Motion of charged particles in EM ¤elds
Jan 07
1
Uniform magnetic ¤eld
The basic motion of a charged particle in a magnetic ¤eld is a helix. Let the
~ be uniform. Then the equation of motion for a particle of
magnetic ¤eld B
charge q and mass m is:
~ = q~v £ B
~ = m~a
F
Thus the acceleration is always perpendicular to the velocity ~v : Compare
with the relation for uniform circular motion:
q ~
~a = ~! £ ~v = ¡ B
£ ~v
m
Thus the angular velocity of the particle is
~! = ¡
q ~
B
m
(1)
The angular speed is !C = qB=m; the cyclotron frequency, and the direction
is along the magnetic ¤eld. The direction of the particlezs rotation depends
on the sign of the particlezs charge. An electron, with q = ¡e; has ~! parallel
~ Ions, with positive charge, rotate in the opposite sense.
to B:
The radius of the helix is given by
v? = !r
where v? is the component of the particlezs velocity perpendicular to the
magnetic ¤eld, and thus
mv?
rL =
(2)
jqj B
This radius is called the Larmor radius. A proton moving at the same speed
as an electron will have a Larmor radius 2000 times bigger than the electronzs
1
Larmor radius. Thus the electrons are more closely tied to the magnetic ¤eld
lines.
The magnetic ¤eld near the surface of a neutron star is about 1012 Gauss,
or 108 T. This corresponds to
!C =
1:6 £ 10¡19 C 8
C
10 T = 1: 8 £ 1019
T
¡31
9 £ 10
kg
kg
Now do the units check out? Since we know that force is qvB; then
C¢T=
and thus
N
m/s
C¢T
kg ¢ m/s2
1
=
=
kg
kg ¢ m/s
s
and so the units check out.
This gyrational motion is the basic motion of charged particles in a
plasma. The center of the circle is called the guiding center. With a uniform
magnetic ¤eld, and no electric ¤eld, the guiding center moves at constant
speed along the ¤eld line.
The circulating charge forms a current loop, and hence has a magnetic
¤eld of its own. This ¤eld produced by the particlezs motion is opposite the
original, uniform ¤eld. Thus the plasma is a diamagnetic material.
2
Uniform magnetic ¤eld plus uniform electric ¤eld
Without a magnetic ¤eld, an electric ¤eld causes a charged particle to ac~ But with a magnetic ¤eld, the particle
celerate along the direction of E:
behaves more like a spinning top, and moves perpendicular to the applied
electric ¤eld. We can understand why by noting that (for positive q) the
~ thus increasing the Larmor raparticle speeds up as it moves parallel to E;
~ v and rL decrease. Thus the
dius (2). As the particle moves opposite E;
circles donzt close, and the particle drifts as shown in the diagram. Here we
~ out of the page and E
~ upward.
show that path of an electron with B
2
~ into components along and perpenNow letzs do the math. Letzs split E
~
dicular to B: Then the equation of motion is:
m
^ we ¤nd
Dotting with b;
³
´
d~v
~ ? + Ek^b + ~v £ B
~
=q E
dt
dvk
= qEk
dt
~ The perpendicular component is:
the same relation as in the absence of B:
m
´
d~v?
q ³~
~
=
E? + ~v? £ B
dt
m
(3)
d (~v? ¡ ~vE )
q
~
= (~v? ¡ ~vE ) £ B
dt
m
(4)
The solution we expect is a combination of a gyration plus a constant drift.
Thus, in the frame moving with the guiding center, we will see only the
gyration, and the equation of motion should look like:
Comparing equations (3) and (4), we have:
~ ? = ¡~vE £ B
~
E
~ to get:
Cross both sides with B
³
´
³
´
~ £B
~ = ¡ ~vE £ B
~ £B
~ = ¡B
~ ~vE ¢ B
~ + ~vE B2
E
3
~ because the cross product selects
We can drop the perpendicular sign on E
only the perpendicular component. And since the drift velocity is perpen~ we have
dicular to B;
~ £B
~
E
~vE =
(5)
B2
The result is independent of both the charge and mass of the particle:
all particles have the same E-cross-B drift velocity. Changing the sign of the
charge changes the direction of gyration, but the particle also accelerates in
the opposite direction, so the direction of the drift is the same. A particle
with smaller mass has a smaller Larmor radius, but gyrates faster. The two
e¤ects combine to create the same drift velocity. Since all particles drift
at the same rate, this drift does not cause any current in the plasma, but
instead causes the whole plasma to drift together.
