Vlasov theory with magnetic effects
S.M.Lea
January 2007
When we include magnetic ®elds in the Vlasov equation several interesting new effects
arise. These include cyclotron damping and waves at harmonics of ! c (the Bernstein
modes).
The Vlasov equation, now with magnetic ®eld included. is:
³
´
@f
~ + q E
~ + ~v £ B
~ ¢ @f = 0
+ ~v ¢ rf
@t
m
@~v
~ 0 6= 0 but E
~ 0 = 0: When we perturb, there are perturbations to both B and E :
Initially B
³
´
@f 1
~ 1+ q E
~ 1 + ~v £ B
~ 1 ¢ @f0 + q ~v £ B
~ 0 ¢ @f 1 = 0
+ ~v ¢ rf
@t
m
@~v
m
@~v
Or, equivalently:
´
df1
q ³~
~ 1 ¢ @f0
=¡
E 1 + ~v £ B
(1)
dt
m
@~v
where df1 =dt is the total time derivative of f 1 taken along the unperturbed orbit in the
~ 0 = B0 ^
magnetic ®eld B
z: We can ®nd a formal solution for f1 by integrating along the
orbit.
Z t
Z t
´
df 1 0
q ³~
~ 1 ¢ @f0 dt0
dt
=
f
(t)
¡
f
(t
)
=
¡
E
+
~
v
£
B
(2)
1
1
0
1
0
m
@~v
t0 dt
t0
The unperturbed trajectory is described by:
and
v z (t0 ) = vz (t) = v k
(3)
vy (t0) = v ? cos [!c (t0 ¡ t) + Á]
(4)
v x (t0 ) = v ? sin [! c (t0 ¡ t) + Á]
(5)
(This works for both signs of charge if we keep the sign of the charge in ! c .) We can
integrate the z-component directly to get
z (t0) ¡ z (t) = vk (t0 ¡ t)
(6)
while for the x and y components we get:
v?
y (t0 ) ¡ y (t) =
fsin [! c (t0 ¡ t) + Á] ¡ sin Ág = rL fsin [(! c (t0 ¡ t) + Á) ¡ sin Á]g
!c
(7)
and similarly for x:
We wonzt have time to explore all the possible waves, so ®rst letzs look at electrostatic
1
~ 1 ´ 0 and ~kkE
~ : We also consider propagation across the
(longitudinal) waves so that B
magnetic ®eld, in the y¡direction, so that
~ = E a exp [iky (t0) ¡ i!t0] y^
E
Then the solution for f1 (2) becomes:
Z
t
q ~ @f0 0
E1 ¢
dt
m
@~v
t0
Z t
0
0
q
~ a eiky(t )¡i!t ¢ @f0 dt0
¡
E
m t0
@~v
f 1 (t) ¡ f 1 (t0 ) =
=
¡
Using expression (7) for y (t0 ), the exponential is:
exp (ik [rL fsin (! c (t0 ¡ t) + Á) ¡ sin Ág + y (t)] ¡ i!t0 )
Now let ¿ = t0 ¡ t:
=
We have:
¡ ikrL sin Á iky( t)
e
e
exp [ikrL fsin (! c ¿ + Á)g ¡ i! (¿ + t)]
¡ ikrL sin Á iky( t)¡i!t
e
e
exp [ikrL fsin (! c ¿ + Á)g ¡ i!¿ ]
If the wave is damped, we can take t0 ! 1; with f1 (t) ! 0 as t ! 1 to get:
Z t
q ~ @f 0 0
f 1 (t) =
¡ E
dt
1¢
m
@~v
1
while for a growing wave, we let t0 ! ¡1; with f 1 (t) ! 0 as t ! ¡1 to get:
Z t
q ~ @f0 0
f 1 (t) =
¡ E
dt
1 ¢
m
@~v
¡1
(8)
(9)
