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Deriving the Energy-Momentum Formula

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the energy-momentum relationship, E^2 = p^2 * c^2 + (m*c^2)^2, follows from the expressions E = (gamma)*m*c and p = (gamma)*m*u

    where

    (gamma) = 1 / sqrt(1 - (u^2)/(c^2)) the lorentz transformation factor.
    m is the rest mass.
    c is the speed of light
    u is the velocity of the particle
    E is the total energy
    p is the momentum

    The book does not teach about relativistic mass, so I think I supposed to derive this without making a distinction between m and m0.

    2. Relevant equations
    (1): E^2 = p^2 * c^2 + (m*c^2)^2
    (2): E = (gamma)*m*c
    (3): p = (gamma)*m*u


    3. The attempt at a solution
    When the chapter introduces the formula E^2 = p^2 * c^2 + (m*c)^2, it does not show how it derived this equation. Instead it says that it just says "By squaring [equations (2) and (3)] and subtracting, we can eliminate u. The result after some algebra is [equation (1)]."

    My first attempt was to start by squaring both sides of equation (3).

    p^2 = (gamma)^2*m^2*u^2

    Then get it in terms of m^2.

    m^2 = (p^2)/(u^2) - (p^2)/(c^2)

    In equation (2), square both sides, then substitute m^2 to get:

    E^2 = (gamma)^2 * [(p^2)/(u^2) - (p^2)/(c^2)] * c^2

    After some algebra I got:

    E^2 = [(gamma)^4 * m^2 * c^4] - [(gamma)^2 * p^2 * c^2]

    This is nearly what I'm trying to derive, however, the (gamma) terms are still there and I don't know how to get rid of them.
     
    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 11, 2011 #2
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