Gradience wrt a Position Vector I found the following equations in a dynamics textbook,
The gravitational potential energy for any two particles in a n-partice system is given by,
$V_{ij} = - \frac {G  m_i  m_j}{r_{ij}}$
where $r_{ij}$ is the distance between $m_i$ and $m_j$. The total potential energy of the system is,
$V = \frac{1}{2} \sum_{i = 1}^{n} \sum_{j = 1}^{n}V_{ij}$ (i $\neq$ j)
If $R_i$ is the postion vector of the $i^{th}$ particle, then
$\frac{\partial{V}} {\partial {\vec{R_{i}}}} = - \frac{\partial{V}}{\partial{\vec{r_{ji}}}} + \frac{\partial{V}}{\partial{\vec{r_{ij}}}} = -2 \frac{\partial{V}}{\partial{\vec{r_{ij}}}}$
What does it mean to take the derivative of a Scalar Function$(V)$ with respect to a vector$(\vec{R_1})$? Is it directional derivative?
I've been trying all day to get the last equation. I would be very grateful if somebody could help me(or mention some reference perhaps). I don't really know which part of math is used to get the last equation.
Thanks in advance.
 A: Let’s do it for two particles – generalization will be straightforward.
The potential energy of two particles is $V_{ij} = - \frac {Gm_i  m_j}{r_{ij}}=-\frac {Gm_i  m_j}{|\vec{r_{ij}}|}=- \frac {Gm_i  m_j}{|\vec{R_{i}}-\vec{R_{j}}|}$,  because $r_{ij} =|\vec{R_{i}}-\vec{R_{j}}|$.
By definition $\frac{\partial}{\partial\vec{R}}V({R_{i}})=\frac{\partial} {\partial{R_x}}\vec{e_{x}}+\frac{\partial}{\partial{R_y}}\vec{e_y}+\frac{\partial} {\partial{R_z}}\vec{e_z}$, where vectors $\vec{e_x}, \vec{e_y}, \vec{e_z}$ are directed along axis X, Y and Z correspondingly and each has a unit length.
Next,
$\frac{\partial}{\partial\vec{R_1}}\frac{1}{|\vec{R_1}-\vec{R_2}|}=(\frac{\partial} {\partial{R_{1x}}}\vec{e_x}+\frac{\partial}{\partial{R_{1y}}}\vec{e_y}$$+\frac{\partial} {\partial{R_{1z}}}\vec{e_z})\frac{1}{\sqrt{(R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2}}$.
Taking derivatives we get explicitly:
$\frac{\partial}{\partial\vec{R_1}}\frac{1}{|\vec{R_1}-\vec{R_2}|}=-\left((R_{1x}- R_{2x})\vec{e_x}+(R_{1y}-R_{2y})\vec{e_y}+(R_{1z}- R_{2z})\vec{e_z}\right)(\frac{1}{\sqrt{(R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2)}})^{3}$$=-\vec{r_{12}}\left((R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2\right)^{-\frac{3}{2}}$
If you take the derivatives with respect to $\vec{R_2}$ you will get the same formula, but with the negative sing.
