plasma physics lecture 4 ian hutchinson

8/3/2019 Plasma Physics Lecture 4 Ian Hutchinson

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Chapter4FluidDescriptionofPlasmaThesingleparticleapproachgetstobehorriblycomplicated,aswehaveseen.Basicallyweneedamorestatisticalapproachbecausewecantfolloweachparticleseparately.If the details of the distribution function in velocity space are important we have to staywiththeBoltzmannequation. Itisakindofparticleconservationequation.

4.1 ParticleConservation(In3-dSpace)

Figure4.1: ElementaryvolumeforparticleconservationNumberofparticles inboxxyz isthevolume, V = xyz, timesthedensityn.Rate of change of number is is equal to the number flowing across the boundary per unittime,theflux. (Inabsenceofsources.)

[xyz n]= FlowOutacrossboundary. (4.1)

t

Takeparticlevelocitytobev(r) [norandomvelocity,onlyflow]andoriginatthecenteroftheboxrefertofluxdensityasnv=J.

FlowOut = [Jz(0,0,z/2)Jz(0,0,z/2)] xy + x + y . (4.2)64


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ExpandasTaylorseries

Jz(0,0,)=Jz(0)+ Jz . (4.3)z

So,

flowout (nvz)zxy + x +y (4.4) z

= V .(nv).HenceParticleConservation

n=.(nv) (4.5)

tNoticewehaveessentialprovedanelementaryformofGaussstheorem

.Ad3r = A.dS. (4.6)v

The expression: FluidDescription refers to any simplified plasma treatment which doesnot keeptrackofv-dependenceoff detail.

1. FluidDescriptionsareessentially3-d(r).2. Dealwithquantitiesaveragedovervelocityspace(e.g. density,meanvelocity,...).3. Omitsomeimportantphysicalprocesses(butdescribeothers).4. Providetractableapproachestomanyproblems.5.

Will

occupy

most

of

the

rest

of

my

lectures.

FluidEquationscanbederivedmathematicallybytakingmoments1 oftheBoltzmannEqua-tion.

0th moment d3v (4.7)1stmoment vd3v (4.8)

2ndmoment vvd3v (4.9)Theselead,respectively,to(0)Particle(1)Momentum(2)Energyconservationequations.We

shall

adopt

amore

direct

physical

approach.

1Theyarethereforesometimescalled MomentEquations.

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4.2 FluidMotionThemotionofafluidisdescribedbyavectorvelocityfieldv(r)(whichisthemeanvelocityof all the individual particles which make up the fluid at r). Also the particle densityn(r) is required. We are here discussing the motion of fluid of a single type of particle ofmass/charge,m/qsothechargeandmassdensityareqnandmnrespectively.Theparticleconservationequationwealreadyknow. ItisalsosometimescalledtheConti-nuityEquation

n+.(nv)=0 (4.10)

tItisalsopossibletoexpandthe.toget:

n+(v.)n+n.v=0 (4.11)

tThesignificance,here,isthatthefirsttwotermsaretheconvectivederivativeofn

D d +v. (4.12)

Dt dt tsothecontinuityequationcanbewritten

DDt

n=n.v (4.13)4.2.1 Lagrangian&EulerianViewpointsThereareessentially2views.

1. Lagrangian. Sitonafluidelementandmovewithitasfluidmoves.

Figure4.2: LagrangeanViewpoint2. Eulerian. Sit at a fixed point in space and watch fluid move through your volume

element: identityoffluidinvolumecontinuallychanging meansrateofchangeatfixed point(Euler).tD d +v.meansrateofchangeatmoving point(Lagrange).Dt dt t

dx +dy +dz : changeduetomotion.v.=t x t y t z

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Figure4.3: EulerianViewpointOurderivationofcontinuitywasEulerian. FromtheLagrangianview

D d N N d 1 dVDt n= dt V =V2 dt V =n V dt (4.14)

since total number of particles in volume element (N) is constant (we are moving withthem). (V =xyz.)

d dx dy dz

Now V = yz+ zx+ yx (4.15)dt dt dt dt1 dx 1 dy 1 dz

= V + + (4.16)x dt y dt x dt

d(x)But = vx(x/2)vx(x/2) (4.17)

dtvx

x etc. ...y ...z (4.18)x

Hence d vx vy vzV =V + + (4.19)dt x y z =V.v

andsoDDt n=n.v (4.20)

Lagrangian Continuity. Naturally, this is the same equation as Eulerian when one putsD = +v..Dt tThequantity.v istherateof(Volume)compressionofelement.

