1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.