Griffiths on Averaging and the Macroscopic Electric Field

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

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


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

    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

    $${\mathbf E}_{\rm aux}({\mathbf r}')=\begin{cases} \displaystyle\frac{\rho_0}{3\varepsilon_0}{\mathbf r'}&\quad{\rm if}\quad r'<R\tag{5}\\ \displaystyle\frac{\rho_0R^3}{3\varepsilon_0}\frac{\mathbf r'}{r'^3}&\quad{\rm if}\quad r'>R. \end{cases}$$

    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 $r'<R$). 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.


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 $(9)$.