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Quiz: E = m * c^2

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  • Quiz: E = m * c^2

    It's quiz time again.

    Derive E = m * c^2

    1st year college level. Less than 1 page. Simplified derivation.

    To the physicists out there who already know the answer, Pls PM me your answer.

    http://aardvark2z.blogspot.com/
    .

  • #2
    Who amougst you is a brave soul. We need a hero

    Here it is,

    Note: the equations are annotated as if in a computer program

    Et = Energy total

    Eo = Energy at rest
    Ev = Energy kinetic

    m = mass
    c = speed of light
    v = speed of object

    Find Eo = m c^2

    Knowing
    Et = Eo + Ev [eqn1]

    and
    Ev = 0.5 m v^2

    also, from special relativity (note 9)

    Et = Eo / ( 1 - (v/c)^2 )^0.5 [eqn2]

    from [eqn1] and [eqn2]
    Eo + Ev = Eo / ( 1 - (v/c)^2 )^0.5

    devide by Eo

    1 + Ev / Eo = 1 / ( 1 - (v/c)^2 )^0.5 [eqn3]

    we know that the following equation has a series equivalent
    (1 - 4 x)^-0.5 = 1 + 2 x + 6 x^2 + ....

    let: 4 x = (v/c)^2 or x = (v/c/2)^2

    just keep the lower terms 1 + 2 x (note

    (1 - 4 x)^-0.5 = 1 + 2 x = 1 + 2 (v/c/2)^2 [eqn4]

    from [eqn3] and [eqn4]
    1 + Ev / Eo = 1 + 2 (v/c/2)^2

    or

    Eo = Ev / 2 / (v/c/2)^2

    Eo = 0.5 m v^2 / 2 / v^2 * c^2 * 4

    Eo = m c^2

    Voila, this deserves a kick in the ass :}

    Appendix:
    ---------

    note 8 : extra points, explain what the ignored higher terms are.

    note 9 : not needed here but remember from special relativity:

    z' = z / ( 1 - (v/c)^2 )^0.5

    t' = t / ( 1 - (v/c)^2 )^0.5

    m' = m / ( 1 - (v/c)^2 )^0.5

    E' = E / ( 1 - (v/c)^2 )^0.5

    ==================================

    from: Checkboard (pprune)

    Einstein's original paper is online here:

    http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

    AA > > thanks for the slightly different topic of translated paper; too bad it's still a pain to read and understand the old terminology. < < AA

    Does the inertia of a body depend upon its energy-content?

    Translated from the original German, it's only three pages and fairly simple to understand. It basically says that, if the speed of light is a constant regardless of frame of reference, then a body emitting light radiation has its kinetic energy change by the amount of mass it loses in radiation emitted.
    .

    Comment


    • #3
      Engineering math
      Classic and Funny from Ma and Pa Kettle (1949)


      Engineering

      Comment


      • #4
        Good stuff

        More more more .....
        .

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