Ch 7 Radiation of Electromagnetic Waves 7.1, potentials of electromagnetic field gauge invariance 7.2 d' Alembert equation and retarded potential 7.3, electric dipole radiation 7.4 EM radiation from arbitrary motion charge
7.1、potentials of electromagnetic field, gauge invariance 7.2、d’Alembert equation and retarded potential 7.3、electric dipole radiation 7.4、EM radiation from arbitrary motion charge Ch 7 Radiation of Electromagnetic Waves
1 What is EM radiation EM field is excited by time-dependent charge and currents. It may propagate in form of waves The problem is usually solved in terms of potentials 特征:与1/r正比的电磁场! 2. It is a boundary value problem Source charge and current excites EMF, EMF in turn affects source distribution -- boundary value problem For convenience our discussions are limited to a simple case Distribution of source is known
1. What is EM radiation EM field is excited by time-dependent charge and currents. It may propagate in form of waves. The problem is usually solved in terms of potentials. 2．It is a boundary value problem Source (charge and current) excites EMF, EMF in turn affects source distribution --- boundary value problem! For convenience, our discussions are limited to a simple case – Distribution of source is known. 特征：与1/r 正比的电磁场！
87. 1 vector potential and scalar potential potentials are slightly different from the static cases 1.a) vector potentia sinceB=0, we can introduce vector potential A as the static field,B=VxA
§7.1 vector potential and scalar potential potentials are slightly different from the static cases since ，we can introduce vector potential as the static field, B = 0 A B A = 1.a）vector potential
= 1.b) scalar potential Since×E=-≠0, scalar potential can not be defined as before aA OA V×E V×A=-V× V×(E+) t Define scalar func⑨ 0A 0A e+ at
Since ，scalar potential can not be defined as before 0 = − t B E 1.b）scalar potential B A = t A A t E = − = − ( ) = 0 + t A E = − + t A E Define scalar func t A E = − −
2). Gauge invariance Potentials are not uniquely determined they differ by a gauge transformation. A>A'=A+Vy q→>9=q at Gauge: Given a set of (A, 9) (A, give identical electric and magnetic fields
2)．Gauge invariance t A A A → = − → = + (A, ) Potentials are not uniquely determined, they differ by a gauge transformation. Gauge: Given a set of give identical electric and magnetic fields ' ' ( , ) A
Prove: since A andA',and p can not change E and B. soB=VxA=V×A+V×Vv=V×A A=A+Vy E=-VP-aA aA OA V(q+) at at at at at aA at ●规范不变性:在规范变换下物理规律满足的动力学方程保持不变 的性质(在微观世界是一条物理学基本原理)
Prove: since and ， and can not change E and B, so A A B A A A = = + = A = A + t A t t t A t A E − = − + − = − − = − − ( ) A t = − − t = − l 规范不变性：在规范变换下物理规律满足的动力学方程保持不变 的性质（在微观世界是一条物理学基本原理）
3. Two typical gauges To reduce arbitrariness of potential, we give some constraint-- Gauge fiⅸing。 Symmetry or explicit physical interpretation ● Coulomb gauge condition V·A=0 transverse (横场),V0 longitudinal(纵场)。 o is determined by instantaneous distribution of charge density (similar to static coulomb field)
l Coulomb gauge condition A = 0 3)．Two typical gauges transverse (横场)， longitudinal (纵场)。 is determined by instantaneous distribution of charge density (similar to static coulomb field) A To reduce arbitrariness of potential, we give some constraint --- Gauge fixing。 Symmetry or explicit physical interpretation
Function satisfies Prove V A'=VA+VVy=oVV=0 ● Lorenz gauge Ludwig Lorenz ondition v4+ a9=o dt Function y satisfies prove:v.A't 1 a =V·4+ V Vu I a1o at t 0 (V·A+ +(V2v c2 at2 satisfy manifest relativistic covariant equations
0 2 Function satisfies = l Lorenz gauge condition 0 1 2 = + c t A A, satisfy manifest relativistic covariant equations Function satisfies 0 1 2 2 2 2 = − c t ) 0 1 ) ( 1 ( 2 2 2 2 2 = + − + c t c t A A = A + = 0 0 2 Prove = 0 1 2 2 2 2 = − c t 2 2 2 2 2 1 1 1 c t c t A c t A − = + + + prove： Ludwig Lorenz
D'Alembert equation 824 10 V(V·A+-x) at Vp+e(VA=_p Prove: substitute E=V×A E=vat Maxwell egs OE V×B=￡μoat ·E And using V×(V×A)=V(V·A)-V2A
2 2 2 2 2 0 2 0 1 1 ( ) ( ) A A A J c t c t A t − − + = − + = − Prove：substitute ， into Maxwell eqs And using B A = t A E = − − 0 0 0 0 , + = = J E t E B A A A 2 ( ) = ( ) − 4)． D’Alembert equation
4. a Under coulomb gauge 02 V c2 at2 So o satisfies Poisson equation as in static case instantaneous interaction? 4. b Under lorenz gauge VA c- at c2 at2
2 2 2 2 2 0 2 0 1 1 A A J c t c t − − = − = − So satisfies Poisson equation as in static case. instantaneous interaction? 4.a) Under coulomb gauge 4.b) Under Lorenz gauge 2 2 2 2 2 2 2 2 0 0 1 1 A A J c t c t − = − − = −