4.2.2 Momentum(Conservation)EquationEachoftheparticlesisactedonbytheLorentzforceq[E+uiB](ui isindividualparticlesvelocity).HencetotalforceonthefluidelementduetoE-Mfieldsis

(q[E+ui B])=N q(E+vB) (4.21)i

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(Usingmean: v= iu/N.)E-MForcedensity (perunitvolume)is:

FEM =nq(E+vB). (4.22)The

total

momentum

of

the

element

is

mui =mNv=V mnv (4.23)

isoMomentumDensity ismnv.Ifnootherforcesareactingthenclearlytheequationofmotionrequiresustosetthetimederivative of mnv equal toFEM. Because we want to retain the identity of the particlesunderconsiderationwewantD/Dti.e. theconvectivederivative(Lagrangianpicture).Ingeneralthereareadditionalforcesacting.

(1)Pressure (2)CollisionalFriction.4.2.3 PressureForceInagasp(=nT) istheforceperunitareaarisingfromthermalmotions. Thesurroundingfluidexertsthisforceontheelement:

Figure4.4: Pressureforcesonoppositefacesofelement.Netforceinxdirectionis

p(x/2)yz+p(x/2)yz (4.24)p p

xyz =V =V (p)x (4.25) x xSo(isotropic)pressureforcedensity (/unitvol)

Fp = (4.26)pHowdoesthisariseinourpictureabove?Answer: Exchangeofmomentumbyparticlethermalmotionacrosstheelementboundary.Although inLagrangianpicturewemovewiththeelement(asdefinedbymeanvelocityv)individualparticlesalsohavethermalvelocitysothattheadditionalvelocitytheyhaveis

wi =ui v peculiarvelocity (4.27)68


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Because of this, some cross the element boundary and exchange momentum with outside.(Eventhough there isnonetchangeofnumberof particles in element.) Rateofexchangeofmomentumduetoparticleswithpeculiarvelocityw,d3wacrossasurfaceelementdsis

f(w)mwd3w

w .ds (4.28)mom

mdensityatw flowrateacrossds

Integrateoverdistribfunctiontoobtainthetotalmomentumexchangerate:ds. mwwf(w)d3w (4.29)

Thethingintheintegralisatensor. Writep = mwwf(w)d3w (PressureTensor) (4.30)

Thenmomentumexchangerateisp.ds (4.31)

Actually,iff(w)isisotropic(e.g. Maxwellian)thenpxy = mwx wy f(w)d3w=0 etc. (4.32)

andpxx = mw2x f(w)d3w nT(=pyy =pzz =p) (4.33)Sothentheexchangerateispds. (ScalarPressure).IntegratedsoverthewholeV thenxcomponentofmomm exchangerateis

px2

yz px2

yz=V (p)x (4.34)

andsoTotalmomentumlossrateduetoexchangeacrosstheboundaryperunitvolumeis

p (=Fp) (4.35)Intermsofthemomentumequation,eitherweputponthemomentumderivativesideorFp onforceside. Theresultisthesame.IgnoringCollisions,MomentumEquationis

D(mnVv)=[FEM +Fp] V (4.36)

DtDRecallthatnV =N ;Dt(N)=0;so

DvL.H.S.=mnV . (4.37)

dtThus,substitutingforFs:

MomentumEquation.Dv v

mn =mn +v.v (4.38)=qn(E+vB) pDt t

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.{ .{

+

4.2.4 MomentumEquation: EulerianViewpointFixedelementinspace. Plasmaflowsthroughit.

1. E.M.forceonelement(perunitvol.)FEM =nq(E+vB) asbefore. (4.39)

2. Momentumfluxacrossboundary(perunitvol)= . m(v+w)(v+w)f(w)d3w (4.40)= m(vv+ vw+wv +ww)f(w)d3w} (4.41)

integratesto0= mnvv+p} (4.42)=

mn(v.)v

+

mv

[ .(nv)]

+

p

(4.43)

(Takeisotropicp.)

