Chapter 10: Potentials and Fields Scalar and Vector Potentials 10.1 The Potential Formulation 10.1.1 Scalar and Vector Potentials In the electrodynamics, In the and , 1 ∂B (i) ∇⋅EE =ε ρ (iii) ∇× =− 1 0 ∂t (i) ∇⋅EE =ε ρ (iii) ∇× =0 0 ∂E (ii) ∇⋅BBJ = 0 (iV) ∇× =000 + (ii) ∇⋅BBJ = 0 (iV) ∇× =µ0 ∂t How do we express the fields in terms of scalar and vector the and can be expressed using potentials? µµε potential: 2 1 E =−∇VV −∇ = ε ρ B remains divergence, so we can still write, BA=∇× 0 Putting this into Faraday’s law (iii) yields, BA=∇× ∇× ( ∇×A ) =µ J 0 ∂∂∂AA 22 ∇×EA =−() ∇× =∇× () − ⇒ ∇× ( E + )0 = ∇×( ∇×A ) =∇ ( ∇⋅ A ) −∇ AJ =µµ00 ⇒ −∇ AJ = ∂∂∂ttt ∂A If A 0. 1 E +=−∇V 2 ∇⋅ = ∂t

Scalar and Vector Potentials Example 10.1 ∂A Find the charge and current distributions that would give rise BA=∇× E =−∇V − ∂t to the potentials. 1 ∂ 1  µ0k (||)ct− x 2 zˆ for |xct |< (i) ∇⋅E = ρ 2 4c ε −∇VA −() ∇ ⋅ = ε ρ V ==0, A  0 ∂t 0  0 for |xct |> ∂E µµε∂∂V µε2A Where k is a constant, and c is the . (iV) ∇×BJ = + ∇×() ∇×AJ = − ∇ () − 000µµε 000 002 ∂ ∂t ∂tt∂ Solution: ρε=−() ∇⋅A 0 ∂t 2 We can further yields. 112 ∂ A JA=− ∇ −00 2 + ∇() ∇⋅ A 2 ∂ 1 ∂t ∇+V () ∇⋅=−A ρ 00 ε µµ ∂t 0 ∂A ∂Ax y µε∂Az 2 ∇⋅A = + + =0 2 ∂∂A V ∂∂∂xyz ∇−AAJµ00εµεµ2 −∇∇⋅+ 00 =− 0 ∂∂tt µε µε 222 ρ = 0  2 ∂∂∂ µ0k ∇=Azz()222 + +Az ˆˆ = These two equations contain all the information in Maxwell’s ∂∂∂x y zcµµ4 J = 0 equations. ∂2A kk 3 −=−=00c2zzˆˆ 4 00∂tcc2 0044 Example 10.1 (ii) 10.1.2 Gauge Transformations Since the volume and are We have succeeded in reducing six components (E and B) both zero, where are the electric and magnetic fields from? down to four (V and A). However, V and A are not uniquely ρ ==0 and J 0 determined. We are free to impose extra conditions on V and A, as long They might originate from surface charge or surface current. as nothing happens to E and B. ∂ µ k EA=−() ∇⋅ =−0 (||)ct − x zˆ Suppose we have two sets of potential (V, A) and (V’, A’), ∂t 2 µµkk∂ which correspond to the same electric and magnetic fields. BA=∇× =−00(||)ct − x2 yˆˆ =± (||) ct − x y 42cx∂ c AA′′=+α and VV =+β BAA=∇× =∇×′ ⇒ ∇×α = 0 ⇒=∇ α λ ∂∂∂AA′ α E =−∇VV′ − =−∇ − − ∇β + ∂∂∂ttt ∂λ ˆ +− β λ ⇒+= ()()β kt There is a surface current K in the yz plane. KnH=×() − H ∂ ∂t 1 µ k ⇒∇ ( + ) = 0 =×nyzˆ 0 ctˆˆ =5 kt ∂t 6 How do we know? µ 0 c