3
Drifts due to other constant, external forces
~ produces a constant drift in
Any constant external force perpendicular to B
a similar way. In general, the equation of motion is
m
d~v
~
= F~ + q~v £ B
dt
and again we look for a solution in which the particle motion perpendicular
~ is gyration plus drift:
to B
d (~v? ¡ ~vD)
q
~
= (~v? ¡ ~vD) £ B
dt
m
Thus
~?
F
q
~
= ¡ ~vD £ B
m
m
~ gives
and crossing with B
1 ~
~ = ~vD
F £B
qB 2 ?
The motion parallel to the ¤eld is uniformly accelerated motion with
ak =
4
Fk
m
(6)
~ =
One particularly important example is the gravitational drift. With F
m~g ; we get
m
~
~vg =
~g £ B
(7)
qB 2
In this case the drift velocity depends on both the mass and the charge of the
particle. Particles with opposite signs of charge drift in opposite directions:
thus this drift creates a current in the plasma. Since the drift is proportional
to mass, the current is carried primarily by the ions.
The result that a plasma move sideways in a gravitational ¤eld is surprising at ¤rst sight. But suppose the plasma is contained in a vessel of ¤nite
size with non-conducting walls. Then the current leads to a charge build-up
on the walls, and thus creates an electric ¤eld. The resulting E-cross-B drift
is downward, and so the plasma falls as expected.
4
4.1
Motion in a non-uniform ¤eld
~
Gradient perpendicular to B:
~ also change the Larmor radius, and so cause
Variations in the magnitude of B
a drift. As B increases, the Larmor radius decreases, and so for a positively
charged particle the path looks as shown in the ¤gure.
5
~ £ rB:
~
The particle drifts in the direction of B
Now for the math. We shall consider the case in which the length scale
over which B varies appreciably is large compared with the Larmor radius.
Then B changes by a small amount over one orbit. Thus to ¤rst order we
may neglect the change in B during one orbit.
~ = B^
~
Let B
z and rB
be in the x¡direction. Then the particlezs unperturbed path may be decribed by:
~r = y^ rL cos ! ct § x^ rL sin ! ct
(8)
where the choice of sign is determined by the sign of the particlezs charge,
and
~v = §v? cos ! ct ^
x ¡ v? sin !ct y
^
The force acting on the particle has components:
³
¯´
¯
~
Fx = qvyB = q (¡v? sin !c t) B0 + ~r ¢ rB ¯
0¯
µ
¶
@B ¯¯
= ¡qv? sin !c t B0 + (§rL sin !ct) ¢
@x ¯ 0
~ (~r) ; and similarly
where we used a Tylor series expansion for B
¯ ¶
µ
@B ¯¯
Fy = ¡qvxB = ¡q (§v? cos ! ct) B0 §rL sin !c t¢
@ x ¯0
Now we average over one orbit. All terms in sin !c t; cos !c t and sin !c t cos ! ct
average to zero. The term in sin2 !c t averages to 1/2. Thus
< Fy > = 0
and
µ
¯ ¶
¯
1
@B ¯¯
1
@B ¯¯
< Fx > = ¡ qv? §rL¢
= ¡ sign (q) qv?rL
2
@ x ¯0
2
@x ¯0
Now we can use this result in equation (6) to get
< Fx > B
y
^
qB 2
¯
1
v?r L @B ¯¯
= sign (q) 2 B
y
^
2
B
@x ¯ 0
v r ~ ~
= sign (q) ? 2L B
£ rB
2B
~vgrad B = ¡
6
(9)
or, using equation (2) for the Larmor radius, we get
2 ~
~
mv?
B £ rB
2 jqj B B 2
~ £ rB
~
mv2? B
=
2qB
B2
~vgrad B = sign (q)
(10)
where in this last expression q is a signed quantity, negative for electrons and
positive for ions.
Since this drift again depends on the charge-to-mass ratio, we see that
electrons and ions drift in opposite directions, giving rise to a current in the
plasma. This current is also carried primarily by the ions.
