# Griffiths on Averaging and the Macroscopic Electric Field

I am self-studying Griffiths's Introduction to Electrodynamics (4th edition) and all has been smooth sailing except for section 4.2.3 in which Griffiths argues for why we can compute (at least at the elementary level of classical electrodynamics that I am currently studying) the field due to a polarized dielectric chunk of material by simply adding up the potential (or field) due to each tiny volume element times the polarization per unit volume $$\mathbf{P}$$, treating each volume element as containing a perfect dipole. I have attached the page and a half of the chapter here, as well as the Eq. 4.8 alluded to there. I'll explain what I understand and where I lose track of his argument, and I'm hoping someone can help me sort things out.

After a very gentle qualitative intro, Griffiths essentially says that we are going to define the macroscopic field $$\mathbf{E(r)}$$ at the point $$\mathbf{r}$$ in space as the average electric field over a sphere of radius a "thousand times the size of a molecule" about the point $$\mathbf{r}$$, which I'll call $$V$$. Fine, this is just a definition, and he says that analogous definitions using ellipsoids etc. yield the same answer, which I'm willing to accept. Now by superposition, the average field on $$V$$ can be found as the average field due to charges inside and outside of the sphere. With (4.17), he gives the average potential over $$V$$ due to all charges (dipoles) outside of $$V$$, and with (4.18) he gives the average field inside $$V$$ due to the charges inside $$V$$. I'm still following up to here, but now he loses me. He says that what is left out of the integral "corresponds" to the field at the centre of a uniformly polarized sphere, but what does this correspondence mean? I see how the integral in (4.17) doesn't include $$V$$, but why does that mean that $$\mathbf{E}_{in}$$ adds back what is missing (what does it even mean for an average field to source potential? We can speak of charge sourcing potential, or taking a line integral of a field giving a potential, but this is just some field value?).

The best I can come up with is to say that if you take (4.19), split the integral out into $$V$$ and outside $$V$$, and then apply the (negative) gradient to the integral, one should be able to show that $$\mathbf{E}_{in}$$ is recovered for the former term, and so that would seem to imply that our coarse graining calculation does indeed give the correct macroscopic field as we have defined it (but I couldn't see how to DO this derivative, so I couldn't prove it).

I've tried to look at other resources on this (eg. Zangwill), but none were digestible for me unfortunately.

Please let me know if this is unclear, and I can try to rephrase. But essentially I'm hoping someone can, in detail, walk me through Griffiths's arguments.

Eq 4.8 (note that Griffiths's script vector r equals $$\mathbf{r} - \mathbf{r'}$$, and hats have their usual unit vector meaning):

Griffiths text:

Since Griffiths' argument relies heavily on two results that are left as an exercise in the book (Exercise 3.47), it's worth proving them first.

## Previous results

1. The first result is to show that the average field inside a sphere of radius $$R$$, due to all the charge within the sphere, is $${\mathbf E}_{\rm ave,in} = \frac{-1}{4\pi\varepsilon_0}\frac{\mathbf p}{R^3},\tag{Griffiths, 3.105}$$ where $$\mathbf p$$ is the total dipole moment with respect to the center of the sphere, which we will take at $$\mathbf r=0$$ for simplicity. To prove this I will try to use Griffiths' notation, except for the script $$\mathbf r$$'s, which are not nicely writable with MathJax.

First, the average of the field is by definition the integral over the sphere divided by the volume $$\cal V$$ of the sphere

$$\mathbf{E}_{\rm ave}=\frac{1}{\cal V}\int_{\rm inside}{\mathbf E}(\mathbf{r})d\tau.\tag{1}$$

In particular, the average field due to all the charge within the sphere is

$$\mathbf{E}_{\rm ave, in}=\frac{1}{\cal V}\int_{\rm inside}{\mathbf E_{\rm in}}(\mathbf{r})d\tau,\tag{2}$$

where

$$\mathbf{E}_{\rm in}(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\rho(\mathbf r')\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau'.\tag{3}$$

