4. Formes différentielles pour l'électromagnétisme
5. Méthode de primitivation de Poincaré
6. Intégration sur un rayon d'espace
7. Intégration sur un rayon d'espace-temps
Dans le cadre de l'électromagnétisme dans le vide, les champs électrique \(\mathbf{\vec {E}}\) et d'induction magnétique \(\mathbf{\vec {B}}\) vérifient
les équations de Maxwell :
\[
\mathbf{\vec {rot} \left (\vec {E} \right ) = - \dfrac{\partial \vec {B}}{\partial t}}\ \ (1.1)
\]
\[
\mathbf{div \left (\vec {B} \right ) = 0}\ \ (1.2)
\]
\[
\mathbf{div \left (\vec {E} \right ) =\dfrac{\rho}{\epsilon _{0}}}\ \ (1.3)
\]
\[
\mathbf{\vec {rot} \left (\vec {B} \right ) = \dfrac{1}{c^{2}} \dfrac{\partial \vec {E}}{\partial t} + \mu _{0} \vec {j}}\ \ (1.4)
\]
Dans ce groupe de quatre équations :
-\(\rho \) et \(\mathbf{\vec {j}}\) sont respectivement la densité de charge électrique et le vecteur densité de courant électrique;
-\(\epsilon _{0}\) et \(\mu _{0}\) sont respectivement la permittivité diélectrique et le perméabilité magnétique.
Ces deux dernières grandeurs citées vérifient la relation de constitution :
\[
\mathbf{\epsilon _{0}\, \mu _{0}\, c^{2} \, = 1}\ \ (1.5)
\]
Les relations (1.1) et (1.2) permettent de démontrer que les champs \(\mathbf{\vec {E}}\) et \(\mathbf{\vec {B}}\) dérivent des potentiels
vecteur \(\mathbf{\vec {A}}\) et scalaire \(\mathbf{V}\) suivant les relations :
\[
\mathbf{\vec {B} = \vec {rot} \left (\vec {A} \right )}\ \ (1.6)
\]
\[
\mathbf{\vec {E} = - \vec {grad}_{\vec {r}} \left ( V \right ) -\dfrac{\partial \vec {A}}{\partial t}}\ \ (1.7)
\]
Pour un même couple \(\mathbf{\left (\vec {E},\vec {B} \right )}\) donné il existe une infinité de choix possibles de couples \(\mathbf{\left (\vec {A},V \right )}\).
On démontre que deux choix possibles de potentiels différent d'un quadrigradient d'une fonction \(\mathbf{\alpha \left (\vec {r} ,t \right )}\) :
\[
\mathbf{\vec {A'} \left ( \vec {r} , t \right ) = \vec {A} \left ( \vec {r} , t \right ) + \vec {grad} _{\vec {r}} \alpha \left (\vec {r} ,t \right )}\ \ (1.8)
\]
\[
\mathbf{V' \left ( \vec {r} , t \right ) = V \left ( \vec {r} , t \right ) - \dfrac{\partial \alpha \left (\vec {r} ,t \right )}{\partial t} }\ \ (1.9)
\]
Dans ce travail on cherche les relations inverses de (1.6) et (1.7) obtenues la première fois par Hertz.
L'idée de Poincaré est de choisir une fonction \(\mathbf{\alpha \left (\vec {r} ,t \right )}\), dite de jauge, particulière qui substituée dans les
relations(1.8) et (1.9) permettra d'obtenir les formules de hertz. Le choix de Poincaré est le suivant :
\[
\mathbf{\alpha \left (\vec {r},t \right )=\alpha \left (\vec {0},t \right ) - \int _{0}^{1} \vec {r} . \vec {A} \left ( \lambda \vec {r} , t \right ) d\lambda}\ \ (2.1)
\]
\[
\mathbf{\dfrac{\partial \alpha \left (\vec {0} ,t \right )}{\partial t} = V \left (\vec {0} ,t \right )}\ \ (2.2)
\]
Développons le produit scalaire spatial présent dans la relation (2.1) :
\[
\begin{align}
\mathbf{\alpha \left (\vec {r},t \right )}
&\mathbf{= \alpha \left (\vec {0},t \right )}\\
&\mathbf{\ \ \ - x \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{\ \ \ - y \int _{0}^{1} A_{y} \left ( \lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{\ \ \ - z \int _{0}^{1} A_{z} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.3)
\end{align}
\]
Commençons par calculer la dérivée partielle de la fonction de jauge suivant la première coordonnée spatiale \(\mathbf{x}\). A partir de (2.1), on obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{- x \int _{0}^{1} \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial x}\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \dfrac{\partial A_{y} \left ( \lambda \vec {r} , t \right )}{\partial y}\, d \lambda}\\
&\mathbf{- z \int _{0}^{1} \dfrac{\partial A_{z} \left ( \lambda \vec {r} , t \right )}{\partial z}\, d \lambda}\ \ (2.4)
\end{align}
\]
L'identité suivante va être utile :
\[
\mathbf{\lambda \, \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda}
= \boxed {x \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial x}}
+ y \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial y}
+ z \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial z}}\ \ (2.5)
\]
Maintenant substituons le terme encadré de la relation (2.5) dans son équivalent de la relation (2.4) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{- \int _{0}^{1}\left (\lambda \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda}
-y \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial y}
-z \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial z} \right )\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \dfrac{\partial A_{y} \left ( \lambda \vec {r} , t \right )}{\partial y}\, d \lambda}\\
&\mathbf{- z \int _{0}^{1} \dfrac{\partial A_{z} \left ( \lambda \vec {r} , t \right )}{\partial z}\, d \lambda}\ \ (2.6)
\end{align}
\]
Effectuons dans l'expression (2.6) le regroupement des termes en \(\mathbf{x}\) puis des termes en \(\mathbf{z}\) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- \int _{0}^{1} A_{x} \left (\lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{- \int _{0}^{1} \lambda \, \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda}\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \lambda \left ( \underbrace {\dfrac{\partial A_{y} \left ( \lambda \vec {r} , t \right )}{\partial \lambda x}
