Download Motion of charged particles in EM fields

Handout
1
Motion of charged particles in EM fields
We consider plasmas in which no collisions take place. The basic equation of motion of a
charged particle in an electromagnetic field is then
m~a = m
0.1
d~v
~ + ~v × B
~
=q E
dt
Energy considerations
d 1 2
d
~ · ~v
mv = W = q E
dt 2
dt
Therefore:
(1)
solution:
vx (t) = A cos(ωg t + φ)
vy (t) = −A sin(ωg t + φ)
(2)
(7)
(8)
where
2
vx2 + vy2 = A2 = v⊥
.
(9)
No spatial change in the magnetic field
results in a change of the total kinetic To get the actual movement of the particle,
integrate the solution:
energy.
Z
v⊥
sin(ωg t + φ)
x(t)
=
v
(τ
)dτ
=
x
+
x
0
Electric fields perpendicular to the diωg
rection of motion do not change the
(10)
Z
total kinetic energy.
v⊥
y(t) = vy (τ )dτ = y0 +
cos(ωg t + φ)
ωg
Only electric fields aligned with the di(11)
rection of motion can change the total
Here, x0 and y0 are constants of integration
and φ is the initial phase. The particle moves
in a circle, perpendicular to the background
0.2 Motion in steady magnetic magnetic field. It "gyrates". The radius of
~ =0
field, E
the gyromotion, the gyroradius, is
v⊥
mv⊥
~ = 0 and B
~ = B0~ez . Then from
Assume E
rg =
=
(12)
ωg
qB0
(1)
kinetic energy.
where ωg is the gyrofrequency:
dvx
q
= vy B0
dt
m
dvy
q
= − vx B0
dt
m
dvz
=0
vz const.
dt
(3)
(4)
(5)
Differentiate (3) and substitute (4):
d2 vx
qB0 dvy
qB0
=
=−
2
dt
m dt
m
2
vx
(6)
This describes a simple harmonic oscillator Figure 1: Sketch of the gyromotion of a posiwith a frequency of ωg = qB0 /m with the tively charged particle.
FYS3610
Lasse Clausen, [email protected]
UiO
Handout
2
Because of the dependence on q, the direction of the gyromotion of positively charged
particles (ions) is opposite that of negatively charged particles (electrons). When
considering the gyromotion of electrons and
ions, for the same perpendicular energy, the
gyrofrequency of electrons is much higher
(me mi ) while the gyroradius is much
larger. (5) shows that vz , i.e., the velocity along the field is constant. Therefore,
for vz 6= 0, the particle trajectory is helical
around the magnetic field line, and the center of the gyromotion is called the "guiding
center".
0.4
Total kinetic energy:
W = W⊥ + Wk
0.3
(14)
~ = 0, dW/dt = dW⊥ /dt + dWk /dt = 0.
For E
Define magnetic moment as µ = W⊥ /B,
then
dW⊥
dµ
dB
=B
+µ
.
(15)
dt
dt
dt
Also
dWk
dB
= −µ
,
(16)
dt
dt
such that
dW⊥ dWk
dµ
+
=B
=0
(17)
dt
dt
dt
and hence dµ/dt = 0, meaning that the magnetic moment µ is conserved as the particle
moves through spatially varying magnetic
fields.
0.5
Figure 2: Sketch of the gyromotion for vk 6= 0.
Magnetic moment
Magnetic mirror
Consider the following situation where the
magnetic field converges:
Pitch angle
The pitch angle α is defined by the ratio of
v⊥ and vk :
α = arctan
v⊥
vk
(13)
Figure 4: Converging magnetic field.
Because
2
W⊥
mv⊥
µ=
=
= const.
