advanced35 min studyLesson 271 in Physics

Macroscopic fields and spatial averaging

Separate averaged electromagnetic fields from microscopic fields and close Maxwell equations with material response.

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What you’ll understand

  • Define polarization and magnetization as dipole moment densities.
  • Derive bound volume and surface charge from polarization.
  • Identify the scale separation required by a local constitutive law.

A voltage probe cannot resolve the electric field around every nucleus. Its reading concerns a field averaged over many microscopic constituents, while retaining variations across the sample. Macroscopic electrodynamics makes that change of description explicit. Maxwell equations still express charge conservation and induction, but part of the microscopic charge and current is represented through polarization and magnetization. The resulting material laws require physical assumptions; they are not additional identities hidden inside Maxwell equations.

Choose an averaging scale

Let aa denote a characteristic molecular spacing and let LL denote the shortest spatial scale of the macroscopic field. Choose a smooth averaging window of width ℓ\ell with many constituents inside it, but small variation of the resolved field across it. The useful hierarchy is a≪ℓ≪La\ll\ell\ll L. Averaging then suppresses atomic fluctuations without erasing the spatial pattern being studied. Near an atomically thin interface, or when the field changes across one unit cell, such a window may not exist.

Linear averaging commutes with spatial derivatives when the window is translation invariant and sufficiently regular. This explains why the averaged Maxwell equations retain their differential structure. Products behave differently: the average of a product need not equal the product of averages. Microscopic energy, force, and local-field corrections therefore require more care than simply replacing every symbol by its mean. A constitutive parameter fitted to one spatial scale should not automatically be trusted for sharp near fields.

a≪ℓ≪L,P=1ΔV∑j∈ΔVpj,M=1ΔV∑j∈ΔVmja\ll\ell\ll L,\qquad \mathbf P=\frac{1}{\Delta V}\sum_{j\in\Delta V}\mathbf p_j,\qquad \mathbf M=\frac{1}{\Delta V}\sum_{j\in\Delta V}\mathbf m_j
The averaging hierarchy is an approximation criterion. Electric dipole moment p has units C m, polarization P has C/m², magnetic dipole moment m has A m², and magnetization M has A/m.↔ Scroll wide equations horizontally.

Move bound sources into material fields

Electric polarization counts dipole moment per volume, not net charge per volume. Neighboring dipoles cancel internally when their density is uniform. Where the polarization varies, incomplete cancellation produces a bound volume charge. A finite uniformly polarized block can therefore have zero bound charge throughout its interior while carrying opposite charges on its faces. This does not mean the material has acquired a net charge: the volume and surface contributions must be counted together.

Define the outward unit normal on a material bordering vacuum. The polarization surface charge equals its normal projection, with its sign fixed by which end of each dipole terminates there. Magnetization similarly describes circulating bound currents, and time-dependent polarization gives an additional bound current. Introducing electric displacement and magnetic intensity packages those contributions so that their Maxwell equations show the chosen free sources explicitly. The free/bound division is a modeling convention; total charge and total current remain the physical sources.

ρb=−∇⋅P,σb=P⋅n^,D=ϵ0E+P,H=Bμ0−M\rho_b=-\nabla\cdot\mathbf P,\quad \sigma_b=\mathbf P\cdot\hat{\mathbf n},\quad \mathbf D=\epsilon_0\mathbf E+\mathbf P,\quad \mathbf H=\frac{\mathbf B}{\mu_0}-\mathbf M
Volume charge density is C/m³ and surface charge density is C/m². The surface formula assumes vacuum outside; at a material interface use the jump in normal polarization.↔ Scroll wide equations horizontally.

Close the description without hiding memory

A local isotropic linear dielectric uses a scalar susceptibility relating polarization to the macroscopic electric field. Locality means the response at one point ignores field gradients and fields at other points; isotropy means rotating the field rotates the response identically. Neither property follows from averaging alone. Crystals may need tensors, and conducting nanostructures may need spatially nonlocal response. The microscopic field that acts on one molecule can differ substantially from the averaged field used in a measured susceptibility.

