Zhang K., Li D. Electromagnetic Theory for Microwaves and Optoelectronics

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1.5 Scalar and Vector Potentials 41

1.5 Scalar and Vector Potentials

Since the current density vector and charge density enter into the inhomo-

geneous wave equations in a rather complicated way, it is diﬃcult to solve

the inhomogeneous vector wave equations (1.128) and (1.129) directly. For

this purpose, the integration of these equations is usually performed by the

introduction of auxiliary potential functions that serve to simplify the math-

ematical analysis. The ﬁrst auxiliary potential functions we will consider are

the scalar and vector potentials or the so-called retarding potentials.

1.5.1 Retarding Potentials, d’Alembert’s Equations

We mention that the magnetic induction B is a solenoidal vector function,

∇ · B = 0 everywhere. In vector analysis, we know that the vector function

formed by the curl of a vector function is a solenoidal vector function, ∇ ·

∇ × A = 0, and hence we may take

B = ∇ × A. (1.214)

Substituting (1.214) into the curl equation for the electric ﬁeld, (1.25), we

obtain

∇ × E = −

∂B

∂t

= −

∂

∂t

(∇ × A) = −∇ ×

∂A

∂t

, (1.215)

and hence

∇ ×

E +

∂A

∂t

= 0. (1.216)

We see that the function E + ∂A/∂t is an irrotational vector function, and

we know that a vector function formed by the gradient of a scalar function

is an irrotational vector function, ∇ × ∇ϕ = 0, and hence we may take

E +

∂A

∂t

= −∇ϕ, or E = −∇ϕ −

∂A

∂t

, (1.217)

where A(x, t) denotes the vector potential function and ϕ(x, t) denotes the

scalar potential function.

Substituting (1.214) and (1.217) into Maxwell’s equations, (1.36) and

(1.37), and assuming σ = 0, we obtain the diﬀerential equations for A and

ϕ in vacuum or nonconducting simple media:

∇ × ∇ × A = −µ²

∂

∂t

∇ϕ +

∂

∂t

+ µJ , (1.218)

∇

ϕ + ∇ ·

∂A

∂t

= −

. (1.219)

Use vector identity (B.45) to give

∇ × ∇ × A = ∇(∇ · A) − ∇

42 1. Basic Electromagnetic Theory

and note that the time derivative and the space derivative are independent

of each other,

∇ ·

∂A

∂t

∂

∂t

∇ · A,

∂

∂t

∇ϕ = ∇

∂ϕ

∂t

Then (1.218) and (1.219) become

∇

A − µ²

∂

∂t

= ∇

∇ · A + µ²

∂ϕ

∂t

− µJ , (1.220)

∇

ϕ +

∂

∂t

∇ · A = −

. (1.221)

According to Helmholtz’s theorem, we know that a vector function is

completely speciﬁed by its divergence and curl. Since (1.214) gives only the

curl of A, we may specify the divergence of A in any way we choose. The

choices are called gauges. The diﬀerential equations as well as the physical

meaning of A and ϕ depend upon the gauge chosen.

(1) The Coulomb Gauge

Choose

∇ · A = 0. (1.222)

This choice is known as the Coulomb gauge, then (1.220) and (1.221) reduce

∇

A − µ²

∂

∂t

= µ²

∂

∂t

∇ϕ − µJ , (1.223)

∇

ϕ = −

. (1.224)

(2) The Lorentz Gauge

We may choose the divergence of A in another way:

∇ · A = −µ²

∂ϕ

∂t

. (1.225)

This choice is known as the Lorentz gauge and (1.220) and (1.221) reduce to

∇

A − µ²

∂

∂t

= −µJ , (1.226)

∇

ϕ − µ²

∂

∂t

= −

. (1.227)

Under the Lorentz gauge, the equations for A and ϕ are of the same kind

and are independent of each other. J is the source of A and % is the source

of ϕ. Nevertheless, A and ϕ themselves are not independent of each other,

they are related to each other through the Lorentz condition (1.225).

