5. Simplified Transport Equations We want to derive two fundamental transport properties, diffusion and viscosity. Unable to handle the 13-moment system of equations, we resort to a simpler description by assuming collision dominance. For such gas the drifting Maxwellian is a good approximation for the velocity distribution function: The ionosphere can be treated as a weakly ionized plasma, meaning that the effects of Coulomb collisions are small compared to electron-neutral and ion-neutral collisions. The opposite is true for a fully ionized plasma (neutrals are still there). We limit our discussion to weakly ionized plasmas. 32 2 2 2 (,) (,) exp ; 3.44 2 2 m mc f t n t c kT kT r r v u
51
Embed
5. Simplified Transport Equations We want to derive two fundamental transport properties, diffusion and viscosity. Unable to handle the 13-moment system.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
5. Simplified Transport Equations
We want to derive two fundamental transport properties, diffusion and viscosity. Unable to handle the 13-moment system of equations, we resort to a simpler description by assuming collision dominance. For such gas the drifting Maxwellian is a good approximation for the velocity distribution function:
The ionosphere can be treated as a weakly ionized plasma, meaning that the effects of Coulomb collisions are small compared to electron-neutral and ion-neutral collisions. The opposite is true for a fully ionized plasma (neutrals are still there). We limit our discussion to weakly ionized plasmas.
3 2 2
22( , ) ( , ) exp ; 3.442 2
m mcf t n t c
kT kT
r r v u
5.1 Basic Transport Properties (1)
To develop the general idea, we assume an isothermal gas with a density n(x) that has a constant gradient in the -x direction, see Fig. 5.1a. We assume that the mean free path length , and that the velocity distribution is given by a non-drifting Maxwellian.
We expect diffusion of particles from higher to lower density, i.e., from left to right in Fig. 5.1a. Let us calculate the net particle flux (# of particles per m2 and s) through the y-z plane at x. Appendix H.26 gives the thermal particle flux at x as [n(x) <c(x)>]/4 (HW#6: derive H.26). If the density n were constant, the net flux through the plane x = const would be zero because of the equal flux from the left and the right. The particles reaching the plane x had their last collision (in the average) in the plane x-x, where x ~ . The density at x-x is
The particles arriving from the right had their last collision at x+x, and
( ) ...dn
n x x n x xdx
( ) ...dn
n x x n x xdx
11 dn
n dx
5.1 Basic Transport Properties (2)The net particle flux crossing the plane at x then becomes
1 1
4 4
5.24 2
Notice - is positive for the density in Fig.5.1a.
For , where is the average collision frequency,
fromleft fromright
collision
c n x x c n x x
c cdn dn dnn x x n x x x
dx dx dx
dn
dxc
x c
2
2
5.32
8Since (App.H.21)
4Fick's Law 5.4
c dn
dxkT
cm
kT dn dnD
m dx dx
5.1 Basic Transport Properties (3)
where
41.3 is the diffusion constant. D is proportional to the temperature
T, and inversely proportional to the mass and the collision frquency.
The net flux vector is
for the six x
kT kTD
m m
n
nu
Γ
Γ u
e
2
2
2
2
tuation in Fig.5.1a. The continuity equation (3.57) is
n for zero production or loss.
t
n
t
nparabolic partial differentialeq. 5.6
t
xx x
n
nu nnu D
x dx x
nD
x
u
e
5.1 Basic Transport Properties (4)
2
4
Solutions of the DE depend on the initial conditions, say the value n(x,0),
and two boundary conditions, say n ,t and n(- ,t). A solution is
, 5.74
This solution has a Gaussian peak at x
x
tNn x t e
Dt
=0, and goes to 0 for x . As
t , n 0. And for t 0, n 0 for all x (except for x=0 where
the function is not defined for t=0). See Fig.5.2.
5.1 Basic Transport Properties (5)Viscosity
The viscosity of a gas determines the transport of momentum perpendicular to the flow
direction. Such transport occurs when there is a velocity gradient like the one in
Fig. 5.1b). In general, the stress tensor is
where is the transfer of x-momentum in the x direction, the transfer of
y-momentum in the x direction, etc. For the conditions in Fig. 5.
xx xy xz
yx yy yz
zx zy zz
xx xy
τ
τ
2yx
1b, only is non-zero.
The viscous stress is the net transfer in the y-direction of x-momentum per m and s
across the plane at y:
1n x c 5.8
4
1n x c
4
yx
yx x x
x xx x
m u y y u y y
u um u y y u y y
y y
2
2
1n x c where /
2
1n x c
2
1 8where c 5.11
2 2
1.3 viscositycoefficient 5.12
x
x
xyx
um y y c
y
um
y
u n kTnm
y
nkT
5.1 Basic Transport Properties (6) Example of Viscosity Effect (1)
A gas flowing in the x direction between two infinite plates at y = 0 and y = a. Assume the plate at y = 0 is fixed and the plate at y = a moves in the x direction with velocity Vo. The particles close to y = a will tend to move with the plate in the x direction, while the particles at y = 0 will be at rest. What is the velocity at y? To find the answer we must solve the momentum equation (equation of motion, or force equation) (3.58):
.