- \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda y}}_{= \,\Large B_{z} \left ( \lambda \vec {r} , t \right )} \right ) d \lambda }\\
&\mathbf{- z \int _{0}^{1} \lambda \left ( \underbrace{\dfrac{\partial A_{z} \left ( \lambda \vec {r} , t \right )}{\partial \lambda x}
- \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda z}}_{= - \,\Large B_{y} \left ( \lambda \vec {r} , t \right )}\right )d \lambda}\ \ (2.7)
\end{align}
\]
La dérivée partielle de la fonction de jauge suivant la première coordonnée spatiale \(\mathbf{x}\) se simplifie un peu en :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- \int _{0}^{1} A_{x} \left (\lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{- \int _{0}^{1} \lambda \, \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda}\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda}\\
&\mathbf{+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.8)
\end{align}
\]
Le deuxième terme du membre de droite de la relation (2.8) peut s'intégrer par parties :
\[
\mathbf{- \int _{0}^{1} \lambda \, \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial \lambda}\, d \lambda =
- \underbrace {\left [ \lambda \, A_{x} \left ( \lambda \vec {r} , t \right )\right ]_{- \infty}^{\infty }}_{\Large = A_{x} \left ( \vec {r} , t\right )}+ \int _{0}^{1} A_{x} \left (\lambda \vec {r} , t \right ) d \lambda}\ \ (2.9)
\]
Finalement il reste :
\[
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =
- A_{x} \left ( \vec {r} , t \right )
- y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda
+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.10)
\]
Par simple permutation circulaire sur les coordonnées spatiales, on obtient :
\[
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial y} =
- A_{y} \left ( \vec {r} , t \right )
- z \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
+ x \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.11)
\]
\[
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial z} =
- A_{z} \left ( \vec {r} , t \right )
- x \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
+ y \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.12)
\]
Il nous faut maintenant calculer la dérivée partielle de la fonction de jauge par rapport à la coordonnée temporelle \(\mathbf{t}\).
A partir de (2.1) on obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{\underbrace{\dfrac{\partial \alpha \left (\vec {0},t \right )}{\partial t}}_{\Large =\, V \left ( \vec {0} ,t \right )}}\\
&\mathbf{- x \int _{0}^{1} \dfrac{\partial A_{x} \left ( \lambda \vec {r} , t \right )}{\partial t}\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \dfrac{\partial A_{y} \left ( \lambda \vec {r} , t \right )}{\partial t}\, d \lambda}\\
&\mathbf{- z \int _{0}^{1} \dfrac{\partial A_{z} \left ( \lambda \vec {r} , t \right )}{\partial t}\, d \lambda}\ \ (2.13)
\end{align}
\]
Dans cette dernière expression (2.13) on peut introduire les coordonnées du champ électrique grâce à (1.7) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{V \left ( \vec {0} ,t \right )}\\
&\mathbf{+ x \int _{0}^{1} \left ( E_{x} \left ( \lambda \vec {r} , t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, x} \right )\, d \lambda}\\
&\mathbf{+ y \int _{0}^{1} \left ( E_{y} \left ( \lambda \vec {r} , t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, y} \right )\, d \lambda}\\
&\mathbf{+ z \int _{0}^{1} \left ( E_{z} \left ( \lambda \vec {r} , t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, z}\right )\, d \lambda\ \ (2.14)}
\end{align}
\]
L'identité suivante va être utile :
\[
\mathbf{ \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda}
= \boxed {x \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, x}
+ y \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, y}
+ z \dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda \, z}}}\ \ (2.15)
\]
Remplaçons le deuxième membre encadré de (2.15) dans (2.14) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{V \left ( \vec {0} ,t \right )}\\
&\mathbf{+ \underbrace {\int _{0}^{1}\dfrac{\partial V \left ( \lambda \vec {r} , t \right )}{\partial \lambda} \, d \lambda }_{\Large = \, V \left ( \vec {r} , t \right ) - V \left ( \vec {0} , t \right )}}\\
&\mathbf{+ x \int _{0}^{1} E_{x} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\\
&\mathbf{+ y \int _{0}^{1} E_{y} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\\
&\mathbf{+ z \int _{0}^{1} E_{z} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\ \ (2.16)
\end{align}
\]
Finalement il reste :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{V \left ( \vec {r} ,t \right )}\\
&\mathbf{+ x \int _{0}^{1} E_{x} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\\
&\mathbf{+ y \int _{0}^{1} E_{y} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\\
&\mathbf{+ z \int _{0}^{1} E_{z} \left ( \lambda \vec {r} , t \right ) \, d \lambda }\ \ (2.17)
\end{align}
\]
Nous sommes désormais en mesure d'appliquer l'idée de Poincaré.