Gauge Transformations 10.1.3 Coulomb Gauge and Lorentz Gauge β There are many famous gauges in the literature. We will α =∇λ =∇λ′ AA′ =+∇λ  show the two most popular ones. ⇒ ∂∂λ λ′  ∂λ ∂ 1 =− +kt() =− VV′ =− 2 ∂tt∂  ∂t ∇+V () ∇⋅=−A ε ρ ∂t 0 ∂∂2A V Conclusion: For any scalar function λ, we can with impunity 2  ∇−AAJµ00εµεµ2 −∇∇⋅+ 00 =− 0 add ∇λ to A, provided we simultaneously subtract ∂λ/∂t to V. ∂∂tt The Coulomb Gauge: ∇⋅A =0 Such changes in V and A do not affect E and B, and are 1 called gauge transformation. 2 ∇=−V ε ρ (Poisson's equation) 0 We have the freedom to choose V and A provided E and B 1(,)r′ t Vt(r , )= d′ (setting V =0 at infinity) do not affect --- gauge freedom. ∫ ρ 4 0πε r V instantaneously reflects all changesτ in ρ. Really? ∂A 7 E =−∇V − unlike electrostatic case. 8 ∂t The Coulomb Gauge The Lorentz Gauge

Advantage: the scalar potential is particularly simple to 2 ∂ 1 ∇+V () ∇⋅=−A ε ρ calculate; ∂t 0 2 1 ∇=−V ρ (Poisson's equation) 2 ε 2 ∂∂A V 0 ∇−AAJµ00εµεµ2 −∇∇⋅+ 00 =− 0 1(,)r′ t ∂∂tt Vt(r , )= ρ d′ (setting V =0 at infinity) 4 ∫ ∂V 0πε r The Lorentz Gauge: ∇⋅A +µε =0 τ 00∂t 2 2 ∂ V 1 Disadvantage: the vector potential is very difficult to ∇−V µ00ερ2 =− ∂t ε0 calculate. 2 2 ∂∂A V 2 ∇=−+AJµµε(()) +∇ µε 2 ∂ A 0002 00 ∇−AJµε µ =− ∂∂tt 00∂t 2 0

2 1 22∂ ,2V =− ρ ∇−µε00 2 ≡, The coulomb gauge is suitable for the static case. ∂t ε0 2 2 9 , : the d'Alembertian , AJ=−µ0 10

The Lorentz Gauge 10.2 Continuous Distributions Advantage: It treat V and A on an equal footing and is 10.2.1 Retarded Potentials particularly nice in the context of special relativity. It can be 2 2 ∂ V 1 regarded as four-dimensional versions of Poisson’s equation. ∇−V µ00ερ =− 2 1 ∂t 2 ε static case ∇=−V ρ 0 ε V and A satisfy the inhomogeneous wave equations, with a ∂2A 0 ∇−2AJµε µ =− ∇=−2AJµ “source” term on the right. 00∂t 2 0 0 1 ,2V =− ρ ε0 Four copies of Poisson's equation ,2 AJ=−µ 1()ρ r′ 0 Vd()r = πε ∫ ′ 4 0 r τ µ Disadvantage: … Jr()′ Ar()= 0 ∫ d ′ 4π r τ We will use the Lorentz gauge exclusively.

11 12 Retarded Potentials Retarded Potentials Satisfy the Lorentz Gauge Condition In the nonstatic case, it is not the status of the source right Show that the retarded scalar potentials satisfy the Lorentz now that matters, but rather its condition at some earlier time gauge condition. πε πε 2 tr when the “message” left. 1 ρ(,)r′ t ∂ V 1 Vt(,)r = r dτ ′ ∇−2V µ ερ =− r ∫ 00∂t 2 ε tt≡− (called the retarded time) 4πε 0 r 0 r c Sol: 11()()ρ(,)r′ t ∇−∇ Retarded potentials: ∇=Vd ∇r τ ′′ = rr dτ ∫∫ 2 1 ρ(,)r′ t 4400rr Vt(,)r = ∫ r dτ ′ Argument: The light we see now 4πε 0 r left each star at the retarded time fgffg∇− ∇ Using quotient rule: ∇= ρρ µ Jr(,)′ t corresponding to that start’s  2 Ar(,)td= 0 r τ ′ gg 4π ∫ distance from the earth. r ∂−ρ 1 ∇=∇ρρ ρ(,)r′ ttrr = ∇= ∇r ∇=ˆ tc r r This heuristic argument sounds reasonable, but is it ∂ r −1 ρ ˆˆρ correct? Yes, we will prove it soon. ∇=Vd[]rr + τ ′ 13 ∫ 2 14 4πε 0 crr

