Potential function for conservative internal forces

Question

In Goldstein's Classical Mechanics book, he considered a system of particles and looked at the conservative internal force between particle $i$ and particle $j$ that satisfy the strong law of action and reaction.

He wrote the potential function for this internal force as $$V_{ij}=V_{ij}(|\vec{r_i}-\vec{r_j}|) .$$

He then said that the forces $\vec{F}_{ji}$ (force particle j exerts on i) and $\vec{F}_{ij}$ (force particle i exerts on j) are automatically equal and opposite:

$$\vec{F}_{ji} = - \nabla_i V_{ij}(|\vec{r_i}-\vec{r_j}|) = + \nabla_j V_{ij}(|\vec{r_i}-\vec{r_j}|) =-\vec{F}_{ij} .$$

I have some problems seeing why $$- \nabla_i V_{ij}(|\vec{r_i}-\vec{r_j}|) = + \nabla_j V_{ij}(|\vec{r_i}-\vec{r_j}|). $$The gradient operator acts on different indices, why does a change of sign makes them equal?

Answer 1

Mathematically is just the chain rule? Since

\begin{align} \dfrac{\partial}{\partial \vec{r}_i}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert)&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}\dfrac{\partial (\vec{r}_i-\vec{r}_j)}{\partial \vec{r}_i}\\ &=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}(1-\delta_{ji}) \end{align}

$$$$

\begin{align} \dfrac{\partial}{\partial \vec{r}_j}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert)&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}\dfrac{\partial (\vec{r}_i-\vec{r}_j)}{\partial \vec{r}_j}\\&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}(\delta_{ij}-1)\\&=-\dfrac{\partial}{\partial \vec{r}_i}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert) \end{align}

As far as $\delta_{ij}=\delta_{ji}.$