plasma physics lecture 2 ian hutchinson

8/3/2019 Plasma Physics Lecture 2 Ian Hutchinson

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Chapter2MotionofChargedParticles inFieldsPlasmasarecomplicatedbecausemotionsofelectronsandionsaredeterminedbytheelectricandmagneticfieldsbutalsochange thefieldsbythecurrentstheycarry.FornowweshallignorethesecondpartoftheproblemandassumethatFieldsarePrescribed.Evenso,calculatingthemotionofachargedparticlecanbequitehard.Equationofmotion:

dvm = q ( E + v B ) (2.1)

dt charge E-field velocityB-fieldRateofchangeofmomentum LorentzForce

Havetosolvethisdifferentialequation,togetpositionrandvelocity(v=r)givenE(r,t),B(r,t).Approach: Startsimple,graduallygeneralize.

2.1 UniformBfield,E= 0.mv =qvB (2.2)

2.1.1 Qualitativelyin the plane perpendicular to B: Accel. is perp tov so particle moves in a circle whoseradiusrL issuchastosatisfy

2v

rListheangular(velocity)frequency

mrL2 =m = q vB (2.3)| |

1stequalityshows2 =v 2L2 /r (rL =v/)

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Figure2.1: Circularorbitinuniformmagneticfield.vHencesecondgivesm 2 = q vB | |

q Bi.e. = | | . (2.4)

mParticlemovesinacircularorbitwith

q Bangularvelocity = | | theCyclotronFrequency (2.5)

mv

andradius rl = theLarmorRadius. (2.6)

2.1.2 ByVectorAlgebra ParticleEnergyisconstant. proof: takev. Eq. ofmotionthen

d 1mv.v = mv2 =qv.(vB)=0. (2.7)dt 2

ParallelandPerpendicularmotionsseparate. v =constantbecauseaccel(vB)isperpendiculartoB.

PerpendicularDynamics:TakeBin z directionandwritecomponents

mvx =qvyB , mvy =qvxB (2.8)Hence

qB qB2vx = vy = vx =2vx (2.9)

m mSolution: vx =vcost (choosezerooftime)Substituteback:

m qvy = vx =|

q|vsint (2.10)

qB18


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Integrate:v

x=x0 + sint , y = y0 + q v cost (2.11) |q|

Figure2.2: Gyrocenter(x0, y0)andorbitThisistheequationofacirclewithcenterr0 =(x0, y0)andradiusrL =v/: GyroRadius.[Angleis=t]Directionofrotationisasindicatedoppositeforoppositesignofcharge:Ionsrotateanticlockwise. Electronsclockwiseaboutthemagneticfield.The current carried by theplasmaalways is in such a direction as toreduce the magneticfield.ThisisthepropertyofamagneticmaterialwhichisDiagmagnetic.Whenv isnon-zerothetotalmotionisalongahelix.

2.2 UniformBandnon-zeroEmv =q(E+vB) (2.12)

Parallel motion: Before,whenE=0thiswasv =const. NowitisclearlyqE

v = (2.13)m

Constantaccelerationalongthefield.Perpendicular MotionQualitatively:Speed of positive particle is greater at top than bottom so radius of curvature is greater.Result is that guiding center moves perpendicular to bothE andB. It drifts across thefield.Algebraically: Itisclearthatifwecanfindaconstantvelocityvd thatsatisfies

E+vd B=0 (2.14)19


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Figure2.3:EBdriftorbitthenthesumofthisdriftvelocityplusthevelocity

dvL = [rLei(tt0)] (2.15)

dtwhichwecalculatedfortheE=0gyrationwillsatisfytheequationofmotion.TakeBtheaboveequation:

0=EB+(vd B)B=EB+(vd.B)BB2vd (2.16)sothat

vd =EB (2.17)B2

doessatisfyit.Hencethefullsolutionis

v= v + vd + vL (2.18)parallel cross-fielddrift Gyration

whereqE

v = (2.19)m

andvd (eq2.17)istheEBdriftofthegyrocenter.CommentsonEBdrift:

1. Itisindependent ofthepropertiesofthedriftingparticle(q,m,v,whatever).2. Henceitisinthesame directionforelectronsandions.3. UnderlyingphysicsforthisisthatintheframemovingattheEBdriftE=0. We

havetransformedawaytheelectricfield.4. Formula given above is exact except for the fact that relativistic effects have been

ignored. Theywouldbeimportantifvd c.20
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2.2.1 DriftduetoGravityorotherForcesSupposeparticleissubjecttosomeotherforce,suchasgravity. WriteitFsothat

1mv =F+qvB=q( F+vB) (2.20)

qThisisjustliketheElectricfieldcaseexceptwithF/qreplacingE.Thedriftistherefore

vd = 1FB (2.21)q B2

Inthiscase,ifforceonelectronsandionsissame,theydriftinopposite directions.Thisgeneralformulacanbeusedtogetthedriftvelocityinsomeothercasesofinterest(seelater).