Wezll consider the damped wave case (8), but we can get a growing wave with only minor
changes to the analysis. Changing variables to ¿ ; the integral becomes
Z 0
q
@f 0
f 1 (t) = ¡ E a e iky(t)¡ i!t
e¡ikrL sin Á exp [ikrL fsin (! c ¿ + Á)g ¡ i!¿ ]
d¿ (10)
m
@vy
1
We can simplify this expression by making use of the generating function for Bessel
functions (see e.g. Lea 8.93 or Jackson problem 3.16c)
µ µ
¶¶
1
X
kr
1
exp
u¡
=
un Jn (kr)
2
u
n= ¡1
Here we let u = ei(! c ¿ +Á) : Then
1
u ¡ = ei! c (! c ¿ +Á) ¡ e ¡i! c (! c ¿ +Á) = 2i sin (! c ¿ + Á)
u
Thus
·
µ
¶¸
+1
´n
X ³
ikrL
1
exp [ikrL sin (!c ¿ + Á)] = exp
u¡
=
ei(! c ¿ +Á) Jn (krL )
2i
u
n=¡ 1
and so (10) becomes
f1 (t) = ¡
q
E a eiky(t)¡ i!t
m
Z
0
+
1
X
1 n= ¡1
³
ei(! c ¿ +Á)
2
´n
J n (krL )
+1
X
m=¡1
¡
e¡iÁ
¢m
Jm (krL ) e¡ i!¿
(11)
@f0
d¿
@v y
Now we expect f0 to be a function of v ? and vk ; so using cylindrical coordinates in the
velocity space, we have
^
@f 0
@f0
@f 0
@f0
^ @f 0 + µ @f0 + ^
+^
y
+^
z
=?
z
@vx
@v y
@v z
@v?
v? @µ
@v z
@f 0
@f 0
= (^
x cos µ + ^
y sin µ)
+0+^
z
@v?
@v z
from the cylindrical symmetry, where µ is the angle that ~v makes with the x-axis. Thus,
(cf eqn 4):
@f 0
v y @f0
@f0
=
= cos (! c ¿ + Á)
@v y
v? @v ?
@v ?
Putting this expression into eqn (11) we have:
q
f1 (t) = ¡ Ea eiky( t)¡i!t £
m
Z 0 +1
1
X +
X
@f 0
ein(! c ¿ +Á) J n (krL ) e¡imÁ Jm (krL ) e¡ i!¿ cos (! c ¿ + Á)
d¿
@v
?
1
~ v f0
r
= x
^
n=¡ 1 m=¡1
q
= ¡
E a e iky(t)¡ i!t £
2m
Z 0 +
1
+1
i
X
X h
@f0
ei(n+ 1)(! c ¿ +Á) + ei(n¡ 1)(! c ¿ + Á) e¡ imÁ Jn (krL ) Jm (krL ) e¡ i!¿
d¿
@v
?
1 n=¡1 m=¡ 1
We can now do the integration over ¿ easily to get:
f1 (t) = ¡
where
Fnm =
+1
X +1
X
q
@f0
E 0 eiky(t)¡ i!t
Jn (krL ) Jm (krL )
F nm
2m
@v
?
n=¡1 m=¡1
¯0
ei(n+1¡m) Á e[i(n+1)! c ¡i!]¿
ei(n¡1¡ m)Á e[i(n¡1)! c ¡i!]¿ ¯¯
+
¯
i [(n + 1) !c ¡ !]
i [(n ¡ 1) ! c ¡ !]
1
(12)
Since we are considering the damped case (! = ! r ¡ i° with ° > 0), then
¡i!¿ = ¡i! r ¿ ¡ °¿ and the integrated term vanishes as ¿ ! 1: (It is straightforward to
repeat the derivation for the growing wave case ° < 0 starting from equation (9) and show
that the end result is the same.) Then:
½
¾
ei(n+1) Á
e i(n¡1)Á
¡ imÁ
F nm = e
+
i [(n + 1) ! c ¡ !] i [(n ¡ 1) !c ¡ !]
3
Then we have:
+1
X
Jn (krL ) F nm
¡ imÁ
= e
n=¡ 1
+1
X
Jn (krL )
n=¡1
= e¡ imÁ
= e¡ imÁ
½
ei(n+1)Á
ei(n¡1)Á
+
i [(n + 1) ! c ¡ !] i [(n ¡ 1) ! c ¡ !]
