University Physics IV · Photons and Matter Waves · 4.2
Radiation Pressure & Photon Flux
Light pushes, and the push is the intensity divided by c — microscopic on a bench, decisive on a solar sail, enormous for a single atom. Learn the factor for an absorber, a mirror and a tilted mirror, then cut the same beam into photons and find out where the two accounts stop agreeing.
Build the model
Connect the measurement to the mechanism.
An electromagnetic wave carries momentum as well as energy, in the fixed ratio 1/c, so a beam of intensity I lands a pressure I/c on a black surface, 2I/c on a mirror that sends it back, and (1 + r)I cos²θ/c on a tilted surface of reflectance r — two cosines, one for the beam a tilted face intercepts and one for the component of momentum a reflection actually reverses. Now cut the same beam into photons of energy hf and momentum hf/c: the arrival rate is R = P/hf, the force is R(hf/c) = P/c, and Planck's constant cancels. The photon picture reproduces the classical mean force exactly, which is why radiation pressure was weighed on a torsion balance in 1901 and settled nothing whatever about quanta.
What the photon picture adds is the fluctuation. A classical wave deposits exactly Iτ of energy in every gate; a coherent beam of photons deposits ⟨N⟩ ± √⟨N⟩ of them, because the arrivals are Poissonian and independent. So the two accounts agree in the mean and only in the mean — and it is in the variance, not in the pressure, that a laser, a lamp and a single atom finally look different.
- Simple definition
- Radiation pressure is the momentum a beam delivers per second per unit area of surface — I/c when the surface absorbs the light head-on, 2I/c when it reflects it, and (1 + r)I cos²θ/c in general.
- Example
- Sunlight above the atmosphere has I = 1361 W m⁻², so a black surface facing the Sun feels 1361 ÷ (2.998 × 10⁸) = 4.54 μPa and a mirror 9.08 μPa — some 4.5 × 10⁻¹¹ of atmospheric pressure, carried by 3.8 × 10²¹ photons per square metre per second.
Total absorption at normal incidence, the floor of every estimate: sunlight's 1361 W m⁻² becomes 4.54 μPa.
I is the time-averaged intensity in W m⁻², c = 2.998 × 10⁸ m s⁻¹, and prad comes out in Pa.
One cosine because a tilted face intercepts less beam, one because only the normal component of momentum is reversed. At 60° the pressure is a quarter of face-on.
r is the power reflectance (0 black, 1 mirror); θ is measured from the surface normal, and the force lies along that normal.
The absorbed fraction keeps the beam's tangential momentum, so a grey sail is dragged sideways as well as pushed — a torque a real sail must trim out.
Same units as p⊥, but acting along the surface. Zero for a perfect mirror, largest at θ = 45°.
1.0 mW at 633 nm is 3.19 × 10¹⁵ photons a second, each carrying 1.05 × 10⁻²⁷ kg m s⁻¹.
hc = 1.986 × 10⁻²⁵ J m = 1240 eV nm; P in W, λ in m, R in s⁻¹.
Why a torsion balance in 1901 could confirm Maxwell and say nothing about quanta: the mean force carries no h.
Normal incidence; P is the incident power in W and F is in N. Planck's constant cancels out entirely.
The shot-noise floor. 1% precision needs 10⁴ counts and 0.1% needs 10⁶, whatever the detector cost.
η is the detection quantum efficiency and τ the counting gate in s. Poisson statistics hold for coherent light.
Where I/c comes from
Maxwell's equations give a plane wave a momentum density g = S/c², where S is the Poynting vector. Put a perfectly absorbing surface of area A in the beam and ask what arrives in a time Δt: the field in a column of length cΔt and cross-section A, carrying momentum (I/c²)(A c Δt) = I A Δt/c. Divide by Δt for the force and by A for the pressure, and radiation pressure is p = I/c — an intensity in W m⁻² divided by a number of order 10⁸, which is why the answer is always microscopic. Sunlight above the atmosphere delivers I = 1361 W m⁻², so a black surface facing the Sun feels 4.54 μPa, about 4.5 × 10⁻¹¹ of atmospheric pressure. Nothing quantum has been used here. This is simply the ratio of momentum to energy carried by a wave, fixed at 1/c.