(1.8) et (2.10) impliquent :
\[
\begin{align}
\mathbf{A'_{x} \left (\vec {r} ,t \right )}
&\mathbf{= A_{x} \left (\vec {r} ,t \right ) + \dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x}}\\
&\mathbf{= \require{cancel}\cancel {A_{x} \left (\vec {r} ,t \right )}+\left ( - \cancel {A_{x} \left ( \vec {r} , t \right )}
- y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda
+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda \right )}\\
&\mathbf{= - y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda
+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda }\ \ (2.18)
\end{align}
\]
Par simple permutation circulaire sur les coordonnées spatiales, on obtient :
\[
\begin{align}
\mathbf{A'_{y} \left (\vec {r} ,t \right )}
&\mathbf{= - z \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
+ x \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , t \right ) d \lambda }\ \ (2.19)
\end{align}
\]
\[
\begin{align}
\mathbf{A'_{z} \left (\vec {r} ,t \right )}
&\mathbf{= - x \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
+ y \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , t \right ) d \lambda }\ \ (2.20)
\end{align}
\]
(1.9) et (2.17) impliquent :
\[
\begin{align}
\mathbf{V' \left ( \vec {r} , t \right )}
&\mathbf{= V \left ( \vec {r} , t \right ) - \dfrac{\partial \alpha \left (\vec {r} ,t \right )}{\partial t} }\\
&\mathbf{\require {cancel} = \cancel {V \left ( \vec {r} , t \right )}
-\left ( \cancel {V \left ( \vec {r} , t \right )}
+ x \int _{0}^{1} E_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
+ y \int _{0}^{1} E_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
+ z \int _{0}^{1} E_{z} \left ( \lambda \vec {r} , t \right ) d \lambda
\right )}\\
&\mathbf{=- x \int _{0}^{1} E_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
- y \int _{0}^{1} E_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
- z \int _{0}^{1} E_{z} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.21)
\end{align}
\]
Les formules de Hertz obtenues peuvent se condenser en :
\[
\boxed {\mathbf{\vec {A'} \left ( \vec {r} , t \right ) = -\vec {r} \times \int _{0}^{1} \lambda \, \vec {B} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.22)\\
\mathbf{V' \left ( \vec {r} , t \right ) = -\vec {r} . \int _{0}^{1} \vec {E} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (2.23)}
\]
L'invariance relativiste des équations de Maxwell invite à l'utilisation des quadrivecteurs position et potentiel :
\[
\mathbf{r \left ( \vec {r} , c\, t \right )}\ \ (3.1)
\]
et
\[
\mathbf{A \left ( \vec {A} , \dfrac{V}{c}\right )}\ \ (3.2)
\]
Les relations (2.1) et (2.2) deviennent :
\[
\mathbf{\alpha \left (\vec {r},t \right )= - \int _{0}^{1} r.A \left ( \lambda \vec {r} , \lambda \,t \right ) d \lambda}\ \ (3.3)
\]
soit
\[
\begin{align}
\mathbf{\alpha \left (\vec {r},t \right ) =}
&\mathbf{-x \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{-y \int _{0}^{1} A_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{-z \int _{0}^{1} A_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{+t \int _{0}^{1} V \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.4)
\end{align}
\]
Commençons par calculer la dérivée partielle de la fonction de jauge suivant la première coordonnée spatiale \(\mathbf{x}\). A partir de (3.4), on obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{- x \int _{0}^{1} \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}\, d \lambda}\\
&\mathbf{- y \int _{0}^{1} \dfrac{\partial A_{y} \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}\, d \lambda}\\
&\mathbf{- z \int _{0}^{1} \dfrac{\partial A_{z} \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}\, d \lambda}\\
&\mathbf{+ t \int _{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}\, d \lambda}\ \ (3.5)
\end{align}
\]
L'identité suivante va être utile :
\[
\mathbf{ \dfrac{\partial A_{x} \left ( \lambda \vec {r} ,\lambda t \right )}{\partial \lambda}
= \boxed {x \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial \lambda x}}
+ y \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial \lambda y}
+ z \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial \lambda z}
+ t \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial \lambda t}}\ \ (3.6)
\]
Maintenant substituons le terme encadré de la relation (3.6) dans son équivalent de la relation (3.5) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{\underbrace {- \int _{0}^{1} A_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- \int_{0}^{1} \lambda \, \dfrac{\partial A_{x} \left ( \lambda \vec {r} ,\lambda t \right )}{\partial \lambda} d \lambda}_{\Large = - A_{x} \left (\vec {r} ,t \right )}}\\
&\mathbf{- y \int_{0}^{1} \left ( \underbrace { \dfrac{\partial A_{y} \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}
- \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial y} }_{= \Large \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right )} \right ) d \lambda}\\
&\mathbf{+ z \int_{0}^{1} \left ( \underbrace { \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial z}
- \dfrac{\partial A_{z} \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}}_{= \Large \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right )} \right ) d \lambda}\\
&\mathbf{+ t \int_{0}^{1} \left ( \underbrace { \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial x}
- \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial t}}_{= \Large - \lambda \, E_{x} \left ( \lambda \vec {r} , \lambda t \right )} \right ) d \lambda}\ \ (3.7)
\end{align}
\]
soit
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x} =}
&\mathbf{- A_{x} \left ( \vec {r}, t \right )}\\
&\mathbf{- y \int_{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{+ z \int_{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{- t \int_{0}^{1} \lambda \, E_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.8)
\end{align}
\]
Par simple permutation circulaire sur les coordonnées spatiales, on obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial y} =}
&\mathbf{- A_{y} \left ( \vec {r}, t \right )}\\
&\mathbf{- z \int_{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{+ x \int_{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{- t \int_{0}^{1} \lambda \, E_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.9)
\end{align}
\]
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial z} =}
&\mathbf{- A_{z} \left ( \vec {r}, t \right )}\\
&\mathbf{- x \int_{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{+ y \int_{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\mathbf{- t \int_{0}^{1} \lambda \, E_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.10)
\end{align}
\]
Il nous faut maintenant calculer la dérivée partielle de la fonction de jauge par rapport à la coordonnée temporelle \(\mathbf{t}\).
