Light reflecting off sphere -- Momentum transfer

[Dazed&Confused]

Homework Statement

Suppose light of momentum $-P\hat{\textbf{k}}$ is shone on a sphere of radius $R$. 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 $(-2P\hat{\textbf{k}} \cdot \hat{\textbf{r}} ) \hat{\textbf{r}}$ 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?

 

[PeroK]

Homework Statement

Suppose light of momentum $-P\hat{\textbf{k}}$ is shone on a sphere of radius $R$. 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 $(-2P\hat{\textbf{k}} \cdot \hat{\textbf{r}} ) \hat{\textbf{r}}$ 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?

Why would it be the same for both shapes?

 

[Dazed&Confused]

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.

 

[Dazed&Confused]

For absorption the answer is the same for both shapes as all that matters is the cross section.

 

[PeroK]

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.

 

[PeroK]

I agree with your answer.

 

[Dazed&Confused]

Thanks for looking at this.

 

[haruspex]

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.