Lecture Notes Em Fields Nov 2012 - Renato Orta

8/16/2019 Lecture Notes Em Fields Nov 2012 - Renato Orta

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Renato Orta

Lecture Noteson

Electromagnetic Field Theory

PRELIMINARY VERSION

November 2012

DEPARTMENT OF ELECTRONICS

AND

TELECOMMUNICATIONS

POLITECNICO DI TORINO


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Contents

Contents 1

1 Fundamental concepts 3

1.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Constitutive relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Waves in homogeneous media 14

2.1 Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Cylindrical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Spherical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 Waves in non homogeneous media . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5 Propagation in good conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Radiation in free space 28

3.1 Green’s functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 Elementary dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3 Radiation of generic sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Antennas 49

4.1 Antenna parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.1.1 Input impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.1.2 Radiation pattern, Directivity and Gain . . . . . . . . . . . . . . . . . . . . 51

4.1.3 Effective area, effective height . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.2 Friis transmission formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.3 Examples of simple antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.3.1 Wire antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.3.2 Aperture antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

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CONTENTS

5 Waveguides 76

5.1 Waveguide modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.2 Equivalent transmission lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3 Rectangular waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.3.1 Design of a single mode rectangular waveguide . . . . . . . . . . . . . . . . 92

5.3.2 Tunneling effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.3.3 Irises and waveguide discontinuities . . . . . . . . . . . . . . . . . . . . . . . 100

A Mathematical Basics 1

A.1 Coordinate systems and algebra of vector fields . . . . . . . . . . . . . . . . . . . . 1

A.2 Calculus of vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

A.3 Multidimensional Dirac delta functions . . . . . . . . . . . . . . . . . . . . . . . . . 17

B Solved Exercises 20

B.1 Polarization and Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

B.2 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

B.3 Antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

B.4 Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Bibliography 45

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Chapter 1

Fundamental concepts

1.1 Maxwell’s Equations

All electromagnetic phenomena of interest in this course can be modeled by means of Maxwell’sequations

∇ × E (r,t) = − ∂

∂tB(r,t) −M(r,t)

∇ ×H(r,t) = ∂ ∂tD(r,t) + J (r,t)

(1.1)

Let us review the meaning of the symbols and the relevant measurement units.

E (r,t) electric field V/m

H(r,t) magnetic field A/m

D(r,t) electric induction C/m2

B(r,t) magnetic induction Wb/m2

J (r,t) electric current density (source) A/m2

M(r,t) magnetic current density (source) [V/m2]

Customarily, only electric currents are introduced; it is in particular stated that magneticcharges and currents do not exist. However, it will be seen in later chapters, that the introductionof fictitious magnetic currents has some advantages:

• The radiation of some antennas, such as loops or horns, is easily obtained• Maxwell’s equations are more symmetric

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Renato Orta - Electromagnetic Field Theory (Nov. 2012) PRELIMINARY VERSION 4

• (surface) magnetic currents are necessary for the formulation of the equivalence theorem, afundamental tool for the rigorous modelling of antennas

In circuit theory, one has two types of ideal generators, i.e. current and voltage ones: likewise, inelectromagnetism one introduces two types of sources.

Concerning the symmetry of Maxwell’s equations, we cite the principle of duality : performingthe exchanges

E ↔HB ↔ −DJ ↔ −M

Maxwell’s equations transform into each others.

Experiments show that the electric charge is a conserved quantity. This implies that electriccurrent density and electric charge (volume density are related by a continuity equation

∇ ·J (r,t) + ∂ ρ(r,t)∂t

= 0 (1.2)

By analogy, we assume that also magnetic charges are conserved, so that a similar continuityequation must be satisfied:

∇ ·M(r,t) + ∂ρm(r,t)∂t

= 0 (1.3)

It can be proved that eqs.(1.1), (1.2) (1.3) imply the well known divergence equations

∇ ·B(r,t) = ρm(r,t) ∇ ·D(r,t) = ρ(r,t) (1.4)Some authors prefer to assume eqs.(1.1), 1.4) as fundamental equations and obtain the conservation

of charge (1.2) (1.3) as a consequence.

