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Four-Velocity & Momentum

  1. Apr 9, 2010 #1
    From Schutz's A First Course in General Relativity

    "A particle of rest mass m moves with velocity v in the x direction of frame O. What are the components of the four-velocity and four-momentum?"

    By definition [tex] \vec{U} = \vec{e}_{\bar{0} [/tex]

    However, I don't see how he gets [tex] U^{\alpha} = \Lambda^{\alpha}_{\bar{\beta}}(\vec{e}_{\bar{0}})^{\bar{\beta}} = \Lambda^{\alpha}_{\bar{0}} [/tex]

    Where [tex] \vec{U} [/tex] is the four-velocity vector, and [tex] U^{\alpha} [/tex] are its components.
     
  2. jcsd
  3. Apr 9, 2010 #2
    Does it help to write it all out explicity as a matrix equation?

    [tex]\begin{bmatrix}
    \gamma & \beta \gamma & 0 & 0\\
    \beta \gamma & \gamma & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
    \end{bmatrix}

    \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix} = \begin{bmatrix}\gamma\\ \beta \gamma \\ 0\\ 0 \end{bmatrix}[/tex]

    where

    [tex]\beta = \frac{v}{c}[/tex]

    and

    [tex]\gamma = \left ( 1 - \beta^2 \right )^{-1/2}.[/tex]

    The components of [itex]U^\alpha[/itex] in reference frame O are the entries of the first column of the transformation matrix.
     
    Last edited: Apr 9, 2010
  4. Apr 9, 2010 #3

    dx

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    In the reference frame in which the mass is at rest, its four-velocity is simply e0, i.e. the components are (1, 0, 0, 0). Now all you have to do is find the components of this same vector in a reference frame which is moving in the opposite direction with speed v. This is done by applying an appropriate Lorentz transformation.
     
    Last edited: Apr 9, 2010
  5. Apr 9, 2010 #4
    Thanks for the help guys! I completely understand it in matrix form - for some reason I struggle with Einstein notation.
     
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