e p ¼
1
γB γ
ð Þ
λ
À3
s f
λ
λ s
:
ð6:34Þ
6.6.2 The Velocity
Also, from Eq. (5.7b) we have the following equation for the particle velocity,
u ¼ AR
À3=2
ϕ η
ð Þ
and using Eq. (5.22), namely, E 0 ¼ ρ 0 A
2 B(γ), and substituting for A yields
u ¼
E
1=2
0
ρ 0 B γ
ð Þ
½
1=2
R
À3=2
ϕ η
ð Þ:
Dividing above and below by p 0 (the ambient air pressure) on the right-hand side
of the latter equation gives,
u ¼
E 0 =p 0
ð
Þ
1=2
ρ 0 B γ
ð Þ
p 0
h
i 1=2 R
À3=2
ϕ η
ð Þ:
As ε
3 ¼ E 0 /p 0 , hence,
u ¼
1
γB γ
ð Þ=
γp 0
ρ 0
h
i 1=2 λ
À3=2
s
ϕ η
ð Þ:
We identify the quantity (γp 0 /ρ 0 )
1/2 as the ambient speed of sound, c 0 , hence,
e u ¼
1
γB γ
ð Þ
½
1=2
λ
À3=2
s
ϕ
λ
λ s
,
ð6:35Þ
which is the initial normalized particle velocity (e u ¼ u=c 0 ) for the numerical
integration.
6.6 Initial Conditions Using the Strong-Shock, Point-Source Solution
293
1
γB γ
ð Þ
λ
À3
s f
λ
λ s
:
ð6:34Þ
6.6.2 The Velocity
Also, from Eq. (5.7b) we have the following equation for the particle velocity,
u ¼ AR
À3=2
ϕ η
ð Þ
and using Eq. (5.22), namely, E 0 ¼ ρ 0 A
2 B(γ), and substituting for A yields
u ¼
E
1=2
0
ρ 0 B γ
ð Þ
½
1=2
R
À3=2
ϕ η
ð Þ:
Dividing above and below by p 0 (the ambient air pressure) on the right-hand side
of the latter equation gives,
u ¼
E 0 =p 0
ð
Þ
1=2
ρ 0 B γ
ð Þ
p 0
h
i 1=2 R
À3=2
ϕ η
ð Þ:
As ε
3 ¼ E 0 /p 0 , hence,
u ¼
1
γB γ
ð Þ=
γp 0
ρ 0
h
i 1=2 λ
À3=2
s
ϕ η
ð Þ:
We identify the quantity (γp 0 /ρ 0 )
1/2 as the ambient speed of sound, c 0 , hence,
e u ¼
1
γB γ
ð Þ
½
1=2
λ
À3=2
s
ϕ
λ
λ s
,
ð6:35Þ
which is the initial normalized particle velocity (e u ¼ u=c 0 ) for the numerical
integration.
6.6 Initial Conditions Using the Strong-Shock, Point-Source Solution
293