# Lorentz transformations: new actual notation for a $4$-vector [duplicate]

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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

• @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 at 10:04