[Physics] Transformation of four-velocity in special relativity

Question

I am revising special relativity introducing more matrix form in the equation. Currently I am reading book in which transformation matrix is defined as
$${\Lambda=
\begin{bmatrix}
\gamma & -v\gamma & 0 & 0 \\
-v\gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
}$$
and the four-velocity as
$${U^{\alpha}=\Lambda^{\alpha}_{\overline{\beta}}(\vec e_{\overline{0}})^{\overline{\beta}}=\Lambda^{\alpha}_{\overline{0}}}$$
The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector
$${\vec U=
\begin{bmatrix}
\gamma \\
-v\gamma \\
0 \\
0
\end{bmatrix}
}$$
And the second thing is that when I apply this rule of transformation twice (for example when I have the following task:Ship is travelling with 0.6c relative to the Earth and another ship is travelling with 0.6c with respect to the first shipt. Find the four-velocity of the second ship relative to the Earth.(it is not homework)) For the four-velocity I get
$${
\vec U=
\begin{bmatrix}
\gamma^{2}+v^{2}\gamma^{2} \\
-v\gamma^{2}-v\gamma^{2} \\
0 \\
0\\
\end{bmatrix}
}$$
The second component is negative which definitely differs from the real vector (I checked this using Lorentz transformation, getting the equations, building the t''-x'' axes of the second ship's frame with respect to Earth). Where am I wrong?
(c=1)

Answer 1

Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a co-vector and my answer below addresses that.

Because boosts are not orthogonal matrices, you have to think about the different way co-vectors and contra-vectors transform. The $\Lambda$ matrix you wrote down is the +v boost for a co-vector. You then applied it to a vector and incorrectly interpreted the result as if it were a contra-vector. The co-vector $U_\alpha$ with +v velocity really does have $-v\gamma$ in it as you found. You must raise U's index with the metric to see what the contra-variant $U^\beta$ is, and you see the +v that you expected.

$$U^\beta = \eta^{\beta \alpha}U_\alpha$$ $$\begin{bmatrix}\gamma \\v\gamma \\0 \\0\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 &-1\end{bmatrix} \begin{bmatrix}\gamma \\-v\gamma \\ 0 \\ 0\end{bmatrix}$$

A contra-vector transforms as $$Unew^\alpha = \Lambda^\alpha _{-\beta} Uold^\beta$$ Then a co-vector transforms like $$Unew_\alpha = Uold_\beta(\Lambda^{-1})^\beta _{-\alpha}$$

This matrix boosts contra-vectors by +v. I have redefined your $\Lambda$ to be the matrix that gives a +v boost to a contra-vector. $${\Lambda=\begin{bmatrix}\gamma & v\gamma & 0 & 0 \\ v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}}$$

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

This matrix boosts co-vectors by +v. $${(\Lambda^{-1})^T=\begin{bmatrix}\gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}}$$

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

Notice that rotations R are orthogonal which means $(R^{-1})^T=R$ and the transformations for co-vectors and contra-vectors are the same. Hence we don't talk about co-vectors and contra-vectors when doing rotations.