Collisions in Plasmas

Chapter 3 Collisions in Plasmas 3.1 Binary collisions between charged particles Reduced-mass for binary collisions: Two particles interacting with each other have forces F12 force on 1 from 2. F21 force on 2 from 1. By Newton’s 3rd law, F12 = −F21 . Equations of motion: m1 r̈1 = F12 ; m2 r̈2 = F21 (3.1) Combine to get r̈1 − r̈2 = F12  1 1 + m1 m2  (3.2) which may be written m1 m2 d2 (r1 − r2 ) = F12 (3.3) m1 + m2 dt2 If F12 depends only on the difference vector r1 − r2 , then this equation is identical to the m2 moving at position r ≡ r1 − r2 with equation of a particle of “Reduced Mass” mr ≡ mm11+m 2 respect to a fixed center of force: mr r̈ = F12 (r) . (3.4) This is the equation we analyse, but actually particle 2 does move. And we need to recognize that when interpreting mathematics. If F21 and r1 − r2 are always parallel, then a general form of the trajectory can be written as an integral. To save time we specialize immediately to the Coulomb force F12 = q1 q2 r 4π0 r3 Solution of this standard (Newton’s) problem: (3.5) 1: projectile v1 χ r b Impact Parameter θ θ 1 2: target Figure 3.1: Geometry of the collision orbit Angular momentum is conserved: mr r2 θ̇ = const. = mr bv1 Substitute u ≡ 1 r then θ̇ = bv1 r2 (θ clockwise from symmetry) (3.6) = u2 bv1 Also ṙ = r̈ = d 1 1 du du = − 2 θ̇ = −bv1 dt u u dθ dθ 2 d2 u d u −bv1 2 θ̇ = −(bv1 )2 u2 2 dθ dθ (3.7) (3.8) Then radial acceleration is 2 2 r̈ − rθ̇ = −(bv1 ) u 2 d2 u +u dθ2 ! = |F12 | mr (3.9) i.e. d2 u q1 q2 1 +u=− 2 dθ 4π0 mr (bv1 )2 This orbit equation has the elementary solution u≡ (3.10) 1 q1 q2 1 = C cos θ − r 4π0 mr (bv1 )2 (3.11) The sin θ term is absent by symmetry. The other constant of integration, C, must be deter1 = 0. So mined by initial condition. At initial (far distant) angle, θ1 , u1 = ∞ 0 = C cos θ1 − There: ṙ1 = −v1 = −bv1 Hence tan θ1 = sin θ1 = cos θ1 q1 q 2 1 4π0 mr (bv1 )2 (3.12) du |1 = +bv1 C sin θ1 dθ −1/Cb q1 q2 1 /C 4π0 mr (bv1 )2 =− b b90 (3.13) (3.14) where b90 ≡ q1 q2 1 4π0 mr v12 . (3.15) Notice that tan θ1 = −1 when b = b90 . This is when θ1 = −45◦ and χ = 90◦ . So particle emerges at 90◦ to initial direction when “90◦ impact parameter00 b = b90 (3.16) Finally: 1 1 b2 C = − cosecθ1 = − 1 + 90 b b b2 3.1.1 !1 2 (3.17) Frames of Reference Key quantity we want is the scattering angle but we need to be careful about reference frames. Most “natural” frame of ref is “Center-of-Mass” frame, in which C of M is stationary. C of M has position: m1 r1 + m2 r2 (3.18) R= m1 + m2 and velocity (in lab frame) m1 v1 + m2 v2 (3.19) V= m1 + m2 Now m2 r1 = R + r (3.20) m1 + m2 m1 r2 = R − r (3.21) m1 + m2 So motion of either particle in C of M frame is a factor times difference vector, r. Velocity in lab frame is obtained by adding V to the C of M velocity, e.g. m2 ṙ m1 +m2 + V. Angles of position vectors and velocity differences are same in all frames. Angles (i.e. directions) of velocities are not same. 3.1.2 Scattering Angle In C of M frame is just the final angle of r. − 2θ1 + χ = π (3.22) (θ1 is negative) χ = π + 2θ1 ; θ1 = χ−π . 2 (3.23) −θ1 χ −θ1 −θ1 −θ1 Figure 3.2: Relation between θ1 and χ. χ π tan θ1 = tan = 2 2   = − cot χ 2 (3.24) So χ b = 2 b90 χ b90 tan = 2 b cot (3.25) (3.26) But scattering angle (defined as exit velocity angle relative to initial velocity) in lab frame is different. Final velocity in CM frame 0 vCM = v1CM (cos χc , sin χC ) = m2 v1 (cos χc , sin χc ) m1 + m2 (3.27) [ χc ≡ χ and v1 is initial relative velocity]. Final velocity in Lab frame vL0 = 0 vCM m2 v1 m2 v1 +V = V + cos χc , sin χc m1 + m2 m1 + m2   (3.28) So angle is given by cot χL = m2 v1 cos χc m1 +m2 m2 v1 sin χx m1 +m2 V + = V m1 + m2 cosecχc + cot χc v1 m2 (3.29) For the specific case when m2 is initially a stationary target in lab frame, then V cot χL m1 v1 and hence m1 + m2 m1 = cosecχc + cot χc m2 = (3.30) (3.31) This is exact. Small angle approximation (cot χ → χ1 , cosecχ → 1 χ gives 1 m1 1 1 m2 = + ⇔ χL = χc χL m2 χc χc m1 + m2 So small angles are proportional, with ratio set by the mass-ratio of particles. (3.32) Center-of-Mass Frame y x χc Particle 1 m _r v m1 0 Particle 2 * b m _r v m2 0 C-of-M Laboratory Frame, Stationary Target χ 1 Particle 1 v0 C-of-M * V b Particle 2 Figure 3.3: Collisions viewed in Center of Mass and Laboratory frame. 3.2 Differential Cross-Section for Scattering by Angle Rutherford Cross-Section By definition the cross-section, σ, for any specified collision process when a particle is passing through a density n2 of targets is such that the number of such collisions per unit path length is n2 σ. Sometimes a continuum of types of collision is considered, e.g. we consider collisions at different angles (χ) to be distinct. In that case we usually discuss differential cross-sections dσ (e.g dχ ) defined such that number of collisions in an (angle) element dχ per unit path length dσ dσ is n2 dχ dx. [Note that dχ is just notation for a number. Some authors just write σ(χ), but I find that less clear.] Normally, for scattering-angle discrimination we discuss the differential cross-section per unit solid angle: dσ . (3.33) dΩs This is related to scattering angle integrated over all azimuthal directions of scattering by: dχ r0 r0 sin χ χ dΩs Figure 3.4: Scattering angle and solid angle relationship. So that since we have dΩs = 2π sin χdχ (3.34) dσ dσ dΩs = dχ dΩs dχ (3.35) dσ 1 dσ = dΩs 2π sin χ dχ (3.36) Now, since χ is a function (only) of the impact parameter, b, we just have to determine the number of collisions per unit length at impact parameter b. 









 















 













 



















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  dl Figure 3.5: Annular volume corresponding to db. Think of the projectile as dragging along an annulus of radius b and thickness db for an elementary distance along its path, d`. It thereby drags through a volume: d`2πbdb . (3.37) Therefore in this distance it has encountered a total number of targets d`2πbdb . n2 (3.38) at impact parameter b(db). By definition this is equal to d` dσ dbn2 . Hence the differential db cross-section for scattering (encounter) at impact parameter b is dσ = 2πb . db (3.39) Again by definition, since χ is a function of b dσ dσ dσ dσ db dx = db ⇒ = = . dχ db dχ db dχ (3.40) [db/dχ is negative but differential cross-sections are positive.] Substitute and we get dσ 1 dσ db b db = = dΩs 2π sin χ db dχ sin dχ . (3.41) [This is a general result for classical collisions.] For Coulomb collisions, in C of M frame, χ b = 2 b90 (3.42) db d χ b90 χ = b90 cot = − cosec2 . dχ dχ 2 2 2 (3.43)   cot ⇒ Hence b90 cot χ2 b90 dσ χ = cosec2 dΩs sin χ 2 2 χ χ 2 1 b cos 2 / sin 2 = 90 χ χ 2 2 sin 2 cos 2 sin2 χ2 b290 = 4 sin4 χ2 (3.44) (3.45) (3.46) This is the Rutherford Cross-Section. dσ b290 = dΩs 4 sin4 χ 2 (3.47) for scattering by Coulomb forces through an angle χ measured in C of M frame. Notice that dσ dΩs → ∞ as χ → 0. This is because of the long-range nature of the Coulomb force. Distant collisions tend to dominate. (χ → 0 ⇔ b → ∞). 