3. Rateofchangeofmomentumwithinelement(perunitvol)

= (mnv) (4.44)t

Hence,totalmomentumbalance:

(mnv)+mn(v.)v+mv[ .(nv)]+p=FEM (4.45)t

Usethecontinuityequation:n

.(nv)=0 , (4.46)t

tocancelthethirdtermandpartofthe1st: n v v

(mnv)+mv(.(nv))=mv{ +.(nv)} +mn =mn (4.47)t t t t

ThentakeptoRHStogetfinalform:MomentumEquation:

vmn +(v.)v =nq(E+vB) p. (4.48)

tAsbefore,viaLagrangianformulation. (Collisionshavebeenignored.)

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4.2.5 EffectofCollisionsFirstnoticethatlikeparticlecollisionsdonotchangethetotalmomentum(whichisaveragedoverallparticlesofthatspecies).Collisionsbetweenunlike particlesdo exchangemomentumbetweenthespecies. Thereforeoncewerealizethatanyquasi-neutralplasmaconsistsofatleasttwodifferentspecies(elec-tronsand ions)andhencetwodifferent interpenetratingfluidswemayneedtoaccount foranothermomentumloss(gain)term.Therateofmomentumdensitylossbyspecies1collidingwithspecies2is:

12n1m1(v1 v2) (4.49)Hencewecanimmediatelygeneralizethemomentumequation to

m1n1 v1t +(v1.)v1 =n1q1(E+v1 B) p1 12n1m1(v1 v2) (4.50)

Withsimilarequationforspecies2.

4.3 TheKeyQuestion forMomentumEquation:Whatdowetakeforp?Basicallyp=nT isdeterminedbyenergybalance,whichwilltellhowT varies. Wecouldwriteanenergyequationinthesamewayasmomentum. However,thiswouldthencontainatermforheatflux,whichwouldbeunknown. Ingeneral,thekth momentequationcontainsatermwhichisa(k+1)th moment.Continuity, 0th equationcontainsvdeterminedbyMomentum,1st equationcontainspdeterminedbyEnergy, 2nd equationcontainsQdeterminedby...Inordertogetasensibleresultwehavetotruncate thishierarchy. Dothisbysomesortofassumptionabouttheheatflux. ThiswillleadtoanEquationofState:

pn =const. (4.51)The value of to be taken depends on the heat flux assumption and on the isotropy (orotherwise)

of

the

energy

distribution.

Examples

1. Isothermal: T =const.: =1.2. Adiabatic/Isotropic: 3degreesoffreedom = 5.

33. Adiabatic/1degreeoffreedom =3.

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+

4. Adiabatic/2degreesoffreedom =2.Ingeneral,n(/2)T =p(V/V) (Adiabaticdegrees)

T V n(4.52)= = +

2 T V nSo

p n T 2 n= + = 1+ , (4.53)

p n T ni.e.

pn(1+2)=const. (4.54)In a normal gas, which holds together by collisions, energy is rapidly shared between 3space-degreesoffreedom. Plasmasareoftenrathercollisionlesssocompressionin1dimensionoften

stays

confined

to

1-degree

of

freedom.

Sometimes

heat

transport

is

so

rapid

that

the

isothermal approach is valid. It depends on the exact situation; so lets leave undefinedfornow.

4.4 SummaryofTwo-FluidEquationsSpeciesjPlasmaResponse

1. Continuity:nj .(njvj )=0 (4.55)t

2. Momentum:vj

mjnj +(vj .)vj jknjmj(vj vk) (4.56)t =njqj(E+vj B) pj

3. Energy/EquationofState:pjn

=const.. (4.57)j(j =electrons,ions).

MaxwellsEquations.B = 0 .E=/o (4.58)

B 1 E B= oj+c2 t E= t (4.59)

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With = qene +qini =e(ne +Zni) (4.60)j = qeneve +qinivi =e(neve +Znivi) (4.61)

= ene(ve vir) (Quasineutral) (4.62)Accounting

Unknowns Equationsne,ni 2 Continuitye,i 2ve,vi 6 Momentume,i 6

pe,pi 2 Statee,i 2E,B 6 Maxwell 8

16 18but

2of

Maxwell

(.

equs)

are

redundant

because

can

be

deduced

from

others:

e.g.

.( E)and.( B)

==

0=t(.B)

0=o.j+ 1c2

t (.E)=

1c2

t

o +.E

(4.63)(4.64)

So16equsfor16unknowns.EquationsstillverydifficultandcomplicatedmostlybecauseitisNonlinearIn some cases can get a tractable problemby linearizing. Thatmeans, takesomeknownequilibrium solution and suppose the deviation (perturbation) from it is small so we canretain

only

the

1st

linear

terms

and

not

the

others.