Absorber, mirror, and the two cosines
Three cases, and the differences are pure geometry. A black surface at normal incidence absorbs the momentum: p = I/c. A perfect mirror reverses it, so twice as much is transferred: p = 2I/c — the mirror feels more force while absorbing no energy at all, which is the first sign that force tracks momentum, not energy. Tilt the surface by θ from its normal and two separate cosines appear. First, a surface of area A now intercepts only a cross-section A cosθ of the beam, so the power landing per unit of surface is I cosθ. Second, specular reflection reverses only the normal component of each photon's momentum and leaves the tangential component untouched, so the transfer per photon is 2(hf/c)cosθ, directed along the surface normal rather than along the beam. Multiply the two: p⊥ = 2I cos²θ/c, half its face-on value at 45° and a quarter at 60°. For reflectance r the normal pressure is (1 + r)I cos²θ/c, and a shear (1 − r)I cosθ sinθ/c appears beside it.
Cut the beam into photons and the force does not move
Now count the same beam. Each photon carries energy hf and momentum hf/c = h/λ, so a power P delivers R = P/hf photons per second. A black surface stops each one: F = R(hf/c) = P/c. A mirror sends each one back: F = 2P/c. Identical to the wave answers, and h has cancelled — it entered once through the energy and once through the momentum. That cancellation is a historical fact worth holding on to. Lebedev in 1900 and Nichols and Hull in 1901 measured radiation pressure on a torsion balance and confirmed Maxwell years before anyone accepted photons, precisely because the mean force cannot tell the two pictures apart. Work one case both ways: a 5.0 mW pointer at 532 nm pushes a black card with P/c = 1.67 × 10⁻¹¹ N; counted, it delivers 1.34 × 10¹⁶ photons per second each carrying 1.25 × 10⁻²⁷ kg m s⁻¹, and the product is the same 1.67 × 10⁻¹¹ N.
Small pressure, large effect: sails, tweezers, cold atoms
Microscopic is not the same as irrelevant; it depends on what the force acts on and for how long. On a bench, 4.5 μPa is hopeless — Nichols and Hull had to work in a partial vacuum because residual gas heated by the beam produced radiometric forces larger than the effect they were after. In orbit nothing competes: a 32 m² aluminised sail at 1 AU feels about 2.8 × 10⁻⁴ N, which on a 5.0 kg spacecraft is 5.5 × 10⁻⁵ m s⁻², adding 4.8 m s⁻¹ every day and roughly 1.7 km s⁻¹ over a year, free. At atomic scale the ratio is better still. A sodium atom absorbing a 589 nm photon takes a velocity kick h/mλ = 2.9 cm s⁻¹, and at saturation it scatters about 3 × 10⁷ photons per second, giving some 9 × 10⁵ m s⁻² — around 90 000 g. The same pressure that cannot lift a speck of dust is what stops an atomic beam and holds a bead in optical tweezers.
Where the accounts part: the count itself fluctuates
So far the two pictures agree exactly. They stop agreeing at the second moment. A classical wave of intensity I deposits exactly I A τ of energy into every gate of length τ: the prediction carries no spread at all. The photon picture says arrivals are discrete events, and for an ideal laser — a coherent state — they are Poissonian and independent, P(N) = e(−⟨N⟩)⟨N⟩N / N!, with Var N = ⟨N⟩ and σN = √⟨N⟩. The absolute noise therefore grows as √N while the relative noise falls as 1/√N. Take 1.0 mW at 633 nm for a full second: ⟨N⟩ = 3.19 × 10¹⁵ and σ = 5.6 × 10⁷, a fractional wobble of 1.8 × 10⁻⁸ that no bench measurement would ever resolve. Attenuate the same beam to 1.0 nW and gate it for 1 μs and the identical law gives ⟨N⟩ = 3186 with σ = 56 — a 1.8% scatter that now dominates the measurement. The graininess was always there; only its fractional size changed.