A partir de (3.4) on obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{- x \int_{0}^{1} \dfrac{\partial A_{x} \left ( \lambda \vec {r} , \lambda t \right )}{\partial t} \, d \lambda}\\
&\mathbf{- y \int_{0}^{1} \dfrac{\partial A_{y} \left ( \lambda \vec {r} , \lambda t \right )}{\partial t} \, d \lambda}\\
&\mathbf{- y \int_{0}^{1} \dfrac{\partial A_{z} \left ( \lambda \vec {r} , \lambda t \right )}{\partial t} \, d \lambda}\\
&\mathbf{+ \int_{0}^{1} V \left ( \lambda \vec {r} , \lambda t \right ) d \lambda + t \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial t}\, d \lambda}\ \ (3.11)
\end{align}
\]
Dans cette dernière expression (3.11) on peut introduire les coordonnées du champ électrique grâce à (1.7) :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{\,\,x \int_{0}^{1} \lambda \left ( E_{x} \left ( \lambda \vec {r} , \lambda t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial x} \right ) d \lambda}\\
&\mathbf{+ y \int_{0}^{1} \lambda \left ( E_{y} \left ( \lambda \vec {r} , \lambda t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial y} \right ) d \lambda}\\
&\mathbf{+ z \int_{0}^{1} \lambda \left ( E_{z} \left ( \lambda \vec {r} , \lambda t \right ) + \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial z} \right ) d \lambda}\\
&\mathbf{+ \int_{0}^{1} V \left ( \lambda \vec {r} , \lambda t \right ) d \lambda + t \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial t}\, d \lambda}\ \ (3.12)
\end{align}
\]
Séparons les termes faisant intervenir le champ électrique d'une part et le potentiel scalaire d'autre part :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{x \int_{0}^{1} \lambda E_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ y \int_{0}^{1} \lambda E_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ z \int_{0}^{1} \lambda E_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\\
&\boxed{
\mathbf{+ x \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial x} d \lambda}\\
\mathbf{+ y \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial y} d \lambda}\\
\mathbf{+ z \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial z} d \lambda}\\
\mathbf{+ t \int_{0}^{1} \dfrac{\partial V \left ( \lambda \vec {r} , \lambda t \right )}{\partial t} d \lambda}}\\
&\mathbf{+ \int_{0}^{1} V \left ( \lambda \vec {r} , \lambda t \right ) d \lambda }\ \ (3.13)
\end{align}
\]
Le terme encadré dans la relation (3.13) ci-dessus vaut :
\[
\mathbf{\int_{0}^{1} \lambda V \left ( \lambda \vec {r} , \lambda t \right ) d \lambda }\ \ (3.14)
\]
Faisons les opérations suivantes :
-Substituons (3.14) dans (3.13);
-Regroupons avec le dernier terme de (3.13);
-Effectuons une intégration par partie;
-Simplifions.
On obtient :
\[
\begin{align}
\mathbf{\dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial t} =}
&\mathbf{V \left ( \vec {r} , t \right )}\\
&\mathbf{+ x \int_{0}^{1} \lambda E_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ y \int_{0}^{1} \lambda E_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ z \int_{0}^{1} \lambda E_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.15)
\end{align}
\]
Nous sommes désormais en mesure d'appliquer la méthode de Poincaré avec la deuxième jauge :
(1.8) et (3.8) impliquent :
\[
\begin{align}
\mathbf{A'_{x} \left (\vec {r} ,t \right )}
&\mathbf{= A_{x} \left (\vec {r} ,t \right ) + \dfrac{\partial \alpha \left (\vec {r},t \right )}{\partial x}}\\
&\mathbf{= \require{cancel}\cancel {A_{x} \left (\vec {r} ,t \right )}+\left ( - \cancel {A_{x} \left ( \vec {r} , t \right )}
- y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- t \int _{0}^{1} \lambda \, E_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda\right )}\\
&\mathbf{= - y \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ z \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- t \int _{0}^{1} \lambda \, E_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda }\ \ (3.16)
\end{align}
\]
Par simple permutation circulaire sur les coordonnées spatiales, on obtient :
\[
\begin{align}
\mathbf{A'_{y} \left (\vec {r} ,t \right )}
&\mathbf{= - z \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ x \int _{0}^{1} \lambda \, B_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- t \int _{0}^{1} \lambda \, E_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda }\ \ (3.17)
\end{align}
\]
\[
\begin{align}
\mathbf{A'_{z} \left (\vec {r} ,t \right )}
&\mathbf{= - x \int _{0}^{1} \lambda \, B_{y} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
+ y \int _{0}^{1} \lambda \, B_{x} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- t \int _{0}^{1} \lambda \, E_{z} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda }\ \ (3.18)
\end{align}
\]
(1.9) et (3.15) impliquent :
\[
\begin{align}
\mathbf{V' \left ( \vec {r} , t \right )}
&\mathbf{= V \left ( \vec {r} , t \right ) - \dfrac{\partial \alpha \left (\vec {r} ,t \right )}{\partial t} }\\
&\mathbf{\require {cancel} = \cancel {V \left ( \vec {r} , t \right )}
-\left ( \cancel {V \left ( \vec {r} , t \right )}
+ x \int _{0}^{1} \lambda E_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
+ y \int _{0}^{1} \lambda E_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
+ z \int _{0}^{1} \lambda E_{z} \left ( \lambda \vec {r} , t \right ) d \lambda
\right )}\\
&\mathbf{=- x \int _{0}^{1} \lambda E_{x} \left ( \lambda \vec {r} , t \right ) d \lambda
- y \int _{0}^{1} \lambda E_{y} \left ( \lambda \vec {r} , t \right ) d \lambda
- z \int _{0}^{1} \lambda E_{z} \left ( \lambda \vec {r} , t \right ) d \lambda}\ \ (3.19)
\end{align}
\]
Les nouvelles formules donnant les potentiels en fonction des champs peuvent se condenser en :
\[
\boxed {\mathbf{\vec {A'} \left ( \vec {r} , t \right ) = -\vec {r} \times \int _{0}^{1} \lambda \, \vec {B} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda
- t \int _{0}^{1} \lambda \, \vec {E} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.20)\\
\mathbf{V' \left ( \vec {r} , t \right ) = -\vec {r} . \int _{0}^{1} \vec {E} \left ( \lambda \vec {r} , \lambda t \right ) d \lambda}\ \ (3.21)}
\]
Remarques
-Les formules obtenues avec la deuxième jauge de Poincaré sont différentes de celles de Hertz.
-Il est très instructif de vérifier (1.6) et (1.7) avec les formules obtenues par la méthode de Poincaré.
