Conversion of $ \nabla $ to $ \nabla' $(please watch video shown, in the time range provided) (doubt explained below video link)

Question

Video: https://youtu.be/JCfhChaJvCo (Doubt is in the time range 26:50 to 28:30 )

A bit of background

I am given to understand that when the lecturer in the video says that [video time: 26:57] $ \nabla $ is the vector differential operator on $ \vec r$ he imples the following:

$ \nabla = \frac \partial {\partial x} \hat \imath+\frac \partial {\partial y}\hat \jmath+\frac \partial {\partial z} \hat k$

$\vec r=x\hat \imath+y\hat\jmath+z\hat k$

$r=\sqrt{x^2+y^2+z^2}$

where $ \hat \imath, \hat \jmath, \hat k $ are the unit vectors in the x,y,z directions respectively.

(Correct me if my understanding is wrong)

Doubts:

1.how does the person convert $ \nabla $ to $ \nabla' $ in his "mathematical trick"? [Video time- 28:30]
2.Does $ \nabla' $ mean something other than $ \nabla $?
3.Is $\nabla'$ the vector differential operatior for $\vec r'$ and if so,what would be the [i,j,k] expanded form of $\nabla'$?
3.is it possible to freely convert between different variations on $\nabla$?
4.What textbook would give me clarity and a deeper look into this $\nabla$ conversion?

Answer 1

The sequence of equalities in question (leaving out the superfluous constants, with the understanding that $E$'s and $r,r'$ are vectors) is as follows: $$ E(r) = \int_V \rho(r') \frac{r - r'}{|r-r'|^3}d^3r'\\ (\nabla_r \cdot E)(r) = \nabla_r\cdot \int_V \rho(r') \frac{r - r'}{|r-r'|^3}d^3r'\\ = \int_V \rho(r') \nabla_r \cdot \left(\frac{r - r'}{|r-r'|^3}\right)d^3r' \\ \overset{!!!}= -\int_V \rho(r') \nabla_{r'} \cdot \left(\frac{r - r'}{|r-r'|^3}\right)d^3r' $$ The fact that we are using in this last equality is as follows:

Claim: For any $f:\Bbb R^3 \to \Bbb R^3$, we have $$ \nabla_r \cdot f(r-r') = -\nabla_{r'} \cdot f(r-r') $$ Or, to write the same thing differently, $$ \nabla_r \cdot f(x-x',y-y',z-z') = -\nabla_{r'} \cdot f(x-x',y-y',z-z') $$

So first, what does the notation mean? Note that $r = (x,y,z)$ is the argument (input) of our function $E$, whereas $r' = (x',y',z')$ is a dummy variable for intgeration.

As an example: if we were to define the single-variable function $g(x) = \int_0^1 t (x-t)^2\,dt$, then $x$ plays the role analogous to $r$, while $t$ plays the role analogous to $r'$.

Now, given an expression involving both $r$ and $r'$, we define $$ \nabla_r \cdot F(x,y,z,x',y',z') = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z}\\ \nabla_{r'} \cdot F(x,y,z,x',y',z') = \frac{\partial F}{\partial x'} + \frac{\partial F}{\partial y'} + \frac{\partial F}{\partial z'} $$ Now, perhaps you can see (as a consequence of the chain rule) that my above claim holds. We can now apply this fact to the integrand in order to reach that last equality.