Impact Force of a Photon

[CycloWizard]

I know that the energy of a photon is E=h*nu, where h is Planck's constant and nu is the frequency. I was wondering if anyone knew what the corresponding force due to impact would be.

The only derivation I can readily find online is an abstract giving the following derivation:
F=dP/dt=d(m*c)/dt for an impulse. Letting E=m*c^2, we get F=1/c*d(m*c^2)/dt=1/c*dE/dt. However, -dE/dt=P. Therefore, |F|=h*f^2/c=hc/lambda^2=E/lambda.

The abstract I linked to is from 2004, though it seems like this must have been derived many years ago. The referenced cite doesn't appear to exist anymore, so I can't really verify any of the references. I thought there were some people here who could tell me whether or not this is really correct.

edit: fixed link

 

[silverpig]

Looks fine to me. The units of force are Newtons, which are Joules/meter. E is in Joules, lambda is in meters. The units work. Also, a more energetic photon (with a shorter lambda) will produce more force. Seems right.

 

[Born2bwire]

Originally posted by: CycloWizard
I know that the energy of a photon is E=h*nu, where h is Planck's constant and nu is the frequency. I was wondering if anyone knew what the corresponding force due to impact would be.

The only derivation I can readily find online[/lhttp://[/L] may not be accessible to everyone, but it goes like this:

F=dP/dt=d(m*c)/dt for an impulse. Letting E=m*c^2, we get F=1/c*d(m*c^2)/dt=1/c*dE/dt. However, -dE/dt=P. Therefore, |F|=h*f^2/c=hc/lambda^2=E/lambda.

The abstract I linked to is from 2004, though it seems like this must have been derived many years ago. The referenced cite doesn't appear to exist anymore, so I can't really verify any of the references. I thought there were some people here who could tell me whether or not this is really correct.

Have you tried looking under "radiation pressure?" I remember doing this for a classical plasma and that was the phrase describing the phenomenon. The only problem is that this was classical electromagnetics.

 

[firewolfsm]

Look under equations for solar sails as well. Not sure if they deal with single photons though.

 

[CycloWizard]

I had read about the pressure and momentum of given photon intensities previously for things like optical tweezers, but didn't remember seeing anything about the contribution of a single photon.

 

[f95toli]

There is a fundamental problem with the question:

A photon carries momentum, with p=hf/c.
Now if a photon hits e.g. a wall and is reflected back it will impart a momentum 2p.
But this is as far as you can get. There is no way to calculate the force unless you know more about the "collision process" (or you are dealing with many photons, i.e a flux, which is a rate).

What you are asking if akin to the following question: a ball with mass m is traveling at speed v and hits a wall, what is the force on the wall?

btw, E=mc^2 is not valid for a massless particles such as the photon.

 

[CycloWizard]

Originally posted by: f95toli
There is a fundamental problem with the question:

A photon carries momentum, with p=hf/c.
Now if a photon hits e.g. a wall and is reflected back it will impart a momentum 2p.
But this is as far as you can get. There is no way to calculate the force unless you know more about the "collision process" (or you are dealing with many photons, i.e a flux, which is a rate).

What you are asking if akin to the following question: a ball with mass m is traveling at speed v and hits a wall, what is the force on the wall?

btw, E=mc^2 is not valid for a massless particles such as the photon.

I see what you're saying. I was just curious why this wasn't readily available information, but now that I think about it that way, it makes perfect sense.

 

[Comdrpopnfresh]

yeh- you can't quantify force without out knowing a lot of contextual settings. In collision dynamics, there is a transfer of momentum based on elastic or inelastic collisions and such. F=ma is a fundamental, so force isn't practical because photons are massless. Even with solar sails, you'd be looking at radiation pressure- and that would once again be a conservation of momentum.
The underlying problem is that light is moving at the same speed from any given reference frame, so if light-comprised of photons- is always traveling at 3x10^8 m/s you have no change in acceleration, and a=0. With a=0, m could be inf or 0 depending up usage and purpose. in either case, F is not a rational concept with photons.

 

[MrDudeMan]

Originally posted by: Comdrpopnfresh
yeh- you can't quantify force without out knowing a lot of contextual settings. In collision dynamics, there is a transfer of momentum based on elastic or inelastic collisions and such. F=ma is a fundamental, so force isn't practical because photons are massless. Even with solar sails, you'd be looking at radiation pressure- and that would once again be a conservation of momentum.
The underlying problem is that light is moving at the same speed from any given reference frame, so if light-comprised of photons- is always traveling at 3x10^8 m/s you have no change in acceleration, and a=0. With a=0, m could be inf or 0 depending up usage and purpose. in either case, F is not a rational concept with photons.

My physics professor in college worked out the force on the earth due to light from the sun before class one day and he showed it to us. IIRC, it was approximately 2 pounds per square mile of unobstructed sunlight.

 

[Comdrpopnfresh]

That's radiation pressure. A photon has no mass, so the force of the impact of one single photon is useless. You could find the flux of that stream of light, and find the energy imparted on the earth, but not the force of a photon impact- because they don't