Final momentum of box with reflective interior hit by ideal laser

[dark_matter_is_neat]

Homework Statement

Consider a square box of mass m whose interior is 100% reflective. Each edge of the box has length L and the box lays on a frictionless surface. One side of the box is removed to allow light into the box. A beam of light from an ideal laser of wavelength $\lambda$ and power P is pointed at an angle of 45 degrees from the horizontal to the opposite corner of the box. The box gradually moves to the right until the laser no longer points into the box. I the box starts at rest determine the final momentum of the box. Ignore the small influence of the doppler effect.

Relevant Equations

Momentum of light = $h\frac{1}{\lambda}$
Pressure on perfectly reflecting surface = $\frac{2I}{c}$

I'm not really sure what to do, clearly I need to get an equation of motion for the box and then solve for the final momentum. I can solve for the rate at which photons hit the box by dividing the power, P, by the energy of photon which would be $\frac{P \lambda}{hc}$ so multiplying this by the momentum of each photon, I would be depositing momentum at a rate of $\frac{P}{c}$, Although this expression considers the photon depositing momentum only once and it assumes the photon is perpendicular to the surface of the cube, both of which isn't actually correct.

Just thinking about the photon hitting the interior of the box and then getting reflected until it exits the box, the photon would get reflected by the box 3 times, so you have 4 photon momentums : The incoming momentum, the first reflected momentum, the second reflected momentum and the outgoing momentum, all of which have the same magnitude of momentum. Following this thinking I believe the photon would effectively deposit momentum into the box 4 times, where all the vertical components of momentum cancel out. The first and second reflected momentums should cancel, so the rate of momentum being deposited would be proportional to $\frac{2P}{c}$.

There is an exception to light getting reflected 3 times in the box which happens when the light hits the corner of the box, where it only gets reflected twice and effectively only one photon momenta is being deposited (so at t = 0, the force should just be $\frac{P}{c}$). Also since at $t = \infty$ there is no light hitting the box, the force is 0 at $t = \infty$.

What I'm struggling to determine it the horizontal component of the momenta deposited by the photons at any given time t. For each point of incidence the force applied is going to be $\frac{P}{c} cos(\theta_{i} (t))$.

 

[hutchphd]

One need not worry about the internal reflections. The cube will act as a "corner reflector"and send most of the light directly back the way it came. So the rate of (vector) momentum transfer to the cube will be twice the incoming (vector) momentum flux. So work out the average value for the incident beam.

 

[dark_matter_is_neat]

One need not worry about the internal reflections. The cube will act as a "corner reflector"and send most of the light directly back the way it came. So the rate of (vector) momentum transfer to the cube will be twice the incoming (vector) momentum flux. So work out the average value for the incident beam.

So, when the light is being reflected by the cube's interior the force is always just $\frac{2Pcos(45)}{c}$ Since $\frac{2P}{c}$ is the rate of photon momentum being deposited, but it is deposited at a angle of 45 degrees from the horizontal so the horizontal force applied is $\frac{2P cos(45)}{c}$.

So considering this, the force is constant until the box moves far enough for the light to miss the box, which would occur after the box moves a distance L. So taking the initial position of the box to be 0, for x<L $\ddot{x} = \frac{2Pcos(45)}{mc}$, so $x(t) = \frac{Pcos(45)}{mc} t^{2}$. So putting in x = L, $t = \sqrt{\frac{Lmc}{Pcos(45)}}$.

So integrating force from t = 0 to this time: p = $\frac{2Pcos(45)}{c} * \sqrt{\frac{Lmc}{Pcos(45)}} = 2\sqrt{\frac{LmPcos(45)}{c}}$ is my thinking right?

 

[haruspex]

the photon would effectively deposit momentum into the box 4 times, where all the vertical components of momentum cancel out. The first and second reflected momentums should cancel

I could not understand this part. As you wrote, there are three reflections. The first and third are off horizontal surfaces, so are not relevant.
And I am unclear what you mean by the photons "depositing" momentum. E.g.

the rate of photon momentum being deposited, but it is deposited at a angle of 45 degrees from the horizontal

The photons transfer momentum, and the momentum transferred at the second reflection is horizontal, not at 45° to it.

But I agree with your final answers.

 

[dark_matter_is_neat]

I could not understand this part. As you wrote, there are three reflections. The first and third are off horizontal surfaces, so are not relevant.
And I am unclear what you mean by the photons "depositing" momentum. E.g.

The photons transfer momentum, and the momentum transferred at the second reflection is horizontal, not at 45° to it.

But I agree with your final answers.

Well since the light source is laser, there is a continuous stream of photons. Since the laser has a power P and the energy of a photon is $\frac{hc}{\lambda}$, the number of incident photons per second is $\frac{P \lambda}{hc}$. The momentum per photon is $\frac{h}{\lambda}$, so multiplying this by the number of photons per second, the momentum per second carried by the beam is $\frac{P}{c}$, this is what I mean by the rate at which the photons "deposit momentum" (which is probably a poor choice of words since this is different to the momentum added per second to the cube).

With photon momentum how I thought about it was momentum every time a photon with momentum p hits the cube, the cube gains momentum $\vec{p}$ and when a photon is reflected from the cube, the cube gains momentum -$\vec{p}$, so I only really need to worry about the ingoing and outgoing photons which have momenta $\vec{p_{in}} = p(cos(45), sin(45), 0)$ and $\vec{p_{out}} = p(-cos(45), sin(45), 0)$ where p is just the magnitude of the photon momentum, so the total momentum added to the block is just $\vec{p_{in}} - \vec{p_{out}} = \vec{p} = (2pcos(45), 0, 0)$