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Light hitting a surface impart a force on the surface, often called "radiation pressure". My question is, given a perfectly reflective surface, if light hits it at 90° to return in the opposite direction, how large is this pressure, expressed as a fraction of the energy in the light?

(Edit: don't take this question as a claim that the answer is one specific number; it could be a function of... not sure what.)

This is not a duplicate of the "will I accelerate?" question. I looked carefully to see if this had been asked before. To my surpise, it had not. To clarify, I'm looking for a dimensionless expression/formula for the fraction of energy turned into kinetic energy, or in other words, efficiency (often denoted η). In contrast, the other question wasn't asking for any expression/formula at all, and none of the formulas/expressions in any of the answers are dimensionless except the one for "number of photons per second" which doesn't apply here. And yes, I am aware that Force = Power/c. I notice that someone deleted the word "fraction", replacing it with "function", in two different places before the question was closed, which must have given people a misleading impression of what the question was about.

A non-relativistic formula will suffice, in the reference frame of a relatively-stationary planet firing a laser at a small reflective object. If it helps, assume the object does not rotate.

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    $\begingroup$ It doesn't make sense to have a ratio of radiation pressure to the energy in light, they don't have the same dimensions. $\endgroup$
    – Triatticus
    Commented Feb 6 at 23:11
  • $\begingroup$ @Triatticus Let's say I shine 1000 joules of light at a perfectly reflective Breakthrough Starshot with starting velocity zero. It will change velocity. That change in velocity can be interpreted as a change in the object's kinetic energy. How big is that energy? Now divide it by 1000 J so it is dimensionless. (My expectation is that less energy will be transferred as its speed nears c, but only because the wavelength of the incoming light will decrease in the reference frame of the Starshot.) $\endgroup$
    – Qwertie
    Commented Feb 6 at 23:18
  • $\begingroup$ @Qwertie The change in kinetic energy is zero. $\endgroup$
    – my2cts
    Commented Feb 6 at 23:21
  • $\begingroup$ Pressure has the same dimensions as energy density: $dU=-PdV$ $\endgroup$
    – mike stone
    Commented Feb 6 at 23:29
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    $\begingroup$ Have you tried calculating this yourself using the equations on Wikipedia for a perfectly reflective surface? Things you'd need to know would be what the irradiance of your source is, and how much of that irradiance actually impacts the surface of the object. If you know the cross sectional area and mass you can convert this to acceleration (which would be almost negligible). $\endgroup$
    – Triatticus
    Commented Feb 7 at 0:03

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"I'm looking for a dimensionless expression/formula for the fraction of energy turned into kinetic energy ..."

Here's a simple treatment... Suppose that the photon of frequency $f_1$ (and energy $hf_1$) hits a small reflective sphere normally, and bounces back with a lower frequency, $f_2$. The frequency is reduced as some kinetic energy will be given to the sphere. Assuming that the sphere's speed, $v\ll c$ we have $$\tfrac 12 mv^2=hf_1-hf_2.$$ But as momentum is a conserved vector, then considering components in the initial direction of motion of the photon:

$$\frac{hf_1}c=mv-\frac{hf_2}c$$

Eliminating $f_2$,

$$\tfrac 12 mv^2=hf_1+hf_1-mcv.$$

Since we're assuming that $v\ll c$ we can neglect $\tfrac 12mv^2$ in comparison with $mcv$ so we have

$$v=\frac{2hf_1}{mc}\ \ \ \ \text{so}\ \ \ \ \ \ \tfrac 12mv^2=\frac{2(hf_1)^2}{mc^2}.$$

The kinetic energy given to the sphere, expressed as a fraction of the initial photon energy, $hf_1$, is therefore

$$\frac{\tfrac 12mv^2}{hf_1}=\frac{2hf_1}{mc^2}.$$

Try putting in some figures, say for a sub-microscopic sphere of mass $1.0 \times 10^{-18}$ kg and an ultraviolet photon of frequency $1.0\times 10^{16}$ Hz. You might not be very impressed with the answer.

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  • $\begingroup$ I specified that the surface is perfectly reflective, so the frequency of the outgoing light should not change. Wikipedia says that reflection yields double the force as compared to a blackbody that absorbs the light. $\endgroup$
    – Qwertie
    Commented Feb 15 at 14:31
  • $\begingroup$ If the frequency didn't change the photon energy wouldn't change, so the body with the reflecting surface couldn't gain any kinetic energy. But a body of finite mass will gain some momentum and some kinetic energy as a result of the impact. The back-travelling photon will have a lower frequency and less energy, not as a result of energy •absorption and dissipation• by the body's surface, but as a result of giving kinetic energy to the body. Another way of understanding the frequency drop is as a Doppler shift due to motion of the reflector. $\endgroup$
    – Philip Wood
    Commented Feb 15 at 14:49
  • $\begingroup$ "Wikipedia says that reflection yields double the force as compared to a blackbody that absorbs the light." And Wiki is right; the factor of 2 appears in my equation $v=2hf_1/mc$. But the equation, as I've said, is very slightly an approximation if the body moves – which it must do if it receives kinetic energy! $\endgroup$
    – Philip Wood
    Commented Feb 15 at 15:07

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