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Photon momentum and lasers transfering energy

  1. Apr 9, 2008 #1
    This is something that I was just playing around with in my head that I was curious about.

    Photons have momentum (p = E/c). That would mean that they could transfer that momentum, and I was curious about collisions between a lot of photons and large objects.

    I was curious if anybody here had any idea how one could calculate the kinetic energy of an object that had a laser pointed at it or something like that.

    So, I guess here's my problem that I'm proposing that I'm curious about.

    You have a mass (not a particle like an electron or something like that, a block or something like that) and you point a laser at it (assuming you could get sufficient power out of this laser to move this object, and you can figure out things like wavelength, frequency, etc about the light emitted), what is the kinetic energy experienced by the mass?

    I appologize if this doesn't make sense. If it does not, please ask me to rephrase.

    Thanks for considering my random thoughts :D
     
  2. jcsd
  3. Apr 9, 2008 #2
    [tex]F = \frac{dp}{dt} = 1/c \frac{dE}{dt} = P/c[/tex]

    Where P is power, p is momentum, t is time, F is force, E is energy, c is speed of light

    So, perhaps I'm just making this up, but could you seriously do F = ma with this when you know the power output of a laser?

    Perhaps my failure in logic here is assuming dE/dt = P, and also that you can't just say that the power that I'm using here is equal the the power output of the laser.

    I'm completely making this up. I honestly have no clue. If I'm being stupid, please correct me. I have no idea what I'm doing.
     
  4. Apr 9, 2008 #3
    Oh! So I read something about radiation pressure.

    I was close for force on the object

    [tex]F = \frac{Q P}{c}[/tex]

    Where Q is the "coefficient of radiation pressure," P is the "effective power" which changes for distance from an object, and c speed of light.

    Now I guess my concern is that I don't really understand this. Could someone explain this all to me?
     
  5. Apr 9, 2008 #4

    dst

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    You are actually correct.

    Yes, you can do the whole F=ma thing - conservation of momentum ALWAYS holds regardless of what function defines that momentum. So even though light has a different form of momentum to ordinary matter, momentum of light in = momentum of mass out.

    And F=p/c is good enough I believe. Use the ridiculously famous mass-energy equivalence. The only issue is that photons don't "accelerate". Are we talking about a mirror? That would make things easier. The photon transfers momentum 2p - once on impact and once on reflection. The small amount of momentum transferred corresponds to some red-shifting of the photons because they lose energy but velocity stays constant.

    Q I have no idea about, P you can take as your laser power since that factor includes silly things like divergence and attenuation, things that don't matter in our hypothetical world.
     
  6. Apr 9, 2008 #5
    Check this out,

    Say you had a mirrored cube of mass m, and you wanted to suspend it via laser :D

    [tex]\frac{2P}{c} = mg[/tex]

    You use 2 because of that coefficient and the fact that there's the whole impact and then reflection thing going on. The 2 implies that the photons are reflecting. 1 means they're being absorbed (only impacting), and 0 means that the photons passed through (no impact).

    So, let's see, that means if you wanted to suspend a 1 kg object, you'd need about 1.5 * 10^9 W of power :D

    Time to break out the 1.5 gigawatt laser.

    Of course there's some terribly flawed logic in there, ignore that.
     
    Last edited: Apr 9, 2008
  7. Apr 9, 2008 #6
    Nice theory, but in reality I dont think the glass block would last very long under that kind of irradiation
     
  8. Apr 9, 2008 #7

    Mapes

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    I did this earlier this week with my own red blood cells (video shows two optical fibers pointing at each other). The laser intensity was 10 GW/m2. So WraithM, your ideas are pretty right-on.
     
  9. Apr 9, 2008 #8

    dst

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    Frankenstein!?!?! Long time no see, damn.
     
  10. Apr 9, 2008 #9
    Gareth, if you had a perfectly reflective mirror, it will hold up under the radiation. That's sort of what I assumed during this whole idea. Of course, there is no such thing as a perfectly reflective mirror, so the specific situation I created is a little nutty. However, the theory still applies to a lot of stuff that isn't exactly in the situation that I proposed. It was more of a proof of concept thing. It's pretty clear that 1.5 gigawatts would destroy pretty much anything, however! Allow me to propose a method to cut the power significantly.

    I was reading a blog about this that I found, and I read about this:

    You can multiply the force tremendously by bouncing the laser beam through two mirrors parallel to eachother at an angle to have multiple points where the laser beam hits the mirrors.

    So, you'd have a situation where you have a 1 kg mass with a mirror on it, and then another mirror that is anchored parallel to the 1 kg mass mirror. You fire the laser at such an angle to make it so the laser has 1000 contact points on the 1 kg mass. For every one of those contact points, it is going to add 2P/c force to the mass.

    So you get this (if you were trying to suspend the block):

    [tex]\frac{2NP}{c} = mg[/tex]

    where N is the number of points of reflection on the mass

    So therefore,

    [tex]P = \frac{mgc}{2N}[/tex]

    So, that's 9.8 x 3 x 10^8 / 2000 = 1.47 * 10^6 W

    A mere 1.47 MW laser, that's a HUGE difference.

    Of course, the laser will begin to fail if you add too many points of reflection.

    But if you did this for a 1 g mass, it'd only be 1,470 W. That's just your plain old kilowatt laser that you could find in pretty much any factory in america. :D

    Of course this is still really difficult to do, because 1.47 kilowatts is still an absurd amount for a laser, but the idea is that there are ways to make this actually practical.

    But any ways, that's just my crazy ranting.
     
  11. Apr 9, 2008 #10
    Were you having some sort of surgery done? Or were you just suspending your red blood cells for good times?
     
  12. Apr 9, 2008 #11

    Mapes

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    Part of my research is on the elastic properties of cells. As I improve this setup, I'll be able to zoom in much closer and look at how the cells deform as they're held in place by the laser beams.

    The easiest way to get cells was just to prick my finger, squeeze a drop of blood into a vial of saline, and put a drop of the diluted mixture on the gap between the optical fibers.

    It was good times!
     
  13. Apr 9, 2008 #12
    That's so awesome!

    Oh man, I wish I could do something sweet like that :(

    Are you a biologist? And where are you doing this research? Just out of curiousity.
     
  14. Apr 9, 2008 #13

    Mapes

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    Graduate student, materials science. Feel free to see my home page for more information on the cell work.
     
  15. Apr 10, 2008 #14
    Wraith M,

    Point taken, this is hypothetical.

    But I can't quite grasp the multiple reflection bit;

    Does any of this conflict with the laws of thermodynamics? (you can't get something for nothing)
     
  16. Apr 10, 2008 #15

    dst

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    You aren't getting anything for nothing, you'll come out with a bunch of slightly redder photons.
     
  17. Apr 10, 2008 #16
    so if I reflect light from a perfect reflector the wavelegth changes?
     
  18. Apr 10, 2008 #17

    dst

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    So says energy/momentum conservation and the constant nature of c.
     
  19. Apr 10, 2008 #18
    I see, but you are assuming that the total momentum is imparted on the glass (X2 to account for absorption and reflection) so by this reasoning would the reflecting particle have any momentum?

    (I know this works, solar sails etc. but I never came across a red shift from plane reflections)
     
  20. Apr 10, 2008 #19

    dst

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    The reflecting particle would still have momentum, yes. The change in momentum corresponds to a tiny amount of energy lost and hence comes red-shift.
     
    Last edited: Apr 10, 2008
  21. Apr 11, 2008 #20
    i see, so the red-shift is negligably small generally
     
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