopes produce the gradient of the scalar function used.

I have never seen the gradient with respect to a position vector. What is this? What is the recipe to calculate it? What does it physically signify?

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    $\begingroup$ Could you write your gradient expressions in MathJax? I believe I can answer this question, but I’m unsure of exactly what your expression/definition of the gradient is. $\endgroup$ Commented Feb 18, 2019 at 17:53
  • $\begingroup$ As far as I remember, Goldstein uses the notation $\nabla_i$ to denote the gradient with respect to the position-vector of the $i$th particle--not with respect to the $i$th component of any position-vector. Extra Point: Nonetheless, whenever it is clear from the context that $\nabla_i$ is supposed to mean gradient with respect to the $i$th component of a position-vector (unlike the case in Goldstein), it simply means the $i$th component of the full gradient with respect to the full position-vector (i.e., simply $\partial_i$). $\endgroup$
    – user87745
    Commented Feb 18, 2019 at 18:14
  • $\begingroup$ I have uploaded the paragraph. It says "derivatives wrt components of ri" $\endgroup$
    – user103515
    Commented Feb 18, 2019 at 18:18
  • $\begingroup$ Yes, exactly! Derivatives with respect to the components of $\vec{r}_i$ means the full gradient with respect to $\vec{r}_i$. It is different from the derivative with respect to the $i$th component of $\vec{r}$. In particular, here, $\vec{r}_i$ is the full position vector of the $i$th particle and the gradient $\nabla_i$ is the full gradient with respect to this full position vector of the $i$th particle. Try to realize how it is different from the derivative, say, $\partial_x$, which is a derivative with respect to a component of the full position vector of some particle. Does this make sense? $\endgroup$
    – user87745
    Commented Feb 18, 2019 at 18:22
  • $\begingroup$ What is the recipe to calculate that? $\endgroup$
    – user103515
    Commented Feb 18, 2019 at 18:40

2 Answers 2

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We have that $V_i$ depends on the position of one particle (the $i$-th particle). In general we could have a potential $V$ that depends on the positions of $N$ particles, so $V=V(\mathbf r_1,...,\mathbf r_N)$. Let's write this out fully for $N=2$: $$V=V(x_1,y_1,z_1,x_2,y_2,z_2).$$ Then $\nabla_1V$ means $$\left(\mathbf i\frac \partial{\partial x_1}+\mathbf j\frac \partial{\partial y_1}+\mathbf k\frac \partial{\partial z_1}\right)V(x_1,y_1,z_1,x_2,y_2,z_2)$$ and similarly you get $\nabla_iV$ by taking the derivative to the coordinates of the $i$'th particle.

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I think, he meant instead of using (x,y,z) in the del- respectively $\nabla$-operator, the author uses $\mathbf{r_i}=(x_i,y_i,z_i)$, i.e. the components of $\mathbf{r}_i$.

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