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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.

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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

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  • $\begingroup$ 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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