(A) EQUATION ONE
∂ρ/∂t = - 1/r^2 ∂/∂r r^2 uρ A
In the steady state, ∂ρ/∂t =0, so that
0 = - 1/r^2 ∂/∂r r^2 uρ Ai
Where - 1/r^2 is smaller entity,
0= ∂/∂r r^2 uρ
Integrating we have,
constant=∫▒∂/∂r r^2 uρ.dr
constant=∫▒2 ruρdr
constant=r^2 uρ
Therefore,
ρ = constant/(r^2 u) Aii
Equation Aii shows that the density ρ is constant only if r^2 u is constant.
(B) Equation number two
∂u/∂t = - u ∂u/∂r - 1/ρ ((∂P_g)/∂r + (∂P_c)/∂r) B
Multiply both sides by ρ:
ρ ∂u/∂t = ρ [–u ∂u/∂r - 1/ρ ((∂P_g)/∂r+ (∂P_c)/∂r) ]
ρ ∂u/∂t = - ρu ∂u/∂r - ((∂P_g)/∂r + (∂P_c)/∂r) Bi
Equation Bi gives the energy equation; where u is the internal heat energy (that is, energy per unit mass).
ρ ∂u/∂t = - u^2 ∂ρ/∂r – ((∂P_g)/∂r + (∂P_c)/∂r)
This is similar to:
ρ ∂u/∂t = -P∇.V + ρg (energy equation)
By comparison, we have:
- u^2 ∂ρ/∂r – ((∂P_g)/∂r + (∂P_c)/∂r) = -P ∇.V + ρg
- u^2 ∂ρ/∂r = -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r)
In quadratic expression, ab = c which means that a = c or b = c. Now, considering only:
∂ρ/∂r = -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r) Bii
Recall that in the steady state,
0= ∂/∂r r^2 uρ
This implies that
∂ρ/∂r =0
Equation Bii becomes;
0= -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r)
((∂P_g)/∂r + (∂P_c)/∂r) -P∇.V +ρg =0
but ρ = nm, where n is the particle density
((∂P_g)/∂r + (∂P_c)/∂r) -∇ .(PV) + nmg =0 Biii
((∂P_g)/∂r + (∂P_c)/∂r) is the ratio of force to volume
That is
force/volume = force/area × 1/length
OR
mass/volume ×acceleration
ρ ×acceleration
ρ × ∂v/∂t = ρ ∂v/∂t
nm ∂v/∂t = ∂(nmv)/∂t
Equation Biii becomes;
∂(nmv)/∂t - ∇.(PV) +nmg =0
∂(nmv)/∂t - ∇.(PV) +nF =0
nm ∂v/∂t =nF - ∇.s +P Biv
Where s is the stress tensor, which is the simple case of an isotropic distribution of the random velocities of the particles.
Equation takes the form:
n_e m_e (∂v_e)/∂t = - n_e e(E + v_e/c × B_0 ) - ∇.s_e + n_e m_e g +P_ei Bv
And
n_i m_i (∂v_i)/∂t = - n_i e(E + v_i/c × B_0 ) - ∇.s_i + n_i m_i g +P_ie Bvi
Adding equations Bv and Bvi gives:
n_e m_e (∂v_e)/∂t +n_i m_i (∂v_i)/∂t = - n_e e ((v_i v_e ) × B_0)/c - ∇P + (n_i m_i + n_e m_e )g
Where the interaction terms Pei and Pie are cancelled by Newton’s 3rd law.
(C) Equation three
(∂P_g)/∂t =-u (∂P_g)/∂r - (γ_g P_g)/r^2 (∂r^2 u)/∂r - (γ_(g )- 1)(w -u) (∂P_c)/∂r C
(∂P_g)/∂t + (u∂P_g)/∂r + (γ_(g )- 1)(w -u) (∂P_c)/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + (∂uP_g)/∂r + (∂(γ_g -1)(w -u).P_c)/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r = - 1/r^2 (∂r^2)/∂r u(P_g γ_g ) Ci
In the steady state, from equation Ai, Ci becomes;
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r =0 Cii
Since ∂ρ/∂t =0, therefore, (∂P_g)/∂t = 0
Equation Cii becomes
P_g ∂u/∂r + P_c ∂( γ_g-1)(w-u)/∂r =0
P_g ∂u/∂r = – P_c ∂( γ_g-1)(w-u)/∂r
This implies that P_g depends on P_c, that is pressure of the gas depends on the pressure of the cosmic ray. Hence if P_g =0 then P_c =0.
P_(g = )the gas pressure.
P_(c =)the cosmic ray pressure.
