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Electromagnetic Waves & Energy Flow

Analytic plane waves in a lossless, nonmagnetic medium: the electric field E, the magnetic field H = (1/η) k̂ × E, and the Poynting vector S = E × H that carries the energy. Add a mirror or a second medium to see how reflection builds a standing wave whose electric and magnetic energy trade places every quarter period.

Sliders have number boxes for exact values • Space on a focused Play button starts/pauses • Every setting is saved in the URL
η₁ 377 Ω Wave impedance
λ₁ 633 nm Medium wavelength
I₀ — Incident ⟨S⟩
R 0 Reflectance
⟨S⟩ — Net flux, medium 1

Fields along the z axis oblique 3-D view, to scale in z

  • E arrows: electric-field vector at points on the z axis
  • η₁H arrows: magnetic field scaled by η₁ (same units, V/m)
  • S arrows (bottom rail): Poynting vector, energy flow along ±z
  • Dotted curves: the line through the arrow tips at this instant. Nothing moves along them; they are not ray paths.
  • Dashed plane: probe zp; the inset shows E and η₁H there, looking back toward the source (x up, y left)

Spatial cross-section at fixed t absolute, V/m

Instantaneous Ex, Ey, η₁Hx, η₁Hy versus z (components that are identically zero are omitted). Thin dashed curves are ±|Ẽ(z)|, the peak envelope. Shaded: z ≥ 0 (medium 2 or conductor).

Temporal cross-section at zp absolute, V/m

The same components versus time over two periods at the probe. The white cursor marks the current instant t.

Energy density and flux vs z normalised

uE/ū₀, uH/ū₀ and Sz/I₀ at the current instant (solid) and their time averages (dashed). ū₀ = ε₁E₀²/2 and I₀ = n₁ε₀cE₀²/2 are the incident wave's averages, so a lone travelling wave has Sz/I₀ = u/ū₀.

Energy exchange at zp vs t normalised

uE/ū₀, uH/ū₀, total u/ū₀ and Sz/I₀ at the probe. In a standing wave, uE and uH peak a quarter period apart and Sz changes sign.

Incident intensity I₀ = n₁ε₀cE₀²/2
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Erms = E₀/√2
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Peak H₀ = E₀/η₁
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Phase velocity c/n₁
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Period T = λ₀/c
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r (source)
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R, T, R + T
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⟨Sz⟩ medium 1 / medium 2
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Standing-wave ratio
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At zp: E·k̂, H·k̂, E·H (forward wave alone)
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At zp, now: uE, uH, Sz
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At zp: ⟨uE⟩/⟨uH⟩
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Poynting check max|∂u/∂t + ∂Sz/∂z|/(ωū₀)
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💡 How to use

Learn with this tool

Learning objectives

  • Construct H from E for a plane wave in any direction with H = (1/η) k̂ × E, and verify that E, H and k are mutually perpendicular.
  • Compute instantaneous and time-averaged Poynting flux and energy densities from a peak amplitude, and convert to RMS without a factor-of-two error.
  • Explain how reflection superposes two waves into a (partial) standing wave with zero or reduced net flux, and how its energy oscillates between electric and magnetic form.

Prerequisites

Model

The medium is homogeneous, isotropic, lossless and nonmagnetic, so ε = n²ε₀ and μ = μ₀. With the suite's convention E(z, t) = Re{Ẽ ei(kz − ωt)} a wave travelling along k̂ satisfies

H = (1/η) k̂ × E,   η = μ₀c/n = η₀/n,   S = E × H,   u = ½εE² + ½μ₀H²

⟨S⟩ = ½ Re(Ẽ × H̃*) → I = n ε₀ c E₀²/2 = Erms²/η,   ⟨uE⟩ = ⟨uH⟩ = εE₀²/4

At normal incidence on z = 0 the reflected wave has amplitude rẼ and travels along −ẑ, so its H has the opposite sign relative to its E. The coefficient r comes from the suite's Fresnel model (fresnel.solve(n₁, n₂, 0).rs, r = (n₁ − n₂)/(n₁ + n₂)); at normal incidence it applies to both Ex and Ey. A perfect electric conductor has r = −1 and no field inside. Medium 1 then contains

Ẽ(z) = E₀ ĵ (eik₁z + r e−ik₁z),   H̃(z) = (E₀/η₁) ẑ × ĵ (eik₁z − r e−ik₁z)

Every number on the page comes from these closed forms. The Poynting-theorem readout differentiates the instantaneous u and Sz numerically (central differences, steps 10⁻⁵λ₀ and 10⁻⁵T) as an independent consistency check.

