Special Relativity – Comprehensive Guide on Deriving the General Lorentz Transformation

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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 $K^\textrm{t}$?

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:

  1. First transformation converts vector to temporary basis so that axis of Lorentz rotation coincides with the axis of the rotation you need, second one returns back to original basis.
  2. Inverse of an orthogonal matrix is equal to its transpose, so it's just easier to use a transpose of transformation to return back to original basis.
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