Chapter 2
Motion of Charged Particles in Fields
Plasmas are complicated because motions of electrons and ions are determined by the electric and magnetic fields but
also change the fields by the currents they carry.
For now we shall ignore the second part of the problem and assume that
Fields are Prescribed.
Even so, calculating the motion of a charged particle can be quite hard.
Equation of motion:

Rate of change of momentum
 = 
q charge  (  E E−field  +  v velocity  ∧  B B−field  ) 
Lorentz Force
 
 (2.1) 
Have to solve this differential equation, to get position
r and velocity (
v=
· r) given
E(
r,t),
B(
r,t).
Approach: Start simple, gradually generalize.
2.1 Uniform B field, E = 0.
2.1.1 Qualitatively
in the plane perpendicular to B:
Figure 2.1: Circular orbit in uniform magnetic field.
Accel. is perp to
v so particle moves in a circle whose radius r
_{L} is such as to satisfy
m r_{L} Ω^{2} = m  v_{⊥}^{2}
r_{L}  = q v_{⊥} B 
 (2.3) 
Ω is the angular (velocity) frequency
1st equality shows Ω
^{2} = v
_{⊥}^{2} /r
_{L}^{2} (r
_{L} = v
_{⊥} / Ω)
Hence second gives m [(v
_{⊥})/(Ω)] Ω
^{2} = q v
_{⊥} B
Particle moves in a circular orbit with
angular velocity Ω =  q B
m  the "Cyclotron Frequency" 
 (2.5) 
and radius r_{l} =  v_{⊥}
Ω  the "Larmor Radius. 
 (2.6) 
2.1.2 By Vector Algebra
 Particle Energy is constant. proof: take v. Eq. of motion then
m v. 
⋅
v
 =  d
dt   ⎛ ⎝  1
2  m v^{2}  ⎞ ⎠  = q v. (v∧B) = 0 . 
 (2.7) 
 Parallel and Perpendicular motions separate. v_{} = constant because accel ( ∝ v∧B) is perpendicular to B.
Perpendicular Dynamics:
Take B in
∧z direction and write components
m 
⋅
v
 x  = q v_{y} B , m 
⋅
v
 y  = − q v_{x} B 
 (2.8) 
Hence

