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2.3. MAGNETOSTATICS
li
Particularlysimpleexpressionsareobtainedforplanarcurrentloops,i.e. incaseswherethe
curveγliesinaplane.Introducingcoordinatessuchthattheunitvectornormaltotheplane
n=e
3
coincideswiththecoordinatevectore
3
,wethenobtainfortheithcomponentofthe
magneticmoment
m
i
=−
I
2c
e
i
·
γ
ds×x=−
I
2c
γ
e
i
·(ds×x)=−
I
2c
γ
ds·(x×de
i
).
ApplicationofStokeslawthenyieldsm
i
=−
I
2c
S(γ)
dσe
3
·(∇×(x×e
i
)). Notingthat
e
3
·(∇×(x×e
i
))=−2δ
3i
,weobtain
m=
AI
c
e
3
,
i.e. amagneticmomentperpendiculartotheplanesupportingthecurrentloopandpropor-
tionaltotheareaoftheloop.
 Imagine e asystem of f point particlesatcoordinates s x
i
whereeachparticle carriescharge
q
i
,isofmassm
i
,andmoveswithvelocityv
i
. Inthiscase,j=
i
v
i
q
i
δ(x−x
i
). Equation
(2.57)reducesto
1
2c
i
q
i
x
i
×v
i
=
1
2c
i
q
i
m
i
l
i
,
(2.61)
wherel =x×(mv) istheangularmomentumcarried by apointparticle. . [Onlyparts s of
themagneticmomentcarriedbygenuineelementaryparticlesareduetotheirorbitalangular
momentuml.Asecondcontributionstemsfromtheir‘intrinsic’angularmomentumorspin.
Speciﬁcally,foranelectronatrest,m=2.002×
e
2cm
S,where|S|=1/2istheelectronspin
andthepre–factor2.002isknownastheg–factoroftheelectron.]
2.3.3 Magneticforces
Tounderstandthe physicalmeaningoftheconnections(currents ﬁelds)derivedabove,letus
exploreanintuitivelyaccessiblequantity,viz.themechanicalforcescreatedbyacurrentdistribution.
Assumingthatthechargedensityofthemobilechargecarriersinthewireisgivenbyρ,andthatthese
chargesmovewithavelocityv,thecurrentdensityinthewireisgivenbyj=ρv.
19
Comparison
withtheLorentzforcelaw(1.6)showsthatinamagneticﬁeldthewirewillbesubjecttoaforce
density = c
−1
ρv×B. Speciﬁcally,forathin n wire of cross section dA, , and d carrying acurrent
I=ρdAv(visthecomponentofv=ve
alongthewire.)theforceactingonasegmentoflength
−1
)×B=c
−1
Ids×dB.Summarizing,theforceactingona
lineelementdsofawirecarryingacurrentIisgivenby
df=
I
c
ds×B.
(2.62)
Inthefollowing,wediscussafewapplicationsoftheprototypicalforce–formula(2.62):
19
TheamountofchargeﬂowingthroughasmallsurfaceelementdσduringatimeintervaldtisdQ=dσρn·vdt.
DividingbydtweﬁndthatthecurrentthroughdσisgivenbyI
=dσn·(ρv)whichmeans thatj=ρvisthe
currentdensity.
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lii
CHAPTER2. THESTATICLIMIT
Forcesoncurrentloops
Considerasingle closed currentloopγ carryingacurrentI. . Thecurrentﬂowwillgeneratea
magneticﬁeld(2.54)whichwillinturnactonthelineelementsoftheloop,asdescribedby(2.62).
Onemaythuswonderwhethertheloopexertsanetforceonitself.Integratingover(2.62),weﬁnd
thatthisforceisgivenby
F=
γ
df=
I
c
γ
ds×B
(2.54)
=
I
2
c
2
γ
γ
ds×(ds
×(x−x
))
|x−x
|
3
=
=
I
2
c
2
γ
γ
ds
(ds·(x−x
))
|x−x
|
3
(x−x
)ds·ds
|x−x
|
3
=
=−
I
2
c
2
γ
ds·
γ
ds
·∇
1
|x−x
|
=0.
Thesecondterminthesecondlinevanishesbecauseitchangessignundercoordinateinterchange
x↔x
. Inthethirdlineweusedthat(x−x
)|x−x
|
−3
=−∇|x−x
|
−1
andthattheintegral
doesnotexertanetforceontoitself.
I
1
I
2
x
y
F
Figure2.2:
Twoinﬁnitewiresexertingaforceoneachother
However,distinctcurrentdistributionsdo,ingeneral,actbymagneticforcesontoeachother.
