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# Mathematical derivation of the Lorentz transformation as an invariant of the d'Alembertian operator .

Alain Guimier

► To cite this version:

Alain Guimier. Mathematical derivation of the Lorentz transformation as an invariant of the d'Alembertian operator .. 2024. hal-04673334v4

HAL Id: hal-04673334

https://hal.science/hal-04673334v4

Preprint submitted on 13 Dec 2024

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Alain Guimier

*Mathematical derivation*
*of the Lorentz Transformation*
*as an invariant of the d'Alembertian operator.*

12/2024
## Introduction :

In special relativity, different methods are known for deriving from the hypothesis of the invariance of the numerical speed of light (along with other hypotheses; see for example [1] p.76) : for two observers O and O' in uniform translation at a constant velocity $v$, the nature of the linear operator $M$ that relates the coordinates $X$ of a point $P$ for O and those $X'$ for O' in spacetime is then characterized by $^tMGM = G$, with $G = (g(i, j))$, $1 \le i, j \le 4$, where $g(1, 1) = 1$ and $g(i, i) = -1$ for $i \ne 1$, and $g(i, j) = 0$ for $i \ne j$: $M$ is named Lorentz transformation. Conversely, from this property of $M$, one can derive a velocity addition formula

(see for example [1] p.68) and show that $c$ is invariant : $\frac{(v+c)}{1+\frac{uv}{c^2}} = c$, for instance.

It has long been shown that the d'Alembertian operator is invariant under an operator $M$ that satisfies $^tMGM = G$ (see for example [1] p.95 and annex at the end). Here, we show the reciprocal : the invariance of the d'Alembertian operator under an operator $M$ twice continuously differentiable such as $M(0) = 0$ implies that $M$ is linear and $^tMGM = G$. In a first approach we suppose $M$ linear.

Since the calculations are very lengthy, they were performed using **Maple 12** software. The advantage of deriving from the invariance of the d'Alembertian operator is that it requires less hypotheses, than the derivation based on the invariance of the numerical speed of light.

Another approach to the problem exists in [5].

Many other derivations of the operator $M$ based on various hypotheses also exist; see [3].

Notice : we can use $(t, x, y, z)$ instead of $(x^1, x^2, x^3, x^4)$ and vice-versa.

## Reminder :

**Definition** : A real square matrix $M$ of order $n > 0$ is called a Lorentz matrix if it satisfies $^tMGM = G$, with $G = (g(i, j))$ where $g(1, 1) = 1$, $g(i, i) = -1$ for $i \ne 1$, and $g(i, j) = 0$ for $i \ne j$.

**Theorem** : The set $L$ of Lorentz matrices $M$ of order $n > 1$ forms a subgroup of $GL_n(\mathbb{R})$ under matrix multiplication due to the following 4 properties :

(1) It is indeed a subset of $GL_n(\mathbb{R})$ because $det(^tMGM) = det(G) = -1 = -(det(M))^2 \ne 0$.

(2) $L \ne \emptyset$ because both $Id$ and $G$ are Lorentz matrices.

(3) Let $M$ and $N$ be two Lorentz matrices : since $^t(MN)G(MN) = ^tN(^tMGM)N = G$, $L$ is therefore closed under multiplication.

(4) Since $^tMGM = G$, we have $G = (^tM)^{-1}GM^{-1} = (^tM^{-1})GM^{-1}$, and $L$ is closed under inversion.

**Corollary** : If $M$ is a Lorentz matrix, then $^tM$ is also a Lorentz one, specifically : $^tMGM = G \Leftrightarrow MG^tM = G$.

Since $^tMGM = G$, we have $(^tMGM)^{-1} = G^{-1} = G = M^{-1}G(^tM)^{-1}$, thus $M^{-1} = G^tMG$, and $^tM = GM^{-1}G$ is a Lorentz matrix as product of matrices of Lorentz.
We can notice for example :

$$
\text{if } M = \begin{bmatrix}
m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\
m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\
m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\
m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4}
\end{bmatrix}
\text{ is a Lorentz matrix then } M^{-1} =
\begin{bmatrix}
m_{1,1} & -m_{2,1} & -m_{3,1} & -m_{4,1} \\
-m_{1,2} & m_{2,2} & m_{3,2} & m_{4,2} \\
-m_{1,3} & m_{2,3} & m_{3,3} & m_{4,3} \\
-m_{1,4} & m_{2,4} & m_{3,4} & m_{4,4}
\end{bmatrix}.
$$

Consequences of (4) on the coefficients of a matrix of Lorentz :

For example if $n=2$, $M = \begin{bmatrix} m_{1,1} & m_{1,2} \\ m_{2,1} & m_{2,2} \end{bmatrix} \quad MG'M = G$ implies :

$$
\begin{bmatrix}
m_{1,1}^2 - m_{1,2}^2 & m_{1,1} m_{2,1} - m_{1,2} m_{2,1} \\
m_{1,1} m_{2,1} - m_{1,2} m_{2,1} & m_{2,1}^2 - m_{2,2}^2
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
$$

and if $n=4$ :

$$
\begin{align*}
& \begin{bmatrix}
m_{1,1}^2 - m_{1,2}^2 - m_{1,3}^2 - m_{1,4}^2 & m_{1,1} m_{2,1} - m_{1,2} m_{2,1} - m_{1,3} m_{2,3} - m_{1,4} m_{2,4} & m_{1,1} m_{3,1} - m_{1,2} m_{3,2} - m_{1,3} m_{3,3} - m_{1,4} m_{3,4} & m_{1,1} m_{4,1} \\
& - m_{1,2} m_{4,2} - m_{1,3} m_{4,3} - m_{1,4} m_{4,4} \end{bmatrix}, \\
& \begin{bmatrix}
m_{2,1} m_{2,1} - m_{2,2} m_{2,2} - m_{2,3} m_{2,3} - m_{2,4} m_{2,4} & m_{2,1}^2 - m_{2,2}^2 - m_{2,3}^2 - m_{2,4}^2 & m_{2,1} m_{3,1} - m_{2,2} m_{3,2} - m_{2,3} m_{3,3} - m_{2,4} m_{3,4} & m_{2,1} m_{4,1} \\
& - m_{2,2} m_{4,2} - m_{2,3} m_{4,3} - m_{2,4} m_{4,4} \end{bmatrix}, \\
& \begin{bmatrix}
m_{3,1} m_{3,1} - m_{3,2} m_{3,2} - m_{3,3} m_{3,3} - m_{3,4} m_{3,4} & m_{3,1} m_{4,1} - m_{3,2} m_{4,2} - m_{3,3} m_{4,3} - m_{3,4} m_{4,4} & m_{3,1}^2 - m_{3,2}^2 - m_{3,3}^2 - m_{3,4}^2 & m_{3,1} m_{4,1} \\
& - m_{3,2} m_{4,2} - m_{3,3} m_{4,3} - m_{3,4} m_{4,4} \end{bmatrix}, \\
& \begin{bmatrix}
m_{4,1} m_{4,1} - m_{4,2} m_{4,2} - m_{4,3} m_{4,3} - m_{4,4} m_{4,4} & m_{4,1} m_{5,1} - m_{4,2} m_{5,2} - m_{4,3} m_{5,3} - m_{4,4} m_{5,4} & m_{4,1}^2 - m_{4,2}^2 - m_{4,3}^2 - m_{4,4}^2 & m_{4,1} m_{5,1} \\
& - m_{4,2} m_{5,2} - m_{4,3} m_{5,3} - m_{4,4} m_{5,4} & m_{4,1} m_{5,1} - m_{4,2} m_{5,2} - m_{4,3} m_{5,3} - m_{4,4} m_{5,4} & m_{4,1}^2 - m_{4,2}^2 - m_{4,3}^2 - m_{4,4}^2 & m_{4,1} m_{5,1} \end{bmatrix} \\
& = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix};
\end{align*}
$$

and vice versa if the coefficients of the matrix *M* verify these relationships,
the matrix *M* is a Lorentz matrix.

Differential calculus reminder :

Subsequently, to simplify the writings, we will assume that the numerical speed of light $c=1$.

