In Carroll's GR book, he develops the general condition rotation matrices satisfy in order to be Lorentz Transformations: \begin{equation}\eta=\Lambda^T\eta\Lambda\end{equation} Which seems natural. Going further into actually writing actual expressions for the transformations, he shows a "boost" in x-direction: \begin{equation}\Lambda^{\mu'}_{\ \ \ \ \nu}=\left[ \begin{array}{cc} \cosh\phi & -\sinh\phi &0 &0 \\ -\sinh\phi & \cosh\phi&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array} \right] \end{equation} Which he just threw in there, and used to latter recover the usual expression for Lorentz Transformations, by representing the coordinates: \begin{align}t'=\ &t\cosh\phi-x\sinh\phi\\ x'=-&t\sinh\phi+x\cosh\phi\end{align} Which I don't really see how it follows from the Rotation Matrix descripted. So: Why does that RM represents a "boost" in x-direction, which makes it something that recovers lorentz transformations in the usual form, and how does that transformation law in $x$ and $t$ follow from the RM?
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$\begingroup$ What you call the usual expression for Lorentz transformation is simply the matrix multiplication of $\Lambda$ with the column vector $(t,x,y,z)^\top$. Was that the question ? $\endgroup$– Kurt G.Oct 4, 2022 at 17:59
1 Answer
First af all, the matrix you just reported isn't a "canonical" rotation matrix, in the sense that the rotation is performed in spacetime and not in the Euclidean space, as we are used to. For this reason the rotation is said to be an hyperbolic rotation and it represents a boost along the $x$ direction because that's simply what boosts are: a rotation in spacetime, where time and spatial coordinates work together. If the form presented isn't satisfactory enough, you can always relate the rapidity $\phi$ to the well-known Lorentz factor $\gamma$ by using the relation (remember that Carroll uses natural units, for which $c=1$): \begin{equation} \phi = \tanh^{-1}{v} \, .\end{equation}
Lastly, the transformations you wrote simply derive from the fact that the 4-vector $x^\mu=(t,x,y,z)$ behaves in the following way under a Lorentz transformation: $$ x'^{\mu}= \Lambda^{\mu}_{\, \,\nu}x^{\nu} $$ so that $$t′= t\cosh{\phi}−x\sinh{\phi}$$ $$x' = -t\sinh{\phi}+x\cosh{\phi} $$ is the immediate resul of matrix multiplication.
