[p. 12 modifica ]
Esiste per l’energia di un campo elettromagnetico un’equazione analoga a quelle che vanno sotto il nome di principio di Hamilton e principio della minima azione nella dinamica ordinaria: si può dimostrare cioè la relazione
δ ∫ t 0 t 1 W e m d t = 0 , {\displaystyle \delta \textstyle \int _{t_{0}}^{t_{1}}\mathrm {W_{em}} \,dt=0,}
quando si suppongano nulle le variazioni ai limiti.
Per vedere questo si scriva:
[1]
x 1 = A ε ∂ X ∂ t − ∂ M ∂ z + ∂ N ∂ y {\displaystyle x_{1}=\mathrm {A} \varepsilon {\frac {\partial \mathrm {X} }{\partial t}}-{\frac {\partial \mathrm {M} }{\partial z}}+{\frac {\partial \mathrm {N} }{\partial y}}}
l 1 = A μ ∂ L ∂ t − ∂ Z ∂ y + ∂ Y ∂ z {\displaystyle l_{1}=\mathrm {A} \mu {\dfrac {\partial \mathrm {L} }{\partial t}}-{\dfrac {\partial \mathrm {Z} }{\partial y}}+{\dfrac {\partial \mathrm {Y} }{\partial z}}}
x 2 = A ε ∂ Y ∂ t − ∂ N ∂ x + ∂ L ∂ z {\displaystyle x_{2}=\mathrm {A} \varepsilon {\frac {\partial \mathrm {Y} }{\partial t}}-{\frac {\partial \mathrm {N} }{\partial x}}+{\frac {\partial \mathrm {L} }{\partial z}}}
l 2 = A μ ∂ M ∂ t − ∂ X ∂ y + ∂ Z ∂ x {\displaystyle l_{2}=\mathrm {A} \mu {\dfrac {\partial \mathrm {M} }{\partial t}}-{\dfrac {\partial \mathrm {X} }{\partial y}}+{\dfrac {\partial \mathrm {Z} }{\partial x}}}
x 3 = A ε ∂ Z ∂ t − ∂ L ∂ y + ∂ M ∂ x {\displaystyle x_{3}=\mathrm {A} \varepsilon {\frac {\partial \mathrm {Z} }{\partial t}}-{\frac {\partial \mathrm {L} }{\partial y}}+{\frac {\partial \mathrm {M} }{\partial x}}}
l 3 = A μ ∂ N ∂ t − ∂ Y ∂ x + ∂ X ∂ x {\displaystyle l_{3}=\mathrm {A} \mu {\dfrac {\partial \mathrm {N} }{\partial t}}-{\dfrac {\partial \mathrm {Y} }{\partial x}}+{\dfrac {\partial \mathrm {X} }{\partial x}}}
[p. 13 modifica ] sarà, per le equazioni di Hertz:
x 1 = x 2 = x 3 = l 1 = l 2 = l 3 = 0 ; {\displaystyle x_{1}=x_{2}=x_{3}=l_{1}=l_{2}=l_{3}=0;}
di più si introducano sei funzioni X . Y . Z . L . M . N {\displaystyle X.Y.Z.L.M.N} delle coordinate e del tempo, che ci riserviamo di determinare.