Let $f$ be a class $C^2$-function (twice continuously differentiable) with real values : $(x^i)_1^4 \rightarrow f((x^i)_1^4)$,

$$
M = \begin{bmatrix}
m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\
m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\
m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\
m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4}
\end{bmatrix}
\text{ a constant matrix, } X \text{ and } X' \text{ such that } X' = MX.
$$
$$ \text{We notice as } X'=MX = \begin{bmatrix} m_{1,1}x^1 + m_{1,2}x^2 + m_{1,3}x^3 + m_{1,4}x^4 \\ m_{2,1}x^1 + m_{2,2}x^2 + m_{2,3}x^3 + m_{2,4}x^4 \\ m_{3,1}x^1 + m_{3,2}x^2 + m_{3,3}x^3 + m_{3,4}x^4 \\ m_{4,1}x^1 + m_{4,2}x^2 + m_{4,3}x^3 + m_{4,4}x^4 \end{bmatrix}. $$

we have $\frac{\partial x'^a}{\partial x^b} = m_{a, b}$ and (cf: [1] p.95 et [4]):

$$ \frac{\partial f(X')}{\partial x^b} = \sum_{a=1}^{4} \frac{\partial f(X')}{\partial x^a} \frac{\partial x'^a}{\partial x^b} = \sum_{a=1}^{4} \frac{\partial f(X')}{\partial x^a} m_{a, b}. \quad (E) $$

Let's evaluate now $\frac{\partial^2 f(X')}{\partial x^{b2}}$:

$$ \frac{\partial^2 f(X')}{\partial x^{b2}} = \frac{\partial}{\partial x^b} \left( \frac{\partial f(X)}{\partial x^b} \right) = \frac{\partial}{\partial x^b} \left( \sum_{a=1}^{4} \frac{\partial f(X')}{\partial x^a} m_{a, b} \right) = \sum_{a=1}^{4} \frac{\partial}{\partial x^b} \frac{\partial f(X')}{\partial x^a} m_{a, b}. $$

Because of (E):

$$ \frac{\partial}{\partial x^b} \frac{\partial f(X')}{\partial x^a} = \sum_{c=1}^{4} \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^a} m_{c, b} \text{ then } \frac{\partial^2 f(X)}{\partial x^{b2}} = \sum_{a=1}^{4} \sum_{c=1}^{4} \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^a} m_{c, b} \cdot m_{a, b}. $$

$$ \frac{\partial^2 f(X')}{\partial x^{12}} - \frac{\partial^2 f(X')}{\partial x^{22}} - \frac{\partial^2 f(X')}{\partial x^{32}} - \frac{\partial^2 f(X')}{\partial x^{42}} = \sum_{a=1}^{4} \sum_{c=1}^{4} \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^a} m_{c, 1} \cdot m_{a, 1} - \sum_{i=2}^{4} \sum_{a=1}^{4} \sum_{c=1}^{4} \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^a} m_{c, i} \cdot m_{a, i}. $$

### Case n=2

We suppose $c=1$. If $t, x$ are the variables used by the observator $O$ and $t', x'$ used by $O'$ the invariance of the d'Alembertian's operator is expressed by:

$$ \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} = \frac{\partial^2}{\partial t'^2} - \frac{\partial^2}{\partial x'^2} $$

$$ \Leftrightarrow \forall (u', v') = U' \in \mathbb{R}^2 \forall f \in C^2 \frac{\partial^2 f(U')}{\partial t'^2} - \frac{\partial^2 f(U')}{\partial x'^2} = \frac{\partial^2 f(U')}{\partial t'^2} - \frac{\partial^2 f(U')}{\partial x'^2} $$

$$ \text{i.e. } \forall (t', x') = X' \quad \forall f \in C^2 \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x'^2} = \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x'^2} $$

$$ \text{or } \forall (t', x') = X' \quad \forall f \in C^2 \frac{\partial^2 f(M \cdot X)}{\partial t'^2} - \frac{\partial^2 f(M \cdot X)}{\partial x'^2} = \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x'^2}. $$

where $M = \begin{bmatrix} m_{1,1} & m_{1,2} \\ m_{2,1} & m_{2,2} \end{bmatrix}$ being an invertible matrix transforming the coordinates

of an event $(t, x)$ of $O$ into the coordinates of an event $(t', x')$ of $O' : X'=MX$.

Now we want to prove: $M$ is a Lorentz' matrix.

In this case we can write:

$$ X := \begin{bmatrix} t \\ x \end{bmatrix} : X' := \begin{bmatrix} t' \\ x' \end{bmatrix}; M := \begin{bmatrix} m_{1,1} & m_{1,2} \\ m_{2,1} & m_{2,2} \end{bmatrix}; M \cdot X = \begin{bmatrix} m_{1,1} t + m_{1,2} x \\ m_{2,1} t + m_{2,2} x \end{bmatrix} = X'; $$

$$ \frac{\partial t'}{\partial t} = m_{1,1}, \quad \frac{\partial t'}{\partial x} = m_{1,2}, \quad \frac{\partial x'}{\partial t} = m_{2,1}, \quad \frac{\partial x'}{\partial x} = m_{2,2} $$
$$ \frac{\partial f(X')}{\partial t} = \frac{\partial f(X')}{\partial t'} \frac{\partial t'}{\partial t} + \frac{\partial f(X')}{\partial x'} \frac{\partial x'}{\partial t} = \frac{\partial f(X')}{\partial t'} m_{1,1} + \frac{\partial f(X')}{\partial x'} m_{2,1} $$

$$ \frac{\partial f(X')}{\partial x} = \frac{\partial f(X')}{\partial t'} \frac{\partial t'}{\partial x} + \frac{\partial f(X')}{\partial x'} \frac{\partial x'}{\partial x} = \frac{\partial f(X')}{\partial t'} m_{1,2} + \frac{\partial f(X')}{\partial x'} m_{2,2} $$

$$ \frac{\partial^2 f(X')}{\partial t^2} = \frac{\partial}{\partial t} \frac{\partial f(X')}{\partial t} = \frac{\partial}{\partial t} \left( \frac{\partial f(X')}{\partial t'} m_{1,1} + \frac{\partial f(X')}{\partial x'} m_{2,1} \right) = \frac{\partial}{\partial t'} \frac{\partial f(X')}{\partial t} m_{1,1} + \frac{\partial}{\partial x'} \frac{\partial f(X')}{\partial t} m_{2,1} $$

$$ = \frac{\partial^2 f(X')}{\partial t'^2} m_{1,1}^2 + \frac{\partial^2 f(X')}{\partial t' \partial x'} m_{1,1} m_{2,1} + \frac{\partial^2 f(X')}{\partial t' \partial x'} m_{1,1} m_{2,1} + \frac{\partial^2 f(X')}{\partial x'^2} m_{2,1}^2 $$

$$ \frac{\partial^2 f(X')}{\partial x^2} = \frac{\partial}{\partial x} \frac{\partial f(X')}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\partial f(X')}{\partial t'} m_{1,2} + \frac{\partial f(X')}{\partial x'} m_{2,2} \right) = \frac{\partial}{\partial t'} \frac{\partial f(X')}{\partial x} m_{1,2} + \frac{\partial}{\partial x'} \frac{\partial f(X')}{\partial x} m_{2,2} $$

$$ = \frac{\partial^2 f(X')}{\partial t'^2} m_{1,2}^2 + \frac{\partial^2 f(X')}{\partial t' \partial x'} m_{2,2} m_{1,2} + \frac{\partial^2 f(X')}{\partial t' \partial x'} m_{1,2} m_{2,2} + \frac{\partial^2 f(X')}{\partial x'^2} m_{2,2}^2 $$

$$ \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x^2} = $$

$$ \frac{\partial^2 f(X')}{\partial t'^2} (m_{1,1}^2 - m_{1,2}^2) + \frac{\partial^2 f(X')}{\partial x'^2} (m_{2,1}^2 - m_{2,2}^2) + 2 \frac{\partial^2 f(X')}{\partial t' \partial x'} (m_{1,1} m_{2,1} - m_{1,2} m_{2,2}) $$

