Estimating Laser Force: The Relationship Between Photon Momentum and Light Power

[azaharak]

Would it be reasonble to estimate the force delivered by a laser in the following way?

I was thinking of demonstrating that photons carry momentum by attempting to make a very light mass displace by imposing laser light on the object.

For a 5mW laser


Power = Force * Velocity

5mW/Speed of light = force that laser light imposes.

I know that its not a correct way of analyzing the physics, but could it crudley be used to give an approximate value?

Thanks

 

[Doc Al]

Yes, that would do OK, assuming that the beam is absorbed by the body (instead of being reflected). But think of it this way. The momentum of a photon is Energy/c. Thus the momentum per second delivered by the beam is Power/c, which would be the effective force of the beam. (For more on this, look up "radiation pressure".)

 

[Dr Lots-o'watts]

If there is reflection, the force is twice as large, so if you don't know the exact absorption or reflection, you still have two strict values between which the actual pressure lies (you know the area of the light beam, so you can relate force and pressure). You can simply see this as an uncertainty.

www.gentec-eo.com

 

[cupid.callin]

@ Dr lots

This won't give correct answer as nothing is completely absorbing or reflecting.

You may assume that it absorbs 50% and reflect 50%.
You can even find the force per photon by calculating energy of each photon (using wavelength)

you can also use this:

power, W=n x E (E= energy of photon)

momentum, p= h/(lambda) = hv/c = E/c
for all photons, p=nE/c = W/c

force, F = dp/dt

use this as you think how many are absorbed or reflected or even reflected at which angles!

* consider momentums perpendicular to surface!

 

[Dr Lots-o'watts]

If there is 100% absorption, the momentum is E(pulse)/c.
If there is 100% reflection, the momentum is 2E(pulse)/c.

In reality, the mass will be partly reflecting, and partly absorbing, so the momentum p transferred will strictly be between both values : E/c < p < 2E/c
(assuming perpendicular incidence of course).

http://maxwell.byu.edu/~spencerr/websumm122/node117.html

Unless you have a well equipped lab, I doubt you'll convincingly push anything with 5 mW though. You'd be better off with high energy laser pulses (on a mirrored membrane perhaps).

 

[kbaumen]

Not very helpful but nonetheless interesting in my opinion:

My high school physics teacher had a small vacuumed glass casing with a tiny metallic axle with three metallic wings, or rather sails on it inside the casing. If the casing was placed in sunlight, the contraption inside would start spinning purely from the momentum of the photons in sunlight. A clear indication of the fact the photons carry momentum.

 

[Doc Al]

Not very helpful but nonetheless interesting in my opinion:

My high school physics teacher had a small vacuumed glass casing with a tiny metallic axle with three metallic wings, or rather sails on it inside the casing. If the casing was placed in sunlight, the contraption inside would start spinning purely from the momentum of the photons in sunlight. A clear indication of the fact the photons carry momentum.

That's a Crookes' radiometer (also called a 'Light Mill'). It's not radiation pressure making it turn, as was first thought, but something more subtle. See: http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html"