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Einstein's potential energy equation

  1. Oct 11, 2003 #1
    What happened to the 1/2 in the transformation of classical potential energy equation to Einstein's potential energy equation. Is it dropped because giving off all rest energy would require annhilation of a particle pair thereof 1/2mv2+1/2mv2=1mv2 gets rid of the half?
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Oct 11, 2003 #2


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    Re: question

    What do you mean by "Einstein's potential energy equation"? I don't understand the question. Do you mean to compare mc^2 with (1/2)mv^2? Apples and oranges?
  4. Oct 11, 2003 #3
    Potential energy doesn't equal 1/2mv^2(that's kinetic energy) it equals mgh

    I think you are asking about the energy E = mc^2 equation relating to classical systems.

    First E = mc^2 is only for stationary objects. The actual equation is E = mc^2/[1-((v^2)/(c^2))]^1/2(you see now why when discussing it in general people assume v = 0)

    Einstein's equation deals with the total energy of an object at any speed. and the derivation can be found in most Modern Physics books( a convineince for me because I don't remember it off the top of my head)
  5. Oct 11, 2003 #4


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    kinetic energy from Einstein

    where m is rest mass.
    Expand Lorentz term in power series:
    Net result:
    The first 2 terms are the energy due to rest mass and the kinetic energy.
  6. Oct 16, 2003 #5
    whats so difficult about my question krab all i was asking was how come there is no half in the rest energy equation.
  7. Oct 16, 2003 #6


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    Well, then it's been competently answered by Mathman and VBPhysics.
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