For a uniform isothermal gas, = 0. Assuming is the dominant force term,
the steady state i.e. 0 solution is obtained from
0.
The divergence of a tenso
Dnm p ne nm
Dt tp
D
Dt
u Mτ E u B G
τ
u
τ
r (L.26) is
.yx xy yy zy yzxx zx xz zzx y zx y z x y z x y z
τ e e e
5.1 Basic Transport Properties (7)Example of Viscosity Effect (2)
For our example, y , and only is nonzero. Therefore
0
0 0 where I used (
x x yx
yx xy yy zy yzxx zx xz zzx y
yxx
yx x
u
x y z x y z x y z
y
u
y y y
u e
τ e e
e
2
2
0
0
x x 0
5.11).
Assuming constant, we can write
0
With the boundary conditions at y = 0 an y = a
0 =0, a = V we get
5.16
The velocity increases linearly from u = 0 to u = V . Thi
xx
x x
x
nkT
uu y Ay B
y
u u
yu y V
a
s means that
viscosity acts to smooth velocity gradients.
5.1 Basic Transport Properties (8)Energy Flow
2
2 3 23 2 3 2
3 2 3 21 2 1 2
3 2 21 2
Thermal energy is / 2 3 2. The energy flux is
3 8 3
2 4 2 4 2 2
3 3 92
22 2 2 2
8
89 9
2 2
x x
x x
m c kT
m c n c kT n kT k nq T T x x T x x
m m
k n T k n TT x T x
x xm m
kTc mx
kTk n T k nmq T T
xm
2
Here is used for the thermal conductivity:
9=
TT
m x
Tq T
x
k nT
m
5.3 Transport in a weakly ionized plasma (1)
Weakly ionized means that Coulomb collisions are of negligible importance in comparison with collisions with neutral particles. We start again with the momentum equation:
' '' ' ' '
' ' '
where the momentum transfer collision integral is given in 4.129b :
.
If we neglect the heat flo
j jj j j j j j j j j
j jj jj jj j jj j j jj j j
j jj j
Dn m p n e n m
Dt t
zn m
t kT
u Mτ E u B G
Mu u q q
w, the momentum equation for the j particles becomes
5.23
where j stands for e (electrons) or i (ions), and n for neutrals.
jj j j j j j j j j j j jn j n
Dn m p n e n m n m
Dt
uτ E u B G u u
5.3 Transport in a weakly ionized plasma (2)Scale Analysis
We start with the so-called diffusion approximation, for which the inertia terms can be neglected. To estimate the importance of the different terms look at the ratios.
2 2 2
2
j
term 2/term 3:
/ /5.24
/ / /
Here u is the average (drift) speed, / thermal speed, and the
Mach number, i.e., the ratio between drift speed and thermal speed.
j
j
j j j j j j j
j j j j j
j j
n m u L n m u L uM
p L n kT L kT m
kT m M
j
This means
the 2nd term in 5.24 can be neglected if M 1, i.e., flow.
Similarly term1/term 3:
/ ' / ' / '' 5.25/
jj j j j
jjj j j j j
j j j j
Lun m u u L L
MkTn kT L kT kT kTm m m m
subsonic
5.3 Transport in a weakly ionized plasma (3)
This means the first term in (5.23) can be neglected if Mj << 1, and ’, the time constant for the plasma processes, is long. Neglecting the time derivative eliminates the description of plasma waves.
Summary:
The diffusion approximation is valid for a slowly varying subsonic flow.
5.3 Transport in a weakly ionized plasma (4)
Consider a special case: no neutral wind (un = 0) and a dominant electric force due to an external field E0.
0
0
j
0
0 0
5.26
For an isothermal plasma (T = const):
, or
5.27
where
is the diffusion constant as de
j j j j j jn j
j j j j j jn j
j j j j j jn j
j j jj j j
j jn j jn
jj
j jn
p n e n m
n kT n e m
kT n n e m
kT n en D n
m m
kTD
m
E u
E Γ
E Γ
Γ E E
0
fined before, and
is called the mobility coefficient.
The +sign in 5.27 is valid for j = i, and the -sign for j = e. In the absence of the
electric field, E 0, (5.27) reduces to Fick's law
j jj
j jn
n e
m
derived from simple mean-free-
path arguments
5.30j jD n Γ
5.3 Transport in a weakly ionized plasma (5)
To analyze the transport in a weakly ionized plasma (Coulomb collisions neglected) we start with the momentum and energy equations
3.58
3 5: 3.59
2 2
With reasonable assumptions valid at least for the height regime 60-160 km
Schunk has shown that one can simplify:
e
s s ss s s s s s s s s
s ss s s s s
Dn m p n e n m
Dt tD E
p pDt t
u Mτ E u B G
q τ u
i
i
2
( ) 5.31m
0 3 ( 5.32
5.31 states that the Lorence force is equal to the friction between ions and neutrals.
