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E = gamma mc^2

  1. Oct 13, 2014 #1
    Hi,
    How exactly does one derive ##E = \gamma mc^2##? Is this an expression for "total energy" contained in an object?

    The velocity of an object differs between frames of reference right? So doesn't that mean that E differs between frames as well? If it varies between frames can it even represent total energy?
     
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  3. Oct 13, 2014 #2

    A.T.

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  4. Oct 13, 2014 #3

    Nugatory

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    Yes. The total energy can be written as the sum of two components: ##m_0c^2## where ##m_0## is the rest mass; and ##(\gamma-1)m_0c^2## which is the kinetic energy. And as you point out:
    Even in classical mechanics, the kinetic energy is different in different frames. It's the same physics when a 1000 kg elephant moving at 1000 m/sec hits a stationary bullet weighing .1kg as when a .1 kg bullet moving at 1000 m/sec hits a stationary elephant; that's just using a frame in which the bullet is at rest instead of the elephant. But the kinetic energy will be different by a factor of 10000.
     
  5. Oct 17, 2014 #4
    Thank you for the replies,

    Nugatory, correct me if I'm wrong. ##E=mc^2## can be derived from an entirely independent thought experiment. Then, using ##E_{tot}=\gamma mc^2## we can find kinetic energy by subtracting rest mass. ##KE =E-E_0=\gamma mc^2 - mc^2 = (\gamma - 1)mc^2##

    So then, the real question is where does ##E_{tot}=\gamma mc^2## come from?

    My book on relativity makes the following argument.
    The square of four momentum is Lorentz invariant
    If you multiply this quantity by ##c^2## we get units of energy squared.
    THUS, ##c## times the magnitude of 4 momentum equals total energy!

    Kinda vague.
     
  6. Oct 17, 2014 #5

    jtbell

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  7. Oct 19, 2014 #6

    stevendaryl

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    There are lots of derivations of [itex]E = mc^2[/itex]. I would say that there are three broad types of derivations:
    1. Mathematical derivation using properties of Lorentz spacetime and/or langrangian mechanics
    2. Using properties of photons.
    3. Using collisions and Newton's laws.
    The first type is a little abstract. It basically amounts to showing that [itex]E = \gamma m c^2[/itex] and [itex]\vec{p} = \gamma m \vec{v}[/itex] are the only possibilities consistent with Lorentz covariance. Those types of derivations can be made perfectly rigorous, and have the advantage of being to state precisely what are the assumptions behind them.

    The second and third derivations are more intuitive, but are a little hand-wavy.

    The photon derivation assumes that you already know that a photon (or pulse of light---it's not quantum-mechanical) has energy [itex]E[/itex] and momentum [itex]p[/itex] related by: [itex]E^2 = p^2 c^2[/itex], and that these quantities transform as a Lorentz 4-vector. Then you can show that if a massive object absorbs a photon of energy [itex]E[/itex], then its rest mass must change by [itex]E/c^2[/itex].

    The collision derivation starts with the assumption that:
    1. The energy of a particle must be some function of velocity and mass and should be proportional to the mass, and should only depend on the magnitude of the velocity, rather than the direction. So there is some function [itex]g[/itex] such that [itex]E = m g(v)[/itex].
    2. The momentum of a particle must be some function of velocity and mass, and should be proportional to the mass. Furthermore, it should be in the same direction as the velocity. So there is some other function [itex]f[/itex] such that [itex]\vec{p} = m \vec{v} f(v)[/itex]
    3. In the limit as [itex]v \rightarrow 0[/itex], the momentum must reduce to the Newtonian case, [itex]\vec{p} = m \vec{v}[/itex], which implies that [itex]f(0) = 0[/itex].
    4. In a collision of two objects, total energy and momentum are conserved.
     
  8. Oct 19, 2014 #7

    robphy

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    It seems to me that the work-energy theorem
    is a relation between the net-work done by the forces on the object
    and the change in Kinetic-Energy (not the change in Energy,
    in spite of this common name for the theorem.).
    In this scheme, one first obtains the kinetic-energy, then constructs the relativistic-energy.
    I made a longer comment in that thread.
     
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