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I know very little about electromagnetic waves and light in general so you are going to have to bear with me. I am attempting to calculate the force on a sphere in a plane wave of light where the light has intensity I. My calculations and research have concluded that, regardless of whether or not the sphere absorbs or reflects the light, that the force can be modeled as: $\frac{\pi{r}^2{I}}{c}$.

This pdf document outlines what I am looking for (with $I$ replaced with $\langle{S}\rangle$) but in equation 24 it seems like a factor of $2\pi\text{r}$ appears out of nowhere.

Another solution I have tried is a double integral in spherical coordinates and exploiting symmetry to find that $\text{Force} = \iint\limits_A \! p \,\,{dA} $ where p is a function describing the pressure on an area element. Every time I calculate this pressure I find a function of one angle rather than of two angles. I find that my integral looks like: $$\int_{0}^{\pi}\!\int_{0}^{\frac{\pi}{2}}\!\frac{4\pi{r}^2{I}}{{c}}cos(\theta)^3sin(\theta) \, d\theta d\phi$$

which evaluates to the correct expression for force (up to the sign) but I have trouble justifying that the pressure does not depend on $\phi$ and only depends on $\theta$. In this instance, I am integrating over half of a hemisphere where $\theta$ is the polar angle and $\phi$ is the azimuthal angle or as described in this image.

I feel like these calculations are incorrect and that the double integral over ${dA}$ should include an element reflecting the influence of $\phi$ as I cannot see any reason why the calculation would be independant of it.

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You will get the right answer if you represent the plane wave by a beam of photons, using the relation between energy and momentum $e=p c$, and setting the density of photons in the beam to match the intensity of the wave. Since reflected photons supply more momentum transfer to the surface than absorbed photons, the pressure force should be larger on reflecting surface. –  Maxim Umansky Jul 17 '13 at 21:32
Not sure about the rest, but reflection or absorption should matter. The change in momentum for a reflected photon is double that of an absorbed photon, so it's an easy factor of 2 (for perfect reflector compared to perfect absorber). If the photons are scattered it starts to get complicated. –  Kyle Oman Jul 17 '13 at 21:34
There is a factor or two included in the integral of pressure. The $4$ comes from multiplying the factor of two from symmetry and the factor of two from the reflection. Despite this factor, when $\theta$ ranges from $ [0,\frac{\pi}{4}]\bigcup[\frac{3\pi}{4},\pi] $ there is a component of reflection pointing in the $+x$ direction or "reflected forward". The net momentum from the reflected photons turns out to be zero. The pdf from the original post outlines this under equation 26. –  user27203 Jul 18 '13 at 15:39

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