ρ τ ρ πρδ τ Retarded Potentials Retarded Potentials Satisfy the Lorentz Gauge Condition (ii) Satisfy the Lorentz Gauge Condition (iii)

2 −1 ρrrˆˆρ ∇⋅∇VV =∇ = ∇⋅[] + dτ ′ 23−11 1ρ ρ (,)r t ∫ 2 ∇=Vdd[4()] − +′′ = − 4πε 0 crr ∫∫22r 44πεπεε000cc ρρrrˆˆ1 r ˆ r ˆ ∇⋅[]()() + = ∇⋅ +∇⋅ 22 22 ρρ ρ22 ρρρ 1 ∂∂ ∂∂VV ccrr r r ddd′′′==== 1 ˆˆˆˆ ∫∫44τττ∂∂tt22 ∫ 4 ∂∂ tt 22 rrrr πε πε πε 00rr 0 r = [][]⋅∇ρρ + ∇⋅ +22 ⋅∇ + ∇⋅ c rrrr 2 ∂ρ−1 ρ −ρ 2 1(,)∂ Vtρ r ∇=∇(,)r′ tt = ∇=ρρ  ρρ ∇=− ˆ and ∇=ρ ˆ ∇=V − rrr r r ct22∂ ∂tccr c ε0 1 ∂2V 1 1(,)∂2Vtr rrˆˆ3 2 2 ρ ∇⋅ =22 and ∇⋅ = 4πδ (r ) ∇−V µ00ερ2 =− ∇−V 22 =− rrr ∂t ε0 ct∂ ε0 ρρrrˆˆ ρρ11 ρ 3 ∇⋅[][][4()] +222 = − + + − + r ccccrr rr r 1 3 =− +4()r 15 16 c2

ρπρδπρδ Retarded Potentials Satisfy the Lorentz Gauge The Principle of Causality Condition Show that the retarded vector potentials satisfy the Lorentz This proof applies equally well to the advanced potentials. gauge condition. Advanced potentials: 2 2 ′ ∂ A 2 ∂ V 1 µ0 Jr(,)tr 2 1 ρ(,)r′ t Ar(,)td= τ ′ ∇−AJµ εµ =− a ∇−V µ00ερ =− ∫ 002 0 Vt(,)r = dτ ′ ∂t 2 ε 4π r ∂t 4πε ∫ 0 0 r 2 2 ∂ A µ0 Jr(,)′ ta ∇−AJµε µ =− Sol: Ar(,)td= ∫ τ ′ 00∂t 2 0 4π r Jr(,)′ t ()∇⋅JJ − ⋅ () ∇ rr- ′ r r r tt≡− rr′ ∇⋅ = 2 r - rr c tta ≡+ c AAAgg()∇⋅ − ⋅ () ∇ The advanced potentials violate the most sacred tenet in Using quotient rule: ∇⋅ = 2 gg all physics: the principle of causality.

