Poynting theorem & electromagnetic energy
Derive the local and fixed-volume Poynting energy balance from Maxwell's curl equations, distinguishing field storage, outward electromagnetic flux and work on material charges. Compute energy transport in vacuum plane waves and energy delivery to a charging capacitor, then identify the limitations of naive stored-energy formulas in lossy, nonlinear and dispersive matter.
- Derive ∇·(E×B) from the double dot product of vacuum Maxwell curl equations using a vector identity.
- Identify vacuum field energy u=½ε₀E²+B²/(2μ₀), Poynting vector S=E×B/μ₀ and work density E·J.
- Obtain ∂t u+∇·S=−E·J and its integral fixed-volume form with signed outward boundary flux.
- Interpret positive outward S as field energy leaving a control volume without confusing it with electric charge current.
- Derive the relation between E, B, energy density and S for a source-free vacuum plane wave and its harmonic time average.
- Check a plane wave at 1 THz and electric amplitude 300 V/m numerically with directions and units.
- Apply Poynting's theorem to an ideal charging parallel-plate capacitor, distinguishing zero conduction through vacuum from incoming field-energy flux.
- Compare Ohmic power E·J=σE² with more general electromagnetic work on moving/bound charges.
- Derive the macroscopic E×H identity and state when u≈½(E·D+B·H) may be interpreted as reversible stored energy.
- Explain why hysteresis, complex dispersive ε(ω) or μ(ω), active response and moving boundaries need a broader material/energy model.
- Solve six worked sign, integration, wave, capacitor, material-limitation and dimensional exercises.
Maxwell's equations predict not only field shapes but the rate at which electromagnetic energy enters, leaves and transfers to material charges. An outward Poynting flux removes FIELD energy from a fixed control volume; positive E·J transfers field energy to charges. We derive this signed balance in vacuum, check a traveling wave and a charging capacitor, and then explain precisely which steps change inside matter.
0105.1 — Derive the Poynting identity from Maxwell's two curl equations
In the vacuum-source formulation, Maxwell gives ∇×E=−∂tB and ∇×B=μ₀J+μ₀ε₀∂tE. Apply the vector identity ∇·(E×B)=B·(∇×E)−E·(∇×B). This identity fixes the sign of the energy flux; switching the cross-product order would reverse it.
Substitute the curl equations. For constant vacuum ε₀ and μ₀, E·∂tE=½∂t(E²) and B·∂tB=½∂t(B²). Rearrange to give local Poynting conservation. J means the TOTAL physical electric-charge current in the vacuum-form equations; if matter is present, all relevant charge-current contributions must be treated consistently.
0205.2 — Define energy density, Poynting vector and work on charges
The local vacuum energy density is u=½ε₀E²+B²/(2μ₀)≥0. The energy-flux vector S=E×B/μ₀ points in the direction of outgoing electromagnetic energy; its SI unit is W/m². The term E·J is the electromagnetic power per volume transferred TO physical charges. It may appear as kinetic, mechanical or heat energy, depending on the matter model.
The magnetic force qv×B does zero instantaneous work on an individual point charge because (qv×B)·v=0. This fact is consistent with Poynting's energy exchange term E·J: only the electric force does direct work on the charged particles, while externally moving conductors and magnetic stresses can redirect mechanical energy.
0305.3 — Integrate over a FIXED control volume and check all signs
Choose a stationary volume V with outward normal n̂. Integrate ∂tu+∇·S=−E·J and apply the divergence theorem. The result is dU_field/dt=−∮S·n̂dS−∫E·Jd³r. A positive outward S-flux REDUCES the stored field energy; a positive E·J also reduces it because energy goes to charges. Incoming energy corresponds to S·n̂<0.
Do not add the mechanical or Joule energy to U_field on the left unless you also formulate a combined matter+field energy balance. For a moving volume, the derivative gains a boundary advection term, just as in EM-03.04; this chapter's displayed integral is specifically for FIXED V.
0405.4 — A vacuum plane wave carries energy along E×B
For the convention cos(kx−ωt), choose k∥+x, E∥+y and B∥+z (out of page). Maxwell–Faraday gives B=(1/ω)k×E, so E×B∥+x and S=E×B/μ0 travels with the wave. The arrows depict one instant when E>0, not a longitudinal electric field or a spatial wave profile. In vacuum |B0|=|E0|/c.
