Post on 06-Apr-2018
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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