“A changing E field makes a B field, and a changing B field makes an E field. So each field should be strongest when the other is changing fastest, which is a quarter cycle (90°) later. Why do the textbook pictures show E and B peaking at the same place? And why are they at right angles?”
Short answer. Maxwell's equations do not say “a changing E makes B a little later”. They say something local and instantaneous: how fast B is changing at a point equals how steeply E slopes in space at that same point, and the same with E and B swapped. Both the slope and the rate are a quarter cycle away from the wave itself. When the wave is moving, those two quarter shifts are the same shift, so they cancel, and E and B peak together. The 90° you expected is real, but it only appears where energy is not travelling: in standing waves and right next to an antenna.
The page proves each piece with something you can drag. Start with section 1, or jump to section 3 for the heart of the argument.
Short answer. Faraday and Ampère–Maxwell are first order in both space and time. For E = x̂ Ex(z,t), B = ŷ By(z,t) in vacuum:
For any travelling solution f(z − ct) we have ∂t = −c∂z, so Faraday becomes ∂z(Ex − cBy) = 0: Ex = cBy pointwise, in phase. With phasors, ikÊ = iωB̂. Each derivative contributes a factor i (a 90° shift), and there is exactly one on each side. The “cause then effect” reading, which would put a single i between E and B, belongs to lumped systems (LC circuits), standing waves and reactive near fields, where ⟨S⟩ = 0.
Red arrows are E, blue arrows are B (drawn as cB so both have the same size). The wave travels to the right. Drag the picture to turn it. The grey square is a probe: the small plot underneath shows E, B and the energy flow at that one point over time.
Try the buttons. A travelling wave has E and B peaking together and the energy flow never goes negative. Put a mirror at the right end and the wave bounces back. The two waves add into a standing wave, and now E and B are a quarter cycle apart, and energy just sloshes back and forth. Force a 90° lag shows what your intuition predicts for a travelling wave, and why nature can't do it: Maxwell's equations fail, and the wave carries no energy on average.
The picture is E = x̂ [cos(kz−ωt) − Γ cos(kz+ωt)], cB = ŷ [cos(kz−ωt+φ) + Γ cos(kz+ωt−φ)], a forward wave plus a reflection of relative size Γ, with B forced to peak φ later than nature allows. With φ = 0 this solves Maxwell exactly for any Γ. At Γ = 1, E = 2 sin kz sin ωt and cB = 2 cos kz cos ωt: 90° apart in time and λ/4 apart in space. The time-averaged Poynting flux is
so the net power carried is (1−Γ²)/2, the part of the field that is travelling. A nonzero φ is not a solution. The residuals below are RMS values of each equation over one λ × one T, normalised to the size of one wave's own curl. Notice also that with φ ≠ 0 and Γ ≠ 0, ⟨S⟩ depends on z: energy would pile up or vanish in empty space.
E cB energy flow S = E × B / μ0 (½ scale)
Three of Maxwell's four equations settle the directions. You can't argue with them, but you can try to break them: move the sliders and watch which law turns red.
The order matters too: E × B points the way the wave travels. Flip B and you get a wave going the other way.
Take a plane wave E = E0ei(k·r−ωt), B = B0ei(k·r−ωt), so ∇ → ik and ∂t → −iω. Maxwell's vacuum equations become algebra:
The first two make both fields transverse. The third makes B0 ⟂ E0 and fixes (E, B, k) as a right-handed set. Feeding the third into the fourth gives k × (k × E0) = −k²E0 = −(ω²/c²)E0, so ω = ck and |E0| = c|B0|. Note that the wave equation □E = 0 alone allows longitudinal solutions. The divergence constraints are what remove them. Note also there is no factor of i in k × E0 = ωB0: that is the phase statement from section 3, in one line.
