The force experienced on a charge, using macroscopic quantities

Question

Zangwill says (Zangwill, Modern Electrodynamics, 2012 edition, page 40, undersection 2.3.1, "Lorentz Averaging")

...An example is the force on the charge density $\rho$ and current density $j$ in a volume $V$ due to electromagnetic fields $E$ and $B$ \begin{equation}F=\iiint (\rho E + j\times B)d\tau \end{equation}(2.40) Interpreted as a microscopic formula, the direct substitution of $\rho$ and $j$ confirms that (2.40) reproduces the Lorentz force law (2.1) for each microscopic particle. However, it is generally the case that $〈\rho E〉 \neq〈\rho 〉〈E〉$ and $〈j × B〉 \neq 〈j〉 × 〈B〉$. How, then, shall we compute the force on a macroscopic body? The answer, not often stated explicitly, is that we simply assume that (2.40) remains valid when all the variables are interpreted macroscopically. No ambiguities arise as long as F is the total force on an isolated sample of macroscopic matter in vacuum.

Let's say we have a point charge $q$ , and I do the spatial averaging of it, and let the resulting macroscopic charge density be $\rho_{macro}$. If I understand Zangwill correctly, the force on this point charge must be equal to $\iiint \rho_{macro}E_{macro}d\tau$ (where $E_{macro}$ is the macroscopic averaged electric field). Is this the correct interpretation of what he has said?

I understand that the force on the point particle is $qE_{micro}$ where $E_{micro}$ is the "true" microscopic electric field. I wonder whether this is equal to the above mentioned integral.

(We are only dealing with electrostatics so we are only bothered with the electric fields)

Answer 1

Yes, this is a correct conclusion, and you're right. Lorentz averaging does not perfectly preserve the system it describes.

To prove this, consider an electric field $\overrightarrow{E}(x,y,z)$ whose value at the origin is different from its average value on the unit sphere (sphere of radius 1 centered at the origin). The field is created by a certain charge distribution $\rho(x,y,z)$. Now consider a point charge $q$ at the origin. It experiences a force from the field equal to $q\overrightarrow{E}(0,0,0)$. The average charge over the unit sphere is $\frac{3}{4\pi}\left(q + \iiint_V \rho(x,y,z)\,dxdydz\right)$.

Note that the electric field difference is independent of $q$, while the average charge is not independent. Let $q_\rho$ be the total charge in the sphere without the point charge $q_0$. Suppose $\left(q_\rho + q_0\right)\overrightarrow{E}_{macro} = q_0\overrightarrow{E}_{micro} = \overrightarrow{F}_0$. Then, necessarily, if we increase $q$ by 1, we get an inequality: $$ \overrightarrow{F}_0 + 1\text{C}\cdot\overrightarrow{E}_{micro} \neq \overrightarrow{F}_0 + 1\text{C}\cdot\overrightarrow{E}_{macro} $$

To complete the proof, here's an example of an electric field which is not the same at the origin as its average over the unit sphere: $$ \overrightarrow{E}(x,y,z) = (x+1)^2\hat{x} $$ $$ \left(\frac{1}{\frac{4}{3}\pi}\iiint_V (x+1)^2\,dxdydz\right)\hat{x} $$ $$ \left(\frac{1}{\frac{4}{3}\pi}\int_{-1}^1 (x+1)^2\cdot \pi(1-x^2)\,dx\right)\hat{x} $$ $$ 1.2\hat{x} $$ Which is inequal to the field at the origin, which is just $\hat{x}$.