B Bateman Waves
213
Thus the vectors u a , e a , b a , p a constitute an orthonormal tetrad. It is useful to note,
using (B.13), that
η abcd u
c B
d
= δ
l
a δ
m
b η lmcd u
c B
d
= (p a e b − p b e a ) p
l η lmcd e
m u
c B
d
= (p a e b − p b e a ) (−E
f E f )
1/2 ,
(B.14)
and thus
η abcd u
c B
d
= p a E b − p b E a .
(B.15)
Similarly
η abcd u
c E
d
= −(p a B b − p b B a ) .
(B.16)
When (B.15) and (B.16) are substituted into (B.5) and (B.6) we find that
F ab = E a k b − E b k a and
∗ F ab = B a k b − B b k a ,
(B.17)
where
k a = u a − p a ⇒ k a k
a
= 0 .
(B.18)
The null vector field k a , which satisfies
F ab k
b
= 0 =
∗ F ab k
b ,
(B.19)
is the degenerate principal null direction associated with the bivector F ab and is thus
the propagation direction in space-time of the electromagnetic waves.
Substituting (B.17) into Maxwell’s equations (B.1), contracting each in turn with
e a and b a and then adding and subtracting the results, we arrive at
k a;b e
a e
b
= k a;b b
a b
b ,
(B.20)
and
(E f E
f ) ,b k
b
+ 2 (E f E
f ){k
b ;b + k a;b e
a e
b
} = 0 ,
(B.21)
from the first of (B.1) and
k a;b e
a b
b
+ k a;b b
a e
b
= 0 ,
(B.22)
213
Thus the vectors u a , e a , b a , p a constitute an orthonormal tetrad. It is useful to note,
using (B.13), that
η abcd u
c B
d
= δ
l
a δ
m
b η lmcd u
c B
d
= (p a e b − p b e a ) p
l η lmcd e
m u
c B
d
= (p a e b − p b e a ) (−E
f E f )
1/2 ,
(B.14)
and thus
η abcd u
c B
d
= p a E b − p b E a .
(B.15)
Similarly
η abcd u
c E
d
= −(p a B b − p b B a ) .
(B.16)
When (B.15) and (B.16) are substituted into (B.5) and (B.6) we find that
F ab = E a k b − E b k a and
∗ F ab = B a k b − B b k a ,
(B.17)
where
k a = u a − p a ⇒ k a k
a
= 0 .
(B.18)
The null vector field k a , which satisfies
F ab k
b
= 0 =
∗ F ab k
b ,
(B.19)
is the degenerate principal null direction associated with the bivector F ab and is thus
the propagation direction in space-time of the electromagnetic waves.
Substituting (B.17) into Maxwell’s equations (B.1), contracting each in turn with
e a and b a and then adding and subtracting the results, we arrive at
k a;b e
a e
b
= k a;b b
a b
b ,
(B.20)
and
(E f E
f ) ,b k
b
+ 2 (E f E
f ){k
b ;b + k a;b e
a e
b
} = 0 ,
(B.21)
from the first of (B.1) and
k a;b e
a b
b
+ k a;b b
a e
b
= 0 ,
(B.22)