It is assumed that the external (superimposed) electric field is
i in i n
n i i i nk T T m
E u B u u
u u
E perpendicular
to the external field, = .B B b E E
5.3 Transport in a weakly ionized plasma (7)
' '2 2 2 2
' i
2
5.35
e, and = is the ion cyclotron frequency (rotations per s).
. 5.363
The relative ion-neutral drift has t
i in cii n
i in ci in ci
n cii
ii n i n i n
Then
e
m
with
B
m
mT T T T
k
u u E E b
E E u B
u u
' '
wo components: the Pederson component, which
is parallel to , and the Hall component which is perpendicular to (and to ).
There is no component parallel to since we had assumed that 0.
The i
E E B
B E
' '2 2 2 2
2' '
2 2 2 2
22'
2 2
on drift is
, and the ion current becomes
5.110
in
i in cii n
i in ci in ci
i i in cii i i n
i in ci in ci
i i ci ini i i n
i in in ci
e
m
n en e
m
n en e
m
u u E E b
J u E E b
J u E
'2 2
2' '
2 2 2 2
2
5.111
with the ion conductivity. 5.112
in
in ci
ci ini i i n i
in ci in ci
i ii
i in
n e
n e
m
E b
J u E E b
5.5 Major Ion Diffusion (1)+Consider a plasma with one major ion, like O in the F region of the ionosphere,
electrons, and one species of neutrals. Let us assume the plasma is stationary, i.e.,
0; the flow is subsonic, i.es t u
e i
// s
., 0; the plasma is neutral, i.e., n =n .
We neglect Coriolis and centrifugul acelleration and the heat flow terms. The momentum
equation 5.50 becomes
p 5.
s s
s s s s s s s s st t ss
n e n m n m
u u
τ E u B G u u
// i // //// // //
// e // //// // //
50 '
Ion and electron momentum equations along the magnetic field:
p 5.51
p 5.52
Note ,
i i i i i i i ie e i i i in n i
e e e e e e e ei i e e e en n e
i e
n e n m n m n m
n e n m n m n m
e e e
τ E G u u u u
τ E G u u u u
// i e //// //
, also from eq. 4.158 , no proof. Add
the two equations:
p p 5.53
i i ie e e ei
i e i i e i i in e en n i
e n m n m
n m m n m m
τ τ G u u
5.5 Major Ion Diffusion (2)
e e
// i e // s s s// //
// i // i // //// //
// //
Neglect terms with m , including which is proportional to m . Then:
p p , or with p =n kT :
n n
e
i i i i i in n i
ii ei n
i i in i i in in i i in
i n
n m n m
kT kT
n m n m n m
τ
τ G u u
τGu u
u u
//// i // //
//// i // //// //
//// i// //
n
We set 2 , then:
2 2 2n. With
n
ii e i e i e
i in i i in i e in i i in
i e p
ip p p pi n a
i in i i in p in i i in i ine
pi n a
i pe
k T T k T T T T
m n m T T n m
T T T
kT kT T kTD
m n m T n m m
T mD
n T
τG
τGu u
u u // // ambipolar diffusion 5.54
2 2ii
p i pkT n kT
τG
5.5 Major Ion Diffusion (3)
// //
If we call r the coordinate along , then 5.54 becomes:
1 1 1
2 2
2Remember , i.e., it becomes large with increasing altitude where 0.
We therefo
pi i i n i
i p p i p a
pa in
i in
Tn m g u u
n r kT T r n kT r D
kTD
m
B
//
re can neglect the last term.
If the magnetic field force is negligible, we can use this equation in any arbitrary
direction, for example the vertical direction. We recall that , i.e.,
g = -g. NergG e
//
//
0p ,
glecting the stress term, we get the classical diffusive equilibrium eq.:
1 1
2
21 1 1, Plasma scale height
If T constant: exp
pi i
i p p
p pip
i p p i
i i op
Tn m g
n r kT T r
T kTnH
n r H T r m g
r rn r n
H
5.11 Electric Currents and Conductivities (1)
e
2' '
2 2 2 2
2
Similarly the elctron current becomes with e , and
5.113
5.115
The total perpendicular current, , is therefo
en
cee
ce ene e n e
en ce en ce
ee
e en
i e
eBe
m
n e
n e
m
J u E E b
J J J
i
2 2' '
2 2 2 2 2 2 2 2
' '
2
2 2
re (for e )
5.116
in en
in
ci in ce eni e n i e i e
in ci en ce in ci en ce
i e n p H
i e n p n H n
p iin ci
e
n n e
n n e
n n e
J u E b E
J u E b E
J u E u B b E u B
2
2 2 2 2 2 2
'
Pederson and Hall conductivity.