Reading the fluctuation: efficiency, shot noise, and g⁽²⁾
That fluctuation belongs to the light, not to the detector, and reading it correctly needs three facts. First, imperfect detection makes it worse, never better: discarding photons at random with probability 1 − η thins a Poisson stream into another Poisson stream of mean ηN, so the relative noise rises to 1/√(ηN). Second, in a photodiode the same statistics show up as current noise, iᵣₘₛ = √(2 e Idc B) — for 1.0 mA in a 1.0 MHz bandwidth that is 18 nA, and no amount of cooling removes it. Third, the variance is diagnostic where the mean is blind. The second-order correlation g⁽²⁾(0) is 1 for a coherent beam, 2 for thermal light of the same mean power, which bunches, and tends to 0 for a single-atom source, which cannot emit two photons at once. Three sources, one radiation pressure, three different noise spectra. In LIGO the same number fluctuation becomes a fluctuating force on the mirrors — quantum radiation-pressure noise, the low-frequency partner of the shot noise on the readout.
Change one variable at a time
Make the relationship visible.
Set r = 1 and slide θ: by 45° the solid curve has already halved while the dashed one-cosine curve has only fallen to 71%, and the gap between the two markers is the second cosine. Then drop r to 0 and the whole picture halves — a black sail is pushed half as hard as a mirror.
NORMAL PRESSURE p⊥6.80 µPa
PHOTON FLUX AT 532 nm3.15 ×10²¹/m²s
COUNTS 1 µm² × 1 µs3154 photons
SHOT NOISE 1/√N1.78 %
Live interpretationNORMAL PRESSURE p⊥: 6.80 µPa. PHOTON FLUX AT 532 nm: 3.15 ×10²¹/m²s. COUNTS 1 µm² × 1 µs: 3154 photons. SHOT NOISE 1/√N: 1.78 %
Catch the common trap
Explain before calculating.
A perfectly reflecting mirror is turned so that the beam now strikes it at 60° from the surface normal. Compared with the same beam at normal incidence, the pressure on the mirror face is:
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 5.0 mW laser pointer at 532 nm shines straight down onto (a) a matt black card and (b) a mirror. Find the force in each case and the rate at which photons arrive. Take c = 2.998 × 10⁸ m s⁻¹ and hc = 1.986 × 10⁻²⁵ J m.
- Normal incidence on a full absorber: every photon's momentum is delivered once and none returns, so F = P/c = 5.0 × 10⁻³ W ÷ 2.998 × 10⁸ m s⁻¹ = 1.67 × 10⁻¹¹ N.
- The mirror sends the light back, doubling the momentum change: F = 2P/c = 3.34 × 10⁻¹¹ N. That is the weight of about 3.4 nanograms — and note that the mirror, which absorbs nothing, feels twice the force of the card, which absorbs everything.
- Photon energy: hf = hc/λ = 1.986 × 10⁻²⁵ J m ÷ 532 × 10⁻⁹ m = 3.73 × 10⁻¹⁹ J, or 2.33 eV.
- Arrival rate: R = P/hf = 5.0 × 10⁻³ ÷ 3.73 × 10⁻¹⁹ = 1.34 × 10¹⁶ photons per second.
- Check the two accounts against each other: each photon carries hf/c = 1.25 × 10⁻²⁷ kg m s⁻¹, and R × hf/c = 1.34 × 10¹⁶ × 1.25 × 10⁻²⁷ = 1.67 × 10⁻¹¹ N — the same as step 1, as it has to be.
AnswerF = 1.7 × 10⁻¹¹ N on the black card and 3.3 × 10⁻¹¹ N on the mirror, carried by R = 1.3 × 10¹⁶ photons per second.
MediumA solar sail of area 32 m² has power reflectance r = 0.90, and the whole spacecraft masses 5.0 kg. At 1 AU the solar intensity is I = 1361 W m⁻². Find the thrust and acceleration with the sail face-on to the Sun, the speed gained in one day, and then the normal force once the sail is tilted 45° to the sunline.