2.3 Non-UniformBFieldIfB-linesarestraightbutthemagnitudeofBvaries inspacewegetorbitsthat lookquali-tativelysimilartotheEBcase:

Figure2.4: B driftorbitCurvatureoforbitisgreaterwhereBisgreatercausinglooptobesmallonthatside. Resultis a drift perpendicular to bothB and B. Notice, though, that electrons and ions go inopposite directions(unlikeEB).AlgebraWe try to find a decomposition of the velocity as before into v = vd +vL where vd isconstant.Weshallfindthatthiscanbedoneonlyapproximately. Alsowemusthaveasimpleexpres-sionforB.ThiswegetbyassumingthattheLarmorradiusismuchsmallerthanthescalelengthofBvariationi.e.,

rL


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| |

in which case we can express the field approximately as the first two terms in a Taylorexpression:

BB0 +(r.)B (2.23)Thensubstitutingthedecomposedvelocityweget:

dvmdt = mvL =q(vB)=q[vL B0 +vd B0 +(vL +vd)(r.)B] (2.24)

or 0 = vd B0 +vL (r.)B+vd (r.)B (2.25)Now we shall find that vd/vL is also small, like r|B /B. Therefore the last term here is|secondorderbutthefirsttwoarefirstorder. Sowedropthelastterm.NowtheawkwardpartisthatvL andrL areperiodic. Substituteforr=r0 +rL so

0=vd B0 +vL (rL.)B+vL (r0.)B (2.26)itWenowaverageoveracyclotronperiod. Thelasttermis soitaveragestozero:e

0=vd B+vL (rL.)B. (2.27)Toperformtheaverageuse

v qsint, cost (2.28)rL =(xL,yL) =

q

cost,qsintvL =(xL, yL) = v (2.29)q| |

d

So [vL (r.)B]x = vyydyB (2.30)d

[vL (r.)B]y = vxydyB (2.31)(TakingB tobeinthey-direction).Then

2

2

v

v

cost

cos

t

costsint =0 (2.32)=vyy

2

1vq

q| |qq| | (2.33)=vxy = 2

So2 B

q 1v

vL (r.)B= (2.34)q 2| |

Substitutein:0=vd B 2 B

2q vq| | (2.35)

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andsolveasbeforetogetv

B B221q||= q v

q| |2

2B B

(2.36)vd =B2 B2

orequivalently 1mvq 2B

2B B (2.37)vd =

B2ThisiscalledtheGradBdrift.

2.4 CurvatureDriftWhentheB-fieldlinesarecurvedandtheparticlehasavelocityv alongthefield,anotherdriftoccurs.

Figure2.5: CurvatureandCentrifugalForceTake|B constant;radiusofcurvatureRe.|To1stordertheparticlejustspiralsalongthefield.Intheframeoftheguidingcenteraforceappearsbecausetheplasmaisrotatingaboutthecenterofcurvature.ThiscentrifugalforceisFcf

Fcf =m 2vRc pointingoutward (2.38)

asavectorRc2Fcf (2.39)=mvR2c

[Thereisalsoacoriolisforce2m( v)butthisaveragestozerooveragyroperiod.]Usethepreviousformulaforaforce

2mv

q B2 qB2 R2c23

1Fcf B Rc B(2.40)vd = =


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ThisistheCurvatureDrift.Itisoftenconvenienttohavethisexpressedintermsofthefieldgradients. SowerelateRctoB etc. asfollows:

Figure2.6: Differentialexpressionofcurvature(Caretsdenoteunitvectors)Fromthediagram

db=b2 b1 =Rc (2.41)and

d= Rc (2.42)So

dbdl =

RcRc =

RcR2c

(2.43)But(bydefinition)

dbdl =(B.)b (2.44)

Sothecurvaturedriftcanbewritten2 2 B ( bb.)Rc Bmv mv (2.45)vd =

R2c =

B2 q B2q2.4.1 VacuumFieldsRelationbetweenB &Rc driftsThe curvature and B are related because of Maxwells equations, their relation dependsonthecurrentdensityj. Aparticularcaseofinterestisj=0: vacuumfields.

