What is the meaning of $\rho _b$ blowing up at the surface?

Question

In Griffith's it has been said that while deriving the equation $\nabla\cdot D=\rho_f$
where $D\equiv\epsilon_0E+P$ and $\rho_f$ is the free charge, there has been no mention of the surface bound charge.
This has been justified by the fact that Gauss' Law cannot be calculated at the surface since the volume bound charge,$\rho_b$ blows up at the surface. By blows up i understand that it goes to infinity, so is it going to infinity at the surface or is it undefined? Why is it undefined?

Answer 1

so is it [i.e., $\rho_b$ ] going to infinity at the surface, or is it undefined?

It is infinite (or undefined depending on your technical definition). Note that by definition $$\rho_b = - \nabla\cdot \vec{P}(\vec{r}) $$

There is an abrupt change in $\vec{P}$ from being non-zero to exactly zero as it crosses the boundary surface. The derivative [i.e., $\nabla\cdot \vec{P}(\vec{r})$] of a function at a point of jump discontinuity is infinite or sometimes described by a Dirac delta function [e.g., $[aH(x)]'=a \delta(x)$ where $H$ is the Heaviside step function and $a$ is the length of jump discontinuity].

Extra: How the surface bound charge changes the picture?

An interesting remark about the bound surface charge density $\sigma_b$ is that in fact, it does not cause a discontinuity in $\nabla \cdot \vec{D}$ as one might expect. Note the change in the perpendicular components of $\vec{D}$ at a boundary is given by

$$\Delta D_{\text{perpend.}}= D^{\text{out}}_{\text{perpend.}}-D^\text{in}_{\text{perpend.}} = P^{\text{out}}_{\text{perpend.}}-P^\text{in}_{\text{perpend.}} + \epsilon_0 \Delta E_{\text{perpend.}} $$

where out and in means outside or inside the dielectric. We know that $P^{\text{out}}_{\text{perpend.}}=0$. Moreover,

$$ \Delta E_{\text{perpend.}}= \dfrac{\sigma_{\text{net}}}{\epsilon_0}=\dfrac{\sigma_{f}+\sigma_{b}}{\epsilon_0} $$

Thus we have

$$\Delta D_{\text{perpend.}} = 0 -P^\text{in}_{\text{perpend.}} + \epsilon_0 \dfrac{\sigma_{f}+\sigma_{b}}{\epsilon_0} $$

Also observe that $P^\text{in}_{\text{perpend.}}= \vec{P}\cdot \hat{\mathbf{n}}=\sigma_{b}$, where $\hat{\mathbf{n}}$ is a unit vector to the normal surface. Using this we get

$$\Delta D_{\text{perpend.}} = 0 -\sigma_{b} + (\sigma_{f}+\sigma_{b})=\sigma_{f} $$

Remarkably, the discontinuity related to $\sigma_b$ from $\Delta P_{\text{perpend.}}$ exactly cancel the one from $\Delta E_{\text{perpend.}}$. It is only $\sigma_f$ that introduces discontinuity in $\vec{D}$. In other words, when you consider surface charge density you get

$$\nabla \cdot \vec{D}= \rho_f +\sigma_f \delta $$

where $\delta$ is the Dirac delta function defined by the coordinates perpendicular to the surface and $\sigma_f$ is the size of the discontinuity jump. When the surface integral is taken on both sides, we get the total free charge due to $\rho_f$ as well as $\sigma_f$

$$\iint_S \vec{D}\cdot d\vec{A}= Q_{f} $$

Answer 2

Surface charge densities or line densities or point charges (doesn't matter if free or bound) always make the charge density blow up, because to describe them mathematically we need to make use of something called the Dirac delta $\delta(\vec{r}-\vec{r}_0)$, which is a "function" that vanishes almost everywhere except on a point ($\vec{r}_0$), where it blows up (i.e. it diverges, $\delta(\vec{r}-\vec{r}_0)\rightarrow\infty$ at $\vec{r}=\vec{r}_0$). Griffiths discusses its properties in the first chapter, so my advice is that you first read about it there.

With this in mind, you can express every surface charge as a 3-dimensional charge density; for example, suppose a uniform surface charge $\sigma$ at the surface $r=a$ in spherical coordinates, then $\rho(\vec{r})=\sigma\delta(r-a)$. When deriving general laws, like $\nabla\cdot\vec{D}=\rho_f$, we don't make distinction between volume charges and surface charges, because both can be considered as included in the term $\rho_f$.

Regarding the application of Gauss's law when dealing with a charged surface, the fact that charge densities can blow up because they are described using Dirac deltas is not really a problem since Dirac deltas are only useful when used inside an integral, and the most important property of a Dirac delta is

$$\int_{\displaystyle V\ni\vec{r}_0}f(\vec{r})\delta(\vec{r}-\vec{r}_0)dV=f(\vec{r}_0)\tag{1}$$

which in simple terms means that if if you integrate a function $f(\vec{r})$ multiplied by a Dirac delta and the point $\vec{r}_0$ where the delta becomes singular lies inside the volume of integration then the result is the function $f(\vec{r})$ evaluated at the singularity: $f(\vec{r}_0)$.

How is this useful in any way? Well, for example, in order to apply Gauss's law in integral form we need first to choose a volume $V$. Then, when calculating the total charge inside the volume it may happen that there is a charged surface or a charged line or a point charge, but then Eq. $(1)$ will come to rescue. Let me put a particular example, suppose that the surface $r=a$ has a charge density $\sigma$ and that there is also a point charge $q$ at the origin. The total 3-dimensional charge density$^1$ will be

$$\rho(\vec{r})=\sigma\delta(r-a)+q\delta(\vec{r}).$$

Let's now find the total charge inside a sphere of radius $r=b>a$

\begin{align} Q&=\int\rho(\vec{r})dV\\ &=\int_{V\ni r=a}\sigma\delta(r-a)\ r^2\ dr\,d\Omega+\int_{V\ni\vec{0}} q\delta(\vec{r})dV\\ &=4\pi a^2\sigma+q \end{align}

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$^1$If you are worried about the dimensions take a look at this other post.