University Physics II · Maxwell's Equations and Electromagnetic Waves · 12.7

Electromagnetic Energy and Momentum

An electromagnetic wave carries energy in both of its fields and moves it at c. Because that energy arrives with momentum, a beam presses on whatever absorbs or reflects it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A wave carries no charge, yet the fields say exactly how much energy is present: ½ε₀E² in the electric field, B²/(2μ₀) in the magnetic. In a plane vacuum wave E = cB makes those two equal at every instant, so u = ε₀E². The transport is the Poynting vector S⃗ = (1/μ₀) E⃗ × B⃗ — power per unit area, pointing along the propagation direction because E⃗, B⃗ and the direction of travel form a right-handed triad.

Its magnitude is uc, which is the energy in a slab sweeping past at speed c. Time-average a sinusoidal wave and ⟨cos²⟩ = ½ gives the intensity I = ½ε₀cE₀². Momentum comes with the energy at the rate p = U/c, so a beam delivers momentum at I/c per unit area: a pressure of I/c on a black surface and 2I/c on a mirror, because the mirror reverses the momentum instead of merely stopping it.

Simple definition: The Poynting vector S⃗ = (1/μ₀) E⃗ × B⃗ gives the power per unit area an electromagnetic field transports and the direction it flows. Its cycle average is the intensity, and I/c is the momentum delivered per second per unit area.
Example: Sunlight above the atmosphere has I = 1.36 × 10³ W m⁻², so E₀ = √(2I/(ε₀c)) = 1.01 × 10³ V m⁻¹ and B₀ = 3.38 μT, and the pressure on a black surface facing it is I/c = 4.5 μPa.

Energy density of the fieldsu = ½ε₀E² + B²/(2μ₀) = ε₀E² = B²/μ₀

The electric and magnetic halves are equal at every instant, not only on average.

J m⁻³ · the last two forms need E = cB, so a plane wave in vacuum

Poynting vectorS⃗ = (1/μ₀) E⃗ × B⃗ · S = EB/μ₀ = ε₀cE² = uc

Power per unit area, pointing the way the energy flows — for a wave, the way it travels.

W m⁻² · E⃗, B⃗ and S⃗ form a right-handed triad

Intensity of a sinusoidal waveI = ⟨S⟩ = ½ε₀cE₀² = E₀B₀/(2μ₀) = cε₀Eᵣₘₛ²

Sunlight at 1.36 × 10³ W m⁻² needs E₀ = 1.01 × 10³ V m⁻¹ and B₀ = 3.38 μT.

W m⁻² · the ½ is ⟨cos²⟩ over one cycle · Eᵣₘₛ = E₀/√2

Momentum carried by the wavep = U/c · momentum density g = u/c = S/c²

Momentum arrives with the energy, along the direction of travel.

g in kg m⁻² s⁻¹ · momentum crosses unit area at the rate S/c

Radiation pressure at normal incidenceP_rad = I/c absorbed · 2I/c reflected · (1 + R)I/c in general

Sunlight gives 4.5 μPa on a black surface and 9.1 μPa on a mirror.

Pa · R is the specular reflectance · I/c is also the mean energy density

Force on an illuminated surfaceF = P_rad A

A 1 km² perfect sail at 1 AU: 2 × 1.36 × 10³ × 10⁶ ÷ (3.00 × 10⁸) = 9.1 N.

N · A is the area projected onto the beam, not the total surface

Both fields hold energy, and in a wave they hold equal shares

The densities carried over from electrostatics and magnetostatics are u_E = ½ε₀E² and u_B = B²/(2μ₀), each in J m⁻³, and both apply point by point to any field, not only inside a capacitor or a solenoid. In a plane wave in vacuum the two are locked together, because E = cB with c = 1/√(ε₀μ₀). Substitute: u_B = B²/(2μ₀) = E²/(2μ₀c²) = ½ε₀E² = u_E. The fields are in phase, so the equality holds at every instant, not merely on average, and the total is u = ε₀E² = B²/μ₀. Sunlight above the atmosphere has E₀ = 1.01 × 10³ V m⁻¹, so the peak density is ε₀E₀² = 9.09 × 10⁻⁶ J m⁻³ and the cycle average is half of it, 4.54 × 10⁻⁶ J m⁻³. That is under 5 μJ in a cubic metre of full sunlight, which is why the energy matters only once you count how fast it moves through.