4.5 Two-FluidEquilibrium:DiamagneticCurrent Slab: =0

y,z =0.x StraightB-field: B=Bz.

Equilibrium:t =0 (E=)

Collisionless: 0.MomentumEquation(s):

mjnj(vj.)vj (4.65)=nj qj(E+vj B) pjDropj suffixfornow. Thentakex,ycomponents:

d dpmnvx vx = nq(Ex +vyB) (4.66)

dx dxd

mnvx vy = nq(0 vxB) (4.67)dx

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Eq 4.67issatisfiedbytakingvx =0.Then4.66dp

nq(Ex +vyB) =0. (4.68)dx

i.e.1 dpvy = Ex + (4.69)B nqB dx

or,invectorform:B

v= EB p (4.70)B2 nq B2

EB drift Diamagnetic DriftNotice:

Inmagneticfield()fluidvelocityisdeterminedbycomponentofmomentumequationorthogonal toit(andtoB).

Additionaldrift(diamagnetic)arisesinstandardFBformfrompressureforce. Diagmagneticdriftisoppositeforoppositesignsofcharge(electronsvs. ions).

Nowrestorespeciesdistinctionsandconsiderelectronsplussingle ionspecies i. Quasineu-tralitysaysniqi =neqe. Henceaddingsolutions

BEB(niqi +neqe) +pi)neqeve +niqivi =

B2 (pe (4.71)B2=0

Hencecurrentdensity: Bj=(pe +pi) (4.72)

B2Thisisthediamagneticcurrent. Theelectricfield,E,disappearsbecauseofquasineutrality.(Generalcase jqjnjvj =( pj)B/B2).

4.6 Reduction ofFluidApproach to the SingleFluidEquations

So far we have been using fluid equations which apply to electrons and ions separately.ThesearecalledTwoFluid equationsbecausewealwayshavetokeeptrackofbothfluidsseparately.AfurthersimplificationispossibleandusefulsometimesbycombiningtheelectronandionequationstogethertoobtainequationsgoverningtheplasmaviewedasaSingleFluid.

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Recall2-fluidequations:nj

Continuity(Cj) +.(njvj )=0. (4.73)t

Momentum(Mj) mjnj +vj . vj =njqj(E+vj B) pj +Fjk (4.74)

t(wherewejustwriteFjk = vjknjmj(vj vk)forshort.)Nowwerearrangethese4equations(2 2species)byaddingandsubtractingappropriatelytogetnewequationsgoverningthenewvariables:

MassDensity m = neme +nimi (4.75)Cof MVelocity V = (nemeve +nimivi)/m (4.76)Chargedensity q = qene +qini (4.77)ElectricCurrentDensity j = qeneve +qinivi (4.78)

= qene(ve vi) byquasineutrality (4.79)TotalPressure p = pe +pi (4.80)

1stequation: takeme Ce +mi Ci m

(1) +.(mV)=0 MassConservation (4.81)t

2ndtakeqe Ce +qI Ci (2)

q+.j=0 ChargeConservation (4.82)t

3rdtakeMe +Mi. Thisisabitmoredifficult. RHSbecomes:=qE+jB (pe +pi) (4.83)njqj(E+vj B) pj +Fjk

(weusethefactthatFei Fie sononet friction). LHSis

mj nj +vj. vj (4.84)tj

Thedifficultyhereisthattheconvectivetermisnon-linearandsodoesnoteasilylenditselftoreexpressionintermsofthenewvariables. Butnotethatsinceme


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so:

(3) m +V. V=q (4.86)t E+jB p

qeFinallywetakemeMe + qi Mi toget:mi

njq2j qj qj+(vj.) (E+vj B) Fjk} (4.87)pj +{njqj vj =t mj mj mjj jAgainseveraldifficultiesarisewhichitisnotveryprofitabletodealwithrigorously. ObservethattheLHScanbewritten(usingquasineutralityniqi +neqe =0)asmt j providedmwe discard the term in (v.)v. (Think of this as a linearization of this question.) [The(v.)vconvectiveterm isatermwhich isnotsatisfactorilydealtwith inthisapproachtothesinglefluidequations.]IntheR.H.S.weusequasineutralityagaintowrite

