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E=mc^2 relation

  1. Jan 17, 2016 #1
    1. From E=mc2 , m= E/c2
    "c" being huge, "m" should be small.

    2. m= mo/√ [1-v2/c2]
    if v= c, then m should reach infinity.

    3. Photons moving with speed of light are massless particles.

    All the above statements are confusing me. (1) doesn't match with (2), (2) doesn't match with (3).
    Please help.
    Thank you.
  2. jcsd
  3. Jan 17, 2016 #2


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    Why do you think (1) doesn't match with (2)? (2) explains why particles with a non-zero rest mass can never reach the speed of light.
    Equations (1) and (2) don't apply to photons.
  4. Jan 17, 2016 #3


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    This is a formula for the rest energy of a particle and its relation to the rest mass. The full dispersion relation reads ##E^2 = m^2 c^4 + p^2 c^2##.

    This is an expression for the "relativistic mass". It is a concept which has largely fallen out of fashion. When physicists talk about "mass", they almost exclusively refer to the invariant mass (or rest mass) of a particle. (See also What is relativistic mass and why it is not used much?)

    Yes, and when you plug that in to the general dispersion relation, you find that the total energy of a photon is given by ##E = pc##.
  5. Jan 17, 2016 #4

    Mister T

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    The problem with saying something is small is that it's meaningless without a comparison to something else. For a slow moving particle most of its energy is its mass, but for a fast moving particle very little of its energy is its mass.

    Photons move with the fastest possible speed. None of their energy is mass.

    Your item 2 is just an arbitrary definition of a quantity called relativistic mass, it applies only to massive (as opposed to massless) particles, offers no advantage to someone trying to learn relativity, is responsible for lots of misconceptions among people trying to learn relativity, and is disappearing from the lexicon. It's best to forget about trying to use it for anything.

    The total energy of a particle is given by ##E=\gamma mc^2##. Choosing to call ##\gamma m## is what you're doing in item 2.
  6. Jan 17, 2016 #5


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    Be aware that the ##m## in that equation is best understood as ##m_0##, the mass the particle has if it is at rest. That way you can use the dispersion relation that Orodruin mentioned, and the same math works for both massive and massless particles.

    And yes, because ##c## is large and ##c^2## is even larger, ##m## can be very small and you still have a lot of energy. There's a recent thread in which someone correctly pointed out the nuclear bomb that wrecked Hiroshima in 1945 converted only about .5 grams (that's less than the mass of a small paper clip) to energy.
    Unless, that is, ##m_0## is zero - as it is for photons and everything else that moves at the speed of light.
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