Facciamo le variazioni di queste funzioni: moltiplichiamo la prima delle [1] per δ X {\displaystyle \delta \,X} , la seconda per δ Y {\displaystyle \delta \,Y} ...... la sesta per δ N {\displaystyle \delta \,N} ; sommiamo membro a membro, moltiplichiamo per d t {\displaystyle dt} e integriamo fra t 0 {\displaystyle t_{0}} e t 1 {\displaystyle t_{1}} otterremo:
∫ t 0 t 1 d t ∭ d v [ A ε ∂ X ∂ t δ X + A ε ∂ Y ∂ t δ Y + ⋯ + A μ ∂ N ∂ t δ N {\displaystyle \int _{t_{0}}^{t_{1}}\,dt\iiint \,dv\left[\mathrm {A} \,\varepsilon {\frac {\partial \mathrm {X} }{\partial t}}\delta \,X+\mathrm {A} \,\varepsilon {\frac {\partial \mathrm {Y} }{\partial t}}\delta \,Y+\cdots +\mathrm {A} \,\mu {\frac {\partial \mathrm {N} }{\partial t}}\delta \,N\right.}
− ∂ M ∂ z δ X − ∂ N ∂ x δ Y − ⋅ ⋅ ⋅ ⋅ ⋅ − ∂ Y ∂ x δ N {\displaystyle -{\frac {\partial \mathrm {M} }{\partial z}}\delta \,X-{\frac {\partial \mathrm {N} }{\partial x}}\delta \,Y-\quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad -{\frac {\partial \mathrm {Y} }{\partial x}}\delta \,N}
+ ∂ N ∂ y δ X + ∂ L ∂ z δ Y + ⋅ ⋅ ⋅ ⋅ ⋅ + ∂ X ∂ y δ N ] = 0. {\displaystyle \left.+{\frac {\partial \mathrm {N} }{\partial y}}\delta \,X+{\frac {\partial \mathrm {L} }{\partial z}}\delta \,Y+\quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad +{\frac {\partial \mathrm {X} }{\partial y}}\delta \,N\right]=0.}
Ora è evidente che si può scrivere:
A ε ∂ X ∂ t δ X = A ε ∂ ∂ t ( X δ X ) − A ε X ∂ ∂ t ( δ X ) {\displaystyle \mathrm {A} \,\varepsilon {\frac {\partial \mathrm {X} }{\partial t}}\delta \,X=\mathrm {A} \,\varepsilon {\frac {\partial }{\partial t}}\left(\mathrm {X} \,\delta \,X\right)-\mathrm {A} \,\varepsilon \,\mathrm {X} {\frac {\partial }{\partial t}}\left(\delta \,X\right)}
e che 5 eguaglianze analoghe si potrebbero formare considerando i termini in ∂ Y ∂ t ⋯ ∂ N ∂ t {\displaystyle {\frac {\partial \mathrm {Y} }{\partial t}}\cdots {\frac {\partial \mathrm {N} }{\partial t}}} , si ha dunque:
t 0 t 1 [ ∭ d v ( A ε X δ X + A ε Y δ Y + ⋯ + A μ N δ N ) ] {\displaystyle \textstyle _{t_{0}}^{t_{1}}\left[\iiint \,dv(\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \,X+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \,Y+\cdots +\mathrm {A} \,\mu \,\mathrm {N} \,\delta \,N)\right]}
− ∫ t 0 t 1 d t ∭ d v [ A ε X δ ⋅ ∂ X ∂ t + A ε Y δ ⋅ ∂ Y ∂ t + ⋯ + A μ N δ ⋅ ∂ Z ∂ t {\displaystyle -\int _{t_{0}}^{t_{1}}\,dt\iiint dv\left[\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \cdot {\frac {\partial X}{\partial t}}+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \cdot {\frac {\partial Y}{\partial t}}+\cdots +\mathrm {A} \,\mu \,\mathrm {N} \,\delta \cdot {\frac {\partial Z}{\partial t}}\right.}
+ ∂ M ∂ z δ X + ∂ N ∂ x δ Y + ⋅ ⋅ ⋅ ⋅ + ∂ Y ∂ x δ N {\displaystyle +{\frac {\partial \mathrm {M} }{\partial z}}\delta \,X+{\frac {\partial \mathrm {N} }{\partial x}}\delta \,Y+\quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad +{\frac {\partial \mathrm {Y} }{\partial x}}\delta \,N}
− ∂ N ∂ y δ X − ∂ L ∂ z δ Y − ⋅ ⋅ ⋅ ⋅ − ∂ X ∂ y δ N ] = 0. {\displaystyle \left.-{\frac {\partial \mathrm {N} }{\partial y}}\delta \,X-{\frac {\partial \mathrm {L} }{\partial z}}\delta \,Y-\quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad -{\frac {\partial \mathrm {X} }{\partial y}}\delta \,N\right]=0.}
Ora si osservi, in modo simile a quanto si è fatto più su, che:
∂ M ∂ z δ X = − M δ ⋅ ∂ X ∂ z + ∂ ∂ z ( M δ X ) {\displaystyle {\frac {\partial \mathrm {M} }{\partial z}}\delta \,X=-\mathrm {M} \,\delta \cdot {\frac {\partial X}{\partial z}}+{\frac {\partial }{\partial z}}\left(\mathrm {M} \,\delta \,X\right)}
