How does author derives $\nabla_i V_{ij} = \nabla_{ij} V_{ij} = - \nabla_j V_{ji}$?

Question

In the book of Classical Mechanics by Golstein, at page 11, it is given that

When the forces are all conservative, the second term in Eq. (1.29) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form $$ -\int_1^2(\nabla_i V_{ij} \cdot d{\bf s}_i + \nabla_j V_{ij} \cdot d{\bf s}_j) $$ If the difference vector ${\bf r}_i - {\bf r}_j$ is denoted by ${\bf r}_{ij}$ and $\nabla_{ij}$ stands for the gradient with respect to ${\bf r}_{ij}$, then $$\nabla_i V_{ij} = \nabla_{ij} V_{ij} = - \nabla_j V_{ji}$$

However, I cannot understand how does the author concludes the equality

$$\nabla_i V_{ij} = \nabla_{ij} V_{ij} = - \nabla_j V_{ji}$$

Answer 1

Because the potential $V_{ij}$ depends on the difference ${\bf x}_{ij} = {\bf x}_i - {\bf x}_j$, so

$$ \frac{\partial V_{ij}}{\partial {\bf x}_j} = \frac{\partial {\bf x}_{ij}}{\partial {\bf x}_j} \frac{\partial V_{ij}}{\partial {\bf x}_{ij}} = \frac{\partial ({\bf x}_i - {\bf x}_j)}{\partial {\bf x}_j} \frac{\partial V_{ij}}{\partial {\bf x}_{ij}} = -\frac{\partial V_{ij}}{\partial {\bf x}_{ij}} $$

Similarly

$$ \frac{\partial V_{ij}}{\partial {\bf x}_i} = +\frac{\partial V_{ij}}{\partial {\bf x}_{ij}} $$

Or in Goldstein's notation

$$ \nabla_i V_{ij} = \nabla_{ij} V_{ij} = -\nabla_j V_{ij} $$