Panofsky and Phillips' Classical Electricity and Magnetism. Chapter 2, Section 3.

The authors show, assuming the microscopic field $\mathbf{\mathcal E}$ and the microscopic charge density $\rho_a$ are related by $\nabla\cdot\mathbf{\mathcal E} = \rho_a/\varepsilon_0$, that the macroscopic field $\mathbf E$ is related by $$ \nabla\cdot\mathbf E = \nabla\cdot\mathbf{\bar{\mathcal E}}\tag{*}, $$ where bar denotes space-time average.

They then claim that this implies that $\mathbf{E} = \bar{\mathcal E}$.

Question: I don't agree that (*) implies $\mathbf{E} = \bar{\mathcal E}$. What if the curls don't match? There can be many functions (which don't differ by just constants) with same divergence, no?

Note that the microscopic fields are not static.


1 Answer 1


If you look at pages 248--258 in Jackson [1999] you will find a detailed explanation of why/how one can derive the macroscopic Maxwell's equations from the microscopic ones. I won't rederive them here but will note a few important points.

First, you need to define macro vs micro. In this context, we assume all spatial scales larger than 10-14 m to be macroscopic, since such scales are large compared to atomic nuclei and electrons. Therefore, we can treat both as point sources/particles. Note that there will be no equivalent $\mathbf{D}$ or $\mathbf{H}$ fields in the microscopic domain because all of the charges and currents are local in the microscopic context.

We also know that all of these point particles are constantly undergoing some form of oscillation due to thermal fluctuations, zero point vibration, or orbital motion. We also know that the variations will occur on spatial scales of 10-10 m or less and temporal scales between 10-13 s and 10-17 s. Almost all instrumentation will average out most of these variations resulting in relatively smooth and slowly varying (compared to the time scales above) macroscopic quantities. This averaging is largely unavoidable for any device seeking to measure macroscopic quantities but it begs the question of what types of averages we should be performing. Spatial or temporal ensemble averages?

It turns out that spatial ensemble averages are the correct answer and one need not worry about temporal. So we define the spatial ensemble average of some function, $F\left( \mathbf{x}, t \right)$, with respect to some test function, $f\left( \mathbf{x} \right)$, as: $$ \langle F\left( \mathbf{x}, t \right) \rangle = \int \ d^{3}x' \ f\left( \mathbf{x}' \right) \ F\left( \mathbf{x} - \mathbf{x}', t \right) \tag{0} $$

So one might assume we need to know $f\left( \mathbf{x} \right)$ ahead of time but we do not need to explicitly define it. We only require that it be continuous such that it has a rapidly converging Taylor series for distances comparable to atomic scales. Without loss of generality, we can also show that: $$ \begin{align} \partial_{j} \langle F\left( \mathbf{x}, t \right) \rangle & = \langle \partial_{j} F\left( \mathbf{x}, t \right) \rangle \tag{1a} \\ \partial_{t} \langle F\left( \mathbf{x}, t \right) \rangle & = \langle \partial_{t} F\left( \mathbf{x}, t \right) \rangle \tag{1b} \end{align} $$ where $\partial_{j} = \tfrac{ \partial }{ \partial x_{j} }$ and $\partial_{t} = \tfrac{ \partial }{ \partial t }$. This results because the partial differential operators commute through the integral and they are of the ordinate $\mathbf{x}$ not $\mathbf{x}'$. Note you can also use a spatial Fourier transform to derive the macroscopic version of Maxwell's equations.

As an aside, this type of averaging is also what lets one derive macroscopic properties like the electric polarization, magnetic moment, and the magnetization of a medium. But recall, the critical piece here is that the averaging be a proper spatial ensemble average, not just the mean of a bunch of measurements or a time average. They are not always equivalent (in fact, they are rarely the same).


  • J.D. Jackson, Classical Electrodynamics, Third Edition, John Wiley & Sons, Inc., New York, NY, 1999.

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