2j 1 1 mini +menenjq qeqi22 22E E=n E= mE, (4.88)+= neq eqe e

mj neme nimi nemenimi memij2 2 2

injq neq niqq eve +vj = vimj me mi

qeqi neqemi niqime= ve + vi}

memi{ qi qeqeqi mi me

= nemeve +nimivi + (qeneve +qinivi)}memi{ qi qe

qeqi mi me(4.89)=

memi{mV qi + qe j}Also,rememberingFei =

j

einemi(ve vi)=Fie,qj

mjFjk = ei

neqe neqimemi

(ve vi)

= ei 1qe

qimemi j (4.90)

Soweget m

tjm =

qeqimemi mE+ mV

miqi +

meqe j B

qe

mepe qi

mipi 1qeqi

memi eij (4.91)

memiRegroupaftermultiplyingbyqeqim:

memi j 1 mi meE+VB = + + jB (4.92)

qeqi t m m qi qeqe qi memi qeme memi

pe pi eijme +mi mqeqi 1 qi mi qeqim

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NoticethatthisisanequationrelatingtheElectricfieldintheframemovingwiththefluid(L.H.S.)tothingsdependingoncurrentji.e. thisisageneralizedformofOhmsLaw.OneessentiallyneverdealswiththisfullgeneralizedOhmslaw. MakesomeapproximationsrecognizingthephysicalsignificanceofthevariousR.H.S.terms.

memi j arisesfromelectroninertia.qeqi t m

itwillbenegligibleforlowenoughfrequency.1 mi me

+ jBiscalledtheHallTerm.m qi qe

andarises because current flow ina B-fieldtendstobedivertedacrossthemagnetic field.Itisalsooftendroppedbutthejustificationfordoingsoislessobviousphysically.

qi q

e

pi term pe forcomparablepressures,mi meandthelatteristheHallterm;soignoreqipi/mi.Lastterminjhasacoefficient,ignoringme/mi c.f. 1whichis

memiei meei= = theresistivity. (4.93)

q2qeqi(nimi) eneHencedroppingelectroninertia,Halltermandpressure,thesimplifiedOhmslawbecomes:

E+VB=j (4.94)Finalequationneeded: state:

pene +pini =constant.e i

Takequasi-neutrality ne ni m. Takee =i,thenp =const. (4.95)m

4.6.1 SummaryofSingleFluidEquations:M.H.D.MassConservation: m

t + (mV)=0 (4.96)ChargeConservation: q

t +.j=0 (4.97)

Momentum: mt+V. V=qE+jB p (4.98)

OhmsLaw: E+VB=j (4.99)Eq.of State: pm =const. (4.100)

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4.6.2 HeuristicDerivation/ExplanationMassCharge: Obvious.

Momm m

t+V. V= qE + jB p (4.101) Electric Magnetic Force Pressure

rate of change of body force on currenttotal momentum density

OhmsLawTheelectricfield seenbyamoving(conducting)fluid isE+VB=EV electricfield inframeinwhichfluidisatrest. Thisisequaltoresistiveelectricfieldj:

EV =E+VB=j (4.102)TheqE term isgenerallydroppedbecause it ismuchsmallerthanthejBterm. Toseethis,takeordersofmagnitude:

.E=q/0 so q E0/L (4.103)1E

+ so E=j B/0L (4.104) B=0jc2 t

Therefore 2 qE 0 B L0 L2/c2

= lighttransittime2

. (4.105)jB L 0L B2 (0L2)2 resistiveskintime

Thisisgenerallyaverysmallnumber. Forexample,evenforasmallcoldplasma,sayTe =1eV

(

2

10

3mho/m),

L

=

1cm,

this

ratio

is

about

10

8.

Conclusion: theqE force ismuch smallerthanthejBforce foressentiallyallpracticalcases. Ignoreit.Normally,also,oneusesMHDonlyforlowfrequencyphenomena,sotheMaxwelldisplace-mentcurrent,E/c2tcanbeignored.AlsoweshallnotneedPoissonsequationbecausethatistakencareofbyquasi-neutrality.4.6.3 MaxwellsEquations forMHDUse

.B=

0

; B

;

.