I have resubstituted .
en ci in ce ene H i e
en ce in ci en ce
n
E E u B
5.11 Electric Currents and Conductivities (2)
2
2 2 2 2
19 56 1
31
For heights above ~100 km
5.120
1.6 10 5 10since . 9 10
9.1 10
1.42
Notice from 5.120 that the electrons contribute
in
en
ci in enp i H i e
in ci in ci ce
ce cee
cece
eBs
m
f MHz
e
only to the Hall current, not to the Pederson current.
Currents along :
To calculate use 5.23 for the electrons which are much more mobile than the ions and neutrals:
ee e e e e e e
Dn m p n e
Dt
B
u
uτ E u
e
5.23
Neglecting the stress term and the terms containing m on the left side, we get
where I used the fact that 0, since is a vector
e e e e en e n
e e e e e en e n
j j
n m n m
p n e n m
B G u u
E u u
u B u B
e
e
to .
When an E exists, u . Therefore
Since and e
n
e e e e e en e
e e e
e e e e e e e en e
u
p n e n m
p n kT e
n k T kT n n e n m
Β
E u
E u
5.11 Electric Currents and Conductivities (3)
Since the electron mobility is much larger than the ion mobility, the current is carried by the electrons. The field-aligned current is therefore
2 2
2
Setting
parallel electric conductivity
current flow conductivity due to thermal
e e e e e ee e e e e e
e en e en e en e en e e en
ee
e en
ee
e en
n ek kT e n e n e kT n ekn e T n n T
m m m m n e m
n e
m
n ek
m
J u E E
n
2 2
2 2 2 2 2 2 2 2
2
gradients
5.124
Recall:
(for u 0) 5.116
;
Notice 5.124 becomes Ohm's law
in en
ee e e e
e
p H
ci in ce enp i e H i e
in ci en ce in ci en ce
ii
i in
kTn T
n e
n e
m
J E
J E b E
when the term dominateseJ E E
6. Wave Phenomena6.1 General Wave Properties(1)
Following Schunk’s notation, we use index 1 to indicate the electric and magnetic wave fields, E1 and B1, and the plasma variations, 1c, caused by the waves.The direction of the propagating wave is given by the propagation constant K.
To find the plasma waves we must solve Maxwell’s differential equations in the plasma environment.
121 1 0 1 s 1 0
s
1
11
711 0 1 0 0 0
1 s 1 1s
1 , e ;charge density disturbance, 8.85 10 ,
2 0
3
4 , 4 10 , .
where e
c c s
s s
n x SI units permittivity
xt
x x SI units permeabilityt
n u
E
B
BE
EB J
J
6.1 General Wave Properties(2)We solve Maxwell’s equations by taking the curl of (3):
1 1
21 1 1
21 1
1 0 0 0 2
22 1 1
1 1 0 0 02
c1 1
But
and using(4): - results in
6.6
This quation can be easily solved for vacuum where = 0, and = 0. In th
xt
x
t t t
t t
E B
E E E
J EB
E JE E
J
c11
0
22 1
1 0 0 2
0 0
22 1
1 2 2
1 10
is case
0 according to 1 , and
0
1or by setting
10 in vacuum. 6.7
Solution: , Re 6.9i t
t
c
c t
t e
K r
E
EE
EE
E r E
6.1 General Wave Properties(3)
1 10 10
10
11
1
, Re cos
where for simplicity I assumed is real. Recall: cos sin .
From Mawell equation 3
we can get . Notice that when the exponential functions are used,
i t
ix
t e t
e x i x
xt
K rE r E E K r
E
BE
B
1 1 1 1
1 1
application of the operators
and simply means multiplication with i and -i , respectively. Therefore:
-i
This means that is to and to . With the help of Maxwell equa
t
i
K
K E B K E B
B K E
1 1 0
1
1 1
1 1 1 1
21 1 1 1 1
1 1
tion 1
it is easy to show that for 0 the electric field is to :
0 .
Further from
.
Summary: In vacuum , , are orthogonal to
c
c
i
KK
E
K
K E K E
K E B K K E K B
K K E E K B E k B
E B K
each other following the right-hand rule.
6.1 General Wave Properties(4)Poynting Vector
1 1 0 1 1
1 1
The flow of energy carried by an electromagnetic wave in the direction is given by
the Poynting vector:
/ 6.13
Since are sinusoidal time varying functions, is a function of t. In
K
S E B E H
E H S
1 1
0
general we are not interested
in the fast in and out energy fluxes, but want to know the time-averaged flux:
1 2dt where is the wave period. It is easy to verify that when using the e
wT
ww
TT
S E H
1
*1
xponential
notation:
1Re time-avearged Poynting vector 6.14
2 S E H
6.2 Plasma Dynamics(1)
The propagation of waves in a plasma is governed by Maxwell’s equations and the transport equations. We assume that the 5-moment simplified continuity, momentum, and energy equations (5.22a-c) can describe the plasma dynamics in the presence of waves. If we neglect gravity and collisions these equations become (Euler equations):
( ) 0 continuity eq. 6.21
[ ] [ ] 0 momentum eq. 6.22
0 energy eq. 6.23
1 16.21 0
Substitute in 6.23 :
ss s
ss s s s s s s s
s ss s
s s s ss s s s s s s
s s s
nn
t
n m p n et
D pp
Dtn n D n
n n nt n t n n Dt
u
uu u E u B
u
u u u u
0 6.25s s s s s
s
D p p D n
Dt n Dt
6.2 Plasma Dynamics (2)
1
This implies that
0, since 6.26a
1 1
1 1
And 6.26a implies that
.
