Suppose light of momentum $-P\hat{\textbf{k}}$ is shone on a sphere. What is the momentum transferred onto the sphere?

I think the transferred momentum upon reflection is given by $(2P\hat{\textbf{k}} \cdot \hat{\textbf{r}} ) \hat{\textbf{r}}$ where $\hat{\textbf{r}}$ is the unit vector perpendicular to the sphere surface. The component of the area parallel to the light is $d\textbf{a} \cdot \hat{\textbf{k}} = R^2 \sin \theta \cos \theta d \phi d \theta$. Thus integrating over one hemisphere $$ \int -2P \cos^2 \theta \sin \theta \ d \phi \ d \theta \ \hat{\textbf{r}}$$

Only the component of $\hat{\textrm{k}}$ 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 $R$. 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?

  • $\begingroup$ Who told you it should be the same? I've done this calculation before and I think your factor of 1/2 is right. $\endgroup$ – knzhou Jan 15 '16 at 0:57
  • $\begingroup$ Unfortunately my lecturer. $\endgroup$ – user110503 Jan 15 '16 at 0:58
  • $\begingroup$ maybe he meant it was the same as an absorbing disc? $\endgroup$ – knzhou Jan 15 '16 at 1:07
  • $\begingroup$ I don't think that can be true because in the solutions there was a factor 2. It was given as 'self evident' it seems. $\endgroup$ – user110503 Jan 15 '16 at 1:12
  • $\begingroup$ Do you know any resource to look into this? $\endgroup$ – user110503 Jan 15 '16 at 1:17

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