E₀
peak phasor amplitude |Ẽ| of the incident wave (V/m); Erms = E₀/√2
ĵ
unit Jones vector (Ex, Ey) of the incident wave
η, η₀
wave impedance η₀/n; η₀ = μ₀c ≈ 376.73 Ω
k₁, λ₁
k₁ = n₁ω/c, λ₁ = λ₀/n₁ (λ₀ is always the vacuum wavelength)
S, I
Poynting vector (W/m²); I = ⟨Sz⟩ of a travelling wave
uE, uH
electric and magnetic energy densities (J/m³)
r, R, T
field reflection coefficient; R = |r|², T = (n₂/n₁)|1 + r|²
Derivation

Transversality and H. For fields ∝ ei(k·r − ωt), ∇ → ik and ∂/∂t → −iω. Gauss's law in a source-free medium gives k·E = 0 and ∇·B = 0 gives k·H = 0. Faraday's law, ∇×E = −μ₀∂H/∂t, becomes ik×E = iωμ₀H, so H = k×E/(ωμ₀) = (n/(μ₀c)) k̂×E = (1/η) k̂×E. Hence E ⟂ H ⟂ k and |H| = |E|/η.

Direction of S. S = E×(k̂×E)/η = [k̂(E·E) − E(E·k̂)]/η = k̂E²/η, which points along k̂ at every instant and equals (c/n)u because ½εE² = ½μ₀H² when |H| = |E|/η.

Time averages. For a = Re{Ae−iωt}, b = Re{Be−iωt}, ⟨ab⟩ = ½Re(AB*). So ⟨Sz⟩ = E₀²/(2η) = nε₀cE₀²/2 for peak E₀. Using RMS amplitudes the ½ disappears: I = Erms²/η.

Standing wave. For r = −1, Ẽ = 2iE₀ĵ sin k₁z and H̃ = (2E₀/η₁) ẑ×ĵ cos k₁z. E and H are 90° out of phase in time and λ₁/4 apart in space, so ⟨Sz⟩ = ½Re(Ẽ×H̃*) = 0 while Sz(t) ∝ sin 2k₁z sin 2ωt sloshes energy back and forth between neighbouring electric and magnetic antinodes. For general real r the cross terms in ½Re(Ẽ×H̃*) are purely imaginary, leaving ⟨Sz⟩ = (1 − |r|²)I₀ at every z.

Poynting's theorem. From Maxwell's curl equations, ∂u/∂t + ∇·S = −J·E = 0 without currents. In one dimension ∂u/∂t + ∂Sz/∂z = 0, which is what the readout checks.

Exercise 1: how steady is the energy flow?

  1. For a linearly polarized wave in vacuum, sketch Sz(t) at a fixed point. What are its maximum, minimum and frequency? What changes for circular polarization at the same E₀?
  2. Load Linear, vacuum and watch the lower-right plot. Then switch the state to circular.
  3. Read the peak Sz/I₀ and count its oscillations in two periods. Compare the I₀ readout with n₁ε₀cE₀²/2 computed by hand.
  4. Explain the result using S = (c/n)u and u ∝ E².
Show answer

Linear: Sz = 2I₀cos²(kz − ωt), so it runs from 0 to 2I₀ at frequency 2ω (four peaks in two periods). Circular: |E| = E₀/√2 at every instant, so Sz = I₀ exactly and uE = uH = ū₀/2 are constant. The average is the same, I₀ = 13.3 W/m² for E₀ = 100 V/m in vacuum.

