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What Happened to Potential Energy in E=m[tex]c^2[tex]

  1. Nov 30, 2005 #1
    I have been wondering why is potential energy not included in the calculation of total energy E=m[tex]c^2[tex]?
    I know that m[tex]c^2[tex] includes the rest energy and the kinetic energy. Suppose the particle or object is charged and is moving in a strong electric field, its potential energy will be sdgnificant wouldn't it?
    If we have to include the PE, how should we do it?
    Thanks
     
  2. jcsd
  3. Dec 18, 2005 #2

    Tom Mattson

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    The energy-momentum relation that you see in a standard general physics textbook is:

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

    It is typically advertised as the energy-momentum relation for a *free* particle (so [itex]V=0[/itex]). If memory serves, Goldstein's "Classical Mechanics" treats relativistic mechanics with a potential, via the Lagrangian formalism.
     
  4. Dec 30, 2005 #3

    krab

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    Properly,
    [tex](E-V)^2=(pc)^2+(mc^2)^2[/tex]
    where V is the potential energy and m, the rest mass, does NOT contain the kinetic energy.
     
  5. Dec 31, 2005 #4

    krab

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    Properly,
    [tex](E-V)^2=(pc)^2+m^2c^4[/tex]
    where m is the rest mass and V the potential energy
     
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