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E = mc^2, so how can anything have zero mass?

  1. Sep 10, 2013 #1
    E = mc^2 results in c^2 = E / m, and division by 0 is undefined.
     
  2. jcsd
  3. Sep 10, 2013 #2
    Oops, my bad. Should be in the Special & General Relativity section.
     
  4. Sep 10, 2013 #3

    Dale

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    Done.
     
  5. Sep 10, 2013 #4

    Nugatory

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    ##E=mc^2## is a special case of the more general ##E^2=(m_0{c}^2)^2+(pc)^2##. The massless particles have ##m_0## equal to zero but ##p## non-zero.
     
  6. Sep 10, 2013 #5

    russ_watters

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    E also equals FD: so how can force or distance ever be zero?
     
  7. Sep 15, 2013 #6
    E = mc^2 only applies in the special case where the particle in question is in its rest frame. Photons do not have a rest frame.
     
  8. Sep 15, 2013 #7
    The 'p' in that is momentum, correct?
    Yet momentum is m * v, mass times velocity. I understand that photons have a velocity of c, but their mass...? If you use the argument that it isn't talking about rest mass, but total mass, then isn't it sort of circular logic?
     
  9. Sep 15, 2013 #8

    ZapperZ

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    Please read the Relativity FAQ subforum.

    https://www.physicsforums.com/forumdisplay.php?f=210 [Broken]

    Zz.
     
    Last edited by a moderator: May 6, 2017
  10. Sep 15, 2013 #9

    Dale

    Staff: Mentor

    Not in general. Photons have momentum but 0 mass, and for massive objects momentum is unbounded as v approaches c.
     
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