Figure2.7: Localpolarcoordinatesinavacuumfield B=0 (staticcase) (2.46)

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Considerthez-component1

0=( B)z = (rB) (Br =0bychoice). (2.47)rr

B B= + (2.48)

r

r

or,inotherwords,B

(B)r = (2.49)Rc

[Notealso0=( B) =B/z :(B)z =0]andhence(B)perp =BRc/R2.cThusthegradBdriftcanbewritten:

mv2 mV2Rc B(2.50)vB = B B =

2q B3

2q R2

B2

candthetotaldriftacrossavacuumfieldbecomes

1 2 1 2vR +vB = mv + mv Rc B . (2.51)q 2 R2B2c

Noticethefollowing:1. Rc &B driftsareinthesame direction.2. Theyareinopposite directionsforoppositecharges.3. Theyareproportionaltoparticleenergies4. CurvatureParallelEnergy(2)

PerpendicularEnergyB5. Asaresultonecanveryquicklycalculatetheaveragedriftoverathermaldistribution

ofparticlesbecause1 2 Tmv = (2.52)2 21 2

mv

=

T

2degrees

of

freedom

(2.53)

2

Therefore b2TRc B 2TB b. (2.54)vR +vB =

q R2B2=

q B2c

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2.5 Interlude: Toroidal Confinement of Single Parti-cles

Since particles can move freely along a magnetic field even if not across it, we cannot ob-viously

confine

the

particles

in

astraight

magnetic

field.

Obvious

idea:

bend

the

field

lines

intocirclessothattheyhavenoends.

Figure2.8: ToroidalfieldgeometryProblemCurvature&B drifts

1 1 RB2 2 (2.55)+vd = mv mv R2B22q

1 1 12 2 (2.56)+= mv|vd| q mv2 BR

Ionsdriftup. Electronsdown. Thereisnoconfinement. Whenthereisfinitedensitythings

Figure2.9: Chargeseparationduetoverticaldriftare

even

worse

because

charge

separation

occurs

E

Outward

Motion.

E

B

2.5.1 Howtosolvethisproblem?Considerabeamofelectronsv =0v =0. Driftis

2mv

q BTR26

1vd = (2.57)


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WhatBz isrequiredtocancelthis?AddingBz givesacompensatingverticalvelocity

Bzv=v forBz


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Inthepoloidalcross-sectionthefielddescribesacircleasitgoesroundin.Equationofmotionofaparticleexactly followingthefieldis:

d B B B Br = v = v = v (2.61)

dt B B B B and

r=constant. (2.62)Nowaddontothismotionthecrossfielddriftinthe zdirection.

Figure2.11: Componentsofvelocityd B

r = v +vdcos (2.63)dt Bdr

= vdsin (2.64)dt

Takeratio,toeliminatetime:1dr udsin

= (2.65)rd Bv +vdcosB

TakeB,B,v,vd tobeconstants,thenwecanintegratethisorbitequation:Bv

[lnr] = [ln +vdcos|]. (2.66)|B

Taker=r0 whencos=0(= 2)thenBvd

r=r0/ 1+ cos (2.67)bv

If Bvd


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Figure2.12: Shifted,approximatelycircularorbitSubstituteforvd

12 + 2)(mvB 1 1mv2 (2.69)r0

B BRq v12

2+1 mv mv rp(2.70)2

qB Rvmv r0 r0If v =0 = (2.71),=rLqB R R

whererL istheLarmorRadiusinafieldB r/R.Providedissmall,particleswillbeconfined. Obviouslytheimportantthingisthepoloidalrotationofthefieldlines: RotationalTransform.RotationalTransform

rotationaltransform

poloidalangle1toroidalrotation (2.72)

(transform/2=) poloidalangletoroidalangle . (2.73)

(Originally,wasusedtodenotethetransform. Sinceabout1990ithasbeenusedtodenotethetransformdividedby2whichistheinverseofthesafetyfactor.)SafetyFactor

1 toroidalanglesq . (2.74)= = poloidalangle

Actuallythevalueoftheseratiosmayvaryasonemovesaroundthemagneticfield. Definitionstrictlyrequiresoneshouldtakethelimitofalargeno. ofrotations.qs isatopologicalnumber: numberofrotationsthelongwayperrotationtheshortway.Cylindricalapprox.:

rBqs = (2.75)

RB

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Intermsofsafetyfactortheorbitshiftcanbewrittenr Br

=rL =rL =rLqs (2.76)||R BR

(assumingB >>B).

