Chapter 5
Ionospheric currents
Agenda:

Based on the momentum equation for ions and electrons we are going to derive expressions for the
Hall- and the Petersen conductivities.


We are going to study how the current density vector j rotates with height.
We are going to discuss the physical limits for electric conductivity at the bottom side E-layer and
topside within the F-layer

The simplest expression for current density is given as:

 
j  ne e vi  ve 
A
 1 m
 m 3 C s  m 2 
(1)

j is proportional with the electron density ne, and the difference in ion and electron velocity. In this lecture
we are going to establish the following equation for the current density
  
j  E
(2)
Momentum equation for ions and electrons

  
 
dvi
ni mi
 N i e E  vi  B  ni mg  ρi  ni mi γi vi  vn  (3)
dt

  
 
dve
ne me
  N e e E  vi  B  ne mg  ρe  ne me γe ve  vn  (4)
dt




For the E-region and F-region of the ionosphere we can neglect:



Pressure force
Gravity force
Acceleration term (*)
______________________________________________________________________________________
(*) Argument
Simple model for the collision frequency:
1
 in  2.6 1015  M  1/ 2 N n 
N n 
 en  5.4 1016 T 1/ 2
nn   5 1017 m 3 Neutral density in the E-region.
T  250K
M  29u Mixture of N2 , O and NO
=>  in  250 s and
1
 en  4300s 1
Assume that the acceleration term can be neglected:

 
dvi
 mi i (vi  vn )
dt
mi
vi
  invi (Neglect the neutral wind)
Ti
Ti 
Te 
1
 in
1
 en
 4 10 -3 s
 2.3 10  4 s
Normally we are not interested in time-scales less than a second, and the acceleration term is negligible.
______________________________________________________________________________________
Simplified momentum equation

vi 
e
mi γin

e
ve  
me γen

E  v  B  v


i
(5)
n
E  v  B  v


e

(6)
n


 
vi ,e  B and vi ,e  B
    
 
E  B  vi  B  B  vn  B
In order to single vi and ve we need to find

 
vi  B 

 
e
mi γin
    

 


e
vi  B 
E  B  vi  B B  B2 vi  v n  B
m i γ in
    
 
 
e
vi  B 
E  B  vi  B  B  vn  B
mi  in
 
e    
vi  B 
E  B  vn  B
mi  in






(7)
 
Insert equation 8 in equation 7:
2
(8)
 
vi  B 
   
e     e     
E  B  u  B  B  B 2 vi   vn  B (9)
 E  B  
mi γin 
 mi γin


Then insert equation 9 in equation 5:

vi 
Isolate
e
mi γin

e
 
E
mi γin


    e     
  
 
2
EB 


E

B

v

B
B

B
v
n
i   vn  B   vn
mγ



 i in





vi on the left hand side:





  
 e2 B 2  
  
e
e2  
e3
e2
e   
1  2 2  vi 
E  2 2 E  B  3 3 E  B B  2 2 vn  B B 
vn  B  vn
mi γin
mi γin
mi γin
mi γin
mi γin
 mi γin 
Introduce the ion gyro-frequency given by
i 
eB
mi





 
  
 
2
3
2 
  ω 2  


ω
ω
ω
ω
v

B
ω
v

E
E

B
E

B
B
i
i
i
i
i
n
i
n B
1     v 
 2
 2
 3 

 vn
i
2
3
2



  γin  
γin B
γin B
γin B
γin  B
γin B




vi 




 
    
   
 ωi   
γin2
ωi2
ωi3
E

v

B

E

B

v

B
B

(
E
 B) B  u 
n
n

ωi2  γin2  γin B
γin2 B 2
γin3 B 3


ω γ
vi  2 i in2
ωi  γin


 
  
    
  
 E  vn  B 
ωi2
E  B  vn  B B
ωi3
γin2 
(E  B) B

 2 2


vn

 ω γ
B
B2
(ωi2  γin2 )γin
B3
ωi2  γin2
i
in


______________________________________________________________________________________
Let us no decompose the electric field in one component along the magnetic field and one component
perpendicular to the magnetic field:

E  E||  E
   

    
E  vn  B  B  E   B  vn  B B  vn B 2
    
    
   
E  B  vn  B B  E  vn  B  B  vn B 2  E   vn  B  B
  
 E   vn  B  B
  

E  B B  E|| B 2














_____________________________________________________________________________
3




 
  
E||
ωi γ in E   v n  B
ωi2
E  vn  B  B
ωi γ in E pa
 i3

vi  2

 2


B
ωi  γ in2
ωi2  γ in2
B2
ωi  γ in2 B
ωi2  γ in2 γ in B

ωi2
ω γ
2
i
2
in


vn 
 
vi  vn 
ν in2
ω γ
2
i
2
in





vn




ωi γin E'
ωi2 E'  B
ωi2 E||
 2 2
 2
ωi2  γin2 B
ωi  γin B 2
γin B
where

E'