⋅⋅
v
 x  =  qB
m  
⋅
v
 y  = −  ⎛ ⎝  qB
m  ⎞ ⎠  2
 v_{x} = − Ω^{2} v_{x} 
 (2.9) 
Solution: v
_{x} = v
_{⊥} cosΩt (choose zero of time)
Substitute back:
v_{y} =  m
qB  
⋅
v
 x  = −  q 
q  v_{⊥} sinΩt 
 (2.10) 
Integrate:
x = x_{0} +  v_{⊥}
Ω  sinΩt , y = y_{0} +  q
q    v_{⊥}
Ω  cosΩt 
 (2.11) 
Figure 2.2: Gyro center (x_{0},y_{0}) and orbit
This is the equation of a circle with center
r_{0} = (x
_{0}, y
_{0}) and radius r
_{L} = v
_{⊥}/ Ω: Gyro Radius. [Angle is θ = Ωt]
Direction of rotation is as indicated opposite for opposite sign of charge:
Ions rotate anticlockwise. Electrons clockwise about the magnetic field.
The current carried by the plasma always is in such a direction as to
reduce the magnetic field.
This is the property of a magnetic material which is
"Diagmagnetic".
When v
_{} is nonzero the total motion is along a helix.
2.2 Uniform B and nonzero E
Parallel motion: Before, when
E = 0 this was v
_{} = const. Now it is clearly
Constant acceleration along the field.
Perpendicular Motion
Qualitatively:
Figure 2.3: E∧B drift orbit
Speed of positive particle is greater at top than bottom so radius of curvature is greater. Result is that guiding center moves perpendicular to both
E and
B. It `drifts' across the field.
Algebraically: It is clear that if we can find a constant velocity
v_{d} that satisfies
then the sum of this drift velocity plus the velocity
v_{L} =  d
dt  [r_{L} e^{iΩ(t−t0)}] 
 (2.15) 
which we calculated for the
E = 0 gyration will satisfy the equation of motion.
Take ∧
B the above equation:
0 = E∧B+ (v_{d} ∧B) ∧B=E∧B+ (v_{d} . B) B− B^{2} v_{d} 
 (2.16) 
so that
does satisfy it.
Hence the full solution is
v=  v_{} parallel  +  v_{d } cross−field drift  +  v_{L } Gyration  
 (2.18) 
where
and
v_{d} (eq
2.17) is the "E × B drift" of the gyrocenter.
Comments on E × B drift:
 It is independent of the properties of the drifting particle (q, m, v, whatever).
 Hence it is in the same direction for electrons and ions.
 Underlying physics for this is that in the frame moving at the E × B drift E = 0. We have `transformed away' the electric field.
 Formula given above is exact except for the fact that relativistic effects have been ignored. They would be important if v_{d} ∼ c.
2.2.1 Drift due to Gravity or other Forces
Suppose particle is subject to some other force, such as gravity. Write it
F so that
m 
⋅
v
 = F + q v∧B = q (  1
q  F+ v∧B) 
 (2.20) 
This is just like the Electric field case except with
F/q replacing
E.
The drift is therefore
In this case, if force on electrons and ions is same, they drift in
opposite directions.
This general formula can be used to get the drift velocity in some other cases of interest (see later).
2.3 NonUniform B Field
If Blines are straight but the magnitude of B varies in space we get orbits that look qualitatively similar to the E ⊥ B case:
Figure 2.4: ∇B drift orbit
Curvature of orbit is greater where B is greater causing loop to be small on that side. Result is a drift perpendicular to both
B and ∇B. Notice, though, that electrons and ions go in
opposite directions (unlike
E∧
B).
Algebra
We try to find a decomposition of the velocity as before into
v =
v_{d} +
v_{L} where
v_{d} is constant.
We shall find that this can be done only approximately. Also we must have a simple expression for B. This we get by assuming that the Larmor radius is much smaller than the scale length of B variation i.e.,
in which case we can express the field approximately as the first two terms in a Taylor expression:
Then substituting the decomposed velocity we get:



m 
⋅
v
 L  = q (v∧B) = q [ v_{L} ∧B_{0} + v_{d} ∧B_{0} + (v_{L} + v_{d}) ∧(r. ∇) B] 
  (2.24) 



v_{d} ∧B_{0} + v_{L} ∧(r. ∇ ) B+ v_{d} ∧(r. ∇) B 
  (2.25) 

Now we shall find that v
_{d}/v
_{L} is also small, like r ∇B  /B. Therefore the last term here is second order but the first two are first order. So we drop the last term.
Now the awkward part is that
v_{L} and
r_{L} are periodic. Substitute for
r =
r_{0} +
r_{L} so
0 = v_{d} ∧B_{0} + v_{L} ∧(r_{L} . ∇ ) B+ v_{L} ∧( r_{0} . ∇ ) B 
 (2.26) 
We now average over a cyclotron period. The last term is ∝ e
^{−iΩt} so it averages to zero:
0 = v_{d} ∧B+ 〈v_{L} ∧(r_{L} . ∇ ) B〉 . 
 (2.27) 
To perform the average use



 v_{⊥}
Ω   ⎛ ⎝  sinΩt ,  q
q   cosΩt  ⎞ ⎠  
  (2.28) 

v_{L} = ( 
⋅
x
 L  , 
⋅
y
 L  ) 


v_{⊥}  ⎛ ⎝  cosΩt ,  −q
q   sinΩt  ⎞ ⎠  
  (2.29) 

So [ v_{L} ∧(r. ∇ ) B]_{x} 


  (2.30) 



  (2.31) 

(Taking ∇B to be in the ydirection).
Then



− 〈cosΩt sinΩt 〉  v_{⊥}^{2}
Ω  = 0 
  (2.32) 



 q
q   〈cosΩt cosΩt 〉  v_{⊥}^{2}
Ω  =  1
2   v_{⊥}^{2}
Ω   q
q   
  (2.33) 