The residuals below are computed numerically from the drawn field with central differences, so they check the picture, not a formula. For α = angle from E to B and longitudinal parts εE, εB: Faraday residual = √(4 sin²((α−90°)/2) + εB²), Ampère = √(4 sin²((α−90°)/2) + εE²).
| law | vacuum form | residual | |
|---|---|---|---|
| Gauss (E) | ∇·E = 0 | | |
| Gauss (B) | ∇·B = 0 | | |
| Faraday | ∇×E = −∂B/∂t | | |
| Ampère–Maxwell | ∇×B = (1/c²)∂E/∂t | | |
This is a snapshot of the wave along its path. Drag the grey probe line along the plot. At the probe you see two things: the slope of E (the red tangent line) and how fast B is changing right now (the blue arrow; the dashed blue curve is B a moment later).
Faraday's law says those two are always equal and opposite, at every point, at every instant. Now follow the chain:
The “cause makes effect a quarter cycle later” idea compares a rate in time with a value. Faraday compares a rate in time with a slope in space. For a moving wave, slope and rate are the same thing seen two ways, and each is a quarter cycle from the value. Press Standing and the chain breaks at step 2: a standing wave doesn't slide, so the steep parts no longer mark the fast-changing parts, and E and B end up 90° apart.
Faraday is ∂zEx = −∂tBy, local in space and time. For Ex = f(z−ct), ∂zEx = f′, hence ∂tBy = −f′, and since B must also be a function of z−ct, By = f/c (plus a static field, zero for a wave). Equivalently, factor the 1-D wave operator:
(add and subtract the two curl equations). So Ex ± cBy are the Riemann invariants: E + cB moves right unchanged and E − cB moves left. A purely right-moving wave has E − cB ≡ 0, which is the in-phase statement; a standing wave has both invariants equal in size, E = ½(R + L), cB = ½(R − L), and their sum and difference are in quadrature.
The intuition “B ∝ ∂E/∂t” is what you get by integrating Ampère over a fixed loop in a quasistatic region, where the spatial structure is set by geometry rather than by propagation (displacement current in a capacitor gap: B ∝ dE/dt, 90° out of phase). That is the near-field limit of section 5.
E now cB now cB a moment later
This is a real simulation of Maxwell's equations on a line (the Yee algorithm used by engineers for antennas and chips). Press Pluck E only: it creates a bump of E with no B at all. Nothing “waits” for B. In the very first step the slope of the E bump starts making B, the curl of that B starts reshaping E, and the bump splits into two half-height bumps, one running each way. Each half already carries its own B, in step with its E. On the right-moving half B points one way, on the left-moving half the other way, because E × B has to point along the travel.
Click anywhere on the line to pluck there. Then try the sine source with the right end open (travelling wave, probe reads 0°) and with a mirror (standing wave, probe reads ±90°).
Initial data E(z,0) = g(z), B(z,0) = 0. Using the invariants of section 3, R = E + cB = g moves right and L = E − cB = g moves left, so
(d'Alembert). The split is immediate. ∂tB = −∂zE gives B a nonzero rate at t = 0⁺, not a delayed value. The scheme is Yee's staggered grid (E at integer, cB at half-integer cells and steps) at Courant number cΔt/Δz = 1, where the 1-D update is exact on the grid (no numerical dispersion), with exact one-way absorbing ends and a total-field/scattered-field source that radiates only to the right. The probe co-locates E and cB at (zp, tn+½), and the phase is a lock-in over the last two periods. With the mirror (a perfect conductor, E = 0), the steady state is E ∝ sin k(L−z) · sin ωt, cB ∝ cos k(L−z) · cos ωt, and the probe reads ±90° at every position; the sign flips each λ/4.
E cB source probe
Your intuition is the right picture for a capacitor and a coil: in an LC circuit the charge (which makes E) is biggest when the current (which makes B) is zero, a quarter cycle apart. Energy is stored, handed back and forth, and not sent anywhere.
A short antenna is exactly that close in. The E field follows the charge piled at its tips and the B field follows the current, so near the antenna they are 90° apart. Move away and the part that escapes as a wave takes over, and the phase difference shrinks to zero. The change happens around r ≈ λ/2π, about a sixth of a wavelength. Click on the map or use the slider to move the probe ring, and watch the phase and the E/B ratio.
