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.


2 Answers 2


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. $\endgroup$ Commented Oct 19, 2016 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
    Commented Oct 19, 2016 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 $\endgroup$ Commented Oct 19, 2016 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 $\endgroup$ Commented Oct 19, 2016 at 10:58

We need to shift from $\mathscr R$ to $r'$ because otherwise the coordinate system would keep changing as we integrate over the whole volume. Now, about that sleight of hand.

Gradient depends upon the coordinate system. By simple definition of gradient we have :-

$dT= \nabla T.\boldsymbol{dl}$, where $\boldsymbol{dl}$ is the change in space of the coordinate system.

Since, $\mathscr R = \boldsymbol r - \boldsymbol r' $ it implies that $d\mathscr R = -d\boldsymbol r'$ as $\boldsymbol r$ is the constant position vector of the point of interest where we wish to calculate electric field by the polarized object, in the source coordinate system.

Now, $d \left( {\frac {1}{\mathscr R}} \right) = \nabla \left( {\frac {1}{\mathscr R}} \right).d\mathscr R = \nabla'\left( {\frac {1}{\mathscr R}} \right).d\boldsymbol r'$, as $d\mathscr R = -d\boldsymbol r'$ this implies

$\nabla' \left( {\frac {1}{\mathscr R}} \right)$ = $-\nabla \left( {\frac {1}{\mathscr R}} \right)$ which simply means that gradient in source coordinate system is just the negative of the gradient in the coordinate system of that differential dipole in consideration, which is what Griffiths touched upon.

PS : When I was having trouble with this, I was incorrectly assuming $d|\mathscr R| \hat{ \mathscr R} = -dr' \hat{r'} $, this was incorrect because I was incorrectly assuming $d\hat{ \mathscr R} = d\hat{r'}= 0$


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