(4.106)

E=

t B=oj

The MHD equations find their major use in studying macroscopic magnetic confinementproblems. InFusionwewantsomehowtoconfinetheplasmapressureaway fromthewallsofthechamber,usingthemagneticfield. InstudyingsuchproblemsMHDisthemajortool.On the other hand if we focus on a small section of the plasma as we do when studyingshort-wavelength waves, other techniques: 2-fluid or kinetic are needed. Also, plasma isapprox. uniform.

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MacroscopicPhenomenaMHDMicroscopicPhenomena2-Fluid/Kinetic

4.7 MHDEquilibriaStudy of how plasma can be held by magnetic field. Equilibrium V = = 0. So

tequationsreduce. MassandFaradayslawareautomatic. Weareleftwith

(Momm) ForceBalance 0=jB p (4.107)Ampere B=oj (4.108)

Plus .B=0,.j=0.NoticethatprovidedwedontaskquestionsaboutOhms law. Edoesntcome intoMHDequilibrium.ThesedeceptivelysimplelookingequationsarethesubjectofmuchofFusionresearch. Thehardpartistakingintoaccountcomplicatedgeometries.Wecandosomeusefulcalculationsonsimplegeometries.4.7.1 -pinch

Figure4.5: -pinchconfiguration.Socalledbecauseplasmacurrentsflowin-direction.UseMHDEquationsTaketobelength,uniforminz-dir.BysymmetryBhasonlyzcomponent.Bysymmetryjhasonly- comp.Bysymmetryphasonlyrcomp.

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Soweonlyneed

Force (jB)r (p)r =0 (4.109)Ampere ( B) = (oj) (4.110)

i.e. jBz rp=0 (4.111)

rBz = oj (4.112)Eliminatej : Bz

oBzr

pr = 0 (4.113)

i.e.

r

B2z2o +p

= 0 (4.114)

Solution B2z2o +p = const. (4.115)

Figure4.6: BalanceofkineticandmagneticpressureB2 B2z z ext+p= (4.116)2o 2o

[RecallSingleParticleProblem]Thinkoftheseasapressureequation. Equilibriumsaystotal pressure=const.

B2z + p =const. (4.117)2o kinetic pressure

magnetic pressure

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Ratioofkinetictomagneticpressureisplasma.2op

=B2z

(4.118)measures efficiency of plasma confinement by B. Want large for fusion but limited byinstabilities,etc.4.7.2 Z-pinch

Figure4.7: Z-pinchconfiguration.so

called

because

j

flows

in

z-direction.

Again

take

to

be

length

and

uniform.

j=jzez B=B e (4.119)

pForce (jB)r (p)r =jzB =0 (4.120)r

1 Ampere = (rB) ojz =0 (4.121)( B)z (oj)z rr

Eliminatej:B p

(rB) =0 (4.122)orr r

or B2 B2 + + p =0 (4.123)or r 2o

Extra Term Magnetic+Kinetic pressureExtratermactslikeamagnetictension force. ArisesbecauseB-fieldlinesarecurved.

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Canintegrateequation b B2 bdr B2 + +p(r) =0 (4.124)a o r 2o a

Ifwechoosebtobeedge(p(b)=0)andseta=rweget

Figure4.8: Radiiofintegrationlimits.B2(b) B2(r) b B2drp(r)= + (4.125)

2o 2o r o rForce balance in z-pinch is somewhat more complicated because of the tension force. Wecantchoosep(r)andj(r)independently;theyhavetobeselfconsistent.Example j=const.

1r

r(rB)=ojz B =

ojz2 r (4.126)

Hencep(r) =

12o

ojz

2 2

{b2

r2 + br 2r

dr

} (4.127)= oj2z

4 {b2 r2} (4.128)

Figure4.9: ParabolicPressureProfile.AlsonoteB(b)= ojzb2 so

p= B2b2o

2b2{b

2 r2} (4.129)

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4.7.3 StabilizedZ-pinchAlsocalledscrewpinch, z pinchorsometimeslooselyjustz-pinch.Z-pinchwithsomeadditionalBz aswellasB

(Force)r jBz jzB =0 (4.130)r

Ampere: Bz =oj (4.131)r

1 (rb)=ojz (4.132)

rrEliminatej:

Bz Bz B p(rB) =0 (4.133)

o r orr ror

B2 B2 + + p =0 (4.134)or r 2o

Mag Tension Mag (+z)+Kinetic pressure only

4.8 SomeGeneralPropertiesofMHDEquilibria4.8.1 Pressure&Tension

jB p=0 : (4.135) B=ojWecaneliminatejinthegeneral casetoget

1p. (4.136)

o ( B)B=Expandthevectortripleproduct:

1 1(4.137)p=

o (B.)B B22oBputb=|B|

sothatB=Bb=Bb+bB. Then1 1

p =o{B

2(b.)b+Bb(b.)B} 2oB

2 (4.138)B2 1

= ( b(b.))B2 (4.139)o (b.)b2oB2 B2

=o (b.)b 2p (4.140)

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Now B2 istheperpendicular(toB)derivativeofmagneticpressureand(b.)bisthe2ocurvature ofthemagneticfieldlinegivingtension.

(b.)b hasvalue 1. R: radiusofcurvature.R

| |

4.8.2 MagneticSurfaces0=B.[jB p] =B.p (4.141)

*Pressureisconstantonafieldline(inMHDsituation).(Similarly,0=j.[jB p] =j.p.)

Figure4.10: Contoursofpressure.Considersomearbitraryvolumeinwhichp=0. Thatis,someplasmaofwhatevershape.Drawcontours(surfacesin3-d)onwhichp=const. Atanypointonsuchanisobericsurfacepisperptothesurface. ButB.p=0impliesthatBisalsoperptop.

Figure4.11:Bisperpendiculartopandsoliesintheisobaricsurface.HenceBliesinthesurfacep=const.Inequilibriumisobaricsurfaces aremagneticsurfaces.[Thisargumentdoesnotwork ifp=const. i.e. p= 0.Thenthereneedbenomagneticsurfaces.]

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4.8.3 CurrentSurfacesSincej.p=0inequilibriumthesameargumentappliestocurrentdensity. Thatisjliesinthesurfacep=const.IsobaricSurfacesareCurrentSurfaces.MoreoveritisclearthatMagneticSurfacesareCurrentSurfaces.(sincebothcoincidewithisobaricsurfaces.)[It is important to note that the existence of magnetic surfaces is guaranteed only in theMHD approximation when p = 0 > Taking account of corrections to MHD we may nothavemagneticsurfacesevenifp=0.]4.8.4 Low equilibria: Force-FreePlasmasInmanycasestheratioofkinetictomagneticpressureissmall,


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Bz = BoJo(or) (4.149)B = BoJ1(or) (4.150)

foror>1stzeroofJo thetoroidalfieldreverses. Thereareplasmaconfinementschemeswithconst. ReversedFieldPinch.

4.9 ToroidalEquilibriumBendaz-pinchintoatorus

Figure4.12: Toroidalz-pinchB fieldsduetocurrentarestrongeratsmallRsidePressure(Magnetic)Forceoutwards.HavetobalancethisbyapplyingaverticalfieldBv topushplasmabackbyj Bv.

Figure4.13: ThefieldofatoroidalloopisnotanMHDequilibrium. Needtoaddaverticalfield.Benda-pinchintoatorus: BisstrongeratsmallRside outwardforce.CannotbebalancedbyBv becausenoj. Noequilibriumforatoroidallysymmetric-pinch.UnderlyingSingleParticlereason:Toroidal-pinchhasB only. Aswehaveseenbefore,curvaturedriftsareuncompensatedinsuchaconfigurationandleadtorapidoutward motion.

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Figure4.14: Charge-separationgivingoutwarddriftisequivalenttothelackofMHDtoroidalforcebalance.Weknowhowtosolvethis: RotationalTransform: getsomeB. Easiestway: addj. FromMHDviewpointthisallowsyoutopushtheplasmabackbyj Bv force. Essentially,thisisTokamak.

4.10Plasma

Dynamics

(MHD)

Whenwewanttoanalyzenon-equilibriumsituationswemustretainthemomentumterms.This will give a dynamic problem. Before doing this, though, let us analyse some purelyKinematicEffects.IdealMHD Seteta=0inOhmsLaw.Agoodapproximationforhighfrequencies,i.e. timesshorterthanresistivedecaytime.

E+VB=0. IdealOhmsLaw. (4.151)Also

B

FaradaysLaw. (4.152) E= tTogetherthesetwoequations implyconstraintsonhowthemagneticfieldcanchangewithtime: EliminateE:

B+ (VB)=+ (4.153)

tThisshowsthatthechangesinBarecompletelydeterminedbytheflow,V.

