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Relativistic momentum and photons

  1. Nov 24, 2008 #1
    I have a question about photons and relativistic momentum. According to my Physics text (Serway & Beichner), the energy of a partle with zero mass, such as a photon, can be related by E[tex]^{}2[/tex]=p[tex]^{}2[/tex]c[tex]^{}2[/tex]+(mc[tex]^{}2[/tex])[tex]^{}2[/tex]. m=0 so the expression becomes E=pc. Since relativistic momentum is p=mu/[tex]\sqrt{}1(-u2^{}[/tex]/c[tex]^{}2[/tex]) and m=0 for a photon, how is there any momentum and thus energy?
  2. jcsd
  3. Nov 24, 2008 #2


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    That definition of momentum only works for things with nonzero rest mass, as you've noticed it becomes undefined (0/0) when you set m=0 and u=c. We can turn to quantum physics for the momentum of a photon--in QM we have E=hf where h is planck's constant and f is the frequency, so you can substitute this value of E into E=pc to find the photon's relativistic momentum.
  4. Nov 24, 2008 #3


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    Also note that in classical electrodynamics, an electromagnetic wave carries momentum in proportion to its energy, via E = pc, same as you get from the relativistic mass-energy-momentum relationship with m = 0.
  5. Nov 24, 2008 #4
    I agree with the posts above. A major trick in physics, that usually comes after studying different pehenomena in different situations, is "which formula(s) applies to the situation I'm working on now"....
  6. Nov 25, 2008 #5
    I would start with the transformation equations for the momentum and the energy of a particle moving with speed u' relative to I' presented as
    p=gp'(1+V/u') (1)
    E=gE'(1+Vu'/cc) (2)
    In the case of a photon (u'=c) (1) and (2) become
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