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Relativistic kinetic energy and force.

  1. Feb 27, 2012 #1
    classically, an electron accelerating from rest in a uniform electric field will have a kinetic energy proportional to the distance 'd' from its point of origin.

    will this continue to hold even when the electron is moving at relativistic velocity?

    I understand that the formula for relativistic kinetic energy is

    656314a4a1ad9593e71227d9c2184c57.png

    so basically kinetic energy is proportional to gamma - 1

    so gamma(d) ≡ d + 1?
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 27, 2012 #2
  4. Feb 29, 2012 #3

    Bill_K

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    granpa, There are several ways of writing the force and acceleration, but one thing you can rely on, even in relativity, is energy conservation. The total energy of a charge particle in an E field is E = γmc2 + eΦ where Φ is the electrostatic potential, and the value of this will be the same both when the particle is at rest and when it is moving relativistically. (This ignores only the radiation it will emit.)
     
  5. Feb 29, 2012 #4
    Thank you. Thats was very helpful.

    I've also been told that teh rate of change of rapidity with respect to proper time will be constant.
    I guess that would be the proper acceleration.

    f3fd7fcce9b254111e10ca5bae382511.png

    gamma(x) = x + 1

    a = dg(x)/dx = 1
     
    Last edited: Feb 29, 2012
  6. Mar 2, 2012 #5
    if instead of a uniform electric field we use an inverse square law

    F = 1/r^2

    KE = integral F dr = 1/r

    KE = gamma(r) - 1 = 1/r

    gamma(r) = 1 + 1/r
     
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