Using equation (9), this drift is approximately:
vgrad B »
v? rL
2 L
¯
¯
¯~ ¯
where L is the scale length over which B changes (L = B= ¯rB ¯). This
expression shows that this is a ¤nite Larmor radius e¤ect. The drift depends
on the fact that over one orbit the particle samples di¤erent values of B:
4.2
Curvature drift
When the ¤eld lines are not straight, but curved, a particle moving along the
¤eld line must be experiencing a force and thus will experience a drift. To
compute the drift, we work in the reference frame moving along the ¤eld line
with the particle. In this accelerated reference frame, the particle experiences
the ¤ctitious centrifugal force
~c =
F
mvk2
mvk2
~
r= 2 R
^
Rc
Rc c
where Rc is the radius of curvature of the ¤eld line, vk is the speed along the
¤eld line, and ^r is the unit vector directed outward. Then, from equation
(6), the drift is:
mv2k
1 ~
~ =
~ £B
~
~vC =
F
£
B
R
(11)
qB2
qB2 R2c c
This curvature drift is also proportional to m=q and so gives rise to a current,
carried primarily by the ions.
7
~ £B
~ = 0; and this means that curvature
Now in vacuum we must have r
is always accompanied by gradients in B: To see why, use cylindrical coords
~ locally be described by B
~ = Bµ:
^ Then r
~ £B
~ has only an r
and let B
component:
~ £B
~ = 1 @ (rB) ^z =0
r
r @r
Thus rB = constant, and thus B _ 1=r: Then
~
~ = ¡ Rc B
rB
R2c
and this gives rise to a gradient drift:
~vgrad B
Ã
!
2 ~
~
~
~
B £ rB
mv? B
Rc
=
=
£ ¡ 2B
2
2
2qB
B
2qB B
Rc
mv2?
=
~c
mv2? R
~
£B
2qB 2 R2c
adding this to the curvature drift, we get
µ 2
¶~
m v?
Rc
2
~
~vtotal drift =
+ vk
£B
2
qB
2
R2c
4.3
(12)
Magnetic mirrors
To round out this discussion, we consider the case where the magnetic ¤eld
~ Then necessarily the ¤eld lines cannot
strength varies in the direction of B:
be straight and parallel, but must converge or diverge. We put the z¡axis
along the "central" ¤eld linew the line along which the guiding center moves.
Then the Lorentz force acting on a particle is:
³
´
~ = q~v £ B
~ = q v µ Bz ½
^ µ B½ ^z
F
^+ [vz B½ ¡ v½Bz ] µ¡v
(13)
The z¡component of the force causes the velocity of the particle along the
¤eld line to change. The ½¡component is the usual centripetal force causing
the particle to gyrate around the ¤eld line. The µ¡component has two parts:
the v½ Bz term changes speed of gyration while the drift due to the vz B½ term
causes the particle to track along the converging ¤eld lines.
~vD »
1
B
~ = vz B½ ^
vz B½^
µ£B
µ £ ^z = vz ½ ½
^
2
B
Bz
Bz
8
With the guiding center on the z¡axis,
~ ¢B
~ = 1 @ (½B½) + @Bz = 0
r
½ @½
@z
Thus
½B½ = ¡
Z
½
0
¯
@Bz
1 2 @Bz ¯¯
½
d½ ' ¡ ½
@z
2
@z ¯½=0
and thus for ½ ¿ L = B= j@Bz =@ zj
¯
1 @Bz ¯¯
B½ ' ¡ ½
2 @z ¯½=0
Then from (13)
¯
q
@Bz ¯¯
Fz = vµ rL
2
@z ¯½=0
Now vµ will be §v? ; depending on the sign of the charge (positive charges
go in the ¡µ direction) and using equation (2) we ¤nd
¯
¯
2
q mv2? @Bz ¯¯
mv?
@Bz ¯¯
Fz = ¨
=¡
2 jqj B @z ¯½=0
2B @z ¯½=0
The quantity
mv2?
¹=
(14)
2B
is the magnetic moment of the particle. To see why, recall that the gyrating
particle forms a current loop with current I = q=T = !q=2¼ and the magnetic
moment of the current loop is
IA =
So
2
¼r2L!q
r v q
mv?