(D) Equation four
∂f/∂t = 1/r^2 ∂/∂r r^2 D(p,r,t) ∂f/∂r –w ∂f/∂r + ∂f/∂p p/(3r^2 ) (∂r^2 w)/∂r + φδ(p-p_inj )/(4πp_inj^2 m) ρ(R+0,t)(R ̇-u(R,0,t) )× δ(r-R_((t) ) ) (D)
If we consider only the f(p,t), momentum distribution function of the particles, according to Ramaty (1986):
D(p)=1/3 (u/lv) p^2 Di
then
∂f/∂t=1/p^2 ∂/∂p [p^2 D(p) ∂f/∂p]
Therefore by the steady state assumption,
∂f/∂t=0
Which gives us;
1/p^2 ∂/∂p [p^2 D(p) ∂f/∂p]=0
This means that
1/p^(2 ) =0
OR
∂/∂p [p^2 D(p) ∂f/∂p]=0 Dii
Using (Di) into (Dii), we have;
∂/∂p [p^2.1/3 (u^2/lv) p^2.∂f/∂p]=0
∂/∂p [p^4.u^2/3lv.∂f/∂p]=0
∂((p^4 u^2)/3lv)/∂p.∂f/∂p=0
∂((p^4 u^2)/3lv)/∂p=0
OR
∂f/∂p=0
From quotient rule,
∂((p^4 u^2)/3lv)/∂p=((3l^4.8p^3 u)-(p^4 u^2.12l^3 ) )/(9l^8 ) Diii
3l^4.8p^3-p^4 u^2.12l^3=0
3l^4.〖8p〗^3 u=p^4 u^2.12l^3
l^4.8p^3 u=p^4 u^2.4l^3
l^4.2p^3 u=p^4 u^2.l^3
l.2p^3 u=p^4 u^2
l.2p^3=p^4 u
l.2=pu
2l=pu
u=2l/p Div
Equation (Div) shows that the internal energy of the electron is inversely proportional to the pressure of the gas particles. If we consider the particle density, we have;
p=ρgh=ρgl
Equation (Div) becomes:
u=2l/ρgl
u=2/ρg
But g=10m⁄s^2
Therefore,
u=1/5ρ
u=1/5.ρ
u∝ρ Dv
u=the internal energy
p=the density of the gas particles.
Equation Dv shows that the internal energy is directly proportional to the density of the gas particles
∂ρ/∂t = - 1/r^2 ∂/∂r r^2 uρ A
In the steady state, ∂ρ/∂t =0, so that
0 = - 1/r^2 ∂/∂r r^2 uρ Ai
Where - 1/r^2 is smaller entity,
0= ∂/∂r r^2 uρ
Integrating we have,
constant=∫▒∂/∂r r^2 uρ.dr
constant=∫▒2 ruρdr
constant=r^2 uρ
Therefore,
ρ = constant/(r^2 u) Aii
Equation Aii shows that the density ρ is constant only if r^2 u is constant.
(B) Equation number two
∂u/∂t = - u ∂u/∂r - 1/ρ ((∂P_g)/∂r + (∂P_c)/∂r) B
Multiply both sides by ρ:
ρ ∂u/∂t = ρ [–u ∂u/∂r - 1/ρ ((∂P_g)/∂r+ (∂P_c)/∂r) ]
ρ ∂u/∂t = - ρu ∂u/∂r - ((∂P_g)/∂r + (∂P_c)/∂r) Bi
Equation Bi gives the energy equation; where u is the internal heat energy (that is, energy per unit mass).
ρ ∂u/∂t = - u^2 ∂ρ/∂r – ((∂P_g)/∂r + (∂P_c)/∂r)
This is similar to:
ρ ∂u/∂t = -P∇.V + ρg (energy equation)
By comparison, we have:
- u^2 ∂ρ/∂r – ((∂P_g)/∂r + (∂P_c)/∂r) = -P ∇.V + ρg
- u^2 ∂ρ/∂r = -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r)
In quadratic expression, ab = c which means that a = c or b = c. Now, considering only:
∂ρ/∂r = -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r) Bii
Recall that in the steady state,
0= ∂/∂r r^2 uρ
This implies that
∂ρ/∂r =0
Equation Bii becomes;
0= -P∇.V + ρg+((∂P_g)/∂r + (∂P_c)/∂r)
((∂P_g)/∂r + (∂P_c)/∂r) -P∇.V +ρg =0
but ρ = nm, where n is the particle density
((∂P_g)/∂r + (∂P_c)/∂r) -∇ .(PV) + nmg =0 Biii
((∂P_g)/∂r + (∂P_c)/∂r) is the ratio of force to volume
That is
force/volume = force/area × 1/length
OR
mass/volume ×acceleration
ρ ×acceleration
ρ × ∂v/∂t = ρ ∂v/∂t
nm ∂v/∂t = ∂(nmv)/∂t
Equation Biii becomes;
∂(nmv)/∂t - ∇.(PV) +nmg =0
∂(nmv)/∂t - ∇.(PV) +nF =0
nm ∂v/∂t =nF - ∇.s +P Biv
Where s is the stress tensor, which is the simple case of an isotropic distribution of the random velocities of the particles.