1.5 Scalar and Vector Potentials 43

Equations (1.226) and (1.227) are known as d’Alembert’s equations. They

are inhomogeneous wave equations. The solutions of both equations are

waves. Substituting A and ϕ into (1.214) and (1.217), we may have E and

H. This is the basic method for the investigation of radiation problems.

In fact, with the Lorentz gauge we can obtain all of the components of

the electric and magnetic ﬁelds from A alone. By using the Lorentz gauge

(1.225) in (1.217), we obtain

E =

µ²

∇(∇ · A)dt −

∂A

∂t

. (1.228)

Under the Coulomb gauge, the equation of the scalar potential ϕ, (1.224),

is Poisson’s equation, which is the same as in the static ﬁeld. The solution of

Poisson’s equation ϕ is determined by the present distribution of the source

%. Does it violate the regulation of wave propagation with ﬁnite velocity

for the time-varying ﬁelds? In fact, the time-varying electric ﬁeld (1.217)

consists of two parts, −∇ϕ and −∂A/∂t. Under the Coulomb gauge, the

ﬁrst part −∇ϕ represents the electric ﬁeld formed by the static source with

the present distribution an is called the Coulomb ﬁeld. The second part

−∂A/∂t is the induction ﬁeld. Both parts together represent the actual

ﬁeld, which propagates with a ﬁnite velocity.

It can be seen that the physical meanings of A and ϕ under diﬀerent

gauges are diﬀerent.

1.5.2 Solution of d’Alembert’s Equations

In simple media, d’Alembert’s equations are linear diﬀerential equations; the

superposition principle is suitable for them. We may ﬁnd the point-source

solution ﬁrst, and the solution of an arbitrary source distribution can be found

by the superposition or integration of the point-source solution. Suppose a

point charge q(t) is placed at the origin of the coordinates. The charge density

is a δ function, %(x, t) = q(t)δ(x). Then the equation for ϕ, (1.227), become

∇

ϕ − µ²

∂

∂t

= −

q(t)δ(x)

. (1.229)

The ﬁeld excited by a point charge at the origin must be spherically

symmetrical, and (1.229) becomes one-dimensional spherical coordinate form,

∂

∂r

∂ϕ

∂r

− µ²

∂

∂t

= −

q(t)δ(x)

. (1.230)

This equation reduces to the following homogeneous equation except at the

origin:

∂

∂r

∂ϕ

∂r

− µ²

∂

∂t

= 0, r 6= 0. (1.231)

44 1. Basic Electromagnetic Theory

Let u(r, t) = rϕ(r, t), and substitute it into (1.231), to give

∂

∂r

= µ²

∂

∂t

. (1.232)

This is a one-dimensional homogeneous wave equation which we have seen

in (1.141). The solution was obtained as (1.149):

u(r, t) = Af

t −

+ Bf

t +

, (1.233)

where f represents an arbitrary function and v

= 1/

√

µ².

So, the solution of the one-dimensional spherical coordinate homogeneous

wave equation (1.231) is

ϕ(r, t) =

Af(t − r/v

)

Bf (t + r/v

)

. (1.234)

The solution includes two spherical waves propagating along +r (outward)

and −r (inward), respectively.

Since the solution of (1.230) is a wave excited by the point charge at the

origin, it must be an outward wave and the second term of (1.234) must be

zero:

ϕ(r, t) =

f(t − r/v

)

. (1.235)

In the static state, ∂/∂t → 0, the wave equation (1.229) reduces to Poisson’s

equation; the solution (1.235) must reduce to

ϕ(r) =

4π²r

. (1.236)

We have enough ground to suppose that the solution (1.235) must be in the

following form:

ϕ(r, t) =

q(t − r/v

)

4π²r

. (1.237)

This solution can be proven by substituting it into the left-hand side of (1.229)

and taking the volume integral inside a small sphere around the origin.

For a point charge located at an arbitrary point x

, ρ(x, t) = q(t)δ(x−x

the solution (1.237) becomes

ϕ(x, t) =

q(x

, t − r/v

)

4π²r

, (1.238)

where r(x, x

) denotes the distance between the source point x

and the ﬁeld

point x, r = |x − x

For an arbitrary volume charge distribution %(x

, t), according to the

principle of superposition we have

ϕ(x, t) =

4π²

%(x

, t − r/v

)

. (1.239)

1.5 Scalar and Vector Potentials 45

The solution of (1.226) excited by an arbitrary volume current J (x

, t) is

A(x, t) =

4π

J (x

, t − r/v

)

, (1.240)

and that excited by an arbitrary line current I(x

, t) is

A(x, t) =

4π

I(x

, t − r/v

)

. (1.241)

The solutions ϕ(x, t) and A(x, t) are known as the retarding potentials.