2.6 TheMirrorEffectofParallelFieldGradients: E=0, BB

Figure2.13: BasisofparallelmirrorforceIntheabovesituationthereisanetforcealongB.Forceis

sin

==

|qvB|Br

B

sin= |q|vBsin (2.77)(2.78)

CalculateBr asfunctionofBz from.B=0.1

.B=rr(rBr)+zBz =0. (2.79)

HencerBr = rBz

z dr (2.80)SupposerL issmallenoughthat Bzz const.

[rBr]rL0

rL0 rdr

Bzz =

12 r2L

Bzz (2.81)

SoBr(rL)=1

2rLBzz (2.82)

sin=BrB =+

rL2

12

Bzz (2.83)

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>=

12

Hence2BzBz mv


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andforceisF=(B.) (constant), (2.94)

We shall show in a moment that || is constant for a circulating particle, regard as anelementarycircuit. Also,foraparticlealwayspoints inthe -Bdirection. [Notethatthismeans

that

the

effect

of

particles

on

the

field

is

to

decrease

it.]

Hence

the

force

may

be

written

F=B (2.95)Thisgivesusboth:

MagneticMirrorForce:F =B (2.96)

andGradBDrift:

1 FB B B. (2.97)vB = =q B2 q B2

2.6.2 isaconstantofthemotionAdiabaticInvariantProof from FParallelequationofmotion

dv

m

=F =

dB(2.98)

dt dz

SodB dB2mv dv = d(1mv)=vz = (2.99) dt dt 2 dz dt

ord 1 2 dB( mv)+ =0 . (2.100)dt 2 dt

ConservationofTotalKEd 1 2 1 2dt(2mv + 2mv)=0 (2.101)d 1= ( mv2 +B)=0 (2.102)dt 2

Combined dB

(B) =0 (2.103)dt dt

d= =0 Asrequired (2.104)

dt32


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AngularMomentumofparticleabouttheguidingcenteris

2mv 2m1mvrLmv = mv = 2 (2.105)

|q|B

q B| |2m= . (2.106)

q| |

Conservationofmagneticmomentisbasicallyconservationofangularmomentumabouttheguidingcenter.Proofdirect fromAngularMomentumConsiderangularmomentumaboutG.C.Because isignorable(locally)Canonicalangularmomentumisconserved.

p= [r (mv+qA)]z conserved. (2.107)HereAisthevectorpotentialsuchthatB= Athedefinitionofthevectorpotentialmeansthat

1 (rA))Bz = (2.108)

r r 2rL mrL rLA(rL) = r.Bzdr=2Bz = q (2.109)0 | |

Hencemp = q rLvm+q (2.110)

q q| |q | |

= m. (2.111)|q|

Sop=const=constant.ConservationofisbasicallyconservationofangularmomentumofparticleaboutG.C.2.6.3 MirrorTrappingF maybeenoughtoreflectparticlesback. Butmaynot!Letscalculatewhetheritwill:Supposereflectionoccurs.Atreflectionpointvr =0.Energyconservation

11 2 2m(v0 +v0)= mv2 (2.112)r2 233


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Figure2.14: MagneticMirrorconservation

1 2 1 2mv2 0 = 2mvr (2.113)

B0 BrHence

2 2 Br 2v0 +v0 = B0v0 (2.114)B0 v2

= 0 (2.115)2Br v0 +v2o2.6.4 PitchAngle

vtan = (2.116)

vB0 v2

= 0 =sin20 (2.117)2Br v0 +v20So,givenapitchangle0,reflectiontakesplacewhereB0/Br =sin20.If0 istoosmallnoreflectioncanoccur.Criticalanglec isobviously

c =sin1(B0/B1)12 (2.118)LossCone isall


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Figure2.15: Criticalanglec dividesvelocityspace intoaloss-coneandaregionofmirror-trapping

22m1mv= 2 (2.120)

q2 B2m=

q2 =constant. (2.121)NotethatifB changessuddenlymightnotbeconserved.