As $\frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x^2} = \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x'^2}$ we have:

$$ \frac{\partial^2 f(X')}{\partial t'^2} (m_{1,1}^2 - m_{1,2}^2 - 1) + \frac{\partial^2 f(X')}{\partial x'^2} (m_{2,1}^2 - m_{2,2}^2 + 1) + 2 \frac{\partial^2 f(X')}{\partial t' \partial x'} (m_{1,1} m_{2,1} - m_{1,2} m_{2,2}) = 0 $$

$$ \text{We can choose } f(X') = \frac{a}{2} \cdot t'^2 + \frac{b}{2} \cdot x'^2 + c \cdot t' \cdot x' $$

$$ \text{then } \frac{\partial f(X')}{\partial t'} = a \cdot t' + c \cdot x', \quad \frac{\partial f(X')}{\partial x'} = a \cdot x' + c \cdot t', $$

$$ \frac{\partial^2 f(X')}{\partial t'^2} = a, \quad \frac{\partial^2 f(X')}{\partial x'^2} = b, \quad \frac{\partial^2 f(X')}{\partial t' \partial x'} = c. $$

$$ \text{thus } a(m_{1,1}^2 - m_{1,2}^2 - 1) + b(m_{2,1}^2 - m_{2,2}^2 + 1) + 2c(m_{1,1} m_{2,1} - m_{1,2} m_{2,2}) = 0 \\ \text{is true for every } a, b, c. $$

$$ \text{Then } m_{1,1}^2 - m_{1,2}^2 - 1 = 0, \ m_{2,1}^2 - m_{2,2}^2 + 1 = 0 \text{ and } m_{1,1} m_{2,1} - m_{1,2} m_{2,2} = 0. $$
$$ \text{We have } \begin{bmatrix} m_{1,1}^2 - m_{1,2}^2 & m_{1,1}m_{2,1} - m_{1,2}m_{2,2} \\ m_{1,1}m_{2,1} - m_{1,2}m_{2,2} & m_{2,1}^2 - m_{2,2}^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$

and thus M is a Lorentz's matrix.

**Reciprocal:**

Consider $\frac{\partial^2 f(M \cdot X)}{\partial t^2} - \frac{\partial^2 f(M \cdot X)}{\partial x^2}$ with M being a Lorentz's matrix and $X'=MX$:

As precedently:

$$ \frac{\partial^2 f(M \cdot X)}{\partial t^2} - \frac{\partial^2 f(M \cdot X)}{\partial x^2} = \frac{\partial^2 f(X')}{\partial t'^2} (m_{1,1}^2 - m_{1,2}^2) + \frac{\partial^2 f(X')}{\partial x'^2} (m_{2,1}^2 - m_{2,2}^2) + 2 \frac{\partial^2 f(X')}{\partial t' \partial x'} (m_{1,1} m_{2,1} - m_{1,2} m_{2,2}) $$

as M being a Lorentz's matrix : $MG'M = G$ and $\frac{\partial^2 f(M \cdot X)}{\partial t^2} - \frac{\partial^2 f(M \cdot X)}{\partial x^2} = \frac{\partial^2 f(X')}{\partial t'^2} - \frac{\partial^2 f(X')}{\partial x'^2}$.

**Case : n = 4**

**Theorem :** Let be $M = \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix}$ a matrix,

$X'=MX$, with $X'=\begin{bmatrix} t' \\ x' \\ y' \\ z' \end{bmatrix}$ and $X=\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$.

Let $f$ be a class $C^2$ function with real values : $f:(t, x, y, z) \rightarrow f(t, x, y, z)$ and $\square$ the operator :

$$ \Box : (f, X) \rightarrow \frac{\partial^2 f}{\partial t^2} - \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} - \frac{\partial^2 f}{\partial z^2} $$

$$ \text{i.e. } \Box(f)(t, x, y, z) = \frac{\partial^2 f(t, x, y, z)}{\partial t^2} - \frac{\partial^2 f(t, x, y, z)}{\partial x^2} - \frac{\partial^2 f(t, x, y, z)}{\partial y^2} - \frac{\partial^2 f(t, x, y, z)}{\partial z^2} $$

If $\forall (t, x, y, z) \ \forall f \in C^2 : \Box(f \circ M)(t, x, y, z) = \Box(f)(t', x', y', z')$ with $\begin{bmatrix} t' \\ x' \\ y' \\ z' \end{bmatrix} = M \begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$

M being a matrix.

then M is a Lorentz matrix and conversely.

This condition can be rewritten as :
$$
\frac{\partial^2 (f \circ M) (t, x, y, z)}{\partial t^2} - \frac{\partial^2 (f \circ M) (t, x, y, z)}{\partial x^2} - \frac{\partial^2 (f \circ M) (t, x, y, z)}{\partial y^2} - \frac{\partial^2 (f \circ M) (t, x, y, z)}{\partial z^2} \\
= \frac{\partial^2 f(t', x', y', z')}{\partial t'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial x'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial y'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial z'^2}
$$

Proof:

[I] First method :

We use **Maple 12**'commands to evaluate :

$$
\frac{\partial^2}{\partial t^2} f(t', x', y', z') - \frac{\partial^2}{\partial x^2} f(t', x', y', z') - \frac{\partial^2}{\partial y^2} f(t', x', y', z') - \frac{\partial^2}{\partial z^2} f(t', x', y', z') : h := M \to (M[1], M[2], M[3], M[4]) :
$$

$$
\text{restart: } g := (t, x, y, z) \rightarrow \left[ \begin{pmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{pmatrix} \cdot \begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix} \right] : h := M \to (M[1], M[2], M[3], M[4]) :
$$

$$
A := \text{simplify} \left( \left( \left( \frac{\partial^2}{\partial t \partial t} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial x \partial x} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial y \partial y} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial z \partial z} f(h(g(t, x, y, z))) \right) \right) :
$$

$$
a := \operatorname{algsubs}(m_{1,1} \cdot t + m_{1,2} \cdot x + m_{1,3} \cdot y + m_{1,4} \cdot z = t_p, A):
$$

$$
b := \operatorname{algsubs}(m_{2,1} \cdot t + m_{2,2} \cdot x + m_{2,3} \cdot y + m_{2,4} \cdot z = x_p, a):
$$

$$
c := \operatorname{algsubs}(m_{3,1} \cdot t + m_{3,2} \cdot x + m_{3,3} \cdot y + m_{3,4} \cdot z = y_p, b):
$$

$$
e1 := \operatorname{algsubs}(m_{4,1} \cdot t + m_{4,2} \cdot x + m_{4,3} \cdot y + m_{4,4} \cdot z = z_p, c):
$$

$$
R := e1:
$$

for i from 1 to 4 do

for j from 1 to 4 do

$$
R := \text{collect}(R, D_{i,j}(f)(t_p, x_p, y_p, z_p)):
$$

end do

end do :

R;

$$
\begin{aligned}
& \left(-m_{4,3}^2 - m_{4,4}^2 - m_{4,2}^2 + m_{4,1}^2\right) D_{4,4}(f)(t_p x_p y_p z_p) + \left(-2 m_{3,4} m_{4,4} - 2 m_{3,2} m_{4,2} - 2 m_{3,3} m_{4,3}\right. \\
& \qquad \left. + 2 m_{3,1} m_{4,1}\right) D_{3,4}(f)(t_p x_p y_p z_p) + \left(m_{3,1}^2 - m_{3,4}^2 - m_{3,3}^2 - m_{3,2}^2\right) D_{3,3}(f)(t_p x_p y_p z_p) \\
& \qquad + \left(2 m_{4,1} m_{2,1} - 2 m_{4,2} m_{2,2} - 2 m_{4,3} m_{2,3} - 2 m_{4,4} m_{2,4}\right) D_{4,4}(f)(t_p x_p y_p z_p) \\
& \qquad + \left(2 m_{3,1} m_{2,1} - 2 m_{3,2} m_{2,2} - 2 m_{3,4} m_{2,4} - 2 m_{3,3} m_{2,3}\right) D_{2,3}(f)(t_p x_p y_p z_p) + \left(-m_{2,3}^2 + \right. \\
& \qquad \left. m_{2,1}^2 - m_{2,4}^2 - m_{2,2}^2\right) D_{2,2}(f)(t_p x_p y_p z_p) + \left(2 m_{4,1} m_{1,1} - 2 m_{4,4} m_{1,4} - 2 m_{4,3} m_{1,3}\right. \\
& \qquad \left. - 2 m_{4,2} m_{1,2}\right) D_{1,4}(f)(t_p x_p y_p z_p) + \left(-m_{1,2}^2 - m_{1,3}^2 - m_{1,4}^2 + m_{1,1}^2\right) D_{1,1}(f)(t_p x_p y_p z_p) \\
& \qquad + \left(2 m_{2,1} m_{1,1} - 2 m_{2,4} m_{1,4} - 2 m_{2,3} m_{1,3} - 2 m_{2,2} m_{1,2}\right) D_{1,2}(f)(t_p x_p y_p z_p) + \left( \right. \\
& \qquad \left. - 2 m_{3,3} m_{1,3} - 2 m_{3,4} m_{1,4} - 2 m_{3,2} m_{1,2} + 2 m_{3,1} m_{1,1}\right) D_{1,3}(f)(t_p x_p y_p z_p)
\end{aligned}
$$

We use $t_p, x_p, y_p, z_p$ instead of $t', x', y', z'$ because of **Maple12** : " ' "is considered as a derivation.