s s
s
s s s s s s s s s ss s
s s s s s
s s s s s s s ss s s
s s s s s
s
s
D p
Dt
D p D p p D D p p D
Dt Dt Dt Dt Dt
D p p D D p p Dn m n
Dt n m Dt Dt n Dt
pconst
1
6.26
Notice that this is the equation of state of a gas. The value for is =3/5 for adiabatic flow,
and =1 for isothermal flow (Can also be written as .)
From 6.26 :s
s s s
b
p V const
b
pp const
1 6.27s s s s ss s s s s
s s s s s
p n kT kT
n m m
6.2 Plasma Dynamics (3)
Substitute in the momentum equation (6.22):
[ ] [ ] 0 6.28
The continuity equation was
( ) 0 6.21
We must solve these equations together with Maxwell's equations
ss s s s s s s s s s
ss s
n m kT n n et
nn
t
uu u E u B
u
s s
0 0 0 0
to find
n , , and (10 unknowns).
:
1. Solve for equilibrium conditions finding n , , , ( I dropped index s on n and u)
that satisfy the differential equations
u E B
Using Perturbation Technique
u B E
0 0
0 1
0 1
0 1
0 1
.
2. Perturb the equilibrium state of the plasma and assume that this will cause small changes in and .
, , 6.31
, , 6.31
, , 6.31
, , 6.31
n t n n t a
t t b
t t c
t t d
B E
r r
u r u u r
E r E E r
B r B B r
6.2 Plasma Dynamics (4)
0 10 1 0 1
0 10 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
Substitute these perturbed functions into the continuity and momentum equations:
( ) 0
[ ] [ ] 0
Carry out the differencia
s s s s
n nn n
t
n n m kT n n n n et
u u
u uu u u u E E u u B B
1 10 1 1 1 0 1 0 1 0 1
tions remembering that all 0-index terms are constants:
0 6.33
where only first order (linear) terms in 1-index functions were kept. The momentum equation be
n nn n n n n
t t
u u u u u
10 0 0 1 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0
s
0 0 0 0
10 0 1
comes
0.
where e = e for ions/electrons. But
0 (equilibrium condition). Therefore
s s s s s s s s s
s
n m n m kT n n e n e n e n e n e n e n et
n e
n mt
uu u E E E u B u B u B u B
E u B
uu u
1 0 1 1 0 0 1 1 0 0 0
10 0 1 1 0 1 1 0 0 1
0.
But the last 0, therefore
0 6.35
s s s s
s s s
kT n n e n e
n m kT n n et
E u B u B E u B
uu u E u B u B
6.2 Plasma Dynamics (5)
( )1 1 1 1
1 0 1 0 1
0 1 0 1
0 1 0 1 1 0 1 1 0 0 1
0 1
We try plane wave solutions for all functions
, , , .
Remember , .t
0
6.37
0
i t
s s s
s
n e
i i
i n n i i n
n n
n m i i kT i n n e
kTi i
k ru E B
K
K u u K
u K K u
u u K u K E u B u B
u K u K
11 1 0 0 1
0
1 1 1
0 6.38
6.37 and 6.38 are 4 algebraic equations that must be satisfied for 6.36 to be solutions.
We also must make use of Maxwell's equations to solve for the unknowns
n , , ,
s sn e
n m m E u B u B
u E B
1 1 1 1 1
22 1 1
1 1 0 0 02
2 2
1 1 0 0 1 0 1 0 0 2
22
1 1 0 12
; n , can have different values for the different species, n , .
From slide 6.1(2):
6.6
1;
6.20
These
s s
t t
i i i i ic
K ic
u u
E JE E
K E K K E E J
E K K E J
1 1 0 1 s 1 1 s 1s 0
1 1 1
11 1 1
are 3 more algebraic equations. from 6.1 1 :
11 , e e . One more algebraic eq.
2 0 0. Only tells that .
3 . Three more algebraic eqs.
c c s ss
n i n
xt
E K E
B K B B K
BE K E B
Electrostatic Waves: B1= 06.3 Electron Plasma Waves (1)
1
i1
We start the discussion with high frequency electron plasma waves assuming that B 0.
The wave frequency is high enough so that the ions cannot follow the motion, i.e., 0.
