Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon's momentum

  1. Sep 1, 2010 #1
    Photon has momentum but is massless.Doesn't that seem strange?
     
  2. jcsd
  3. Sep 1, 2010 #2
    If you write down the fully relativistic equation for momentum (in pc units) for any particle (including photons) of kinetic energy T, the quation is

    (pc)2 = (T + mc2)2 - (mc2)2

    = T2 + 2(mc2)T

    So, for a bullet use the second term; for a photon use the first (for mass = 0). So a massless photon has momentum.

    Bob S
     
  4. Sep 1, 2010 #3

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c can only have zero mass.
     
  5. Sep 1, 2010 #4
    No it's not strange because photon has energy. So its momentum is [tex]p=E/c=h/\lambda[/tex]

    (ok it is little bit strange...)
     
  6. Sep 1, 2010 #5
    No.

    "Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c [STRIKE]can only have[/STRIKE] may have zero mass".

    Also, in Newtonian mechanics, p = sqrt(2mE) where E = ½mv2

    Bob S
     
  7. Sep 1, 2010 #6

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    No, not 'may', 'must'. The four-momentum of a particle is tangent to it's wordline for a particle travelling at c this means it's four-momentum is always null. The mass of particle (in the absence of a rest frame to define it) can be taken as the norm of it's four-momentum which is zero for a particle with null four-momentum. Hence all particles travelling at c MUST have zero mass (unless we're going to allow particles to have undefined four-momentum and hence undefined momentum and energy).

    In Newtonian mechanics a particles momentum is defined as p(t) = mv(t).

    Your definition is clearly incorrect as momentum is a vector quantity whereas your definition defines a scalar quantity.
     
  8. Sep 1, 2010 #7

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    Exactly.

    [tex]v = c \sqrt{1-\frac{m^2 c^4}{E^2}}[/tex]
     
  9. Sep 1, 2010 #8
    You are correct.

    My pocket calculator shows that a 3.5 TeV proton in LEP has a v/c =

    β = 1-1/2γ2 = 1-3.6·10-8 = 0.999 999 964

    For the 1020 eV Oh My God cosmic ray proton

    β = 1-1/2γ2 = 1-5·10-23 = 0.999 999 999 999 999 999 999 9 (more or less)

    So any particle with mass >0 cannot have a velocity of exactly c. I was not careful enough in rounding off.

    Bob S
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook