206
8 Reissner–Nordström Particle
Since a 3 (u) = 0 and v 4 (u) = 0 it is straightforward to show from (8.128)–(8.132)
that
v
1 (u) = 0 = v
2 (u) ,
(8.134)
for all values of u. Hence in addition to (8.133) we are left with (8.124) which now
reads
− (a
3 )
2
+ (a
4 )
2
= −B
2 e
2 μ u
+
8
5
e
2 B
4 e
4 μ u
−
8
5
e
2 B
4 e
2 μ u .
(8.135)
But this equation is a consequence of (8.133) (neglecting O 2 -terms) since
−(v
3 )
2
+ (v
4 )
2
= 1 and − a
3 v
3
+ a
4 v
4
= 0 ⇒ (a
3 )
2
− (a
4 )
2
= (a
3 v
4
− a
4 v
3 )
2 .
(8.136)
To solve (8.133) it is useful to define
k 1 = v
4
+ v
3 and k 2 = v
4
− v
3
⇒ k 1 k 2 = 1 ,
(8.137)
then
d
du
(log k 1 ) = a
3 v
4
− a
4 v
3
= B e
μ u
−
4
5
e
2 B
3 e
3 μ u
+
4
5
e
2 B
3 e
μ u .
(8.138)
Integrating and using the initial condition that k 1 = 1 when u = 0 we arrive at
k 1 = v
4
+ v
3
= exp
B
μ
e
μ u
− 1
−
4 e 2 B 3
15 μ
e
3 μ u
+
4 e 2 B 3
5 μ
e
μ u
−
8 e 2 B 3
15 μ
,
(8.139)
We can simplify this to read
v
4
+ v
3
= exp
B
μ
(e
μ u
− 1)
1 −
4 e 2 B 3
15 μ
(e
μ u
− 1)
2 (e
μ u
+ 2) + O 2
.
(8.140)
Since v 4 − v 3 = k 2 = (k 1 ) −1 we have
v
4
− v
3
= exp
−
B
μ
(e
μ u
− 1)
1 +
4 e 2 B 3
15 μ
(e
μ u
− 1)
2 (e
μ u
+ 2) + O 2
.
(8.141)