3.3 Relaxation Processes There are 2 (main) different types of collisional relaxation process we need to discuss for a test particle moving through a background of scatterers: 1. Energy Loss (or equilibrium) 2. Momentum Loss (or angular scattering) The distinction may be illustrated by a large angle (90◦ ) scatter from a heavy (stationary) target. If the target is fixed, no energy is transferred to it. So the energy loss is zero (or small if scatterer is just ‘heavy’). However, the momentum in the x direction is completely ‘lost’ in this 90◦ scatter. This shows that the timescales for Energy loss and momentum loss may be very different. 3.3.1 Energy Loss For an initially stationary target, the final velocity in lab frame of the projectile is vL0 m 2 v1 m2 v1 ml v1 + cos χc , sin χc = m1 + m2 m1 + m2 m1 + m2   (3.48) So the final kinetic energy is K 0 1 1 = m1 vL02 = m1 v12 2 2 ( m1 m1 + m2 2 + 2m1 m2 cos χc (m1 + m2 )2   m22 2 2 + cos χ + sin χ c c (m1 + m2 )2 ( ) 1 2m1 m2 2 = m 1 v1 1 + (cos χc − 1) 2 (m1 + m2 )2 ( ) 1 2m1 m2 2 2 χc = m 1 v1 1 + 2 sin 2 2 (m1 + m2 )2 (3.49) ) (3.50) (3.51) (3.52) Hence the kinetic energy lost is ∆K = K − K 0 1 4m1 m2 2 χc m1 v2 2 sin 2 2 (m1 + m2 ) 1 4m1 m2 1 m1 v12 = 2 2  b 2 (m1 + m2 ) +1 b90 (3.53) = [using cot χc b = ] 2 b90 (3.54) (exact). For small angles χ  1 i.e. b/b90  1 this energy lost in a single collision is approximately !2   1 4m1 m2 b90 2 m 1 v1 (3.55) 2 b (m1 + m2 )2 If what we are asking is: how fast does the projectile lose energy? Then we need add up the effects of all collisions in an elemental length d` at all relevant impact parameters. The contribution from impact parameter range db at b will equal the number of targets encountered times ∆K: n d`2πbdb} | 2 {z encounters 1 4m1 m2 m1 v12 2 (m1 + m2 )2 | b90 b !2 {z (3.56) } Loss per encounter (∆K) This must be integrated over all b to get total energy loss. Z 4m1 m2 dK = n2 d`K (m1 + m2 )2 so b90 b !2 2πbdb (3.57) dK m1 m2 = K n2 8πb290 [ln |b|]max min d` (m1 + m2 )2 (3.58) We see there is a problem both limits of the integral (b → 0, b → ∞) diverge logarithmically. That is because the formulas we are integrating are approximate. 1. We are using small-angle approx for ∆K. 2. We are assuming the Coulomb force applies but this is a plasma so there is screening. 3.3.2 Cut-offs Estimates 1. Small-angle approx breaks down around b = b90 . Just truncate the integral there; ignore contributions from b < b90 . 2. Debye Shielding says really the potential varies as φ∝ exp  −r λD  instead of ∝ r 1 r (3.59) so approximate this by cutting off integral at b = λD equivalent to bmin = b90 . dK d` = Kn2 bmax = λD . (3.60) m1 m2 2 2 8πb90 ln |Λ| (m1 + m2 ) λD Λ = = b90  0 Te ne2  12 , q1 q 2 4π0 mr v12 (3.61) ! (3.62) So Coulomb Logarithm is ‘ln Λ’ λD Λ = = b90  0 Te ne2  12 , q1 q2 4π0 mr v12 ! (3.63) Because these cut-offs are in ln term result is not sensitive to their exact values. One commonly uses Collision Frequency. Energy Loss Collision Frequency is νK ≡ v1 m1 m2 1 dK 8πb290 ln |Λ| = n2 v1 K dL (m1 + m2 )2 (3.64) Substitute for b90 and mr (in b90 ) " νK Collision time q1 q2 m1 m2 = n2 v1 m2 2 2 8π 4π0 mm11+m v (m1 + m2 ) 2 1 q12 q22 8π ln Λ = n2 2 (4π0 ) m1 m2 v13 ln Λ (3.65) (3.66) τK ≡ 1/νK Effective (Energy Loss) Cross-section h 1 dK K d` σK = νK /n2 v1 = 3.3.3 #2 = σK n2 i 8π q12 q22 ln Λ 2 (4π0 ) m1 m2 v14 (3.67) Momentum Loss Loss of x-momentum in 1 collision is 0 ) ∆px = m1 (v1 − vLx m1 m2 = m 1 v1 1 − + cos χc m1 + m2 m1 + m2 m2 = px (1 − cos χc ) m1 + m2 m2 χ2c m2 2b290 ' px = px m1 + m2 2 m1 + m2 b2   (3.68)  (3.69) (3.70) (3.71) (small angle approx). Hence rate of momentum loss can be obtained using an integral identical to the energy loss but with the above parameters: Z bmax dp m2 2b290 = n2 p 2πbdb d` m1 + m2 b2 bmin m2 = n2 p 4π b290 ln Λ m1 + m2 (3.72) (3.73) Note for the future reference: dp dp m1 m2 =v = n2 v 2 4πb290 ln Λ. dt d` m1 + m2 (3.74) Therefore Momentum Loss. Collision Frequency νp = v1 1 dp m2 = n 2 v1 4πb290 ln Λ p d` m1 + m2 " m2 q 1 q2 = n 2 v1 4π m2 2 m1 + m2 4π0 mm11+m v 2 1 = n2 Collision Time (3.75) #2 ln Λ q12 q22 4π (m1 + m2 ) ln Λ m2 m21 v13 (4π0 )2 (3.76) (3.77) τp = 1/νp Cross-Section (effective) σ = νp /n2 v1 Notice ratio Energy Loss νK 2 = Momentum loss νp m1 m2 , m1 + m2 2m1 = 2 m2 m1 m1 + m2 (3.78) This is ' 2 = 1  1 if if if m1 >> m2 m1 = m2 m1  m2 . (3.79) (3.80) (3.81) Third case, e.g. electrons → shows that mostly the angle of velocity scatters. Therefore Momentum ‘Scattering’ time is often called ‘90◦ scattering’ time to ‘diffuse’ through 90◦ in angle. 3.3.4 ‘Random Walk’ in angle When m1 << m2 energy loss << momentum loss. Hence |vL0 ' v1 . All that matters is the scattering angle: χL ' χc ' 2b90 /b. Mean angle of deviation in length L is zero because all directions are equally likely. But: Mean square angle is ∆α2 = n2 L Z bmax χ2 2πbdb (3.82) bmin = Ln2 8π b290 ln Λ (3.83) Spread is ‘all round’ when ∆α2 ' 1. This is roughly when a particle has scattered 90◦ on average. It requires Ln2 8πb290 ln Λ = 1 . (3.84) So can think of a kind of ‘cross-section’ for ‘σ90 ’ 90◦ scattering as such that n2 L‘σ90 ’ = 1 when Ln2 8π b290 ln Λ = 1 i.e. ‘σ90 ’ = 8π b290 ln Λ (= 2σp ) (3.85) (3.86) This is 8 ln Λ larger than cross-section for 90◦ scattering in single collision. Be Careful! ‘σ90 ’ is not a usual type of cross-section because the whole process is really diffusive in angle. Actually all collision processes due to coulomb force are best treated (in a Mathematical way) as a diffusion in velocity space → Fokker-Planck equation. 3.3.5 Summary of different types of collision The Energy Loss collision frequency is to do with slowing down to rest and exchanging energy. It is required for calculating Equilibration Times (of Temperatures) Energy Transfer between species. The Momentum Loss frequency is to do with loss of directed velocity. It is required for calculating Mobility: Conductivity/Resistivity Viscosity Particle Diffusion Energy (Thermal) Diffusion Usually we distinguish between electrons and ions because of their very different mass: Energy Loss [Stationary Targets] Momentum Loss K νee = K νei = k K νii = νie = e4 ne (4π0 )2 Z 2 e4 ni (4π0 )2 Z 2 e4 ni (4πeo )2 Z 2 e4 ne e 2 (4π0 ) 8π ln Λ m2e ve3 8π ln Λ me mi ve3 8π ln Λ m2i vi3 8π ln Λ mi me vi3 me + me νee = νee × =1 2me   me + mi mi p νei =K νei × ' 2me 2me   mi + mi p K νii = νii × =1 2mi   me + mi 1 p K νie = νie = × ' 2mii 2 p  K  (3.87) Sometimes one distinguishes between ‘transverse diffusion’ of velocity and ‘momentum loss’. The ratio of these two is ∆p21 p2 ∆L , ∆pk p∆L dχ2L = dL  = , 1 dp p dL 2 m2 χ m1 +m2 c χ2c m2 m1 +m2 2 = (3.88) 2m2 . m1 + m2 (3.89) So 2m2 ‘σ90 ’ = ‘σp ’ m1 + m2 =1 like particles '2 m1 << m2 2m2 ' m2 << m1 . m1 (3.90) (3.91) (3.92) Hence ⊥ νee = ⊥ νei ⊥ ⊥ νii νie p νee = K νee (= ‘νee ’!!) ni = 2p νei = K νee Z 2 (= Zνee ) (= ‘νei ’) ne p K = νii = νii (= νii !!) (Like Ions) 2me p me K = νie = νie = K νii = νii mi mi (3.93) (3.94) (3.95) (3.96) [But note: ions are slowed down by electrons long before being angle scattered.] 3.4 Thermal Distribution Collisions So far we have calculated collision frequencies with stationary targets and single-velocity projectiles but generally we shall care about thermal (Maxwellian) distributions (or nearly thermal) of both species. This is harder to calculate and we shall resort to some heuristic calculations. 