Présentation des formes de base
Le potentiel \(\mathbf{A}\) est une 1-forme tandis que le courant \(\mathbf{J}\) est une 3-forme.
\[
\mathbf{A = A_{x}\, dx + A_{y}\, dy + A_{z}\, dz - V\, dt}\ \ (4.1)
\]
\[
\mathbf{J = i\,c\, J_{x}\, dy\, \wedge\, dz\, \wedge \, dt + i\,c\, J_{y}\, dz\, \wedge\, dx\, \wedge \, dt + i\,c\, J_{z}\, dx\, \wedge\, dy\, \wedge \, dt +
\dfrac{c\, \rho}{i}\, dx\, \wedge \, dy\, \wedge \, dz}\ \ (4.2)
\]
Calcul du champ
La 2-forme champ \(\mathbf{F}\) est égale à la différentielle extérieure de la 1-forme potentiel \(\mathbf{A}\) :
\[
\begin{align}
\mathbf{F}
&\mathbf{=\, dA}\\
&\mathbf{=\left (\dfrac{\partial A_{x}}{\partial x}\, dx + \dfrac{\partial A_{x}}{\partial y}\, dy +\dfrac{\partial A_{x}}{\partial z}\, dz + \dfrac{\partial A_{x}}{\partial t}\, dt \right )\wedge dx}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \left (\dfrac{\partial A_{y}}{\partial x}\, dx + \dfrac{\partial A_{y}}{\partial y}\, dy +\dfrac{\partial A_{y}}{\partial z}\, dz + \dfrac{\partial A_{y}}{\partial t}\, dt\right )\wedge dy}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \left (\dfrac{\partial A_{z}}{\partial x}\, dx + \dfrac{\partial A_{z}}{\partial y}\, dy +\dfrac{\partial A_{z}}{\partial z}\, dz + \dfrac{\partial A_{z}}{\partial t}\, dt\right )\wedge dz}\\
&\mathbf{\ \ \ \ \ \ \ -}\\
&\mathbf{\ \ \ \ \left (\dfrac{\partial V}{\partial x}\, dx + \dfrac{\partial V}{\partial y}\, dy + \dfrac{\partial V}{\partial z}\, dz + \dfrac{\partial V}{\partial t}\, dt \right )\wedge dt}\\
&\mathbf{=\,\left (-\dfrac{\partial V}{\partial x}-\dfrac{\partial A_{x}}{\partial t} \right)\, dx\wedge \,dt
+\left (-\dfrac{\partial V}{\partial y}-\dfrac{\partial A_{y}}{\partial t} \right)\, dy\wedge \,dt
+\left (-\dfrac{\partial V}{\partial z}-\dfrac{\partial A_{z}}{\partial t} \right)\, dz\wedge \,dt}\\
&\mathbf{\ \ +\,\left (\dfrac{\partial A_{y}}{\partial x}-\dfrac{\partial A_{x}}{\partial y} \right)\, dx\wedge \,dy
+\left (\dfrac{\partial A_{z}}{\partial y}-\dfrac{\partial A_{y}}{\partial z} \right)\, dy\wedge \,dz
+\left (\dfrac{\partial A_{x}}{\partial z}-\dfrac{\partial A_{z}}{\partial x} \right)\, dz\wedge \,dx}
\end{align}
\]
On obtient :
\[
\mathbf{F = E_{x}\, dx\, \wedge \, dt + E_{y}\, dy \, \wedge \, dt + E_{z}\, dz\, \wedge \, dt
+ B_{z}\, dx\, \wedge \, dy + B_{x}\, dy\, \wedge \, dz + B_{y}\, dz\, \wedge \, dx}\ \ (4.3)
\]
Calcul du champ dual
La 2-forme champ dual \(\mathbf{^{\star} F}\) est la conjuguée de Hodge de la 2-forme \(\mathbf{F}\).
Afin d'éviter le tenseur métrique et le tenseur de Levi-Civita; Nous allons simplifier en disant que la conjugaison de Hodge agit
linéairement sur les formes \(\mathbf{\omega}\) telle que :
\[
\mathbf{\omega \, \wedge \, ^{\star} \, \omega \, = dx \, \wedge \, dy \, \wedge \, dz \, \wedge i\, c\, dt}\ \ (4.4)
\]
Appliquons la définition précédente.
\[
\mathbf{^{\star}F = E_{x}\, ^{\star} \left (dx\, \wedge \, dt \right ) + E_{y}\, ^{\star} \left (dy \, \wedge \, dt\, \right ) + E_{z}\, ^{\star} \left (dz\, \wedge \, dt \right)
+ B_{z}\, ^{\star} \left (dx\, \wedge \, dy \right ) + B_{x}\, ^{\star} \left (dy\, \wedge \, dz \right ) + B_{y}\, ^{\star} \left (dz\, \wedge \, dx \right )}
\]
On obtient :
\[
\mathbf{^{\star}F = \dfrac{E_{x}}{i\, c}\, dy\, \wedge \, dz + \dfrac{E_{y}}{i\, c}\, dz \, \wedge \, dx + \dfrac{E_{z}}{i\, c}\, dx\, \wedge \, dy
+ i\, c\, B_{z}\, dz\, \wedge \, dt + i\, c\, B_{x}\, dx\, \wedge \, dt + i\, c \, B_{y}\, dy\, \wedge \, dt}\ \ (4.5)
\]
Equation du champ
\[ \begin{align} \mathbf{dF} &\mathbf{=\left (\dfrac{\partial E_{x}}{\partial x}\, dx + \dfrac{\partial E_{x}}{\partial y}\, dy +\dfrac{\partial E_{x}}{\partial z}\, dz + \dfrac{\partial E_{x}}{\partial t}\, dt \right )\wedge dx \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial E_{y}}{\partial x}\, dx + \dfrac{\partial E_{y}}{\partial y}\, dy +\dfrac{\partial E_{y}}{\partial z}\, dz + \dfrac{\partial E_{y}}{\partial t}\, dt\right )\wedge dy \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial E_{z}}{\partial x}\, dx + \dfrac{\partial E_{z}}{\partial y}\, dy +\dfrac{\partial E_{z}}{\partial z}\, dz + \dfrac{\partial E_{z}}{\partial t}\, dt\right )\wedge dz \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial B_{z}}{\partial x}\, dx + \dfrac{\partial B_{z}}{\partial y}\, dy +\dfrac{\partial B_{z}}{\partial z}\, dz + \dfrac{\partial B_{z}}{\partial t}\, dt \right )\wedge dx \, \wedge dy}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial B_{x}}{\partial x}\, dx + \dfrac{\partial B_{x}}{\partial y}\, dy +\dfrac{\partial B_{x}}{\partial z}\, dz + \dfrac{\partial B_{x}}{\partial t}\, dt\right )\wedge dy \, \wedge dz}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial B_{y}}{\partial x}\, dx + \dfrac{\partial B_{y}}{\partial y}\, dy +\dfrac{\partial B_{y}}{\partial z}\, dz + \dfrac{\partial B_{y}}{\partial t}\, dt\right )\wedge dz \, \wedge dx}\\ &\mathbf{=\left (\dfrac{\partial E_{x}}{\partial z}-\dfrac{\partial E_{z}}{\partial x}+\dfrac{\partial B_{z}}{\partial t} \right )\, dz\, \wedge dx \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial E_{y}}{\partial x}-\dfrac{\partial E_{x}}{\partial y}+\dfrac{\partial B_{x}}{\partial t} \right )\, dx\, \wedge dy \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial E_{z}}{\partial y}-\dfrac{\partial E_{y}}{\partial z}+\dfrac{\partial B_{y}}{\partial t} \right )\, dy\, \wedge dz \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \left (\dfrac{\partial B_{x}}{\partial x}+\dfrac{\partial B_{y}}{\partial y}+\dfrac{\partial B_{z}}{\partial z} \right )\, dx\, \wedge dy \, \wedge \, dz}\ \ (4.6) \end{align} \] On retrouve \(\mathbf{dF = 0}\) car \(\mathbf{\vec {rot}\, \left (\vec {E}\, \right ) = -\, \dfrac{\partial \vec {B}}{\partial t}}\) et \(\mathbf{div\, \left (\vec {B}\right ) = 0}\).