Oscillating point dipole p ẑ e−iωt, with x = kr (Gaussian-style, c = 1, fields ÷ pk³):
For x ≪ 1, E ∝ p/r³ tracks the charge and B ∝ ṗ/r² tracks the current I = ṗ: quadrature, wave impedance |E/cB| ≈ 1/kr ≫ 1 (capacitive), and the 1/r³·1/r² product carries no net power. At x = 1 the ratio is (1+i)/2: exactly 45° and 1/√2. Radiated power ∮⟨S⟩·dA is the same through every sphere; only the in-phase 1/r · 1/r parts contribute to it, so the quadrature near field is purely reactive energy sloshing in and out each half cycle.
Eθ cBφ (traces each scaled to their own peak; map scaled by r to show the far field)
Light is not special here. Many waves are carried by a pair of quantities: a push (like voltage or pressure) and a flow (like current or air velocity), and the power is push × flow. They are linked the same way E and B are: how fast one changes in time is set by how steeply the other slopes in space. So the same rule holds. When the wave travels, push and flow are in phase and power flows. When it bounces back and forms a standing wave, they are 90° apart and energy only sloshes. Pick a system below. Red is always the push and blue the flow, like E and B above.
Each system is a first-order hyperbolic pair with the same structure as the two curl equations:
A single travelling wave has P = ±Z F with Z real, so there is no phase between them. A termination with impedance ZL reflects Γ = (ZL − Z)/(ZL + Z), and ⟨P F⟩ = (1 − |Γ|²)/2 in units of the incident power, the same at every z. The page draws P̂ = eikz + Γe−ikz, ZF̂ = eikz − Γe−ikz, exactly the §1 model with ΓE = −Γ.
A coax cable or ladder line carries a voltage between its conductors and a current along them. They are really E and B: the voltage is the E field across the gap, and the current makes the B field around the wire. Into a matched load (a resistor equal to the line's characteristic impedance, such as 50 Ω), all the power is absorbed and nothing comes back. The wave only travels, and V and I are in phase all along the line. Leave the end open or short it and all the power bounces back. V and I form standing waves a quarter wavelength apart, 90° apart in time, and the SWR is infinite. A mismatched resistor reflects part of the power, and the phase between V and I then changes along the line: zero at the voltage peaks and dips, largest in between. Add reactance X and the load itself puts V and I out of phase. This is the pattern an SWR meter and an antenna tuner work with.
Telegrapher's equations for a lossless line with inductance L′ and capacitance C′ per metre:
With d the distance back from the load, V(d) = V⁺(ejβd + Γe−jβd), Z0I(d) = V⁺(ejβd − Γe−jβd), SWR = (1+|Γ|)/(1−|Γ|), and the input impedance is Z(d) = Z0(ZL + jZ0 tan βd)/(Z0 + jZL tan βd). The V–I phase at d is arg Z(d). Z is real, so V and I are in phase, at every voltage maximum (Z = Z0·SWR) and minimum (Z = Z0/SWR). A stub or tuner cancels the reactive part at one point and leaves the line between tuner and antenna still standing-wave. The load sliders use the engineering ejωt convention, so X > 0 is inductive and current lags voltage by atan(X/R) at the load.
Sound is a pressure wave. The push is the pressure and the flow is the air's back-and-forth velocity. Watch the dots, which are bits of air. With an absorber at the far end, the sound only travels. The air moves forward exactly where it is squeezed and backward where it is thinned out, so pressure and velocity are in phase and energy flows away from the speaker. Cap the end with a closed wall and the air at the wall can't move. Velocity has a node there and pressure a peak. The reflection builds a standing wave, and pressure and velocity are 90° apart. An open end does the opposite: pressure there is held at room pressure, so pressure has the node. This is how organ pipes, flutes and clarinets pick their notes. The same 90° appears right next to a small loudspeaker, which mostly shoves air back and forth instead of sending it away. That is why small speakers make so little bass.