3.4.1 e→i Very rare for thermal ion velocity to be ∼ electron. So ignore ion motion. Average over electron distribution. Momentum loss to ions from (assumed) drifting Maxwellian electron distribution:  fe (v) = ne me 2πTe 3 2 m (v − vd )2 exp − 2T " # (3.97) Each electron in this distribution is losing momentum to the ions at a rate given by the collision frequency q 2 q 2 4π (me + mi ) ln Λ (3.98) νp = ni e i 2 mi m2e v 3 (4π0 ) so total rate of loss of momentum is given by (per unit volume) dp Z = fe (v) νp (v) me v d3 v dt To evaluate this integral approximately we adopt the following simplifications. − (3.99) 1. Ignore variations of ln Λ with v and just replace a typical thermal value in Λ = λD /b90 (v1 ). 2. Suppose q that drift velocity vd is small relative to the typical thermal velocity, written ve ≡ te /me and express fe in terms of u ≡ vve to first order in ud ≡ vved : −1 exp (u − ud )2 2 1 fe = ne  3  (3.100) (2π) 2 ve3 " # ne −u2 ' (1 + u.ud ) exp = (1 + ux ud ) fo 3 2 (2π) 2 ve3 (3.101) taking x-axis along ud and denoting by fo the unshifted Maxwellian. Then momentum loss rate per unit volume Z dpx − = fe νp me vx d3 v dt = νp (vt )me Z (1 + ux ud )fo = νp (vt )me vd Z ve3 vx d3 v v3 (3.102) u2x fo d3 v u3 To evaluate this integral, use the spherical symmetry of fo to see that: Z u2x 1 Z u2x + u2y + u3z 1 Z u2 3 3 f d v = f d v = fo d3 v o o u3 3Z u3 3 u3 1 α ve = fo 4πv 2 dv 3 0 v 2π Z α = ve fo 2vdv 3 0 ! Z α −v 2 2π ne exp dv 2 = ve 2ve2 3 (2π) 32 v 3 0 e 2π ne 2 = 3 2 = 1 ne . 3 (2π) 2 3 (2π) 2 Thus the Maxwell-averaged momentum-loss frequency is 1 dp 2 − ≡ ν ei = 1 νp (vt ) p dt 3 (2π) 2 (3.103) (3.104) (where p = me vd ne is the momentum per unit volume attributable to drift). ν ei = = 2 1 3 (2π) 2 2 3 (2π) 1 2 ni ni qe2 qi2 4π (me + mi ) ln Λe mi m2e ve3 (4π0 )2 ze2 4π0 !2 4π 1 2 3 2 ln Λe me Te (substituting for thermal electron velocity, ve , and dropping This is the standard form of electron collision frequency. me mi (3.105) (3.106) order term), where Ze = qi . 3.4.2 i→e Ion momentum loss to electrons can be treated by a simple Galilean transformation of the e → i case because it is still the electron thermal motions that matter. Ions −→ Electrons Ions ←− Electrons ≡ vi vi = 0 0 vd = −vi Figure 3.6: Ion-electron collisions are equivalent to electron-ion collisions in a moving reference frame. Rate of momentum transfer, dp , dt is same in both cases: Hence pe νei = pi νie or ν ie = dp = −pν dt (3.107) pe ne me ν ei = ν ei pi ni mi (3.108) (since drift velocities are the same). Ion momentum loss to electrons is much lower collision frequency than e → i because ions possess so much more momentum for the same velocity. 3.4.3 i→i Ion-ion collisions can be treated somewhat like e → i collisions except that we have to account for moving targets i.e. their thermal motion. Consider two different ion species moving relative to each other with drift velocity vd ; the targets’ thermal motion affects the momentum transfer cross-section. Using our previous expression for momentum transfer, we can write the average rate of transfer per unit volume as: [see 3.74 “note for future reference”] − m1 m2 dp Z Z = vr vr 4π b290 ln Λ f1 f2 d2 v1 d3 v2 dt m1 + m2 (3.109) where vr is the relative velocity (v1 − v2 ) and b90 is expressed b90 = q1 q2 1 4π0 mr vr2 (3.110) and mr is the reduced mass m1 m2 . m1 +m2 Since everything in the integral apart from f1 f2 depends only on the relative velocity, we proceed by transforming the velocity coordinates from v1 , v2 to being expressed in terms of relative (vr ) and average (V say) vr ≡ v1 − v2 V≡ ; m1 v1 + m2 v2 . m1 + m2 (3.111) Take f1 and f2 to be shifted Maxwellians in the overall C of M frame:  fj = nj mj 2πT 3 2 mj (vj − vdj )2 exp − 2T " # (j = 1, 2) (3.112) where m1 vd1 + m2 vd2 = 0. Then f1 f2 3  3 m2 2 m1 v12 m2 v22 exp − − = n1 n2 2πT 2T 2T   v1 .m1 vd1 v2 .m2 vd2 × 1+ + T T  m1 2πT 2 " # (3.113) to first order in vd . Convert CM coordinates and find (after algebra)  f1 f2 = n1 n2 M 2πT   32  × 1+ mr 2πT 3 mr vd .vr T 2 M V 2 mr vr2 exp − − 2T 2T " #  (3.114) where M = m1 + m2 . Note also that (it can be shown) d3 v1 d3 v2 = d3 vr d3 V . Hence dp Z Z = − dt 3  3 M 2 mr 2 vr mr vr 4π ln Λn1 n2 2πT! 2πT !  