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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.

Qmechanic
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Qwertie
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1 Answers1

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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.

Philip Wood
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