Inserting $$(3)$$ in $$(2)$$ we get

\begin{align} \mathbf{E}_{\rm ave, in}&=\frac{1}{\cal V}\int_{\rm inside}\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\rho(\mathbf r')\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau'\right]d\tau\\ &=\frac{1}{\cal V}\int_{\rm inside}\rho(\mathbf r')\underbrace{\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau\right]}_{\displaystyle{\mathbf E}_{\rm aux}({\mathbf r}')}d\tau'.\tag{4} \end{align}

This last step may seem tricky, but the only thing we did is changing the order of integration. Instead of integrating first over the primed variables and then over the unprimed ones, we do it the other way around, first over the unprimed variables and then over the primed ones. Doing this, we are left with an expression for $$\mathbf{E}_{\rm ave, in}$$ that depends on an auxiliary field which I have called $${\mathbf E}_{\rm aux}({\mathbf r}')$$ because we are going to calculate it aside.

We can rewrite $${\mathbf E}_{\rm aux}({\mathbf r}')$$ as

\begin{align} {\mathbf E}_{\rm aux}({\mathbf r}')&=\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau\\ &=\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}(-1)\frac{{\mathbf r'}-{\mathbf r}}{|{\mathbf r'}-{\mathbf r}|^3}d\tau \end{align}

and notice that the field $${\mathbf E}_{\rm aux}({\mathbf r}')$$ is the field that a sphere with uniform charge density $$\rho_0=-1$$ would create at a point $$\mathbf r'$$, and this is something we know how to solve with Gauss' Law

$${\mathbf E}_{\rm aux}({\mathbf r}')=\begin{cases} \displaystyle\frac{\rho_0}{3\varepsilon_0}{\mathbf r'}&\quad{\rm if}\quad r'R. \end{cases}$$

Now, back to the integral $$(4)$$:

$$\mathbf{E}_{\rm ave, in}=\frac{1}{\cal V}\int_{\rm inside}\rho(\mathbf r'){\mathbf E}_{\rm aux}({\mathbf r}')d\tau'$$

Since we want to integrate inside the sphere substitute $${\mathbf E}_{\rm aux}({\mathbf r}')$$ with the appropriate case (i.e. the one for $$r'). Finally, remember the expression for the total dipole moment

$$\mathbf{p}=\int \mathbf{r}'\rho(\mathbf{r}')d\tau'$$

and you should arrive to $$({\rm Griffiths,\ 3.105})$$.

2. The second result we need to show is that the average field over the volume of a sphere, due to all the charges outside ($$\mathbf{E}_{\rm ave,out}$$), is the same as the field they produce at the center. This result is obtained in a rather similar way. Start again from $$(1)$$, but now average the field $$\mathbf{E}_{\rm out}$$ produced by charges outside

$$\mathbf{E}_{\rm ave, out}=\frac{1}{\cal V}\int_{\rm inside}{\mathbf E_{\rm out}}(\mathbf{r})d\tau.\tag{6}$$

The integral is still over the inside, though, because the average is over the inside points. On the other hand, $$\mathbf E_{\rm out}(\mathbf{r})$$ is the field produced by charges outside, so

$$\mathbf{E}_{\rm out}(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\int_{\rm outside}\rho(\mathbf r')\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau',$$

and then

\begin{align} \mathbf{E}_{\rm ave, out}&=\frac{1}{\cal V}\int_{\rm inside}\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm ouside}\rho(\mathbf r')\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau'\right]d\tau\\ &=\frac{1}{\cal V}\int_{\rm outside}\rho(\mathbf r')\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{{\mathbf r}-{\mathbf r}'}{|{\mathbf r}-{\mathbf r}'|^3}d\tau\right]d\tau'\\ &=\frac{1}{\cal V}\int_{\rm outside}\rho(\mathbf r'){\mathbf E}_{\rm aux}({\mathbf r}')d\tau'\\ &=\frac{1}{\cal V}\int_{\rm outside}\rho(\mathbf r')\frac{-R^3}{3\varepsilon_0}\frac{\mathbf r'}{r'^3}d\tau'\\ &=\frac{-1}{4\pi\varepsilon_0}\int_{\rm outside}\rho(\mathbf r')\frac{\mathbf r'}{r'^3}d\tau'. \end{align}

