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Light reflecting off sphere -- Momentum transfer

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    3. 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?
     
    Last edited: Jan 15, 2016
  2. jcsd
  3. Jan 15, 2016 #2

    PeroK

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    Why would it be the same for both shapes?
     
  4. Jan 15, 2016 #3
    I personally do not think it would be, but that's what my lecturer has said. I'm asking here if the calculation I've done makes sense.
     
  5. Jan 15, 2016 #4
    For absorption the answer is the same for both shapes as all that matters is the cross section.
     
  6. Jan 15, 2016 #5

    PeroK

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    Unless all the light reflects vertically backwards, then it can't be the same.

    I can't quite check your integration in my head. It looks okay. I can check when I get back to my desk, if no one else has by then.
     
  7. Jan 15, 2016 #6

    PeroK

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    I agree with your answer.
     
  8. Jan 15, 2016 #7
    Thanks for looking at this.
     
  9. Jan 15, 2016 #8

    haruspex

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    There's a cute shortcut.
    Slicing the sphere into thin equal width sections perpendicular to the incident light, each section has the same surface area (Archimedes) so gets the same light. At a latitude of 45 degrees, the light reflects perpendicularly to its original direction, giving the same force as for the absorption case. At two latitudes equally either side of that, i.e. 45+/-theta, you can average out the reflections to get the same. Hence for the hemisphere it is equivalent to absorption.
    Well, maybe not that short, but it avoids the hazards of integration.
     
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