1.5.3 Complex d’Alembert Equations

For steady-state sinusoidal sources,

%(x

, t) = =[ρ(x

j ωt

], J (x

, t) = =[J(x

j ωt

the potentials may also be written in complex form,

ϕ(x, t) = =[ϕ(x)e

j ωt

], A(x, t) = =[A(x)e

j ωt

D’Alembert’s equations (1.226) and (1.227) become the following inho-

mogeneous Helmholtz equations,

∇

A + k

A = −µJ , (1.242)

∇

ϕ + k

ϕ = −

. (1.243)

The Lorentz gauge (1.225) becomes

∇ · A = −j ωµ²ϕ. (1.244)

The solution of (1.243) for a point charge located at x

ϕ(x, t) =

q e

−jkr

4π²r

jωt

, where r = |x − x

|. (1.245)

The sinusoidal solutions of (1.242) and (1.243) are

ϕ(x, t) =

4π²

ρ(x

j(ωt−kr)

, (1.246)

A(x, t) =

4π

J(x

j(ωt−kr)

, (1.247)

A(x, t) =

4π

I(x

j(ωt−kr)

. (1.248)

46 1. Basic Electromagnetic Theory

The magnetic and electric ﬁelds are found by using (1.214) and (1.217):

∇ × A, (1.249)

E=−∇ϕ − j ωA. (1.250)

Rewrite the expression for E (1.250) by substituting the Lorentz equation

(1.244) into it:

E = −j ω

∇(∇ · A)

+ A

. (1.251)

In a source-free region, from Maxwell’s equation (1.76),

∇ × H = j ω²E,

we have

E = −

ω²

∇ × H = −

j ω

∇ × ∇ × A. (1.252)

In a static or low-frequency state, ∂/∂t → 0, jω → 0, the Lorentz gauge

and the Coulomb gauge are no longer diﬀerent: the solutions for A and H

become the Biot–savart law and the solutions for ϕ and E become Coulomb’s

law.

1.6 Hertz Vectors

The second auxiliary potential functions we will consider are the Hertz vectors

or polarization potentials [24, 96, 101]. It is possible under certain conditions

to deﬁne an electromagnetic ﬁeld in terms of a set of vector functions named

electric Hertz vector and magnetic Hertz vector.

1.6.1 Instantaneous Hertz Vectors

The current density as the source includes an irrotational component and

a solenoidal component. We may express the irrotational component of the

current density by means of an equivalent polarization current,

J =

∂P

∂t

where P denotes the equivalent polarization vector. Using the equation of

continuity (1.6), we get

% = −∇ · P,

where % is the equivalent polarization charge density. Hence the d’Alembert’s

equation of vector potential A (1.226) become

∇

A − µ²

∂

∂t

= −µ

∂P

∂t

. (1.253)

1.6 Hertz Vectors 47

Deﬁne a vector function

, π

, called the electric Hertz vector, which

is related to A as follows:

A = µ²

∂

∂t

. (1.254)

Substituting it into (1.253) and integrating, we have the inhomogeneous wave

equation of

∇

− µ²

∂

∂t

= −

. (1.255)

Thus the electric Hertz vector

satisﬁes the inhomogeneous wave equation,

with the equivalent polarization vector P as its source.

Substituting (1.254) into the Lorentz equation (1.225), gives

ϕ = −∇ ·

. (1.256)

Substituting (1.254) and (1.256) into (1.217) and (1.214), we have

E = ∇(∇ ·

) − µ²

∂

∂t

, (1.257)

H = ²∇ ×

∂

∂t

. (1.258)

The electric Hertz vector

is suitable for solving problems in which the

source is an irrotational current.

Furthermore, we may express the solenoidal component of the current

density by means of an equivalent magnetization current,

J = ∇ × M,

where M denotes the equivalent magnetization vector. Then d’Alembert’s

equation (1.226) becomes

∇

A − µ²

∂

∂t

= −µ∇ × M. (1.259)

Deﬁne another vector function

, called the magnetic Hertz vector or

Fitzgerald vector, which is related to A as follows:

A = µ∇ ×

. (1.260)

Substituting it into (1.259), we have the inhomogeneous wave equation of

∇

− µ²

∂

∂t

= −M. (1.261)

We can see in (1.260) that for this choice ∇ · A = 0. From the Lorentz

condition (1.225), the time-varying component of ϕ must be zero, and as we

48 1. Basic Electromagnetic Theory

are not interested in the d-c component ϕ = 0. Substituting (1.260) into

(1.217) and (1.214), we have

E = −µ∇ ×

∂

∂t

, (1.262)

H = ∇ × ∇ ×

. (1.263)

The magnetic Hertz vector

is suitable for solving problems in which the

source is a solenoidal current.