Finally, the last step can be rewritten to show Griffiths' second result

$$\mathbf{E}_{\rm ave, out}=\frac{1}{4\pi\varepsilon_0}\int_{\rm outside}\rho(\mathbf r')\frac{\mathbf 0-\mathbf r'}{(0-r')^3}d\tau'=\mathbf{E}_{\rm out}(\mathbf{0})$$

namely that the average field over the volume of a sphere, due to all the charges outside, is the same as the field they produce at the center.

## Problem

Griffiths tells us in an earlier section that we can calculate the potential produced by a polarized dielectric as

$$V(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\int_{\cal V}\frac{{\mathbf P}(\mathbf r')\cdot (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'\tag{7}$$

where the integration is over the entire dielectric. But now he makes you wonder whether this might not be strictly true, because when we choose $$\mathbf r$$ to be inside the dielectric, the molecules close to $$\mathbf r$$ create a field for which the dipole approximation might not be very good.

Then, a distinction is made between the real microscopic electric field, and the macroscopic electric field, the latter defined as an average over regions that contain many molecules. The microscopic is impossible to calculate, but the macroscopic, Griffiths claims, can be calculated via $$(7)$$, and is what we refer to as the "Electric field inside matter".

## Main argument

With the two previous results in mind, Griffiths' argument to prove that $$(7)$$ yields the correct macroscopic potential is:

1. The macroscopic electric field at a point $$\mathbf r$$ is the average of the microscopic field over a sphere of radius $$R$$ centered at $$\mathbf r$$ (cf. Eq. $$(1)$$).
2. The real microscopic field has two contributions, the field created by charges outside the sphere $$\mathbf E_{\rm out}$$ and the one created by charges within the sphere $$\mathbf E_{\rm in}$$. Therefore, the macroscopic field also has two contributions (cf. Eqs. $$(2)$$ and $$(6)$$)

\begin{align}{\mathbf E}_{\rm macro}(\mathbf r)&\equiv{\mathbf E}_{\rm ave}(\mathbf r)\\ &={\mathbf E}_{\rm ave,out}(\mathbf r)+{\mathbf E}_{\rm ave,in}(\mathbf r) \end{align}

1. The average field created by charges outside is just the real microscopic field that they create at the center (we proved it before). Since all those molecules are far away from the center of the sphere I can calculate it with the dipole approximation

$${\mathbf E}_{\rm ave,out}(\mathbf r)={\mathbf E}_{\rm out}(\mathbf r)=-\nabla V_{\rm out}=-\nabla\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm outside}\frac{{\mathbf P}(\mathbf r')\cdot (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'\right]$$

2. I would very much like to be able to express the average field due to the internal charges as

$${\mathbf E}_{\rm ave,in}(\mathbf r)=-\nabla\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{{\mathbf P}(\mathbf r')\cdot (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'\right]\tag{8}$$

because then both contributions would be symmetrical and I would be able to calculate $$\mathbf E(\mathbf r)=-\nabla V$$ using Griffits' earlier proposal (Eq. $$(7)$$) without having to worry about what charges are inside or outside or whatever. However, we showed that the correct expression for $${\mathbf E}_{\rm ave,in}$$ is

$${\mathbf E}_{\rm ave,in} = \frac{-1}{4\pi\varepsilon_0}\frac{\mathbf p}{R^3},\tag{Griffiths, 3.105}$$

which can be written using $$\mathbf p={\cal V}\mathbf P=(4/3)\pi R^3\mathbf P$$ as

$${\mathbf E}_{\rm ave,in} = \frac{-\mathbf P}{3\varepsilon_0}.\tag{9}$$

3. The only way out is: are the expressions $$(8)$$ and $$(9)$$ equivalent (the same)? The answer is YES. If we assume the polarization $$\mathbf P$$ is constant within the sphere then we can get it out of the integral in $$(8)$$:

