verificação da equação da onda por transformação de Galileu e Lorentz

Prévia do material em texto

Mostre que a equação 
𝛻2𝛹 −
1
𝑐2
𝜕2
𝜕𝑡2
𝛹 = 0 , 
É invariante sob transformação de Lorentz, mas não sob uma transformação de Galileu. 
Variância sob transformação de Galileu 
Da transformação de Galileu, temos 
𝑥′ = 𝑥 − 𝑣𝑥𝑡 
𝑦′ = 𝑦 − 𝑣𝑦𝑡 
𝑧′ = 𝑧 − 𝑣𝑧𝑡 
𝑡′ = 𝑡 
Da equação de onda de onda 
𝛻2𝛹 −
1
𝑐2
𝜕2
𝜕𝑡2
𝛹 = 0 
[𝛻2 −
1
𝑐2
𝜕2
2𝜕𝑡2
]𝛹 = 0 
Do operador de d’Alembert 
⎕2 = 𝛻2 −
1
𝑐2
𝜕2
2𝜕𝑡2
 
⎕2 = (
𝜕2
𝜕𝑥2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
) −
1
𝑐2
𝜕2
𝜕𝑡2
 
Usando a regra da cadeia para x 
𝜕
𝜕𝑥
=
𝜕𝑥′
𝜕𝑥
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑥
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑥
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑥
𝜕
𝜕𝑡′
 
Substituindo 
𝜕
𝜕𝑥
=
𝜕
𝜕𝑥
(𝑥 − 𝑣𝑥𝑡)
𝜕
𝜕𝑥′
+
𝜕
𝜕𝑥
(𝑦 − 𝑣𝑦𝑡)
𝜕
𝜕𝑦′
+
𝜕
𝜕𝑥
(𝑧 − 𝑣𝑧𝑡)
𝜕
𝜕𝑧′
+
𝜕𝑡
𝜕𝑥
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑥
= (
𝜕𝑥
𝜕𝑥
−
𝜕(𝑣𝑥𝑡)
𝜕𝑥
)
𝜕
𝜕𝑥′
+ 0 + 0 + 0 
𝜕
𝜕𝑥
= (1 − 0)
𝜕
𝜕𝑥′
 
𝜕
𝜕𝑥
=
𝜕
𝜕𝑥′
 
Para y 
𝜕
𝜕𝑦
=
𝜕𝑥′
𝜕𝑦
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑦
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑦
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑦
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑦
=
𝜕
𝜕𝑦
(𝑥 − 𝑣𝑥𝑡)
𝜕
𝜕𝑥′
+
𝜕
𝜕𝑦
(𝑦 − 𝑣𝑦𝑡)
𝜕
𝜕𝑦′
+
𝜕
𝜕𝑦
(𝑧 − 𝑣𝑧𝑡)
𝜕
𝜕𝑧′
+
𝜕𝑡
𝜕𝑦
𝜕
𝜕𝑡′
 
 
𝜕
𝜕𝑦
= 0 + (
𝜕𝑦
𝜕𝑦
−
𝜕(𝑣𝑦𝑡)
𝜕𝑦
)
𝜕
𝜕𝑦′
+ 0 + 0 
𝜕
𝜕𝑦
= (1 − 0) 
𝜕
𝜕𝑦′
 
𝜕
𝜕𝑦
=
𝜕
𝜕𝑦′
 
Agora para z 
𝜕
𝜕𝑧
=
𝜕𝑥′
𝜕𝑧
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑧
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑧
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑧
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑧
=
𝜕
𝜕𝑧
(𝑥 − 𝑣𝑥𝑡)
𝜕
𝜕𝑥′
+
𝜕
𝜕𝑧
(𝑦 − 𝑣𝑦𝑡)
𝜕
𝜕𝑦′
+
𝜕
𝜕𝑧
(𝑧 − 𝑣𝑧𝑡)
𝜕
𝜕𝑧′
+
𝜕𝑡
𝜕𝑧
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑧
= 0 + 0 + (
𝜕𝑧
𝜕𝑧
−
𝜕(𝑣𝑧𝑡)
𝜕𝑧
)
𝜕
𝜕𝑧′
+ 0 
𝜕
𝜕𝑧
= (1 − 0)
𝜕
𝜕𝑧′
 
