special relativity - Derivation of the general Lorentz transformation

The standard Lorentz transformation or boost with velocity $u$ is given by $$\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = \left(\begin{matrix} \gamma & \gamma u/c & 0 & 0 \\ \gamma u/c & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right) \, \left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right) = L_u \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)$$ where $\gamma = \gamma(u) = 1/\sqrt{1-u^2/c^2}$. In the standard Lorentz transformation, it is assumed that the $x$ and $x^\prime$ axes coincide, and that $O^\prime$ is moving directly away from $O$.

If we drop the first condition, allowing the inertial frames to have arbitrary orientations, then "we must combine [the standard Lorentz transformation] with an orthogonal transformation of the $x$, $y$, $z$ coordinates and an orthogonal transformation of the $x^\prime$, $y^\prime$, $z^\prime$ coordinates. The result is $$\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = L \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)$$ with $$L = \left(\begin{matrix} 1 & 0 \\ 0 & H \end{matrix}\right)\, L_u \,\left(\begin{matrix} 1 & 0 \\ 0 & K^\textrm{t} \end{matrix}\right)$$ where $H$ and $K$ are $3 \times 3$ proper orthogonal matrices, $L_u$ is the standard Lorentz transformation matrix with velocity $u$, for some $u < c$, [and 't' denotes matrix transpose]."

I have two questions:

1. Why are two orthogonal transformations, for both the unprimed and primed spatial coordinates, necessary? That is, why isn't one orthogonal transformation sufficient to align the axes of the inertial frames?


2. Why does the first orthogonal transformation use the transposed orthogonal matrix $K^\textrm{t}$?

Answer

It's actually very simple. The general Lorentz transformation can be rewritten as
$$\left(\begin{matrix} 1 & 0 \\ 0 & H^\textrm{t} \end{matrix}\right)\,\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = L_u \,\left(\begin{matrix} 1 & 0 \\ 0 & K^\textrm{t} \end{matrix}\right) \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)\,.$$ This corresponds to aligning the $x$ and $x^\prime$ axes with the direction of the relative velocity, and then applying the standard Lorentz transformation.