= L ? =
= ¹:
2¼
2
2B
¯
@Bz ¯¯
Fz = ¡¹
@z ¯½=0
or, in a coordinate-free notation,
Fk = ¡¹r kB
9
Thus
dvk
= ¡¹rkB
dt
Now let s be the distance travelled along the ¤eld line by the guiding center,
so vk = ds=dt: Multiply both sides of the equation by vk to get:
m
dv
dB ds
vk m k = ¡¹
dt¶
ds dt
µ
d 1 2
dB
mv
= ¡¹
dt 2 k
dt
(15)
The magnetic force does no work, so as the particle moves along the ¤eld
line its energy must remain constant. As vk decreases, v? must increase to
compensate.
µ
¶
µ
¶
d 1 2
d 1
1 2
2
mv =
mv + mv
=0
dt 2
dt 2 k 2 ?
Then, using the de¤nition of ¹ and equation (15), we have
¡¹
dB
d
d¹
+ (¹B) = B
=0
dt
dt
dt
Thus ¹ does not change as the particle moves along the ¤eld line: it is an
invariant of the motion.
Strictly, ¹ is an adiabatic invariant. It remains constant only so long
as the system changes slowly compared with the gyrational period of the
particle. Or, equivalently, rL ¿ L; the length scale over which B changes.
Invariance of ¹ shows that the particle gyrates faster as the magnetic
¤eld strength increases: the parallel velocity decreases. Eventually, if B gets
strong enough, the parallel velocity is reduced to zero. Since Fk does not
depend on the parallel velocity, it continues to act in the same direction,
and accelerates the particle back in the opposite direction: the particle is
re¢ected. This phenomenon is called the magnetic mirror e¤ect.
Not all particles are re¢ected. Obviously a particle with v? = 0 is not
re¢ected. We can ¤nd v? at the turning point from conservation of energy,
and the fact that vk = 0 there. Suppose a particle starts at a point where the
magnetic ¤eld strength is B0 with gyrational velocity v?0; and total speed v0 :
At the turning point v? = v0; the particlezs initial speed. It will be re¢ected
where
2
1 v2?0
1 v?
1
v02
m
= m
= m
2 B0
2 B
2 BTP
10
Thus
B0
v2
= ?0
= sin2 µ
2
BTP
v0
where µ is the pitch angle, the angle between the particlezs velocity vector
and the ¤eld line at the starting point. Since BTP · Bm; the maximum value
of B; this equation shows that particles with small pitch angles will not be
re¢ected. The continual loss of particles with small pitch angles causes the
particlesz velocity distribution to become anisotropic, leaving a hole called
the loss cone. Collisions repopulate the loss cone.
Example: the Van Allen belts
The earthzs ¤eld is essentially a dipole, with B » B0 (r0=r)3 (ignoring the
angular dependence for now). For a particle that starts at 5 earth radii and
reaches the top of the atmosphere (essentially 1 earth radius) then particles
that re¢ect have pitch angle given by
sin2 µ ¸ (r=r0 )3 =
sin µ ¸ 0:09
µ ¸ 5±
1
125
Thus the loss cone is very small. The particles in the loss cone are responsible
for the aurora. I have asked you to investigate this system further in Problem
set 3.
5
Non-uniform electric ¤eld
One of the most important examples of a non-uniform electric ¤eld is the
¤eld produced by a wave, which has the form
~=E
~ 0 cos kx
E
Using Fourier transforms, we can reduce a general variation to a sum of terms
of this form.