So
〈v_{L} ∧(r. ∇ ) B〉 = −  q
q    1
2   v_{⊥}^{2}
Ω  ∇ B 
 (2.34) 
Substitute in:
0 = v_{d} ∧B−  q
q    v_{⊥}^{2}
2Ω  ∇ B 
 (2.35) 
and solve as before to get
v_{d} = 
 ⎛ ⎝  −1
q    v_{⊥}^{2}
2 Ω  ∇ B  ⎞ ⎠  ∧B 
B^{2}  =  q
q    v_{⊥}^{2}
2 Ω   B∧∇ B
B^{2}  
 (2.36) 
or equivalently
v_{d} =  1
q   mv_{⊥}^{2}
2 B   B∧∇ B
B^{2}  
 (2.37) 
This is called the `Grad B drift'.
2.4 Curvature Drift
When the Bfield lines are curved and the particle has a velocity v
_{} along the field, another drift occurs.
Figure 2.5: Curvature and Centrifugal Force
Take B constant; radius of curvature R
_{e}.
To 1st order the particle just spirals along the field.
In the frame of the guiding center a force appears because the plasma is rotating about the center of curvature.
This centrifugal force is F
_{cf}
F_{cf} = m  v_{}^{2}
R_{c}  pointing outward 
 (2.38) 
as a vector
F_{cf} = m v_{}^{2}  R_{c}
R_{c}^{2}  
 (2.39) 
[There is also a coriolis force 2m(ω ∧
v) but this averages to zero over a gyroperiod.]
Use the previous formula for a force
v_{d} =  1
q   F_{cf} ∧B
B^{2}  =  m v_{}^{2}
q B^{2}   R_{c} ∧B
R_{c}^{2}  
 (2.40) 
This is the "Curvature Drift".
It is often convenient to have this expressed in terms of the field gradients. So we relate
R_{c} to ∇B etc. as follows:
Figure 2.6: Differential expression of curvature
(Carets denote unit vectors)
From the diagram
db = 
^
b
 2  − 
^
b
 1  = − 
^
R
 c  α 
 (2.41) 
and
So
 d b
d l  = − 
R_{c}  = −  R_{c}
R_{c}^{2}  
 (2.43) 
But (by definition)
 d b
d l  = ( 
^
B
 . ∇ ) 
^
b
 
 (2.44) 
So the curvature drift can be written
v_{d} =  m v_{}^{2}
q   R_{c}
R_{c}^{2}  ∧  B
B^{2}  =  mv_{}^{2}
q  
B^{2}  
 (2.45) 
2.4.1 Vacuum Fields
Relation between ∇ B &
R_{c} drifts
The curvature and ∇B are related because of Maxwell's equations, their relation depends on the current density
j. A particular case of interest is
j=0: vacuum fields.
Figure 2.7: Local polar coordinates in a vacuum field
Consider the zcomponent



 1
r   ∂
∂r  (r B_{θ} ) (B_{r} = 0 by choice). 
  (2.47) 



  (2.48) 

or, in other words,
[Note also 0 = (∇ ∧
B)
_{θ} = ∂B
_{θ}/∂z : (∇ B )
_{z} = 0]
and hence (∇ B)
_{perp} = − B
R_{c}/R
_{c}^{2}.
Thus the grad B drift can be written:
v_{∇B} =  mv_{⊥}^{2}
2 q   B∧∇ B
B^{3}  =  mv_{⊥}^{2}
2 q   R_{c} ∧B
R_{c}^{2} B^{2}  
 (2.50) 
and the total drift across a vacuum field becomes
v_{R} + v_{∇B} =  1
q   ⎛ ⎝  mv_{}^{2} +  1
2  mv_{⊥}^{2}  ⎞ ⎠   R_{c} ∧B
R_{c}^{2} B^{2}  . 
 (2.51) 
Notice the following:
 R_{c} & ∇B drifts are in the same direction.
 They are in opposite directions for opposite charges.
 They are proportional to particle energies
 Curvature ↔ Parallel Energy (× 2)
∇B ↔ Perpendicular Energy
 As a result one can very quickly calculate the average drift over a thermal distribution of particles because
Therefore
〈v_{R} + v_{∇B} 〉 =  2T
q   R_{c} ∧B
R_{c}^{2} B^{2}   ⎛ ⎜ ⎝  =  2T
q  
B^{2}  ⎞ ⎟ ⎠  
 (2.54) 
2.5 Interlude: Toroidal Confinement of Single Particles
Since particles can move freely along a magnetic field even if not across it, we cannot obviously confine the particles in a straight magnetic field. Obvious idea: bend the field lines into circles so that they have no ends.
Figure 2.8: Toroidal field geometry
Problem
Curvature & ∇B drifts



