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Derivation of E=mc^2

  1. May 3, 2014 #1
    Reading into some special relativity, I have seen E=mc^2 proposed from the assumption of four momentum conservation and the fact that the 'mass' component varies with velocity with the gamma factor, like a kinetic energy.

    This seems a bit of a leap of faith to me so I was wondering if there are more rigorous derivations preferably only assuming special relativity?
  2. jcsd
  3. May 3, 2014 #2


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  4. May 3, 2014 #3
    As is the case with most things in physics, seemingly arbitary quantities arise naturally from a lagrangian approach.
    If you start with the relavistic free lagrangian, you will see that the famous (and somewhat misunderstood) term pops out of the hamiltonian upon a power expansion.
  5. May 3, 2014 #4

    Jano L.

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  6. May 3, 2014 #5


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  7. May 8, 2014 #6
    World's Quickest Derivation of E = mc2 (not written to satisfy classical physicists like Newton)
    Adapted from a derivation posted on the internet by John D. Norton
    Department of History and Philosophy of Science, University of Pittsburgh

    Consider a body of mass m moving at a velocity very close to c. A constant force F acts on the body in the same direction as its motion for a duration set at unit time interval delta T (=1), and as a result, the force increases both the energy and momentum of the body. The force cannot increase its speed because it is already c, so all of the increase of momentum = mass x velocity of the body manifests here mathematically as an increase in mass. This thought experiment dispenses with any references to center of mass for Newton's benefit.

    Side note to particle physicists: Of course, mass = rest mass + kinetic energy, expressed in units of energy. No one expects that a particle changing relativistic speed actually gains or loses inertial mass. The Higgs field pushes back to keep speeds below c for objects that have mass, including itself (by giving mass to them). But it would be impractical to use (delta P) = m x (delta v) because c can't change, and (delta v) would then be zero. By the same token, a single photon has no defined energy or even existence relative to an observer who knows nothing of its origin, but a pair of photons do. This is the raison d'etre for any observer-based version of relativity, expressed without the obligatory math. This principle works for non-virtual particles also.
    A basic rule in quantum physics is: If you look, it's a particle; if you don't look, it's a wave.

    Note the choice of unit time in this derivation (as opposed to unit c, popular among classical physicists including Einstein himself).

    We want to show that in unit time interval delta T, the energy E gained due to the action of the Force applied is equal to mc^2, where m is the mass gained by the relativistic projectile, and c is the speed of light. For this example, E = delta E, and m = delta m.

    We have two relations involving Energy, Force, and Momentum over the unit time interval delta T (=1):

    The first relation is about the change in energy in unit time interval delta T, (or 'work' in classical physics):

    delta E = E = Force x (distance through which force acts) = Force x (c x delta T), or E = Force x c (1-1)

    The second relation is about the change in momentum P in unit time interval delta T (or 'impulse' in classical physics):

    delta P = (delta m) x c = m x c = (Force) x (delta T), or m x c = Force (1-2)

    Which means: delta E = E = Force x c = (m x c) x c

    In other words (QED): E = mc^2 (0-0)

    We now see that c^2 = c x c derives from relating energy to distance and momentum to time in the same expression.
    Last edited: May 8, 2014
  8. May 9, 2014 #7
    My favourite derivation follows. There are only 2 assumptions.

    1. Let's assume that time is the fourth dimension, much like the length, width and breadth. Time is perpendicular to other dimensions.

    The particle trajectory in the 4 dimensional spacetime is called a world line.

    The immediate collorary is that speed is related to an angle between the world line and the "space" part. Particles at rest have their world line ideally perpendicular to the space (zero angle to the time axis). Moving particle's world line has a greater angle relative to the time axis and smaller to the space.

    2. The second assumption: the speed of light is exactly the angle of 45 degrees between time and space.

    This assumption lets us to relate meter and second. If time is the same as space then interval is the same as length so they should be measured in the same units. Time could be measured in meters, length in seconds. The speed of light allows us to find a conversion factor from time to length and it is exactly:

    [m] = c

    "Meter" is a "second" times the dimensionless numerical value "c", equal to the speed of light in meters per second.

    Now consider the joule definition.

    [J] = [kg] [m]^2 / ^2

    If a meter is just a scaled second then a joule is just a scaled kilogram.

    [J] = [kg] c^2

    Energy is the same thing as mass, just in different units. The scaling factor is the dimensionless constant c^2, equal to the squared value of the speed of light in meters per second.

    The identification of energy and mass comes from identification of time and space.

    Similar identifications can be done for mass and momentum, momentum and energy, electric current and magnetic field, and between mass density, pressure and tension.
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