Exercise 2: energy sloshing in front of a mirror

  1. In front of a perfect conductor, where are the nodes of E and of H? What is the net time-averaged flux?
  2. Load PEC mirror. Drag the probe between zp = −0.25λ₁ and −0.125λ₁ and play the animation.
  3. Measure the E-node spacing on the spatial plot (in nm) and read ⟨Sz⟩. At zp = −λ₁/8, find the times where uE and uH peak.
  4. Where does the energy go when uE falls to zero at an electric antinode?
Show answer

E nodes sit at z = 0, −λ₁/2, −λ₁, … (spacing 316.5 nm at λ₀ = 633 nm in vacuum); H nodes are shifted by λ₁/4. ⟨Sz⟩ = 0 to round-off. At z = −λ₁/8, uE and uH peak T/4 apart and their sum is constant there. Energy flows (Sz ≠ 0) from electric antinodes to the magnetic antinodes λ₁/4 away and back, twice per period, so the energy is stored locally with no net transport.

Exercise 3 (limiting cases): from no reflection to a perfect mirror

  1. Set a second dielectric with n₁ = 1. What happens as n₂ → n₁? As n₂ grows, which pattern should the fields in medium 1 approach? Predict the standing-wave ratio for n₂ = 1.5.
  2. Load Air → glass, then sweep n₂ from 1.00 to 4.00. Compare with PEC mirror.
  3. Read R, T, R + T and the standing-wave ratio; check the ratio against the max/min of the dashed envelope.
  4. Why is ⟨Sz⟩ in medium 1 independent of z even though the fields are not?
Show answer

n₂ = n₁ gives r = 0: one travelling wave and no boundary. For n₂ = 1.5, r = −0.2, R = 0.04, T = 0.96 and SWR = (1 + |r|)/(1 − |r|) = 1.5. As n₂ → ∞, r → −1 and the pattern approaches the PEC standing wave (already R = 0.36 at n₂ = 4). ⟨Sz⟩ = (1 − R)I₀ everywhere in medium 1 because the forward–backward interference terms carry no average flux; energy conservation then fixes ⟨Sz⟩ = T·I₀ in medium 2.

Worked example: sunlight at 1 kW/m²

Treat full sunlight at normal incidence as a plane wave with I = 1.00 kW/m² in air (n = 1). The peak field is E₀ = √(2I/(ε₀c)) = √(2000 / 2.654×10⁻³) ≈ 868 V/m, so Erms ≈ 614 V/m and H₀ = E₀/η₀ ≈ 2.30 A/m. The mean energy density is ⟨u⟩ = I/c ≈ 3.34 µJ/m³, split equally between electric and magnetic parts. On a perfect mirror the standing wave has zero net flux, and the reflected momentum gives radiation pressure 2I/c ≈ 6.7 µPa. If the same 1 kW/m² travels inside glass (n = 1.5), E₀ = √(2I/(nε₀c)) ≈ 709 V/m: a denser medium needs a smaller field for the same flux, because η is smaller. Load Sunlight 1 kW/m² (E₀ = 868 V/m) and confirm that I₀ reads ≈ 1.00 kW/m². (Real sunlight is broadband and partially coherent; this example treats only the power balance.)

When the model fails

  • Absorbing media and real metals: n and η become complex, E and H are no longer in phase, and ⟨S⟩ decays with depth. Optical metals are not perfect conductors: silver reflects about 95–98 % in the visible and the field penetrates a skin depth of tens of nanometres.
  • Magnetic or anisotropic media: μ ≠ μ₀ changes η = √(μ/ε). In crystals S is generally not parallel to k (walk-off).
  • Dispersion: with n(ω) the energy moves at the group velocity, not c/n, and u needs dispersive corrections. The model uses a single frequency.
  • Finite beams and oblique incidence: real beams have longitudinal field components and transverse energy flow near their edges. This page models normal incidence only; use the Fresnel tool for angles.
  • Very high intensity or very low photon flux: nonlinear response invalidates the linear medium, and at a few photons the classical energy density describes only detection probabilities.

References

  • D. J. Griffiths, Introduction to Electrodynamics, 5th ed., Ch. 8–9 (Poynting's theorem, plane waves, reflection at normal incidence).
  • J. D. Jackson, Classical Electrodynamics, 3rd ed., §6.7–6.8 and §7.1–7.3.
  • E. Hecht, Optics, 5th ed., §3.2–3.3 (energy and momentum of light).
  • M. Born and E. Wolf, Principles of Optics, 7th ed., §1.1–1.5.
  • B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed., Ch. 5.