The Poynting vector gives the rate and the direction

Define S⃗ = (1/μ₀) E⃗ × B⃗, in W m⁻²: the power crossing unit area, pointing along the flow. For a wave travelling along x̂ with E⃗ = E ŷ and B⃗ = B ẑ, the cross product gives S⃗ = (EB/μ₀) x̂, so the energy travels the way the wave does and the triad E⃗, B⃗, propagation is right-handed. Its size follows from the energy density: with B = E/c and c² = 1/(ε₀μ₀), S = EB/μ₀ = E²/(μ₀c) = ε₀cE² = uc, which is what you get by pushing a slab of density u past a plane at speed c. S⃗ is not limited to waves. Take a wire of radius a and length L carrying a steady current i with V across it: at the surface E = V/L runs along the wire and B = μ₀i/(2πa) wraps around it, so S⃗ points radially inward with magnitude Vi/(2πaL). Multiply by the side area 2πaL and the inflow is Vi — the dissipated power enters through the surface of the wire, not along the copper.

Intensity is the cycle average, and that is where the ½ comes from

For E = E₀cos(kx − ωt), S = ε₀cE₀²cos²(kx − ωt), so S oscillates at twice the field frequency and never goes negative. No detector follows 5 × 10¹⁴ Hz, so what is measured is the average, and ⟨cos²⟩ = ½ gives I = ⟨S⟩ = ½ε₀cE₀² = E₀B₀/(2μ₀) = cε₀Eᵣₘₛ², with Eᵣₘₛ = E₀/√2. Run it backwards to size the fields behind a familiar intensity. The Sun radiates 3.85 × 10²⁶ W, and at r = 1.50 × 10¹¹ m the inverse-square spread gives I = 3.85 × 10²⁶ ÷ (4π(1.50 × 10¹¹)²) = 1.36 × 10³ W m⁻². Then E₀ = √(2I/(ε₀c)) = 1.01 × 10³ V m⁻¹ and B₀ = E₀/c = 3.38 μT — an order of magnitude below the Earth's static field, and an electric field far below the 3 × 10⁶ V m⁻¹ that breaks down air. A 5.0 mW pointer with a 1.0 mm beam concentrates more: I = 5.0 × 10⁻³ ÷ 7.85 × 10⁻⁷ = 6.4 × 10³ W m⁻², giving E₀ = 2.2 × 10³ V m⁻¹.

The wave carries momentum as well as energy

Maxwell's equations give a wave that brings momentum p = U/c along its direction with every energy U it delivers, so the momentum density is g = u/c = S/c², in kg m⁻² s⁻¹, and momentum crosses unit area at the rate S/c. The mechanism shows up in the force on a charge sitting in the beam. The electric field drives the charge along E⃗, giving it a velocity component parallel to E⃗; the magnetic force qv⃗ × B⃗ acting on that motion then points along the propagation direction, and it keeps pointing that way through both halves of the cycle, because v⃗ and B⃗ reverse together. Absorbing a wave therefore leaves the absorber with net forward momentum, not only with heat. For sunlight, g = 1.36 × 10³ ÷ (3.00 × 10⁸)² = 1.5 × 10⁻¹⁴ kg m⁻² s⁻¹ — nothing to a wall, but decisive for a micrometre dust grain, whose area-to-mass ratio lets radiation pressure compete with the Sun's gravity.