2 2  MV mr vr mr exp − 1+ vd .vr d3 vr d3 V exp − 2T 2T T  b290 (3.115) and since nothing except the exponential depends on V , that integral can be done: dp Z mr − = vr mr vr 4π ln Λ n1 n2 dt 2πT  3 2 −mr vr2 exp 2π ! mr 1+ vd .vr d3 vr T  (3.116) This integral is of just the same type as for e − i collisions, i.e. − Z 2 dp ux ˆ = vd vrt mr 4π b290 (vrt ) ln Λt n1 n2 fo (vr )d3 vr dt u3 2 = vd vrt mr 4 π b290 (vrt ) ln Λt n1 n2 2 3 (2π) 2 (3.117) q where vrt ≡ mTr , b290 (vrt ) is the ninety degree impact parameter evaluated at velocity vtr , and fˆo is the normalized Maxwellian. dp 2 − = 1 dt 3 (2π) 2  q 1 q2 4π0 2 4π ln Λt n1 n2 mr vd 3 m2r vrt (3.118) This is the general result for momentum exchange rate between two Maxwellians drifting at small relative velocity vd . To get a collision frequency is a matter of deciding which species is stationary and so what the momentum density of the moving species is. Suppose we regard 2 as targets then momentum density is n1 m1 vd so 2 1 q 1 q2 dp 2 = = 1 n2 n1 m1 vd dt 4π0 3 (2π) 2  ν 12 4π ln Λt . 3 mr vrt m1 (3.119) This expression works immediately for electron-ion collisions substituting mr ' me , recovering previous. For equal-mass ions mr = m2i mi +mi = 12 mi and vrt = q T mr = q 2T . mi Substituting, we get ν ii = 1 3π  1 2 ni q1 q2 4π0 2 4π 1 2 3 2 ln Λ (3.120) mi Ti √1 2 that is, times the e − i expression but with ion parameters substituted. [Note, however, that we have considered the ion species to be different.] 3.4.4 e→e Electron-electron collisions are covered by the same formalism, so ν ee = 1 3π 1 2 ne e2 4π0 !2 4π 1 3 ln Λ . (3.121) me2 Te2 However, the physical case under discussion is not so obvious; since electrons are indistiguishable how do we define two different “drifting maxwellian” electron populations? A more specific discussion would be needed to make this rigorous. √ Generally νee ∼ νei / 2 : electron-electron collision frequency ∼ electron-ion (for momentum loss). 3.4.5 Summary of Thermal Collision Frequencies For momentum loss: ν ei √ !2 2 Ze2 4π = √ ni ln Λe . 3 π 4π0 me12 Te32 (3.122) 1 ν ee ' √ ν ei . (electron parameters) 2 ne me ν ie = ν ei . ni mi √    1 2 2 qi qi0 2 4π mi0 ν ii0 = √ ni0 ln Λi 1 3 3 π 4π0 m 2 T 2 mi + mi0 i Energy loss K (3.123) (3.124) (3.125) i ν related to the above (p ν) by K ν= 2mi p ν . m1 + m2 (3.126) Transverse ‘diffusion’ of momentum ⊥ ν, related to the above by: ⊥ 3.5 3.5.1 ν= 2m2 p ν . m1 + m2 (3.127) Applications of Collision Analysis Energetic (‘Runaway’) Electrons Consider an energetic ( 21 me v12 >> T ) electron travelling through a plasma. It is slowed down (loses momentum) by collisions with electrons and ions (Z), with collision frequency: p νee = νee = ne p νei = e4 8π ln Λ 2 (4π0 ) m2e v13 1 Z νee 2 (3.128) (3.129) Hence (in the absence of other forces) d (me v) = − (p νee + p νei ) mp v dt   Z = − 1+ νee me v 2 (3.130) (3.131) This is equivalent to saying that the electron experiences an effective ‘Frictional’ force d Z (me v) = − 1 + νee me v = dt 2   Z e4 8π ln Λ = − 1+ ne 2 2 (4π0 ) me v 2  Ff Ff  (3.132) (3.133) Notice 1. for Z = 1 slowing down is 2 3 on electrons 1 3 ions 2. Ff decreases with v