𝜕
𝜕𝑧
=
𝜕
𝜕𝑧′
 
Também para t 
𝜕
𝜕𝑡
=
𝜕𝑥′
𝜕𝑡
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑡
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑡
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑡
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
=
𝜕
𝜕𝑡
(𝑥 − 𝑣𝑥𝑡)
𝜕
𝜕𝑥′
+
𝜕
𝜕𝑡
(𝑦 − 𝑣𝑦𝑡)
𝜕
𝜕𝑦′
+
𝜕
𝜕𝑡
(𝑧 − 𝑣𝑧𝑡)
𝜕
𝜕𝑧′
+
𝜕𝑡
𝜕𝑡
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= (
𝜕𝑥
𝜕𝑡
−
𝜕(𝑣𝑥𝑡)
𝜕𝑡
)
𝜕
𝜕𝑥′
+ (
𝜕𝑦
𝜕𝑡
−
𝜕(𝑣𝑦𝑡)
𝜕𝑡
)
𝜕
𝜕𝑦′
+ (
𝜕𝑧
𝜕𝑡
−
𝜕(𝑣𝑧𝑡)
𝜕𝑡
)
𝜕
𝜕𝑧′
+
𝜕𝑡
𝜕𝑡
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= −𝑣𝑥
𝜕𝑡
𝜕𝑡
𝜕
𝜕𝑥′
− 𝑣𝑦
𝜕𝑡
𝜕𝑡
𝜕
𝜕𝑦′
− 𝑣𝑧
𝜕𝑡
𝜕𝑡
𝜕
𝜕𝑦′
+
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= −𝑣𝑥
𝜕
𝜕𝑥′
− 𝑣𝑦
𝜕
𝜕𝑦′
− 𝑣𝑧
𝜕
𝜕𝑧′
+
𝜕
𝜕𝑡′
 
Para as coordenadas espaciais 
𝜕2
𝜕𝑥2
=
𝜕2
𝜕𝑥′2
 
𝜕2
𝜕𝑦2
=
𝜕2
𝜕𝑦′2
 
𝜕2
𝜕𝑧2
=
𝜕2
𝜕𝑧′2
 
Para t, temos 
𝜕2
𝜕𝑡2
=
𝜕
𝜕𝑡
(
𝜕
𝜕𝑡
) 
Lembrando que 
𝜕
𝜕𝑡
= −𝑣𝑥
𝜕
𝜕𝑥′
− 𝑣𝑦
𝜕
𝜕𝑦′
− 𝑣𝑧
𝜕
𝜕𝑧′
+
𝜕
𝜕𝑡′
 
𝜕2
𝜕𝑡2
= (−𝑣𝑥
𝜕
𝜕𝑥′
− 𝑣𝑦
𝜕
𝜕𝑦′
− 𝑣𝑧
𝜕
𝜕𝑧′
+
𝜕
𝜕𝑡′
) (−𝑣𝑥
𝜕
𝜕𝑥′
− 𝑣𝑦
𝜕
𝜕𝑦′
− 𝑣𝑧
𝜕
𝜕𝑧′
+
𝜕
𝜕𝑡′
) 
𝜕2
𝜕𝑡2
= 𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
− 𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝑣𝑦𝑣𝑥
𝜕2
𝜕𝑦′𝜕𝑥′
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
− 𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
+ 𝑣𝑧𝑣𝑥
𝜕2
𝜕𝑧′𝜕𝑥′
+ 𝑣𝑧𝑣𝑦
𝜕2
𝜕𝑧′𝜕𝑦′
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
− 𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
− 𝑣𝑥
𝜕2
𝜕𝑡′𝜕𝑥′
− 𝑣𝑦
𝜕2
𝜕𝑡′𝜕𝑦′
− 𝑣𝑧
𝜕2
𝜕𝑡′𝜕𝑧′
+
𝜕2
𝜕𝑡′2
 
𝜕2
𝜕𝑡2
= (𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
) + 2𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 2𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
+ 2𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
− 2𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
− 2𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
− 2𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
+
𝜕2
𝜕𝑡′2
 
𝜕2
𝜕𝑡2
= (𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
) + 2(𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
+ 𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
)
− 2(𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
+ 𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
) +
𝜕2
𝜕𝑡′2
 
