In July 2020 comet NEOWISE could be seen from India with the naked eye, low in the north-west after sunset. Its long, curved tail pointed upwards, away from the Sun that had just set below the horizon. Photographs taken over the following days showed the tail always pointing away from the Sun. It did not trail behind the comet the way smoke trails behind a train.
Momentum delivered by a beam
A photon carries momentum \(h/\lambda = E/c\). When a surface absorbs the photon, that momentum goes into the surface. A beam delivers a steady stream of photons, so it delivers a steady stream of momentum. From the chapter on Newton's laws, momentum delivered per second is a force. So light pushes on everything it strikes.
Let a beam of power \(P\) fall straight onto a black surface that absorbs it all. Every second the beam brings energy \(P\), carried by photons of energy \(E\) each. So \(P/E\) photons arrive each second, each bringing momentum \(E/c\). The momentum delivered each second is \((P/E)(E/c) = P/c\). The size of the photons cancels out.
\[ n = \dfrac{P}{E} \]
Why. the beam brings energy \(P\) each second, in photons of energy \(E\) each (lesson 5).
\[ \Delta p = \dfrac{E}{c} \text{ per photon} \]
Why. a black surface absorbs each photon and takes all its momentum \(p = E/c\).
\[ F = n\,\Delta p = \dfrac{P}{E} \cdot \dfrac{E}{c} = \dfrac{P}{c} \]
Why. force is the momentum delivered per second, from the impulse-momentum theorem.
\[ F = \dfrac{IA}{c}, \qquad \dfrac{F}{A} = \dfrac{I}{c} \]
Read it. a beam of intensity \(I\) covering an area \(A\) carries power \(P = IA\).
A perfect mirror does more. It sends each photon straight back. The photon's momentum changes from \(+E/c\) to \(-E/c\), a change of \(2E/c\). By the third law, the mirror receives \(2E/c\) from every photon, twice as much as the black surface. A ball bouncing off a wall pushes it harder than a ball of wet clay that sticks.
The force per unit area is the radiation pressure. It is very small. Sunlight above the Earth's air has an intensity of about \(1360\,\text{W/m}^2\). On one square metre of black card it pushes with \(1360/3 \times 10^{8} = 4.5 \times 10^{-6}\,\text{N}\), the weight of half a milligram. On a square metre of mirror the push is twice this, \(9.1 \times 10^{-6}\,\text{N}\).
Two identical beams fall straight onto two plates, one black and one a perfect mirror. Which plate feels the larger force?
Sunlight of intensity \(900\,\text{W/m}^2\) falls straight onto a flat mirror of area \(2\,\text{m}^2\) that reflects it all. What force does the light exert? Answer in micronewtons, \(10^{-6}\,\text{N}\).
Light that arrives at an angle
Now tilt the surface, so the beam arrives at an angle \(\theta\) to the normal. A patch of area \(A\) now catches less of the beam. Seen along the beam, the patch looks narrower, with a cross-section \(A\cos\theta\). So the power it receives is \(IA\cos\theta\), and fewer photons land on it each second.
A black patch absorbs every photon, momentum and all. The momentum arrives along the beam, so the force points along the beam, with size \(IA\cos\theta/c\). Split it into two parts. The part along the normal presses on the surface: \(IA\cos^2\theta/c\). The part along the surface drags it sideways: \(IA\cos\theta\sin\theta/c\).
A mirror treats the two parts of each photon's momentum differently. The part along the surface is unchanged by the reflection. The part along the normal, \((E/c)\cos\theta\), is reversed, a change of \(2(E/c)\cos\theta\). So the mirror feels no sideways force at all. Its whole force is along the normal, \(2IA\cos^2\theta/c\).
Why does \(\cos\theta\) appear twice? One factor is geometric: the tilted patch catches a thinner beam. The other factor picks out the normal part of each photon's momentum. For a mirror at \(60^\circ\) the two factors give \(\cos^2 60^\circ = 1/4\) of the force at normal incidence.
Sunlight of \(1360\,\text{W/m}^2\) falls on a flat mirror of area \(1\,\text{m}^2\) at \(60^\circ\) to its normal. Find the force on the mirror. Then find the force on a black plate of the same size at the same angle.