Maxwell’s equations can be written in differential form as above, so that they are assumed tohold in every point of a domain, or in integral form, so that they refer to a finite volume. Theintegral form can be obtained by integrating eq.(1.1) over an open surface Σo with boundary Γand applying Stokes theorem:

Γ

E · τ̂ ds = − ddt

Σo

B · ν̂ dΣo − Σo

M · ν̂ dΣo

Γ

H · τ̂ ds = ddt

Σo

D · ν̂ dΣo + Σo

J · ν̂ dΣo(1.5)

In words, the first equation says that the line integral of the electric field, i.e. the sum of all voltagedrops along a closed loop, equals the time rate of change of the magnetic induction flux plus thetotal magnetic current. The second equation says that the line integral of the magnetic field alonga closed loop equals the time rate of change of the electric induction flux plus the total electriccurrent.

Then we integrate eq.(1.4) over a volume V with surface Σ and apply the divergence theorem:

Σ

B · ν̂ dΣ = V

ρm dV

Σ

D · ν̂ dΣ = V

ρ dV (1.6)


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Figure 1.1. Open surface Σo for the application of Stokes theorem. Notice that the orien-tations of ν̂ and τ̂ are related by the right-hand-rule: if the thumb points in the direction

of ν̂ , the fingers point in that of τ̂ .

Figure 1.2. Closed surface Σ for the application of the divergence theorem.

which is the usual formulation of Gauss theorem.

The same procedure on eq.(1.2) produces: Σ

J · ν̂ dΣ + ddt

V

ρ dV = 0 (1.7)

This says that the total current out of a volume equals the time rate of decrease of the internal

charge.

1.2 Phasors

Field variables can have any time dependence but a particularly important one is the so calledtime-harmonic regime. Consider a general time-harmonic electric field in a particular point:

E (t) = E x0 cos(ω0t + ϕx)x̂ + E y0 cos(ω0t + ϕy)ŷ + E z0 cos(ω0t + ϕz)ẑ


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The three cartesian components are sinusoidal functions of time with different amplitudes andphases, but the same frequency. This equation can be transformed in the following way.

E (t) = RE x0e j(ω0t+ϕx) x̂ + RE y0e j(ω0t+ϕy) ŷ + RE z0e j(ω0t+ϕz) ẑ

= RE x0e jϕx x̂ + E y0e jϕy ŷ + E z0e jϕz ẑ e jω0t

= R

E e jω0t (1.8)

The complex vector E is called the phasor of the time-harmonic vector E (t) and is measured inthe same units. It can be decomposed into real and imaginary parts:

E = E + jE

withE = E x0 cos ϕxx̂ + E y0 cos ϕyŷ + E z0 cos ϕz ẑ

and

E

= E x0 sin ϕxx̂ + E y0 sin ϕyŷ + E z0 sin ϕz ẑConsider again eq.(1.8):

E (t) = R

E e jω0t

= R

(E + jE) e jω0t

= R {(E + jE) (cos ω0t + j sin ω0t)}= E cos ω0t − E sin ω0t

We have obtained a representation of the time-harmonic electric field as the sum of two vectors,with arbitrary directions, in time quadrature one with respect to the other: in other words boththese vectors are sinusoidal functions of time, with the same frequency but with a relative delay of a quarter of a period. This representation is useful to study the polarization of the time-harmonic

vector, i.e. the form of the curve traced by the vector E (t) as a function of time. It can be shown

t E ( )

E"

E'

Figure 1.3. Elliptically polarized field

that, in general, this curve is an ellipse inscribed in a parallelogram that has E and E as half medians, as shown in Fig. 1.3. We see easily from the previous equation that


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• for t = 0, E = E

• for t = T /4, E =

−E

• for t = T /2, E = −E

• for t = 3T /4, E = E

where T = 2π/ω is the period. Hence the sense of rotation is from E to −E. In this case thefield is said to be elliptically polarized .

There are particular cases. When E and E are parallel or one of the two is zero, the paral-lelogram degenerates into a line and the polarization is linear . The two cases can be condensed inthe single condition (cross product, i.e. vector product):

E × E = 0

The other particular case is that in which

|E| = |E| and E · E = 0

The parallelogram becomes a square and the ellipse a circle: the field is circularly polarized .

The plane defined by the two vectors E and E is called polarization plane . Suppose weintroduce a cartesian reference in this plane with the z axis orthogonal to it. The phasor E hasonly x and y components,

E = E xx̂ + E yŷ

where E x and E y are complex numbers. This means that the original time-harmonic field isrepresented as the sum of two sinusoidally varying orthogonal vectors, with arbitrary magnitudesand phases. On this basis, the type of polarization is ascertained with the following rules:

• if the phase difference between the two components δ = arg E y − arg E x is 0 or π thepolarization is linear, along a line that forms an angle ψ = arctan(|E y|/|E x|) (if δ = 0) orψ = − arctan(|E y|/|E x|) (if δ = π)

• if δ = ±π/2 and |E y| = |E x| the polarization is circular, clockwise if δ = π/2, counterclock-wise if δ = −π/2

• in the other cases, the polarization is elliptical

To illustrate these concepts consider the following example.