Substituindo as derivadas espaciais, temos 
⎕2 = (
𝜕2
𝜕𝑥2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
) −
1
𝑐2
𝜕2
𝜕𝑡2
 
⎕2 = (
𝜕2
𝜕𝑥′2
+
𝜕2
𝜕𝑦′2
+
𝜕2
𝜕𝑧′2
)
−
1
𝑐2
{(𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
)
+ 2(𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
+ 𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
)
− 2(𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
+ 𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
) +
𝜕2
𝜕𝑡′2
} 
Arrumando, temos 
⎕2 = (
𝜕2
𝜕𝑥′2
+
𝜕2
𝜕𝑦′2
+
𝜕2
𝜕𝑧′2
−
1
𝑐2
𝜕2
𝜕𝑡′2
)
−
1
𝑐2
{(𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
)
+ 2(𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
+ 𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
)
− 2(𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
+ 𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
)} 
Logo 
⎕2 = ⎕′2 −
1
𝑐2
{(𝑣𝑥
2
𝜕2
𝜕𝑥′2
+ 𝑣𝑦
2
𝜕2
𝜕𝑦′2
+ 𝑣𝑧
2
𝜕2
𝜕𝑧′2
)
+ 2(𝑣𝑥𝑣𝑦
𝜕2
𝜕𝑥′𝜕𝑦′
+ 𝑣𝑥𝑣𝑧
𝜕2
𝜕𝑥′𝜕𝑧′
+ 𝑣𝑦𝑣𝑧
𝜕2
𝜕𝑦′𝜕𝑧′
)
− 2(𝑣𝑥
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝑣𝑦
𝜕2
𝜕𝑦′𝜕𝑡′
+ 𝑣𝑧
𝜕2
𝜕𝑧′𝜕𝑡′
)} 
Assim 
⎕2𝛹 ≠ ⎕′2𝛹 
Percebesse que quando troca de referencial, a equação de onda não é invariante por 
transformação de Galileu 
 
Invariância sob Transformação de Lorentz 
Da equação de onda 
𝛻2𝛹 −
1
𝑐2
𝜕2
𝜕𝑡2
𝛹 = 0 
Do operador de d’Alembert 
⎕2 = (
𝜕2
𝜕𝑥2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
) −
1
𝑐2
𝜕2
𝜕𝑡2
 
As equações de transformação de Lorentz 
𝑥′ = 𝛾(𝑥 − 𝑣𝑡) 
𝑦′ = 𝑦 
𝑧′ = 𝑧 
𝑡′ = 𝛾 (𝑡 −
𝑣𝑥
𝑐2
) 
Lembrando que 𝛾 =
1
√1−
𝑣2
𝑐2
, 
Usando a regra da cadeia para x 
𝜕
𝜕𝑥
=
𝜕𝑥′
𝜕𝑥
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑥
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑥
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑥
𝜕
𝜕𝑡′
 
Substituindo 
𝜕
𝜕𝑥
= {
𝜕
𝜕𝑥
[𝛾(𝑥 − 𝑣𝑡)]}
𝜕
𝜕𝑥′
+ 0 + 0 + {
𝜕
𝜕𝑥
[𝛾 (𝑡 −
𝑣𝑥
𝑐2
)]}
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑥
= 𝛾 [
𝜕𝑥
𝜕𝑥
−
𝜕(𝑣𝑡)
𝜕𝑥
]
𝜕
𝜕𝑥′
+ 𝛾 [
𝜕𝑡
𝜕𝑥
−
𝜕
𝜕𝑥
(
𝑣𝑥
𝑐2
)]
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑥
= 𝛾
𝜕
𝜕𝑥′
− 𝛾
𝑣𝑥
𝑐2
𝜕
𝜕𝑡′
 
Ainda 
𝜕2
𝜕𝑥2
=
𝜕
𝜕𝑥
(
𝜕
𝜕𝑥
) 
𝜕2
𝜕𝑥2
= (𝛾
𝜕
𝜕𝑥′
− 𝛾
𝑣𝑥
𝑐2
𝜕
𝜕𝑡′
) (𝛾
𝜕
𝜕𝑥′
− 𝛾
𝑣
𝑐2
𝜕
𝜕𝑡′
) 
𝜕2
𝜕𝑥2
= 𝛾2
𝜕2
𝜕𝑥′2
− 𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑥′𝜕𝑡′
− 𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑡′𝜕𝑥′
+ 𝛾2
𝑣2
𝑐4
𝜕2
𝜕𝑡′2
 
