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:
- 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?
- Why does the first orthogonal transformation use the transposed orthogonal matrix $K^\textrm{t}$?
Best Answer
To apply the Lorentz transformation to some vector $\vec v$, having a $L_x$ matrix, but doing it along another axis $\vec q$, you can temporarily change coordinates so that the vector is parallel to $\vec e_x$:
$$\vec v'=U\vec v.$$
Now your intermediate result would be
$$L_x\vec v'=L_x U\vec v.$$
But it's still in the temporary basis. Let's now go back to original basis. As $U$ is orthogonal, its inverse equals its transpose, so we get:
$$L_q\vec v=U^TL_xU\vec v.$$
Thus,
$$L_q=U^TL_xU.$$
So, the answers are: