# Différentes méthodes pour obtenir la forme polaire d'une matrice de Lorentz

Alain Guimier

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## To cite this version:

Alain Guimier. Différentes méthodes pour obtenir la forme polaire d'une matrice de Lorentz. 2023. hal-04188756

HAL Id: hal-04188756

https://amu.hal.science/hal-04188756v1

Preprint submitted on 27 Aug 2023

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Distributed under a Creative Commons CC0 1.0 - Universal - International License
A. Guimier

Different methods for obtaining
the polar form of a Lorentz matrix M under the form :

$$
M = \begin{bmatrix}
1 + \frac{\gamma^2}{(1+\gamma)} \beta_1^2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_3 \\
\frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_1 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_2^2 & \frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_3 \\
\frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_1 & \frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_2 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_3^2
\end{bmatrix} \cdot \Omega
$$

Summary : We present three different methods
to obtain the polar form of a Lorentz matrix.
We provide references for two additional methods.

Date : 08/23
If the matrix $M = (m_i^j)_{0 \le i \le n, 0 \le j \le n}$ we note $^t M = (m_j^i)_{0 \le i \le n, 0 \le j \le n}$.

A Lorentz matrix $M$ is defined as a square matrix satisfying $^t MGM = G$, where
$G = (g_j^i)$ with $g_j^i = 0$ if $i \ne j$, $g_j^i = 1$, $g_i^j = -1$ for $i > 1$.

If $M$ is a 4-th order matrix, it represents the transformation matrix
from a spacetime basis of an observer $O$ to another spacetime basis associated
with another observer $O'$, moving relative to $O$ with a uniform rectilinear motion
of velocity $\vec{V}$. We aim to identify the terms of this matrix $M$ as a function of $\vec{V}$ or
$\vec{\mathbf{\beta}} = \frac{\vec{V}}{c}$, $c$ the lightspeed.

**First method :**

The first method is a consequence of the theorems found in:
J-M. Souriau. "Calcul Linéaire" * PUF 1964 * p.378.

**Theorem 1 :**

Let $X \in \mathbb{R}^n$ such that $^t XX = I$.

Let $N$ be the matrix defined by:

$$
\mathcal{O}_{n-1} \text{ being the zero matrix of } \mathbf{M}_{n-1}(\mathbb{R}), N = \begin{bmatrix} \mathbf{0} & {}^{t}\mathbf{X} \\ \mathbf{X} & \mathcal{O}_{n-1} \end{bmatrix}
$$

Then, $\forall \mathbf{\alpha} \in \mathbb{R}$:

$$
M = \exp(\alpha N) = \begin{bmatrix} \text{ch}(\alpha) & \sh(\alpha)^t X \\ \sh(\alpha)X & (\text{Id}_{\mathbb{R}^n} - 1 + (\text{ch}(\alpha) - 1)X^t X) \end{bmatrix}
$$

is a Lorentz matrix of determinant equal to 1.

**Proof:**

We have $N^2 = \begin{bmatrix} 1 & 0 \\ 0^n - 1 & X^t X \end{bmatrix}$ by noting $\mathbf{0}^n - \mathbf{1}$ as the zero column of order $n-1$,

since $^tXX = I$ and then we have $N^3 = N$.

By noting that for any square matrix $A$ and $\forall \mathbf{\alpha} \in \mathbb{R}$,

$$
\exp(A) = \sum_{n=0}^{+\infty} \frac{A^n}{n!} , \quad \sh(\alpha) = \sum_{n=0}^{+\infty} \frac{\alpha^{2n+1}}{(2n+1)!} , \quad \text{ch}(\alpha) = \sum_{n=0}^{+\infty} \frac{\alpha^{2n}}{(2n)!}
$$

therefore

$$
\begin{align*}
\exp(\alpha N) &= \mathrm{Id}_{\mathbb{R}^n} + \alpha N + \frac{\alpha^2 N^2}{2!} + \frac{\alpha^3 N}{3!} + \frac{\alpha^4 N^2}{4!} + \dots + \\
&\quad + \frac{\alpha^{2p-1} N}{(2p-1)!} + \frac{\alpha^{2p} N^2}{(2p)!} + \text{....................................}
\end{align*}
$$
$$ = \mathrm{Id}_{\mathbb{R}^n} + \mathrm{sh}(\boldsymbol{\alpha})N + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)N^2. $$

Now we can verify if $M$ is indeed a Lorentz transformation : $^tMGM = G$?

It is worth noting that $M$ is symmetric: $^tM = M$.

$$ MG = \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) & \mathrm{sh}(\boldsymbol{\alpha})^t X \\ \mathrm{sh}(\boldsymbol{\alpha})X & (\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) \end{bmatrix} \begin{bmatrix} 1 & {}^{t0} \\ 0 & -\mathrm{Id}_{\mathbb{R}^n - 1} \end{bmatrix} $$

$$ = \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) & -\mathrm{sh}(\boldsymbol{\alpha})^t X \\ \mathrm{sh}(\boldsymbol{\alpha})X & -(\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) \end{bmatrix} \quad \text{and} $$

$$ \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) & -\mathrm{sh}(\boldsymbol{\alpha})^t X \\ \mathrm{sh}(\boldsymbol{\alpha})X & -(\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) \end{bmatrix} \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) & \mathrm{sh}(\boldsymbol{\alpha})^t X \\ -(\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) & \mathrm{sh}(\boldsymbol{\alpha})^t X \end{bmatrix}, $$

$$ = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad : \text{Let's calculate } a, b, c, \text{ and } d. $$

$$ a = \mathrm{ch}^2(\boldsymbol{\alpha}) - \mathrm{sh}^2(\boldsymbol{\alpha})^t XX = 1, \text{ because } ^tXX = 1. $$

$$ b = \mathrm{ch}(\boldsymbol{\alpha}) \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha})^t X (\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) $$

$$ = \mathrm{ch}(\boldsymbol{\alpha}) \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha}) (\mathrm{ch}(\boldsymbol{\alpha}) - 1)^t XX^t X $$

$$ = \mathrm{ch}(\boldsymbol{\alpha}) \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha}) \mathrm{ch}(\boldsymbol{\alpha})^t X + \mathrm{sh}(\boldsymbol{\alpha})^t X = 0, $$

$$ c = \mathrm{ch}(\boldsymbol{\alpha}) \mathrm{sh}(\boldsymbol{\alpha})^t X - (\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) \mathrm{sh}(\boldsymbol{\alpha})^t X $$

$$ = \mathrm{ch}(\boldsymbol{\alpha}) \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha})^t X - \mathrm{sh}(\boldsymbol{\alpha}) (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X = 0, $$

$$ d = \mathrm{sh}^2(\boldsymbol{\alpha})X^t X - (\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X)^2 \quad \text{but:} $$

$$ (\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X)(\mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X) $$

$$ = \mathrm{Id}_{\mathbb{R}^n} - 1 + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)^2 X^t XX^t X, $$

(It can be observed that as $^tXX = 1 : X(^tXX)^t X = X$ (using matrix multiplication associativity), and $(\mathrm{ch}(\boldsymbol{\alpha}) - 1)^2 = \mathrm{ch}^2(\boldsymbol{\alpha}) - 2\mathrm{ch}(\boldsymbol{\alpha}) + 1$)

$$ = \mathrm{Id}_{\mathbb{R}^n} - 1 + (2(\mathrm{ch}(\boldsymbol{\alpha}) - 1) + \mathrm{ch}^2(\boldsymbol{\alpha}) - 2\mathrm{ch}(\boldsymbol{\alpha}) + 1)X^t X $$
$$ = \mathbf{Id}_{\mathbb{R}^1-1} + (ch^2(\boldsymbol{\alpha}) - 1)X^tX = \mathbf{Id}_{\mathbb{R}^1-1} + \mathrm{sh}^2(\boldsymbol{\alpha})X^tX $$

hence :

$$ \mathrm{sh}^2(\boldsymbol{\alpha})X^tX - (\mathbf{Id}_{\mathbb{R}^1-1} + (ch(\boldsymbol{\alpha}) - 1)X^tX)^2 = -\mathbf{Id}_{\mathbb{R}^1-1}. $$

$$ \text{Consequently} \quad \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & {}^{t}0 \\ 0 & -\mathbf{Id}_{\mathbb{R}^1-1} \end{bmatrix} $$

and $M$ is a Lorentz matrix.

**Another proof :**

$$ \begin{align*} \exp(\alpha N) G &= \begin{bmatrix} \mathrm{ch}(\alpha) & \mathrm{sh}(\alpha)^t X \\ \mathrm{sh}(\alpha)X & (\mathrm{Id}_{\mathbb{R}^1-1} + (\mathrm{ch}(\alpha) - 1)X^tX)^t \end{bmatrix} \begin{bmatrix} 1 & {}^t0 \\ 0 & -\mathrm{Id}_{\mathbb{R}^1-1} \end{bmatrix} \\ &= \begin{bmatrix} \mathrm{ch}(\alpha) & -\mathrm{sh}(\alpha)^t X \\ \mathrm{sh}(\alpha)X & -(\mathrm{Id}_{\mathbb{R}^1-1} + (\mathrm{ch}(\alpha) - 1)X^tX)^t \end{bmatrix} \quad \text{and} \end{align*} $$

$$ \begin{align*} G\exp(-\alpha N) &= \begin{bmatrix} 1 & {}^t0 \\ 0 & -\mathrm{Id}_{\mathbb{R}^1-1} \end{bmatrix} \begin{bmatrix} \mathrm{ch}(\alpha) & -\mathrm{sh}(\alpha)^t X \\ -\mathrm{sh}(\alpha)X & (\mathrm{Id}_{\mathbb{R}^1-1} + (\mathrm{ch}(\alpha) - 1)X^tX)^t \end{bmatrix} \\ &= \begin{bmatrix} \mathrm{ch}(\alpha) & -\mathrm{sh}(\alpha)^t X \\ \mathrm{sh}(\alpha)X & -(\mathrm{Id}_{\mathbb{R}^1-1} + (\mathrm{ch}(\alpha) - 1)X^tX)^t \end{bmatrix} \end{align*} $$

$$ \text{hence } \exp(\alpha N) G = G\exp(-\alpha N) \Rightarrow \exp(\alpha N) G\exp(\alpha N) = G $$

as $M = \exp(\alpha N)$ is symmetric : $^tMGM = G$ and $M$ is a Lorentz matrix.

**Lemma :**

If $M$ is a Lorentz matrix and its first column $K_1$ has the form : $K_1 = {}^t(\alpha, 0, \dots, 0)$, then $M$ has the form :

$$ M = \begin{bmatrix} \varepsilon & {}^tQ \\ Q & \Omega \end{bmatrix} \quad \text{où } \varepsilon = \pm 1, {}^tQ = (0, 0, 0), {}^t\Omega\Omega = \mathbf{Id}_{\mathbb{R}^1-1} \quad (1). $$

Conversely, any matrix of this form (1) is a Lorentz matrix.

**Proof:**

$$ \text{Let } M = \begin{bmatrix} \alpha & {}^tL \\ 0 & C \end{bmatrix} \text{ with } C \in \mathbf{M}_3(\mathbb{R}), {}^tL = (l_1, l_2, l_3). $$

Since $(^tMG)M = G$, we have :
$$
G = \begin{bmatrix} \alpha & 0 \\ L & -tC \end{bmatrix} \begin{bmatrix} \alpha & tL \\ 0 & C \end{bmatrix} = \begin{bmatrix} \alpha^2 & \alpha^t L \\ \alpha L & L^t L - tCC \end{bmatrix}.
$$

By identification, we obtain $\alpha=\pm 1$, $L=0$, $tCC=Id_{\mathbb{R}^1-i}$

If $\mathbf{M}$ is of the form (1), an immediate calculation shows that $^t\mathbf{MGM} = \mathbf{G}$.

**Theorem 2 :**

Any Lorentz matrix $M$ can be written in the form :

$$
M = \exp(\alpha N) - \begin{bmatrix} \varepsilon & 0 \\ 0 & \Omega \end{bmatrix} \quad \text{where} \quad \alpha \in \mathbb{R}, N = \begin{bmatrix} 0 & tX \\ X & 0 \end{bmatrix}, X \in \mathbb{R}^{n-1}, \varepsilon = \pm 1,
$$

such as : $^tXX=1$ , $^t\Omega\Omega=Id_{\mathbb{R}^1-1}$ .

Proof:

Let $\mathbf{M}$ be a Lorentz matrix $\mathbf{M} = (\mathbf{m}[i, j])$

and $\mathbf{M}[1] = (\mathbf{m}[i, 1])$ be the first column of $\mathbf{M}$ then:

$$
{}^t M_1 G M_1 = m_{1,1}^2 - m_{1,2}^2 - \dots - m_{1,n}^2 = G_{1,1} = 1
$$

We can set $^tM_1 = (\beta, Y)$ avec $\beta = m_{1,1}$ and $^tY = (m_{1,2}, \dots, m_{1,n})$. Thus, we have $\beta^2 - tYY = 1$, so $|\beta| \ge 1$.

We can set $\beta = \varepsilon \cdot ch(\alpha)$, where $\varepsilon = \pm 1$ and $\alpha \in \mathbb{R}$.

Additionally, $^tYY = \beta^2 - 1 = sh^2(\alpha)$.

If $\alpha \neq 0$, let's set $X = \frac{Y}{\varepsilon \cdot sh(\alpha)}$, then we have $^tXX = 1$.

If $\alpha=0$, we have $^tM_1 = (\varepsilon, 0, 0, 0)$.

In all cases, $^tM_1 = \varepsilon(ch(\alpha), sh(\alpha)X)$ with $^tXX=1$.

$$
According to lemma, $M[1]$ is the first column of $\varepsilon \cdot \exp(\alpha N)$ where $N = \begin{bmatrix} 0 & tX \\ X & 0 \end{bmatrix}$.
$$

We notice that $\exp(\alpha N) \cdot M$ is a Lorentz matrix since $M$ and $\exp(-\alpha N)$ are.

Let's calculate the first column $K[1]$ of this product :

$$
K[1] = \begin{bmatrix} ch(\alpha) & (sh(\alpha))^{tX} \\ (sh(\alpha))X & \mathrm{Id}_{\mathbb{R}^3} + (ch(\alpha) - 1)X^{tX} \end{bmatrix} \cdot M[1]
$$

By noting that $^tXX=1$, $ch^2(\alpha) - sh^2(\alpha) = 1$, and

$$
(Id_{\mathbb{R}^3} + (ch(\alpha) - 1)X^t X) \cdot X = ch(\alpha) \cdot X,
$$

we obtain : $^tK_1 = (\varepsilon, 0, \dots, 0)$.

We can then apply the lemma, and the theorem is proven.

**Unicity of the decomposition :**
The decomposition of the previous theorem is unique.

**Proof:** It follows from the classical results that follow :

**Lemma :**

Any square symmetric matrix $\mathbf{A}$ of order $n \ge 1$, with real coefficients, is diagonalizable.

Moreover the eigenspaces of $\mathbf{A}$ are pairwise orthogonal. Therefore, we can write $\mathbf{A} = \mathbf{P}\mathbf{D}^t\mathbf{P}$,
where $\mathbf{P}$ is an orthogonal matrix and $\mathbf{D}$ is a diagonal matrix.

**Lemma :**

Let $\mathbf{M}$ be an invertible matrix with real coefficients

. Then there exists a unique pair of matrices $\mathbf{O}$ (orthogonal)

and $\mathbf{S}$ (symmetric positive definite) such that $\mathbf{M} = \mathbf{S}\mathbf{O}$.

*Proof:* The proof of this lemma can be found at the following link : https://en.wikipedia.org/wiki/Polar_decomposition.

This lemma implies that the decomposition in the theorem is unique.

For more details, refer to the following link : https://en.wikipedia.org/wiki/Matrix_exponential

Now let's apply these lemmas :

It is sufficient to verify that $exp(\alpha\mathbf{N})$ is positive definite. Since $\alpha\mathbf{N}$ is a real symmetric matrix,
there exists a real orthogonal matrix $\mathbf{U}$ and a real diagonal matrix $\mathbf{D} = (d_{i, i})$ such that $\alpha\mathbf{N} = t\mathbf{U}\mathbf{D}\mathbf{U}$.

As $\exp(\alpha\mathbf{N}) = \exp(t\mathbf{U}\mathbf{D}\mathbf{U}) = t\mathbf{U}\exp(\mathbf{D}\mathbf{U})\mathbf{U}$, the eigenvalues of $\exp(\alpha\mathbf{N})$ are strictly positive,
and $\exp(\alpha\mathbf{N})$ is positive definite. From the formula $\det(e^A) = e^{\mathrm{Tr}(A)}$, we deduce that $\det(\exp(\alpha\mathbf{N})) = 1$,

and $\det(\mathbf{M}) = \varepsilon \cdot \det(\Omega) = \pm 1$.

Application :

Let $\mathbf{M}$ be a Lorentz matrix, we have the following equivalences :

$$ \mathbf{M} \text{ is symmetric} \Leftrightarrow \mathbf{M} = \exp(\alpha \mathbf{N}) $$

$$ \mathbf{M} \text{ is orthogonal} \Leftrightarrow \mathbf{M} = \begin{bmatrix} \varepsilon & 0 \\ 0 & \Omega \end{bmatrix} $$

For the direct parts, it is sufficient to note that :

$$ \mathbf{M} = \exp(\alpha\mathbf{N}) \cdot \begin{bmatrix} \varepsilon & 0 \\ 0 & \Omega \end{bmatrix} \quad \text{and} \quad \mathbf{M} = \mathbf{M} \cdot \mathbf{Id} \text{ or } \mathbf{M} = \mathbf{Id} \cdot \mathbf{M}. $$

**Consequence :**

We can now specify the polar form. By taking the previous notations, we will show that :

If we define $\vec{V}$ as the velocity of the observer $\mathbf{O}'$ relative to the observer $\mathbf{O}$,

$c$ as the speed of light, and $\vec{\mathbf{\beta}} = \vec{V}/c$. We have if $\boldsymbol{\gamma} = \frac{\mathbf{1}}{\sqrt{1 - \mathbf{\beta}^2}}$ :
$$ \exp(\alpha N) = \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix}^{t[\vec{\gamma\beta}]} \begin{bmatrix} \vec{\gamma\beta} \\ C \end{bmatrix} \quad \text{with} \quad [\vec{\gamma\beta}] \text{ the coordinates of } \vec{\gamma\beta} \text{ in the frame of the observer } \mathbf{O}, $$

$$ \vec{\beta} = th(\alpha)\vec{X}, N = \begin{bmatrix} 0 & X \\ X & \mathcal{O}_{n-1} \end{bmatrix}^{t'X}, \gamma = ch(\alpha), $$

$$ C = \begin{bmatrix} 1 + \frac{\gamma^2}{(1+\gamma)} \beta_1^2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_3 \\ \frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_1 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_2^2 & \frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_3 \\ \frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_1 & \frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_2 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_3^2 \end{bmatrix} $$

**Proof :**

Let us consider $W' = \begin{bmatrix} ct' \\ 0 \\ 0 \\ 0 \end{bmatrix}$ as the coordinates of $O'$ in the frame of the observer $O'$.

Also, let $W = MW' = \exp(\alpha N) \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} W'$ be the coordinates of $O'$ in the frame of the observer $O$,
since $M$ is the transformation matrix from the frame of the observer $O$ to the frame of observer $O'$,

which are also equal to $\begin{bmatrix} ct \\ c\beta_1 \\ c\beta_2 \\ c\beta_3 \end{bmatrix}$. We first observe that

$$ \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} W' = W' \text{ thus } W = \exp(\alpha N) W'. \text{ We are led to :} $$
$$ \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) & & & & \mathrm{sh}(\boldsymbol{\alpha})^t X \\ & \mathrm{sh}(\boldsymbol{\alpha})X & & & \\ & & \mathrm{Id}_{\mathbb{R}^3} + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X^t X & \\ & & & t' = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} & t \end{bmatrix}, \text{ and} $$

$$ \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) \\ \mathrm{sh}(\boldsymbol{\alpha})X_1 \\ \mathrm{sh}(\boldsymbol{\alpha})X_2 \\ \mathrm{sh}(\boldsymbol{\alpha})X_3 \end{bmatrix}_{t'=1} \begin{bmatrix} 1 \\ \boldsymbol{\beta}_1 \\ \boldsymbol{\beta}_2 \\ \boldsymbol{\beta}_3 \end{bmatrix}_t \quad \text{and} \quad t = \mathrm{ch}(\boldsymbol{\alpha})t' \quad \text{and consequently:} $$

$$ \mathrm{ch}(\boldsymbol{\alpha}) \begin{bmatrix} 1 \\ \boldsymbol{\beta}_1 \\ \boldsymbol{\beta}_2 \\ \boldsymbol{\beta}_3 \end{bmatrix} = \begin{bmatrix} \mathrm{ch}(\boldsymbol{\alpha}) \\ \mathrm{sh}(\boldsymbol{\alpha})X_1 \\ \mathrm{sh}(\boldsymbol{\alpha})X_2 \\ \mathrm{sh}(\boldsymbol{\alpha})X_3 \end{bmatrix} \quad \text{for } t' \neq 0 \Rightarrow \mathrm{ch}(\boldsymbol{\alpha})\boldsymbol{\beta} = \mathrm{sh}(\boldsymbol{\alpha})\bar{X}, $$

$$ \text{thus } \vec{\boldsymbol{\beta}} = \mathrm{th}(\boldsymbol{\alpha})\vec{X} \Rightarrow \vec{\boldsymbol{\beta}}^2 = \mathrm{th}^2(\boldsymbol{\alpha}) \text{ since } \vec{X}^2 = 1. $$

$$ \text{We put } \boldsymbol{\beta} = \sqrt{\vec{\boldsymbol{\beta}}^2} \text{ et } \boldsymbol{\gamma} = \frac{1}{\sqrt{1 - \vec{\boldsymbol{\beta}}^2}}. $$

$$ \text{As } 1 - \mathrm{th}^2(\boldsymbol{\alpha}) = \frac{1}{\mathrm{ch}^2(\boldsymbol{\alpha})} \Rightarrow \mathrm{ch}^2(\boldsymbol{\alpha}) = \frac{1}{1 - \mathrm{th}^2(\boldsymbol{\alpha})} = \frac{1}{1 - \vec{\boldsymbol{\beta}}} = \gamma^2, $$

$$ \begin{align*} & \text{as } \mathrm{ch}(\boldsymbol{\alpha}) \ge 1 \quad \gamma = \mathrm{ch}(\boldsymbol{\alpha}); \text{ as } \mathrm{sh}^2(\boldsymbol{\alpha}) = \mathrm{ch}^2(\boldsymbol{\alpha}) - 1 = \gamma^2 - 1 \\ & = \frac{1}{1 - \vec{\boldsymbol{\beta}}} - 1 = \frac{\vec{\boldsymbol{\beta}}^2}{1 - \vec{\boldsymbol{\beta}}} = \gamma^2 \vec{\boldsymbol{\beta}}^2. \end{align*} $$

$$ \text{For short we have } \gamma = \mathrm{ch}(\boldsymbol{\alpha}), \gamma^2 \vec{\boldsymbol{\beta}}^2 = \mathrm{sh}^2(\boldsymbol{\alpha}), \vec{\boldsymbol{\beta}}^2 = \mathrm{th}^2(\boldsymbol{\alpha}). $$

Furthermore:

$$ \gamma^2 \vec{\boldsymbol{\beta}}^2 = \gamma^2 - 1 = (\gamma + 1)(\gamma - 1) \Rightarrow \frac{\gamma^2 \vec{\boldsymbol{\beta}}^2}{(1 + \gamma)} = (\gamma - 1) = \mathrm{ch}(\boldsymbol{\alpha}) - 1 $$

$$ \text{et } X_i X_j = \frac{(\boldsymbol{\beta}_i \boldsymbol{\beta}_j)}{\mathrm{th}^2(\boldsymbol{\alpha})} = \frac{(\boldsymbol{\beta}_i \boldsymbol{\beta}_j)}{\vec{\boldsymbol{\beta}}^2} \quad \text{consequently} $$

$$ (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X_i X_j = \frac{\gamma^2 \vec{\boldsymbol{\beta}}^2}{(1 + \gamma)} \frac{(\boldsymbol{\beta}_i \boldsymbol{\beta}_j)}{\vec{\boldsymbol{\beta}}^2} = \frac{\gamma^2}{(1 + \gamma)} \boldsymbol{\beta}_i \boldsymbol{\beta}_j, \text{ that implies:} $$
$$ \mathrm{Id}_{\mathbb{R}^3} + (\mathrm{ch}(\boldsymbol{\alpha}) - 1)X'X = \begin{bmatrix} 1 + \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_1^2 & \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_1 \boldsymbol{\beta}_2 & \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_1 \boldsymbol{\beta}_3 \\ \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_2 \boldsymbol{\beta}_1 & 1 + \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_2^2 & \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_2 \boldsymbol{\beta}_3 \\ \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_3 \boldsymbol{\beta}_1 & \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_3 \boldsymbol{\beta}_2 & 1 + \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_3^2 \end{bmatrix} = C $$

As $\mathrm{sh}(\boldsymbol{\alpha})X_i = \frac{\mathrm{sh}(\boldsymbol{\alpha})\boldsymbol{\beta}_i}{\mathrm{th}(\boldsymbol{\alpha})} = \mathrm{ch}(\boldsymbol{\alpha})\boldsymbol{\beta}_i = \boldsymbol{\gamma}\boldsymbol{\beta}_i$, we have finally:

$$ \exp(\alpha N) = \begin{bmatrix} \boldsymbol{\gamma} & t[\boldsymbol{\gamma}\boldsymbol{\beta}] \\ [\boldsymbol{\gamma}\boldsymbol{\beta}] & C \end{bmatrix} \quad \text{with} \quad [\boldsymbol{\gamma}\boldsymbol{\beta}] \text{ the coordinates of } \boldsymbol{\gamma}\boldsymbol{\beta}. $$

We notice that the inverse of $\exp(\alpha N)$ is $\exp(-\alpha N)$, simply by replacing $\boldsymbol{\gamma}\boldsymbol{\beta}$ with $\boldsymbol{\gamma}\boldsymbol{\beta}$ to obtain the inverse while noting that $\boldsymbol{\gamma}$ and $C$ remain unchanged, since: $\exp(\alpha N) \cdot \exp(-\alpha N) = \exp(0) = \mathrm{Id}$

if $\boldsymbol{\gamma}\boldsymbol{\beta}$ is parallel to the space coordinate of $\boldsymbol{i}$ then only the diagonal term of $C$ are nonzero : 1

$$ + \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_p^2, 1, 1 \text{ and} $$

$$ 1 + \frac{\boldsymbol{\gamma}^2}{(1+\boldsymbol{\gamma})} \boldsymbol{\beta}_1^2 = \frac{(1+\boldsymbol{\gamma}) + \boldsymbol{\gamma}^2 \boldsymbol{\beta}^2}{(1+\boldsymbol{\gamma})} = \frac{1+\boldsymbol{\gamma}+\boldsymbol{\gamma}^2-1}{(1+\boldsymbol{\gamma})} = \boldsymbol{\gamma} \quad \text{because} $$

$$ \boldsymbol{\gamma}^2 - 1 = \frac{1}{1-\boldsymbol{\beta}} - 1 = \frac{\boldsymbol{\beta}^2}{1-\boldsymbol{\beta}} = \boldsymbol{\gamma}^2 \boldsymbol{\beta}^2. $$

**Other forms of expression of matrix M:**

We can also write after the above;

$$ M = \begin{bmatrix} \boldsymbol{\varepsilon} \cdot \boldsymbol{\gamma} & t[\boldsymbol{\gamma}\boldsymbol{\beta}] \cdot \boldsymbol{\Omega} \\ \boldsymbol{\varepsilon} \cdot [\boldsymbol{\gamma}\boldsymbol{\beta}] & C \cdot \boldsymbol{\Omega} \end{bmatrix} = \exp \left( \mathrm{Argch}(\boldsymbol{\gamma}) \begin{bmatrix} 0 & \frac{t[\boldsymbol{\gamma}\boldsymbol{\beta}]}{\boldsymbol{\beta}} \\ \frac{[\boldsymbol{\gamma}\boldsymbol{\beta}]}{\boldsymbol{\beta}} & 0 \end{bmatrix} \right). \begin{bmatrix} \boldsymbol{\varepsilon} & 0 \\ 0 & \boldsymbol{\Omega} \end{bmatrix}. $$

**Another form of the polar form :**

Furthermore if $\boldsymbol{\varepsilon}=1$:
$$
\text{Let's write: } \Lambda(\vec{\beta}) = \begin{bmatrix} \gamma & {}^{t}[\vec{\gamma\beta}] \\ [\vec{\gamma\beta}] & C([\vec{\beta}]) \end{bmatrix} \quad \text{and} \quad \tilde{\Omega} = \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix}.
$$

We have

$$
M = \Lambda(\vec{\beta}). \tilde{\Omega} = \tilde{\Omega} \cdot (\tilde{\Omega} \cdot \Lambda(\vec{\beta}) \cdot \tilde{\Omega}) = \tilde{\Omega} \cdot \Lambda({}^t[\Omega]\vec{\beta}) \text{ because :}
$$

$$
\begin{align*}
& \tilde{\Omega} \cdot \Lambda(\vec{\beta}) \cdot \tilde{\Omega} = \begin{bmatrix} 1 & 0 \\ 0 & {}^t \Omega \end{bmatrix} \cdot \begin{bmatrix} \gamma & {}^{t}[\vec{\gamma\beta}] \\ [\vec{\gamma\beta}] & C([\vec{\beta}]) \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} \\
&= \begin{bmatrix} \gamma & {}^{t}[\vec{\gamma\beta}] \\ {}^t \Omega \cdot [\vec{\gamma\beta}] & {}^t \Omega \cdot C \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} = \begin{bmatrix} \gamma & {}^{t}[\vec{\gamma\beta}] \cdot \Omega \\ {}^t \Omega \cdot [\vec{\gamma\beta}] & {}^t \Omega \cdot C \cdot \Omega \end{bmatrix} = \begin{bmatrix} \gamma & \gamma \cdot {}^t({}^t[\Omega]\vec{\beta}) \\ {}^t \gamma \cdot {}^t \Omega \cdot [\vec{\gamma\beta}] & {}^t \gamma \cdot {}^t \Omega \cdot C \cdot \Omega \end{bmatrix}
\end{align*}
$$

as $C = \text{Id} + \frac{\gamma^2}{(1+\gamma)} [\vec{\beta}] \cdot [\vec{\beta}]^{\text{t}}$ we have :

$$
{}^t \Omega \cdot C \cdot \Omega = \mathrm{Id} + \frac{\gamma^2}{(1+\gamma)} {}^t \Omega \cdot [\vec{\beta}] \cdot [\vec{\beta}]^{\mathrm{t}} \cdot \Omega = \mathrm{Id} + \frac{\gamma^2}{(1+\gamma)} ({}^t \Omega [\vec{\beta}]^{\mathrm{t}}) \cdot ({}^t \Omega [\vec{\beta}]^{\mathrm{t}}).
$$

**Independence of Ω with respect to β → when ε = +1 :**

Now let's show that Ω does not depend on **β**. We assume ε = +1 (non – time inversion).

Let **M** be a Lorentz matrix of order 4; it represents the transformation from a spacetime basis of an observer **O** to another spacetime basis associated with another observer **O'**, who is in relative motion with respect to **O**, moving with a uniform velocity **v**.

More precisely : Let $\mathcal{B}(c\vec{\tau}, i, j, k)$ be the basis associated with observer **O**,
and $\mathcal{B}'(c\vec{\tau}', i', j', k')$ be the basis associated with observer **O'**.

$$
M = \mathcal{P}(\mathcal{B}(c\vec{\tau}, i, j, k), \mathcal{B}'(c\vec{\tau}', i', j', k')) \text{ is the change of basis matrix.}
$$

A change of – basis matrix $\mathcal{P}(\mathcal{B}(c\vec{\tau}, i, j, k), \mathcal{B}'(c\vec{\tau}', i', j', k'))$ is the representation
of vectors $(c\vec{\tau}', i', j', k')$ in the old basis $\mathcal{B}(c\vec{\tau}, i, j, k)$

For observer O, we consider the basis $\mathcal{B}_I(c\vec{\tau}, i_I, j_I, k_I)$ such that the change – of – basis matrix

$$
A = \mathcal{P} \left( \mathcal{B}(c\vec{\tau}, i, j, k), \mathcal{B}_I(c\vec{\tau}, i_I, j_I, k_I) \right) \text{ is an orthogonal matrix of the form}
$$

$$
A = \begin{bmatrix} 1 & 0 \\ 0 & R_I \end{bmatrix}_{\rightarrow} \text{ where } R_I \text{ is an orthogonal operator, transforms the spatial components of } i_{\rightarrow} \\
\text{to those of } i_I \text{ such that they are parallel and have the same direction as } \vec{\beta}_{\rightarrow}.
$$
For the observer $O'$, we consider the basis $\mathcal{B}'_l(c\tau', i'_p, j'_p, k'_l)$ such that the tchange of
-basis matrix

$$
{}^t\mathbf{B} = \mathcal{P} \left( \mathcal{B}'(c\tau', i', j', k') , \mathcal{B}'_l(c\tau', i'_p, j'_p, k'_l) \right) \text{ is an orthogonal matrix of the form}
$$

$$
{}^t \mathbf{B} = \begin{bmatrix} 1 & 0 \\ 0 & R_2 \end{bmatrix}_{\substack{\rightarrow \\ \rightarrow}} \quad \text{où } R_2, \text{ an orthogonal operator, transforms the spatial components of} \\
\text{i' to those of } i'_1 \text{ such that it is parallel and has the same direction as } \vec{\beta}, \text{ and } j'_1 \text{ and } \vec{k}'_1 \text{ are parallel} \\
\text{and have the same direction as } j'_1, \text{ and } \vec{k}'_1 \text{ respectively, (rotation around the first axis).}
$$

Let's calculate the operator $N$ that allows us to transform the basis

$$
\mathcal{B}(c\vec{\tau}, i_1, j_1, k_1) \text{ to the basis } \mathcal{B}'(c\vec{\tau}', i'_p, j'_p, k'_1):
$$

$$
N = \mathcal{P} \left( \mathcal{B}(c\vec{\tau}, i_1, j_1, k_1), \mathcal{B}'(c\vec{\tau}', i'_p, j'_p, k'_1) \right).
\xrightarrow{\rightarrow} \rightarrow
$$

Since $\mathbf{\beta}$ is perpendicular to the spatial components of $\mathbf{j}_1$ and $\mathbf{k}_1$, we have:

$$
N = \begin{bmatrix} \gamma & \gamma - \beta & 0 & 0 \\ \gamma - \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \omega_{1,1} & \omega_{1,2} & \omega_{1,3} \\ 0 & \omega_{2,1} & \omega_{2,2} & \omega_{2,3} \\ 0 & \omega_{3,1} & \omega_{3,2} & \omega_{3,3} \end{bmatrix} = \begin{bmatrix} \gamma & \gamma\beta\omega_{1,1} & \gamma\beta\omega_{1,2} & \gamma\beta\omega_{1,3} \\ \gamma\beta & \gamma\omega_{1,1} & \gamma\omega_{1,2} & \gamma\omega_{1,3} \\ 0 & \omega_{2,1} & \omega_{2,2} & \omega_{2,3} \\ 0 & \omega_{3,1} & \omega_{3,2} & \omega_{3,3} \end{bmatrix}.
$$

As the spatial components of the new spatial vectors are parallel and in the same direction as
those of the old one, we necessarily have :

$$
\omega_{2,1} = \omega_{3,1} = \omega_{3,2} = \omega_{1,2} = \omega_{2,3} = \omega_{1,3} = 0 \text{ et } \omega_{1,1} > 0 \text{ then :}
$$

$$
N = \begin{bmatrix} \gamma & \gamma\beta\omega_{1,1} & 0 & 0 \\ \gamma\beta & \gamma\omega_{1,1} & 0 & 0 \\ 0 & 0 & \omega_{2,2} & 0 \\ 0 & 0 & 0 & \omega_{3,3} \end{bmatrix},
$$

$$
\text{as } \gamma^2 - \gamma^2 \beta^2 = 1 \text{ and } {}^t N \cdot G - N = G \text{ imply that } \gamma^2 \beta^2 \omega_{1,1}^2 - \gamma^2 \omega_{1,1} = -1,
$$

$$
\text{and we have } \omega_{1,1} = \omega_{2,2} = \omega_{3,3} = 1 \text{ et } N = \begin{bmatrix} \gamma & \gamma\beta & 0 & 0 \\ \gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.
$$

Now we have :

$$
\begin{align*}
M &= \mathcal{P}(\mathcal{B}(c\vec{\tau}, \vec{i}, \vec{j}, \vec{k}), \mathcal{B}'(c\vec{\tau}', \vec{i}', \vec{j}', \vec{k}')) \\
  &= \mathcal{P}(\mathcal{B}(c\vec{\tau}, \vec{i}, \vec{j}, \vec{k}), \mathcal{B}(c\vec{\tau}, \vec{i}', \vec{j}', \vec{k}')) \mathcal{P}(\mathcal{B}(c\vec{\tau}', \vec{i}', \vec{j}', \vec{k}')) \mathcal{P}(\mathcal{B}'(c\vec{\tau}', \vec{i}', \vec{j}', \vec{k}'), \mathcal{B}'(c\vec{\tau}', \vec{i}', \vec{j}', \vec{k}'))
\end{align*}
$$
$$ =A \cdot N \cdot B = (A \cdot N \cdot {}^t A) \cdot (A \cdot B) = \begin{bmatrix} \gamma & {}^t[\vec{\gamma\beta}] \\ [\vec{\gamma\beta}] & C \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix}. $$

By the unicity of the decomposition, we have
$$ \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} = A \cdot B $$

$$ \text{and} \quad \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix} \cdot {}^t[\vec{\gamma\beta}] = A \cdot \begin{bmatrix} \gamma & \gamma\beta & 0 & 0 \\ \gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot {}^t A. $$

$\Omega$ is therefore independent of $\vec{\beta}$ since A and B are independent of $\vec{\beta}$

$$ \text{As } \lim_{\vec{\beta} \to 0} \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix} \cdot {}^t[\vec{\gamma\beta}] \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix}, $$

$$ \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} \text{ is the Lorentz matrix } M \text{ for } \vec{\beta} = \vec{0}. $$

### Transformation of Galileo:

This also allows us to show that the Galileo transformation is the limit as $c \to +\infty$ of the Lorentz transformation. We notice that if we write the Lorentz matrix in the form:

$$ M = \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix} + \text{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)} [\vec{\beta}] \cdot {}^t[\vec{\beta}] I \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} $$

$$ = \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix} + \begin{bmatrix} \text{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)} [\vec{\beta}] \cdot {}^t[\vec{\beta}] \Omega \\ [\vec{\gamma\beta}] \end{bmatrix} $$

On the other hand:

$$ MY' = \begin{bmatrix} \gamma \\ [\vec{\gamma\beta}] \end{bmatrix} + \begin{bmatrix} \text{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)} [\vec{\beta}] \cdot {}^t[\vec{\beta}] \Omega \\ [\vec{\gamma\beta}] \end{bmatrix} \cdot \begin{bmatrix} {}^t[\vec{\gamma\beta}] \Omega \\ [ct'] \\ [X'] \end{bmatrix} $$
$$ = \begin{bmatrix} c\gamma & t[\vec{\gamma\beta}]\Omega \\ c[\vec{\gamma\beta}] & (\mathrm{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)}[\vec{\beta}] \cdot t[\vec{\beta}])\Omega \end{bmatrix} \begin{bmatrix} t' \\ X' \end{bmatrix} = \begin{bmatrix} ct \\ X \end{bmatrix} $$

and then :

$$ \begin{bmatrix} \gamma \\ c[\vec{\gamma\beta}] \end{bmatrix} \begin{bmatrix} t[\vec{\gamma\beta}] \frac{\Omega}{c} \\ (\mathrm{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)}[\vec{\beta}] \cdot t[\vec{\beta}])\Omega \end{bmatrix} \begin{bmatrix} t' \\ X' \end{bmatrix} = \begin{bmatrix} t \\ X \end{bmatrix} $$

since $[\vec{\beta}] = [\frac{\vec{V}}{c}]$ and $\gamma = \frac{1}{1 - (\frac{\vec{V}}{c})^2}$ hence:

$$ \begin{bmatrix} \gamma \\ [\vec{\gamma\mathcal{W}}] \end{bmatrix} \begin{bmatrix} t[\vec{\gamma\mathcal{V}}] \frac{\Omega}{c} \\ (\mathrm{Id}_{\mathbb{R}^3} + \frac{\gamma^2}{(1+\gamma)}[\vec{\mathcal{V}}] \cdot t[\vec{\mathcal{V}}])\Omega \end{bmatrix} \begin{bmatrix} t' \\ X' \end{bmatrix} = \begin{bmatrix} t \\ X \end{bmatrix} \quad \text{then} $$

For a given velocity $\vec{V}$, $\Omega$ is independent of $c$,

$$ \text{if } c \to +\infty \text{, we have } \begin{bmatrix} 1 & 0 \\ [\vec{V}] & \Omega \end{bmatrix} \begin{bmatrix} t' \\ X' \end{bmatrix} = \begin{bmatrix} t \\ X \end{bmatrix}, $$

that is the Galileo transformation.

## Calculation of eigenvalues;

We can determine the eigenvalues directly :

$$ \text{As } \Lambda = \begin{bmatrix} 1 + \frac{\gamma^2}{(1+\gamma)} \beta_1^2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_2 & \frac{\gamma^2}{(1+\gamma)} \beta_1 \beta_3 \\ \frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_1 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_2^2 & \frac{\gamma^2}{(1+\gamma)} \beta_2 \beta_3 \\ \frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_1 & \frac{\gamma^2}{(1+\gamma)} \beta_3 \beta_2 & 1 + \frac{\gamma^2}{(1+\gamma)} \beta_3^2 \end{bmatrix} = A \begin{bmatrix} \gamma & \gamma\beta & 0 & 0 \\ \gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. $$

$^T A$, where $A$ est orthogonal : $A.A^{-1} = \mathrm{Id}$.
$$ \text{we note that } \det(\Lambda - \lambda Id) = \det A \cdot {}^T A - \lambda \cdot A \cdot {}^T A $$
$$ \begin{bmatrix} \gamma & \gamma-\beta & 0 & 0 \\ \gamma-\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

$$ = \det(A) \det \left( \begin{bmatrix} \gamma & \gamma-\beta & 0 & 0 \\ \gamma-\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} - \lambda \cdot \mathrm{Id} \right) \det({}^T A). $$

and

$$ \begin{vmatrix} \gamma & \gamma-\beta & 0 & 0 \\ \gamma-\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 & 0 \\ +\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix} \begin{vmatrix} \gamma-(1+\beta) & 0 & 0 & 0 \\ 0 & \gamma-(1-\beta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix} \begin{vmatrix} \frac{\sqrt{2}}{2} & +\frac{\sqrt{2}}{2} & 0 & 0 \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix} $$

In the general case, it is shown that every Lorentz matrix admits at least one real eigenvalue : (cf: E • GOURGOULON : Relativité restreinte EDP Sciences).

As it is of order 4, it has 2 eigenvalues. This is the maximum number for the general case : consider the case where M is orthogonal to Ω, a rotation matrix around the Ox axis.

## Another expression of Ω :

We can directly calculate Ω because

$$ \text{if } M = \Lambda(\vec{\beta}).\tilde{\Omega} \Leftrightarrow \Omega = \Lambda(-\vec{\beta}) \cdot M. $$

It is possible to have a simpler calculation :

$$ \text{We have } (\vec{\beta} \leftrightarrow \beta) (\vec{\beta} \leftrightarrow \beta) = \vec{\beta} (\leftrightarrow \beta\beta) \leftrightarrow \beta = \beta^2 (\leftrightarrow \beta\beta) \text{ and } \vec{\gamma}^2 \vec{\beta}^2 = \vec{\gamma}^2 - 1, \\ \text{that imply that :} $$

$$ \begin{aligned} & \left( \mathrm{Id}_{\mathbb{R}^3} + \frac{[\vec{\gamma\beta}]^t [\vec{\gamma\beta}]}{(1+\gamma)} \right) \cdot \left( \mathrm{Id}_{\mathbb{R}^3} - \gamma \frac{\vec{\beta} \leftrightarrow \beta}{(1+\gamma)} \right) \\ &= \mathrm{Id}_{\mathbb{R}^3} - \gamma \frac{\vec{\beta} \leftrightarrow \beta}{(1+\gamma)} + \gamma^2 \frac{\vec{\beta} \leftrightarrow \beta}{(1+\gamma)} - \gamma^3 \frac{\vec{\beta}^2 (\vec{\beta} \leftrightarrow \beta)}{(1+\gamma)^2} \\ &= \mathrm{Id}_{\mathbb{R}^3} + \frac{\vec{\beta} \leftrightarrow \beta}{(1+\gamma)^2} \left( -\gamma(1+\gamma) + \gamma^2(1+\gamma) - \gamma(\gamma^2-1) \right) = \mathrm{Id}_{\mathbb{R}^3}. \end{aligned} $$
Then $C^{-1} = \left( \mathrm{Id}_{\mathbb{R}^3} + \frac{[\vec{\gamma\beta]^t}[\vec{\gamma\beta]^t}}{(1+\gamma)} \right)^{-1} = \mathrm{Id}_{\mathbb{R}^3} - \gamma \frac{\vec{\gamma\beta} \leftrightarrow \beta}{(1+\gamma)}$

As $M = \begin{bmatrix} \gamma & t[\vec{\gamma\beta}] \\ [\vec{\gamma\beta}] & C \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} = \begin{bmatrix} \gamma & t[\vec{\gamma\beta}]\Omega \\ [\vec{\gamma\beta}] & C\Omega \end{bmatrix}$

Now if we write $M = \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix}$.

We have $\Omega = \begin{bmatrix} \mathrm{Id}_{\mathbb{R}^3} - \gamma \frac{\vec{\gamma\beta} \leftrightarrow \beta}{(1+\gamma)} \end{bmatrix} \begin{bmatrix} m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,2} & m_{3,3} & m_{3,4} \\ m_{3,2} & m_{4,3} & m_{4,4} \end{bmatrix}$.

**A final remark :**

We can note :

$M$ symmetric $\Leftrightarrow$ the spatial components of the spatial vectors of $\mathcal{B}(\vec{c\tau}, \vec{i}, \vec{j}, \vec{k})$ and $\mathcal{B}'(\vec{c\tau'}, \vec{i'}, \vec{j'}, \vec{k'})$ are parallel.

Now we propose two other ways to calculate the terms of $\exp(\alpha N)$ :

**Second method :**

The second method starts with a demonstration by J. Parizet in "La géométrie de la relativité restreinte (Ellipses) p.89."

Let's Ma Lorentz matrix : $G = tMGM$ that implies

$GM^{-1} = tMG$ and $M^{-1} = G'MG$.

We can write $M$ as $\begin{bmatrix} \gamma & tU \\ V & A \end{bmatrix}$ où $\gamma \in \mathbb{R}$, $U$ et $V$ 2 vecteurs de $\mathbb{R}^3$ et $A$ une matrice 3 x 3.

from $M^{-1} = G'MG$ we have $M^{-1} = \begin{bmatrix} \gamma & -tV \\ -U & tA \end{bmatrix}$ and then :

$$M \cdot M^{-1} = \begin{bmatrix} \gamma^2 - tV \cdot V & \gamma \cdot tU - tV \cdot A \\ -\gamma U + tA \cdot V & tA \cdot A - U \cdot tU \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \mathrm{Id}_3 \end{bmatrix} \quad \text{that implies :}$$

$\gamma^2 - tV \cdot V = 1, -\gamma U + tA \cdot V = \gamma \cdot tU - tV \cdot A = 0, tA \cdot A - U \cdot tU = \mathrm{Id}_3$.
We note $\gamma^2 \ge 1$ since $\gamma^2 - 1 = {}^tV \cdot V \ge 0$, ${}^tA \cdot V = \gamma U$ and ${}^tV \cdot A = \gamma \cdot {}^tU$.

We have $V=0 \Leftrightarrow \gamma^2=1$, $V=0 \Leftrightarrow U=0$, $V=0 \Leftrightarrow {}^tA \cdot A = \text{Id}_3$.

Now if $V=0$ then $M = \begin{bmatrix} \pm 1 & 0 \\ 0 & A \end{bmatrix}$ with ${}^tA \cdot A = \text{Id}_3$.

Consider the case $V \neq 0$ and then $\gamma^2 > 1$.

We then consider $\Omega = A - \frac{V \cdot {}^tU}{1 + \gamma}$.

$$ \begin{aligned} \text{Let' calculate } {}^t\Omega \cdot \Omega &= \left({}^tA - \frac{U \cdot {}^tV}{1 + \gamma}\right) \left(A - \frac{V \cdot {}^tU}{1 + \gamma}\right) \\ &= {}^tA \cdot A - \frac{{}^tAV \cdot {}^tU}{1 + \gamma} - \frac{U \cdot VA}{1 + \gamma} + \frac{U \cdot VV \cdot U}{(1 + \gamma)^2}, \\ &= \text{Id}_3 + U \cdot {}^tU - \frac{2 \cdot \gamma U \cdot U}{1 + \gamma} + \frac{(\gamma^2 - 1)U \cdot U}{(\gamma^2 - 1)} \\ &= \text{Id}_3 + U \cdot {}^tU \left(1 - \frac{2 \cdot \gamma}{1 + \gamma} + \frac{(\gamma^2 - 1)}{(1 + \gamma)^2}\right) \\ &= \text{Id}_3 + U \cdot {}^tU \left(1 + \gamma - 2 \cdot \gamma + \gamma - 1\right) = \text{Id}_3. \end{aligned} $$

We conclude that $\Omega$ is an orthogonal matrix.

Now we can consider:

$$ {}^tV \cdot \Omega = {}^tV \cdot \left(A - \frac{V \cdot {}^tU}{1 + \gamma}\right) = {}^tV \cdot A - \frac{{}^tV \cdot V \cdot {}^tU}{1 + \gamma} = \gamma \cdot {}^tU - \frac{(\gamma^2 - 1) \cdot {}^tU}{1 + \gamma} = {}^tU. $$

Then $\Omega \cdot U = \Omega^t \cdot \Omega \cdot V = V \cdot U$. Because $\Omega = A - \frac{V \cdot {}^tU}{1 + \gamma}$

$$ M = \begin{bmatrix} \gamma & {}^tU \\ V & A \end{bmatrix} = \begin{bmatrix} \gamma & {}^tU \\ V & \Omega + \frac{V \cdot {}^tU}{1 + \gamma} \end{bmatrix} = \begin{bmatrix} \gamma & {}^tU \cdot {}^tO \\ V & \text{Id} + \frac{V \cdot {}^tU \cdot {}^t\Omega}{1 + \gamma} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} \quad \text{and then:} $$

$$ M = \begin{bmatrix} \gamma & {}^tV \\ V & \text{Id} + \frac{V^tV}{1 + \gamma} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix}. $$

If we consider $O$ and $O'$, $O'$ with a constant speed $\vec{v}$ with respect to $O$:

$O'$ has the coordinate $\begin{bmatrix} c \cdot t' \\ 0 \end{bmatrix}_{\mathbb{R}^3}$ in the basis associated with $O'$ and

$\begin{bmatrix} c \cdot t \\ [\vec{v}] \cdot t \end{bmatrix}$ in the basis associated with $O$.
$$ \text{Then : } \begin{bmatrix} c \cdot t \\ \vec{\nu} \end{bmatrix} \cdot t = \begin{bmatrix} \gamma & {}^{t}V \\ V & \text{Id} + \frac{V^{t}V}{1+\gamma} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} \begin{bmatrix} c \cdot t' \\ 0 \end{bmatrix} $$

we deduce from it : $t = \gamma \cdot t$ and $\vec{\nu} \cdot t = c \cdot t' \cdot V$ then $V = \frac{\gamma[\vec{\nu}]}{c} = \gamma[\vec{\beta}]$

Finally :

$$ M = \begin{bmatrix} \gamma & \gamma \cdot [\vec{\beta}] \\ \gamma[\vec{\beta}] & \text{Id} + \frac{\gamma^2[\vec{\beta}]^t[\vec{\beta}]}{1+\gamma} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix} $$

We can also write $M = \begin{bmatrix} -\gamma & -\gamma[\vec{\beta}] \\ -\gamma[\vec{\beta}] & -\left(\text{Id} + \frac{\gamma^2[\vec{\beta}]^t[\vec{\beta}]}{1+\gamma}\right) \end{bmatrix} \cdot \begin{bmatrix} -1 & 0 \\ 0 & -\Omega \end{bmatrix}$ and

In a general way $M$ can be written as :

$$ M = \begin{bmatrix} \gamma & \gamma \cdot [\vec{\beta}] \\ \gamma[\vec{\beta}] & \text{Id} + \frac{\gamma^2[\vec{\beta}]^t[\vec{\beta}]}{1+\gamma} \end{bmatrix} \cdot \begin{bmatrix} \epsilon & 0 \\ 0 & \Omega \end{bmatrix} \quad \text{with } \epsilon = \pm 1, \gamma \ge 1, \gamma = \frac{1}{\sqrt{1 - \vec{\beta} \cdot \vec{\beta}}} $$

with $\Omega$ an orthogonal matrix.

### Third method :

We start from: Special relativity by Ulrich E. Schroder, page 92, World Scientific, modifying his demonstration.

(The book can be consulted at : https://archive.org/details/specialrelativit0000schr/page/92/mode/2 up"?view=theater).

Following Schroder, let us consider a symmetric Lorentz matrix $M$ of the form :

$$ M = A \cdot \begin{bmatrix} \gamma & \gamma \cdot \beta & 0 & 0 \\ \gamma \cdot \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot A^T \quad \text{with } \beta = \|\vec{\beta}\|, A = \begin{bmatrix} 1 & 0 \\ 0 & \Omega \end{bmatrix}, \Omega \text{ an orthogonal matrix 3 x 3.} $$

$M$ is indeed a Lorentz matrix, since it can be expressed as the product of 3 Lorentz matrices.

Since $\Omega$ is a 3 x 3 rotation matrix,

it can be decomposed into a product of 3 elementary rotations (Euler angles) :

see, for example, https://en.wikipedia.org/wiki/Rotation_formalisms_in three\_dimensions •
$$
\text{If we define: } \mathbf{Rx} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos(\theta) & -\sin(\theta) \\ 0 & 0 & \sin(\theta) & \cos(\theta) \end{bmatrix}, \quad \mathbf{Ry} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\varphi) & 0 & \sin(\varphi) \\ 0 & 0 & 1 & 0 \\ 0 & -\sin(\varphi) & 0 & \cos(\varphi) \end{bmatrix}, \quad \mathbf{Rz} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\psi) & -\sin(\psi) & 0 \\ 0 & \sin(\psi) & \cos(\psi) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
$$

We can write $\mathbf{A} = \mathbf{Rz} \cdot \mathbf{Ry} \cdot \mathbf{Rx}$.

This order of composition is chosen to simplify the calculations.