As $D_{i,j} = D_{j,i}$ their multiplying factors are gathered together.

$$
\text{then } (E) \quad R = \frac{\partial^2 f(X')}{\partial x^{12}} - \frac{\partial^2 f(X')}{\partial x^{22}} - \frac{\partial^2 f(X')}{\partial x^{32}} - \frac{\partial^2 f(X')}{\partial x^{42}} =
$$
Given 10 reals ($u_i$) $1 \le i \le 10$ from the above there exists at least one function $f_u$ of class $C^2$ such that
$Z1 = {}^t[u_p, \dots, u_{10}]$. So $\sum {}^tZ2.Z1 = 0$ for all $u$ then $Z2 = 0$.

Returning to $M$, remembering that :

$$
\left[
\begin{aligned}
& \left[ m_{1,1}^2 - m_{1,2}^2 - m_{1,3}^2 - m_{1,4}^2, m_{1,1} m_{2,1} - m_{1,2} m_{2,2} - m_{1,3} m_{2,3} - m_{1,4} m_{2,4}, m_{1,1} m_{3,1} - m_{1,2} m_{3,2} - m_{1,3} m_{3,3} - m_{1,4} m_{3,4}, m_{1,1} m_{4,1} \right. \\
& \qquad \left. - m_{1,2} m_{4,2} - m_{1,3} m_{4,3} - m_{1,4} m_{4,4} \right], \\
& \left[ m_{1,1} m_{2,1} - m_{1,2} m_{2,2} - m_{1,3} m_{2,3} - m_{1,4} m_{2,4}, m_{2,1}^2 - m_{2,2}^2 - m_{2,3}^2 - m_{2,4}^2, m_{2,1} m_{3,1} - m_{2,2} m_{3,2} - m_{2,3} m_{3,3} - m_{2,4} m_{3,4}, m_{2,1} m_{4,1} \right. \\
& \qquad \left. - m_{2,2} m_{4,2} - m_{2,3} m_{4,3} - m_{2,4} m_{4,4} \right], \\
& \left[ m_{1,1} m_{3,1} - m_{1,2} m_{3,2} - m_{1,3} m_{3,3} - m_{1,4} m_{3,4}, m_{2,1} m_{3,1} - m_{2,2} m_{3,2} - m_{2,3} m_{3,3} - m_{2,4} m_{3,4}, m_{3,1}^2 - m_{3,2}^2 - m_{3,3}^2 - m_{3,4}^2, m_{3,1} m_{4,1} \right. \\
& \qquad \left. - m_{3,2} m_{4,2} - m_{3,3} m_{4,3} - m_{3,4} m_{4,4} \right], \\
& \left[ m_{1,1} m_{4,1} - m_{1,2} m_{4,2} - m_{1,3} m_{4,3} - m_{1,4} m_{4,4}, m_{2,1} m_{4,1} - m_{2,2} m_{4,2} - m_{2,3} m_{4,3} - m_{2,4} m_{4,4}, m_{3,1} m_{4,1} - m_{3,2} m_{4,2} - m_{3,3} m_{4,3} \right. \\
& \qquad \left. - m_{3,4} m_{4,4}, m_{4,1}^2 - m_{4,2}^2 - m_{4,3}^2 - m_{4,4}^2 \right]
\end{aligned}
\right]
$$

$$
= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}
$$

characterizes a Lorentz matrix, as $Z2=0$ we conclude that $M$ is a Lorentz matrix.

[II] Second method :

We start from the formula proved in the remainder of differential calculation :

$$
\frac{\partial^2 f(X')}{\partial x^{12}} - \frac{\partial^2 f(X')}{\partial x^{22}} - \frac{\partial^2 f(X')}{\partial x^{32}} - \frac{\partial^2 f(X')}{\partial x^{42}} = \sum_{a=1}^{4} \sum_{c=1}^{4} \left( \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^u} m_{c,1} \cdot m_{a,1} - \sum_{i=2}^{4} \sum_{a=1}^{4} \sum_{c=1}^{4} \left( \frac{\partial}{\partial x'^c} \frac{\partial f(X')}{\partial x'^u} m_{c,i} \cdot m_{a,i} \right) \right)
$$

To do the calculation with Maple 12 we have the commands :

$$
M := \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix}; F := \begin{bmatrix} D_{1,1}(f)(t,x,y,z_p) & D_{1,2}(f)(t,x,y,z_p) & D_{1,3}(f)(t,x,y,z_p) & D_{1,4}(f)(t,x,y,z_p) \\ D_{1,2}(f)(t,x,y,z_p) & D_{2,2}(f)(t,x,y,z_p) & D_{2,3}(f)(t,x,y,z_p) & D_{2,4}(f)(t,x,y,z_p) \\ D_{1,3}(f)(t,x,y,z_p) & D_{2,3}(f)(t,x,y,z_p) & D_{3,3}(f)(t,x,y,z_p) & D_{3,4}(f)(t,x,y,z_p) \\ D_{1,4}(f)(t,x,y,z_p) & D_{2,4}(f)(t,x,y,z_p) & D_{3,4}(f)(t,x,y,z_p) & D_{4,4}(f)(t,x,y,z_p) \end{bmatrix}.
$$

for k from 1 to 4 do

X1 := 0 :

for i from 1 to 4 do

for j from 1 to 4 do

X1 := X1 + F[j, i] · M[j, k] · M[i, k] :

end do;

end do;

X[k] := X1 :

end do:

X1 := X[1] :

for k from 2 to 4 do

X1 := X1 - X[k];

end do:

X1;

for i from 1 to 4 do

for j from 1 to 4 do

X1 := collect(X1, F[j, i]):
end do : end do :

XI;

We obtain :

$$
\begin{align*}
& (m_{4,1}^2 - m_{4,3}^2 - m_{4,2}^2 - m_{4,4}^2) D_{4,4}(f) (t_p x_p y_p z_p) + (-2 m_{4,3} m_{3,3} - 2 m_{4,2} m_{3,2} + 2 m_{4,1} m_{3,1} \\
& \quad - 2 m_{4,4} m_{3,4}) D_{3,4}(f) (t_p x_p y_p z_p) + (-m_{3,4}^2 - m_{3,2}^2 + m_{3,1}^2 - m_{3,3}^2) D_{3,3}(f) (t_p x_p y_p z_p) \\
& \quad + (-2 m_{4,2} m_{2,2} + 2 m_{4,1} m_{2,1} - 2 m_{4,3} m_{2,3} - 2 m_{4,4} m_{2,4}) D_{2,4}(f) (t_p x_p y_p z_p) \\
& \quad + (2 m_{3,1} m_{2,1} - 2 m_{3,3} m_{2,3} - 2 m_{3,4} m_{2,4} - 2 m_{3,2} m_{2,2}) D_{2,3}(f) (t_p x_p y_p z_p) \\
& \quad + (2 m_{2,1} m_{1,1} - 2 m_{2,4} m_{1,4} - 2 m_{2,3} m_{1,3} - 2 m_{2,2} m_{1,2}) D_{1,2}(f) (t_p x_p y_p z_p) + ( \\
& \quad -2 m_{3,4} m_{1,4} + 2 m_{3,1} m_{1,1} - 2 m_{3,3} m_{1,3} - 2 m_{3,2} m_{1,2}) D_{1,3}(f) (t_p x_p y_p z_p) + (m_{2,1}^2 - m_{2,3}^2 \\
& \quad - m_{2,2}^2 - m_{2,4}^2) D_{2,2}(f) (t_p x_p y_p z_p) + (-m_{1,4}^2 - m_{1,3}^2 - m_{1,2}^2 + m_{1,1}^2) D_{1,1}(f) (t_p x_p y_p z_p) \\
& \quad + (-2 m_{4,3} m_{1,3} - 2 m_{4,2} m_{1,2} - 2 m_{4,4} m_{1,4} + 2 m_{4,1} m_{1,1}) D_{1,4}(f) (t_p x_p y_p z_p).
\end{align*}
$$

We find the same calculated result in the first part and we finish in the same way.

The reciprocal is obvious :

As previously,

$$
\begin{align*}
& \frac{\partial^2}{\partial t^2} f(t', x', y', z') - \frac{\partial^2}{\partial x^2} f(t', x', y', z') - \frac{\partial^2}{\partial y^2} f(t', x', y', z') - \frac{\partial^2}{\partial z^2} f(t', x', y', z') \\
&= (m_{4,1}^2 - m_{4,3}^2 - m_{4,2}^2 - m_{4,4}^2) D_{4,4}(f) (t_p x_p y_p z_p) + (-2 m_{4,3} m_{3,3} - 2 m_{4,2} m_{3,2} + 2 m_{4,1} m_{3,1} \\
& \quad - 2 m_{4,4} m_{3,4}) D_{3,4}(f) (t_p x_p y_p z_p) + (-m_{3,4}^2 - m_{3,2}^2 + m_{3,1}^2 - m_{3,3}^2) D_{3,3}(f) (t_p x_p y_p z_p) \\
& \quad + (-2 m_{4,2} m_{2,2} + 2 m_{4,1} m_{2,1} - 2 m_{4,3} m_{2,3} - 2 m_{4,4} m_{2,4}) D_{2,4}(f) (t_p x_p y_p z_p) \\
& \quad + (2 m_{3,1} m_{2,1} - 2 m_{3,3} m_{2,3} - 2 m_{3,4} m_{2,4} - 2 m_{3,2} m_{2,2}) D_{2,3}(f) (t_p x_p y_p z_p) \\
& \quad + (2 m_{2,1} m_{1,1} - 2 m_{2,4} m_{1,4} - 2 m_{2,3} m_{1,3} - 2 m_{2,2} m_{1,2}) D_{1,2}(f) (t_p x_p y_p z_p) + ( \\
& \quad -2 m_{3,4} m_{1,4} + 2 m_{3,1} m_{1,1} - 2 m_{3,3} m_{1,3} - 2 m_{3,2} m_{1,2}) D_{1,3}(f) (t_p x_p y_p z_p) + (m_{2,1}^2 - m_{2,3}^2 \\
& \quad - m_{2,2}^2 - m_{2,4}^2) D_{2,2}(f) (t_p x_p y_p z_p) + (-m_{1,4}^2 - m_{1,3}^2 - m_{1,2}^2 + m_{1,1}^2) D_{1,1}(f) (t_p x_p y_p z_p) \\
& \quad + (-2 m_{4,3} m_{1,3} - 2 m_{4,2} m_{1,2} - 2 m_{4,4} m_{1,4} + 2 m_{4,1} m_{1,1}) D_{1,4}(f) (t_p x_p y_p z_p) \\
&= \frac{\partial^2 f(t', x', y', z')}{\partial t'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial x'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial y'^2} - \frac{\partial^2 f(t', x', y', z')}{\partial z'^2}
\end{align*}
$$

if M is a matrix of Lorentz.

**Numerical examples :** (with Maple 12)

$$
M := \frac{1}{\sqrt{2}} \cdot \begin{bmatrix} -2 & 1 & 0 & -1 \\ -1 & 1 & 1 & -1 \\ 1 & -1 & 1 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix} : G := \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} : M^{\%T} \cdot G \cdot M = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix},
$$
$$
M.G.M^T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}, \quad IM := G.M^T.G, \quad IM = \begin{bmatrix} -\sqrt{2} & \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} & 0 \\ -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \\ 0 & \frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} & 0 \\ \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \end{bmatrix},
$$

$$
IM.M = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad g := (t, x, y, z) \to \begin{bmatrix} \begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix} \end{bmatrix} : h := M \to (M[1], M[2], M[3], M[4]) :
$$

$$
h(g(t, x, y, z)); -\sqrt{2} t + \frac{1}{2} \sqrt{2} x - \frac{1}{2} \sqrt{2} z, -\frac{1}{2} \sqrt{2} t + \frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} y - \frac{1}{2} \sqrt{2} z, \frac{1}{2} \sqrt{2} t - \frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} y \\
+ \frac{1}{2} \sqrt{2} z,
$$

$$
\frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} z
$$

$$
f := (t, x, y, z) \to y - x^3 + 3 \cdot t^2 + x \cdot t + x \cdot y + z^4 + \sin(x):
$$

$$
A := \text{simplify} \left( \left( \frac{\partial^2}{\partial t \partial t} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial x \partial x} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial y \partial y} f(h(g(t, x, y, z))) \right) - \left( \frac{\partial^2}{\partial z \partial z} f(h(g(t, x, y, z))) \right) \right);
$$

$$
6 - 6x^2 - 12x z - 6z^2 - 3 \sqrt{2} t + 3 \sqrt{2} x - 3 \sqrt{2} z + 3 \sqrt{2} y - \sin \left( \frac{1}{2} \sqrt{2} t - \frac{1}{2} \sqrt{2} x - \frac{1}{2} \sqrt{2} y + \frac{1}{2} \sqrt{2} z \right)
$$

$$
B := \text{simplify} \left( \left( \frac{\partial^2}{\partial t \partial t} f(t, x, y, z) \right) - \left( \frac{\partial^2}{\partial x \partial x} f(t, x, y, z) \right) - \left( \frac{\partial^2}{\partial y \partial y} f(t, x, y, z) \right) - \left( \frac{\partial^2}{\partial z \partial z} f(t, x, y, z) \right) \right);
$$

$$
6 + 6x + \sin(x) - 12z^2
$$

If you replace *t* by *t*', ..., ...

$$
6 + 6 \cdot \left( -\frac{1}{2} \sqrt{2} t + \frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} y - \frac{1}{2} \sqrt{2} z \right) + \sin \left( -\frac{1}{2} \sqrt{2} t + \frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} y - \frac{1}{2} \sqrt{2} z \right) \\
- 12 \cdot \left( \frac{1}{2} \sqrt{2} x + \frac{1}{2} \sqrt{2} y \right)^2 :
$$

simplify(%);

$$
6 - 6x^2 - 12xz - 6z^2 - 3\sqrt{2}t + 3\sqrt{2}x - 3\sqrt{2}z + 3\sqrt{2}y - \sin\left(\frac{1}{2}\sqrt{2}t - \frac{1}{2}\sqrt{2}x - \frac{1}{2}\sqrt{2}y + \frac{1}{2}\sqrt{2}z\right)
$$

**Case : M is not linear :**

Now we assume only that **M** is an operator twice continuously differentiable such that
**M(0, 0) = (0, 0)** we can prove that is matrix of Lorentz.
To do this and to limit the length of the formulas we limit our proof to $n=2$.