11 equazioni analoghe si potrebbero scrivere, sicchè, sostituendo, si ottiene:
t 0 t 1 [ ∭ d v ( A ε X δ X + A ε Y δ Y + ⋯ + A μ N δ N ) {\displaystyle _{t_{0}}^{t_{1}}\textstyle [\iiint \,dv(\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \,X+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \,Y+\cdots +\mathrm {A} \,\mu \,\mathrm {N} \,\delta \,N)}
− ∫ t 0 t 1 d t ∭ d v [ A ε X δ ⋅ ∂ X ∂ t + A ε Y δ ⋅ ∂ Y ∂ t + ⋯ + A μ N δ ⋅ ∂ Z ∂ t {\displaystyle -\int _{t_{0}}^{t_{1}}\,dt\iiint dv\left[\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \cdot {\frac {\partial X}{\partial t}}+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \cdot {\frac {\partial Y}{\partial t}}+\cdots +\mathrm {A} \,\mu \,\mathrm {N} \,\delta \cdot {\frac {\partial Z}{\partial t}}\right.}
[p. 14 modifica ]
− M δ ⋅ ∂ X ∂ z − N δ ⋅ ∂ Y ∂ x − ⋯ − Y δ ⋅ ∂ N ∂ x {\displaystyle -\mathrm {M} \,\delta \cdot {\frac {\partial X}{\partial z}}-\mathrm {N} \,\delta \cdot {\frac {\partial Y}{\partial x}}-\cdots -\mathrm {Y} \,\delta \cdot {\frac {\partial N}{\partial x}}}
+ ∂ ∂ z ( M δ X ) + ∂ ∂ x ( N δ Y ) + ⋯ + ∂ ∂ x ( Y δ N ) {\displaystyle +{\frac {\partial }{\partial z}}\left(\mathrm {M} \,\delta \,X\right)+{\frac {\partial }{\partial x}}\left(\mathrm {N} \,\delta \,Y\right)+\cdots +{\frac {\partial }{\partial x}}\left(\mathrm {Y} \,\delta \,N\right)}
+ N δ ⋅ ∂ X ∂ y + L δ ⋅ ∂ Y ∂ z + ⋯ + X δ ⋅ ∂ N ∂ y {\displaystyle +\mathrm {N} \,\delta \cdot {\frac {\partial X}{\partial y}}+\mathrm {L} \,\delta \cdot {\frac {\partial Y}{\partial z}}+\cdots +\mathrm {X} \,\delta \cdot {\frac {\partial N}{\partial y}}}
− ∂ ∂ y ( N δ X ) − ∂ ∂ z ( L δ Y ) − ⋯ − ∂ ∂ y ( X δ N ) ] = 0. {\displaystyle \left.-{\frac {\partial }{\partial y}}\left(\mathrm {N} \,\delta \,X\right)-{\frac {\partial }{\partial z}}\left(\mathrm {L} \,\delta \,Y\right)-\cdots -{\frac {\partial }{\partial y}}\left(\mathrm {X} \,\delta \,N\right)\right]=0.}
Se si ammette che X . Y . . . N {\displaystyle X.Y...N} siano nulle sulla superficie che racchiude il campo si possono cancellare i 12 termini dell’integrale che sono derivate parziali rispetto alle coordinate.
Ordinando ciò che resta rispetto ad X . Y . . . N {\displaystyle X.Y...N} si trova:
t 0 t 1 [ ∭ d v ( A ε X δ X + A ε Y δ Y + ⋯ + A μ Z δ Z ) ] {\displaystyle _{t_{0}}^{t_{1}}\textstyle [\iiint \,dv(\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \,X+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \,Y+\cdots +\mathrm {A} \,\mu \,\mathrm {Z} \,\delta \,Z)]}
− ∫ t 0 t 1 d t ∭ d v [ X δ ⋅ ( A ε ∂ X ∂ t − ∂ M ∂ z + ∂ N ∂ y ) {\displaystyle -\int _{t_{0}}^{t_{1}}\,dt\iiint dv\left[\mathrm {X} \,\delta \cdot \left(\mathrm {A} \,\varepsilon {\frac {\partial X}{\partial t}}-{\frac {\partial M}{\partial z}}+{\frac {\partial N}{\partial y}}\right)\right.}
+ Y δ ⋅ ( A ε ∂ Y ∂ t − ∂ N ∂ x + ∂ L ∂ z ) {\displaystyle +\mathrm {Y} \,\delta \cdot \left(\mathrm {A} \,\varepsilon {\frac {\partial Y}{\partial t}}-{\frac {\partial N}{\partial x}}+{\frac {\partial L}{\partial z}}\right)}
(*)
+ ⋯ ⋯ {\displaystyle +\cdots \cdots }
+ N δ ⋅ ( A μ ∂ L ∂ t − ∂ Y ∂ x + ∂ X ∂ z ) ] = 0. {\displaystyle \left.+\mathrm {N} \,\delta \cdot \left(\mathrm {A} \,\mu {\frac {\partial L}{\partial t}}-{\frac {\partial Y}{\partial x}}+{\frac {\partial X}{\partial z}}\right)\right]=0.}
Le X . Y . . . N {\displaystyle X.Y...N} sono ancora in nostro arbitrio, le assoggetteremo alle equazioni seguenti.