𝜕2
𝜕𝑥2
= 𝛾2
𝜕2
𝜕𝑥′2
− 2𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝛾2
𝑣2
𝑐4
𝜕2
𝜕𝑡′2
 
 
Para y 
𝜕
𝜕𝑦
=
𝜕𝑥′
𝜕𝑦
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑦
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑦
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑦
𝜕
𝜕𝑡′
 
Substituindo 
𝜕
𝜕𝑦
= {
𝜕
𝜕𝑦
[𝛾(𝑥 − 𝑣𝑡)]}
𝜕
𝜕𝑥′
+
𝜕𝑦
𝜕𝑦
𝜕
𝜕𝑦′
+
𝜕𝑧
𝜕𝑦
𝜕
𝜕𝑧′
+ {
𝜕
𝜕𝑦
[𝛾 (𝑡 −
𝑣𝑥
𝑐2
)]}
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑦
=
𝜕
𝜕𝑦′
 
logo 
𝜕2
𝜕𝑦2
=
𝜕
𝜕𝑦′
(
𝜕
𝜕𝑦′
) 
𝜕2
𝜕𝑦2
=
𝜕2
𝜕𝑦′2
 
Para z 
𝜕
𝜕𝑧
=
𝜕𝑥′
𝜕𝑧
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑧
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑧
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑧
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑧
= {
𝜕
𝜕𝑧
[𝛾(𝑥 − 𝑣𝑡)]}
𝜕
𝜕𝑥′
+
𝜕𝑦
𝜕𝑧
𝜕
𝜕𝑦′
+
𝜕𝑧
𝜕𝑧
𝜕
𝜕𝑧′
+ {
𝜕
𝜕𝑧
[𝛾 (𝑡 −
𝑣𝑥
𝑐2
)]}
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑧
=
𝜕
𝜕𝑧′
 
Também temos 
𝜕2
𝜕𝑧2
=
𝜕
𝜕𝑧′
(
𝜕
𝜕𝑧′
) 
 
𝜕2
𝜕𝑧2
=
𝜕2
𝜕𝑧′2
 
Também para t 
𝜕
𝜕𝑡
=
𝜕𝑥′
𝜕𝑡
𝜕
𝜕𝑥′
+
𝜕𝑦′
𝜕𝑡
𝜕
𝜕𝑦′
+
𝜕𝑧′
𝜕𝑡
𝜕
𝜕𝑧′
+
𝜕𝑡′
𝜕𝑡
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= {
𝜕
𝜕𝑡
[𝛾(𝑥 − 𝑣𝑡)]}
𝜕
𝜕𝑥′
+
𝜕𝑦
𝜕𝑡
𝜕
𝜕𝑦′
+
𝜕𝑧
𝜕𝑡
𝜕
𝜕𝑧′
+ {
𝜕
𝜕𝑡
[𝛾 (𝑡 −
𝑣𝑥
𝑐2
)]}
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= {
𝜕
𝜕𝑡
[𝛾(𝑥 − 𝑣𝑡)]}
𝜕
𝜕𝑥′
+ 0 + 0 + {
𝜕
𝜕𝑡
[𝛾 (𝑡 −
𝑣𝑥
𝑐2
)]}
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= 𝛾 [
𝜕𝑥
𝜕𝑡
−
𝜕(𝑣𝑡)
𝜕𝑡
]
𝜕
𝜕𝑥′
+ 𝛾 [
𝜕𝑡
𝜕𝑡
−
𝜕
𝜕𝑡
(
𝑣𝑥
𝑐2
)]
𝜕
𝜕𝑡′
 