Equation takes the form:
n_e m_e (∂v_e)/∂t = - n_e e(E + v_e/c × B_0 ) - ∇.s_e + n_e m_e g +P_ei Bv
And
n_i m_i (∂v_i)/∂t = - n_i e(E + v_i/c × B_0 ) - ∇.s_i + n_i m_i g +P_ie Bvi
Adding equations Bv and Bvi gives:
n_e m_e (∂v_e)/∂t +n_i m_i (∂v_i)/∂t = - n_e e ((v_i v_e ) × B_0)/c - ∇P + (n_i m_i + n_e m_e )g
Where the interaction terms Pei and Pie are cancelled by Newton’s 3rd law.
(C) Equation three
(∂P_g)/∂t =-u (∂P_g)/∂r - (γ_g P_g)/r^2 (∂r^2 u)/∂r - (γ_(g )- 1)(w -u) (∂P_c)/∂r C
(∂P_g)/∂t + (u∂P_g)/∂r + (γ_(g )- 1)(w -u) (∂P_c)/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + (∂uP_g)/∂r + (∂(γ_g -1)(w -u).P_c)/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r = - (γ_g P_g)/r^2 (∂r^2 u)/∂r
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r = - 1/r^2 (∂r^2)/∂r u(P_g γ_g ) Ci
In the steady state, from equation Ai, Ci becomes;
(∂P_g)/∂t + ∂(uP_g + (γ_g-1)(w -u).P_c )/∂r =0 Cii
Since ∂ρ/∂t =0, therefore, (∂P_g)/∂t = 0
Equation Cii becomes
P_g ∂u/∂r + P_c ∂( γ_g-1)(w-u)/∂r =0
P_g ∂u/∂r = – P_c ∂( γ_g-1)(w-u)/∂r
This implies that P_g depends on P_c, that is pressure of the gas depends on the pressure of the cosmic ray. Hence if P_g =0 then P_c =0.
P_(g = )the gas pressure.
P_(c =)the cosmic ray pressure.
(D) Equation four
∂f/∂t = 1/r^2 ∂/∂r r^2 D(p,r,t) ∂f/∂r –w ∂f/∂r + ∂f/∂p p/(3r^2 ) (∂r^2 w)/∂r + φδ(p-p_inj )/(4πp_inj^2 m) ρ(R+0,t)(R ̇-u(R,0,t) )× δ(r-R_((t) ) ) (D)
If we consider only the f(p,t), momentum distribution function of the particles, according to Ramaty (1986):
D(p)=1/3 (u/lv) p^2 Di
then
∂f/∂t=1/p^2 ∂/∂p [p^2 D(p) ∂f/∂p]
Therefore by the steady state assumption,
∂f/∂t=0
Which gives us;
1/p^2 ∂/∂p [p^2 D(p) ∂f/∂p]=0
This means that
1/p^(2 ) =0
OR
∂/∂p [p^2 D(p) ∂f/∂p]=0 Dii
Using (Di) into (Dii), we have;
∂/∂p [p^2.1/3 (u^2/lv) p^2.∂f/∂p]=0
∂/∂p [p^4.u^2/3lv.∂f/∂p]=0
∂((p^4 u^2)/3lv)/∂p.∂f/∂p=0
∂((p^4 u^2)/3lv)/∂p=0
OR
∂f/∂p=0
From quotient rule,
∂((p^4 u^2)/3lv)/∂p=((3l^4.8p^3 u)-(p^4 u^2.12l^3 ) )/(9l^8 ) Diii
3l^4.8p^3-p^4 u^2.12l^3=0
3l^4.〖8p〗^3 u=p^4 u^2.12l^3
l^4.8p^3 u=p^4 u^2.4l^3
l^4.2p^3 u=p^4 u^2.l^3
l.2p^3 u=p^4 u^2
l.2p^3=p^4 u
l.2=pu
2l=pu
u=2l/p Div
Equation (Div) shows that the internal energy of the electron is inversely proportional to the pressure of the gas particles. If we consider the particle density, we have;
p=ρgh=ρgl
Equation (Div) becomes:
u=2l/ρgl
u=2/ρg
But g=10m⁄s^2
Therefore,
u=1/5ρ
u=1/5.ρ
u∝ρ Dv
u=the internal energy
p=the density of the gas particles.
Equation Dv shows that the internal energy is directly proportional to the density of the gas particles