8.5 Energy Radiation Rate and Run-Away Motion
205
From this we derive the following two equations:
μ a i a
i
=
1
2
d
du
(a i a
i ) −
4
5
e
2 a k a
k d
du
(a i a
i ) ,
(8.122)
μ (a
i v
j
− a
j v
i ) =
d
du
(a
i v
j
− a
j v
i ) −
8
5
e
2 (a k a
k )
d
du
(a
i v
j
− a
j v
i ) ,
(8.123)
with an O 2 -error understood in each case from now on. Solving the first of these
with the initial condition (8.119) results in
a i a
i
= −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 .
(8.124)
The general solution of (8.123) reads
a
i v
j
− a
j v
i
= K
ij e
μ u
−
4
5
e
2 B
2 K
ij e
3 μ u
+
ij
1
e
μ u ,
(8.125)
where K ij = −K ji = O 0 are constants and ij
1
= − ji
1
= O 1 are constants, all
to be determined. Writing out (8.125) for i, j = 1, 2, 3, 4 and imposing the initial
conditions (8.119) we find that
K
ij
= 0 except K
34
= −K
43
= B ,
(8.126)
and
ij
1
= 0 except
34
1
= −
43
1
=
4
5
e
2 B
3
= O 1 .
(8.127)
Now writing out (8.125) explicitly we have
a
1 v
2
− a
2 v
1
= 0 ,
(8.128)
a
1 v
3
− a
3 v
1
= 0 ,
(8.129)
a
1 v
4
− a
4 v
1
= 0 ,
(8.130)
a
2 v
3
− a
3 v
2
= 0 ,
(8.131)
a
2 v
4
− a
4 v
2
= 0 ,
(8.132)
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 .
(8.133)