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Momentum and Light

  1. Aug 4, 2004 #1
    In classical mech since light has no mass it had no p cause p=mv with m at 0 you have no momenum nor a force correct, by take the der.

    But in GR, how is light considered to have momentum, I read somehting about Einstein saying that in the GR field EM spectrum could now be called matter. Wow interesting but how... just becuase EM now has momentum.

    Also found

    p = E/c which holds for a light-speed massless particle. is that E energy and if so pot or K or both just take - of the other.??

    can someone derive this?

    also in GR since the definition of momentum changed did also the defintion of a force. That would include all Em spectrum to apply a force also light shined from a flashlight.

    I know of radiation pressure felt by such things is this kind of like the force that is felt by a burst of photons falling on an object.

    By no means am I qualified in any of this stuff just trying to learn, so feel free to hammer anything I have messed up in assuming or if my definitons are lacking horribly. But if anyone could answer a little of this I would be greatly appreciative.
  2. jcsd
  3. Aug 4, 2004 #2
    I think this is how its derived...
    [tex]E = mc^2[/tex]
    [tex]E/c = mc[/tex] (mc is mass times the velocity of light, which is momentum)
    [tex]E/c = p[/tex]

    I think theres another way to derive it, by using the total energy equation.
    [tex]E^2 = m^2c^4 + p^2v^2[/tex]
    Since the photon is moving at velocity c, and has mass 0, we plug those in...
    [tex]E^2 = 0^2c^4 + p^2c^2[/tex]
    [tex]E^2 = p^2c^2[/tex]
    [tex]E^2/c^2 = p^2[/tex]
    [tex]E/c = p[/tex]
  4. Aug 4, 2004 #3
    wow, that was nicely done, thanx alot.
  5. Aug 4, 2004 #4
    E/c = P

    but what is E/c are we calling that light

    and what are the units for E/c
  6. Aug 4, 2004 #5


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    This is incorrect and a common misconception.

    If you open a good E&M text, (example: Jackson's Classical Electrodynamics), you will encounter a section on "radiation pressure" exerted by light. This came about even with a purely classical treatment of light (i.e. not as photons) - a lot of this work was attributed to P.N. Lebedev. Since "pressure" implies a "change in momentum", this clearly shows that even the classical version of light also contains a description of light having a momentum.

  7. Aug 4, 2004 #6


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    The E part of E/c is energy. The c part is the speed of light.

    Energy is measured in Joules (Newton-meters or kg*m^2/sec^2).
    The speed of light is measured in meters/sec.

    That puts your momentum into kg*m/sec.

    The E is found by multiplying the frequency times Planck's constant.

    Planck's constant is about 6.626 x 10^-34 Joule seconds.
    Frequency is measured in Hertz (cycles per second - the cycles are unitless, just as radians are).

    Since blue light has a higher frequency than red light, blue light has more energy than red light, etc.
  8. Aug 4, 2004 #7
    zapper you didint' read my whole post?
  9. Aug 4, 2004 #8
    change in momentum with respect to what..

    force is with respect to time right??? so is pressure the change in momentum with respect to position or what.
  10. Aug 4, 2004 #9


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    I did! You equate "radiation pressure" with photons, which isn't part of the classical picture. Thus, you stated that the classical idea contains no concept of momentum of light, since light has no mass. I disagreed with this by pointing out that with JUST using classical E&M, you can still show that light has a momentum without having to invoke modern physics into it.

  11. Aug 4, 2004 #10
    ok i'm sorry, your right

  12. Aug 4, 2004 #11
    what was it then?? just light, did they not know of photons, and is what is radiation pressure caused from if not by photons, i thought it would be the change in the momentum of photons with respect to position.
  13. Aug 4, 2004 #12


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    Er... you do know that F = dp/dt, ya? Furthermore, "pressure" is also proportional to this rate of change of momentum. When light impinges on a surface, if it is either absorbed, or reflected, it has undergone a change in momentum (simple classical mechanics). That is the "change in momentum" that I talked about.

  14. Aug 4, 2004 #13
    Im sorry zapperz,, really i don't wanna upset you, let me look around and learn some more before I ask questions, k, so no one gets frusterated, thanx though.
  15. Aug 4, 2004 #14


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    The treatment isn't trivial to describe on here if you haven't had basic E&M. Suffice to say that the E-field vector in light plays a significant role in generating a momentum effect when it impinges upon a surface. That is why this momentum transfer (from light to a surface) is most efficient when the surface is metallic (which is why most solar sails in Sci-Fi books are made of mylar). A metallic surface has "free" conduction electrons, and these electrons are the most easily effected by the oscillating E-field that's present in EM radiation. This interaction between the oscillating E-field in the EM radiation and the surface electrons imparts a "recoil" effect onto the surface. This is what produces the apparent momentum in the classical picture of light, without having to invoke any photon picture.

  16. Aug 4, 2004 #15
    The sad thing is I have had basic EM
  17. Aug 4, 2004 #16
    btw very good definition I see now.
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