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A proper Lorentz transformation of a vector $\bf{x}$ is given by

$$\bf{x}\to \bf{x}'=\Lambda\cdot\bf{x}$$

where $\Lambda$ is a matrix with the properties

$$\Lambda^T\cdot\eta\cdot\Lambda=\eta~~~,~~~\det\Lambda=1,$$

where $\eta=\text{diag}(-1,1,1,1)$ is the Minkowski metric.

What is a convenient parametrization of all the individual components $\Lambda_{ij}$ of matrix $\Lambda$ for a generic proper Lorentz transformation continuously connected to the identity?

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The usual six parameters for proper Lorentz transformations are a three-component axis-angle vector $\vec\theta$ for the rotation part and a three-component rapidity vector $\vec\zeta$ for the boost part, as described in this Wikipedia article.

If $\vec J$ are the rotation generators and $\vec K$ the boost generators, then

$$\Lambda(\vec\theta,\vec\zeta)=e^{\vec\theta\cdot\vec J-\vec\zeta\cdot\vec K}.$$

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  • $\begingroup$ The exponential of matrices is defined through an infinite sum. I'm curious how this expression looks like in terms of a single matrix (not in exponential) where the infinite sum has been performed? $\endgroup$ – Kagaratsch Sep 19 '19 at 19:31
  • $\begingroup$ It’s something ugly that I can’t help you with. The exponential, by contrast, is beautiful, because it builds up a finite transformation from an infinitesimal one. $\endgroup$ – G. Smith Sep 19 '19 at 19:33
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    $\begingroup$ I shouldn’t have said that it is something ugly. I have never seen even an ugly expression for this exponential, so I doubt that one exists that doesn’t involve an infinite sum. However, for a pure rotation or a pure boost the sum can be done and a nice result obtained. $\endgroup$ – G. Smith Sep 19 '19 at 21:30

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