I am reading the Feynman lectures and at this point http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S3 it says as follows:

The time derivate of the potential energy is

$\begin{equation} \dfrac{d}{dt}\sum\limits_{pairs}-\frac{Gm_{i}m_{j}}{r_{ij}} = \sum\limits_{pairs} \left( +\frac{Gm_{i}m_{j}}{r^2_{ij}} \right) \left( \dfrac{dr_{ij}}{dt} \right) \end{equation}$


$\begin{equation} r_{ij}=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}, \end{equation}$

so that

$\begin{equation} \begin{split} \frac{dr_{ij}}{dt} = \frac{1}{2r_{ij}} \biggl[ & 2 \left( x_{i}-x_{j} \right) \left( \dfrac{dx_{i}}{dt} - \dfrac{dx_{j}}{dt} \right) \\ +& 2 \left( y_{i}-y_{j} \right) \left( \dfrac{dy_{i}}{dt} - \dfrac{dy_{j}}{dt} \right) \\ +& 2 \left( z_{i}-z_{j} \right) \left( \dfrac{dz_{i}}{dt} - \dfrac{dz_{j}}{dt} \right) \biggr] \end{split} \end{equation}$


What happens after "so that"? I don't understand what is being done mathematically?

  • $\begingroup$ He took a derivative. $\endgroup$ – DumpsterDoofus Apr 21 '14 at 20:31
  • 2
    $\begingroup$ By the way, I never visited the online edition of the Feynmann lectures until I read this question; the site is typeset beautifully (even the Table of Contents are beautiful). Thanks for posting the link. $\endgroup$ – DumpsterDoofus Apr 21 '14 at 20:35

It's the chain rule. We have

$$ r_{ij} = \sqrt{w} $$

where I've defined $w$ as all those terms underneath the square root. So

$$ \frac{d r_{ij}}{dt} =\frac{d r_{ij}}{dw} \frac{dw}{dt} = \frac{1}{2 \sqrt{w}} \frac{dw}{dt} = \frac{1}{2 r_{ij}} \frac{dw}{dt} \,, $$

where to go to the second expression I've used the chain rule (the rest is just manipulation). To calculate $dw/dt$, you need to apply the chain rule again. Hopefully you can see how this will work. If not, consider each term under the square root seperately, and make substitutions such as $u = x_i - x_j$.


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.