 1
q   ⎛ ⎝  mv_{}^{2} +  1
2  mv_{⊥}^{2}  ⎞ ⎠   R ∧B
R^{2} B^{2}  
  (2.55) 



 1
q   ⎛ ⎝  mv_{}^{2} +  1
2  mv_{⊥}^{2}  ⎞ ⎠   1
BR  
  (2.56) 

Figure 2.9: Charge separation due to vertical drift
Ions drift
up. Electrons down. There is no confinement. When there is finite density things are even worse because charge separation occurs → E → E ∧B → Outward Motion.
2.5.1 How to solve this problem?
Consider a beam of electrons v
_{} ≠ 0 v
_{⊥} = 0. Drift is
v_{d} =  mv_{}^{2}
q   1
B_{T}R  
 (2.57) 
What B
_{z} is required to cancel this?
Adding B
_{z} gives a compensating vertical velocity
v = v_{}  B_{z}
B_{T}  for B_{z} << B_{T} 
 (2.58) 
We want total
v_{z} = 0 = v_{}  B_{z}
B_{T}  +  mv_{}^{2}
q   q
B_{T}R  
 (2.59) 
So B
_{z} = − mv
_{}/ Rq is the right amount of field.
Note that this is such as to make
r_{L} (B_{z}) =  mv_{} 
q B_{z}   = R . 
 (2.60) 
But B
_{z} required depends on v
_{} and q so we can't compensate for all particles simultaneously.
Vertical field along cannot do it.
2.5.2 The Solution: Rotational Transform
Figure 2.10: Tokamak field lines with rotational transform
Toroidal Coordinate system (r, θ, ϕ) (minor radius, poloidal angle, toroidal angle), see figure
2.8.
Suppose we have a
poloidal field B
_{θ}
Field Lines become helical and wind around the torus: figure
2.10.
In the poloidal crosssection the field describes a circle as it goes round in ϕ.
Equation of motion of a particle
exactly following the field is:
r  d θ
dt  =  B_{θ}
B_{ϕ}  v_{ϕ} =  B_{θ}
B_{ϕ}   B_{ϕ}
B  v_{} =  B_{θ}
B  v_{} 
 (2.61) 
and
Now add on to this motion the cross field drift in the
∧z direction.
Figure 2.11: Components of velocity



 B_{θ}
B  v_{} + v_{d} cosθ 
  (2.63) 



  (2.64) 

Take ratio, to eliminate time:
 1
r   dr
dθ  =  u_{d} sinθ
B_{θ}
B  v_{} +v_{d} cosθ 
 
 (2.65) 
Take B
_{θ}, B, v
_{}, v
_{d} to be constants, then we can integrate this orbit equation:
[ lnr ] = [ − ln  B_{θ} v_{}
B  + v_{d} cosθ] . 
 (2.66) 
Take r = r
_{0} when cosθ = 0 (θ = [( π)/2]) then
r = r_{0} /  ⎡ ⎣  1 +  Bv_{d}
b_{θ} v_{}  cosθ  ⎤ ⎦  
 (2.67) 
If [( Bv
_{d})/(B
_{θ} v
_{} )] << 1 this is approximately
where ∆ = [(Bv
_{d})/(B
_{θ} v
_{} )] r
_{0}
This is approximately a circular orbit shifted by a distance ∆:
Figure 2.12: Shifted, approximately circular orbit
Substitute for v
_{d}



r_{0}  B
B_{θ}   1
q  
(mv_{}^{2} +  1
2  mv_{⊥}^{2}) 
v_{}   1
B_{ϕ} R  
  (2.69) 



