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:
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.
Figure 2.1: Circular orbit in uniform magnetic field.
Accel. is perp to v so particle moves in a circle whose radius rL is such as to satisfy
Ω is the angular (velocity) frequency
1st equality shows Ω2 = v⊥2 /rL2 (rL = v⊥ / Ω)
Hence second gives m [(v⊥)/(Ω)] Ω2 = |q |v⊥ B
Particle moves in a circular orbit with
Take B in ∧z direction and write components
Solution: vx = v⊥ cosΩt (choose zero of time)
Figure 2.2: Gyro center (x0,y0) and orbit
This is the equation of a circle with center r0 = (x0, y0) and radius rL = 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 non-zero the total motion is along a helix.
Parallel motion: Before, when E = 0 this was v|| = const. Now it is clearly
Constant acceleration along the field.
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 vd that satisfies
then the sum of this drift velocity plus the velocity
which we calculated for the E = 0 gyration will satisfy the equation of motion.
Take ∧B the above equation:
does satisfy it.
Hence the full solution is
vd (eq 2.17) is the "E × B drift" of the gyrocenter.
Comments on E × B drift:
Suppose particle is subject to some other force, such as gravity. Write it F so that
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).
If B-lines 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).
We try to find a decomposition of the velocity as before into v = vd + vL where vd 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:
Now we shall find that vd/vL 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 vL and rL are periodic. Substitute for r = r0 + rL so
We now average over a cyclotron period. The last term is ∝ e−iΩt so it averages to zero:
To perform the average use
(Taking ∇B to be in the y-direction).
and solve as before to get
This is called the `Grad B drift'.
When the B-field 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 Re.
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 Fcf
as a vector
[There is also a coriolis force 2m(ω ∧v) but this averages to zero over a gyroperiod.]
Use the previous formula for a force
This is the "Curvature Drift".
It is often convenient to have this expressed in terms of the field gradients. So we relate Rc to ∇B etc. as follows:
Figure 2.6: Differential expression of curvature
(Carets denote unit vectors)
From the diagram
But (by definition)
So the curvature drift can be written
Relation between ∇ B & Rc 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 z-component
or, in other words,
[Note also 0 = (∇ ∧B)θ = ∂Bθ/∂z : (∇ B )z = 0]
and hence (∇ B)perp = − B Rc/Rc2.
Thus the grad B drift can be written:
and the total drift across a vacuum field becomes
Notice the following:
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
Curvature & ∇B drifts
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.
Consider a beam of electrons v|| ≠ 0 v⊥ = 0. Drift is
What Bz is required to cancel this?
Adding Bz gives a compensating vertical velocity
We want total
So Bz = − mv||/ Rq is the right amount of field.
Note that this is such as to make
But Bz required depends on v|| and q so we can't compensate for all particles simultaneously.
Vertical field along cannot do it.
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 Bθ
Field Lines become helical and wind around the torus: figure 2.10.
In the poloidal cross-section the field describes a circle as it goes round in ϕ.
Equation of motion of a particle exactly following the field is:
Now add on to this motion the cross field drift in the ∧z direction.
Figure 2.11: Components of velocity
Take ratio, to eliminate time:
Take Bθ, B, v||, vd to be constants, then we can integrate this orbit equation:
Take r = r0 when cosθ = 0 (θ = [( π)/2]) then
If [( Bvd)/(Bθ v|| )] << 1 this is approximately
where ∆ = [(Bvd)/(Bθ v|| )] r0
This is approximately a circular orbit shifted by a distance ∆:
Figure 2.12: Shifted, approximately circular orbit
Substitute for vd
where rLθ is the Larmor Radius in a field Bθ ×r / R.
Provided ∆ is small, particles will be confined. Obviously the important thing is the poloidal rotation of the field lines: Rotational Transform.
(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.)
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.
qs is a topological number: number of rotations the long way per rotation the short way.
In terms of safety factor the orbit shift can be written
(assuming Bϕ >> Bθ).
Figure 2.13: Basis of parallel mirror force
In the above situation there is a net force along B.
Calculate Br as function of Bz from ∇ . B = 0.
Suppose rL is small enough that [(∂Bz)/(∂z)] ≅ const.
As particle enters increasing field region it experiences a net parallel retarding force.
Define Magnetic Moment
Note this is consistent with loop current definition
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.
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
(Using ∇ . B = 0).
