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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?

Sebastiano
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    See this (and its keywords) you are good to go. https://en.wikipedia.org/wiki/Wick_rotation – Paradoxy Jul 14 '19 at 09:48
  • @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 $ds^2=dx^2+dy^2+dz^2+c^2dt^2$. – Sebastiano Jul 14 '19 at 10:04
  • Possible duplicates: https://physics.stackexchange.com/q/168532/2451 , https://physics.stackexchange.com/q/107443/2451 and links therein. – Qmechanic Jul 14 '19 at 10:12
  • @ThomasFritsch Excuse me, did you vote against my question because it is a duplicate that I did not know about at all? If it were really so then on TeX.SE. there would be an infinity of negative, unjust votes. A question is always appreciated unless it is off-topic or ramshackle. – Sebastiano Jul 14 '19 at 12:19
  • @Sebastiano No, I didn't downvote your question. It is a legitimate question. – Thomas Fritsch Jul 14 '19 at 12:24
  • @ThomasFritsch Then I'm so sorry. I have another way of understanding the negative vote of some other user who is having fun. Excuse me again. Thank you very much. – Sebastiano Jul 14 '19 at 12:26
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    The question has been closed as a duplicate, but there is a better duplicate (https://physics.stackexchange.com/questions/121380/special-relativity-and-imaginary-coefficient-of-the-time-coordinate/121403#121403) 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. – G. Smith Jul 14 '19 at 19:18
  • @G.Smith Thank you for your interest. I have compensated you for the votes lost after the cancellation of my application this morning. – Sebastiano Jul 14 '19 at 19:27

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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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    I am voting in favour of the help offered. Is he Kip Thorne peraphs? – Sebastiano Jul 14 '19 at 10:43
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    See https://physics.stackexchange.com/questions/121380/special-relativity-and-imaginary-coefficient-of-the-time-coordinate/121403#121403 – G. Smith Jul 14 '19 at 19:21