Producing & Propagating EM Waves

A charge at rest sits inside its own Coulomb field: energy is stored, none leaves. Set it moving at constant velocity and the whole pattern simply travels with it, still radiating nothing. Now jerk it. News of the new motion spreads outward at c, so far away the field lines still aim at where the charge was headed while close in they aim at where it actually is, and something has to join the two. That joint is a transverse kink in the field, and it falls off as 1/r rather than 1/r². The exponent is the whole story: energy flow goes as field squared times area, (1/r)² × 4πr², which is independent of r, so the kink never comes back. The Larmor formula P = q²a²/(6πε₀c³) prices it, and it is worth noticing what sits in the formula and what does not — the acceleration, not the velocity and not the position. An electron accelerating at 1.0 × 10¹⁴ m s⁻² radiates 5.7 × 10⁻²⁶ W. Tiny alone, which is why transmitting is an engineering problem about making many charges accelerate together.

Nothing accelerates charge as conveniently as an alternating current. Drive charge along a rod as x(t) = x₀ sin ωt and its acceleration amplitude is ω²x₀, so the radiated power, which goes as a², scales as ω⁴. Multiply the frequency by ten and the same charge moving through the same distance radiates ten thousand times as much. That single exponent is why a mains-frequency circuit leaks essentially nothing while a 100 MHz oscillator is a broadcast hazard. Geometry then sets the length. Current has to vanish at an open end, so the standing wave on a centre-fed pair of rods fits half a wavelength across the whole antenna: at 98.5 MHz, λ = c/f = 3.04 m and the dipole is 1.52 m. A car's roof whip is the quarter-wave half of that, about 0.76 m, using the metal body beneath it in place of the missing arm — either way the aerial is cut from the wavelength of the station it listens to. The pattern is not uniform either. A dipole radiates most strongly broadside and not at all along its own axis, with intensity going as sin²θ, because a charge accelerating straight towards you shows you no transverse kink.

Take a flat wavefront sweeping along +x at speed v, leaving uniform E along +y and B along +z behind it and nothing ahead. Lay a rectangle of height a in the xy-plane, straddling the front. In a time dt the front advances v dt, so the magnetic flux through the rectangle grows by B a v dt. Faraday's law equates that rate to the circulation of E around the rectangle, which is E a because only the trailing side sits in the field: E a = B a v, so E = vB. Now turn the rectangle into the xz-plane. No conduction current threads it, but the electric flux grows by E a v dt, and the Ampere-Maxwell law gives B a = ε₀μ₀ E a v, so B = ε₀μ₀ v E. Substitute the second result into the first: E = v(ε₀μ₀vE), which cancels to ε₀μ₀v² = 1. The speed is not chosen, it is forced: v = 1/√(8.854 × 10⁻¹² × 1.2566 × 10⁻⁶) = 2.998 × 10⁸ m s⁻¹. Nothing in that argument mentions the antenna. Once the pair exists, it keeps itself alive.

Three more features fall out of the same laws, and the ratio E = cB comes free with them. First, no longitudinal component: for a plane wave whose fields depend only on x and t, wrap a closed box around a slab of empty space. Gauss's law makes the flux of E through it zero, which forces Eₓ to take the same value on both faces — a constant, so a static background rather than part of the wave, and the magnetic Gauss law does the same to Bₓ. Second, E ⟂ B: the two rectangles of the previous section only closed because the fields sat in perpendicular planes. Third, in phase: both relations carry the same sinusoid, so E and B reach their peaks at the same place at the same instant and cross zero together. The sense follows from the signs, with E × B pointing along the travel direction, since +y × +z gives +x. Last, that ratio in numbers. E₀ = 100 V m⁻¹ goes with B₀ = E₀/c = 3.34 × 10⁻⁷ T. B looks negligible only because of the SI units: for a charge crossing the wave at speed u, the magnetic force quB is smaller than the electric force qE by the factor u/c.

Write the wave as E_y(x, t) = E₀ cos(kx − ωt). A point of fixed phase needs kx − ωt to stay constant, so x = (ω/k)t + constant and the whole pattern slides along +x at the phase speed ω/k. Setting that speed to c is what makes λf = c, since ω/k = 2πf ÷ (2π/λ) = λf. Two independent periodicities live in the one expression: freeze time and the snapshot repeats every λ in space, stand at one point and the field repeats every T = 1/f in time. For a satellite television downlink at 12.0 GHz, λ = c/f = 2.50 cm, k = 2π/λ = 251 rad m⁻¹, T = 83.3 ps and ω = 2πf = 7.54 × 10¹⁰ rad s⁻¹. Two edits are worth knowing. Adding a constant φ inside the bracket only slides the snapshot sideways and changes no physics. Flipping the sign to cos(kx + ωt) sends the wave along −x instead, and then B must flip too, because E × B has to keep pointing the way the wave goes.

Replace ε₀ and μ₀ by the material's ε and μ and the derivation runs unchanged: v = 1/√(εμ) = c/√(K Kₘ), with K = ε/ε₀ and Kₘ = μ/μ₀. Non-magnetic matter has Kₘ ≈ 1, so the refractive index n = c/v is simply √K. What cannot change at a boundary is the frequency: the fields on the two sides stay joined cycle for cycle, and a surface delivering fewer crests than it received would have to store the difference forever. So f is whatever the source made it, and λ = v/f = λ₀/n absorbs the whole slowdown. A sodium lamp's 589 nm line entering water with n = 1.333 keeps its 5.090 × 10¹⁴ Hz, slows to 2.249 × 10⁸ m s⁻¹, and shortens to 441.9 nm. The field ratio shifts too: inside, E = vB, so a given E rides with a larger B. One warning about K — it must be the value at the wave's own frequency. The handbook K ≈ 80 for water belongs to kilohertz, where whole molecules have time to turn and face the field, and it would predict n = 8.94; at 10¹⁴ Hz only the electron clouds can follow, the working K is n² = 1.777, and the index is 1.333. That frequency dependence is dispersion, and it is why a prism spreads white light.