To simplify the discussion
u
i0 e0 0
1 0 1
11 1
0
we also assume 0, and 0, then the algebraic
electron transport equations 6.37 and 6.38 become with - :
6.39
0 6.39
From Gauss's law (see slide 6
o
s
e e e
e e ee
e e e
e e
n n a
kT n ei i b
n m m
u u E B
K u
u K E
1 1 0
1 1
11
.2(5)):
i / 6.39
Our immediate goal is to find the dispersion relation that relates K and .
Muliply 6.39 with and use 6.39 and 6.39 to substitute for and :
e
e
e e ee
en c
b a b
kT ni i
K E
K K u K E
K u K K
10
21 11 0
0 0
22 2 0
10
0 6.40
/ 0
0 6.41
e e e
e e e ee
e e e e
s s ee
e e
e
n m m
i n kT n eiK en
n n m im
kT e nn K
m m
K E
6.3 Electron Plasma Waves (2) (B1=0)
22 2 0
0
22 20
0
2 2 2 2e
20
0
This gives the dispersion relation
0, or
, or
; usually is set equal to 3. 6.42
plasma frequency; electron thermal s
e e e
e e
e e e
e e
p e e
e ep e
e e
kT e nK
m m
e n kTK
m m
V K
e n kTV
m m
2 2
peed.
The dispersion relation 6.42 relates with the wavelength (=2 /K). Notice there is no
propagating wave in a cold plasma where 0. In the cold plasma
plasma oscillation 6.45
e
p
T
6.4 Ion-Acoustic Waves (1) (B1=0)
1 0 1
11 1
0
We now consider the low frequency waves for which the ion motion must be included. The
ion transport equations are, similar to 6.39 , with :
6.46
0 6.46
In G
s
i i i
i i ii
i i i
e e
n n a
kT n ei i b
n m m
K u
u K E
1 1 1 0
2 2 1
0
auss's law (see slide 6.2(5)) we must include positive and negative charges:
i ( ) / . 6.46
Looking back at slide 6.3 (1), equation (6.40) for the electron motion:i e
e e e
e e
e n n c
kT n ei K
m n m
K E
1
2 2 01 1
2 2 01 1
0 0
0 or 6.40
0 6.48
Similarly from the ion transport equations we get from 6.46b
0 6.49
Assuming = (neutral plasma), and adding 6.48 and 6.48 gives
e
ee e e e
ii i i i
e i
enm K kT n
i
enm K kT n
in n
K E
K E
K E
2 21 1 1 1
2 2 21 1
( ) ( ) 0. Since :
( ) 0 6.50
i e e i i e e e i ei
i i i e e e
mn m n K kTn kT n m m
m K kT n K kT n
6.4 Ion-Acoustic Waves (2) (B1=0)
1 1
1 1 1 0
2 2 01 1
2 21 1 0
0
How do and relate? we can combine Gauss's law and the momentum equation for the electrons
given above as
i ( ) / . 6.46
0 6.48
1( ) /
i e
i e
ee e e e
i e e e ee
n n
e n n c
enm K kT n
i
e n n m K kTen
K E
K E
1
2
20
1 1 20
1 01 D2 2 2
0
2 2
6.51
If we neglect the electron inertia term again for low frequencies:
1 0
with , the Debye length 6.521
Substitute into 6.50
( )
e
e
e ei e
e
i ee
e D e
i i
n
m
K kTn n
e n
n kTn
K e n
m K kT n
21 1
2 2
2 2
22 2 D
0,
gives the dispersion relation for ion plasma waves:
6.531
2For very long waves = 1. We get the dispersion relation for ion acoustic wa
i e e e
i e e
i i e D
D
K kT n
kT kTK
m m K
K
2 2 2 2
ves, or ion sound waves:
ion acoustic waves 6.55
ion acoustic speed 6.56
i e es
i
i e es
i
kT kTK K V
m
kT kTV
m
6.5 Upper Hybrid Oscillations
0 0
e1 0 1
1 1
Assume 0, and 0.Upper hybrid oscillations are directed .
We start again with the electron continuity 6.37 and momentum 6.38 equations:
n 657e e
e e
n a
ei
m
0B E high frequency oscillations B
K u
u E u
1 0
1 1 0
1 1 1
1 1
0 6.57
We also use two Maxwell equations:
i / Gauss's Law 6.57
i where 0 Faraday's Law 6.57
0 6.57
The dispersion relation for the upper hybrid oscillation becomes
e
b
en c
i d
e
B
K E
K E B B
K E K E
2 2 2
2 22 20 0
0
:
6.62
,
and K are not related, so there is no wave velocity defined. .
pe ce
ce pee e
eB n ewhere
m m
There is no wave
6.6 Lower Hybrid Oscillations
0 0Assume again 0, and 0. Lower hybrid oscillations are directed .
