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

[schwarzschild]

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.

 

[Rasalhague]

Does it help to write it all out explicity as a matrix equation?

$$\begin{bmatrix} \gamma & \beta \gamma & 0 & 0\\ \beta \gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix} = \begin{bmatrix}\gamma\\ \beta \gamma \\ 0\\ 0 \end{bmatrix}$$

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.

 

[dx]

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.

 

[schwarzschild]

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