\begin{align} {\mathbf E}_{\rm ave,in}(\mathbf r)&=-\nabla\left[\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{{\mathbf P}(\mathbf r')\cdot (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'\right]\\ &=-\nabla\left[{\mathbf P}\cdot\left(\frac{1}{4\pi\varepsilon_0}\int_{\rm inside}\frac{ (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'\right)\right]\\ &=-\nabla\left[{\mathbf P}\cdot\left(-\mathbf E_{\rm aux}(\mathbf r)\right)\right]\\ &=-\nabla\left[{\mathbf P}\cdot\left(\frac{\mathbf r}{3\varepsilon_0}\right)\right]\\ &=-\nabla\left[\frac{1}{3\varepsilon_0}(xP_x+yP_y+zP_z)\right]\\ &=\frac{-\mathbf P}{3\varepsilon_0} \end{align}

hence recovering $$(9)$$.

• This answer is superb, thank you! Before I accept it, I'm just wondering why, in your last line, you can take $\mathbf{P}$ as a constant in evaluating the gradient? $\mathbf{P}=\mathbf{P(r)}$, right?
– EE18
Oct 31, 2020 at 1:08
• First, this is a very nice answer. Second, @1729_SR, the key step is that you can take any $\mathbf{P}$ inside of $\mathcal{V}$, so long as it produces the same average $\mathbf{p}$. The true $\mathbf{P}$ is a horrible mess. However, it's also equivalent (and much easier) to use a constant $\mathbf{P}$ which does not depend on $\mathbf{r}$. In Urb's answer, before Eq. 9, this is implicitly done by writing $\mathbf{p}=\mathcal{V}\mathbf{P}$. This is not true of the exact microscopic $\mathbf{P}$, but this is where we sneak in the idea of a macroscopic, average $\mathbf{P}$. Oct 31, 2020 at 2:31
• @Andrew, thanks for the response, and for the very insightful answer below as well. I think I follow your argument regarding constant $\mathbf{P}$ on $V$, but what I don't understand still is why the $\mathbf{P}$ is constant with respect to the gradient at the end (with respect to the integral, fair enough, that's over a small $V$, but with respect to the gradient? The macroscopic field should change as $f(\mathbf{r})$ after all.).
– EE18
Oct 31, 2020 at 11:36
• Hi @1729_SR, of course we have to take into account the dependence of $\mathbf P(\mathbf r)$, i.e. the macroscopic $\mathbf P$ changes from one point in the dielectric to another. But in $(8)$ we don't have the entire $\mathbf r$-dependence. Think of it this way: we want to calculate $V$ at $\mathbf r$ with Eq. $(7)$, and we divide the integral into the part of the dielectric that is far away from $\mathbf r$ and the part that is close to $\mathbf r$.
– Urb
Oct 31, 2020 at 13:51
• $\frac{1}{4\pi\varepsilon_0}\int_{all\ dielectric}\frac{\mathbf P(\mathbf r')\cdot (\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}d\tau'=\int_{inside}+\int_{outside}$. The integral over the inside (over $\cal V$) turns out to be $\mathbf P_0\cdot \mathbf r/(3\varepsilon_0)$ where $\mathbf P_0$ is just the value of $\mathbf P$ at the neighborhood of the point where we are calculating the potential.
– Urb
Oct 31, 2020 at 13:52

A note about notation: for clarity, I will use $$V$$ to refer to the volume of the small region, and $$\phi$$ to refer to the electric potential (as opposed to $$\mathcal{V}$$ and $$V$$, which I find unnecessarily confusing).

Griffiths's argument has 4 main steps:

1. The average field $$\vec{E}$$ over the volume $$V$$ can be broken up, via the superposition principle, into two pieces: $$\vec{E}=\vec{E}_{\rm out} + \vec{E}_{\rm in}$$, where $$\vec{E}_{\rm out}$$ is the field due to charges outside of $$V$$ and $$\vec{E}_{\rm in}$$ is the field due to charges inside of $$V$$.
2. The field $$\vec{E}_{\rm out}$$ can be estimated using the leading non-vanishing term in the multipole expansion, since the sources are all assumed to be far away.
3. The field $$\vec{E}_{\rm in}$$ can be estimated as the field due to a sphere with a uniform polarization density with the same average polarization as the real volume. Assuming a uniform polarization density is an approximation which is fine so long as $$V$$ is a small volume.
4. Combining steps 2 and 3 leads to Eq 4.19.