𝜕
𝜕𝑡
= −𝛾𝑣
𝜕
𝜕𝑥′
+ 𝛾
𝜕
𝜕𝑡′
 
Ainda 
𝜕
𝜕𝑡2
=
𝜕
𝜕𝑡
(
𝜕
𝜕𝑡
) 
𝜕
𝜕𝑡2
= (−𝛾𝑣
𝜕
𝜕𝑥′
+ 𝛾
𝜕
𝜕𝑡′
) (−𝛾𝑣
𝜕
𝜕𝑥′
+ 𝛾
𝜕
𝜕𝑡′
) 
𝜕
𝜕𝑡2
= 𝜸𝟐𝒗𝟐
𝜕2
𝜕𝑥′2
− 𝛾2𝑣
𝜕2
𝜕𝑥′𝜕𝑡′
− 𝛾2𝑣
𝜕2
𝜕𝑡′𝜕𝑥′
+ 𝛾2
𝜕2
𝜕𝑡′2
 
𝜕
𝜕𝑡2
= 𝜸𝟐𝒗𝟐
𝜕2
𝜕𝑥′2
− 2𝛾2𝑣
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝛾2
𝜕2
𝜕𝑡′2
 
Substituindo as derivadas espaciais na equação de onda 
⎕2 = (
𝜕2
𝜕𝑥2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
) −
1
𝑐2
𝜕2
𝜕𝑡2
 
⎕2 = [(𝛾2
𝜕2
𝜕𝑥′2
− 2𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝛾2
𝑣2
𝑐4
𝜕2
𝜕𝑡′2
) +
𝜕𝟐
𝜕𝑦′𝟐
+
𝜕𝟐
𝜕𝑧′𝟐
]
−
1
𝒄𝟐
(𝜸𝟐𝒗𝟐
𝜕2
𝜕𝑥′2
− 2𝛾2𝑣
𝜕2
𝜕𝑥′𝜕𝑡′
+ 𝛾2
𝜕2
𝜕𝑡′2
) 
⎕2 = (𝛾2
𝜕2
𝜕𝑥′2−
𝛾𝟐𝒗𝟐
𝑐2
𝜕2
𝜕𝑥′2
) + (−2𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑥′𝜕𝑡′
+ 2𝛾2
𝑣
𝑐2
𝜕2
𝜕𝑥′𝜕𝑡′
) +
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
+ (𝛾2
𝑣2
𝑐4
𝜕2
𝜕𝑡′2
−
𝛾2
𝑐2
𝜕2
𝜕𝑡′2
) 
⎕2 = 𝛾2
𝜕2
𝜕𝑥′2
(1 −
𝑣2
𝑐2
) +
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
+
𝛾2
𝑐2
𝜕2
𝜕𝑡′2
(
𝑣2
𝑐2
− 1) 
⎕2 = 𝛾2
𝜕2
𝜕𝑥′2
(1 −
𝑣2
𝑐2
) +
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
−
𝛾2
𝑐2
𝜕2
𝜕𝑡′2
(1 −
𝑣2
𝑐2
) 
Lembrando que 𝛾 =
1
√1−
𝑣2
𝑐2
 
⎕2 =
(
 
1
√1 −
𝑣2
𝑐2)
 
2
𝜕2
𝜕𝑥′2
(1 −
𝑣2
𝑐2
) +
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
−
1
𝑐2
(
 
1
√1 −
𝑣2
𝑐2)
 
2
𝜕2
𝜕𝑡′2
(1 −
𝑣2
𝑐2
) 
⎕2 = (
1
1 −
𝑣2
𝑐2
)
𝜕2
𝜕𝑥′2
(1 −
𝑣2
𝑐2
) +
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
−
1
𝑐2
(
1
1 −
𝑣2
𝑐2
)
𝜕2
𝜕𝑡′2
(1 −
𝑣2
𝑐2
) 
⎕2 = (
1 −
𝑣2
𝑐2
1 −
𝑣2
𝑐2
)
𝜕2
𝜕𝑥′2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
−
1
𝑐2
(
1 −
𝑣2
𝑐2
1 −
𝑣2
𝑐2
)
𝜕2
𝜕𝑡′2
 
 
⎕2 =
𝜕2
𝜕𝑥′2
+
𝜕2
𝜕𝑦2
+
𝜕2
𝜕𝑧2
−
1
𝑐2
𝜕2
𝜕𝑡′2
 
⎕2 = ∇′
2
−
1
𝑐2
𝜕2
𝜕𝑡′2
 
⎕2𝛹 = ⎕′2𝛹 
Quando muda de referencial a equação de onda é a mesma sob Transformação de Lorentz.