**Hypothesis :**

$$ \forall (t, x) \quad \forall f \in \mathcal{C}^2 : \square(f \circ M)(t, x) = \square(f)(t', x') \quad \text{with} \quad \begin{bmatrix} t' \\ x' \end{bmatrix} = M \begin{bmatrix} t \\ x \end{bmatrix}. $$

This condition can be rewritten as :

$$ \frac{\partial^2 (f \circ M)(t, x)}{\partial t^2} - \frac{\partial^2 (f \circ M)(t, x)}{\partial x^2} = \frac{\partial^2 f(t', x')}{\partial t'^2} - \frac{\partial^2 f(t', x')}{\partial x'^2} \quad \text{with} \quad (t', x') = M(t, x). $$

**Proof :**

As previously let's use Maple 12 with the commands :

**restart :**

$$ h := M \rightarrow (M[1], M[2]) : $$

$$ A := \text{simplify} \left( \left( \frac{\partial}{\partial t} \frac{\partial}{\partial t} f(h(M(t, x))) \right) - \left( \frac{\partial}{\partial x} \frac{\partial}{\partial x} f(h(M(t, x))) \right) \right) ; $$

$$ R := A : $$

for i from 1 to 2 do

for j from 1 to 2 do

$$ R := \text{collect}(R, D_{i,j}(f)(M(t,x)_p M(t,x)_q)) : $$

end do

end do :

$$ R; $$

We obtain :

$$ \frac{\partial^2 f(t', x')}{\partial t^2} - \frac{\partial^2 f(t', x')}{\partial x'^2} = \frac{\partial^2}{\partial t^2} f(M(t, x)_p, M(t, x)_q) - \frac{\partial^2}{\partial x^2} f(M(t, x)_p, M(t, x)_q) = \\ \left( \left( \frac{\partial}{\partial t} M(t, x)_2 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_2 \right)^2 \right) D_{2,2}(f) \left( M(t, x)_p M(t, x)_q \right) + \left( 2 \left( \frac{\partial}{\partial t} M(t, x)_1 \right) \left( \frac{\partial}{\partial t} M(t, x)_2 \right) - 2 \left( \frac{\partial}{\partial x} M(t, x)_1 \right) \left( \frac{\partial}{\partial x} M(t, x)_2 \right) \right) D_{1,2}(f) \left( M(t, x)_p M(t, x)_q \right) \\ + \left( \left( \frac{\partial}{\partial t} M(t, x)_1 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_1 \right)^2 \right) D_{1,1}(f) \left( M(t, x)_p M(t, x)_q \right) + D_{2,1}(f) \left( M(t, x)_p M(t, x)_q \right) \\ \left( \left( \frac{\partial^2}{\partial t^2} M(t, x)_2 \right) - \left( \frac{\partial^2}{\partial x^2} M(t, x)_2 \right) \right) + D_{1,1}(f) \left( M(t, x)_p M(t, x)_q \right) \\ \left( \left( \frac{\partial^2}{\partial t^2} M(t, x)_1 \right) - \left( \frac{\partial^2}{\partial x^2} M(t, x)_1 \right) \right) $$

$$ \text{Notice : } D_{1,2}(f) = D_{2,1}(f) . $$

Let's consider the functions $f_{\alpha, \beta}$ of the form $f_{\alpha, \beta}(t, x) = \alpha \cdot t + \beta \cdot x$ we have :

$$ \frac{\partial^2}{\partial t^2} f_{\alpha, \beta}(t', x') - \frac{\partial^2}{\partial x^2} f_{\alpha, \beta}(t', x') = 0, \quad D_{i,j}(f_{\alpha, \beta}) = 0, \quad D_{1}(f_{\alpha, \beta}) = \alpha \quad \text{and} \quad D_{2}(f_{\alpha, \beta}) = \beta $$

Now we have :

$$ 0 = \beta \left( \frac{\partial^2}{\partial t^2} M(t, x)_2 - \frac{\partial^2}{\partial x^2} M(t, x)_2 \right) + \alpha \left( \frac{\partial^2}{\partial t^2} M(t, x)_1 - \frac{\partial^2}{\partial x^2} M(t, x)_1 \right) \quad \forall \alpha \text{ and } \beta. $$

Thus :
$$
\frac{\partial^2}{\partial t^2} M(t, x)_1 - \frac{\partial^2}{\partial x^2} M(t, x)_1 = 0 \quad \text{et} \quad \frac{\partial^2}{\partial t^2} M(t, x)_2 - \frac{\partial^2}{\partial x^2} M(t, x)_2 = 0
$$

Then (Wave equation) :

$$
M(t, x)_1 = F1(t+x) + F2(t-x) \quad \text{et} \quad M(t, x)_2 = G1(t+x) + G2(t-x) \quad (E)
$$

with arbitry functions F1, F2, G1, G2 of class C² .

If $\mathbf{M}$ is a matrix, for example:

$$
M(t, x)_1 = m_{1,1} t + m_{1,2} x = \frac{(m_{1,1} + m_{1,2})}{2} \cdot (t+x) + \frac{(m_{1,1} - m_{1,2})}{2} \cdot (t-x).
$$

$$
M(t, x)_2 = m_{2,1} t + m_{2,2} x = \frac{(m_{2,1} + m_{2,2})}{2} \cdot (t+x) + \frac{(m_{2,1} - m_{2,2})}{2} \cdot (t-x)
$$

Now if we restrict to the $M(t, x)$ of the form above (E) we have :

$$
\frac{\partial^2}{\partial t^2} f(M(t, x)_p, M(t, x)_{p2}) - \frac{\partial^2}{\partial x^2} f(M(t, x)_p, M(t, x)_{p2}) =
$$

$$
\left( \left( \frac{\partial}{\partial t} M(t, x)_2 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_2 \right)^2 \right) D_{2,2}(f) (M(t, x)_p, M(t, x)_{p2}) + \left( 2 \left( \frac{\partial}{\partial t} M(t, x)_1 \right) \left( \frac{\partial}{\partial t} M(t, x)_2 \right) - 2 \left( \frac{\partial}{\partial x} M(t, x)_1 \right) \left( \frac{\partial}{\partial x} M(t, x)_2 \right) \right) D_{1,2}(f) (M(t, x)_p, M(t, x)_{p2}) \\
+ \left( \left( \frac{\partial}{\partial t} M(t, x)_1 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_1 \right)^2 \right) D_{1,1}(f) (M(t, x)_p, M(t, x)_{p2}).
$$

$$
\text{Let's consider the functions } f_{\alpha, \beta, \gamma}(t, x) = \frac{1}{2} \cdot \alpha \cdot t^2 + \beta \cdot t \cdot x + \frac{1}{2} \cdot \gamma \cdot x^2 :
$$

We have :

$$
\frac{\partial^2}{\partial t^2} f_{\alpha, \beta, \gamma}(t', x') - \frac{\partial^2}{\partial x^2} f_{\alpha, \beta, \gamma}(t', x') = \alpha - \gamma
$$

$$
= \frac{\partial^2}{\partial t^2} f_{\alpha, \beta, \gamma} (M(t, x)_p, M(t, x)_{p2}) - \frac{\partial^2}{\partial x^2} f_{\alpha, \beta, \gamma} (M(t, x)_p, M(t, x)_{p2})
$$

$$
= \left( \left( \frac{\partial}{\partial t} M(t, x)_2 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_2 \right)^2 \right) \gamma + \left( 2 \left( \frac{\partial}{\partial t} M(t, x)_1 \right) \left( \frac{\partial}{\partial t} M(t, x)_2 \right) - 2 \left( \frac{\partial}{\partial x} M(t, x)_1 \right) \left( \frac{\partial}{\partial x} M(t, x)_2 \right) \right) \beta + \left( \left( \frac{\partial}{\partial t} M(t, x)_1 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_1 \right)^2 \right) \alpha.
$$

Hence :

$$
\left( \left( \frac{\partial}{\partial t} M(t, x)_2 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_2 \right)^2 + 1 \right) \gamma + \left( 2 \left( \frac{\partial}{\partial t} M(t, x)_1 \right) \left( \frac{\partial}{\partial t} M(t, x)_2 \right) - 2 \left( \frac{\partial}{\partial x} M(t, x)_1 \right) \left( \frac{\partial}{\partial x} M(t, x)_2 \right) \right) \beta + \left( \left( \frac{\partial}{\partial t} M(t, x)_1 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_1 \right)^2 - 1 \right) \alpha = 0 \quad \forall (\alpha, \beta, \gamma)
$$