We must use the electron and ion continuity 6.37 and momentum 6.38 equations together with
0B E lowfrequency electrostatic oscillations B
0 0
2
the two
Maxwell equations. This gives the algebraic equation:
6.68
where
,
are the electron-cyclotron and ion-cyclotron frequencies. Again, .
ce cie i
ce ci
eB eB
m m
no waves
6.7 Ion-Cyclotron Waves
0 0
0
2 2 2
Assume again 0, and 0.The ion-cyclotron waves are
that propagate in a direction to . The algebraic dispersion relation becomes:
ci K
B E low frequency electrostatic waves
almost perpendicular B
2
2
6.74
where
s
e es
i
V
kTV
m
6.8 Electromagnetic Waves in a Plasma (1)
Now we consider the case where E1 and B1are non-zero. We start with the general wave equation (6.20) assuming again a plane wave solution:
22
1 1 0 12
1 1
22
1 0 12
6.20
Let's look first for transverse waves, i.e., forsolutions for which 0 :
6.75
In a two-component plasma (electrons and one ion species) th
K ic
K ic
E K K E J
K E K E
E J
0 1
1
0 1
0 0
0 0 1 1 1
e current density is given by
6.76
Perturbation and linearization:
/ for ions/electrons. Charge neutrality:
i i e e
s so s io eo o
s s s
i i e e
i e
i e i
en en
n n n s i e n n n
en en
en
en en
J u u
J J J
u u u
J u u
J u u
J J u u u
0 1 0
1 0 1 1 1 0 1 0
, . .,
6.80
i e e
i e i i e e
en i e
en en en
u
J u u u u
6.8 Electromagnetic Waves in a Plasma (2)
0 0 0 0
1 0 1 1
1
1 0 1
10 1 1 1 0 0 1
0
To simplify the notation, we assume 0.Then
High frequency approximation: 0
6.81
The electron momentum equation 6.38
i e i e
i e
i
e
s s s
T T
en
en
kT n ei i
n m m
E B u u
J u u
u
J u
u K u K E u B u B
1 1
20
1 1
2 22 0
1 0 12
1
2 22 0
1 02
2 2 2 2
0 becomes
0, therefore 6.82
Substitute into (6.75):
6.83
Since 0, this gives the dispersion relation
, or
6.84
ee
e
e
e
pe
ei
m
n ei
m
n eK
c m
n eK
c m
c K
u E
J E
E E
E
6.8 Electromagnetic Waves in a Plasma (3)
2 22
2 2 2
The phase velocity of high frequency EM waves is
1 .
The group velocity is
./
pe peph
gph
v c c cK K c K
c cv c c c
K K v
6.9 Ordinary and Extraordinary Waves (1)
0
0 0
0 0 0
We now look for a high frequency EM wave solutions in a plasma with 0
and begin with waves perpendicular to , i.e., . Again we assume
E 0, , and we neglect ion motion. We will get die e iT n n
B
B K B
1 0 1 0 1 0
0
1 0
fferent solutions
depending on whether or . The wave with is called the
ordinary wave since has no effect on the wave propagation. The wave with
is called the extraordinary wa
E B E B E B
B
E B
ve.
: in ionospheric radio science the terms ordinary and extraordinary waves are used for waves with left-hand and right hand elliptical polarizations.Note
6.9 Ordinary and Extraordinary Waves (2)
We can use the following equations:
22
1 1 0 12
1 0 1
1 1 1 0
1 0 1
0 1 1 0 1
1 0 1
6.20
6.39
0 from 6.38
6.81
For the ordinary wave since 0.
Since accelarates the electron along , then
e e e
e ee
e
e
K ic
n n a
ei
m
en
E K K E J
K u
u E u B
J u
B E K E K B K E
E B u B
0 1 0
2 2 2 2
1 0 1
1 1 1
=0
The equation are therefore the same as for the case , and the dispersion relation relation is
again given by 6.84 :
For the extraordinary wave 0.
Set = and start
e
pec K
u B
E B K E
E E E
2 2
2 2 2 2
2 2 2
cranking. The result is
6.96pepe
pe ce
c K
6.10 L and R Waves
0
22
1 1 0 12
1 0 1
1 1 1 0
1 0 1
1
Consider high frequency transverse EM waves propagating parallel to . Using the equations
6.20
6.39
0 from 6.38
6.81
0.
This sy
e e e
e ee
e
K ic
n n a
ei
m
en
B
E K K E J
K u
u E u B
J u
K E
0
( )1 10
stem of equations has two solutions. If we orient the coordinate system such that and
point in the direction of the the z-axis, then
right-hand circularly polarized wave R wi Kz tR E x iy e
B K
E
( )1 10
22 2 2
22 2 2
ave, also e-wave
left-hand circularly polarized wave (L wave, also i-wave)
And the respective dispersion relations are:
R wave 6.1021 /
L wa1 /
i Kz tR
pe
ce
pe
ce
E x iy e
c K a
c K
E
ve 6.102b
6.11 Alfvén and Magnetosonic Waves
Low frequency transverse (i.e. )electromagnetic waves are called:
Alfvén waves, if
magnetosonic waves, if
The dispersion relations are, respectively:
0K B
0K B
1 E K
2 2 2
2 22 2
2 2
2 0
0
6.103
6.1041 /
where
A
s A
A
Aio i
K V
V VK
V c
BV
n m
9. Ionization and Energy Exchange Processes1. Solar extreme ultraviolet (EUV) is the major source of energy
input into the thermospheres/ionospheres in the solar system.