~ be uniform, B
~ = B0 ^
Now let B
z; and let
~ = E 0^
E
x cos kx
(a longitudal wave). Then the particle equation of motion is:
³
´
d~v
~ + ~v £ B
~
m
=q E
dt
11
or, in components,
dvx
q
q
= E0 cos kx + vy B0
dt
m
m
dvy
q
= ¡ v x B0
dt
m
and
dvz
=0
dt
Now we take a second derivative, to eliminate vx from the vy equation:
and similarly
³q
´
d2 vy
q
dvx
= ¡ B0
= ¡!c
E0 cos kx + !cvy
dt2
m
dt
m
d2vx
q
dx
dvy
= ¡k E0 sin kx
+ !c
2
dt
m
dt
dt
E0
= ¡k!c vx
sin kx ¡ ! 2c vx
B0
Next we average over a gyrational period to isolate the drift. We assume
~ (krL ¿ 1) and use the unperturbed orbit position (8)
slow variations of E
in our expressions for the acceleration:
sin kx = sin k (x0 § rL sin !t)
(cf section 4.1). Expanding the sine gives:
sin kx = sin kx0 cos (krL sin !t) § cos kx0 sin (krL sin !t)
and with krL ¿ 1;
sin kx = sin kx0
·
¸
1
2
1 ¡ (krL sin !t) § cos kx0 (krL sin !t)
2
to second order in small quantities. When we time average, only the squares
of sin !t; cos !t do not average to zero, and so
µ
¶
1
2
< sin kx > = 1 ¡ (krL ) sin kx0
4
12
and similarly
so
and
µ
¶
1
2
< cos kx > = 1 ¡ (krL ) cos kx0
4
µ
¶
d2vy
1
2
2
2 E0
< 2 > = ¡! c < vy > ¡!c
1 ¡ (krL) cos kx0
dt
B0
4
µ
µ
¶
¶
d2 vx
E0
1
2
< 2 > = ¡ < v x > !c k
1 ¡ (krL) sin kx0 + !c
dt
B0
4
From past experience, we expect the drift to be constant, so <
2
< ddtv2x > = 0; and thus
< vx > = 0
d2 vy
dt2
> =
and
µ
¶
E0
1
2
< vy >= ¡
1 ¡ (krL) cos kx0
B0
4
Wezd like to write this in a coordinate-independent way. Remember that if
~ were constant, we would get
E
~vE =
~ £B
~
E
E0 cos kx
=
¡
y
^
B2
B0
which is the ¤rst term of our result. The second term arises from the non~ Thus:
uniformity of E:
µ
¶ ~
~
1 2 2 E
£B
~vD = 1 + rLr
4
B2
This is another example of a ¤nite Larmor radius e¤ect. The sign of the
charge does not e¤ect the result, but the mass of the particle enters through
rL: A larger mass particle has a larger Larmor radius and thus samples more
of the gradient of E:
6
~
Time-varying, uniform E:
Here we investigate the particle motion in an electric ¤eld of the form
~ = E0ei!tx
E
^;
13
~ = B0^z
B
Again, this might be one Fourier component of a more general time variation.
We may neglect induced magnetic ¤elds provided that the variations are slow.
(Bind » ¹ 0"0 !LE » !L
E where L is the length scale for variation of B: Thus
c2
the magnetic force due to induced B is of order v!L=c2 compared with the
electric force. ) Then the Lorentz force is
³
´
~=q E
~ + ~v £ B
~ = m d~v
F
dt
Thus
¢
dvx
q ¡
=
E0 ei!t + vyB0
dt
m
and
dvy
q
= ¡ vxB0 = ¡!cvx
dt
m
Di¤erentiating again, and eliminating vx; we get
µ
¶
d2vy
dvx
2 E 0 i!t
= ¡!c
= ¡!c
e + vy
dt2
dt
B0
and similarly
µ
¶
d2vx
E0 i!t
= ! c i! e ¡ ! cvx
dt2
B0
~ £B
~ drift, ~vE = ¡ (E0ei!t =B0) ^
Once again we can recognize the E
y = vE ^
y:
Letzs also de¤ne
i! E0 i!t
vP =
e
!c B0
Then
d 2vx
= ¡! 2c (vx ¡ vP )
dt2
and
d2vy
= ¡!2c (vx ¡ vE )
dt2
Here we have to be a little more careful, because both vE and vP are time
dependent. However,
d2vP
= ¡!2 vP
dt2
and provided that ! ¿ !c ;
d 2 (vx ¡ vP )
= ¡!2c (vx ¡ vP ) + ! 2vP ' ¡! 2c (vx ¡ vP )
dt2
14
so we have a gyration in the frame drifting with velocity
~
1 @E
~vP =
(16)
!c B @t
~ £B
~ drift, which is in the y
We also have the E
^ direction. The new drift
with velocity ~vP (equation 16) is called the polarization drift. It depends on
both the charge and mass of the particle, since both appear in !c ; and thus
there is a polarization current :
~
~jP = ne (~vP + ¡ ~vP ¡) = ne [M ¡ (¡m)] @ E
eB 2
@t
~
½ @E
=
(17)
B2 @t
where ½ = n (M + m) is the plasma mass density. Once again the current
is carried primarily by the ions.
This drift is a |start-up e¤ect}. As the electric ¤eld is applied, the
~ and only
particles begin to move along (+ charge) or opposite (¡ charge) E;
then start to gyrate.
