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Derivation of E=mc^2 in Wikipedia

  1. Jan 11, 2009 #1
    Wikipedia derives the kinetic energy of a rigid body at relativistic speed to be

    [tex]E_k = m\gamma c^2 - m c^2[/tex]​

    The continue to say:
    Can anybody explain this reasoning? Just because the zero value of the kinetic energy at zero speed has the representation [itex]x - x[/itex] does not mean that the rest energy must be [itex]x[/itex], right? Or is that the reasoning?

  2. jcsd
  3. Jan 11, 2009 #2


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    The reasoning is just that the energy required to accelerate a particle from 0 to v is equal to the change in the quantity [itex]\gamma m c^2[/itex], so it makes sense to think of that quantity as representing a kind of energy.

    I think you have to look at conservation of four-momentum to really justify the equivalence between mass and energy.
  4. Jan 12, 2009 #3
    There's an important distinction here; Wikipedia calculates the kinetic energy as what you listed above, however, the rest energy is not the kinetic energy, obviously, because the kinetic energy is obviously zero when v = 0.

    So what they mean is that
    [tex]E_t = E_k + E_m = m\gamma c^2 - m c^2 + m c^2 = m\gamma c^2[/tex]

    Or, equivalently,

    [tex]E_r = E_m = E_t - E_k = (m\gamma c^2) - (m\gamma c^2 - m c^2) = m c^2[/tex]

    Where [tex]E_m[/tex] is the mass energy, [tex]E_t[/tex] is the total energy, and [tex]E_r[/tex] is the rest energy.

    Sorry if that was confusing.
  5. Jan 12, 2009 #4
    Don't worry, this is not confusing. But it somehow gets me where I started. I was after a simple derivation for E=mc^2, found a link to Wikipedia and now I am back with [itex]E_t=m\gamma c^2[/itex] asking where this comes. Hmm, yes, trivially it is [itex]E_t = E_k + E_m[/itex]:rolleyes: So I need to look closer to other recommendations in this other thread, I am afraid.

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