Equation du champ dual
\[ \begin{align} \mathbf{d\, ^{ \star} F} &\mathbf{=\, \dfrac{1}{i\, c}\,\left (\dfrac{\partial E_{x}}{\partial x}\, dx + \dfrac{\partial E_{x}}{\partial y}\, dy +\dfrac{\partial E_{x}}{\partial z}\, dz + \dfrac{\partial E_{x}}{\partial t}\, dt \right )\wedge dy \, \wedge dz}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, \dfrac{1}{i\, c}\,\left (\dfrac{\partial E_{y}}{\partial x}\, dx + \dfrac{\partial E_{y}}{\partial y}\, dy +\dfrac{\partial E_{y}}{\partial z}\, dz + \dfrac{\partial E_{y}}{\partial t}\, dt\right )\wedge dz \, \wedge dx}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, \dfrac{1}{i\, c}\,\left (\dfrac{\partial E_{z}}{\partial x}\, dx + \dfrac{\partial E_{z}}{\partial y}\, dy +\dfrac{\partial E_{z}}{\partial z}\, dz + \dfrac{\partial E_{z}}{\partial t}\, dt\right )\wedge dx \, \wedge dy}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ i\, c\,\left (\dfrac{\partial B_{z}}{\partial x}\, dx + \dfrac{\partial B_{z}}{\partial y}\, dy +\dfrac{\partial B_{z}}{\partial z}\, dz + \dfrac{\partial B_{z}}{\partial t}\, dt \right )\wedge dz \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, i\, c\, \left (\dfrac{\partial B_{x}}{\partial x}\, dx + \dfrac{\partial B_{x}}{\partial y}\, dy +\dfrac{\partial B_{x}}{\partial z}\, dz + \dfrac{\partial B_{x}}{\partial t}\, dt \right )\wedge dx \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ i\, c\,\left (\dfrac{\partial B_{y}}{\partial x}\, dx + \dfrac{\partial B_{y}}{\partial y}\, dy +\dfrac{\partial B_{y}}{\partial z}\, dz + \dfrac{\partial B_{y}}{\partial t}\, dt\right )\wedge dy \, \wedge dt}\\ &\mathbf{=\, \dfrac{c}{i}\,\left (-\dfrac{\partial B_{z}}{\partial y}+\dfrac{\partial B_{y}}{\partial z}+\dfrac{\partial E_{x}}{c^{2}\, \partial t} \right )\, dy\, \wedge dz \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, \dfrac{c}{i}\,\left (-\dfrac{\partial B_{x}}{\partial z}+\dfrac{\partial B_{z}}{\partial x}+\dfrac{\partial E_{y}}{c^{2}\, \partial t} \right )\, dz\, \wedge dx \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, \dfrac{c}{i}\,\left (-\dfrac{\partial B_{y}}{\partial x}+\dfrac{\partial B_{x}}{\partial y}+\dfrac{\partial E_{z}}{c^{2}\,\partial t} \right )\, dx\, \wedge dy \, \wedge \, dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, \dfrac{1}{i\, c}\,\left (\dfrac{\partial E_{x}}{\partial x}+\dfrac{\partial E_{y}}{\partial y}+\dfrac{\partial E_{z}}{\partial z} \right )\, dx\, \wedge dy \, \wedge \, dz}\ \ (4.7) \end{align}\] On retrouve \(\mathbf{d\, ^{\star} F = \mu _{0} \, J}\) car \(\mathbf{\vec {rot}\, \left (\vec {B}\, \right ) = -\, \dfrac{\partial \vec {E}}{c^{2}\,\partial t} + \mu _{0}\, \vec {J}}\) et \(\mathbf{div\, \left (\vec {E}\right ) =\, \dfrac{\rho}{\epsilon _{0}}}\).
Equation de continuité
\[ \begin{align} \mathbf{dJ} &\mathbf{=\, i\, c\,\left (\dfrac{\partial J_{x}}{\partial x}\, dx + \dfrac{\partial J_{x}}{\partial y}\, dy +\dfrac{\partial J_{x}}{\partial z}\, dz + \dfrac{\partial J_{x}}{\partial t}\, dt \right )\wedge dy \, \wedge dz \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, i\, c\,\left (\dfrac{\partial J_{y}}{\partial x}\, dx + \dfrac{\partial J_{y}}{\partial y}\, dy +\dfrac{\partial J_{y}}{\partial z}\, dz + \dfrac{\partial J_{y}}{\partial t}\, dt\right )\wedge dz \, \wedge dx \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \, i\, c\,\left (\dfrac{\partial J_{z}}{\partial x}\, dx + \dfrac{\partial J_{z}}{\partial y}\, dy +\dfrac{\partial J_{z}}{\partial z}\, dz + \dfrac{\partial J_{z}}{\partial t}\, dt\right )\wedge dx \, \wedge dy \, \wedge dt}\\ &\mathbf{\ \ \ \ \ \ \ +}\\ &\mathbf{\ \ \ \ \dfrac{c}{i}\,\left (\dfrac{\partial \rho}{\partial x}\, dx + \dfrac{\partial \rho}{\partial y}\, dy +\dfrac{\partial \rho}{\partial z}\, dz + \dfrac{\partial \rho}{\partial t}\, dt\right )\wedge dx \, \wedge dy \, \wedge dz}\\ &\mathbf{=\, i\, c \,\left (\dfrac{\partial J_{x}}{\partial x}+\dfrac{\partial J_{y}}{\partial y}+\dfrac{\partial J_{z}}{\partial z} + \frac{\partial \rho}{\partial t} \right )\, dx\, \wedge dy \, \wedge \, dz \, \wedge dt}\ \ (4.8) \end{align}\] On retrouve \(\mathbf{dJ = 0}\) car \(\mathbf{div\, \left (\vec {J} \, \right ) + \dfrac{\partial \rho}{\partial t}\, = 0}\).