 1
q B_{θ}  
mv_{}^{2} +  1
2  mv_{⊥}^{2} 
v_{}   r_{p}
R  
  (2.70) 



∆ =  mv_{}
q B_{θ}   r_{0}
R  = r_{Lθ}  r_{0}
R  , 
  (2.71) 

where r
_{Lθ} is the Larmor Radius in a field B
_{θ} ×r / R.
Provided ∆ is small, particles
will be confined. Obviously the important thing is the poloidal rotation of the field lines: Rotational Transform.
Rotational Transform



 poloidal angle
1 toroidal rotation  
  (2.72) 



 poloidal angle
toroidal angle  . 
  (2.73) 

(Originally, ι was used to denote the transform. Since about 1990 it has been used to denote the transform divided by 2π which is the inverse of the safety factor.)
`Safety Factor'
`q_{s}′ =  1
ι  =  toroidal angle
poloidal angle  . 
 (2.74) 
Actually the value of these ratios may vary as one moves around the magnetic field. Definition strictly requires one should take the limit of a large no. of rotations.
q
_{s} is a topological number: number of rotations the long way per rotation the short way.
Cylindrical approx.:
In terms of safety factor the orbit shift can be written
∆ = r_{L θ}  r
R  = r_{L ϕ}  B_{ϕ} r
B_{θ} R  = r_{L} q_{s} 
 (2.76) 
(assuming B
_{ϕ} >> B
_{θ}).
2.6 The Mirror Effect of Parallel Field Gradients: E = 0, ∇B B
Figure 2.13: Basis of parallel mirror force
In the above situation there is a net force along
B.
Force is



− q v∧B sinα = − q v_{⊥} B sinα 
  (2.77) 



  (2.78) 

Calculate B
_{r} as function of B
_{z} from ∇ .
B = 0.
∇ . B=  1
r   ∂
∂r  (rB_{r}) +  ∂
∂z  B_{z} = 0 . 
 (2.79) 
Hence
rB_{r} = −  ⌠ ⌡  r  ∂B_{z}
∂z  dr 
 (2.80) 
Suppose r
_{L} is small enough that [(∂B
_{z})/(∂z)] ≅ const.
[ r B_{r}]_{0}^{rL} ≅  ⌠ ⌡  r_{L}
0  r dr  ∂B_{z}
∂z  = −  1
2  r_{L}^{2}  ∂B_{z}
∂z  
 (2.81) 
So
B_{r}(r_{L}) = −  1
2  r_{L}  ∂B_{z}
∂z  
 (2.82) 
sinα = −  B_{r}
B  = +  r_{L}
2   1
2   ∂B_{z}
∂z  
 (2.83) 
Hence
< F_{} > = − q   v_{⊥} r_{L}
2   ∂B_{z}
∂z  = − 
B   ∂B_{z}
∂z  . 
 (2.84) 
As particle enters increasing field region it experiences a net parallel
retarding force.
Define
Magnetic Moment
Note this is consistent with loop current definition
μ = A I = πr_{L}^{2} .  q v_{⊥}
2 πr_{L}  =  q r_{L} v_{⊥}
2  
 (2.86) 
Force is F
_{} = μ . ∇
_{} B
This is force on a `magnetic dipole' of moment μ.
Our μ always points along
B but in opposite direction.
2.6.1 Force on an Elementary Magnetic Moment Circuit
Consider a plane rectangular circuit carrying current I having elementary area dxdy = dA. Regard this as a vector pointing in the
z direction
dA. The force on this circuit in a field
B(
r) is
F such that



Idy [B_{z} (x+dx) − B_{z}(x)] = I dydx  ∂B_{z}
∂x  
  (2.88) 



−I dx [B_{z} (y+dy) − B_{z}(y)] = I dydx  ∂B_{z}
∂y  
  (2.89) 



− I dx [B_{y} (y+dy) − B_{y}(y)] − I dy[B_{x}(x+dx) − B_{x}(x)] 
  (2.90) 



