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Kinetic energy

  1. Oct 4, 2005 #1
    How do you prove that (1/2)*(gamma)mv^2 doen't equal the kinetic energy of a particle moving at relativistic speeds?
  2. jcsd
  3. Oct 4, 2005 #2


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    Simply because

    [tex] E_{kin}=m_{0}c^{2}\left(\gamma-1\right) [/tex]

    ,which is different from your [itex] \frac{1}{2} \gamma m_{0}v^{2} [/itex]...?

  4. Oct 4, 2005 #3
    ... yet another example why relativistic mass was a bad idea.

    Mass doesn't change with speed.
    [tex]E_{K} = (\gamma -1)mc^{2}[/tex]
  5. Oct 4, 2005 #4
    How so mass does not change with speed? I thought that as a particle approaches the speed of light then it must lose mass. What is it i got wrong here?
  6. Oct 4, 2005 #5


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    There are two sorts of mass. One of them, called invariant mass, stays constant regardless of velocity and is a property of the particle itself (it doesn't depend on the particles state of motion).

    This is preferred by many, probably even most, people, but there are a few vocal people who prefer the other sort of mass, relativistic mass.

    Relativistic mass _increases_ with velocity according to the formula

    [itex]m_r = \gamma m_0[/itex]

    where [itex]m_r[/itex] is the relativistic mass, [itex]m_0[/itex] is the invariant mass, and [itex]\gamma = \frac{1}{\sqrt{1-(v/c)^2}}[/itex] depends on the velocity of the particle.

    For some more information, see for instance the sci.physics.faq "Does mass change with velocity".

    Last edited: Oct 4, 2005
  7. Oct 4, 2005 #6
    This question has absolutely nothing to do with the great idea of relativistic mass.

    asdf1 - Its a matter of calculation. Simply calculate the kinetic energy and you'll obtain

    [tex]K = (\gamma - 1)m_0 c^2[/tex]

    I worked out the calculation based on the work-energy theorem and placed them online at - http://www.geocities.com/physics_world/sr/work_energy.htm

  8. Oct 6, 2005 #7
    wow! thank you very much! :)
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