danshawen
Valued Senior Member
And your answer is, as per usual, perfectly consistent. Thank you.The analogy between hyperbolic rotations and rotations in three dimensions in space is an analogy between two types transformations between geometrically equivalent coordinate systems. There is no concept of motion or rate with the latter and therefore no analogy with rate with the former.
So your demand seems ignorant and specious.
$$\vec{\zeta} \to A(\vec{\zeta}) = \begin{pmatrix}0 & \zeta_x & \zeta_y & \zeta_z \\ \zeta_x & 0 & 0 & 0 \\ \zeta_y & 0 & 0 & 0 \\ \zeta_z & 0 & 0 & 0 \end{pmatrix}
\\ A(\vec{\zeta})^2 = \begin{pmatrix}\zeta^2 & 0 & 0 & 0 \\ 0 & \zeta_x^2 & \zeta_x \zeta_y & \zeta_x \zeta_z \\ 0 & \zeta_x \zeta_y & \zeta_y^2 & \zeta_y \zeta_z \\ 0 & \zeta_x \zeta_z & \zeta_y \zeta_z & \zeta_z^2 \end{pmatrix}
\\ A(\vec{\zeta})^3 = \zeta^2 A(\vec{\zeta})
\\ A(\vec{\zeta})^4 = \zeta^2 A(\vec{\zeta})^2
\\ \Lambda(\vec{\zeta}) = e^{A(\vec{\zeta})} = I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2
\\ \Lambda(-\vec{\zeta}) = I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2
\\ \Lambda(\vec{\zeta}) \Lambda(-\vec{\zeta}) = \left( I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right) \left( I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right)
\\ \quad \quad \quad = I + \frac{ 2 \cosh \sqrt{ \zeta^2 } - 2 - \sinh^2 \sqrt{ \zeta^2 } + \cosh^2 \sqrt{ \zeta^2 } - 2 \cosh \sqrt{ \zeta^2 } + 1}{ \zeta^2} A(\vec{\zeta})^2
\\ \quad \quad \quad = I
$$
$$\vec{\theta} \to B(\vec{\theta}) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta_z & \theta_y \\ 0 & \theta_z & 0 & - \theta_x \\ 0 & -\theta_y & \theta_x & 0 \end{pmatrix}
\\ B(\vec{\theta})^2 = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & \theta_x^2 - \theta^2 & \theta_x \theta_y & \theta_x \theta_z \\ 0 & \theta_x \theta_y & \theta_y^2 - \theta^2 & \theta_y \theta_z \\ 0 & \theta_x \theta_z & \theta_y \theta_z & \theta_z^2 - \theta^2\end{pmatrix}
\\ B(\vec{\theta})^3 = -\theta^2 B(\vec{\theta})
\\ B(\vec{\theta})^4 = -\theta^2 B(\vec{\theta})^2
\\ R(\vec{\theta}) = e^{B(\vec{\theta})} = I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2
\\ R(-\vec{\theta}) = I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2
\\ R(\vec{\theta}) R(-\vec{\theta}) = \left( I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right) \left( I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right)
\\ \quad \quad \quad = I + \frac{ 2 - 2 \cos \sqrt{ \theta^2 } - \sin^ \sqrt{ \theta^2 } - 1 + 2 \cos \sqrt{ \theta^2 } - \cos^2 \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2
\\ \quad \quad \quad = I
$$
$$\Lambda(\rho \hat{x})$$ "rotates" $$(ct = \textrm{sech} \, \rho, x=0, y=0, z=0)$$ to $$(ct' = 1, x'= \tanh \rho, y'=0, z'=0)$$ in a manner to how $$R(t \hat{z})$$ rotates $$(ct = 0, x= \sec t, y=0, z=0)$$ to $$(ct' = 0, x'= 1, y'= \tan t, z'=0)$$. Both are linear transformations, preserve colinearity of trajectories and preserve $$(ct)^2 - x^2 -y^2 -z^2 = (ct')^2 - x'^2 -y'^2 -z'^2$$.
Translation: "it doesn't matter, because it doesn't matter."