- #1
Dazed&Confused
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Homework Statement
Suppose light of momentum [itex]-P\hat{\textbf{k}}[/itex] is shone on a sphere of radius [itex]R[/itex]. What is the momentum transferred onto the sphere?
Homework Equations
The Attempt at a Solution
I think the transferred momentum upon reflection is given by [itex](-2P\hat{\textbf{k}} \cdot \hat{\textbf{r}} ) \hat{\textbf{r}}[/itex] where [itex]\hat{\textbf{r}}[/itex] is the unit vector perpendicular to the sphere surface. The component of the area parallel to the light is [itex]d\textbf{a} \cdot \hat{\textbf{k}} = R^2 \sin \theta \cos \theta d \phi d \theta[/itex]. Thus integrating over one hemisphere $$ \int -2P \cos^2 \theta \sin \theta \ d \phi \ d \theta \ \hat{\textbf{r}}$$
Only the component of [itex]\hat{\textrm{k}}[/itex] remains and so this becomes $$ \int -2P \cos^3 \theta \sin \theta \ d \phi \ d \theta \ \hat{\textbf{k}} = 4 \pi R^2 P \hat{\textbf{k}} \left [ \frac{\cos^4 \theta}{4} \right ]_{0}^{\pi/2} = -\pi R^2 P \hat{\textbf{k}}$$
which is half the result of the reflection of a disc of radius [itex]R[/itex]. I've told that it should be the same answer for both shapes, but this does not make sense to me. Where have I gone wrong?
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