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Light mass?

  1. May 26, 2007 #1
    Wikipedia says that photons have momentum. How can that be possible since p=mv?
  2. jcsd
  3. May 26, 2007 #2

    The Poynting vector,S, gives momentum of light, and is given by:

    S = E X B........E= electric field
    B= magnetic field
    X is cross product

    However, this is the classical answer....and maybe this answer should be in classical section.:smile:
    Last edited: May 26, 2007
  4. May 26, 2007 #3

    George Jones

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    According to relativity, the spatial momentum of a "particle" that has rest mass [itex]m[/itex] is

    [tex]p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}.[/tex]

    To consider a photon, let [itex]m \rightarrow 0[/itex] and [itex]v \rightarrow c.[/itex] This gives zero over zero, which is indeterminant, and can be anything (including non-zero values), depending on how the limits are taken.

    Also, relaltivity says that for any particle,

    [tex]E^2 - (cp)^2 = \left(mc^2\right)^2.[/tex]

    Even if [itex]m=0[/itex], [itex]p[/itex] is non-zero when [itex]E[/itex] is non-zero. In fact,

    [tex]E = cp[/tex]

    for a photon.
  5. May 26, 2007 #4
    i wish that i was in a higher physics class so that i could understand you.
  6. May 26, 2007 #5
    The inertial mass M (aka relativistic mass) of a photon is defined just like the mass of all particles, as M/i] = p/v = p/c where p is the magnitude of the momentom and v is the particle's speed. I believe that you're thinking about a photon's proper mass m, which is zero for all photons.

    The definition of the mass of a particle allows the following relation to be derived (let m = proper mass)

    p = sqrt[1-(v/c)sup2]/c

    This can be solved for m which in turn will be a function of the energy and momentum and when E = pc me will always turn out to be 0.

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