E 0 ¼ ρ 0 A
2 2π
Z 1
0
ψϕ
2
η
2 dη þ
4π
γ γ À 1
ð
Þ
Z 1
0
f η
2 dη
2
4
3
5 :
The quantity in square brackets in this latter equation is just a function of γ, hence
we may write it as
B γ
ð Þ ¼ 2π
Z 1
0
ψϕ
2
η
2 dη þ
4π
γ γ À 1
ð
Þ
Z 1
0
f η
2 dη
2
4
3
5
so that
E 0 ¼ ρ 0 A
2 B γ
ð Þ:
ð5:22Þ
Numerical integration gives the following results for γ ¼ 1.4;
Z 1
0
ψϕ
2
η
2 dη ¼ 0:185 and
Z 1
0
f η
2 dη ¼ 0:185,
hence,
B γ
ð Þ ¼ 5:31,
ð5:23Þ
and we can identify A according to Eq. (5.22) as,
A ¼
1
ffiffiffiffiffiffiffiffiffi
B γ
ð Þ
p
E 0
ρ 0
1
2
However, Eq. (5.3) gave the following relationship between η 0 and A,
A ¼
2
5
η
5
2
0
E 0
ρ 0
1
2
,
hence,
2
5
η
5
2
0 ¼
1
ffiffiffiffiffiffiffiffiffi
B γ
ð Þ
p
and therefore, η 0 ¼ 1.033 when γ ¼ 1.4. The radius of the shock front according to
Eq. (5.1) can now be written as
230
5 Spherical Shock Waves: The Self-similar Solution