− I dxdy  ⎡ ⎣  ∂B_{x}
∂x  +  ∂B_{y}
∂y  ⎤ ⎦  = I dydx  ∂B_{z}
∂z  
  (2.91) 

(Using ∇ .
B = 0).
Hence, summarizing:
F = I dydx ∇ B
_{z}. Now define μ = I
dA = I dydx
∧z and take it constant. Then clearly the force can be written
F = ∇ (B. μ ) [ Strictly = (∇ B) . μ] 
 (2.92) 
μ is the (vector) magnetic moment of the circuit.
The shape of the circuit does not matter since any circuit can be considered to be composed of the sum of many rectangular circuits. So in general
and force is
F = ∇ (B. μ) (μ constant), 
 (2.94) 
We shall show in a moment that μ  is constant for a circulating particle, regard as an elementary circuit. Also, μ for a particle always points in the 
B direction. [Note that this means that the effect of particles on the field is to
decrease it.] Hence the force may be written
This gives us both:
 Magnetic Mirror Force:
and
 Grad B Drift:
v_{∇B} =  1
q   F ∧B
B^{2}  =  μ
q   B∧∇ B
B^{2}  . 
 (2.97) 
2.6.2 μ is a constant of the motion
`Adiabatic Invariant'
Proof from F_{}
Parallel equation of motion
m  dv_{}
dt  = F_{} = − μ  dB
dz  
 (2.98) 
So
m v_{}  dv_{}
dt  =  d
dt  (  1
2  m v_{}^{2} ) = − μv_{z}  dB
dz  = − μ  dB
dt  
 (2.99) 
or
 d
dt  (  1
2  m v_{}^{2}) + μ  dB
dt  = 0 . 
 (2.100) 
Conservation of Total KE



 d
dt  (  1
2  mv_{}^{2} +  1
2  mv_{⊥}^{2} ) = 0 
  (2.101) 



 d
dt  (  1
2  mv_{}^{2} + μB) = 0 
  (2.102) 

Combine
Angular Momentum
of particle about the guiding center is



 mv_{⊥}
q B  m v_{⊥} =  2m
q   
B  
  (2.105) 



  (2.106) 

Conservation of magnetic moment is basically conservation of angular momentum about the guiding center.
Proof direct from Angular Momentum
Consider angular momentum about G.C. Because θ is ignorable (locally) Canonical angular momentum is conserved.
p = [ r ∧( m v+ q A) ]_{z} conserved. 
 (2.107) 
Here
A is the vector potential such that
B = ∇ ∧
A
the definition of the vector potential means that



  (2.108) 



 ⌠ ⌡  r_{L}
0  r . B_{z} dr =  r_{L}^{2}
2  B_{z} =  μm
q   
  (2.109) 

Hence



 − q
q   r_{L} v_{⊥} m + q  mμ
q  
  (2.110) 



  (2.111) 

So p = const ↔ μ = constant.
Conservation of μ is basically conservation of angular momentum of particle about G.C.
2.6.3 Mirror Trapping
Figure 2.14: Magnetic Mirror
F
_{} may be enough to reflect particles back. But may not!
Let's calculate whether it will:
Suppose reflection occurs.
At reflection point v
_{r} = 0.
Energy conservation
 1
2  m (v_{⊥0}^{2} + v_{0}^{2}) =  1
2  m v_{⊥r}^{2} 
 (2.112) 
μ conservation
Hence



  (2.114) 



 v_{⊥0}^{2}
v_{⊥0}^{2} + v_{o}^{2}  
  (2.115) 

2.6.4 Pitch Angle θ



  (2.116) 



 v_{⊥0}^{2}
v_{⊥0}^{2} +v_{0}^{2}  = sin^{2} θ_{0} 
  (2.117) 

So, given a pitch angle θ
_{0}, reflection takes place where B
_{0}/B
_{r} = sin
^{2}θ
_{0}.
If θ
_{0} is too small no reflection can occur.
Critical angle θ
_{c} is obviously
θ_{c} = sin^{−1} (B_{0} / B_{1})^{1/2} 
 (2.118) 
Loss Cone is all θ < θ
_{c}.
Figure 2.15: Critical angle θ_{c} divides velocity space into a losscone and a region of mirrortrapping
Importance of Mirror Ratio: R
_{m} = B
_{1} / B
_{0}.
2.6.5 Other Features of Mirror Motions
Flux enclosed by gyro orbit is constant.