See Prob. 10.8… No direct physical significance. 17 18

Example 10.2 µ I ()ct22− s 1 00ˆ 0for t ≤ 0 Az(,)st= ( ) 22 dz  ∫−−()ct s 22 An infinite straight wire carries the current It()=  4π sz+ It0 for > 0 22 How?  µI ()ct− s Find the resulting electric and magnetic fields. =++()ln()00zˆ szz22 2 π 0 Sol: The wire is electrically neutral, so the retarded scalar 22 00I µ ct+−() ct s potential is zero. = ()ln(zˆ ) 2 s π µ Jr(,)′ tItτ µ ∞ () Ar(,)tst== A (,) 00rr d′ = zˆ dz 44π ∫∫π −∞ rr ∂A µ Ic For ts/c, only the segment z ≤− () ct s contributes. ∂A µ I ct BA=∇× =−z φφˆˆ = 00 ∂ss2π ()ct22− s

19 20 επ πε ρ Retarded Fields? 10.2.2 Jefimenko’s Equations Can we express the electric field and magnetic field using the Retarded potentials: concept of the retarded potentials? No. 1 (,)rJr′′ttµ (,) Vt(,)rAr==∫∫rr dτ ′′ and (,) t0 dτ 440 rr Retarded potentials: Retarded fields: (wrong) 1 ρrrˆˆρ 1 ρ(,)r′ t 1 ρ(,r′ t ) −∇Vd =[]ππ + τ ′ Vt(,)r = r dτ ′ Er(,)td≠ r rˆ τ ′ 4πε ∫ c 2 ∫ 4πε ∫ 2 ∂A 0 rr 4πε 0 r 0 r E =−∇V − ∂∂AJµµJr(,)′ tt∂  µ Jr(,)′ t 1 Jr(,)′ tr ×rˆ ∂t 00rr Ar(,)td= 0 r τ ′ Br(,)td≠ τ ′ −=−()ddτ ′′ =− τ ∫ ∫ 2 π πε ρρ∂∂tt44∫∫ ∂ t 4π r 4πε 0 r r rr 1  ρρˆˆ µ J How to correct this problem? E =+−πε[]rrddτ ′′0 τ ∫∫2 440 crr r Jefimenko’s equations. 1 ˆˆ J τ =+−[]rr d ′ ∫ 22 4 0 rrrcc 21 The time-dependent generalization of Coulomb’s law. 22

επ πε ρ 10.3 Point Charges Jefimenko’s Equations (ii) 10.3.1 Lienard-Wiechert Potentials Retarded potentials:ππ 1 (,)rJr′′ttµ (,) What are the retarded potentials of a moving point charge q? Vt(,)rAr==rr dτ ′′ and (,) t0 dτ 44∫∫ 0 rr Consider a point charge q that is moving on a specified µ Jr(,)′ t µ 11 trajectory BA=∇× =00∫∫ ∇×r ddτ ′′ =[] ∇× JJ − ×∇ τ W(tqt )≡ position of at time . 44rrr rw()t 11rˆ - r  The retarded time is: ttr ≡− ∇×JJ = ×rˆ and ∇ ( ) =− 2 c rr c W(tr ) the retarded position of the charge. µ J 1 The time-dependent generalization BJ=+×0 [] ˆdτ ′ ∫ 2 r The separation vector r is the vector from the retarded 4π rrc of the Biot-Savart law. position to the field point r

These two equations are of limited utility, but they provide r =−rW()tr a satisfying sense of closure to the theory. 23 24 επε πε Communication Total Charge

Is it possible that more than one point on the trajectory are 111ρ(,)r′ t Vt(,)rr==r dτ ′′′ρτ ( , td ) “in communication” with r at any particular time t? ∫∫r 4()4()00rw−−ttrr rw  No, one and only one will contribute. ≠ q

Suppose there are two such points, with retarded time t and The retardation obliges us to evaluate ρ at different times for 1 different parts of the configuration. t2: rr1122=−ct( t ) and =− ct ( t ) rr12− =−ct() 12 t This means the average velocity of the particle in the The source in motion lead to a distorted picture of the total direction of r would have to be c. Í violate special relativity. charge. q ρτ(,)r′′td = No matter how small the ∫ r Only one retarded point contributes to the potentials at any 1/−⋅rˆ v c charge is. given moment. To be proved.

