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  ∂B  ∇ × E = −  ∂t  ∂D  ∇ × H = + J ∂t c   ∇ · D = ρ   ∇ · B = 0 Since B field is non-divergent ( ∇ · B = 0 ), so we can define

B = ∇ × 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 ∇ × E = − ∂t ∂A = −∇ × ∂t

 ∂A ⇐⇒ ∇ × E + = 0 ∂t ∂A ⇐⇒ E + is curl-less ∂t ∂A ⇐⇒ E + = −∇φ ∂t ∂A E = −∇φ − ∂t Thus ∂A B = ∇ × 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 = ∇ × AE = −∇φ − ∂t Now consider another paris of potentials A0 and φ0

A0 = A + ∇ψ

∂ψ φ0 = φ − ∂t Then because curl grad = 0

∇ × A0 = ∇ × A + ∇ × ∇ψ = B | {z } 0 ∂f ∂f And because of mixed derivative equality = ∂x1∂x2 ∂x2∂x1 ∂A0  ∂ψ  ∂A ∂  −∇φ0 − = −∇φ + ∇ − + ∇ψ ∂t ∂t ∂t ∂t    ∂A ∂ψ ∂ = −∇φ − + ∇ − ∇ψ ∂t  ∂t ∂t  | {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 ∇·A , thus we have many different ways to define ∇·A. Different limitation on ∇ · 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 ∇ · A = 0 and the Lorenz gauge ∇ · A + = 0. c2 ∂t 3 d’Alembert equations and the Gauges

With these equations  ∂B  ∇ × E = −  ∂t   ∂D B = ∇ × A  ∇ × H = + J  ∂t c ∂A   E = −∇φ −  ∇ · D = ρ ∂t   ∇ · B = 0 The Ampere’s Law thus becomes

2 ∂D ∇ × H = + J ∂t c ∂E ∇ × B = µ ε + µ J 0 0 ∂t 0 c ∂  ∂A ∇ × ∇ × A = µ ε −∇φ − + µ J 0 0 ∂t ∂t 0 c ∂ ∂2A ∇(∇ · A) − ∇2A = −µ ε ∇φ − µ ε + µ J 0 0 ∂t 0 0 ∂t2 0 c

Rearrange

∂2A  ∂φ ∇2A − µ ε − ∇ ∇ · A + µ ε = −µ J 0 0 ∂t2 0 0 ∂t 0 Next, consider the Gauss’s Law for electric field

∇ · D = ρ ρ ∇ · E = ε  ∂A ρ ∇ · −∇φ − = ∂t ε ∂ ρ −∇2φ − ∇ · A = ∂t ε

Thus we have the following 2 d’alembert equations

 ∂2A  ∂φ  ∇2A − µ ε − ∇ ∇ · A + µ ε = −µ J  0 0 ∂t2 0 0 ∂t 0 ∂ ρ  −∇2φ − ∇ · A = ∂t ε Just by looking at the second equation, if ∇ · A = 0 ( Coulomb gauge ) :

 2 2 ∂ A ∂  ∇ A − µ0ε0 2 − µ0ε0 ∇φ = −µ0J ∂t ρ∂t  ∇2φ = − ε The second equation is the same Laplace equation in electrostatics. ∂φ If Lorenz gauge is used ∇ · A + µ ε = 0, the d’Alembert equations become a symmetric form : 0 0 ∂t  ∂2A  ∇2A − µ ε = −µ J Current generated  0 0 ∂t2 0 2 2 ∂ φ ρ  ∇ φ − µ0ε0 = − Charge generated ∂t ε −END−

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