+1
X
eiºÁ
[J º¡1 (krL ) + Jº+1 (krL )]
i [º ! c ¡ !]
º=¡ 1
+1
X
eiºÁ
2º
Jº (krL )
i
[º
!
¡
!]
kr
c
L
º=¡ 1
where we have made use of a recursion relation for the Bessel functions (eg equation 3.87 in
Jackson or Lea 8.89). Finally then (12) is:
+1
1
X +
X
q
ei(n¡ m)Á n!c
@f0
iky( t)¡i!t
f1 (t) = ¡ Ea e
Jn (krL ) J m (krL )
m
i
[n!
¡
!]
kv
@v
c
?
?
n=¡ 1 m=¡1
(13)
Next we use Poissonzs equation:
Z
n1 q
q
ikE 1 =
=
f1 dv k v? dv? dÁ
"0
"0
The integration over the azimuthal angle Á is easy:
Z 2¼
ei(n¡m)Á dÁ = 2¼±nm
0
and thus the double sum collapses, leaving:
Z +
· µ
¶¸2
1
X
q2
2¼
n! c
kv?
@f0
ik = ¡
Jn
dv k v? dv?
m"0
i
(n!
¡
!)
kv
!
@v
c
?
c
?
n=¡1
Z · µ
¶¸2
+1
q2 X
n! c
kv?
@f 0
k2 = 2¼
Jn
dv dv ?
m"0
(n! c ¡ !)
!c
@v? k
n= ¡1
Here we wrote the Larmor radius in terms of v? explicitly, since we are integrating over v ? :
Now we can factor the plasma density from the distribution function to get:
¶¸2 ^
Z · µ
+1
X
n! c
kv?
@ f0
2
2
k = 2¼! p
Jn
dvk dv ?
(n!
¡
!)
!
@v
c
c
?
n=¡1
For a Maxwellian, the integral over v k is straightforward, leaving
f^0;?
@ f^0;?
@v ?
2
¯ ¡ ¯v?
=2
e
2¼
=
= ¡¯v? f^0;?
4
¾
where ¯ = m=kT ; and then:
k 2 = ¡!2p
+1
X
n!c
(n! c ¡ !)
n=¡ 1
Z
1
0
·
Jn
µ
kv?
!c
¶¸2
2
¯ 2 v? e¡ ¯v? =2 dv?
where the n = 0 term gives zero. To do the integral we can use the result (G&R 6.633#2)
valid for n > ¡1
Z 1
¡ ¢
2
2
2
Jn2 (x) e ¡x =2¾ xdx = ¾ 2 e¡ ¾ In ¾ 2
0
n
2
where In is a modi®ed Bessel function. Since J¡ n = (¡1) J n ; (J ¡n ) = Jn2 ;
and I¡n = In ; the same result holds
for the negative n terms. . Here we have
p
x = krL = kv ? =! c ; and x=¾ = ¯v ? ; so
¾=p
Thus:
k
2
=
=
x
kv ? =! c
k
= p
=p
¯ v?
¯ v?
¯ !c
Z
n!c
¯! 2c 1 2
2
2
Jn (x) e¡x =2¾ x dx
2
[n!
¡
!]
k
c
0
n=¡1
µ
¶ µ 2 ¶
+1
1 X
n! c
k2
k
¡ 2
exp ¡ 2 In
¯! c
¯! 2c
¸ D n=¡ 1 [n!c ¡ !]
¡! 2p ¯
+1
X
where we used the fact that ¸ 2D = 1=! 2p ¯: Notice that we can write ¾ as:
¾= p
k
kv th
=
= krth
!c
¯! c
p
where rth is the Larmor radius of a particle moving at the thermal speed kT =m:Thus the
dispersion relation is:
1
¡
¢ ¡
¢
1 X
n! c
2
k =
exp ¡k 2 r2th In k 2 r2th
2
[!
¡
n!