I
1
andI
2
,resp. Choosingcoordinatesasshownintheﬁgureandparameterizingthelineelements
asds
1,2
=dx
1,2
e
x
,weobtaintheforceF
12
wire#2exertsonwire#1as
F
12
=
I
1
I
2
c
2
dx
1
dx
2
e
x
×e
x
×(e
x
(x
1
−x
2
)−e
y
d)
|(x
1
−x
2
)
2
+d
2
|
3
)=
=e
y
I
1
I
2
d
c
2
dx
1
dx
2
1
(x
2
2
+d
2
)
3/2
=e
y
2I
1
I
2
c
2
d
dx
1
,
whereinthesecondlineweused
−∞
dx(x
2
+d
2
)
−3/2
=2/d
2
. Reﬂectingtheirinﬁnitelength,the
wiresexertaninﬁniteforceoneachother.However,theforceperunitlength
dF
dx
=
2I
1
I
2
c2d
e
y
isﬁnite
forcurrentsﬂowinginthesame/oppositedirection. (Oneofthemoreprominentmanifestationsof
theattractivemagneticforcesbetweenparallelcurrentﬂowsisthephenomenonofcurrentimplosion
inhollowconductors.)
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2.3. MAGNETOSTATICS
liii
Forcesonlocalcurrentdistributions
Consideraspatiallylocalizedcurrentdistributioninanexternalmag-
B
m
neticﬁeld. Wewishtocomputetotalforceactingonthedistribution
F=
d
3
xf=
1
c
d
3
xj×B.
(2.63)
Choosingtheoriginofthecoordinatesystemsomewhereinsidethecurrent
distribution,andassumingthatthemagneticﬁeldvariesslowlyacrossthe
extentofthecurrentﬂow,onemayTaylorexpandBintheLorentzforce
formula:
F=
1
c
d
3
xj(x)×[B(0)+x·∇B(0)+...].
(x·∇Bisavectorwithcomponentsx
i
i
B
j
.) Theauxiliaryidentity(2.56)impliesthattheﬁrst
contributiontotheexpansionvanish. Theithcomponentoftheforceisthusgivenby
F
i
1
c
d
3
x
ijk
j
j
x
l
l
B
k
(2.56)
=
1
2c
d
3
x
ijk
mjl
(j×x)
m
l
B
k
=
=
1
2c
d
3
x(δ
kl
δ
im
−δ
km
δ
il
)(j×x)
m
l
B
k
=
1
2c
d
3
x((j×x)
i
k
B
k
−(j×x)
k
i
B
k
)=
=∂
i
|
x=0
(m·B(x)),
or
F=∇|
x=0
(m·B(x)),
(2.64)
whereinthesecondlineweused∇·B=∂
l
B
l
=0Thus,(a)theforceisstrongestifthemagnetic
momentisalignedwiththemagneticﬁeld,and(b)proportionaltotherateatwhichtheexternal
magneticﬁeldvaries.Speciﬁcally,forauniformﬁeld,noforcesact.
ﬁelddoesexertaﬁnitetorque
N=
d
3
xx×f=
1
c
d
3
xx×(j×B).
(2.65)
Approximatingtheﬁeldbyitsvalueattheorigin,
N
i
=
1
c
d
3
x
ijk
klm
x
j
j
l
B(0)
m
=
1
c
d
3
x(δ
il
δ
jm
−δ
im
δ
lj
)x
j
j
l
B(0)
m
=
=
1
c
d
3
x(x
j
j
i
B(0)
j
−x
j
j
j
B(0)
i
)
(2.56)
=
1
2c
d
3
x
lji
(x×j)
l
B(0)
j
=
=(m×B(0))
i
.
Thus,thetorqueactingonthemoment,
N=m×B(0),
(2.66)
isperpendiculartoboththeexternalﬁeldandthemomentvector. Itactssoastoaligntheﬁeld
andthemoment.
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liv
CHAPTER2. THESTATICLIMIT
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Chapter3
Electrodynamics
3.1 Magneticﬁeldenergy
magneticﬁeld? In n section2.2.2, , the e analogous questionforthe electricﬁeld was answered in a
constructivemanner:wecomputedthemechanicalenergyrequiredtobuildupasystemofcharges.
Itturnedoutthatthe answercould be formulated entirelyin termsoftheelectric ﬁeld,without
explicitreferencetothechargedistribution creatingit. . Bysymmetry,onemightexpectasimilar
currentdistributionagainstthemagneticﬁeldcreatedbythoseelementsofthecurrentdistribution
quite asstraightforwardlyimplementedasintheelectric case: : nomatterhowslowly y wemove a
‘currentloop’in anmagneticﬁeld,anelectricﬁeldactingonthechargecarriersmaintainingthe
currentintheloopwillbeinduced—theinductionlaw. Workwillhavetobedonetomaintaina
constantcurrentanditisthisworkfunctionthatessentiallyenterstheenergybalanceofthecurrent
distribution.