Retour vers le haut de page
Considérons une 2-forme \(\mathbf{F}\) construite sur un espace de dimension 3 dont le triplet de coordonnées est \(\mathbf{(x,y,z)}\) :
\[
\mathbf{F = p(x,y,z)\, dx\, \wedge\,dy + q(x,y,z)\, dy\, \wedge \, dz + r(x,y,z)\, dz\, \wedge \,dx}\ \ (5.1)
\]
Pour garantir l'existence d'une primitive \(\mathbf{A}\) il faut :
\[
\mathbf{\dfrac{\partial p(x,y,z)}{\partial z} + \dfrac{\partial q(x,y,z)}{\partial x} + \dfrac{\partial r(x,y,z)}{\partial y} = 0}\ \ (5.2)
\]
La proposition de Poincaré est :
\[
\begin{align}
\mathbf{A}
&\mathbf{=\, \int _{0}^{1}\, \lambda \, \underbrace{p(\lambda \, x,\lambda \,y,\lambda \,z)}_{{\Large P(x,y,z,\lambda)}}\, d\lambda\, \left(x\, dy - y\, dx\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, \underbrace{q(\lambda \, x,\lambda \,y,\lambda \,z)}_{{\Large Q(x,y,z,\lambda)}}\, d\lambda\, \left(y\, dz - z\, dy\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, \underbrace{r(\lambda \, x,\lambda \,y,\lambda \,z)}_{{\Large R(x,y,z,\lambda)}}\, d\lambda\, \left(y\, dz - z\, dy\, \right )}\ \ (5.3)
\end{align}
\]
Dans l'élément différentiel (idem pour les autres) \(\mathbf{dx\, \wedge \, dy}\),parcouru de gauche à droite :
-On omet \(\mathbf{dx}\) que l'on remplace par \(\mathbf{x}\) et on conserve \(\mathbf{dy}\);
-On omet \(\mathbf{dy}\) que l'on remplace par \(\mathbf{-y}\) et on conserve \(\mathbf{dx}\);
-Le signe est fonction de la position dans l'élément différentiel \(\mathbf{(+-+-+-etc)}\);
-On additionne pour obtenir \(\mathbf{\left(x\, dy - y\, dx\, \right )}\).
Vérifions le bien-fondé de cette méthode :
\[
\begin{align}
\mathbf{dA}
&\mathbf{=\, \int _{0}^{1}\,\lambda \, \left (\dfrac{\partial P(x,y,z,\lambda)}{\partial x}\, dx + \dfrac{\partial P(x,y,z,\lambda)}{\partial y}\, dy + \dfrac{\partial P(x,y,z,\lambda)}{\partial z}\, dz \right )\, d\lambda \, (x\,dy - y\, dx)}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{2\, \int _{0}^{1}\, \lambda \, P(x,y,z,\lambda)\, d\, \lambda \ \ dx\, \wedge \, dy}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\, \int _{0}^{1}\,\lambda \, \left (\dfrac{\partial Q(x,y,z,\lambda)}{\partial x}\, dx + \dfrac{\partial Q(x,y,z,\lambda)}{\partial y}\, dy + \dfrac{\partial Q(x,y,z,\lambda)}{\partial z}\, dz \right )\, d\lambda \, (y\,dz - z\, dy)}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{2\, \int _{0}^{1}\, \lambda \, Q(x,y,z,\lambda)\, d\, \lambda \ \ dy\, \wedge \, dz}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\, \int _{0}^{1}\,\lambda \, \left (\dfrac{\partial R(x,y,z,\lambda)}{\partial x}\, dx + \dfrac{\partial R(x,y,z,\lambda)}{\partial y}\, dy + \dfrac{\partial R(x,y,z,\lambda)}{\partial z}\, dz \right )\, d\lambda \, (z\,dx - x\, dz)}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{2\, \int _{0}^{1}\, \lambda \, R(x,y,z,\lambda)\, d\, \lambda \ \ dz\, \wedge \, dx}\ \ (5.4)
\end{align}
\]
Dans l'expression (5.4) calculons le coefficient de \(\mathbf{dx\, \wedge \, dy}\) :
\[
\mathbf{\int _{0}^{1}\, \lambda \, \left [x\, \dfrac{\partial P(x,y,z,\lambda)}{\partial x} + y\, \dfrac{\partial P(x,y,z,\lambda)}{\partial y}
-z\, \dfrac{\partial Q(x,y,z,\lambda)}{\partial x} - z\, \dfrac{\partial R(x,y,z,\lambda)}{\partial y}\right ]\, d\lambda
+ 2\, \int _{0}^{1}\, \lambda \, P(x,y,z,\lambda)\, d \lambda }\ \ (5.5)
\]
Soit :
\[
\mathbf{\int _{0}^{1}\, {\lambda}^{2} \, \left [x\, \dfrac{\partial p(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial x} + y\, \dfrac{\partial p(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial y}
-z\, \dfrac{\partial q(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial x} - z\, \dfrac{\partial r(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial y}\right ]\, d\lambda
+ 2\, \int _{0}^{1}\, \lambda \, p(\lambda \,x,\lambda \,y,\lambda \,z)\, d \lambda }\ \ (5.6)
\]
En utilisant la condition (5.2) l'expression (5.6) ci-dessus devient :
\[
\mathbf{\int _{0}^{1}\, {\lambda}^{2} \, \left [x\, \dfrac{\partial p(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial x} + y\, \dfrac{\partial p(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial y}
+z\, \dfrac{\partial p(\lambda \,x,\lambda \,y,\lambda \,z)}{\partial z} \right ]\, d\lambda
+ 2\, \int _{0}^{1}\, \lambda \, p(\lambda \,x,\lambda \,y,\lambda \,z)\, d \lambda }\ \ (5.7)
\]
Intégrons le premier terme de (5.7) par parties :
\[
\mathbf{\left [{\lambda}^{2}\,p(\lambda \,x,\lambda \,y,\lambda \,z) \right ]_{0}^{1} - 2\, \int _{0}^{1}\, \lambda \, \require {cancel} \cancel {p(\lambda \,x,\lambda \,y,\lambda \,z)}\, d \lambda
+ 2\, \int _{0}^{1}\, \lambda \, \require {cancel} \cancel {p(\lambda \,x,\lambda \,y,\lambda \,z)}\, d \lambda }
\]
Finalement on obtient \(\mathbf{p(x,y,z)}\). En étudiant les trois termes de \(\mathbf{dA}\) on rtrouve bien l'expression de \(\mathbf{F}\).
