Calculating Four-Velocity and Four-Momentum Using Schutz's Method

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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 \vec{U} = \vec{e}_{\bar{0}

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

Where \vec{U} is the four-velocity vector, and U^{\alpha} are its components.
 
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Does it help to write it all out explicity as a matrix equation?

\begin{bmatrix}<br /> \gamma &amp; \beta \gamma &amp; 0 &amp; 0\\ <br /> \beta \gamma &amp; \gamma &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix} = \begin{bmatrix}\gamma\\ \beta \gamma \\ 0\\ 0 \end{bmatrix}

where

\beta = \frac{v}{c}

and

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

The components of U^\alpha in reference frame O are the entries of the first column of the transformation matrix.
 
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schwarzschild said:
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 \vec{U} = \vec{e}_{\bar{0}

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

Where \vec{U} is the four-velocity vector, and U^{\alpha} are its components.

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.
 
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Thanks for the help guys! I completely understand it in matrix form - for some reason I struggle with Einstein notation.
 
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