In Griffith's Electrodynamics, in the section 4.2, just after the equation 4.9, he writes "sleight-of-hand casts this integral into a much more illuminating form"... I have a doubt in that. If the Gradient (or differentiation if carried out) is with respect to primed coordinates, how can variable r be differentiated as r' ? It would be of great help if someone clarifies this point.


The point is that the 'scripty r' (i dont know how to write it here) depends only on the difference between the coordinates; note that ($\frac{\partial}{\partial x}$) $f(x- x')$ = -($\frac{\partial}{\partial x'}$)$f(x- x')$. What i am trying to say is that the 'scripty r' is a vector joining the primed co-ordinates to the unprimed co-ordinates.....hence the gradient is taken wrt either of the co-ordinates....ie, gradient wrt unprimed is negative of gradient wrt primed co-ordinates

  • $\begingroup$ Sure. it was a little confusing to me too in the beginning. $\endgroup$ – Prasad Mani Oct 19 '16 at 9:08
  • $\begingroup$ But can you tell me why we shift from normal r to r' ? [read r as that that scripty r] $\endgroup$ – Paul Oct 19 '16 at 10:17
  • $\begingroup$ Because after some steps, you will see that he invoked a product rule where $P$. $\triangledown '(\frac{1}{scripty r})$ = $\triangledown ' (\frac{P}{scripty r})$ - $\frac{1}{scripty r}(\triangledown'.P)$. For this to happen, the $\triangledown'$ has to be applied to both P(which is a source co-ordinate..primed) and the scripty r (which is flexible)...which is why he shifted to the primed co-ordinates $\endgroup$ – Prasad Mani Oct 19 '16 at 10:50
  • $\begingroup$ The P is strictly a source co-ordinate...so we have to accomodate the $\frac{1}{scriptyr}$ in such a way that it can also be written as a function of source co-ordinates...which is why the shift from unprimed to prime $\endgroup$ – Prasad Mani Oct 19 '16 at 10:58

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