25 26

Total Charge: a Geometrical Effect Total Charge: a Geometrical Effect (ii) A train coming towards you looks a little longer than it really In general, if the train’s velocity makes an angle θ with your is, because the light you receive from the caboose left earlier line of sight, the extra distance light from the caboose must than the light you receive simultaneously from the engine. cover is L ′ cos θ .

LLL′′cosθ − L LLL′′− L =⇒= L′ =⇒= L′ cv1cos/− vcθ cv1/− vc This effect does not distort the dimensions perpendicular to L L′ = Approaching train appears longer. the motion. 1/− vc τ L The apparent volume τ’ of the train is τ ′ = L′ = A train going away from you looks shorter. related to the actual volume τ by . 1/−⋅rˆ v c 1/+ vc 27 28 Lienard-Wiechert Potentials Example 10.3 It follows that Find the potentials of a point charge moving with constant επε πε velocity. Assume the particle passes through the origin at time 11ρ(,)r′ t q Vt(,)r ==ππr dτ ′ , t =0. 44(/)∫ −⋅v c µ 00rrr Sol: The trajectory is: Wv()tt=

π First compute the retarded time: rW−=−=−()ttcttrrr rv ( ) µ00ρ(,)()r′ tvtrrµ vt () r Ar(,)tdtd==τ ′′′ρτ ( r ,r ) 44∫∫ 222222 rr rtvtctttt−⋅2(2)rvrr + = − rr + 0 qvv ==Vt(,)r 222 2 222 2 ()2()()0cvt−+⋅−+−=rv cttctr 4(r −⋅r v /)cc rr ()()()()ct22222222−⋅rv ± rv ⋅− ct − c − v ct − r Which sign where ρ(rrr′′ ,tq )=−δ ( , t ) tr = is correct? rr ()cv22−

2222 The famous Lienard-Wiechert potentials for a moving point Consider v =0 ttr =± t −(/)/ trc − =± trc charge. 29 We want the negative sign 30

Cont’: ()()()()ct22222222−⋅rv − rv ⋅− ct − c − v ct − r t = 10.3.2 The Fields of a Moving Point Charge r ()cv22− Using the Lienard-Wiechert potentials we can calculate the rv− tr rctt=−(r ), and rˆ = fields of a moving point charge. ct()− tr vvrrv− t ⋅ v2 1 q v −⋅v /()1-cctt = − ⋅r = ctt () − − − t Vt(,)r = and Ar ( ,tVt )= ( r , ) r r rrr 4(πε −⋅v /)c c2 cctt()− r c c 0 r r 1 222 ∂A =−⋅−−()()ctrv c v tr Find: E =−∇V − and BA=∇× c ∂t 1 22 2 2 22 2 =⋅−−−−()()()rv ct c v ct r ′  c The separation vector: r =−rr =− rW()ttrr and v = W ()

1 qc The retarded time t : rW−=−()tctt ( ) Vt(,)r = r rr 22 2 2 22 2 4πε 0 ()()()rv⋅−ct − c − v ct − r

µ qcv tr is a function of r and t. Ar(,)t = 0 4π 22 2 2 22 2 ()()()rv⋅−ct − c − v ct − r 31 32 ∂∂∂ #2 ()()()()(vvrvW⋅∇r = ⋅∇ − ⋅∇tvvvrxyz = − + + )() W t r Gradient of the Scalar Potential ∂∂∂xyz

