Eletromagnetismo II Aula 21 Os potenciais de Lienard-Wiechert - IFSC
Eletromagnetismo II Aula 21 Os potenciais de Lienard-Wiechert - IFSC
Eletromagnetismo II Aula 21 Os potenciais de Lienard-Wiechert - IFSC
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<strong>Eletromagnetismo</strong> <strong>II</strong><br />
<strong>Aula</strong> <strong>21</strong><br />
<strong>Os</strong> <strong>potenciais</strong> <strong>de</strong> <strong>Lienard</strong>-<strong>Wiechert</strong> (continuação)<br />
Na aula passada, obtivemos os <strong>potenciais</strong> <strong>de</strong> <strong>Lienard</strong>-<strong>Wiechert</strong>, que po<strong>de</strong>m ser<br />
escritos como<br />
φ (r, t) = 1<br />
q<br />
4πε 0 |r − r 0 (t R )| − [r − r 0 (t R )] · v(t R)<br />
c<br />
e, analogamente,<br />
on<strong>de</strong><br />
A (r, t) = µ 0<br />
4π<br />
qv (t R )<br />
|r − r 0 (t R )| − [r − r 0 (t R )] · v(t R)<br />
c<br />
t R = t − |r − r 0 (t R )|<br />
.<br />
c<br />
Para simplificar a notação, <strong>de</strong>finimos<br />
R = r − r 0 (t R ) ,<br />
R = |R| ,<br />
,<br />
ˆR = R R ,<br />
e<br />
v = v (t R )<br />
= dr 0 (t R )<br />
dt R<br />
= ṙ 0 (t R )<br />
β = v c .<br />
Com essas <strong>de</strong>finições, po<strong>de</strong>mos simplificar as expressões dos <strong>potenciais</strong> assim:<br />
φ (r, t) = 1 ( )<br />
q<br />
,<br />
4πε 0 R − R · β<br />
A (r, t) = µ ( )<br />
0 qv<br />
4π R − R · β<br />
1
e<br />
t R = t − R c .<br />
Vamos agora calcular os campos B e E. Começamos com o cálculo <strong>de</strong> B:<br />
ou, em termos <strong>de</strong> componentes,<br />
B i =<br />
B = ∇ × A<br />
3∑<br />
j=1<br />
3∑<br />
ε ijk ∂ j A k ,<br />
on<strong>de</strong> ε ijk é o tensor <strong>de</strong> Levi-Civita, que é dado por<br />
⎧<br />
⎪⎨ 1, se (i, j, k) for uma permutação par <strong>de</strong> (1, 2, 3) ,<br />
ε ijk = 0, se pelo menos dois dos índices i, j, k forem iguais e<br />
⎪⎩<br />
−1, se (i, j, k) for uma permutação ímpar <strong>de</strong> (1, 2, 3) .<br />
Também utilizamos a notação<br />
k=1<br />
∂ j = ∂<br />
∂x j<br />
,<br />
para j = 1, 2, 3. A convenção <strong>de</strong> Einstein para somas permite que escrevamos<br />
B i = ε ijk ∂ j A k ,<br />
on<strong>de</strong> subenten<strong>de</strong>mos que os índices j e k estão somados <strong>de</strong> 1 a 3, porque aparecem<br />
repetidos no mesmo termo. Temos, assim,<br />
∂ j A k = µ ( )<br />