Radiation pressure: absorb once, reflect twice

Momentum arriving per second per unit area is a pressure. A perfectly absorbing surface facing the beam takes I/c, and since I/c is also the mean energy density, the absorbing pressure equals ⟨u⟩. A perfect mirror at normal incidence returns the wave with reversed momentum, so it takes twice as much, 2I/c. Sunlight at Earth gives 1.36 × 10³ ÷ (3.00 × 10⁸) = 4.5 μPa absorbed and 9.1 μPa reflected, against 1.0 × 10⁵ Pa of atmosphere — smaller by a factor of 2 × 10¹⁰, which is why the effect was not measured until Lebedev, then Nichols and Hull, worked in evacuated chambers around 1900. Scale the area and it stops being negligible. A square-kilometre sail, perfectly reflecting and face-on at 1 AU, feels 2IA/c = 9.1 N. At an areal density of 10 g m⁻² that sail masses 1.0 × 10⁴ kg, so its acceleration is 9.1 × 10⁻⁴ m s⁻² — about 79 m s⁻¹ of speed gained per day, with no propellant spent.

What the ideal-wave assumptions are doing

Every relation above assumes a monochromatic plane wave in vacuum, with E = cB in phase and both fields transverse, meeting a flat target head-on. Loosen them one at a time. Turn the mirror through θ and it both intercepts less power and returns the momentum at an angle, so the pressure on its surface falls to 2I cos²θ/c — a quarter of the head-on value at θ = 60°. Real surfaces are neither black nor perfect mirrors: with specular reflectance R the pressure is (1 + R)I/c, and diffuse scattering redistributes the returned momentum rather than reversing it, so the factor sits below 2. Time-averaging needs many cycles inside the measurement window, which is free at 5 × 10¹⁴ Hz but not for a femtosecond pulse. A finite beam is only locally plane, so I varies across the profile. And inside a dielectric the split of energy and momentum between field and medium is still argued over — Abraham and Minkowski wrote different answers — which is why these results are quoted for vacuum.

Change one variable at a time

Make the relationship visible.

Interactive model

Peak electric field E₀1000 V m⁻¹

Reflectance R of the surface0.0

Raise E₀ and both pressure arrows grow as E₀², while the wave amplitude beside them grows only as E₀, because I = ½ε₀cE₀²; then slide R from 0 to 1 and watch the push arrow stretch to exactly twice the dashed tick — a mirror reverses the momentum instead of merely stopping it.

INTENSITY I = ½ε₀cE₀²1327 W m⁻²

PEAK B FIELD B₀ = E₀/c3.34 µT

MEAN ENERGY DENSITY ⟨u⟩4.43 µJ m⁻³

RADIATION PRESSURE (1+R)I/c4.43 µPa

Live interpretationINTENSITY I = ½ε₀cE₀²: 1327 W m⁻². PEAK B FIELD B₀ = E₀/c: 3.34 µT. MEAN ENERGY DENSITY ⟨u⟩: 4.43 µJ m⁻³. RADIATION PRESSURE (1+R)I/c: 4.43 µPa

Catch the common trap

Explain before calculating.

A plane electromagnetic wave in vacuum has peak electric field E₀ = 1.0 × 10³ V m⁻¹ and strikes a perfectly reflecting flat surface head-on. What is the radiation pressure on that surface? Take ε₀ = 8.85 × 10⁻¹² F m⁻¹ and c = 3.00 × 10⁸ m s⁻¹.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 4.0 mW laser pointer puts its whole beam, 2.0 mm across, onto a matt black card. Find the intensity in the beam, the radiation pressure on the card, and the force the beam exerts. Take c = 3.00 × 10⁸ m s⁻¹.