A ε ∂ X ∂ t − ∂ M ∂ z + ∂ N ∂ y = − ε 4 π X {\displaystyle \mathrm {A} \,\varepsilon {\frac {\partial X}{\partial t}}-{\frac {\partial M}{\partial z}}+{\frac {\partial N}{\partial y}}=-{\frac {\varepsilon }{4\pi }}\mathrm {X} }
A ε ∂ Y ∂ t − ∂ N ∂ x + ∂ L ∂ z = − ε 4 π X {\displaystyle \mathrm {A} \,\varepsilon {\frac {\partial Y}{\partial t}}-{\frac {\partial N}{\partial x}}+{\frac {\partial L}{\partial z}}=-{\frac {\varepsilon }{4\pi }}\mathrm {X} }
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ {\displaystyle \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot }
A μ ∂ N ∂ t − ∂ Y ∂ x + ∂ X ∂ y = − μ 4 π Z . {\displaystyle \mathrm {A} \,\mu {\frac {\partial N}{\partial t}}-{\frac {\partial Y}{\partial x}}+{\frac {\partial X}{\partial y}}=-{\frac {\mu }{4\pi }}\mathrm {Z} .}
Ciò posto l'equazione (*) diventa:
t 0 t 1 [ ∭ d v ( A ε X δ X + A ε Y δ Y + ⋯ + A μ Z δ Z ) ] {\displaystyle _{t_{0}}^{t_{1}}\textstyle [\iiint \,dv(\mathrm {A} \,\varepsilon \,\mathrm {X} \,\delta \,X+\mathrm {A} \,\varepsilon \,\mathrm {Y} \,\delta \,Y+\cdots +\mathrm {A} \,\mu \,\mathrm {Z} \,\delta \,Z)]}
+ δ ∫ t 0 t 1 d t ∭ d v [ ε 8 π ( X 2 + Y 2 + Z 2 ) + μ 8 π ( L 2 + M 2 + N 2 ) ] = 0 ; {\displaystyle +\delta \int _{t_{0}}^{t_{1}}\,dt\iiint dv\left[{\frac {\varepsilon }{8\pi }}\left(\mathrm {X^{2}+Y^{2}+Z^{2}} \right)+{\frac {\mu }{8\pi }}\left(\mathrm {L^{2}+M^{2}+N^{2}} \right)\right]=0;}
se le variazioni ai limiti sono nulle la parte integrata rispetto al tempo si annulla da sè, quindi bisogna che sia
[p. 15 modifica ]
δ ∫ t 0 t 1 d t ∭ d v [ ε 8 π ( X 2 + Y 2 + Z 2 ) + μ 8 π ( L 2 + M 2 + N 2 ) ] = 0 ; {\displaystyle \delta \int _{t_{0}}^{t_{1}}\,dt\iiint dv\left[{\frac {\varepsilon }{8\pi }}\left(\mathrm {X^{2}+Y^{2}+Z^{2}} \right)+{\frac {\mu }{8\pi }}\left(\mathrm {L^{2}+M^{2}+N^{2}} \right)\right]=0;}
vale a dire
δ ∫ t 0 t 1 W e m d t = 0 , {\displaystyle \delta \,\int _{t_{0}}^{t_{1}}\mathrm {W_{\mathrm {em} }} \,dt=0,}
ciò che si voleva dimostrare.
Questo teorema si deve al prof. Vito Volterra 1 .