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Radiation pressure

Did you know that light has momentum? Even though it doesn't have mass? This allows light to exert a pressure on a surface, known as radiation pressure:

\[P = \frac{\text{Flux density}}{c}\]

Where \(c\) is the speed of light. Flux density will be written in full to prevent confusion with Force which is also denoted \(F\). Flux density is also sometimes called "intensity" and represented with \(I\), but this is much too confusing since intensity can mean many different things (see this note).

Not all light is absorbed

If the surface perfectly reflects all the light, the radiation pressure actually doubles since light being reflected back also gives additional pressure on the surface:

\[P = \frac{2\cdot\text{Flux density}}{c}\]

We know that \(I\) depends on the Luminosity and Area of the object, so we can rewrite it as such:

\[P = \frac{\text{Flux density}}{c} = \frac{L}{4\pi D^2}\frac{1}{c} = \frac{L}{4\pi c D^2}\]

We also know that \(P=\frac{F}{A}\), so we can equate the force as such:

\[\begin{align*} P&=\frac{\text{Flux density}}{c} \\ \frac{F}{A} &= \frac{\text{Flux density}}{c} \\ F &= \frac{\text{Flux density}\cdot A}{c} \\ &= \frac{LA}{4\pi c D^2}\end{align*}\]

where \(A\) is the area of the surface being hit by light.

Why is there a mirror in space

There is a mirror 1 AU from the sun that is a 100% reflective surface. What is the pressure exerted on the surface? What is the force exerted on the surface if the mirror is a circle with radius of \(1m\)? What is it instead is a square of sides 1m? (\(1AU = 1.496\times10^{11}m\), \(L_\odot = 3.828\times 10^{26}W\), \(c=2.998\times10^8 m/s\))

We first find the pressure:

\[\begin{align*}P &= \frac{2\cdot\text{Flux density}}{c} \\ &= \frac{2L}{4\pi c D^2} \\ &= \frac{2\times3.828\times 10^{26}}{4\pi \times 2.998\times10^8 \times (1.496\times10^{11})^2} \\ &= 9.080 \times 10^{-6}Pa\end{align*}\]

We can then find the force \(F\) using \(P=\frac{F}{A}\)

For a circle, \(A = \pi r^2 = 3.1416\)

\[\begin{align*}F &= PA \\ &= 9.080 \times 10^{-6} \times 3.1416 \\ &= 2.853 \times 10^{-5} N \end{align*}\]

For a square \(A = a^2 = 1\), hence \(F = 9.080 \times 10^{-6} N\)

Note that this is an approximation because we are assuming the rays are parallel, when the source isn't actually at infinity. However since 1AU is much bigger than 1m, we can effectively treat it like the rays are parallel.