πr_{L}^{2} B =  πm^{2} v_{⊥}^{2}
q^{2} B^{2}  B 
  (2.119) 



  (2.120) 



  (2.121) 

Figure 2.16: Flux tube described by orbit
Note that if B changes `suddenly' μ might not be conserved.
Basic requirement
Slow variation of B (relative to r
_{L}).
2.7 Time Varying B Field (E inductive)
Figure 2.17: Particle orbits round B so as to perform a line integral of the Electric field
Particle can gain energy from the inductive
E field



  (2.123) 



−  ⌠ ⌡ 
s  
⋅
B
 . ds = −  d Φ
d t  
  (2.124) 

Hence work done on particle in 1 revolution is
δw = −  ⌠ (⎜) ⌡  qE. dl = + q   ⌠ ⌡ 
s  
⋅
B
 . ds = + q   d Φ
dt  = q  
⋅
B
 πr_{L}^{2} 
 (2.125) 
(
dl and
v_{⊥} q are in opposition directions).



q  
⋅
B
 πr _{L}^{2} = 
q B  
B  
  (2.126) 



  (2.127) 

Hence
 d
dt   ⎛ ⎝  1
2  m v_{⊥}^{2}  ⎞ ⎠  =  Ω
2 π  δ  ⎛ ⎝  1
2  mv_{⊥}^{2}  ⎞ ⎠  = μ  db
dt  
 (2.128) 
but also
 d
dt   ⎛ ⎝  1
2  mv_{⊥}^{2}  ⎞ ⎠  =  d
dt  ( μB ) . 
 (2.129) 
Hence
Notice that since Φ = [(2 πm)/(q
^{2} )] μ, this is just another way of saying that the flux through the gyro orbit is conserved.
Notice also
energy increase. Method of `heating'. Adiabatic Compression.
2.8 Time Varying Efield (E, B uniform)
Recall the
E∧
B drift:
when E varies so does
v_{E ∧B}. Thus the guiding centre experiences an acceleration

⋅
v
 E ∧B  =  d
dt   ⎛ ⎝  E∧B
B^{2}  ⎞ ⎠  
 (2.132) 
In the frame of the guiding centre which is accelerating, a force is felt.
F_{a} = − m  d
dt   ⎛ ⎝  E∧B
B^{2}  ⎞ ⎠  (Pushed back into seat! −ve.) 
 (2.133) 
This force produces another drift



 1
q   F_{a} ∧B
B^{2}  =  m
qB^{2}   d
dt   ⎛ ⎝  E∧B
B^{2}  ⎞ ⎠  ∧B 
  (2.134) 



−  m
qB   d
dt  ( ( E. B) B− B^{2}E) 
  (2.135) 



  (2.136) 

This is called the
`polarization drift'.



v_{p} =  E∧B
B^{2}  +  m
q B^{2}  
⋅
E
 ⊥  
  (2.137) 



  (2.138) 