]
¸D n= ¡1
c
k 2 ¸2D
=
2
1
X
2
¡
¢ ¡
¢
(n! c )
exp ¡k2 rth2 In k2 r2th
2
2
2
! ¡ n !c
n= 1
(14)
Now if ¾ is large (wavelength small compared with the thermal gyro-radius) we may use the
large argument expansion of the Bessel function
2
¡ ¢
e¾
In ¾ 2 ¼ p
2¼ ¾
and then the dispersion relation simpli®es:
1
X
(n! c )2
1
k2 ¸ 2D = 2
p
2
2
2
! ¡ n !c 2¼krth
n= 1
r
1
X (n!c )2
¼ 3 2
k ¸ D rth =
2
! 2 ¡ n2 ! 2c
n=1
5
(15)
so as k ! 1; ! ! n! c ; the harmonics of the cyclotron frequency. For k large but not
in®nite and ! near n! c ; one term dominates and we have:
r
2
2 (n! c )
2
2
! ¡ (n! c ) =
¼ k3 ¸ 2D rth
The right hand side is positive, so the limit is approached from above.
Now ³
as k !
´ 0; we should use the small argument approximation to the Bessel function:
k 2r2
n
1
th
In ¼ n!
(Lea eqn 8.101). The exponential in eqn (14) is approximately 1 in this
2
case. Then (14) becomes:
µ
¶n
1
2
X
¡
¢
(n! c )
1 k 2 r2th
k2 ¸ 2D = 2
1 ¡ k 2 r2th + ¢ ¢ ¢
(16)
2
2
2
! ¡ n !c n!
2
n= 1
The sum is dominated by the n = 1 term, which gives:
¸ 2D
r2th
1
!2p
! 2 ¡ ! 2c
!2
=
=
¯! 2c
!2
= 2 c 2
2
¯! p
! ¡ !c
1
!2 ¡ ! 2c
=
! 2p
=
! 2p + ! 2c = ! 2UH
the upper hybrid frequency.
However, if the frequency is very close to one of the
resonances at ! = n! c (other than n = 1), then we cannot ignore the resonant term. So
again we get ! ¼ n!c :
In more detail, for n 6= 1 but ! ¼ n! c ; we have, including the n = 1 term,
µ
¶n
! 2c
2 (n! c )2
1 k2 r2th
2 2
i
k ¸D =
+h
2
! 2 ¡ !2c
2
! 2 ¡ (n! c ) n!
h
i
µ 2 2 ¶ n !2 ! 2 ¡ (n! )2
2
c
c
2 (n! c ) 1 k rth
2
! 2 ¡ (n! c )
=
+
2
k2 ¸ 2D n!
(! 2 ¡ ! 2c ) k2 ¸ 2D
£
¤
µ
¶n
2
!2c ! 2 ¡ ! 2c ¡ (n ¡ 1) ! 2c
2 (n! c ) 1 k2 rth2
=
+
2
k2 ¸ 2D n!
(! 2 ¡ ! 2c ) k 2 ¸2D
·
µ
¶
¸
n
! 2c
2n
k2 r2th
(n ¡ 1) ! 2c
=
1
+
¡
(17)
(n ¡ 1)!
2
!2 ¡ !2c
k2 ¸ 2D
Generally if ! » n! c ; the last term in [] is small (» (n + 1)¡1 ), the right hand side is
positive, and so the limit is approached from above.
With ! p = !c ; the plot of ! versus k looks like this:
6
5
4
3
w/wc
2
1
0
1
2
krth 3
4
5
These are the Bernstein modes.
Note: to get the plot we used eqn( 14) with the Bessel function series evaluated to 10
terms.
¡
¢
First write it in terms of dimensionless variables. yn = !2 ¡ n2 ! 2c =n2 ! 2c and
k¸ D = x; so that
2 !2
r2
2 vth p
k2 r2th = k 2 ¸2D th
=
x
= x2 ® 2
!2c vth2
¸2D
where ® = !p =! c : Then for ! ' n!c ; (14) becomes:
¡
¢ ¡
¢
¡
¢ ¡
¢
1
X
1 exp ¡x 2 ® 2 In x 2 ®2
2 exp ¡x 2 ®2 In x2 ® 2
1=2
'
y
x2
yn
x2
n= 1 n
The small xplimit is slightly different if ! UH > 2!c : To investigate this, let
!UH =! c = ° = 1 + ® 2
7
Letzs look at (16) keeping one more term in the exponential.
