176
7 Small Magnetic Black Hole
Next we require E (1)(1) − E (2)(2) + 2 i E (1)(2) . To aid the calculation of this complex
variable we first note from (7.161)–(7.164) that we can write
F (1)(3) = F
0
(1)(3) + O(r) , F (2)(3) = F
0
(2)(3) + O(r) ,
(7.179)
and
F (1)(4) =
F 2
(1)(4)
r 2 +
F 1
(1)(4)
r
+ F
0
(1)(4) + O(r) ,
(7.180)
F (2)(4) =
F 2
(2)(4)
r 2 +
F 1
(2)(4)
r
+ F
0
(1)(4) + O(r) ,
(7.181)
with
F
0
(1)(3) + i F
0
(2)(3) = 2 i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
− 2 i P 0 (m 2 − i l 2 ) + O 2 ,
(7.182)
F
2
(1)(4) + i F
2
(2)(4) = −5 i g
2 P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ O 3 ,
(7.183)
F
1
(1)(4) + i F
1
(2)(4) = −2 i g P 0 a i
∂k i
∂ ¯
ζ
+ 2 m i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ O 2 , (7.184)
F
0
14 + i F
0
(2)(4) = i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ 2 i P 0
∗ F ij
∂k i
∂ ¯
ζ
v
j
+ O 1 .
(7.185)
Now
E (1)(1) − E (2)(2) + 2 i E (1)(2) =
T 2
r 2 +
T 1
r
+ O(r
0 ) ,
(7.186)
with (neglecting O 3 -terms in order to simplify the presentation at this stage)
T 2 = 2 (F
0
(1)(3) + i F
0
(2)(3) )(F
2
(1)(4) + i F
2
(2)(4) )
= 20 g
2
P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
2
=
∂
∂ ¯
ζ
−20 g
2 P
2
0
∗ F
p
i
∗ F pj k
i ∂k j
∂ ¯
ζ
− 40 g
2 P
2
0
∗ F ij k
i ∂k j
∂ ¯
ζ
∗ F pq k
p v
q
.
(7.187)
7 Small Magnetic Black Hole
Next we require E (1)(1) − E (2)(2) + 2 i E (1)(2) . To aid the calculation of this complex
variable we first note from (7.161)–(7.164) that we can write
F (1)(3) = F
0
(1)(3) + O(r) , F (2)(3) = F
0
(2)(3) + O(r) ,
(7.179)
and
F (1)(4) =
F 2
(1)(4)
r 2 +
F 1
(1)(4)
r
+ F
0
(1)(4) + O(r) ,
(7.180)
F (2)(4) =
F 2
(2)(4)
r 2 +
F 1
(2)(4)
r
+ F
0
(1)(4) + O(r) ,
(7.181)
with
F
0
(1)(3) + i F
0
(2)(3) = 2 i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
− 2 i P 0 (m 2 − i l 2 ) + O 2 ,
(7.182)
F
2
(1)(4) + i F
2
(2)(4) = −5 i g
2 P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ O 3 ,
(7.183)
F
1
(1)(4) + i F
1
(2)(4) = −2 i g P 0 a i
∂k i
∂ ¯
ζ
+ 2 m i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ O 2 , (7.184)
F
0
14 + i F
0
(2)(4) = i P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
+ 2 i P 0
∗ F ij
∂k i
∂ ¯
ζ
v
j
+ O 1 .
(7.185)
Now
E (1)(1) − E (2)(2) + 2 i E (1)(2) =
T 2
r 2 +
T 1
r
+ O(r
0 ) ,
(7.186)
with (neglecting O 3 -terms in order to simplify the presentation at this stage)
T 2 = 2 (F
0
(1)(3) + i F
0
(2)(3) )(F
2
(1)(4) + i F
2
(2)(4) )
= 20 g
2
P 0
∗ F ij k
i ∂k j
∂ ¯
ζ
2
=
∂
∂ ¯
ζ
−20 g
2 P
2
0
∗ F
p
i
∗ F pj k
i ∂k j
∂ ¯
ζ
− 40 g
2 P
2
0
∗ F ij k
i ∂k j
∂ ¯
ζ
∗ F pq k
p v
q
.
(7.187)
