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# There exists a Lorentz transform between any two unit vector

Theorem : For any two unit vectors of the same sign, ie for $u, v$ such that $\langle u,u \rangle = \langle v, v \rangle = \pm 1$, there exists a Lorentz transformation $\Lambda$ such that $\Lambda u = v$.

Proof : First let's show that a linear transformation exists for such a thing. Consider the following linear map :

$$T^\pm_{u,v} w = w - \frac{\langle (u + v), w \rangle}{\pm 1 + \langle u, v \rangle} (u + v) \pm 2 \langle u, w \rangle v$$

$T^+$ is for spacelike vectors, $T^-$ for timelike vectors. If we apply our operator to $u$ :

\begin{eqnarray} T^\pm_{u,v} u &=& u - \frac{\langle (u + v), u \rangle}{\pm 1 + \langle u, v \rangle} (u + v) \pm 2 \langle u, u \rangle v\\ &=& u - \frac{\langle u, u \rangle + \langle v, u \rangle}{\pm 1 + \langle u, v \rangle} (u + v) + 2 v\\ &=& u - \frac{\pm 1 + \langle v, u \rangle}{\pm 1 + \langle u, v \rangle} (u + v) + 2 v\\ &=& u - (u + v) + 2 v\\ &=& v \end{eqnarray}

This indeed transforms $u$ to $v$. Now let's show that this is a Lorentz transformation, by considering the inner product

\begin{eqnarray} \langle T^\pm_{u,v} x, T^\pm_{u,v} y\rangle &=& \langle x - \frac{\langle (u + v), x \rangle}{\pm 1 + \langle u, v \rangle} (u + v) \pm 2 \langle u, x \rangle v, y - \frac{\langle (u + v), y \rangle}{\pm 1 + \langle u, v \rangle} (u + v) \pm 2 \langle u, y \rangle v \rangle\\ &=& \langle x, y \rangle - \frac{\langle (u + v), y \rangle}{\pm 1 + \langle u, v \rangle} \langle x, (u + v) \rangle \pm 2 \langle u, y \rangle \langle x, v \rangle\\ && - \frac{\langle (u + v), x \rangle}{\pm 1 + \langle u, v \rangle} \langle (u + v), y \rangle + \frac{\langle (u + v), x \rangle}{\pm 1 + \langle u, v \rangle} \frac{\langle (u + v), y \rangle}{\pm 1 + \langle u, v \rangle} \langle (u + v), (u + v) \rangle\\ && \mp \frac{\langle (u + v), x \rangle}{\pm 1 + \langle u, v \rangle} 2 \langle u, y \rangle \langle (u + v), v \rangle \pm 2 \langle u, x \rangle \langle v, y \rangle\\ &&\mp 2 \langle u, x \rangle \frac{\langle (u + v), y \rangle}{\pm 1 + \langle u, v \rangle} \langle v, (u + v) \rangle + 4 \langle u, x \rangle \langle u, y \rangle \langle v, v \rangle\\ &=& \langle x, y \rangle - \frac{\langle (u + v), y \rangle}{\pm 1 + \langle u, v \rangle} \langle x, (u + v) \rangle + \frac{\langle (u + v), x \rangle}{\pm 1 + \langle u, v \rangle} \langle (u + v), y \rangle\\ &&\pm 2 \langle u, y \rangle \langle x, v \rangle \mp 2 \langle u, x \rangle \langle u, y \rangle \mp 2 \langle v, x \rangle \langle u, y \rangle \pm 2 \langle u, x \rangle \langle v, y \rangle\\ &&\mp 2 \langle u, x \rangle \langle u, y \rangle \mp 2 \langle u, x \rangle \langle v, y \rangle \pm 4 \langle u, x \rangle \langle u, y \rangle \\ &=& \langle x, y \rangle + \frac{1}{\pm 1 + \langle u, v \rangle}\left( \langle (u + v), x \rangle \langle (u + v), y \rangle - \langle (u + v), y \rangle \right \langle x, (u + v) \rangle) \\ &=& \langle x, y \rangle \end{eqnarray}

The transformation does preserve the inner product, and is therefore a Lorentz transform.

Corollary : Any timelike (resp. spacelike) vector $u$ can be Lorentz transformed to a vector colinear to any other timelike (resp. spacelike) vector. To show this simply take the normalized vector $u' = u / \| u \|$ and $v' = v / \| v \|$, so that

\begin{eqnarray} T_{u', v'} u &=& T_{u', v'} u' \| u \|\\ &=& \| u \| T_{u', v'} u'\\ &=& \| u \| v'\\ &=& \frac{\| u \|}{\| v \|} v \end{eqnarray}