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For the Lorentz trasformations I use this notation

\begin{equation*} \left\{\begin{aligned} x&=\gamma (x'+\beta ct')\\ y&=y'\\ z&=z'\\ ct&=\gamma (ct'+\beta x')\\ \end{aligned}\right. \end{equation*}

with this matrix

$$L^*=\begin{pmatrix}\gamma & 0 & 0 & \beta\gamma\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \beta \gamma & 0 & 0 & \gamma\end{pmatrix}$$ Introducing the imaginary unit $i=\sqrt{-1}$, the Lorentz transformations will allow you to switch from an orthogonal Cartesian coordinate system to an orthogonal one. Hence I, actually, use $L$ that is an orthogonal matrix. $$L=L(\beta)=\begin{pmatrix}\gamma & 0 & 0 & -i\beta\gamma\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ i\beta \gamma & 0 & 0 & \gamma \end{pmatrix}$$

My usual notation that I use is the following to define a quadrivector $\boldsymbol{\mathcal{X}}=(x,y,z,ict)$, or even better is:

$$\boldsymbol{\mathcal{X}}^\intercal=\begin{pmatrix} x \\ y \\ z \\ ict \end{pmatrix}$$ Why most physicists now use $(ct,x,y,z)$ instead of $(x,y,z,ict)$ (or $(ict, x,y,z)$) and let the electromagnetic field tensor have real components?


marked as duplicate by AccidentalFourierTransform, Thomas Fritsch, Community Jul 14 at 12:20

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    $\begingroup$ See this (and its keywords) you are good to go. en.wikipedia.org/wiki/Wick_rotation $\endgroup$ – Paradoxy Jul 14 at 9:48
  • $\begingroup$ @Paradoxy I have seen your link in the first note. It's the same for me. The elimination of the imaginary unit by placing $t=i\tau$ simply serves to make the Euclidian metric. That is, starting from $ds^2=dx^2+dy^2+dz^2-c^2d\tau^2$ I can have $ds^2=dx^2+dy^2+dz^2+c^2dt^2$. $\endgroup$ – Sebastiano Jul 14 at 10:04
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/168532/2451 , physics.stackexchange.com/q/107443/2451 and links therein. $\endgroup$ – Qmechanic Jul 14 at 10:12
  • $\begingroup$ Possible duplicate of Minkowski Metric Signature $\endgroup$ – Thomas Fritsch Jul 14 at 12:10
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    $\begingroup$ The question has been closed as a duplicate, but there is a better duplicate (physics.stackexchange.com/questions/121380/…) with what I consider to be the definitive answer, by Misner, Thorne, and Wheeler in Gravitation. Sebastiano, please read what they had to say in their “Farewell to ict”. This book was the magnum opus of GR and has had tremendous influence over the last 50 years. $\endgroup$ – G. Smith Jul 14 at 19:18

I had the same question before. I think there's a paragraph in Kip Thorne's Modern Classical Physics specifically pointed out that the imaginary number could not capture the total spacial/geometrical aspects of the GR (could not recall the detail), therefore I guess, people slowly used the real number and differential metric instead of the imaginary number (you don't actually need to write it in metric form if you use imaginary number, not necessarily). It's useful, though, to notice that imaginary number was a very “cheap”/neat solution, which was used in many old books. Like @bolbteppa has mentioned.


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