0q<br />
4π ∂ v k<br />
j<br />
R − R · β<br />
= µ [<br />
( )]<br />
0q ∂ j v k<br />
4π R − R · β + v 1<br />
k∂ j<br />
R − R · β<br />
= µ [<br />
]<br />
0q ∂ j v k<br />
4π R − R · β − v k<br />
(R − R · β) 2 (∂ jR − β · ∂ j R − R · ∂ j β) .<br />
Também,<br />
∂ j v k = dv k<br />
dt R<br />
∂ j t R<br />
= a k ∂ j t R ,<br />
2
Façamos agora o cálculo <strong>de</strong> ∂ j t R :<br />
∂ j t R = ∂ j<br />
(<br />
t − R c<br />
= − 1 c ∂ jR<br />
)<br />
= − 1 c ˆR · ˆx j + ˆR · β∂ j t R ,<br />
ou seja,<br />
( )<br />
1 − ˆR · β ∂ j t R = − 1 c ˆR · ˆx j ,<br />
resultando em<br />
∂ j t R<br />
= − 1 ˆR · ˆx<br />
( j<br />
).<br />
c 1 − ˆR · β<br />
Portanto,<br />
∂ j v k<br />
= − a k<br />
c<br />
ˆR · ˆx<br />
( j<br />
),<br />
1 − ˆR · β<br />
∂ j R = ∂ j r − ∂ j r 0 (t R )<br />
= ˆx j − dr 0 (t R )<br />
∂ j t R<br />
dt R<br />
= ˆx j + v 1 ˆR · ˆx<br />
( j<br />
)<br />
c 1 − ˆR · β<br />
ˆR · ˆx j<br />
= ˆx j + β(<br />
),<br />
1 − ˆR · β<br />
3
∂ j R = 1<br />
2R ∂ jR 2<br />
= 1<br />
2R ∂ j (R · R)<br />
= R R · ∂ jR<br />
ˆR · ˆx j<br />
= ˆR · ˆx j + ˆR · β(<br />
)<br />
1 − ˆR · β<br />
)<br />
ˆR · ˆx j<br />
(1 − ˆR · β + ˆR · β ˆR · ˆx j<br />
=<br />
( )<br />
1 − ˆR · β<br />
= ˆR · ˆx j − ˆR · ˆx j ˆR · β + ˆR · β ˆR · ˆx<br />
( )<br />
j<br />
1 − ˆR · β<br />
e<br />
on<strong>de</strong> <strong>de</strong>notamos<br />
e<br />
=<br />
ˆR · ˆx<br />
( j<br />
)<br />
1 − ˆR · β<br />
∂ j β =<br />
= 1 c<br />
1 c ∂ jv<br />
dv<br />
dt R<br />
∂ j t R<br />
= − a ˆR · ˆx<br />
( j<br />
),<br />
c 2 1 − ˆR · β<br />
a k = dv k<br />
dt R<br />
a =<br />
dv<br />
dt R<br />
.<br />
4
Assim,<br />
∂ j A k = µ 0q<br />
4π<br />
= µ 0q<br />
4π<br />
= µ 0q<br />
4π<br />
[<br />
∂ j v k<br />
R − R · β −<br />
⎡<br />
⎢<br />
⎣<br />
R<br />
⎡<br />
⎢<br />
⎣<br />
R<br />
∂ j v<br />
( k<br />
1 − ˆR · β<br />
∂ j v<br />
( k<br />
1 − ˆR · β<br />
Substituindo as <strong>de</strong>rivadas parciais, temos<br />
⎡<br />
on<strong>de</strong><br />
]<br />
v k<br />
(R − R · β) 2 (∂ jR − β · ∂ j R − R · ∂ j β)<br />
)<br />
⎤<br />
v k<br />
(∂ j R − β · ∂ j R − R ˆR · ∂ j β<br />
)<br />
⎥<br />
−<br />
) 2 ⎦<br />
R<br />
(1 2 − ˆR · β<br />
⎤<br />
v<br />
) k ˆR · ∂ j β<br />
+ ( ) 2<br />
− v k (∂ j R − β · ∂ j R) ⎥<br />
) 2 ⎦ .<br />
R 1 − ˆR · β R<br />
(1 2 − ˆR · β<br />
∂ j A k = µ 0q ⎢ −a k ˆR · ˆx j<br />
⎣ ( )<br />
4π<br />
2<br />
− v ˆR k · a ˆR · ˆx j<br />