7
7.1
More on adiabatic invariance
~
Time-varying B
We know that the magnetic force does no work, and thus does not change the
~ changes in time there is an induced electric
particlezs energy. But when B
¤eld that does do work. From Faradayzs law
~
~ £E
~ = ¡ @B
r
@t
~ is perpendicular to B:
~ The equation of motion is:
and thus the induced E
³
´
d~v
~ + ~v £ B
~
m ?=q E
dt
Now dot this equation with ~v? to get:
³
´
d~v?
~ + ~v £ B
~
m~v? ¢
= q~v? ¢ E
dt¶
µ
d 1 2
~
mv
= q~v? ¢ E
dt 2 ?
15
Now we express v? ; the gyrational velocity, as
~v? =
where
d~`
dt
d~` = rLdµ ^
µ
is a vector element along the Larmor orbit. Thus
µ
¶
~
d 1 2
~ ¢ d`
mv? = q E
dt 2
dt
Integrating over one period, we get
µ
¶ I
1 2
~ ¢ d~`
±
mv
=
qE
2 ?
orbit
~ changes slowly compared with the gyrational period, we
where, provided B
may integrate over the unperturbed orbit. Then applying Stokesz theorem,
we get:
µ
¶
Z ³
Z ~
´
1 2
@B
~
~
±
mv? = q
r £ E ¢ ^n dA = ¡q
¢ ^n dA
2
@t
Note that there is a connection between the directions of d~` and n
^ through the
right hand rule. For positively charged particles that gyrate with ~! opposite
~ ^
~ For negatively charged particles, ^
~
B;
n is also opposite B:
n is parallel to B:
Thus
µ
¶
1 2
@B 2
±
mv?
= ¡q (¨)
¼r
2
@t L
µ
¶
@B
mv? 2
= §
¼q
@t
qB
2
@B 1 mv? 2¼m
= §
@t 2 B qB
@B 2¼
=
¹
@t !c
16
where ! c = jqj B=m is the cyclotron frequency, and ¹ is the magnetic moment
(equation (14)). Thus the change in gyrational energy over one period is
µ
¶
1 2
±
mv
= ± (¹B) = ¹±B
2 ?
where ±B is the change in B over one period. Thus ±¹ = 0; and the magnetic
moment does not change. This is the same result that we obtained in §4.3.
This analysis corresponds to the situation wezd see in the reference frame
moving with the guiding center.
This analysis also shows that the magnetic moment is proportional to the
¢ux enclosed within one Larmor orbit. Thus the particle moves so as to keep
the magnetic ¢ux within its orbit a constant.
The magnetic moment is an adiabtic invariantw it remains constant only
so long as the time scale for variation of B satis¤es ! cT > 2¼: (i.e.!c =! > 1)
|Much greater than 1} is not required (although we have not proved that
here.)
7.2
The second adiabatic invariant.
A particle trapped between two magnetic mirrors, such as a particle in the
earthzs radiation belts, oscillates along the ¤eld lines at the |bounce frequency}. It also drifts around the earth due to grad-B and curvature drift.
If P1 and P2 are the points at which the particle re¢ects at the two mirrors,
the integral
Z
P2
J=
P1
vk ds
remains invariant as the particle drifts. This is a second adiabatic invariant.
~ are slow compared with the bounce
It is invariant so long as changes in B
period. Since this period is much longer than the gyrational period, J is less
likely to be invariant than ¹:
Invariance of J guarantees that a drifting particle returns to the original
¤eld line after drifting through 2¼ radians.
Compare two neighboring ¤eld lines as shown in the diagram.
17
Then
±s
±s0
= 0 = ±µ
Rc
Rc
Thus
±s0 ¡ ±s R0c ¡ Rc
=
¢t±s
¢tRc
The guiding center drift is a combination of grad-B (equation 10) and curvature drift (equation 12). Over most of the bounce period grad-B drift
dominates (vk ¿ v?): The radial component is
~vD ¢ r^ =
~ £ rB
~
~c
mv2? B
R
R0c ¡ Rc
¢
=
2qB
B2
Rc
¢t
Thus
~
±s0 ¡ ±s
mv2? B
Rc =
¢t±s
2qB
~
1 d±s
mv2? B
=
±s dt
2qB
~
£ rB
B2
~
£ rB
B2
~c
R
Rc
~c
R
¢ 2
Rc
¢
We also need to know how vk varies. The total kinetic energy is
1
1
1
W = mvk2 + mv2? = mvk2 + ¹B
2
2
2
18
and thus
r
2
(W ¡ ¹B)
m
where both W and ¹ are constants in this expression. Thus
r
dvk
2 1 ¡¹dB=dt
p
=
dt
m 2 W ¡ ¹B
vk =
and
1 dvk
1 ¡¹dB=dt
¡¹dB=dt
=
=
vk dt
2 (W ¡ ¹B)
mvk2
The magnetic ¤eld B at the particlezs position changes because of the guiding
center motion, so
"
#
~
dB
m
R
c
~ =
~ ¢ rB
~
= ~vgc ¢ rB
v2
£B
dt
qB 2 k R2c
and rearranging the triple scalar product,
Thus
´
~c ³
1 dvk
¡¹ m 2 R
~
~
=
v
¢ B £ rB
vk dt
mvk2 qB2 k R2c
´
2 ~
mv?