In the general case, the source current density J includes both the irro-

tational component and the solenoidal component:

J =

∂P

∂t

+ ∇ × M. (1.264)

According to the theorem of superposition,

A = µ²

∂

∂t

+ µ∇ ×

, (1.265)

E = ∇(∇ ·

) − µ²

∂

∂t

− µ∇ ×

∂

∂t

, (1.266)

H = ∇ × ∇ ×

+ ²∇ ×

∂

∂t

. (1.267)

In the source-free region, P = 0 and M = 0, the Hertz vectors satisfy

homogeneous wave equations:

∇

− µ²

∂

∂t

= 0, (1.268)

∇

− µ²

∂

∂t

= 0. (1.269)

Applying these two equations and the vector identity (B.45), we ﬁnd that

the expressions of E and H in the source-free region become

E = ∇(∇ ·

) − µ²

∂

∂t

− µ∇ ×

∂

∂t

, (1.270)

H = ∇(∇ ·

) − µ²

∂

∂t

+ ²∇ ×

∂

∂t

(1.271)

E = ∇ × ∇×

− µ∇ ×

∂

∂t

, (1.272)

H = ∇ × ∇ ×

+ ²∇ ×

∂

∂t

. (1.273)

1.6 Hertz Vectors 49

1.6.2 Complex Hertz Vectors

For the steady-state sinusoidal time-varying ﬁelds,

= Π

j ωt

, π

= Π

j ωt

where Π

and Π

denote the complex amplitudes of the Hertz vectors. Then

the equations (1.255) and (1.261) become

∇

+ k

= −

, (1.274)

∇

+ k

= −M. (1.275)

The expressions of the complex amplitudes of potentials and ﬁelds become

A = j ωµ² Π

+ µ∇ × Π

, (1.276)

ϕ = −∇ · Π

, (1.277)

E = ∇(∇ · Π

) − k

− j ωµ∇ × Π

, (1.278)

H = ∇ × ∇× Π

+ j ω²∇ × Π

. (1.279)

In the source-free region, the equations of the complex amplitudes of the

Hertz vectors become Helmholtz’s equations:

∇

+ k

= 0, (1.280)

∇

+ k

= 0. (1.281)

The expressions for the ﬁelds become

E = ∇(∇ · Π

) + k

− j ωµ∇ × Π

, (1.282)

H = ∇(∇ · Π

) + k

+ j ω²∇ × Π

(1.283)

E = ∇ × ∇ × Π

− j ωµ∇ × Π

, (1.284)

H = ∇ × ∇× Π

+ j ω²∇ × Π

. (1.285)

The method for the solution of vector Helmholtz’s equations using Hertz

vector functions is given in Section 4.3.2.

50 1. Basic Electromagnetic Theory

1.7 Duality

In Section 1.1.6 we have seen that by introducing the equivalent magnetic

charge and magnetic current we convert Maxwell’s equations into dual equa-

tions [37]. The electromagnetic ﬁelds excited by an electric source and those

excited by a magnetic source are dual to each other.

∇ × E = −j ωµH ∇ × H = j ω²E

∇ × H = j ω²E + J ∇ × E = −j ωµH − J

∇ · ²E = ρ ∇ · µH = ρ

∇ · µH = 0 ∇·²E = 0

These two sets of equations take the same mathematical form. A systematic

interchange of symbols changes the ﬁrst set of equations into the second set.

E → H

H → −E

J → J

ρ → ρ

µ → ²

² → µ

If we have the solutions to one problem, we can obtain the solutions to

the dual problem by means of interchange of symbols. For example, the

electromagnetic ﬁeld of an a-c electric dipole and that of a small a-c current

loop are dual problems because the later can be seen as an a-c magnetic

dipole.

The boundary with a surface electric charge and current and the boundary

with a surface magnetic charge and current are dual boundary conditions.

n × (E

− E

) = 0 n × (H

− H

) = 0

n × (H

− H

) = J

n × (E

− E

) = −J

n · (D

− D

) = ρ

n · (B

− B

) = ρ

n · (B

− B

) = 0 n · (D

− D

) = 0

So, the short-circuit surface or electric wall and the open-circuit surface or

magnetic wall are dual boundary conditions.

short-circuit surface open-circuit surface

n × E = 0 n ×H = 0

n × H = J

n × E = −J

n · D = ρ

n · B = ρ

n · B = 0 n · D = 0

Note that the problems with not only dual equations but also dual boundary

conditions are dual problems. For example, an electric dipole beside a perfect

conductor surface and a magnetic dipole beside the same perfect conductor

surface are not dual problems. An electric dipole beside an electric wall

and a magnetic dipole beside a magnetic wall with the same shape are dual

problems.