It is observed that due to rotational invariance around the x-axis.

$$
\mathbf{Rx.} \quad \begin{bmatrix} \gamma & \gamma-\beta & 0 & 0 \\ \gamma-\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot' \mathbf{Rx} = \begin{bmatrix} \gamma & \gamma-\beta & 0 & 0 \\ \gamma-\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
$$

After calculation (with Maple) :

$$
A = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & \cos(\psi) \cos(\varphi) & -\sin(\psi) \cos(\theta) + \cos(\psi) \sin(\varphi) \sin(\theta) & \sin(\psi) \sin(\theta) + \cos(\psi) \sin(\varphi) \cos(\theta) \\
0 & \sin(\psi) \cos(\varphi) & \cos(\psi) \cos(\theta) + \sin(\psi) \sin(\varphi) \sin(\theta) & -\cos(\psi) \sin(\theta) + \sin(\psi) \sin(\varphi) \cos(\theta) \\
0 & -\sin(\varphi) & \cos(\varphi) \sin(\theta) & \cos(\varphi) \cos(\theta)
\end{bmatrix}
$$

We have since the $\theta$ variable is eliminated :

$$
\mathbf{M} = \begin{bmatrix}
\gamma & \cos(\psi) \cos(\varphi) \gamma\beta & \sin(\psi) \cos(\varphi) \gamma\beta & -\sin(\varphi) \gamma\beta \\
\cos(\psi) \cos(\varphi) \gamma\beta & 1 + \cos(\psi)^2 (-1 + \gamma) \cos(\varphi)^2 & \sin(\psi) \cos(\varphi) \cos(\varphi)^2 (-1 + \gamma) & -\cos(\varphi) \cos(\psi) \sin(\varphi) (-1 + \gamma) \\
\sin(\psi) \cos(\varphi) \gamma\beta & \sin(\psi) \cos(\varphi) \cos(\varphi)^2 (-1 + \gamma) & 1 + ((1 - \gamma) \cos(\psi)^2 - 1 + \gamma) \cos(\varphi)^2 & -\cos(\varphi) \sin(\psi) \sin(\varphi) (-1 + \gamma) \\
-\sin(\varphi) \gamma\beta & -\cos(\varphi) \cos(\psi) \sin(\varphi) (-1 + \gamma) & -\cos(\varphi) \sin(\psi) \sin(\varphi) (-1 + \gamma) & (1 - \gamma) \cos(\varphi)^2 + \gamma
\end{bmatrix}
$$

$\rightarrow$

One can still directly evaluate the first column of a general Lorentz matrix with $\beta$ by considering
the coordinates of the observer $O'$ in the two bases associated with $O$ and $O'$:

$$
\text{If we write } \mathbf{M} = \begin{bmatrix} \mathbf{m}_1, 1 & \mathbf{m}_1, 2 & \mathbf{m}_1, 3 & \mathbf{m}_1, 4 \\ \mathbf{m}_2, 1 & \mathbf{m}_2, 2 & \mathbf{m}_2, 3 & \mathbf{m}_2, 4 \\ \mathbf{m}_3, 1 & \mathbf{m}_3, 2 & \mathbf{m}_3, 3 & \mathbf{m}_3, 4 \\ \mathbf{m}_4, 1 & \mathbf{m}_4, 2 & \mathbf{m}_4, 3 & \mathbf{m}_4, 4 \end{bmatrix} \quad \text{then } \mathbf{M} \cdot \begin{bmatrix} \mathbf{c} \cdot \mathbf{t}' \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}_{O'} = \begin{bmatrix} \mathbf{c} \cdot \mathbf{t} \\ \mathbf{c} \cdot \mathbf{\beta}_1 \cdot \mathbf{t} \\ \mathbf{c} \cdot \mathbf{\beta}_2 \cdot \mathbf{t} \\ \mathbf{c} \cdot \mathbf{\beta}_3 \cdot \mathbf{t} \end{bmatrix}_{O} \quad \text{hence :}
$$

if we write $\boldsymbol{\gamma}=\mathbf{m}_{1,1}$ then $\boldsymbol{\gamma}-\boldsymbol{t}'=\boldsymbol{t}$ moreover $\boldsymbol{\gamma}>\mathbf{0}$. (Principle of causality).

From this, we deduce that the first column of $\mathbf{M}$ is written as
$$ \begin{bmatrix} \gamma \\ \gamma \cdot \boldsymbol{\beta}_1 \\ \gamma \cdot \boldsymbol{\beta}_2 \\ \gamma \cdot \boldsymbol{\beta}_3 \end{bmatrix}. $$
By identification : $\cos(\psi) \cos(\varphi) = \frac{\beta_1}{\beta}$ , $\sin(\psi) \cos(\varphi) = \frac{\beta_2}{\beta}$ , $\sin(\varphi) = -\frac{\beta_3}{\beta}$ .

We recall that $\gamma^2 \cdot \beta^2 = \gamma^2 - 1$ since $1 = \gamma^2(1 - \beta^2)$ since $^tMGM = G$.

Calculation of the $m_{i,j}$:

$$ 
\begin{aligned}
m_{2,2} &= 1 + \cos(\psi)^2 (-1 + \gamma) \cos(\varphi)^2 = 1 + \cos(\psi)^2 \cos(\varphi)^2 \frac{\gamma^2 \cdot \beta^2}{1 + \gamma} \\
&= 1 + \left(\frac{\beta_1}{\beta}\right)^2 \frac{\gamma^2 \cdot \beta^2}{1 + \gamma} = 1 + \frac{\gamma^2}{1 + \gamma} \beta_1^2.
\end{aligned}
$$

$$ 
\begin{aligned}
m_{3,3} &= 1 + ((1 - \gamma) \cos(\psi)^2 - 1 + \gamma) \cos(\varphi)^2 = 1 + (1 - \gamma) (\cos(\psi)^2 \cos(\varphi)^2 - \cos(\varphi)^2) \\
&= 1 + (1 - \gamma) (\cos(\psi)^2 \cos(\varphi)^2 - 1 + \sin(\varphi)^2) = 1 + (1 - \gamma) \left( \frac{\beta_1^2 - \beta^2 + \beta_3^2}{\beta^2} \right) \\
&= 1 + \frac{\gamma^2}{1 + \gamma} \cdot \beta_2^2.
\end{aligned}
$$

$$ m_{4,4} = (1 - \gamma) \cos(\varphi)^2 + \gamma = 1 + (\gamma - 1)(1 - \cos(\varphi)^2) = 1 + \frac{\gamma^2 \cdot \beta^2}{1 + \gamma} \sin(\varphi)^2 = 1 + \frac{\gamma^2}{1 + \gamma} \beta_3^2. $$

$$ 
\begin{aligned}
m_{3,2} &= \sin(\psi) \cos(\psi) \cos(\varphi)^2 (-1 + \gamma) = \sin(\psi) \cos(\varphi) \cos(\psi) \cos(\varphi) (-1 + \gamma) = \frac{\gamma^2 \cdot \beta^2}{1 + \gamma} \frac{\beta_2}{\beta} \frac{\beta_1}{\beta} \\
&= \frac{\gamma^2}{1 + \gamma} \beta_1 \beta_2.
\end{aligned}
$$

$$ m_{4,3} = -\cos(\varphi) \sin(\psi) \sin(\varphi) (-1 + \gamma) = \frac{\gamma^2}{1 + \gamma} \beta_2 \beta_3 . $$

$$ m_{4,2} = -\cos(\varphi) \cos(\psi) \sin(\varphi) (-1 + \gamma) = \frac{\gamma^2}{1 + \gamma} \beta_1 \beta_3 . $$

We complete by symmetry .

**Other methods :**

See also

**E • GOURGOULON** : Relativité restreinte EDP Sciences p.203 •

a geometric method based on the existence of an eingenvalue,

and for a heuristic approach :

**C • Semay B • Silvestre – Brac** Relativité Restreinte Dunod p.120 .
**Bibliography :**

Annequin et Boutigny. "Mécanique relativiste, Exercices ". Vuibert 1978.

R.G. Bartle. "Modern theory of integration ". AMS 2001.

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P. Boyer. "Algèbre et Géométries ". C&M 2015.

P. Brousse. "Mécanique • Armand Colin 1968.

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F. R. Gantmacher. "Théorie des matrices ". Edition J. Gabay 1990.

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E. Gourgoulhon. "Relativité restreinte ". EDP Sciences 2010.

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J. B. Hiriart - Urruty, Y. Plusquellec. "Exercices Algèbre linéaire ". Cepadues éditions 1988.

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J. M. Lévy - Leblond. "One more derivation of the Lorentz transformation."

American Journal of Physics, vol 44, n°3, March 1976 p271-277

J.R. Lucas, P.E. Hodgson. "Spacetime and electromagnetism ". Clarendon Press 1990.

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N.M.J. Woodhouse. "Special Relativity ". Springer 2002.

**Preliminary works :**

https://archive.org/details/version-1a_202006/mode/2 up

https://archive.org/details/matricesdelorentz

https://archive.org/details/p_20220209_202202/mode/2 up

https://archive.org/details/une-nouvelle-approche-axiomatique-de-la-theorie-de-la-relativite-restreinte-docu/mode/2 up

https://hal-amu.archives-ouvertes.fr/hal-02965773/document
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