Tomakethispicturequantitative,consideracurrentloopcarry-
I
B
ingacurrentI. Wemaythinkofthisloopasbeingconsecutively
builtupbyimportingsmallcurrentloops(carryingcurrentI)from
inﬁnity(see theﬁgure.) ) The e currentsﬂowingalong adjacentseg-
ments of f these loops s will eventually y cancel l so o that t only y the net
currentI ﬂowingaroundtheboundaryremains.Letus,then,com-
putetheworkthatneedstobedonetobringinoneoftheseloops
frominﬁnity.
Consider,ﬁrst,anordinarypointparticlekeptata(mechanical)
potentialU.Therateatwhichthispotentialchangesiftheparticle
changesitspositionisd
t
U=d
t
U(x(t))=∇U·d
t
x=−F·v,where
Fistheforceactingontheparticle. Speciﬁcally,forthecharged
particlesmovinginourprototypicalcurrentloops,F=qE,where
Eistheelectricﬁeldinducedbyvariationofthemagneticﬂuxthroughtheloopasitapproaches
frominﬁnity.
lv
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lvi
CHAPTER3. ELECTRODYNAMICS
Nextconsideraninﬁnitesimalvolumeelementd
3
xinsidetheloop. Assumingthatthecharge
carriers move at avelocity v, , the e charge ofthevolume elementisgiven by q q = = ρd
3
x and rate
ofitspotentialchange by d
t
φ= −ρd
3
xv·E= −d
3
xj·E. Wemaynowusetheinductionlaw
to express theelectric ﬁeld in terms ofmagnetic quantities:
ds·E = = −c
−1
δS
dσn·d
t
B =
−c
−1
d
t
δS
dσn·d
t
(∇×A) = −c
−1
ds·d
t
A. Since e this holds regardless of f the geometry
of the e loop, we have E = −c
−1
d
t
A, where e d
t
A is the change e in vector potential l due e to the
movementoftheloop.Thus,therateatwhichthepotentialofavolumeelementchangesisgiven
byd
t
φ=c
−1
d
3
xj·d
t
A. Integratingovertimeandspace,weﬁndthatthetotalpotentialenergyof
theloopduetothepresenceofavectorpotentialisgivenby
E=
1
c
d
3
xj·A.
(3.1)
Althoughderivedforthespeciﬁccaseofacurrentloop,Eq.(3.1)holdsforgeneralcurrentdistribu-
tionssubjecttoamagneticﬁeld. (Forexample,forthecurrentdensitycarriedbyapointparticle
at x(t), j = = qδ(x−x(t))v(t), , we e obtain E E = = qv(t)·A(x(t)), , i.e. . the e familiar r Lorentz–force
contributiontotheLagrangianofachargedparticle.)
Now,assumethatweshifttheloopatﬁxedcurrentagainstthemagneticﬁeld. Thechangein
potentialenergycorrespondingtoasmallshiftisgivenbyδE=c
−1
d
3
xj·δA,where∇×δA=δB
denotesthechangeintheﬁeldstrength.Usingthat∇×H=4πc
−1
j,werepresentδEas
δE=
1
d
3
x(∇×H)·δA=
1
d
3
x
ijk
(∂
j
H
k
)δA
i
=
=−
1
d
3
xH
k
ijk
j
δA
i
=
1
d
3
xH·δB,
whereintheintegrationbypartswe notedthatdue tothespatialdecay ofthe ﬁeldsnosurface
termsatinﬁniteyarise.DuetothelinearrelationH=µ
−1
0
B,wemaywriteH·δB=δ(H·B)/2,
i.e. δE=
1
δ
B·E. Finally,summingoverallshiftsrequiredtobringthecurrentloopinfrom
inﬁnity,weobtain
E=
1
d
3
xH·B
(3.2)
forthemagneticﬁeldenergy. Notice(a)thatwehaveagainmanagedtoexpresstheenergyof
thesystementirelyintermsoftheﬁelds,i.e.withoutexplicitreferencetothesourcescreatingthese
ﬁeldsand(b)thestructuralsimilaritytotheelectricﬁeldenergy(2.20).
3.2 Electromagneticgaugeﬁeld
ConsiderthefullsetofMaxwellequationsinvacuum(E=DandB=H),
∇·E = = 4πρ,
∇×B−
1
c
∂t
E =
c
j,
∇×E+
1
c
∂t
B = = 0,
∇·B = = 0.