increasing. Suppose now there is an electric field, E. The electron experiences an accelerating Force. Total force d Z e4 8π ln Λ (mv) = −eE + Ff = −eE − 1 + ne 2 dt 2 (4π0 ) me v 2  F =  (3.134) Two Cases (When E is accelerating) 1. |eE| < |Ff |: Electron Slows Down 2. |eE| > |Ff |: Electron Speeds Up! Once the electron energy exceeds a certain value its velocity increases continuously and the friction force becomes less and less effective. The electron is then said to ahve become a ‘runaway’. Condition: 1 Z e4 8π ln Λ me v 2 > 1 + ne 2 2 2 2eE (4π0 )  3.5.2  (3.135) Plasma Resistivity (DC) Consider a bulk distribution of electrons in an electric field. They tend to be accelerated by E and decelerated by collisions. In this case, considering the electrons as a whole, no loss of total electron momentum by e − e collisions. Hence the friction force we need is just that due to ν ei . If the electrons have a mean drift velocity vd (<< vthe ) then d (me vd ) = −eE − ν ei me vd dt (3.136) Hence in steady state vd = −eE . me ν ei (3.137) The current is then ne e2 E (3.138) me ν ei Now generally, for a conducting medium we define the conductivity, σ, or resistivity, η, by j = −ne evd = j = σE ; ηj = E 1 σ= η ! (3.139) Therefore, for a plasma, σ= ne e2 1 = η me ν ei (3.140) Substitute the value of ν ei and we get 1 ni Z 2 e2 me2 8π ln Λ · η ' ne (4π )2 3√2π Te32 0 (3.141) 1 Ze2 me2 8π ln Λ = 3 √ (4π0 )2 3 2π Te2 (for a single ion species). (3.142) Notice 1. Density cancels out because more electrons means (a) more carriers but (b) more collisions. 2. Main dependence is η ∝ Te−3/2 . High electron temperature implies low resistivity (high conductivity). 3. This expression is only approximate because the current tends to be carried by the more energetic electrons, which have smaller νei ; thus if we had done a proper average over f (ve ) we expect a lower numerical value. Detailed calculations give ln Λ η = 5.2 × 10−5 3 Ωm (3.143) (Te /eV ) 2 for Z = 1 (vs. ' 10−4 in our expression). This is ‘Spitzer’ resistivity. The detailed calculation value is roughly a factor of two smaller than our calculation, which is not a negligible correction! 3.5.3 Diffusion For motion parallel to a magnetic field if we take a typical electron, with velocity vk ' vte it will travel a distance approximately `e = vte /ν ei (3.144) before being pitch-angle scattered enough to have its velocity randomised. [This is an orderof-magnitude calculation so we ignore ν̄ee .] ` is the mean free path. Roughly speaking, any electron does a random walk along the field with step size ` and step frequency ν ei . Thus the diffusion coefficient of this process is Dek ' `2e 2 vte ν ei ' . ν ei (3.145) Similarly for ions Dik ' `2i ν ii ' vti2 ν ii (3.146) Notice ν ii /ν ei ' Hence  me mi 1 2 ' vti vte (if Te ' Ti ) (3.147) `e ' `i Mean free paths for electrons and ions are ∼ same. The diffusion coefficients are in the ratio Di me ' De mi  1 2 : Ions diffuse slower in parallel direction. (3.148) Diffusion Perpendicular to Mag. Field is different GC rL GC rL Figure 3.7: Cross-field diffusion by collisions causing a jump in the gyrocenter (GC) position. Roughly speaking, if electron direction is changed by ∼ 90◦ the Guiding Centre moves by a distance ∼ rL . Hence we may think of this as a random walk with step size ∼ rL and frequency ν ei . Hence v2 2 De⊥ ' rLe ν ei ' te2 ν ei (3.149) Ωe Ion transport is similar but requires a discussion of the effects of like and unlike collisions. Particle transport occurs only via unlike collisions. To show this we consider in more detail the change in guiding center position at a collision. Recall mv̇ = qv ∧ B which leads to v⊥ = q rL ∧ B m (perp. velocity only). (3.150) This gives rL = B ∧ mv⊥ qB 2 (3.151) At a collision the particle position does not change (instantaneously) but the guiding center position (r0 ) does. r00 + r0L = r0 + rL ⇒ ∆r0 ≡ r00 − r0 = −(r0L − rL ) (3.152) Change in rL is due to the momentum change caused by the collision: r0L − rL = B B 0 ∧ m(v − v) ≡ ∧ ∆(mv) qB 2 qB 2 So ∆r0 = − B ∧ ∆(mv). qB 2 (3.153) (3.154) The total momentum conservation means that ∆(mv) for the two particles colliding is equal and opposite. Hence, from our equation, for like particles, ∆r0 is equal and opposite. The mean position of guiding centers of two colliding like particles (r01 + r02 )/2 does not change. No net cross field particle (guiding center) shift. Unlike collisions (between particles of different charge q) do produce net transport of particles of either type. And indeed may move r01 and r02 in same direction if they have opposite charge. v2 2 p ν ie ' ti2 p ν ie (3.155) Di⊥ ' rLi Ωi 2 2 Notice that rLi /rLe ' mi /me ; pν ie /ν ei ' me mi So Di⊥ /De⊥ ' 1 (for equal temperatures). Collisional diffusion rates of particles are same for ions and electrons. However energy transport is different because it can occur by like-like collisins. Thermal Diffusivity: 2 2 χe ∼ rLe (ν ei + ν ee ) ∼ rLe ν ei 2 p 2 χi ∼ rLi ( ν ie + ν ii ) ' rLi ν ii 1 mi me2 mi r2 ν ii ∼ 2Li ' 1 = rLe ν ei me m 2 me i  χi /χe (ν ei ∼ ν ee ) (ν ii >> ν ie ) (3.156) (3.157) 1 2 (equal T) (3.158) 1 Collisional Thermal transport by Ions is greater than by electrons [factor ∼ (mi /me ) 2 ]. 3.5.4 Energy Equilibration If Te 6= Ti then there is an exchange of enegy between electrons and ions tending to make Te = Ti . As we saw earlier 2me p me ⊥ K νei = νei = νei (3.159) mi mi So applying this to averages. Kν ei ' 2me ν ei mi (' ν ie ) (3.160) Thermal energy exchange occurs ∼ me /mi slower than momentum exchange. (Allows Te 6= Ti ). So dTe dTi (3.161) =− = −K ν ei (Te − Ti ) dt dt From this one can obtain the heat exchange rate (per unit volume), Hei , say: d 3 d ne Te = dt 2 dt 3 d = − n (Te − Ti ) = 4 dt Hei = −    3 ni Ti 2  (3.162) 3 K n ν ei (Te − Ti ) 2 (3.163) Important point: Kν ei 1 me ' Zνee ' 2 mi Z  Me mi  12 ν ii . (3.164) ‘Electrons and Ions equilibrate among themselves much faster than with each other’. 3.6 Some Orders of Magnitude 1. ln Λ is very slowly varying. Typically has value ∼ 12 to 16 for laboratory plasmas. 3 2. ν ei ≈ 6 × 10−11 (ni /m3 ) / (Te /eV ) 2 (ln Λ = 15, Z = 1). 5 −1 20 −3 e.g. = 2 × 10 s (when n = 10 m and Te = 1keV.) For phenomena which happen much faster than this, i.e. τ  1/νei ∼ 5µs, collisions can be ignored. Examples: Electromagnetic Waves with high frequency. 3. Resistivity. Because most of the energy of a current carrying plasma is in the B field not the K.E. of electrons. Resistive decay of current can be much slower than ν ei . E.g. Coaxial Plasma: (Unit length) Inductance L = µo ln ab Resistance R = η 1/πa2 L/R decay time µo πa2 b ne e2 b ln ' µo πa2 ln η a me ν ei a 2 2 2 2 ω a ne e a 1 1 1 ∼ = p2 · >> . 2 me 0 c ν ei c ν ei ν ei τR ∼ (3.165) Comparison 1 keV temperature plasma has ∼ same (conductivity/) resistivity as a slab of copper (∼ 2 × 10−8 Ωm). Ohmic Heating Because η ∝ Te−3/2 , if we try to heat a plasma Ohmically, i.e. by simply passing a current through it, this works well at low temperatures but its effectiveness falls off rapidly at high temperature. Result for most Fusion schemes it looks as if Ohmic heating does not quite yet get us to the required ignition temperature. We need auxilliary heating, e.g. Neutral Beams. (These slow down by collisions.)