Conclusion :

$$
\left( \frac{\partial}{\partial t} M(t, x)_1 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_1 \right)^2 = 1,
$$

$$
\left( \frac{\partial}{\partial t} M(t, x)_2 \right)^2 - \left( \frac{\partial}{\partial x} M(t, x)_2 \right)^2 = -1 \quad \text{et}
$$

$$
\left( \frac{\partial}{\partial t} M(t, x)_1 \right) \left( \frac{\partial}{\partial t} M(t, x)_2 \right) - \left( \frac{\partial}{\partial x} M(t, x)_1 \right) \left( \frac{\partial}{\partial x} M(t, x)_2 \right) = 0.
$$
As $M(t, x)_1 = F1(t+x) + F2(t-x)$ et $M(t, x)_2 = G1(t+x) + G2(t-x)$ :

$$ \left(\frac{\partial}{\partial t} M(t, x)_1\right)^2 - \left(\frac{\partial}{\partial x} M(t, x)_1\right)^2 = 4 F1'(t+x) \cdot F2'(t-x), $$

and

$$ \left(\frac{\partial}{\partial t} M(t, x)_2\right)^2 - \left(\frac{\partial}{\partial x} M(t, x)_2\right)^2 = 4 G1'(t+x) \cdot G2'(t-x). $$

Notice : When M is linear,

$$ 4 F1'(t+x) \cdot F2'(t-x) = 4 \cdot \frac{(m_{1,1} + m_{1,2})}{2} \cdot \frac{(m_{1,1} - m_{1,2})}{2} = m_{1,1}^2 - m_{1,2}^2 $$

$$ 4 G1'(t+x) \cdot G2'(t-x) = 4 \cdot \frac{(m_{2,1} + m_{2,2})}{2} \cdot \frac{(m_{2,1} - m_{2,2})}{2} = m_{1,2}^2 - m_{2,2}^2 $$

And

$$ \begin{aligned} & \left(\frac{\partial}{\partial t} M(t, x)_1\right) \left(\frac{\partial}{\partial t} M(t, x)_2\right) - \left(\frac{\partial}{\partial x} M(t, x)_1\right) \left(\frac{\partial}{\partial x} M(t, x)_2\right) \\ &= \left(\frac{\partial}{\partial t} (F1(t+x) + F2(t-x))\right) \left(\frac{\partial}{\partial x} (G1(t+x) + G2(t-x))\right) \\ &\quad - \left(\frac{\partial}{\partial x} (F1(t+x) + F2(t-x))\right) \left(\frac{\partial}{\partial x} (G1(t+x) + G2(t-x))\right); \\ &= 2 (F1'(t+x)G2'(t-x) + F2'(t-x)G1'(t+x)) \end{aligned} $$

Notice : When M is linear,

$$ 2 \cdot \left( \frac{(m_{1,1} + m_{1,2})}{2} \cdot \frac{(m_{2,1} - m_{2,2})}{2} + \frac{(m_{1,1} - m_{1,2})}{2} \cdot \frac{(m_{2,1} + m_{2,2})}{2} \right) = m_{1,1} m_{2,1} - m_{1,2} m_{2,2}. $$

We can conclude :

$$ \text{As } M(t, x) \text{ can be written as } M(t, x) = \begin{bmatrix} M(t, x)_1 \\ M(t, x)_2 \end{bmatrix} = \begin{bmatrix} F1(t+x) + F2(t-x) \\ G1(t+x) + G2(t-x) \end{bmatrix} $$

where $F1, F2, G1$ et $G2$ are arbitrary $C^2$-functions with the conditions :

$$ 4 F1'(t+x) \cdot F2'(t-x) = 1, \quad (1) $$

$$ 4 G1'(t+x) \cdot G2'(t-x) = -1 \quad (2) $$

$$ F1'(t+x)G2'(t-x) + F2'(t-x)G1'(t+x) = 0. \quad (3) $$

If in (1) and (2) we put $t=x$ and $t=-x$ we can see $F1', F2', G1'$ et $G2'$ are constant, then $F1, F2, G1$ et $G2$ are affinities.

If $M(0, 0) = (0, 0)$ M is linear and a matrix of Lorentz.

**Notice :**

Why $MG'M = G$ is the key point?

That comes from chain rule :
$$ \text{If } X \triangleq MX = \begin{bmatrix} m_{1,1} x^1 + m_{1,2} x^2 + m_{1,3} x^3 + m_{1,4} x^4 \\ m_{2,1} x^1 + m_{2,2} x^2 + m_{2,3} x^3 + m_{2,4} x^4 \\ m_{3,1} x^1 + m_{3,2} x^2 + m_{3,3} x^3 + m_{3,4} x^4 \\ m_{4,1} x^1 + m_{4,2} x^2 + m_{4,3} x^3 + m_{4,4} x^4 \end{bmatrix}. $$

we have $\frac{\partial x'^a}{\partial x^b} = m_{a,b}$ and (cf: [1] p.95 et [4]):

$$ \frac{\partial f(X')}{\partial x^b} = \sum_{a=1}^{4} \frac{\partial f(X')}{\partial x^a} \frac{\partial x'^a}{\partial x^b} = \sum_{a=1}^{4} \frac{\partial f(X')}{\partial x^a} m_{a,b} $$

$$ \text{then } \frac{\partial f(X')}{\partial x^b} = \frac{\partial f(X')}{\partial x'^1} m_{1,b} + \frac{\partial f(X')}{\partial x'^2} m_{2,b} + \frac{\partial f(X')}{\partial x'^3} m_{3,b} + \frac{\partial f(X')}{\partial x'^4} m_{4,b} $$

$$ \text{and } \left[ \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^3}, \frac{\partial}{\partial x^4} \right] f(X') = \left[ \frac{\partial}{\partial x'^1}, \frac{\partial}{\partial x'^2}, \frac{\partial}{\partial x'^3}, \frac{\partial}{\partial x'^4} \right] \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix} f(X'), $$

that implies:

$$ \left[ \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^3}, \frac{\partial}{\partial x^4} \right] \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \frac{\partial}{\partial x^1} \frac{\partial}{\partial x^2} \frac{\partial}{\partial x^3} \frac{\partial}{\partial x^4} f(X') $$

$$ = \left[ \frac{\partial}{\partial x'^1}, \frac{\partial}{\partial x'^2}, \frac{\partial}{\partial x'^3}, \frac{\partial}{\partial x'^4} \right] \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix} \frac{\partial}{\partial x'^1} \frac{\partial}{\partial x'^2} \frac{\partial}{\partial x'^3} \frac{\partial}{\partial x'^4} f(X') $$

if M is a matrix of Lorentz :

$$ \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} m_{1,1} & m_{1,2} & m_{1,3} & m_{1,4} \\ m_{2,1} & m_{2,2} & m_{2,3} & m_{2,4} \\ m_{3,1} & m_{3,2} & m_{3,3} & m_{3,4} \\ m_{4,1} & m_{4,2} & m_{4,3} & m_{4,4} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \quad \text{and then:} $$
$$ \frac{\partial^2 f(X')}{\partial x^{12}} - \frac{\partial^2 f(X')}{\partial x^{22}} - \frac{\partial^2 f(X')}{\partial x^{32}} - \frac{\partial^2 f(X')}{\partial x^{42}} = \frac{\partial^2 f(X')}{\partial x'^{12}} - \frac{\partial^2 f(X')}{\partial x'^{22}} - \frac{\partial^2 f(X')}{\partial x'^{32}} - \frac{\partial^2 f(X')}{\partial x'^{42}} $$

Annex:

[I] The D'Alembertien as consequence of the Maxwell equations.

[1] p.39, 40, 41

Consider the Maxwell's equations of the electric and magnetic fields $\vec{E}$ et $\vec{B}$ in a region of empty space isolated from charges generating electric and magnetic fields :

$$ \mathrm{div} \, \vec{E} = 0, \mathrm{div} \, \vec{B} = 0, \mathrm{curl} \, \vec{B} = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} \quad \text{and} \quad \mathrm{curl} \, \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$

From these equations :

$$ \mathrm{curl} \, \vec{B} = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} \Rightarrow \mathrm{curl} \, \mathrm{curl} \, \vec{B} = \frac{1}{c^2} \frac{\partial \mathrm{curl} \, \vec{E}}{\partial t} $$

$$ \begin{align*} \text{as } \mathrm{curl} \, \mathrm{curl} \, \vec{B} &= \mathrm{grad} \, \mathrm{div} \, \vec{B} - \nabla^2 \vec{B} && (\text{three vector identity}) \\ &= - \nabla^2 \vec{B} \end{align*} $$

$$ \text{Hence } \mathrm{curl} \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} = - \nabla^2 \vec{B} \Rightarrow - \frac{1}{c^2} \frac{\partial^2 \vec{B}}{\partial t^2} = - \nabla^2 \vec{B} \quad \text{and} $$

then the three components of $\vec{B}$ in empty space satisfy

$$ \text{the (scalar) wave equation: } \Box \mathbf{u} = \mathbf{0}. $$

The invariance of the Maxwell's equations implies this of the D'Alembertian.

