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How do you verify that v/c=pc/E?

  1. Oct 4, 2005 #1
    How do you verify that v/c=pc/E?
     
  2. jcsd
  3. Oct 4, 2005 #2

    dextercioby

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    HINT:

    [tex] p=\gamma m_{0} v [/tex]

    [tex] E=\gamma m_{0} c^{2} [/tex]

    Daniel.
     
  4. Oct 4, 2005 #3

    krab

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    Your question is too short, and BTW only applies to free particles. I don't know what you are starting with. Or even whether you mean experimental verification. As Dextrecioby points out, if you already know expressions for p and E, the result is trivial. Another possible starting point is that you know only
    [tex]E^2=p^2c^2-m_0^2c^4[/tex]
    and Hamilton's equations. Then use H=E(p,x), and
    [tex]v=dx/dt=\partial H/\partial p[/tex]
    or the vector generalization.
     
  5. Oct 6, 2005 #4
    @@a
    why does it only apply to free particles?
     
  6. Oct 6, 2005 #5

    krab

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    If a particle is in a potential field [itex]\Phi(x,y,z)[/itex], then
    [tex](E-\Phi)^2=p^2c^2-m_0^2c^4[/tex]
    so v/c is not equal to pc/E. One can define "kinetic energy" as [itex]K=E-\Phi(x,y,z)[/itex], then v/c=pc/K. That's just one example of non-free.
     
  7. Oct 7, 2005 #6
    i see... thank you very much!
     
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