4.11 FluxConservationConsideranarbitraryclosedcontourC andspawningsurfaceS inthefluid.FluxlinkedbyC is

= B.ds (4.154)S

LetC andS movewithfluid:Totalrateofchangeofisgivenbytwoterms:

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Figure4.15: Motionofcontourwithfluidgivesconvectivefluxderivativeterm.

B = .ds + B.(V dl) (4.155)

CS

t

DuetochangesinB DuetomotionofC

= (VB).dl (4.156)S E.ds C

= (E+VB).dl=0 byIdealOhmsLaw. (4.157)C

Fluxthroughanysurfacemovingwithfluidisconserved.

4.12 FieldLineMotionThinkofafieldlineastheintersectionoftwosurfacesbothtangentialtothefieldeverywhere:

Figure4.16: Fieldlinedefinedbyintersectionoftwofluxsurfacestangentialtofield.Letsurfacesmovewithfluid.Sinceallpartsofsurfaceshadzerofluxcrossingatstart,theyalsohavezeroafter,(byfluxconserv.).Surfacesaretangentaftermotion

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Theirintersectiondefinesafieldlineafter.Wethinkofthenewfieldlineasthesamelineastheoldone(onlymoved).Thus:

1. Numberoffieldlines(flux)throughanysurfaceisconstant. (FluxCons.)2. Alineoffluidthatstartsasafieldlineremainsone.

4.13 MHDStabilityThe factthatonecanfindanMHDequilibrium (e.g. z-pinch)doesnotguaranteeausefulconfinementschemebecausetheequil. mightbeunstable. Ballonhillanalogies:

Figure4.17: PotentialenergycurvesAn equilibrium is unstable if the curvature of the Potential energy surface is downward

d2awayfromequil. Thatisifdx2{Wpot}


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Figure4.19: Skin-current,sharpboundarypinch.Attheplacewhereitpinchesin (A)B and 1 increase Mag. pressure& tension increase inward forceno longer balancerbypperturbationgrows.Atplacewhereitbulgesout (B)B & 1 decrease Pressure&tensionperturbationgrows.rConclusionasmallperturbationinducesaforcetendingtoincreaseitself.Unstable(W


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Figure4.24: Invertedwaterglassanalogy. RayleighTaylorinstability.

2B2AB B

2o 2oPerturbationGrows.


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Figure4.26: Examplesofmagneticconfigurationswithgoodandbadcurvature.

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|

4.16 QuickandSimpleAnalysisofPinches-pinch B|=const. outsidepinch|Nofieldlinecurvature. Neutralstabilityz-pinch B|awayfromplasmaoutsideBadCurvature(Towardsplasma)Instability.Generally it is difficult to get the curvature to be good everywhere. Often it is sufficientto make it good on average on a field line. This is referred to as Average Minimum B.Tokamakhasthis.Generalideaisthatiffieldlineisonlyinbadcurvatureoverpartofitslengththentoperturbinthatregionandnotinthegoodregionrequiresfieldlinebending:

Figure4.27: Parallellocalizationofperturbationrequiresbending.Butbendingisnot preferred. Sothismaystabilize.Possiblewaytostabilizeconfigurationwithbadcurvature: ShearShearofFieldLines

Figure4.28: Depictionoffieldshear.DirectionofBchanges. AperturbationalongB atz3 isnot alongBatz2 orz1 soitwouldhavetobend fieldthereStabilizingeffect.GeneralPrinciple: Fieldlinebendingisstabilizing.Example: Stabilizedz-pinchPerturbations(e.g. sausageorkink)bendBz sothetension inBz actsasarestoring forcetoprevent instability. Ifwave lengthvery long bending is less. Leaststabletendstobelongestwavelength.Example: CylindricalTokamak

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Tokamakisinsomewayslikeaperiodiccylindricalstabilizedpinch. Longestallowablewavelength=1turnroundtorusthelongway,i.e.

kR=1: =2R. (4.159)Express this in terms of a toroidal mode number, n (s.t. perturbation expi(n+m):= z n=kR.

RMostunstablemodetends toben=1.[Careful! Tokamak has important toroidal effects and some modes can be localized in thebadcurvatureregion(n=1).

Figure4.29: Ballooningmodesarelocalizedintheoutboard,badcurvatureregion.

plasma physics lecture 4 ian hutchinson

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