Appendix E. Gradient B Drift
Appendix E.
181
Gradient B Drift
Let us consider a charge particle moving in a system with non-uniform magnetic field B(r) .
If the non-uniformity of the magnetic field is small that we can approximately use Taylor
expansion to estimate the magnetic field seen by the charge particle along its gyro orbit in
terms of the magnetic field observed at its guiding center location. Namely, for r = rg.c. + rgyro ,
we have
B(r) = B(rg.c. ) + rgyro ⋅ ∇B + ⋅ ⋅ ⋅ ⋅
(E.1)
If this particle has no velocity component parallel to the local magnetic field and magnetic
moment of this particle is conserved then we can decompose its velocity v into
v = v gyro + v drift
where vgyro is the high frequency gyro motion velocity and vdrift is a time independent
guiding center drift velocity.
m
The equation of motion of this charge particle is
dv
= qv × B ≈ q(v gyro + v drift ) × [B(rg.c. ) + rgyro ⋅ ∇B]
dt
(E.2)
The low frequency guiding-center equation of motion becomes
v drift × B(rg.c. ) + v gyro × (rgyro ⋅ ∇B) = 0
where the notation
f
denotes the time average value of f .
(E.3)
Now let us consider a
simplified case.
Let B = B(x,y) zˆ , then we have
∇B = ∇(Bzˆ ) =
∂ B(x, y)
∂ B(x, y)
xˆzˆ +
yˆ zˆ
∂x
∂y
Equation of gyro motion component with respect to guiding center is
m
dv gyro
= qv gyro × B
dt
(E.4)
drgyro
= v gyro
dt
(E.5)
Solution of gyro motion Eq. (E.4) and Eq. (E.5) can be written as
v gyro = v gyro [ xˆ cos(Ωt + ϕ ) −
rgyro =
Thus,
v gyro
Ω
[ xˆ sin(Ωt + ϕ ) +
q
yˆ sin(Ωt + ϕ )]
q
(E.6)
q
yˆ cos(Ωt + ϕ )]
q
(E.7)
182
Appendix E. Gradient B Drift
rgyro ⋅ ∇B =
v gyro
v
q
∂ B(x, y)
∂ B(x, y)
sin(Ωt + ϕ )
zˆ + gyro cos(Ωt + ϕ )]
zˆ
Ω
∂x
Ω q
∂y
v gyro × rgyro ⋅ ∇B
2
2
v gyro
v gyro
q 2
∂ B(x, y)
∂ B(x, y)
=
cos(Ωt + ϕ )sin(Ωt + ϕ )
( xˆ × zˆ ) −
sin (Ωt + ϕ )
( yˆ × zˆ )
Ω
∂x
Ω q
∂x
+
2
v gyro
v2
q
∂ B(x, y)
∂ B(x, y)
cos 2 (Ωt + ϕ )
( xˆ × zˆ ) − gyro sin(Ωt + ϕ )cos(Ωt + ϕ )
( yˆ × zˆ )
Ω q
∂y
Ω
∂y
where Ω = q B /m , thus
2
2
v gyro
q mv gyro
=
Ω q
qB
Since
sin(Ωt + ϕ )cos(Ωt + ϕ ) = 0
cos 2 (Ωt + ϕ ) = sin 2 (Ωt + ϕ ) =
1
2
(E.8)
we have
2
2
mv gyro
mv gyro
∂ B(x, y)
∂ B(x, y)
v gyro × rgyro ⋅ ∇B = −
( yˆ × zˆ ) +
( xˆ × zˆ )
2qB
∂x
2qB
∂y
=−
2
2
mv gyro
mv gyro
∂ B(x, y)
∂ B(x, y)
[
xˆ +
yˆ ] =
(−∇ ⊥ B)
2qB
∂x
∂y
2qB
(E.9)
Substituting Eq. (E.9) into Eq. (E.3), yields
v drift × B(rg.c. ) +
2
mv gyro
(−∇ ⊥ B) = 0
2qB
(E.10)
Solution of v drift in Eq. (E.10) can be written as
v drift
2
mv gyro
(−∇ ⊥ B) × B
=
2qB
B2
(E.11)
For v drift << v gyro , the perpendicular speed, v ⊥ , of the charge particle is approximately equal
to v gyro .
v drift =
Thus, it is commonly using the following expression to denote gradient-B drift
mv ⊥2 (−∇ ⊥ B) × B
2qB
B2
(E.12)