In a source-free monochromatic vacuum plane wave propagating along +z, choose E=E₀ cos(kz−ωt)x̂ and B=(E₀/c)cos(kz−ωt)ŷ. Their cross product points +z, while u_E=ε₀E²/2 equals u_B=B²/(2μ₀) instantaneously. Thus total u=ε₀E² and S=cu ẑ, which is positive during either sign of the field oscillation.
For a sinusoid, ⟨cos²⟩=½. The cycle-averaged intensity is ⟨S⟩=½ε₀cE₀² ẑ. Reversing the propagation direction reverses the field orientation appropriate for a traveling wave and reverses S; reversing E alone while keeping an independently fixed B is not the same wave.
0505.5 — Numerical vacuum-wave intensity with coherent field amplitudes
For f=1.00 THz and peak E₀=300 V/m in vacuum, λ₀=c/f≈300 μm and B₀=E₀/c≈1.00 μT, consistent with EM-03.03. Substituting ε₀≈8.854×10⁻¹² F/m and exact c gives ⟨S⟩≈119 W/m² along the wave's propagation direction.
The cycle-averaged total energy density is ⟨u⟩=ε₀E₀²/2≈3.98×10⁻⁷ J/m³, so c⟨u⟩≈119 W/m². These values apply to a single traveling plane wave in VACUUM; standing waves, near fields and lossy/dispersive materials need their actual fields and energy model.
0605.6 — Charging capacitor: field energy arrives through the GAP SIDE
Take z UP, x RIGHT, y INTO THE PAGE. A charging lower positive plate gives E∥+z and Bφ∥+y at the right gap side; consequently S=E×B/μ0∥−x, inward. Field energy increases at dU/dt≈VI, supplied by the inward lateral Poynting flux. No conduction electron travels across the ideal vacuum gap; edge-field and high-frequency corrections are omitted.
For ideal parallel vacuum plates of area A, separation d and V(t)>0, take the bottom plate positive and E=+E ẑ in the gap, E≈V/d. Increasing positive plate charge gives axial displacement-current flux +z and approximately azimuthal B=+Bφφ̂. Consequently S=(E ẑ)×(Bφφ̂)/μ₀=−EBφr̂/μ₀: energy flows RADially inward from the lateral boundary of the gap even though no conduction charge crosses the vacuum gap.
In the quasistatic uniform-gap model, C≈ε₀A/d and U_E≈½ε₀E²Ad=½CV². Differentiation gives dU_E/dt=V I where I=dQ/dt=C dV/dt at fixed C. The inward lateral Poynting flux has the same magnitude: at an ideal circular gap radius a, Bφ(a)≈μ₀I/(2πa), giving −∮side S·n̂dS≈[E I/(2πa)](2πa d)=EId=VI.
The uniform interior field, negligible fringing and quasistatic symmetric feed are MODEL assumptions. A real feed has nonuniform fields and may store magnetic energy or radiate; the exact Poynting theorem still holds for the actual fields even when the idealized side-only integral does not.
0705.7 — Distinguish work on charges from simple Joule heating
For an isotropic OHMIC medium with real nonnegative static conductivity σ and J=σE, the local field-to-charge power is E·J=σE²≥0. This is Joule heating in a fixed passive resistive model. Example: E=10.0 V/m and σ=2.00 S/m give J=20.0 A/m² and p=200 W/m³.
In general E·J includes reversible kinetic/mechanical transfer as well as possibly irreversible processes. For a perfectly insulating vacuum capacitor gap J_conduction=0 and no local Joule heating follows from a nonzero S or displacement current. Do not relabel all E·J as heat without a constitutive and mechanical account.
0805.8 — Material macroscopic balance and the limits of ½(E·D+B·H)
For macroscopic fields define D=ε₀E+P and H=B/μ₀−M. Using ∇×H=J_free+∂tD and ∇×E=−∂tB in ∇·(E×H)=H·(∇×E)−E·(∇×H) gives ∇·(E×H)+E·J_free+E·∂tD+H·∂tB=0. This identity is an electromagnetic power balance, but the two time-derivative products are NOT universally equal to the derivative of a simple stored-energy function.