From your question, I think you follow steps 1 and 2, so I'll focus on steps 3 and 4. (Actually step 4 is basically trivial if you follow everything else so this is mostly about step 3).

Step 3 is actually derived in Griffifths in an earlier section, the punchline is given by Eq 3.105. The point is that the average field inside the sphere, regardless of the charge distribution, is proportional to the average polarization of the sphere, $$\vec{p}$$. Then here is a tricky step: we can use any polarization density $$\vec{P}$$ inside the sphere that gives us the same average polarization $$\vec{p}$$. The reason is that the field $$\vec{E}_{\rm in}$$ only depends on the average polarization $$\vec{p}$$. So, we assume a uniform polarization density, since (a) is the easiest thing to do, and (b) our goal is to "smooth" over the volume $$V$$ and assign a single macroscopic polarization density and field to this region.

I said above that $$V$$ should be a small volume for the uniform polarization density to be a good approximation. This statement requires stepping back a bit to think about what we are doing. If all we want is the average field at the center of some arbitrary volume, then mathematically we could use a uniform polarization density that gives the same average polarization as the real volume, for any size $$V$$. However, what we really want is a reasonable description of the macroscopic field everywhere in the solid. If $$V$$ is too big, we will "coarse grain" too much and lose track of large scale variations in the polarization that are significant. Therefore, quantifying exactly what we mean by "a small volume" is important, but problem-dependent.

Step 4 is just the mathematical statement that the sum of two integrals over two disjoint volumes is the integral over the combined volume.

Finally, you mention that you should be able to differentiate the potential to recover the total field. This certainly has to be true. However note this gives you new information, in a sense -- you can take the derivative but you won't be able to compare it to something else you already knew in this problem. Anyway here's what you get if you compute the field:

$$\begin{eqnarray} E_i=\partial_i \phi &=& \frac{1}{4\pi \epsilon_0} \int {\rm d}^3 x' P_j (\vec{r}') \partial_i \frac{r_j - r'_j}{|\vec{r}-\vec{r}'|^3} \\ &=& \frac{1}{4\pi \epsilon_0} \int {\rm d}^3 x' P_j (\vec{r}') \partial_i \partial_j' \frac{1}{2 |\vec{r}-\vec{r'}|^2} \\ &=& \frac{1}{4\pi \epsilon_0} \left[\int_S {\rm d^2 S} n_j P_j \frac{r_i - r_i'}{|\vec{r}-\vec{r}'|^3} + \int_V {\rm d}^3 x \partial_j' P_j \frac{r_i-r'_i}{|\vec{r}-\vec{r}'|^3}\right] \\ &=& \frac{1}{4\pi \epsilon_0}\left[\int_S {\rm d}^2 x \sigma_b \frac{\scr{r}_i}{|\scr{r}|^3} + \int_V {\rm d}^3 x \rho_b \frac{\scr{r}_i}{|\scr{r}|^3} \right] \end{eqnarray}$$

where I defined $$\vec{\scr{r}} = \vec{r}-\vec{r}'$$, the unit normal to the surface $$n_i$$, and the surface bound charge $$\sigma_b = \hat{n} \cdot (\vec{P}_{\rm out} - \vec{P}_{\rm in})$$ and the volume bound charge $$\rho_b = - \nabla \cdot \vec{P}$$. Here $$V$$ and $$S$$ are the volume and surface of the actual material, not the little averaging volume. This should make a lot of sense -- the electric field is sourced by the bound volume and surface charges. I am assuming there are no free charges around here. It's also possible I got some signs wrong so please be careful using this, I tried to be careful but my main aim was to show conceptually what was going on.