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4 Momentum and 4 velocity relationship

  1. May 23, 2012 #1
    [tex]
    P = \left( \begin{array}{c}
    E/c
    \\ \bar{p}
    \end{array}\right)
    [/tex]

    and

    [tex]
    U = \left( \begin{array}{c}
    \gamma c
    \\ \gamma \bar{v}
    \end{array}\right)
    [/tex]

    right? But I frequently see in textbooks that [itex] P = m_0 U [/itex]. Surely,
    [tex]m_0 U =
    \left( \begin{array}{c}
    \gamma m_0 c
    \\ \gamma m_0 \bar{v}
    \end{array}\right)
    =
    \left( \begin{array}{c}
    E/c
    \\ \gamma \bar{p}
    \end{array}\right)
    \neq
    \left( \begin{array}{c}
    E/c
    \\ \bar{p}
    \end{array}\right)
    [/tex]

    So how does this work?
    Yours confusedly
     
  2. jcsd
  3. May 23, 2012 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    The p in (E/c,p) is three momentum but it is still relativistic three momentum not Newtonian 3-momentum. Thus p=γmv, and your confusion is resolved.
     
  4. May 23, 2012 #3
    Ah, thanks
     
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