# What is gradient with respect to components of a position vector?

I 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?

• 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. Commented Feb 18, 2019 at 17:53
• 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 $\partial_i$).
– user87745
Commented Feb 18, 2019 at 18:14
• I have uploaded the paragraph. It says "derivatives wrt components of ri" Commented Feb 18, 2019 at 18:18
• 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?
– user87745
Commented Feb 18, 2019 at 18:22
• What is the recipe to calculate that? Commented Feb 18, 2019 at 18:40

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.
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 $$\mathbf{r}_i$$.