(3.3)
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3.2. ELECTROMAGNETICGAUGEFIELD
lvii
Asinprevioussectionswewilltrytouseconstraintsinherenttotheseequationstocompactifythem
toasmallersetofequations.Asinsection2.3,theequation∇·B=0implies
B=∇×A.
(3.4)
(However,wemaynolongerexpectAtobetimeindependent.)Now,substitutethisrepresentation
intothelawofinduction: ∇×(E+c
−1
t
A)=0. ThisimpliesthatE+c
−1
t
A=−∇φcanbe
E=−∇φ−
1
c
t
A.
(3.5)
We have, , thus, managed d to represent the electromagnetic ﬁelds as in terms of derivatives of a
generalized four–componentpotential A= {A
µ
}= (φ,−A). . (The e negative sign multiplying A
hasbeenintroducedforlaterreference.) SubstitutingEqs.(3.4)and(3.5)intotheinhomogeneous
Maxwellequations,weobtain
−∆φ−
1
c
t
∇·A=4πρ,
−∆A+
1
c2
2
t
A+∇(∇·A)+
1
c
t
∇φ=
c
j
(3.6)
Theseequationsdonotlookparticularlyinviting. However,asinsection2.3wemayobservethat
thechoiceofthegeneralizedvectorpotentialAisnotunique;thisfreedomcanbeusedtotransform
Eq.(3.6)intoamoremanageableform:Foranarbitraryfunctionf:R
3
×R→R,(x,t)→f(x,t).
The transformation A A → → A+∇f f leaves s the magnetic ﬁeld unchanged while the electric ﬁeld
changesaccordingtoE→E−c
−1
t
∇f.If,however,wesynchronouslyredeﬁnethescalarpotential
asφ→φ−c
−1
t
f,theelectricﬁeld,too,willnotbeaﬀectedbythetransformation.Summarizing,
thegeneralizedgaugetransformation
A →
A+∇f,
φ →
φ−
1
c
t
f,
(3.7)
leaves the electromagnetic ﬁelds (3.4) and (3.5) unchanged. . (In n the fourcomponent shorthand
notationintroducedabove,thegaugetransformationassumeseasy–to–memorize theformA
µ
A
µ
−∂
µ
f,wherethefour–derivativeoperator{∂
µ
}=(∂
0
,∇)andx
0
=ctasbefore.)
Thegaugefreedommaybeusedtotransformthevectorpotentialintooneofseveralconvenient
representations.OfparticularrelevancetothesolutionofthetimedependentMaxwellequationsis
theso–calledLorentzgauge
∇·A+
1
c
t
φ=0.
(3.8)
Thisequation,too,aﬀordsacompactfour–vectornotation:forageneralvector{v
µ
}=(v
0
,v)we
deﬁneavector{v
µ
}≡(v
0
,−v),i.e. aan n objectwith ‘raised components’thatdiﬀersfromthe
onewith‘loweredcomponents’byasignchangeinthespace–likesector. Usingthisnotation,the
Lorentzconditionassumestheform∂
µ
A
µ
=0. Itisalwayspossibletosatisfythisconditionbya
suitablegaugetransformation. Indeed,ifA
does notobey theLorentzcondition,wemaydeﬁne
A
µ
=A
µ
−∂
µ
f toobtain0
!
=∂
µ
A
µ
=∂
µ
A
µ
−∂
µ
µ
f.Ifwechosefsoastosatisfytheequation
lviii
CHAPTER3. ELECTRODYNAMICS
µ
f
µ
=∂
µ
A
µ
,theLorentzequationissatisﬁed.Expressedintermsofspaceandtimecomponents,
thislatterequationassumestheform
∆−
1
c
2
2
t
f=−
∇·A
+
1
c
t
φ
,
i.e. awaveequationwithinhomogeneity−(∇·A+c
−1
t
φ). Weshallseemomentarilythatsuch
equationscanalwaysbesolved,i.e. animplementationoftheLorentzgaugeconditionispossible.
IntheLorentzgauge,theMaxwellequationsassumethesimpliﬁedform
∆−
1
c
2
2
t
φ = = −4πρ,
∆−
1
c
2
2
t
A = = −
c
j.
(3.9)
Incombinationwiththegaugecondition(3.8),Eqs.(3.9)arefullyequivalenttothesetofMaxwell
equations(3.3). Again,thefour–vectornotationmaybeemployedtocompactifythenotationstill
further.With∂
µ
µ
=−∆+c
−2
2
t
andj
µ
=(cρ,−j),thepotentialequationsassumetheform
ν
ν
A
µ
=
c
j
µ
,
µ
A
µ
=0.