Throughout this module real harmonic fields are the real parts of complex amplitudes multiplied by e−iωte^{-i\omega t}, with positive angular frequency. Time differentiation therefore multiplies an amplitude by −iω-i\omega. A causal material can retain memory, so its frequency-domain susceptibility is generally complex and frequency dependent. For a passive local scalar dielectric, the imaginary part of permittivity is nonnegative at positive frequency. Fixing this convention now prevents later loss terms from accidentally describing amplification.

Jb=∂P∂t+∇×M,P(ω)=ϵ0χe(ω)E(ω),ϵr=1+χe\mathbf J_b=\frac{\partial\mathbf P}{\partial t}+\nabla\times\mathbf M,\qquad \mathbf P(\omega)=\epsilon_0\chi_e(\omega)\mathbf E(\omega),\qquad \epsilon_r=1+\chi_e
Bound current is A/m². The scalar constitutive relation assumes linear, local, isotropic response; εr and χe are dimensionless and may depend on angular frequency in rad/s.↔ Scroll wide equations horizontally.

Charges on a polarized slab

A slab of thickness d=2.0 mmd=2.0\,\mathrm{mm} and face area A=3.0 cm2A=3.0\,\mathrm{cm^2} has uniform polarization P=4.0 μC/m2 z^\mathbf P=4.0\,\mu\mathrm{C/m^2}\,\hat{\mathbf z} normal to its faces. Find the bound charge on each broad face and in the interior.

  1. Uniform polarization has zero divergence, so the interior bound volume charge density vanishes. Side faces have normals perpendicular to the polarization and carry no bound surface charge in this ideal geometry.

  2. The upper face has outward normal +z^+\hat{\mathbf z}, giving σb=+4.0 μC/m2\sigma_b=+4.0\,\mu\mathrm{C/m^2}; the lower face has the opposite normal and opposite surface charge density.

  3. Convert the area to 3.0×10−4 m23.0\times10^{-4}\,\mathrm{m^2} before multiplying. Each face has charge magnitude 4.0×10−6×3.0×10−4=1.2×10−9 C4.0\times10^{-6}\times3.0\times10^{-4}=1.2\times10^{-9}\,\mathrm C.

The result

The top and bottom charges are +1.2 nC and −1.2 nC; the interior bound charge is zero and the slab is neutral.

Explore the model

Interactive lab

Interactive lab · vectors

Surface charge is a projection

For each fresh comparison, Reset and set BOTH controls. First set Angle = 0° and Magnitude = 1; then Reset and set Angle = 90° and Magnitude = 1. Regard the x direction as a face normal and predict the normal projection changing from one to zero. The instrument shows a vector and components in arbitrary units; it does not average microscopic fields or solve a polarized slab.

Surface charge is a projectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe normal component controls surface charge even when the polarization magnitude is unchanged.

Check your understanding

A finite uniformly polarized slab has no bound volume charge. What can still occur?

Try it yourself

Inside a slab 0<z<d0<z<d, take Pz=αzP_z=\alpha z with α=2.0×10−3 C/m3\alpha=2.0\times10^{-3}\,\mathrm{C/m^3}, d=1.0 mmd=1.0\,\mathrm{mm} and area A=1.0 cm2A=1.0\,\mathrm{cm^2}. Find the total bound volume charge and the two face charges.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate P inside the slab, then include the jump to zero polarization outside each face.

Compare with the solution

The interior density is ρb=−α\rho_b=-\alpha. Its integral is −αAd=−2.0×10−10 C-\alpha Ad=-2.0\times10^{-10}\,\mathrm C. At the top, Pz=αd=2.0×10−6 C/m2P_z=\alpha d=2.0\times10^{-6}\,\mathrm{C/m^2}, so the top charge is +2.0×10−10 C+2.0\times10^{-10}\,\mathrm C. At the bottom the polarization is zero, hence its surface charge is zero. Adding all three terms gives zero total charge, as required for a dipole distribution with no added free charge.

Take it with you

  • A macroscopic field requires a window between microscopic and resolved spatial scales.
  • Bound surface charge can remain when bound volume charge vanishes.
  • Constitutive locality, isotropy and linearity are separate assumptions.

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