Laser beams interacting with a screen | AM and momentum conservation

[JD_PM]

Homework Statement

A screen is tied to an axis that can rotate, as in the below figure. On one side the screen is black and absorbs photons completely.On the other side there is a perfect mirror. An identical laser appears on both sides of the plates as shown in the figure.

a) Given that photons are massless but do have momentum in which direction will the axis turn? Or will not turn? Explain.

b) What if both sides were to be mirrors? and what if both sides were to be black absorbing screens?

c) Suppose that the system is invariant under rotations around the $z$-axis (which is the rotational axis) and the $x$-axis (note that the screen lies on the $xz$ plane). State all quantities that are conserved over time while dealing with the screen as originally settled (black and mirror sides).

Relevant Equations

Please see attached image.

a)

On the black part, all incoming light is absorbed. This means that the momentum of the left-light beam doesn't change (i.e. momentum before hitting the black screen is $\vec p_0$ and after hitting it is zero. Thus $\Delta \vec p = \vec p_0$). If momentum doesn't change, we get no impulse. So we expect no rotation in the counterclockwise direction.

On the mirror part we deal with light reflection. The magnitude of the momentum of the right-light beam doesn't change but its direction does. Thus the momentum of the right-light beam changes after the reflection and there is an impulse. So we expect no rotation in the clockwise direction.

The overall effect is that the screen rotates clockwise.

b)

Based on a) reasoning we expect no rotation in neither case.

c)

I guess here we are asked to check conservation of angular momentum and momentum.

I think neither momentum nor angular momentum are conserved due to the presence of an external force and an external torque respectively.

I would like to discuss whether you are with my answers or not.

Thank you.

 

[Cutter Ketch]

You are on the right track for part a) except your idea that the momentum on the black side doesn’t change. I presume you mean the momentum of the light. The very next thing you write is the change in momentum! So why would you say it doesn’t change and there is no impulse?

In the second paragraph I think you have a cut and paste error in that as soon as you correctly reason that there is a change in momentum on the mirror side you say “so we expect no rotation in the clockwise direction” which you clearly don’t mean based on your very next sentence “the overall effect is the screen rotates clockwise”

b) all of the confusion in a has led to bad reasoning here.

Try again on a). You wrote an expression for $\Delta$p on the black side. Can you write a similar expression for the mirror side?

 

[haruspex]

Just pointing out that the question ought to state that this is in a vacuum. Otherwise the rotation can go the other way! See https://en.m.wikipedia.org/wiki/Crookes_radiometer#Movement_with_black-body_absorption.

 

[JD_PM]

You are on the right track for part a) except your idea that the momentum on the black side doesn’t change. I presume you mean the momentum of the light. The very next thing you write is the change in momentum! So why would you say it doesn’t change and there is no impulse?

In the second paragraph I think you have a cut and paste error in that as soon as you correctly reason that there is a change in momentum on the mirror side you say “so we expect no rotation in the clockwise direction” which you clearly don’t mean based on your very next sentence “the overall effect is the screen rotates clockwise”

My bad, I did not explain myself properly. I will argue a) again

a)

On the black part, all incoming light is absorbed. This means that the momentum of the left-light is $\vec p_{0.left}$ before the collision and zero after the collision. This means that the momentum does change and thus there is an impulse. So we expect rotation in the counterclockwise direction.

On the mirror part we deal with light reflection. The magnitude of the momentum of the right-light beam doesn't change but its direction does. Thus the momentum of the right-light beam changes after the reflection and there is an impulse. So we expect rotation in the clockwise direction.

We notice that overall there are two impulse vectors with opposite directions. Will they cancel each other out and then no rotation will occur? or one impulse vector overcomes the other?

You wrote an expression for $\Delta$p on the black side. Can you write a similar expression for the mirror side?

Change of momentum of the left-light beam:

$$\Delta p_{left} = p_{f.left} - p_{0.left} = -p_{0.left}$$

Change of momentum of the right-light beam:

We know that:

$$p_{f.right} = - p_{0.right}$$

So:

$$\Delta p_{right} = p_{f.right} - p_{0.right} = 2p_{f.right}$$

 

[Cutter Ketch]

My bad, I did not explain myself properly. I will argue a) again

a)

On the black part, all incoming light is absorbed. This means that the momentum of the left-light is $\vec p_{0.left}$ before the collision and zero after the collision. This means that the momentum does change and thus there is an impulse. So we expect rotation in the counterclockwise direction.

On the mirror part we deal with light reflection. The magnitude of the momentum of the right-light beam doesn't change but its direction does. Thus the momentum of the right-light beam changes after the reflection and there is an impulse. So we expect rotation in the clockwise direction.

We notice that overall there are two impulse vectors with opposite directions. Will they cancel each other out and then no rotation will occur? or one impulse vector overcomes the other?
Change of momentum of the left-light beam:

$$\Delta p_{left} = p_{f.left} - p_{0.left} = -p_{0.left}$$

Change of momentum of the right-light beam:

We know that:

$$p_{f.right} = - p_{0.right}$$

So:

$$\Delta p_{right} = p_{f.right} - p_{0.right} = 2p_{f.right}$$

The notation is throwing me a little, but I think you’ve got it. However, you haven’t reached a conclusion yet. Which is bigger, left or right?

 

[JD_PM]

The notation is throwing me a little, but I think you’ve got it. However, you haven’t reached a conclusion yet. Which is bigger, left or right?

The right is the bigger. Thus we would expect the screen to turn clockwise. At this point, we are assuming that the force that causes screen motion is due to the pressure exerted by the light on the screen (which is in vacuum).

There are two important ideas to point out (which can be found in the link haruspex provided):

- If light pressure were the cause of the rotation, then the better the vacuum, the less air resistance to movement, and the faster the screen should spin clockwise.

- If light pressure were the motive force the screen would spin clockwise, as the photons on the mirror side being reflected would deposit more momentum than on the black side where the photons are absorbed. This results from conservation of momentum - the momentum of the reflected photon exiting on the light side must be matched by a reaction on the side that reflected it.

However, it turns out that if you were to do the experiment (in vacuum) you wouldn't observe clockwise rotation but counterclockwise rotation.

Thus light pressure cannot be the driving force.

The currently accepted explanation is based on Thermodynamics (which can also be found in the link haruspex provided):

The screen behaves as a heat engine. The operation of a heat engine is based on a difference in temperature that is converted to a mechanical output. In this case, the black side of the screen becomes hotter than the other side, as radiant energy from a light source warms the black side by absorption faster than the mirror side. The internal air molecules are heated up when they touch the black side of the screen.

The details Wikipedia provides on exactly how this moves the warmer side of the screen counterclockwise do not look clear to me.

But this goes beyond the question I posted.