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(
)
(
)
ar
*
(18)
FF F
=
F
/2
F
+
F
/2
F
ia
rb
1
bi
2
bi
(
)
(
)
*
*
ar
*
*
FF F
=
F
/4
F
−
F
/4
F
ia
rb
2
bi
1
bi
From Eq. (13a), it is simple to show that
,
(
)
(
)
2
(
)(
)
2
ab
(19a)
det
F
=
1/16
F
=
E B
⋅
2
which is a particular case of the following theorem, see Drazin [21]:
“The expression
det
F
, with antisymmetric
F
×
and even
n
, is the
square of a rational polynomial in
i
F
” (19b)
Now, we mention the interesting and useful Stachel theorem [2] page
1261:
“If
F
satisfi es
nn
FFF
++=
0
(
)
FF
=−
det
F
=
0
-
,
,
(20a)
ar c
,
rc a
,
ca r
,
ar
ra
ab
then there exist functions
β
and
ψ
such that
F
=
βψ
−
βψ
(20b)
ar
,
a
,
r
,
r
,
a
That is, the conditions (20a) reduce
F
to an antisymmetric product of gra-
dients. If we extend the Stachel result to the Maxwell field, then the first
two conditions Eq. (20a) will be immediately verified, thereby
“If the Faraday tensor fulfi lls
2
F
=
0
, then it has the form Eq. (20b).”
(20c)
The Liénard-Wiechert solution satisfi es Eq. (20c), and thus permits
us to write
F
in the form of Plebañski [6]. In general, an electromagnetic
fi eld with a different type of A has the structure Eq. (20b). Results Eq. (20)
are valid in the presence of curvature because its differential expressions
remain undisturbed if covariant derivatives are used instead of partial ones:
ββ
=
*
FF
ar
=
*
ar
=
0
,
,
r
;
r
,
r
;
r
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