) 3<br />
Rc 1 − ˆR · β Rc<br />
(1 2 − ˆR · β<br />
⎛<br />
v k<br />
−<br />
) ⎝( ) ⎞⎤<br />
1 − β<br />
2 ˆR · ˆx j<br />
2<br />
( ) − β j<br />
⎠⎥<br />
⎦<br />
R<br />
(1 2 − ˆR · β 1 − ˆR · β<br />
β j = v j<br />
c .<br />
Em termos vetoriais, po<strong>de</strong>mos escrever<br />
⎡<br />
B =<br />
−<br />
µ 0q<br />
4π<br />
⎢<br />
⎣<br />
Rc<br />
− ˆR × a<br />
(<br />
1 − ˆR · β<br />
⎤<br />
⎥<br />
) 3 ⎦ ,<br />
(<br />
1 − β<br />
2 ) ˆR × v<br />
R 2 (1 − ˆR · β<br />
ˆR · a ˆR × v<br />
) 2<br />
−<br />
Rc<br />
(1 2 − ˆR · β<br />
on<strong>de</strong> o termo proporcional a β × v se anula. Po<strong>de</strong>mos ainda escrever<br />
⎡ ( )<br />
B =<br />
µ (<br />
0q<br />
4π ˆR ⎢<br />
−a 1 − ˆR · β − ˆR · aβ<br />
) ⎤<br />
1 − β<br />
2<br />
v ⎥<br />
× ⎣ ( ) 3<br />
−<br />
) 3 ⎦<br />
Rc 1 − ˆR · β R<br />
(1 2 − ˆR · β<br />
⎡ ( ) ( )<br />
= µ (<br />
0q<br />
4π ˆR ⎢<br />
−a ˆR · ˆR − β + ˆR · a ˆR − β<br />
) ⎤<br />
1 − β<br />
2<br />
v ⎥<br />
× ⎣<br />
( ) 3<br />
−<br />
) 3 ⎦ ,<br />
Rc 1 − ˆR · β<br />
R<br />
(1 2 − ˆR · β<br />
) 3<br />
5
on<strong>de</strong> adicionamos um termo proporcional a ˆR entre colchetes, que não contribui<br />
para B, pois é multiplicado vetorialmente pelo ˆR que aparece fora dos colchetes.<br />
Notamos agora que<br />
( )<br />
−a ˆR · ˆR − β<br />
( )<br />
+ ˆR · a ˆR − β<br />
[( ) ]<br />
= ˆR × ˆR − β × a .<br />
Logo,<br />
B =<br />
µ 0q<br />
4π ˆR ×<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
[( )<br />
ˆR × ˆR − β<br />
Rc<br />
(<br />
1 − ˆR · β<br />
]<br />
× a<br />
) 3<br />
−<br />
( ) ⎫<br />
1 − β<br />
2 ⎪⎬<br />
v<br />
) 3<br />
.<br />
R<br />
(1 2 ⎪<br />
− ˆR · β ⎭<br />
Exercício proposto<br />
Mostre que<br />
⎡( (1<br />
E =<br />
q ⎢ ˆR − β) ) ( ) ⎤<br />
− β<br />
2 ˆR − β ˆR · a<br />
a ⎥<br />
⎣<br />
)<br />
4πε 3<br />
+<br />
) 3<br />
−<br />
) 2 ⎦<br />
0<br />
R<br />
(1 2 − ˆR · β Rc<br />
(1 2 − ˆR · β Rc<br />
(1 2 − ˆR · β<br />
⎡( (1<br />
= q ⎢ ˆR − β) ) ( ) ( ) ⎤<br />
− β<br />
2 ˆR − β ˆR · a − a 1 − ˆR · β<br />
⎥<br />
⎣<br />
)<br />
4πε 3<br />
+<br />
) 3 ⎦<br />
0<br />
R<br />
(1 2 − ˆR · β<br />
Rc<br />
(1 2 − ˆR · β<br />
⎡( (1<br />
= q ⎢ ˆR − β) ) [( ) ] ⎤<br />
− β<br />
2 ˆR × ˆR − β × a<br />
⎥<br />
⎣<br />
)<br />
4πε 3<br />
+<br />
) 3 ⎦<br />
0<br />
R<br />
(1 2 − ˆR · β Rc<br />
(1 2 − ˆR · β<br />
e que, portanto,<br />
B =<br />
ˆR<br />
c × E.<br />
6