retarded是什么意思arded在线翻译读音例

RetardedPotentialsandRadiation

Initialquestion:Whyareparticlecollidersthatinvolveprotonslargecircles,whereas

thosethatinvolveelectronsarestraightlines?

No,thisisn’taboutpotentialsthatwereheldbackagrade:).Retardedpotentialsare

neededbecauseatagivenlocationinspace,aparticle“feels”thefieldsorpotentialsof

othercharges,notwherethosechargesarenow,butwheretheywerealighttraveltime

cturesagowetalkedabouthowelectromagnetismcanbephrasedintermsof

potentialsratherthanfiss:ifyouhaveasinglechargeeadistanceraway

fromagivenpoint,whatistheelectrostaticpotentialthere,ignoringlighttraveltimes?

It’sjust=e/,generalizing,supposeonehadlotsof(static!)chargesatdifferent

ss:howthenwouldyoufindthepotential?It’sadditive,sothetotal

potentialwouldbe

=i

e

i

/r

i

,(1)

whereeachsourcehaschargee

i

andisadistancer

i

,

supposethatyouhadabunchofmovingchargese

i

.Ifyoupickaparticulartimetandyou

knowthatforeachparticlethedistancewasrret

i

(t)alighttraveltimeago,Askclass:what

willthepotentialbethen?

=i

e

i

/rret

i

(t).(2)

Notethatmoredistantchargeswillhavehadalongerlighttraveltimethannearercharges,

kethisformulaand

writeitforacontinuouschargedensity,thenthescalarpotentialatatimetandlocation

risthen

(r,t)=[]d3r′

|r−r′|

(3)

wherethebracketsmeantoevaluatethequantityattheretardedtime:

[Q]=Qr′,t−

1

c

|r−r′|.(4)

Notethatherewe’veshiftedthingsslightly:insteadoffollowingindividualcharges,we’re

evaluatingthepotentialfromfixedpointsinspace,butallowingthechargedensityto

ntstothesamething.

Onecangothroughanidenticalprocedureforthevectorpotential:

A(r,t)=

1

c[j]d3r′

|r−r′|

.(5)

Rememberthatthepotentialshavegaugefreedom,soinwritingandAthiswaywe’ve

chosenaparticulargauge,inthiscasetheLorentzgauge,inwhich

∇A+

1

c

∂/∂t=0.(6)

That’snobigdealinmostcircumstance程门立雪的道理 s,butit’sgoodtobeclear.

Aspecificcharge“distribution”thatonecanimagineisthatofasinglecharge(!).If

ithaschargeqandmovesalongatrajectoryr0(t),sothatitsvelocityisu(t)=

r(t),then

wecancongratulateourselvesonbeingcoolbywritingthechargeandcurrentdensitiesin

termsofDiracdeltafunctions:

(r,t)=q(r−r0(t)),

j(r,t)=qu(t)(r−r0(t)).

(7)

ankabit(see3.1ofRybickiandLightman)wegettheLienard-Wiechart

potentials

=q

R

A=qu

cR.

(8)

Herethebracketsmeananevaluationattheretardedtime,and(t′)≡1−n(t′)u(t′)/c,

wheren=R/factorismighty

hargeisn’tmoving,u=0,thewholethingisjust1andyougetthe

potentialsyouexpectfromelectrostatics(A=0,=q/R).Ontheotherhand,forspeeds

closetothespeedoflightthisfactorproducesstron霏组词 gintensityinthedirectionofmotionof

,abeamingeffect.

rememberthatthePoyntingfluxcarried

byanelectromagneticfieldisS=c

4

Eorget,rememberthat(asidefromthe

1/4factor)youcangetthisjustfromunits,ifyourememberthatB2andE2haveunits

,ss:whatare

andA?isjustthesumofq

i

/R

i

evaluatedatthecharges’currentposition(sincethey

aren’tmoving),andA=0becausethechargesarefixed(nocurrent).Askclass:so,what

isSinthatcase?Zero,cchargedistributionhasnoflux.

lessclearisthatit’sthe

differentiatethepotentialstogetthefields(RL

saythatthisis“straightforwardbutlengthy”,alwaysawarningthatyoumaybeinfor

analgebraicchallenge).Onethenendsup(3.2inRL)withanelectricfieldthatcanbe

writtenasthesumoftwoterms:onethatdoesnotdependontheacceleration

(where

=u/c),ss:whichofthesetermsdotheyexpectwillproduce

radiation?Fromtheargumentsweusedafewclassesago,itmustbetheoneproportional

cchargedistributionemitsnoradiation,soachargeatconstant

velocitycan’teither(sinceyoucouldtransformtoaframeinwhichitisstatic).