Rc ³ ~
1 d±s
~
= ¡
¢
B
£
rB
=¡
3
2
2qB Rc
±s dt
¢
d ¡
vk ±s = 0
dt
Can we now conclude that J is invariant? What if the turning points P1 and
P2 are not quite at the same point on a neighboring ¤eld line? It doesnzt
matter because vk ! 0 at the turning points, and so the contribution to the
integral due to a small change in P is negligible. Thus
Z P2
J=
vkds is an adiabatic invariant.
P1
19
2nd proof of invariance of J: (Boyd andSanderson pg 29) (See also Sturrock, Plasma Physics Ch 4)
The invariant is
I
Z P2
Z P1
Z P2
J = vkds =
vkds +
vkds = 2
vk ds
P1
We begin by writing
vk =
P2
r
P1
2
(W ¡ ¹B)
m
¡
¢
where W = 12 m v2? + vk is the energy, ¹ (14) is the magnetic moment, and
looking at
Z sr
2
J (W; s; t) =
(W ¡ ¹B)ds
m
s1
where s is a coordinate measured along the ¤eld line. Then
dJ
@J
@J dW
@J ds
=
+
+
dt
@t
@W dt
@s dt
where
and
@J
=¡
@t
Z
s
¹ @B
ds
2
m @t
s1
(W
¡
¹B)
m
½
¾
Z s
@vk
@J dW
1
¹ @B
¹ @B
=¡
q
ds vk
+
+ vk
@W dt
@t
m @t
m @s
2
s1
(W
¡
¹B)
m
@J ds
=
@s dt
r
1
q
2
(W ¡ ¹B)vk ¡ vk
m
Now we want to evaluate
Z
s
s1
q
1
¹ @B
ds
2
m @s
(W
¡
¹B)
m
dJ (W; s1; t)
dt
and we note that at s = s1; vk = 0 since this is a turning point. Thus many
20
of these terms vanish. We are left with
Z s
dJ (W; s1; t)
1
¹ @B
q
= ¡
ds
dt
2
m @t
s1
(W
¡
¹B)
m
Z s
1
¹ @B
q
¡
ds
2
s1
(W ¡ ¹B) m @t
m
Z T =2
Z T=2
¹ @B
¹ @B
= ¡
dt ¡
(s 1)
dt
m @t
m @t
0
0
where we used the fact that ds=vk = dt; and T is the bounce period. Evaluating the integrals, we get
·
µ ¶
µ ¶¸
dJ (W; s 1; t)
¹
T
T @B T
=
B (0) ¡ B
¡
dt
m
2
2 @t 2
Now we may expand B in a Taylor series, to get
µ ¶
µ ¶
µ ¶
T
T @B T
1 T 2 @2 B
B (0) = B
¡
+
+¢¢¢
2
2 @t 2
2 2
@t2
Thus
and is of order
·
µ ¶
¸
dJ (W; s 1; t)
¹
@B T
=
¡T
+ ¢¢¢
dt
m
@t 2
¹B
m
µ ¶
T
¿
where ¿ is the time-scale for change in B: Thus J is invariant provided that
T ¿ ¿:
7.3
The third adiabatic invariant
There are three adiabatic invariants because the particle has three degrees of
freedom. These are the integrals of the motion (cf your classical mechanics
course). The third invariant is ©; the ¢ux enclosed by the orbit of the
guiding center as it drifts. This is invariant when system changes are slow
compared with the guiding center orbit period. Since this is quite long, this
third invariant is much less useful than the ¤rst two.
21