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Relativistic Kinetic Energy

  1. Jul 28, 2008 #1
  2. jcsd
  3. Jul 28, 2008 #2
    I haven't read the first link, but from what I know I'd say that the second definition is the right one. [itex]E = \gamma mc^{2}[/itex] is the total energy, not the kinetic energy. You can Taylor expand the first expression [itex]E = \gamma mc^{2} - mc^{2}[/itex] with respect to the variable v/c. The first term in the expression, [itex]\frac{1}{2}mv^{2}[/itex] represents the Newtonian kinetic energy, which will be the dominant term if v/c is small.
  4. Jul 28, 2008 #3
    Where does the first link say that? All I can see under "conservation of energy" is:

    E_final = γmc^2 + E_L

  5. Jul 28, 2008 #4


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  6. Jul 28, 2008 #5
    Your expressions are better written as:

    see : Conservation of Energy
    TE = gamma*m*c^2

    While here :
    see : Relativistic kinetic energy of rigid bodies
    KE = gamma*m*c^2 - m*c^2

    where TE is Total Energy and KE is Kinetic Energy.
    The second expression can also be written as:

    KE = TE - RE

    where RE is Rest Energy or rest mass energy.

    Total Energy can can also be found from this relationship:

    [itex]TE = \sqrt{(gamma*m*v*c)^2+(m*c^2)^2} = \sqrt{(pc)^2+(mc^2)^2}[/itex]

    which can be written as:

    [itex]TE = \sqrt{ME^2+RE^2}[/itex]

    where ME is Momentum Energy.

    By rearranging this becomes :

    [itex]RE = \sqrt{TE^2-ME^2}[/itex]

    Since rest energy is usually an invariant, the quantity [itex]\sqrt{TE^2-ME^2}[/itex] is the same when switching from one reference frame to another. In fact, in a perfectly elastic collision, the quantity [itex]\sqrt{TE^2-ME^2}[/itex] is the same for a given particle before and after the collision.

    Hope that helps.
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