 
E  vn  B

We have now obtained the following expressions for the electron and ion velocity:

vi

ve




ωi γin E' 
ωi2 E'   B
ωi E||

 vn  2
 2

γin B
ωi  γin2 B
ωi  γin2 B 2




ωe γen E'
ωe2 E'  B
ωe E||
 vn  2 2
 2 2

ωe  γen B
ωe  γen B 2
γen B
(10)
(11)
Inserting Eq. 10 and 11 into Eq. 1 we get

j 


 
2
2


E






ω
γ
ω
γ
ω
ω
ω
ω
E
'
B

E
'

|| 

ne e  2 e en 2  2 i in 2     2 e 2  2 i 2 
  e  i  
2
νin  B 
ωi  γin  B
ωi  γin  B

 γen
 ωe  γen
 ωe  γen

which can be written on the form

j

 


B  E' 
σ p E'   σ H
 σ|| E||
B
(12)
where
σP
σH

ωγ 
e  ωe γen
ne  2
 2i i2 
2
B  ωe  γen
ωi  γin 
e
ne αe  αi 
B
(13)

ωi2 
e  ωe2
ne 
 2 2 
B  ωe2  γen2
ωi  γin 
e
ne  βe  βi 
B
(14)
σ P and σ H are the Pedersen and Hall conductivities, respectively. The Pedersen conductivity is
associated with the Pedersen current along the electric field but perpendicular to the magnetic field. The
Hall conductivity is associated with the Hall current perpendicular to the magnetic and the electric field.
The  e,i and  e,i on the right are the Pedersen and Hall mobility coefficients for electrons and ions:
4
e

i

e

i

 e  en
 e2   en2
 i  in
 i2   in2
 e2
 e2   en2
 i2
 i2   in2
Figure 5.2 shows the altitude variation of these coefficients. By inspection of figure 5.2 we can easily
quantify the Hall and Pedersen mobility coefficients, k H   e  i and k P   e   i , at three different
altitudes:
90km
125km
180km
e
i
e
i
kP  e  i
k H   e  i
0
0
0
0
0.5
0
1
1
1
0
0.5
1
0
0.5
0
1
0.5
0
Table 5.1
Let us now consider how do


vi and ve rotates with height.
Assume that E|| = 0, vn = 0.
Equations (10) and (11) then become

vi


ve




ωi γin E 
ωi2 E   B
 2
ωi2  γin2 B
ωi  γin2 B 2



ωe γen E 
ωe2 E   B
 2
 2
ωe  γen2 B
ωe  γen2 B 2
tan  i 
i
 in
(15)
tan  e 
e
 en
(16)
Figure 5.1 shows the electron-neutral and ion-neutral collision frequency and the ion and electron gyro
frequency versus height. By reading out values from this figure 5.1 we easily find the angle θe and θi for
three different altitudes as shown in Table 5.2 below.
5
80 km
γen = ωe
γen >> ωi
=>θe=45°
=>θi=0°
125 km
γen << ωe
γen = ωe
=> θe=90°
=> θi=45°
200 km
γen << ωe
γen << ωi
=> θe=90°
=> θe=90°
Table 5.2: The ion vector rotates anti-clockwise and is indicated by a dashed arrow. The electron vector is
fill line arrow rotating clock-wise.
Please notice that


vi and ve are in the same direction and have the same magnitude above 200 km.
Therefore now current according to Eq. 1. Also according to Table 5.1, the electrons and ions have no
mobility across the magnetic field at this altitude. At 80 km there will be practically now current because
the electron density is very low.
Let us now organize Eq. 12 on a matrix form
 j1    P
  
 j2    0
j   0
 3 
0
H
0
 E1 
 
 E 2 
 pa  E3 
0
0
 



B  E 
where E1  E , E2 
, E3  E||
B
However, this is not a convenient co-ordinate system as the direction of the E-field is a subject to rapid
variations. Let us therefore introduce a co-ordinate system where x is magnetic north, y is negative east and
z is always the magnetic field.

B 

E 

E 
Bzˆ
E x xˆ  E y yˆ  E z zˆ
E x xˆ  E y yˆ
6
Inserted in equation 12

Bẑ  (E x x̂  E y ŷ)
j  σ P (E x x̂  E y ŷ)  σ H
 σ || E z ẑ  (σ P E x  σ H E y )x̂  (σ p E y  σ H E x )ŷ  σ || E z ẑ
B
 jx    P
  
 j y    H
j   0
 z 
 H
P
0
Version JM 10.09.2008
Figure 5.1
7
 E x 
 
 E y 
 ||  E z 
0
0
Figure 5.2
8