- Face-on, the absorbed 10% delivers I/c and the reflected 90% delivers a further 0.90 I/c, so p = (1 + r)I/c = 1.90 × 1361 ÷ 2.998 × 10⁸ = 8.63 × 10⁻⁶ Pa.
- Thrust: F = pA = 8.63 × 10⁻⁶ Pa × 32 m² = 2.76 × 10⁻⁴ N.
- Acceleration: a = F/m = 2.76 × 10⁻⁴ ÷ 5.0 = 5.52 × 10⁻⁵ m s⁻². Over one day, Δv = a t = 5.52 × 10⁻⁵ × 86400 = 4.8 m s⁻¹ — about 1.7 km s⁻¹ over a year, if it stayed at 1 AU and face-on.
- Tilt to θ = 45° and apply both cosines: p⊥ = (1 + r)I cos²45°/c = 8.63 × 10⁻⁶ × 0.500 = 4.31 × 10⁻⁶ Pa, giving F⊥ = 1.38 × 10⁻⁴ N directed along the sail's own normal, not along the sunlight.
- Resolve that normal force about the sunline: 1.38 × 10⁻⁴ × cos45° = 9.8 × 10⁻⁵ N radially outward and the same 9.8 × 10⁻⁵ N transverse. The transverse component is what raises or lowers the orbit; a face-on sail can only push outward.
AnswerFace-on: 2.8 × 10⁻⁴ N, a = 5.5 × 10⁻⁵ m s⁻², 4.8 m s⁻¹ per day. At 45° the normal force halves to 1.4 × 10⁻⁴ N but gains a 9.8 × 10⁻⁵ N transverse component — the steering thrust. The absorbed 10% adds a 7 × 10⁻⁶ N shear.
HardA 632.8 nm helium–neon beam is attenuated to 1.20 nW and falls on a photon counter of quantum efficiency η = 0.65 with negligible dark counts. Find the photon arrival rate, the mean count and its standard deviation in a 10.0 μs gate, and the gate length needed to fix the power to 0.10%. Then say what a classical wave of the same intensity would predict for the spread.
- Photon energy: hf = hc/λ = 1.986 × 10⁻²⁵ ÷ 632.8 × 10⁻⁹ = 3.139 × 10⁻¹⁹ J, or 1.96 eV.
- Arrival rate R = P/hf = 1.20 × 10⁻⁹ ÷ 3.139 × 10⁻¹⁹ = 3.82 × 10⁹ s⁻¹. The counter registers only ηR = 0.65 × 3.82 × 10⁹ = 2.48 × 10⁹ s⁻¹.
- Mean count in the gate: ⟨N⟩ = ηRτ = 2.48 × 10⁹ × 10.0 × 10⁻⁶ = 2.48 × 10⁴. Coherent light is Poissonian, so Var N = ⟨N⟩ and σ = √(2.48 × 10⁴) = 158 counts — a relative spread of 0.63%.
- Relative noise is 1/√N, so 0.10% needs N = (1/0.0010)² = 1.0 × 10⁶ counts, i.e. τ = 1.0 × 10⁶ ÷ 2.48 × 10⁹ = 4.0 × 10⁻⁴ s = 402 μs. Six times the precision costs forty times the time.
- The lost 35% is not free: at η = 1 the same 10⁶ counts would arrive in 262 μs. Discarding photons at random thins one Poisson stream into another of smaller mean, so it raises the relative noise rather than leaving it untouched.
- A classical wave of the same intensity delivers exactly Pτ into every gate, so it predicts σ = 0. Mean force and mean count agree with the photon picture; the spread does not. Thermal light of the same mean power would give Var N > ⟨N⟩, with g⁽²⁾(0) = 2 instead of 1 — so only the variance identifies the source.
AnswerR = 3.82 × 10⁹ s⁻¹; ⟨N⟩ = 2.48 × 10⁴ ± 158 counts (0.63%) in 10.0 μs; 402 μs is needed for 0.10%. Classically σ = 0 — the mean is blind to the source, the variance is not.