2. Electron precipitation contributes near the magnetic poles.
3. > 90 nm causes dissociation (O2 O+O)
4. nm causes ionization
5. Energy losses (sinks), from the ionosphere point of view, are airglow and the heating of the neutral atmosphere (thermosphere)
6. Energy flow diagram in Fig. 9.1
9.1 Absorption of Solar Radiation in the Thermosphere (1)
The photon flux I(s) decreases along its path s through the atmosphere. The flux change dI due to path
element ds is
9.1
9.1
where , the constant of proportionality, is usual
a
a
dI s I s n s ds a
dI s I s n s ds b
ly called the absorption cross section. Of course,
depends on the wavelength . To understand the principle of "layer formation" we assume monochromatic
radiation for our dicussion. The positive di
a
rection for ds is from the sun towards the earth. If the solar rays
have an angle with the vertical (solar zenith angle, Fig. 9.2), then
sec 9.2
sec 9.1
sec , or after integr
a
a
ds dz
dI z I z n z dz c
dIn dz
I
ation from z to :
Iln sec 9.3
I z
I z exp sec
a
z
a
z
n z dz
I n z dz
9.1 Absorption of Solar Radiation in the Thermosphere (2)
00
To get an estimate for the variation of I with z, we assume an exponential decrease of n with height:
exp 9.4
and assume , H, and are constant. Then the or a
z zn z n z
H
optical depth optical thi
a
becomes:
sec sec
, exp sec 9.5
In general:
, , exp , , 9.9
Here is added to the variables because I (I at top of thermosphere) and are functions o
a a
z
a
ckness
n z dz n z H
I z I n z H
I z I z
0 0
0 00 0
f .
The simple realtion 9.4 holds only for an isothermal atmosphere. If T is not constant we can write
exp exp ./
is the neutral gas scale hei
z z
z z
T z T zdz dzn z n z n z
T z kT mg T z H
kTH
mg
ght.
9.1 Absorption of Solar Radiation in the Thermosphere (3)
0
0 0 0
0 0 0
0
0 00 0 0
00
0
can be calculated by setting / . Then
exp exp
z
z
z
n z dz d dz H
kTn z dz n H d n d
mg
T z kT zkTn z d d n z d
T z mg mg
kT zn z
mg
n z dz n z
The vertical columnar content
a
0
9.16
To derive this equation equations we only had assumed that g does not vary with height.
If we assume that is height independent, the optical depth for plane geometry becomes
z
n z dz n z H z
z
H z
a a a
as
, , sec sec sec .
, , sec . 9.17
For a gas with different constituents:
s s
z z
z n z H z
n z dz n z dz n z H z
9.3 Photoionization (1)Photons that exceed the ionization energy threshold can produce electron-ion pairs when
absorbed by neutral particles. Th excess energy is transfered to the kinetic energy of the
electron, or used to
a
excite the ion. If the probability for an absorbed photon to produce an
electron-ion pair is , the production rate is
, , 9.21
with
, , sec .
Sidney Chapman (1932) was th
a aCP z I z n z I n z e
z n z H z
e first to derive this . The two
factors in the product of opposite trends: with decreasing height, n increases while
decreases because increases. At what height
Chapman production function
n z e e
a
a
C
C
secC
sec a
amax
will P be a maximum?
Answer: at the height where dP / 0.
dP d d0
For an isothermal gas (H = const):
1 sec 0
1 sec 0
n z Ha a
n z H
dz
I n z e I n z nedz dz dz
dnI n z e n z H
dz
n z H n z
1amax
amax max
sec
Notice at the height of max production sec 1
H
n z H
9.3 Photoionization (2)
max 0
max 0
max
max 0
a0 max 0 0a
11
max max max a
max 1
But for constant H:
n z
1ln sec 9.22
sec
From (9.21):
,sec
cos, 9.23
The maximum production
z z
H
z z
H
aa a
C
C
n e
n e z z H n HH
I eP z I n z e I n z e
H
IP z
He
rate occurs at local noon.
The height of max production increases with according to 9.22 . This has ben observed for the
E and F1 layers in Earth's ionosphere
11.4 Ionospheric LayersEarth's ionosphere is usually stratified in D, E, F1, and F2 layers during the daytime. At night,
only the F2 layer survives. When transport processes can be neglected, the ion continuity equation
bec
e
e
2
omes
dn11.49
dtIn equilibrium, i.e., during most of the daytime
The loss is caused by recombination of electrons with ions. If we assume one dominant ion, then