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Photon Momentum

  1. Dec 27, 2004 #1
    Hi all,

    When thinking about equations of momentum for photons and i came up with this problem.

    Say you take the equation E=mc^2 and re-arrange it so m is the subject, m=E/(c^2)

    Now the Energy of a photon is also given by the equation Ep=hf and the 'mass' can therefore be said as m=hf/(c^2)

    Now momentum is said to = mv so the momentum of the photon should =hf/c.

    My question is this, how is it this equation works when it relies on the fact that E=mc^2; as photon has no mass should it not have any energy? (Which it clearly does). I am also confused by the fact that is combines to separate energy equations i.e. Ep=hf which is designed to overcome the problems that occur when calculating the energy of a photon with E=mc^2.

    Is this due to the fact that the mass is equivalent to the energy of the photon as in Relativity?

    PS: Does this forum support LaTeX, and if so any resources on using it would be much appreciated.

    Thank you for any help in advance :smile:
    Last edited: Dec 27, 2004
  2. jcsd
  3. Dec 27, 2004 #2
    Your question is a fairly common one, and stems from a misunderstanding of the relativistic energy-momentum relation. The full equation is

    [tex] E^2 = (pc)^2 + (m_0 c^2)^2 [/tex]

    For a photon the rest mass term vanishes, and you get E=pc.

    This being said, it begs the question "How can something that is massless have momentum?" The answer to that is that momentum in relativity is a more general concept than momentum in classical mechanics, and in this more general form it can be carried by massless particles. Relativity itself cannot differentiate photons of different momentum/energy (i.e. it offers no property that can differentiate a radio wave photon from a gamma ray); this task belongs to relativistic quantum mechanics.

    In regards to the other question, this forum has extensive Latex support and you can read all about it in this thread: https://www.physicsforums.com/showthread.php?t=8997
  4. Dec 27, 2004 #3
    Thank you that help alot :biggrin:
  5. Dec 28, 2004 #4


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    Do recall that classically, the E&M field has momentum -- the Poynting vector E x B gives the momentum density -- and energy -- the energy density is 1/2(E*E + B*B). Most any E&M text will walk you through the Fourier transform analysis of the above. This analysis is but a stone's throw from the picture shown by photons.

    From experiments like Compton scattering, and photoproduction of mesons, and so forth show that, indeed, photons carry momentum and energy, just as the theory explains.

    Reilly Atkinson
  6. Jan 1, 2005 #5
    Right. That's what many relativistists such as A.P. French calls the inertial mass of the photon, e.g. Special Relativity footnote page 16
    Yes. That is correct as long as you're referring to the inertial mass of the photon defined as m = p/v.
    So long as m = inertial mass and v = 3-velocity then yes, hf/c = 3-momentum.
    You're using the term "mass" in two different contexts and thus you mean two different things when you use them. Its parallel to the difference between proper time and time. Particle physicists will often use the term "lifetime" of a particle when they really mean "proper lifetime". In this case the "proper mass" of the photon is zero while its mass is not zero

    In special relativity inertial mass, m is defined such that mv is a conserved quantity. mv is then called the momentum of the particle.

  7. Jan 2, 2005 #6
    Actually, in a sense it is. (As alluded to by Reilly), classically Maxwell's E&M says the total momentum, p, imparted to an object resulting from incident light which is totally absorbed is directly proportional to the total energy, E, absorbed during a period of time, and given by:

    P = E/c

    Creator :biggrin:
    Last edited: Jan 2, 2005
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