(18)
B
2B
as B increases v⊥ must also increase. However, the total kinetic energy must not
change, such that vk must decrease. That
increases the pitch angle (see (13)), meaning that the movement along the field line
slows down. Once the field-aligned move~ = −µ∇B
Figure 3: Pitch angle for the case for vk < v⊥ . ment stops, there is still a force F
FYS3610
Lasse Clausen, [email protected]
UiO
Handout
3
on the gyrating particle, pushing it back out equatorial plane and the Earth’s surface. For
of the region of higher B.
a field lines that crosses the equatorial plane
2
2
2
Define total speed v as v = vk + v⊥ , then at a distance of 4 Earth radii (Re )
s
v⊥ = v sin α and µ = mv 2 sin2 α/2B. Then
Beq
because µ0 = µ1 :
α = arcsin
' 5.5◦
(21)
Bsf
sin2 α0
B0
=
(19) If α is smaller the particle will hit the Earth’s
2
sin α1
B1
surface before it has a chance to mirror,
Therefore, all particles with α < 5.5◦ are
0.6 Trapped particles in a lost; this defines the loss cone (for that pardipole field
ticular field line at 4 Re .
Earth’s magnetic field is essentially a dipole
~ ×B
~ drift
field where B(r) ∝ 1/r3 . As particles move 0.8 E
along field lines toward the magnetic poles,
~
~
they experience stronger magnetic fields and Now assume E 6= 0 and B = B0~ez . Then
are eventually mirrored. As they gyrate they from (1)
are said to "bounce" between hemispheres;
q
dvx
= (Ex + vy B0 )
(22)
they are trapped in the terrestrial magnetic
dt
m
field.
dvy
q
= (Ey − vx B0 )
(23)
dt
m
dvz
q
= Ez
(24)
dt
m
(24) describes simple accelerated motion
~ (change in total kinetic energy!).
along B
~ = Ey~ey :
For simplicity, assume E
Figure 5: Terrestrial dipolar field lines.
dvx
q
= vy B0
dt
m
q
dvy
= (Ey − vx B0 )
dt
m
(25)
(26)
Differentiate and substitute:
0.7
Loss cone
d2 v x
qB0 2
Ey
=
−
vx −
2
dt
m
B0
2
2
d vy
qB0
=−
vy ,
2
dt
m
At the mirror point, where Wk = 0, the pitch
angle becomes αm = 90◦ . Hence from (19)
we see that
(27)
(28)
B
sin2 α
=
1
Bm
(20) where the additional Ey /B0 term indicates
a drift velocity in the +x direction. A more
general consideration gives
which is true anywhere along the trajectory
(field line). For a terrestrial dipole field line,
~ ×B
~
E
~vE =
.
(29)
the maximum ratio of B/Bm is between the
B2
FYS3610
Lasse Clausen, [email protected]
UiO
Handout
4
~vE is superposed on the gyromotion and in- Formally
dependent of charge; i.e., electrons and pro2 ~
mv⊥
B × ∇B
~
v
=
,
(30)
tons drift in the same direction
no net
G
2
qB03
current.
the drift velocity is charge dependent, i.e.,
the gradient drift causes a current. The gradient drift velocity depends on the perpendicular kinetic energy.
The second magnetic drift is the curvature
drift. It is due to centrifugal forces experienced by a particle as it gyrates along a
curved field line:
~ ×B
~ drift motion for a positively
Figure 6: E
charged particle.
0.9
Magnetic drifts
Two magnetic drifts exists, the gradient drift
and the curvature drift. For the first, assume
~ = 0 and B
~ is spatially inhomogeneous,
E
~ the
i.e., ∇B 6= 0. In the region of large B
gyroradius is smaller than in the region of
Figure 7: Illustration of the curvature drift.
~ From this results a drift:
smaller B.
The drift velocity is given by
~c × B
~
mvk2 2R
~vR =
,
(31)
2 qRc2 B02
Again, the curvature drift is charge dependent, i.e., positively and negatively charged
particles drift in opposite directions, creating a current. The velocity of the curvature
drift depends on the parallel kinetic energy
0.10
General force drifts
~ in (29) with E
~ = F~ /q one obReplacing E
tains:


~
1  F~
B
~v =
× 
(32)
Figure 7: Illustration of the gradient drift for
ωg m B
positively and negatively charged particles.
FYS3610
Lasse Clausen, [email protected]
UiO