\(\theta = 60^\circ\), so \(\cos\theta = 0.5\) and \(\sin\theta = 0.866\).
Mirror: \(F = 2IA\cos^2\theta/c\), along the normal. Black plate: \(F = IA\cos\theta/c\), along the beam.
Mirror: \(F = 2 \times 1360 \times 0.25 / 3 \times 10^{8} = 2.3 \times 10^{-6}\,\text{N}\). Black plate: \(F = 1360 \times 0.5 / 3 \times 10^{8} = 2.3 \times 10^{-6}\,\text{N}\), made of \(1.13 \times 10^{-6}\,\text{N}\) along the normal and \(1.96 \times 10^{-6}\,\text{N}\) along the surface.
The mirror is pushed straight away from its face. The black plate is pushed along the beam, partly sideways.
Sunlight of \(1360\,\text{W/m}^2\) falls on a black plate of area \(1\,\text{m}^2\) at \(30^\circ\) to its normal. What is the part of the force along the normal? Answer in micronewtons.
A beam strikes a perfect mirror at \(45^\circ\) to its normal. In which direction is the mirror pushed?
Surfaces in between
Real surfaces are neither perfectly black nor perfect mirrors. Suppose a surface reflects a fraction \(r\) of the light, its reflectivity, and absorbs the rest, \(1 - r\). At normal incidence the absorbed part pushes with \((1 - r)P/c\), and the reflected part with \(2rP/c\). The total is \((1 + r)P/c\).
The formula checks itself at both ends. With \(r = 0\) it gives \(IA/c\), the black surface. With \(r = 1\) it gives \(2IA/c\), the mirror. At an angle the two parts behave as before. The normal force is \((1 + r)IA\cos^2\theta/c\), and only the absorbed part, \((1 - r)IA\cos\theta\sin\theta/c\), drags along the surface.
Light of intensity \(1200\,\text{W/m}^2\) falls straight onto a plate of area \(0.5\,\text{m}^2\) that reflects a quarter of it and absorbs the rest. What force does the light exert? Answer in micronewtons.
A comet swings round the Sun and heads back out. As it moves away from the Sun, which way does its dust tail point?
The comet's head releases fine dust as the Sun warms it. For grains about a micrometre across, the push of sunlight is comparable to the pull of the Sun's gravity. Both forces fall off as \(1/r^2\), so the ratio stays the same all along the orbit. The dust is pushed outwards, and the tail points away from the Sun, ahead of the comet on its way out.
Treat the Earth as a black disc of radius \(6.4 \times 10^{6}\,\text{m}\) in sunlight of \(1360\,\text{W/m}^2\). Estimate the force of sunlight on the Earth, in newtons.
In 1873 James Clerk Maxwell predicted from his theory of light that a beam must press on what it strikes, with a pressure \(I/c\). The pressure was so small that nobody could measure it for almost thirty years. In 1900 the Russian physicist Pyotr Lebedev hung thin metal vanes, some black and some shiny, on a fine thread inside an evacuated glass bulb. Light made the thread twist by a tiny angle. The shiny vanes were pushed harder than the black ones, just as Maxwell's pressure required.
ISRO's solar observatory Aditya-L1 has watched the Sun since early 2024 from near the Lagrange point L1, about \(1.5\) million km from the Earth. There the push of sunlight never stops. On a sunlit face of, say, \(10\,\text{m}^2\) it is roughly \(4.5 \times 10^{-5}\,\text{N}\). Over a year that force would change the velocity of a \(1500\,\text{kg}\) craft by about \(1\,\text{m/s}\). That is enough to drift it off its path, so the small corrective firings of its thrusters must allow for it.
- Giving a mirror the same force as a black surface.A mirror reverses each photon's momentum, a change of \(2E/c\), so it feels twice the force.
- Using a single \(\cos\theta\) at oblique incidence.One \(\cos\theta\) for the thinner beam caught, another for the normal part of the momentum: \(\cos^2\theta\).
- Giving a mirror a sideways force at an angle.Reflection keeps the momentum along the surface. The mirror's force is purely normal.
- Measuring \(\theta\) from the surface instead of the normal.\(\theta\) is the angle between the beam and the normal. At \(\theta = 0\) the light falls straight on.