Scalar and Vector Potentials, Gauge Conditions Andersen Ang First created: 2013. Last updated: 2017-Feb-1 1 The vector potential A Consider EM fields in free space 8 @B > r × E = − > @t > @D < r × H = + J @t c > > r · D = ρ > : r · B = 0 Since B field is non-divergent ( r · B = 0 ), so we can define B = r × A What A means : The close loop line integral of A = flux pass through that loop. Unlike electrostatics that E is created by charges, in electrodynamics E is created by charges and induced by B, thus a single potential is not sufficient to describe E @B r × E = − @t @A = −∇ × @t @A () r × E + = 0 @t @A () E + is curl-less @t @A () E + = −∇φ @t @A E = −∇φ − @t Thus @A B = r × AE = −∇φ − @t Vector potential A and scalar potential φ now descripe B and E. * The φ now is not the electric potential. Thus when frequency is high, the electric potential in electrostatic case is not valid anymore. 1 2 Gauge It is necessary to force some constraints on the definition of A such that the potentials can uniquely describe E and B. Without such constraint, the description is not unique. For example, consider potentials A and φ @A B = r × AE = −∇φ − @t Now consider another paris of potentials A0 and φ0 A0 = A + r @ φ0 = φ − @t Then because curl grad = 0 r × A0 = r × A + r × r = B | {z } 0 @f @f And because of mixed derivative equality = @x1@x2 @x2@x1 @A0 @ @A @ −∇φ0 − = −∇φ + r − + r @t @t @t @t 0 1 @A @ @ = −∇φ − + Br − r C @t @ @t @t A | {z } 0 = E Thus two potentials are describing the same thing and there is no way to differentiate them. From observation, such non-uniqueness exists because there is no limitation on the divergence of A. Since B and E do not enforce any limitation on r·A , thus we have many different ways to define r·A. Different limitation on r · A fits different problem. Thus , a Gauge , is a imposed limitations that cope with the redundant extra degrees of freedom. There are 2 famous gauge in electromagnetics, 1 @φ the Coulomb gauge r · A = 0 and the Lorenz gauge r · A + = 0. c2 @t 3 d’Alembert equations and the Gauges With these equations 8 @B > r × E = − > @t 8 > @D B = r × A < r × H = + J < @t c @A > : E = −∇φ − > r · D = ρ @t > : r · B = 0 The Ampere’s Law thus becomes 2 @D r × H = + J @t c @E r × B = µ " + µ J 0 0 @t 0 c @ @A r × r × A = µ " −∇φ − + µ J 0 0 @t @t 0 c @ @2A r(r · A) − r2A = −µ " rφ − µ " + µ J 0 0 @t 0 0 @t2 0 c Rearrange @2A @φ r2A − µ " − r r · A + µ " = −µ J 0 0 @t2 0 0 @t 0 Next, consider the Gauss’s Law for electric field r · D = ρ ρ r · E = " @A ρ r · −∇φ − = @t " @ ρ −∇2φ − r · A = @t " Thus we have the following 2 d’alembert equations 8 @2A @φ > r2A − µ " − r r · A + µ " = −µ J < 0 0 @t2 0 0 @t 0 @ ρ :> −∇2φ − r · A = @t " Just by looking at the second equation, if r · A = 0 ( Coulomb gauge ) : 8 2 2 @ A @ < r A − µ0"0 2 − µ0"0 rφ = −µ0J @t ρ@t : r2φ = − " The second equation is the same Laplace equation in electrostatics. @φ If Lorenz gauge is used r · A + µ " = 0, the d’Alembert equations become a symmetric form : 0 0 @t 8 @2A > r2A − µ " = −µ J Current generated < 0 0 @t2 0 2 2 @ φ ρ :> r φ − µ0"0 = − Charge generated @t " −END− 3.