Nowaninterestingthingistolookattheradiusdependencesofthetwotermsforthe

electricfield(andthecorrespondingmagneticfield,whichisjustB(r,t)=[nE(r,t)]).

The“velocityfield”termisproportionalto1/R2,whereasthe“radiationfield”termis

proportionalto1/nsiderjustthevelocityfieldthenthefluxS∝EB∝1/R4,

meaningthatwhenoneintegratedoverasphereatlargeradiusthenetfluxwouldbe

∝1/R2,sothesystemwouldn’tradiatetoinfiiationfieldisnecessaryfor

radiation,record,theradiationfieldis

Erad(r,t)=

q

cn

3R

(n−)

.(9)

NotethatE⊥n,soBrad=nrtherecord,the

velocityfieldis

Evel(r,t)=q(n−)(1−2)

3R2.(10)

Thereasontowritetheseissothatwecanunderstandtheminsomelimits.

Let’strythelimitofsmall,≪ss:at

largedistances,whichcomponentofEdotheyexpectwilldominate?Theradiationfield.

Askclass:bywhatpowerofR?’sseeifthisisvalidbycomparing

themagnitudesofthevelocityandradiationfiss:ignoringcrossproductsand

othercomplications,whatisthemagnitudeofthevelocityfieldtolowestorderin?It’s

Evel≈q/(3R2).Notethatfor≪1,≈1,sothisreducestothestandardelectricfield

ss:tolowestorderin,whatisthemagnitudeof

theradiationfield?It’sErad≈(q/c)

/(3R).Takingtheratioandwriting

=u/c,we

have

Erad/Evel≈Ru/c2.(11)

,theratioincreaseslikethefirstpowerof

R,,whenthereisnoaccelerationthereisnoradiation(we’veseenthis

severaltimesbefore,butit’salwaysgoodtocheck).Third,ifthespeedoflightwereinfinite

inforcesthatradiationisfundamentallyarelativistic

effect,etsomefurtherunderstandingbyassuming

thattheparticlehassomecharacteristicfrequencyofoscillation,sothatu≈u=uc/.

Inthatcase,

Erad/Evel≈(u/c)(R/).(12)

It’softengoodtoexpressadimensionlessratioastheproductofdimensionlessratios,so

ysthatinthe“nearzone”,R<,

thevelocityfielddominatesbyafactor>c/“farzone”,R≫(c/u),theradiation

fielddominates.

Wecannowusethistogetahandleonthesimplesttypeofradiation:Larmor’sformula

imit≪1,anangle

fromthedirectionofaccelerationthemagnitudesoftheelectricandmagneticradiation

fieldsareErad=Brad=(qu/Rc2)sin.Askclass:howcanweusethistocomputethe

energyflux?ThePoyntingfluxisS=(c/4)E2

rad

inthiscase(becausetheelectricand

magneticfieldmagnitudesareequal).Ifweconsidertheenergypertimeemittedintoa

solidangledΩatradiusR,thentheareaatthatsolidangleisR2dΩ,sowehave

dW

dtdΩ

=

q2u2

4c3

sin2.(13)

Togetthetotalpowerweintegrateoversolidangles,whichgivesustheLarmorformula:

P=2q2u2/3c3.(14)

Again,let’slookatthisequationtoseeifitsatisfiermustbe

proportionaltoanevenpowerofthecharge,sincethesigndoesn’not

dependonthevelocity,butrathertheacceleration,andagainthesigncan’tmattersoit

(asalways!)thisisarelativistic

effect,ifc→∞seintuitiveconditionsaresatisfiedbythis

,theradiationisinatypicaldipole

pattern,proportionaltosin2;thismeansthatnoradiationisemittedalongthedirection

,theinstantaneousdirectionofEradisdeterminedbyboth

uand

icular,forlinearaccelerationtheradiationwillbe100%linearlypolarizedinthe

planeof

uandn(theselasttwopointsaretakendirectlyfromRybickiandLightman).

Bytheway,ingtothis

formula,anelectro三秋桂子 十里荷花 ncirclingaroundaprotonwouldemitenergycontinuouslyandinan

acceleratingway,sincesoneofthecontradictions

thathelpedspurthedevelopmentofquantummechanics.

