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Chapter 10: Potentials and Fields Example 10.1

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Chapter 10: Potentials and Fields Example 10.1 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 electrostatics and magnetostatics, 1 ∂B (i) ∇⋅EE =ρ (iii) ∇× =− 1 ε ∂t (i) ∇⋅EE =ρ (iii) ∇× =0 0 ε0 ∂E (ii) ∇⋅BBJ =0 (iV) ∇× =µµε000 + (ii) ∇⋅BBJ =0 (iV) ∇× =µ0 ∂t How do we express the fields in terms of scalar and vector the electric field and magnetic field can be expressed using potentials? potential: 1 E =−∇VV −∇2 = ρ B remains divergence, so we can still write, BA=∇× ε0 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 −∇2VA −() ∇ ⋅ = ρ V ==0, A 4c ε 0 ∂t ε0 0 for |xct |> ∂E ∂∂V 2A Where k is a constant, and c is the speed of light. (iV) ∇×BJ =µµε + ∇×() ∇×AJ =µµε − ∇ () − µε 000 000 002 ∂ ∂t ∂tt∂ Solution: ρε=−() ∇⋅A 0 ∂t 2 We can further yields. 112 ∂ A JA=− ∇ −µε00 2 + ∇() ∇⋅ A 2 ∂ 1 µµ∂t ∇+V () ∇⋅=−A ρ 00 ∂t ε0 ∂A ∂Ax y ∂Az 2 ∇⋅A = + + =0 2 ∂∂A V ∂∂∂xyz ∇−AAJµ00εµεµ2 −∇∇⋅+ 00 =− 0 ∂∂tt ∂∂∂222 µ k ρ = 0 ∇=2Azz() + +A ˆˆ =0 These two equations contain all the information in Maxwell’s ∂∂∂x222y zcz 4 J = 0 equations. ∂2A µµkk 3 −=−=µε µε 00c2zzˆˆ 4 00∂tcc2 0044 Example 10.1 (ii) 10.1.2 Gauge Transformations Since the volume charge density and current density 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 kt There is a surface current K in the yz plane. KnH=×() − H ∂λ ∂t 1 µ k ⇒∇ (β + ) = 0 How do we know? =×nyzˆ 0 ctˆˆ =5 kt ∂t 6 µ0 c Gauge Transformations 10.1.3 Coulomb Gauge and Lorentz Gauge There are many famous gauges in the literature. We will α =∇λ =∇λ′ AA′ =+∇λ show the two most popular ones. ∂∂λ λ′ ⇒ ∂λ β =− +kt() =− VV′ =− 2 ∂ 1 ∂tt∂ ∂t ∇+V () ∇⋅=−A ρ ∂t ε0 ∂∂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. ∇=−2V ρ (Poisson's equation) ε0 We have the freedom to choose V and A provided E and B 1(,)ρ r′ t Vt(r , )= dτ ′ (setting V=0 at infinity) do not affect --- gauge freedom. ∫ 4πε 0 r V instantaneously reflects all changes in ρ. Really? ∂A 7 E =−∇V − unlike electrostatic case. 8 ∂t The Coulomb Gauge The Lorentz Gauge ∂ 1 Advantage: the scalar potential is particularly simple to ∇+2V () ∇⋅=−A ρ calculate; ∂t ε0 2 1 ∇=−V ρ (Poisson's equation) 2 2 ∂∂A V ε0 ∇−AAJµ00εµεµ2 −∇∇⋅+ 00 =− 0 1(,)ρ r′ t ∂∂tt Vt(r , )= dτ ′ (setting V=0 at infinity) 4πε ∫ ∂V 0 r The Lorentz Gauge: ∇⋅A +µε =0 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 = ∫ τ ′ 4πε 0 r Disadvantage: … µ Jr()′ Ar()= 0 ∫ dτ ′ 4π r 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. 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 ∇−∇ρρ Retarded potentials: ∇=Vd ∇r τ ′′ = rr dτ ∫∫ 2 1 ρ(,)r′ t 44πε00rr πε Vt(,)r = ∫ r dτ ′ Argument: The light we see now 4πε 0 r left each star at the retarded time fgffg∇− ∇ Using quotient rule: ∇= µ Jr(,)′ t corresponding to that start’s 2 Ar(,)td= 0 r τ ′ gg 4π ∫ distance from the earth. r ∂−ρ 1 ∇=∇ρρ(,)r′ ttrr = ∇= ρ ∇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 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()] −ρ + πρδ τ′′ = τ − 4πε 0 crr ∫∫22r 44πεπεε000cc ρρrrˆˆ1 r ˆ r ˆ ∇⋅[]()() + = ∇⋅ρρ +∇⋅ 22 22 22 ρρρ 1 ∂∂ ∂∂VV ccrr r r dddτττ′′′==== 1 ˆˆˆˆ ∫∫44πε πε∂∂tt22 ∫ 4 πε ∂∂ tt 22 rrrr 00rr 0 r = [][]⋅∇ρρ + ∇⋅ +22 ⋅∇ ρρ + ∇⋅ c rrrr 2 ∂ρ−1 ρ −ρ 2 1(,)∂ Vtρ r ∇=∇ρρ(,)r′ ttrr = ∇= ρ ∇=−r rˆ and ∇=ρ rˆ ∇=V 22 − ∂tccr c ct∂ ε0 2 2 rrˆˆ1 3 2 ∂ V 1 2 1(,)∂ Vtρ r ∇⋅ =22 and ∇⋅ =4πδ (r ) ∇−V µ00ερ2 =− ∇−V 22 =− rrr ∂t ε0 ct∂ ε0 ρρrrˆˆ11 ρρ ρ 3 ∇⋅[][][4()] +222 = − + + − + πρδ r ccccrr rr r 1 3 =−ρπρδ +4()r 15 16 c2 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? µ I ()ct− s Find the resulting electric and magnetic fields. =++()ln()00zˆ szz22 2π 0 Sol: The wire is electrically neutral, so the retarded scalar µ I ct+−() ct22 s potential is zero. = ()ln(00zˆ ) 2π s µ Jr(,)′ tItµ ∞ () Ar(,)tst== A (,) 00rr dτ ′ = zˆ dz 44π ∫∫π −∞ rr ∂A µ Ic For t<s/c, the “news” has not yet reached P, and the Ez=− =− 00 ˆ 22 potential is zero. ∂t 2()π ct− s 22 For t>s/c, only the segment z ≤− () ct s contributes. ∂A µ I ct BA=∇× =−z φφˆˆ = 00 ∂ss2π ()ct22− s 19 20 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 ρ(,)rJr′′ttµ (,) Vt(,)rAr==∫∫rr dτ ′′ and (,) t0 dτ 44πε0 rr π Retarded potentials: Retarded fields: (wrong) 1 ρrrˆˆρ 1 ρ(,)r′ t 1 ρ(,r′ t ) −∇Vd =[] + 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 ∂∂tt44ππ∫∫ ∂ t 4π r 4πε 0 r r rr 1 ρρ ˆˆ µ J How to correct this problem? E =+−[]rrddτ ′′0 τ ∫∫2 44πε0 crr π r Jefimenko’s equations. 1 ρρˆˆ J =+−[]rr dτ ′ ∫ 22 4πε 0 rrrcc 21 The time-dependent generalization of Coulomb’s law.
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