Linearised momentum and continuity for air at rest (density ρ0, sound speed c):
A rigid cap forces u = 0, so Γp = +1, the acoustic open circuit. An ideal open end forces p ≈ 0, so Γp = −1, the acoustic short. A real open end radiates a little and acts about 0.6a longer, where a is the pipe radius. The dots are drawn at displacement ξ with u = ∂tξ and p = −ρ0c² ∂zξ, so bunching is pressure. Near a pulsating sphere of radius a, the radiation impedance is Z = ρ0c · jka/(1 + jka). For ka ≪ 1 this is almost purely reactive, a mass load, which is the acoustic twin of the dipole near field in §5.
A tide is a very long water wave. It repeats every 12.4 hours and is hundreds to thousands of kilometres long. The push is the height of the water and the flow is the current. Along an open coast, where the tide travels past and never comes back, the flood current (toward the land) is strongest at high water and the ebb is strongest at low water. That is in phase. In a closed bay the tide reflects off the head of the bay and makes a standing wave. Then the water is slack at high and at low tide, and the current is strongest in between, about 3 hours before high water. That is 90°. Sailors use both rules: “slack at high water” is a standing-tide rule. The Bay of Fundy is nearly a quarter wavelength long, so it resonates like an organ pipe, which is why its tides are the biggest on Earth. The tide gauge marks the harbour. Real bays are shorter than a quarter wavelength; the picture shows a longer channel so you can see the whole pattern, and the gauge starts in the last quarter wave before the bay head, where a real harbour would be. Move it further out and the standing pattern repeats, with the sign of the lead flipping every quarter wavelength.
Long-wave (shallow-water) equations for depth h, surface height η, depth-averaged current u:
In the deep ocean (h ≈ 4 km), c ≈ 200 m/s and the M2 tide has λ ≈ 8900 km. A closed channel of length L has η ∝ cos k(L−z) cos ωt and u ∝ sin k(L−z) sin ωt: current leads height by a quarter period seaward of the head. The channel resonates when L ≈ λ/4 = T√(gh)/4. The Fundy–Gulf of Maine system has a natural period near 13 h, close to the 12.42 h forcing. Real tides also feel Earth's rotation (Kelvin waves, amphidromic points), but along a narrow channel the height–current phase follows this 1-D picture. The “partly reflecting” slider sets a real Γ between the two limits.
| situation | E vs B in time | energy | see |
|---|---|---|---|
| travelling wave in free space | in phase, |E| = c|B| | flows forward, S ≥ 0 always | §1, §3, §4 |
| standing wave (wave + mirror) | 90°, and λ/4 apart in space | sloshes, ⟨S⟩ = 0 | §1, §4 |
| part reflected (0 < Γ < 1) | between, depends on position | net (1−Γ²) forward | §1 |
| near an antenna (r ≪ λ/2π) | 90°, E ≫ cB | stored, reactive | §5 |
| far from an antenna (r ≫ λ/2π) | in phase, E = cB | radiated | §5 |
| LC circuit | 90° (charge vs current) | stored | §5 |
| coax into a matched load | in phase, V = Z₀I | delivered to the load | §6 |
| coax, open or shorted end | 90°, SWR ∞ | all reflected | §6 |
| sound into an absorber / closed pipe | in phase / 90° (p vs u) | radiated / stored | §6 |
| tide on an open coast / in a closed bay | current max at high water / slack at high water | travels / sloshes | §6 |
The rule of thumb: fields that carry energy are in phase; fields that store energy are 90° apart. A travelling wave is the first kind. The “changing E makes B” story is not wrong, it is just about a slope in space, not a value, and for a moving wave that slope is already in step with the other field.
One line: in k × E0 = ωB0 the curl and the time derivative each bring one factor of i, so no net phase survives. A relative phase between E and B is the signature of reactive power, Im(E × H*), which requires counter-propagating or evanescent components (standing waves, near fields, waveguides below cutoff). For a single plane wave in a lossless medium the impedance η = √(μ/ε) is real; in a lossy medium η is complex and B lags E by arg η, up to 45° in a good conductor.