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B Confusion on Momentum of Light

  1. Aug 11, 2016 #1
    In E2=m2c4+p2c2
    how do you prove light has momentum? I've tried but my answer comes to be undefined. Light has m=0, so you're left with p2c2. p=(mv)/sqrt(1-v^2/c^2) When substituting values of light you get that p=0/0. How do you prove light has momentum using math and equations?
     
  2. jcsd
  3. Aug 12, 2016 #2
    The expression for momentum for photons in QM is ##p=\hbar k##. Where ##k## is the wave number (##k=\frac{2\pi}{\lambda}##) and ##\hbar## is the Planck constant divided by ##2\pi## . Or equivalently, ##p=\frac{h}{\lambda}## . Your equation for ##p## is for massive particles with non zero rest mass.
    Edit: Also note that in your expression for p if you put v=c (photon) then you get a division by zero! It is good to check for these stuff when you are doing calculations.
    Edit 2: Sorry I thought that this was posted in QM section. Apologies.
     
  4. Aug 12, 2016 #3

    andrewkirk

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    In relativity, p is not necessarily obtained by multiplying mass by velocity.

    For a photon, instead we work with the four-momentum, the first component ##p^0## of which, in any reference frame, is the energy of the photon in that frame. The other three components ##p^1,p^2,p^3## are momenta in the directions of the three spatial axes of the frame. We want the four-momentum to be a null vector (vector of zero magnitude), because light travels along null geodesics. So that means that not all the other three components of the four momentum can be zero - since the first component (Energy) ##p^0## is nonzero. If we rotate the frame so that the x spatial axis points in the direction of the photon's travel, we know that the other two momentum components ##p^2,p^3## must be zero, and we conclude that ##p^1=p^0## where ##p^0=E=h\nu## is the photon's energy.
     
  5. Aug 12, 2016 #4

    Orodruin

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    This is a relation that only holds fot massive particles. You cannot use it for light.

    Yes, and so you are done. If light carries energy, then it carries momentum p=E/c.

    You don't. This is not how physics works, it is an empirical science. Your theory makes a prediction and you test it in the lab.
     
  6. Aug 12, 2016 #5

    Orodruin

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    I strongly suspect that this will pass over the head of the OP in a B level thread ...
     
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