Rappelons les expressions des formes \(\mathbf{A}\) et \(\mathbf{F}\) :
\[
\mathbf{A = A_{x}\, dx + A_{y}\, dy + A_{z}\, dz - V\, dt}\ \ (6.1)
\]
\[
\mathbf{F = E_{x}\, dx\, \wedge \, dt + E_{y}\, dy \, \wedge \, dt + E_{z}\, dz\, \wedge \, dt
+ B_{z}\, dx\, \wedge \, dy + B_{x}\, dy\, \wedge \, dz + B_{y}\, dz\, \wedge \, dx}\ \ (6.2)
\]
On applique la technique de Poincaré en intégrant suivant un rayon d'espace seulement.
Il s'ensuit une disymétrie entre la partie magnétique (2-forme à intégrer) et la partie électrique (1-forme à intégrer car dt reste fixe).
On obtient :
\[
\begin{align}
\mathbf{A}
&\mathbf{=\, \int _{0}^{1}\, \lambda \, B_{z}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(x\, dy - y\, dx\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, B_{x}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(y\, dz - z\, dy\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, B_{y}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(z\, dx - x\, dz\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, E_{x}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(x\, dt \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, E_{y}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(y\, dt \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, E_{z}(\lambda \, x,\lambda \,y,\lambda \,z,t)\, d\lambda\, \left(z\, dt \right )}\ \ (6.3)
\end{align}
\]
Le terme en \(\mathbf{\lambda}\) devant la composante du champ dans la relation (6.3) est du type \(\mathbf{\lambda ^{r-1}}\) ou
\(\mathbf{r}\) est le grade de la forme à intégrer (2 pour la partie magnétique et 1 pour la partie électrique); Ce qui explique une
différence de traitement dans (6.3).En identifiant avec (6.1) on retrouve les relations (2.22) et (2.23).
Champ magnétique constant
\[ \mathbf{\vec {B} = B_{x}\, \vec {\imath} + B_{y}\, \vec {\jmath} + B_{z}\, \vec {k}} \] \[ \mathbf{\vec {A} = \dfrac{\vec {B} \times \vec {r}}{2}} \] \[ \mathbf{V = 0} \]
Champ électrique constant
\[ \mathbf{\vec {E} = E_{x}\, \vec {\imath} + E_{y}\, \vec {\jmath} + E_{z}\, \vec {k}} \] \[ \mathbf{\vec {A} = \vec {0} } \] \[ \mathbf{V = -\vec {E}\, . \vec {r}} \]
Retour vers le haut de page
On applique la technique de Poincaré en intégrant suivant un rayon d'espace-temps comme prescrit par la relativité restreinte.
On obtient :
\[
\begin{align}
\mathbf{A}
&\mathbf{=\, \int _{0}^{1}\, \lambda \, B_{z}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(x\, dy - y\, dx\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, B_{x}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(y\, dz - z\, dy\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, B_{y}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(z\, dx - x\, dz\, \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda _, E_{x}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(x\, dt- t\, dx \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, E_{y}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(y\, dt - t\, dy \right )}\\
&\mathbf{\ \ \ \ \ \ \ +}\\
&\mathbf{\ \ \ \ \, \int _{0}^{1}\, \lambda \, E_{z}(\lambda \, x,\lambda \,y,\lambda \,z,\lambda \, t)\, d\lambda\, \left(z\, dt - t\, dz\right )}\ \ (7.1)
\end{align}
\]
En identifiant ave (6.1) on retrouve les relations (3.20) et (3.21)
Champ magnétique constant
\[ \mathbf{\vec {B} = B_{x}\, \vec {\imath} + B_{y}\, \vec {\jmath} + B_{z}\, \vec {k}} \] \[ \mathbf{\vec {A} = \dfrac{\vec {B} \times \vec {r}}{2}} \] \[ \mathbf{V = 0} \]
Champ électrique constant
\[ \mathbf{\vec {E} = E_{x}\, \vec {\imath} + E_{y}\, \vec {\jmath} + E_{z}\, \vec {k}} \] \[ \mathbf{\vec {A} = \dfrac{-\vec {E} \, t}{2}} \] \[ \mathbf{V = -\dfrac{\vec {E}\, . \vec {r}}{2}} \]
Remarques
-Les jauges de Poincaré permettent un petit miracle qui mériterait une explication.
-La méthode de primitivation de Poincaré est une parmi d'autres, il faudrait pouvoir en donner une caractéristique.
-L'intégration suivant l'espace ou suivant l'espace-temps donne deux résultats satisfaisants. A priori seule l'intégration
suivant un rayon d'espace-temps aurait du donner un bon résultat; Il faudrait éclaircir la situation.