[II] The d'Alembertian is an invariant operator under the Lorentz transformation [1] p.95, 96, 97

Now consider two observators $O$ and $O'$ in uniform translation at a constant velocity $v$ having each a coordinates system

$X = (t, x, y, z)$ and $X' = (t', x', y', z')$ and we suppose : there is a matrix $M$ such as $X = MX'$.

If we suppose the Maxwell's equations are invariant for the two observators

$$ \begin{align*} \Box \text{ is also invariant: } \Box (\mathbf{u}) (t, x, y, z) &= \Box (\mathbf{u}) (t', x', y', z') \quad \forall \mathbf{u} \\ \Leftrightarrow \Box (\mathbf{u} \circ \mathbf{M}) (t', x', y', z') &= \Box (\mathbf{u}) (t', x', y', z') \quad \forall \mathbf{u}. \end{align*} $$

the problem is to characterize $M$.

If $M$ is a matrix de Lorentz it is known that :

$$ \Box (\mathbf{u} \circ \mathbf{M}) (t', x', y', z') = \Box (\mathbf{u}) (t', x', y', z') \quad \forall \mathbf{u}. $$

Proof:

We put :

$$ (ct, x, y, z) = (x^0, x^1, x^2, x^3), (ct', x', y', z') = (x'^0, x'^1, x'^2, x'^3) $$

$$ \partial_{\alpha} = \frac{\partial}{\partial x^{\alpha}}, \quad \partial'_{\alpha} = \frac{\partial}{\partial x'^{\alpha}} $$

$$ \partial f = (\partial_0 f, \partial_1 f, \partial_2 f, \partial_3 f), \partial' f = (\partial'_0 f, \partial'_1 f, \partial'_2 f, \partial'_3 f) $$
Lemma : $\partial'f = \partial fM$.

We have $\frac{\partial x^a}{\partial x^b} = M_b^a$.

By the chain rule :

$$ \partial'_b f = \frac{\partial f(X)}{\partial x^b} = \sum_{a=0}^{3} \frac{\partial f(X)}{\partial x^a} \frac{\partial x^a}{\partial x^b} = \sum_{a=0}^{3} \frac{\partial f(X)}{\partial x^a} M_b^a $$

$$ \text{as } \partial fM = (\partial_0 f, \partial_1 f, \partial_2 f, \partial_3 f) \quad \left[ \begin{array}{cccc} M_0^0 & M_1^0 & M_2^0 & M_3^0 \\ M_0^1 & M_1^1 & M_2^1 & M_3^1 \\ M_0^2 & M_1^2 & M_2^2 & M_3^2 \\ M_0^3 & M_1^3 & M_2^3 & M_3^3 \end{array} \right] \quad \text{we can conclude.} $$

As $M^{-1} = G^t MG$ and $\partial'f = \partial fM \Rightarrow \partial'fM^{-1} = \partial f \Rightarrow \partial'fG^t MG = \partial f \Rightarrow \partial'fG^t M = \partial fG \Rightarrow {}^t(\partial fG) = MG^t(\partial'f) = M^t(\partial'fG)$

hence ${}^t(\partial fG) = M^t(\partial'fG) : {}^t(\partial fG)$ is a four-vector.

We can put $\text{Grad } f = {}^t(\partial fG) = \left( \frac{1}{c} \frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, -\frac{\partial f}{\partial z} \right)$.

We can also define for X a four-vector field the four-divergence :

$$ \text{Div } X = \frac{1}{c} \frac{\partial X^0}{\partial t} + \frac{\partial X^1}{\partial x} + \frac{\partial X^2}{\partial y} + \frac{\partial X^3}{\partial z} $$

Theorem :

If $X = MX'$ then $\text{Div } X = \text{Div } X'$.

Proof :

We note $X = (X^0, X^1, X^2, X^3)$, $X' = (X'^0, X'^1, X'^2, X'^3)$.

Then $\text{Div}(X) = \sum_{i=0}^{3} \frac{\partial X^i}{\partial x^i}$ and $\text{Div}(X') = \sum_{i=0}^{3} \frac{\partial X^i}{\partial x'^i}$.

We put $\partial_i = \frac{\partial}{\partial x^i}, \partial'_i = \frac{\partial}{\partial x'^i}$.

We have : $\partial'_i = \frac{\partial}{\partial x'^i} = \sum_{j=0}^{3} \frac{\partial x^j}{\partial x'^i} \frac{\partial}{\partial x^j} = \sum_{i=0}^{3} M_j^i \partial_i$ and

$$ \partial_i = \frac{\partial}{\partial x^i} = \sum_{j=0}^{3} \frac{\partial x^j}{\partial x^i} \frac{\partial}{\partial x'^j} = \sum_{i=0}^{3} (M^{-1})^i_j \partial'_i $$

Then $\text{Div}(X) = \sum_{i=0}^{3} \frac{\partial X^i}{\partial x^i} = \sum_{i=0}^{3} \sum_{j=0}^{3} (M^{-1})^j_i \partial'_j X^i$

as $X^i = \sum_{j=0}^{3} M_j^i X^j$ :
$$ \mathbf{Div}(X) = \sum_{i=0}^{3} \sum_{j=0}^{3} (M^{-1})_{i}^{j} \partial'_{j} \left( \sum_{k=0}^{3} M_{k}^{i} X^{k} \right) $$

since the components of M are constant :

$$ \partial'_{j} \left( \sum_{k=0}^{3} M_{k}^{i} X^{k} \right) = \sum_{k=0}^{3} M_{k}^{i} \partial'_{j} X^{k} $$

$$ \begin{aligned} \text{and } \mathbf{Div}(X) &= \sum_{i=0}^{3} \sum_{j=0}^{3} (M^{-1})_{i}^{j} \sum_{k=0}^{3} M_{k}^{i} \partial'_{j} X^{k} \\ &= \sum_{j=0}^{3} \sum_{k=0}^{3} \left( \sum_{i=0}^{3} (M^{-1})_{i}^{j} M_{k}^{i} \right) \partial'_{j} X^{k} \\ &= \sum_{k=0}^{3} \sum_{j=0}^{3} \delta_{k}^{j} \partial'_{j} X^{k} = \mathbf{Div}(X'). \end{aligned} $$

**Theorem :** $\mathbf{Div Grad} = \square$

Proof:

$$ \mathbf{Grad} f = {}^{t}(\partial f G) = \left( \frac{1}{c} \frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, -\frac{\partial f}{\partial z} \right). $$

$$ \mathbf{Div} \mathbf{Grad} f = \mathbf{Div} \left( \frac{1}{c} \frac{\partial f}{\partial t}, \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \square f $$

**Theorem :**

$\square$ is an invariant operator under the Lorentz transformation.

Proof:

$$ \square f = \mathbf{Div} \mathbf{Grad} f = \mathbf{Div} M \mathbf{Grad} f' = \mathbf{Div} \mathbf{Grad} f' = \square f' $$

**Bibliography :**

[1] N.M.J Woodhouse : Special Relativity 2003 Springer.

https://archive.org/details/specialrelativit0000wood/mode/2 up

[2] A. Guimier https://amu.hal.science/hal-02965773.

[3] J.R Lucas and P.E Hodgson : Spacetime and electromagnetism 1990 Clarendon Press Oxford.

https://archive.org/details/spacetimeelectro0000luca

[4] https://en.wikipedia.org/wiki/Chain_rule.

[5] Ricaro Heras : Lorentz transformation and the wave equation

March 2016 European Journal of Physics

https://www.researchgate.net/publication/292339127_Lorentz_transform