Whatifwehaveabunchofparticles?Ingeneral,itbecomesalottougherbecauseall

thedifferentparticleswillhavedifftherway,ifwe’reworried

aboutaparticularfrequencycomponentoftheradiation,wehavetokeeptrackofphase

ss:cantheythinkofacircumstancein

whichthephaserelationsareclos红楼风月梦 eenoughtoconstantthatthiscanbesimplified?One

wayisifthefrequencyofinterestismuchlowerthanc/L,whereListhecharacteristic

,thedifferencesinretardedtimeamounttojustasmall

ataparticularexampleofthisisthatiftheparticlesaremoving

nonrelativistically,u≪c,thenthecharacteristicfrequencyoftheirmovementacrossthe

region,u/L,ismuchlessthanc/ansthatfornonrelativisticmotionwecan

,theradiationfieldis

Erad=i

q

i

c2

n(n

u

i

)

R

i

,(15)

inkaboutanobservationpointveryfarfromthesources,

thenthedifferencesindistancesarenegligibleandwecantakesomeR0ascharacteristicof

ad≈n(n

d)/c2R0,wherewehavedefinedthedipolemoment

asd≡i

q

i

r

i

.InananalogouswaytotheLarmorformulaforasinglecharge’spower,the

radiationpowerisP=2

d2/3c3inthisapproximation.

example,inNewtoniangravityyouca沁园春雪课件ppt nestimatethepotentialaroundamassdistribution

byexpandingitinpowersofthedistance,wherethefirsttermisthatofapointmassatthe

centerofmass,’ssimilarhere,withthecaveat

kgoesintoa

littlemoredetailaboutgeneralmultipolarexpansions.

nacceleratedparticle

radiates,itcarriesawayenergy,linearmomentum,ans

thatthemotionoftheparticleitselfmustbemodifietanapproximateideaof

whatthismodificationdoesbytreatingitasanextraforce,theforceofradiationreaction.

First,though,let’sfigureoutunderwhatcircumstancestheforcecanbetreatedasa

etheparticlehasaspeed书院二小松阅读答案 u,soitskineticenergyis∼om

theLarmorformulathetimeinwhichthekineticenergyischangedsubstantiallyis

T∼mu2/P∼(3mc3/2e2)(u/u)2.(16)

Let’sestimateatypicalo在线电子新华字典 rbitaltimefortheparticleoft

p

∼u/rtheenergylostin

anorbitalperiodtobesmall,T/t

p

≫1,ort

p

≫≡2e2/3mc3≈10−23s(!).That’samighty

outthetimenecessaryforlighttotraveladistanceequaltotheclassical

electronradius2.810−,tofigureouttheforceourfirstinclinationwouldbe

tosettheforcetimesthevelocityequaltothepowerradiated,Fradu=−2e2u2/

problemisthat(1)Fradcan’tdependonusincethatwouldimplyapreferredframe,so

(2)onesideofthisequationdependsonuwhiletheotherdoesn’t!d,wecan

seeifthiscanbesatisfiedinatime-averagedsense.

−t

2

t

1

Fradudt=(2e2/3c3)t

2

t

1

u

udt

=(2e2/3c3)

uu

t

2

t

1

−t

2

t

1

uudt.

(17)

sumethatthemotionorperiodic,or

atleastthat

uuisthesameatt2asatt1,thenthefirsttermontherighthandside

vanishesandwefindthatFrad=m

’sallverywell,butas

ss:whatdoesthissayabout

aparticlewithconstantlinearacceleration?Then

u=0,soitwouldimplynoradiation

reactionforce,eventhoughtheparticledoesradiate(sinceitisaccelerated).Theproblemis

thatthentheexpressionatthelimitsdoesn’tvanish,

mostcases,though,itisifyouaverageoveralongenoughtimeandthemotionisbounded.

Ifyouputthisintoagrandequationofmotionitreads

m(

u−

u)=F(18)

mallydoesn’tencounter

emonecanencounterinsuchcasesisspurious

mple,supposeF==constantisobviouslyasolution,but

soisu=u0exp(t/),

casesonemus

mathematicalreasonisthat

uu(t1)=

uu(t2),soinfacttheradiationreactionforce

wouldhaveadiffetobecarefulincaseslikethis.

RecommendedRybickiandLightmanproblem:3.1