Figure 2.18: Suddenly turning on an electric field causes a shift of the gyrocenter in the direction of force. This is the polarization drift.
Startup effect: When we `switch on' an electric field the average position (gyro center) of an initially stationary particle shifts over by ∼
^{1}/
_{2} the orbit size. The polarization drift is this polarization effect on the medium.
Total
shift due to
v_{p} is
∆r  ⌠ ⌡  v_{p} dt =  m
q B^{2}   ⌠ ⌡  
^
E
 ⊥  dt =  m
q B^{2}  [ ∆E_{⊥} ] . 
 (2.139) 
2.8.1 Direct Derivation of [(dE)/dt] effect: `Polarization Drift'
Consider an oscillatory field
E =
Ee
^{−iωt} (⊥r
_{0}B)
Try for a solution in the form
v= v_{D} e^{−iωt} + v_{L} 
 (2.142) 
where, as usual,
v_{L} satisfies m
· v_{L} = q
v_{L} ∧
B
Then
(1) m(−i ωv_{D} = q ( E+ v_{D} ∧B) xl^{−i ωt} 
 (2.143) 
Solve for
v_{D}: Take ∧
B this equation:
(2) − m i ω( v_{D} ∧B) = q ( E∧B+ ( B^{2} . vD ) B−B^{2} v_{D} ) 
 (2.144) 
add m i ω×(1) to q ×(2) to eliminate
v_{D} ∧
B.
m^{2} ω^{2} v_{D} + q^{2} ( E∧B− B^{2} v_{D} ) = mi ωq E 
 (2.145) 
or: v_{D}  ⎡ ⎣  1 −  m^{2} ω^{2}
q^{2} B^{2}  ⎤ ⎦  = −  miω
qB^{2}  E+  E∧B
B^{2}  
 (2.146) 
i.e. v_{D}  ⎡ ⎣  1 −  ω^{2}
Ω^{2}  ⎤ ⎦  = −  iωq
ΩB q   E+  E∧ B
B^{2}  
 (2.147) 
Since −iω↔ [(∂)/(∂t)] this is the same formula as we had before: the sum of polarization and
E∧
B drifts
except for the [1 − ω
^{2}Ω
^{2}] term.
This term comes from the change in
v_{D} with time (accel).
Thus our earlier expression was only approximate. A good approx if ω << Ω.
2.9 Non Uniform E (Finite Larmor Radius)
m  dv
dt  = q ( E(r) + v∧B) 
 (2.148) 
Seek the usual soltuion
v =
v_{D} +
v_{g}.
Then average out over a gyro orbit
Hence drift is obviously
So we just need to find the
average E field experienced.
Expand
E as a Taylor series about the G.C.
E(r) = E_{0} +( r. ∇) E+  ⎛ ⎝  x^{2}∂^{2}
2!∂x^{2}  +  y^{2}
2!   ∂^{2}
∂y^{2}  ⎞ ⎠  E+ cross terms + . 
 (2.152) 
(E.g. cross terms are x y [(∂
^{2})/(∂x ∂y)]
E).
Average over a gyro orbit:
r = r
_{L} (cosθ, sinθ,0).
Average of cross terms = 0.
Then
〈E(r) 〉 = E+ ( 〈r_{L} 〉. ∇) E+  〈r_{L}^{2}〉
2!  ∇^{2} E. 
 (2.153) 
linear term 〈r
_{L}〉 = 0. So
〈E(r)〉 ≅ E+  r_{L}^{2}
4  ∇^{2} E 
 (2.154) 
Hence
E∧
B with 1st finiteLarmorradius correction is
v_{E ∧B} =  ⎛ ⎝  1 +  r_{L}^{2}
r  ∇^{2}  ⎞ ⎠   E∧B
B^{2}  . 
 (2.155) 
[Note: Grad B drift is a finite Larmor effect already.]
Second and Third Adiabatic Invariants
There are additional approximately conserved quantities like μ in some geometries.
2.10 Summary of Drifts



  (2.156) 



 1
q   F ∧B
B^{2}  General Force 
  (2.157) 



 ⎛ ⎝  1 +  r_{L}^{2}
4  ∇^{2}  ⎞ ⎠   E∧B
B^{2}  Nonuniform E 
  (2.158) 



 mv_{⊥}^{2}
2q   B∧∇ B
B^{3}  Grad B 
  (2.159) 



 mv_{}^{2}
q   R_{c} ∧B
R_{c}^{2} B^{2}  Curvature 
  (2.160) 



 1
q   ⎛ ⎝  mv_{}^{2} +  1
2  m v_{⊥}^{2}  ⎞ ⎠   R_{c} ∧B
R_{c}^{2} B^{2}  Vacuum Fields. 
  (2.161) 



  (2.162) 

Mirror Motion
μ ≡  mv_{⊥}^{2}
2B  is constant 
 (2.163) 
Force is
F = − μ∇B.