Figure2.16: FluxtubedescribedbyorbitBasicrequirement

rL


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Figure2.17: ParticleorbitsroundBsoastoperformalineintegraloftheElectricfield(dandvqareinoppositiondirections).

12mv2

= |q|Br2L = 2Bm|q|B

12mv2

B2B

(2.126)=

|| . (2.127)Hence

ddt

12mv2

= ||

2

12mv2

=db

dt (2.128)butalso

ddt

12mv2

= d

dt(B) . (2.129)Hence

d=0. (2.130)

dt

Noticethatsince= 2m,thisisjustanotherwayofsayingthatthefluxthroughthegyroq2

orbitisconserved.Noticealsoenergy increase. Methodofheating. AdiabaticCompression.

2.8 TimeVaryingE-field (E,Buniform)RecalltheEBdrift:

vEB =EB (2.131)B2

whenE variessodoesvEB. ThustheguidingcentreexperiencesanaccelerationvEB = d EB (2.132)

dt B2Intheframeoftheguidingcentrewhichisaccelerating,aforceisfelt.

Fa =m d EB (Pushedbackintoseat! ve.) (2.133)dt B2

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Thisforceproducesanotherdrift1Fa B m d EB

(2.134)vD = =q B2 qB2dt B2 B

m d

= (E.B)BB2E (2.135)qBdtm = E (2.136)

qB2

Thisiscalledthepolarizationdrift.m vD =vEB + vp =EB +

qB2 E (2.137)B2 1 = EB +

BE (2.138)B2

Figure 2.18: Suddenly turning on an electric field causes a shift of the gyrocenter in thedirectionofforce. Thisisthepolarizationdrift.Start-up effect: Whenwe switchon anelectric fieldtheaverage position(gyrocenter)ofaninitiallystationaryparticleshiftsoverby 1 theorbitsize. Thepolarizationdriftisthis2polarizationeffectonthemedium.Totalshift duetovp is

r vpdt= mqB2 Edt=

mqB2 [E] . (2.139)

2.8.1 DirectDerivationof dE effect: PolarizationDriftdt

ConsideranoscillatoryfieldE=Eeit (r0B)dv

m = q(E+vB) (2.140)dt

= q Eeit +vB (2.141)Tryforasolutionintheform

v=vDeit +vL (2.142)37


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where,asusual,vL satisfiesmvL =qvL BThen

(1) m(ivD =q(E+vD B) xit (2.143)SolveforvD : TakeBthisequation:

(2) mi(vD B)=q EB+ B2.v D B B2vD (2.144)|addmi (1)toq (2)toeliminatevD B.

2m22vD +q (EB B2vD)=miqE (2.145)m22 mi EB

(2.146)or: vD = E+1q2B2 qB2 B22 iq

i.e. vD 12 =B|q|E+EB

(2.147)B2Since i this is the same formula as we had before: the sum of polarization and

tEBdriftsexcept forthe[1 22]term.ThistermcomesfromthechangeinvD withtime(accel).Thusourearlierexpressionwasonlyapproximate. Agoodapproxif


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2(E.g. crosstermsarexyxyE).

Averageoveragyroorbit: r=rL(cos,sin,0).Averageofcrossterms=0.Then

2

rE(r) =E+(rL.)E+ L (2.153)2! 2E.lineartermrL =0. So

2(2.154)E(r) E+r

4L2EHenceEBwith1stfinite-Larmor-radiuscorrectionis

2 EBvEB = 1+rL 2 . (2.155)

r B2[Note: GradBdriftisafiniteLarmoreffectalready.]SecondandThirdAdiabaticInvariantsThereareadditionalapproximatelyconservedquantitieslikeinsomegeometries.

2.10 SummaryofDriftsvE = EB

B2 ElectricField (2.156)vFvE

==

1qFB

B21+r2L

42EB

B2

GeneralForceNonuniformE

(2.157)(2.158)

2B Bmv GradB (2.159)vB =

B32q2 Rc Bmv

Curvature (2.160)vR =q R2B2c

1 1 Rc B2 mv2 VacuumFields. (2.161)vR +v += mvB R2B2c2qEq

q |B| | |Polarization (2.162)vp =

MirrorMotionmv22B isconstant (2.163)

ForceisF=B.39

plasma physics lecture 2 ian hutchinson

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