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Kinetic energy transformation

  1. Mar 31, 2007 #1
    Please let me know a reference where the transformation of the kinetic energy is performed.
  2. jcsd
  3. Apr 2, 2007 #2
    Since the kinetic energy is the total energy minus the rest energy, and since the rest energy is invariant, won't any reference suffice that gives the transformation of total energy? That is, K = E - mc^2, m is invariant, and E transforms in the usual way.
  4. Apr 2, 2007 #3
    kinetic energy

    Thanks. There is a poblem. K is a physical quantity for which we do not define a proper value whereas for E we do.
  5. Apr 2, 2007 #4


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    If by "proper" you mean the value in an object's own rest frame, then the "proper kinetic energy" of any object is zero! :biggrin:
  6. Apr 2, 2007 #5
    I suggest that you will need to calculate K in each frame after transforming E. Kinetic energy is not a component of a 4-tensor, since it is just the additional energy imparted by motion. However, it can be written

    K = mc^2 [(1/sqrt(1-(v/c)^2))-1]
  7. Apr 2, 2007 #6


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    Geometrically, the relativistic kinetic energy is difference between the projections of two 4-momenta-related-by-a-boost onto one of those 4-momenta.
  8. Apr 2, 2007 #7
    Thanks. For a less sophisticated audience I would present the problem as:
    Start with the expression of the kinetic energy in I
    K=mcc[(1/g(u))-1] (1)
    where g(u) stands for gamma as a function of the speed of the bullet in I,
    Express the right side of (1) as a function of u' the speed of the bullet in I' via the composition law of parallel speeds in order to obtain
    K=mcc[(1+u'V/cc)/g(V)g(u'))-1] (2)
    The transformation equation (2) leads to the following consequences
    a. For u'=0
    b.For V=0 (u'=u)
    whereas for u'=0 and V=0
    all in good accordance with phyhsical reality!
    Is there some thing wrong in my derivation?
    Use please soft words and hard arguments:rofl:
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