For a stationary local LINEAR, reciprocal, lossless, nondispersive medium with constant real symmetric ε and μ, D=εE and B=μH, and E·∂tD+H·∂tB=∂t[½(E·D+B·H)]. Nonlinear reversible systems require integrating along the material state path; hysteresis and frequency-dependent complex ε(ω), μ(ω) introduce loss, memory and internal degrees of freedom that invalidate a naive static substitution at THz.
0905.9 — Numerical capacitor and conservation checklist
For an ideal vacuum capacitor A=1.00×10⁻² m², d=1.00 mm and V=10.0 V, C≈ε₀A/d≈8.85×10⁻¹¹ F, E≈1.00×10⁴ V/m and U_E≈½CV²≈4.43×10⁻⁹ J. At a charging instant when I=+1.00 μA, field energy rises at VI=10.0 μW; the signed OUTWARD lateral Poynting flux is approximately −10.0 μW.
Always declare the control volume fixed, its outward normal, which physical current enters E·J, whether the field is real instantaneous or a complex harmonic amplitude, and whether energy of the material is separately included. For a harmonic vacuum wave use ½ for peak-amplitude averaging. For THz conducting or dispersive interfaces, calculate actual Poynting flux and a specified material loss model rather than interpreting 1−R−T as local stored energy without additional assumptions.
05.10 — Six exercises with worked solutions
- Starting from vacuum Maxwell curl equations, derive ∂tu+∇·S=−E·J. Identify the vector identity and the SIGN of each term.
- A fixed source-free volume has an outward Poynting flux +2.00 W and no field work on charges. Find dU_field/dt. Repeat if a net field-to-charge work rate +0.500 W occurs.
- For a +z traveling vacuum plane wave with peak E₀=300 x̂ V/m and f=1.00 THz, compute B₀ vector and cycle-averaged S vector, including direction and unit.
- An ideal capacitor with C=88.5 pF has V=10.0 V and charges at I=+1.00 μA. Compute U and its rate of change. Which boundary carries energy into its ideal vacuum gap when E is axial?
- For a passive Ohmic region with E=10.0 V/m and σ=2.00 S/m, calculate J and E·J. Is the same Joule conclusion valid for displacement current in a vacuum gap?
- Write the macroscopic E×H identity and state the conditions needed to identify ½(E·D+B·H) as local reversible stored energy. Explain why substituting a complex THz permittivity is not generally sufficient.
Worked solutions
1. Use ∇·(E×B)=B·(∇×E)−E·(∇×B). Maxwell gives −B·∂tB−μ₀E·J−μ₀ε₀E·∂tE. Divide by μ₀ and identify u and S; outward flux and charge work both reduce field energy.
2. dU/dt=−2.00 W in the first case. With additional +0.500 W field-to-charge work, dU/dt=−2.50 W. A negative outward flux would instead contribute energy TO the volume.
3. B₀=(E₀/c)ŷ≈+1.00 μT ŷ. ⟨S⟩=½ε₀cE₀² ẑ≈+119 W/m² ẑ. The field vectors and energy flux form the +x,+y,+z right-handed triad.
4. U=½CV²≈4.425×10⁻⁹ J≈4.43 nJ. dU/dt=VI=+10.0 μW. With axial E and azimuthal B, E×B points inward radially through the gap's lateral boundary.
5. J=σE=20.0 A/m² and E·J=σE²=200 W/m³. No: a changing E in an ideal vacuum gap is not an Ohmic carrier current; J_conduction=0 there.
6. ∇·(E×H)+E·J_free+E·∂tD+H·∂tB=0. If response is stationary, local, linear, real, reciprocal, lossless and nondispersive, the derivative products yield ∂t½(E·D+B·H). Complex dispersive ε(ω) includes memory/loss, and a static quadratic is not generically the stored-energy functional.
EM-03 now closes the electrodynamics foundation: Faraday induction, displacement current, the Maxwell system, charge continuity and Poynting energy. The natural next step is wave propagation: plane-wave solutions, polarization, interfaces and the way fields transport energy through space and matter.