498
G. Kamath
The last term in (35) will be zero by current conservation and the momentum
integration yields
Ψ (x) =
2i
c
z
− →
j ×( − → x − − → z )
|
− → x − − → z |
2
−
j 0
2 ∂ x log
− → x − − → z
2 −
j 0
2 ∂ y log
− → x − − → z
2 0
T
≡
B + iH E 1 + iD 1 E 2 + iD 2 0
T
(36)
The second equality in (36) yields
− →
E + i
− →
D = −i grad
log
− → x −
− →
R
2
(37)
when j 0 = cδ
− → z −
− →
R
reflecting an unit charge at − → z =
− →
R . By extension one
also has
B + iH =
2i
c
z
− →
j ×
− → x − − → z
− → x − − → z
2
(38)
The relation between
B,
− →
E
and
H,
− →
D
follows from
f a0 = g ab g 0c f
ab , f 21 = g 2μ g 1ν f
μν
(39)
as D a = f 0a , H = f 21 , E a = f a0 , B = f 21 vide Eq. (27) above. To elaborate one has for
Eq. (39)
f a0 = (g a0 g 01 − g a1 g 00 ) f 01 + (g a0 g 02 − g a2 g 00 ) f 02 + (g a2 g 01 − g a1 g 20 ) f 21
f 21 = (g 20 g 11 − g 21 g 10 ) f 01 + (g 20 g 12 − g 22 g 10 ) f 02 + (g 22 g 11 − g 21 g 12 ) f 21
(40)
and with a little effort one gets from Eqs. (40)
⎛
⎝
B + iH
E 1 + iD 1
E 2 + iD 2
⎞
⎠ = K
⎛
⎝
f 01
f 02
f 21
⎞
⎠
(41)
where
K =
⎛
⎝
(g 20 g 11 − g 21 g 10 )
(g 20 g 12 − g 22 g 10 ) (g 22 g 11 − g 21 g 12 + i)
(g 10 g 01 − g 11 g 00 + i) (g 10 g 02 − g 12 g 00 )
(g 12 g 01 − g 11 g 20 )
(g 20 g 01 − g 21 g 00 ) (g 20 g 02 − g 22 g 00 + i) (g 22 g 01 − g 21 g 20 )
⎞
⎠
(42)