Bound Charges in Polarized Material

Question

I'm currently working my way through chapter 4 of Griffith's Introduction to Electrodynamics (4th ed) and I ran into an issue with his notation.

In chapter 4 sectin two, when calculating the electric potential of a polarized object with polarization $\mathbf{P}$, he eventually gets to this expression

$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\oint_S \frac{1}{\mathscr{R}} \mathbf{P}\cdot d\mathbf{a}' - \frac{1}{4\pi\epsilon_0}\int_V\frac{1}{\mathscr{R}}\left(\nabla' \cdot \mathbf{P}\right)d\tau'$

where $\mathscr{R} = |\mathbf{r} - \mathbf{r}'|$. I understand the derivation up until here. In the next part he claims we can define a bound volume charge density $\rho_b = -\nabla \cdot \mathbf{P}$. This is where I get confused. Where did the prime on the del operator go? At first I though this was a typo, but then in section 3 he defines Gauss' law inside a dielectric as

$\epsilon_0 \nabla \cdot \mathbf{E} = \rho_b + \rho_f = -\nabla \cdot \mathbf{P} + \rho_f$

This is then used to group $\mathbf{E}$ and $\mathbf{P}$ inside the same del operator when defining the electric displacement. So it seems like for the rest of the chapter to work out, it must be the case that $\rho_b = -\nabla \cdot \mathbf{P}$ and not $\rho_b = -\nabla' \cdot \mathbf{P}$. Is it the case that $\nabla \cdot \mathbf{P} = \nabla' \cdot \mathbf{P}$?

Answer 1

In chapter 4 lesson 2.1 you will see an image(FIGURE 8) of an object with irregular shape. The prime notation indicates that the differentiation or integration is being done with respect to the source coordinates. The equation $$ V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \oint_S \frac{\mathbf{P} \cdot d\mathbf{a}'}{R} - \frac{1}{4\pi\epsilon_0} \int_V \frac{\nabla' \cdot \mathbf{P}}{R} \, d\tau' $$ where $$R=|r−r′|$$ was derived considering source coordinate but you can change the coordinate however you want but the outcome will be same that's why they removed the prime. Yes its true in special coordinate taking the source as coordinate calculation more easier.