Hence, summarizing: F = I dydx ∇ Bz. Now define μ = I dA = I dydx ∧z and take it constant. Then clearly the force can be written
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
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:
Parallel equation of motion
Conservation of Total KE
of particle about the guiding center is
Conservation of magnetic moment is basically conservation of angular momentum about the guiding center.
Consider angular momentum about G.C. Because θ is ignorable (locally) Canonical angular momentum is conserved.
Here A is the vector potential such that B = ∇ ∧A
the definition of the vector potential means that
So p = const ↔ μ = constant.
Conservation of μ is basically conservation of angular momentum of particle about G.C.
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.
So, given a pitch angle θ0, reflection takes place where B0/Br = sin2θ0.
If θ0 is too small no reflection can occur.
Critical angle θc is obviously
Loss Cone is all θ < θc.
Figure 2.15: Critical angle θc divides velocity space into a loss-cone and a region of mirror-trapping
Importance of Mirror Ratio: Rm = B1 / B0.
Flux enclosed by gyro orbit is constant.
Figure 2.16: Flux tube described by orbit
Note that if B changes `suddenly' μ might not be conserved.
Slow variation of B (relative to rL).
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
Hence work done on particle in 1 revolution is
(dl and v⊥ q are in opposition directions).
Notice that since Φ = [(2 πm)/(q2 )] 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.
Recall the E∧B drift:
when E varies so does vE ∧B. Thus the guiding centre experiences an acceleration
In the frame of the guiding centre which is accelerating, a force is felt.
This force produces another drift
This is called the `polarization drift'.
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.
Start-up 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 vp is
Consider an oscillatory field E = Ee−iωt (⊥r0B)
Try for a solution in the form
where, as usual, vL satisfies m · vL = q vL ∧B
Solve for vD: Take ∧B this equation:
add m i ω×(1) to q ×(2) to eliminate vD ∧B.
Since −iω↔ [(∂)/(∂t)] this is the same formula as we had before: the sum of polarization and E∧B drifts except for the [1 − ω2Ω2] term.
This term comes from the change in vD with time (accel).
Thus our earlier expression was only approximate. A good approx if ω << Ω.
Seek the usual soltuion v = vD + vg.
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.g. cross terms are x y [(∂2)/(∂x ∂y)]E).
Average over a gyro orbit: r = rL (cosθ, sinθ,0).
Average of cross terms = 0.
linear term 〈rL〉 = 0. So
Hence E∧B with 1st finite-Larmor-radius correction is
[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.
Force is F = − μ∇B.
Motion of Charged Particles in Fields
2.1 Uniform B field, E = 0.
2.1.1 Qualitativelyin the plane perpendicular to B:
2.1.2 By Vector Algebra
- Particle Energy is constant. proof: take v. Eq. of motion then
m v.⋅v = d
m v2 ⎞
= q v. (v∧B) = 0 . (2.7)
- Parallel and Perpendicular motions separate. v|| = constant because accel ( ∝ v∧B) is perpendicular to B.
2.2 Uniform B and non-zero E
- 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 vd ∼ c.
2.2.1 Drift due to Gravity or other Forces
2.3 Non-Uniform B Field
2.4 Curvature Drift
2.4.1 Vacuum Fields
- Rc & ∇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)
- As a result one can very quickly calculate the average drift over a thermal distribution of particles because
m v||2 〉 = T
(2.52) 〈 1
m v⊥2 〉 = T 2 degrees of freedom (2.53) 〈vR + v∇B 〉 = 2T
. ∇ ⎞
2.5 Interlude: Toroidal Confinement of Single Particles
2.5.1 How to solve this problem?
2.5.2 The Solution: Rotational Transform
2.6 The Mirror Effect of Parallel Field Gradients: E = 0, ∇B ||B
2.6.1 Force on an Elementary Magnetic Moment Circuit
- Magnetic Mirror Force:
F|| = −μ∇||B (2.96)
- Grad B Drift:
v∇B = 1
2.6.2 μ is a constant of the motion`Adiabatic Invariant'
Proof from F||
Proof direct from Angular Momentum
2.6.3 Mirror Trapping
2.6.4 Pitch Angle θ
2.6.5 Other Features of Mirror Motions
2.7 Time Varying B Field (E inductive)
2.8 Time Varying E-field (E, B uniform)
2.8.1 Direct Derivation of [(dE)/dt] effect: `Polarization Drift'
2.9 Non Uniform E (Finite Larmor Radius)
2.10 Summary of Drifts