µ
¶n
1
2
X
¡
¢
(n! c )
1 k2 rth2
2 2
k ¸D = 2
1 ¡ k2 rth2 + ¢ ¢ ¢
! 2 ¡ n2 ! 2c n!
2
n=1
"
¡
¢ µ
¶ n¡ 1 #
1
X
¡
¢
®2 ! 2c
n2 ! 2 ¡ ! 2c 1 k 2rth2
1 =
1+
1 ¡ k2 r2th + ¢ ¢ ¢
2
2
2
2
2
! ¡ !c
! ¡ n ! c n!
2
n=2
" 1 ¡
¢
¡
¢
µ 2 2 ¶ n¡ 1 #
m
1
X ¡k2 r 2
X
n ! 2 ¡ ! 2c
1
k rth
2
2
2 2
2 2
th
! ¡ !c ¡ ® ! c = ® ! c
+
2
2
2
m!
! ¡ n ! c (n ¡ 1)!
2
m=1
n=2
"
¡ 2 2 ¢m
¡ 2
¢
µ
¶ n¡ 1 #
1
1
X ¡k rth
X n ! ¡ ! 2c
1
k 2 r2th
2
2
2 2
! ¡ ! UH = ® ! c
+
m!
! 2 ¡ n2 ! 2c (n ¡ 1)!
2
m=1
n=2
where we will keep the same order of terms in the exponential as in the Bessel function.
2
! ¡°
2
! 2c
2
=®
! 2c
"
¢m
¡
¢
µ 2 2 ¶ n¡ 1 #
1 ¡
1
X
X
¡k2 rth2
n ! 2 ¡ ! 2c
1
k rth
+
2
2
2
m!
! ¡ n ! c (n ¡ 1)!
2
m=1
n=2
If 2 < ° < 3; then we keep the n = 2 and n = 3 terms, and truncate the ®rst sum at m = 2:
"
#
¡ 2
¢
¡
¢
! ¡ ! 2c
3 !2 ¡ ! 2c 1 k2 rth2
k2 r2th
2
2 2
2 2 2 2
! ¡ ° ! c = ® ! c k rth ¡1 +
+ 2
+
2
! ¡ 4!2c
! 2 ¡ 9! 2c 2! 4
"
#
¡ 2
¢
2
2
2 2
2 2
3
!
¡
!
3!
k
r
1
k
r
c
c
th
th
= ®2 ! 2c k2 r2th
+
+
!2 ¡ 4! 2c
2
! 2 ¡ 9!2c 2! 4
Since the ®rst term is positive, we still approach the limit from above.
We also have, including the n = 1 and 2 terms,
"
#
¡ 2
¢
! ¡ !2c 2 2
2
2
2 2
2 2
! ¡ ! c = ® ! c 1 ¡ k rth + 2
k rth
! ¡ 4! 2c
½
¾
3!2c
2 2
2 2
' ® !c 1 + 2
k rth + ¢ ¢ ¢
! ¡ 4! 2c
which does not go to zero as k ! 0: So there is no limit at ! = ! c in this case. However,
"
#
¡
¢
3 ! 2 ¡ !2c 1 k 2 rth2
3!2c
k 2 r2th
2 2 2 2
= ® ! c k rth
+
+
!2 ¡ 4! 2c
2
! 2 ¡ 9!2c 2! 4
¡ 2
¢
3 ! ¡ ! 2c 1 k2 r2th
3! 2c
! 2 ¡ ° 2 !2c
k2 r2th
=
¡
+
! 2 ¡ 4! 2c
®2 ! 2c k2 r2th
2
!2 ¡ 9! 2c 2! 4
! 1 as k ! 0
Thus, since ° > 2;
! 2 ¡ 4! 2c
® 2 ! 2c k2 r2th
!
! 0 from below
3! 2c
! 2 ¡ °2 ! 2c
Thus the plot now looks like this:
8
Reference: Boyd and Sanderson pg 281
See Chen pg 281 for the curves for 3 < ° < 4:
9