1 −qc dddWWW∂∂∂tttrrr ∇=Vc ∇−⋅(/)v =−v ()vvvxyz + + 2 r r dt∂∂∂ x dt y dt z 4(πε 0 r −⋅r v /)c rrr =−⋅∇vv(1 (tr )) ∇=∇−ct() t =−∇ c t ∂∂vv r rr ∂∂vvzzyy∂∂vvxx #3 rr×∇×()vxyz = × ( − )(ˆˆˆ + − )( + − ) ∂∂yz ∂∂ zx ∂∂ xy ∇()rr ⋅vvv = ⋅∇ + ⋅∇ rr + × () ∇× vv + × () ∇× r N N   ∂∂vv∂∂vvyy∂∂vv #1 #2 zzxx #3 #4 =×r ()()() −xyzˆˆˆ + − + − ∂∂yz ∂∂ zx ∂∂ xy ∂∂∂ #1 ()(rrrr⋅∇vv =xyz + + ) =×−×∇r ()a tr ∂∂∂xyz ()yW ∂−()zWz ∂−y ∂−()xWx dddvvv∂∂∂tttrrr #4 vv×∇×()r = × ( − )( xˆ + =++()rrrxyz ∂∂yz ∂ z dtrrr∂∂∂ x dt y dt z  ∂−()zW ∂−()yW ∂−()xW =⋅∇a()t z y x r r −+−)(yzˆˆ ) ∂∂∂xxy acceleration 33 34 =×−×∇vv()tr

∇() ⋅vvv = ⋅∇ + ⋅∇ + × () ∇× vv + × () ∇× rrN N rr  r 1 qc 22 #1 #2 ∇=Vccv()() −⋅vv − − +⋅ a #3 #4 3 r rrr 4(πε 0 rc −⋅r v ) =⋅∇+−⋅∇−××∇+××∇avvavv()(1())rrttttrrrr()( ) 2 =+va()r ⋅−vt ∇r Similar calculations 11 1 r 1/2 ()(/)rcc−⋅rrvva −+ ∇=−∇=−∇=−∇⋅tr r ()rr =−1/2 ∇ () rr ⋅ ∂A 1 qc cc c2( crr⋅ ) =  3 +−+⋅r ()cv22 av 1 ∂−⋅tc4(πε 0 r r v )r c = −1/2 []rrrr ×∇×+⋅∇ ( ) ( ) 2(c rr⋅ )

 rr×∇×() = ∂A q 22 Euua=−∇Vcv − =r () − + × () × where  3 r ()()(())(rrr⋅∇ = ⋅∇rW −ttrr = r − v r ⋅∇ ) ∂⋅t 4()πε 0 r u 1 ∇=−tttrrr1/2 []rrr ×()vv ×∇+− () ⋅∇ 2(c rr⋅ ) where uv≡−crˆ 1 −r =− 2[]rr − ( ⋅v ) ∇ttrr ) ⇒∇ = 2ccrr−⋅r v 35 36 Curl of the Vector Potential Generalized Coulomb Field 11 ∇×Av =22 ∇×()VV =() ( ∇× vv ) − ×∇ V cc  1 q r 22  =−3 rrr ×()()()cv −vavua + ⋅ − ⋅ q r 22 c 4()u  Euua=−+××()cv () πε 0 r ⋅ 3  r 4()πε 0 r ⋅u  11q 22 velocity field acceleration field =×−+××=×r ()()cvuuaEˆ  3 rrr radiation field cc4()πε0 r ⋅u where rr×=−×vu. v ==0 and a 0

qqr 3 1 The magnetic field of a point charge is always E ==()c rrˆˆ 1 πε4()πε c 32 4 BE=×rˆ perpendicular to the electric field, and to the 00rr c vector from the retarded point.

37 38

Example 10.4 Fields of a Moving Point Charge Calculate the electric and magnetic fields of a point charge moving with constant velocity.

Solution: qvc1-22 / Rˆ E = 22 23/22, q r 22 θ4(1sin/)πε 0 − vcR Eua=−=3 (cv ) , since 0. 4()πε 0 r ⋅u where Rrv≡−t uv=−c rˆ θ ⇒=−=−−−=− rruvrvvrvcctcttctr ()()();rr 22 2 ⇒⋅=−⋅= rrruvcRcvc 1 − sinθ / (Prob. 10.14) where is the angle between Rv and . 11 BEvE=×=×()rˆ () cc2 qvc1-22 / Rˆ ERrv= 22 23/22, where ≡−t εθ4(1sin/)πε 0 − vcR 39 40 Homework of Chap.10

Prob. 4, 9, 12, 13, 23, 24

41