(3.10)
Beforeturningto the discussion of thesolution of these equationsa few generalremarks are in
order:
 TheLorentzconditionistheprevalentgaugechoiceinelectrodynamicsbecause(a)itbrings
theMaxwellequationsintoamaximallysimpleformand(b)willturnoutbelowtobeinvariant
underthemostgeneralclassofcoordinatetransformations,theLorentztransformationssee
below. (ItisworthwhiletonotethattheLorentzconditiondoesnotunambiguouslyﬁxthe
µ
afunctionA
µ
→ A
µ
+f
µ
satisfying the homogeneous waveequation∂
µ
µ
f = = 0without
alteringthecondition∂
µ
A
µ
=0.) Othergaugeconditionsfrequentlyemployedinclude
 theCoulomb b gaugeorradiationgauge∇·A=0(employedearlierinsection2.3.) ) The
Poissonequation∆φ(x,t)=4πρ(x,t)whichissolvedby φ(x,t)=
d
3
x
ρ(x,t)/|x−x
|,
i.e. byaninstantaneous’Coulombpotential’(hencethenameCoulombgauge.) Thisgauge
However,wewon’tdiscussitanyfurtherinthistext.
3.3 Electromagneticwavesinvacuum
AsawarmuptoourdiscussionofthefullproblemposedbythesolutionofEqs.(3.9),weconsider
thevacuumproblem,i.e. asituationwherenosourcesarepresent,j
µ
=0. Takingthecurlofthe
secondofEqs.(3.9)andusing(3.4)wethenﬁnd
∆−
1
c
2
2
t
B=0.
3.3. ELECTROMAGNETICWAVESINVACUUM
lix
−1
timesthetimederivativeofthesecond,
andusingEq.(3.5),weobtain
∆−
1
c
2
2
t
E=0,
i.e.invacuumboththeelectricﬁeldandthemagneticﬁeldobeyhomogeneouswaveequations.
3.3.1 Solutionofthehomogeneouswaveequations
ThehomogeneouswaveequationsareconvenientlysolvedbyFouriertransformation. Tothisend,
wedeﬁneafour–dimensionalvariantofEq.(2.10),
˜
f(ω,k)=
1
(2π)
4
d
3
xdtf(t,x)e
−ik·x+iωt
,
(3.11)
f(t,x)=
d
3
kdωf(ω,k)e
ik·x−iωt
,
The onediﬀerencetoEq.(2.10)is thatthe sign–convention in the exponentofthe(ω/t)–sector
ofthetransformdiﬀersfromthatin the (k,x)–sector. . We e nextsubjectthe homogeneous wave
equation
∆−
1
c
2
2
t
ψ(t,x)=0,
(whereψismeanttorepresentanyofthecomponentsoftheE–orB–ﬁeld)tothistransformation
andobtain
k
2
ω
c
2
ψ(ω,k)=0.
Evidently,thesolutionψ mustvanish h forallvalues (ω,k),exceptforthoseforwhich the factor
k
2
−(ω/c)
2
=0. Wemaythuswrite
ψ(ω,k)=c
+
(k)δ(ω−kc)+c
(k)δ(ω+kc),
wherec
±
∈Carearbitrarycomplexfunctionsofthewavevectork.Substitutingthisrepresentation
into the inverse transform,we obtain the general solution n of the scalar r homogeneous wave
equation
ψ(t,x)=
d
3
k
c
+
(k)e
i(k·x−ckt)
+c
(k)e
i(k·x+ckt)
.
(3.12)
The general solution is obtained by y linearsuperposition n of elementary plane e waves, e
i(k·x∓ckt)
,
whereeachwaveisweightedwithanarbitrarycoeﬃcientc
±
(k). Theelementaryconstituentsare
called waves becauseforanyﬁxedinstanceofspace,x/time,tthey harmonicallydepend on the
complementaryargumenttime,t/positionvector,x. Thewavesareplanarinthesensethatforall
points in theplaneﬁxedbytheconditionx·k= = const. thephase e ofthewaveisidentical,i.e.
thesetofpoints k·x= = const. deﬁnesa‘wavefront’perpendiculartothewavevectork. . The
spacingbetweenconsequtivewavefrontswiththesamephasearg(exp(i(k·x−ckt))isgivenby
∆x=2π
k≡λ
,whereλisthewavelengthofthewaveandλ
−1
=2π/kitswavenumber.Thetemporal
oscillationperiodofthewavefrontsissetby2π/ck.