Beam area: A = πd²/4 = π(2.0 × 10⁻³ m)²/4 = 3.14 × 10⁻⁶ m².
Intensity is beam power per unit area: I = 4.0 × 10⁻³ W ÷ 3.14 × 10⁻⁶ m² = 1.27 × 10³ W m⁻².
Matt black absorbs, so the surface takes the momentum without returning it: P_rad = I/c = 1.27 × 10³ ÷ (3.00 × 10⁸) = 4.24 × 10⁻⁶ Pa.
F = P_rad A = 4.24 × 10⁻⁶ Pa × 3.14 × 10⁻⁶ m² = 1.3 × 10⁻¹¹ N — which is just (beam power)/c = 4.0 × 10⁻³ ÷ 3.00 × 10⁸, since the area cancels. Focusing the beam raises the pressure but never the force.

AnswerI = 1.3 × 10³ W m⁻², P_rad = 4.2 μPa, F = 1.3 × 10⁻¹¹ N

MediumA plane wave in vacuum has intensity I = 5.0 × 10² W m⁻². Find E₀ and B₀, show that the two fields hold equal energy, and find the pressure on a perfect mirror at normal incidence. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹, μ₀ = 4π × 10⁻⁷ H m⁻¹, c = 3.00 × 10⁸ m s⁻¹.

I = ½ε₀cE₀² is already a cycle average, so invert it directly: E₀ = √(2I/(ε₀c)) = √(2 × 5.0 × 10² ÷ (8.85 × 10⁻¹² × 3.00 × 10⁸)) = √(3.77 × 10⁵) = 6.14 × 10² V m⁻¹.
B₀ = E₀/c = 613.7 ÷ (3.00 × 10⁸) = 2.05 × 10⁻⁶ T — microtesla-scale, far below the Earth's static field.
Peak electric share: ½ε₀E₀² = 0.5 × 8.85 × 10⁻¹² × 3.77 × 10⁵ = 1.67 × 10⁻⁶ J m⁻³. Peak magnetic share: B₀²/(2μ₀) = (2.05 × 10⁻⁶)² ÷ (2 × 1.257 × 10⁻⁶) = 1.67 × 10⁻⁶ J m⁻³. Equal, so B is not the junior partner.
A perfect mirror reverses the momentum rather than stopping it, so it takes twice the absorbing value: P_rad = 2I/c = 2 × 5.0 × 10² ÷ (3.00 × 10⁸) = 3.3 × 10⁻⁶ Pa.

AnswerE₀ = 6.1 × 10² V m⁻¹, B₀ = 2.05 × 10⁻⁶ T, each field holds 1.67 × 10⁻⁶ J m⁻³ at the crest, and P_rad = 3.3 μPa

HardA solar sail of area 2.5 × 10⁴ m² and specular reflectance R = 0.88 faces the Sun square-on at 1 AU, where I = 1.36 × 10³ W m⁻². Sail plus payload mass 250 kg. Find the radiation pressure, the force, the acceleration, and the speed gained in 30 days.

A real surface returns only the fraction R of the arriving momentum, so P_rad = (1 + R)I/c = 1.88 × 1.36 × 10³ ÷ (3.00 × 10⁸) = 8.52 × 10⁻⁶ Pa — between the absorbing I/c and the perfect-mirror 2I/c.
F = P_rad A = 8.52 × 10⁻⁶ Pa × 2.5 × 10⁴ m² = 0.213 N, about the weight of a 22 g mass on Earth.
a = F/m = 0.213 N ÷ 250 kg = 8.52 × 10⁻⁴ m s⁻². The areal density is 250 ÷ 2.5 × 10⁴ = 10 g m⁻², which is what makes the number this large.
Δv = at with t = 30 × 86 400 s = 2.59 × 10⁶ s: Δv = 8.52 × 10⁻⁴ × 2.59 × 10⁶ = 2.2 × 10³ m s⁻¹, holding I fixed — honest only while the sail has not receded far from 1 AU.

AnswerP_rad = 8.5 μPa, F = 0.21 N, a = 8.5 × 10⁻⁴ m s⁻², Δv ≈ 2.2 km s⁻¹ in 30 days