Example

The phasor of a magnetic field is H = (1 + j)x̂ + (1 − 3 j)ŷ. Determine the type of polarization,write the expression of the time varying field H(t) and draw the polarization curve defined by thisvector.

Solution Find real and imaginary part of the phasor

H = x̂ + ŷ H = x̂ − 3ŷ


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Compute

H × H = (x̂ + ŷ) × (x̂ − 3ŷ) = (−3 − 1)ẑ = 0

H · H = (x̂ + ŷ) · (x̂ − 3ŷ) = 1 − 3 = −2 = 0The polarization is neither linear nor circular, hence it is elliptical counterclockwise (H(t) goesfrom H to −H).The time varying field is

H(t) = (x̂ + ŷ)cos ω0t − (x̂ − 3ŷ)sin ω0t

The plot is shown in Fig. 1.4

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

y

x

Figure 1.4. Polarization curve defined by H(t) above

The time-harmonic regime is important because of the property

d

dtE (r,t) =

d

dtR

E(r) e jω0t

= R jω0E(r) e

jω0t

(1.9)

so that time derivatives become algebraic operations. If we substitute the representation (1.8) into(1.1) we obtain, after canceling common factors exp( jω0t):

∇ × E(r) = − jω0B(r) − M(r)

∇ × H(r) = jω0D(r) + J(r)(1.10)

If the time dependence is of general type, eq. (1.8) is generalized by the spectral representation(inverse Fourier transform)

E (r,t) = 1

2π

∞−∞

E(r,ω) e jωtdω (1.11)


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In words, a generic time varying electric field is represented as a sum of an infinite number of harmonic components of all frequencies and amplitude E(r,ω)dω/(2π), where

E(r,ω) = ∞

−∞

E (r,t) e− jωtdt (1.12)

(direct Fourier transform). E(r,ω) is a spectral density of electric field, hence it is measured inV/(m Hz). Due to the fact that E (r,t) is real, E(r, − ω) = E∗(r,ω), so that the previous equationcan also be written

E (r,t) = 2R

1

2π

∞0

E(r,ω) e jωtdω

in terms of positive frequencies only.

The importance of the spectral representation is related to the property

d

dtE (r,t) =

d

dt

1

2π ∞−∞ E(r,ω) e

jωt

dω =

1

2π ∞−∞ jωE(r,ω) e

jωt

dω

If we take the Fourier transform of (1.1), we get

∇ × E(r,ω) = − jωB(r,ω) − M(r,ω)

∇ × H(r,ω) = jωD(r,ω) + J(r,ω)(1.13)

While ω0 is a specific frequency value, (1.13) must be satisfied for all frequencies. We refer tothese system as Maxwell’s equations in the frequency domain. The variables will be interpreted asphasors or Fourier transforms, depending on the application.

1.3 Constitutive relations

Clearly Maxwell’s equations as written in the previous section form an underdetermined system:indeed there are only two equations but four unknowns, the two fields and the two inductions. Itis necessary to introduce the constitutive relations, i.e. the equations linking the inductions to thefields. From a general point of view, matter becomes polarized when it is introduced into a regionin which there is an electromagnetic field, that is, the electric charges at molecular and atomiclevel are set in motion by the field and produce an additional field that is summed to the originalone. The inductions describe the electromagnetic behaviour of matter.

The simplest case is that of free space in which

B(r,t) = µ0H(r,t)

D(r,t) = 0 E (r,t)(1.14)

where 0, dielectric permittivity , and µ0 magnetic permeability , have the values

µ0 = 4π · 10−7 H/m0 =

1

µ0c2 ≈ 1

36π · 10−9 F/m


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and the speed of light in free space c has the value

c = 2.99792458

·108 m/s.

In the case of linear, isotropic, dielectrics, the constitutive relations (1.14) are substituted by

B(r,ω) = µ(ω) H(r,ω)

D(r,ω) = ε(ω) E(r,ω)(1.15)

where

µ(ω) = µ0µr(ω)

(ω) = 0r(ω)

and µr(ω), r(ω) (pure numbers) are the relative permittivity and permeabilities. All non fer-romagnetic materials have values of µr very close to 1. Notice that since molecular and atomiccharges have some inertia, they cannot respond instantaneously to the applied field, so that theresponse depends on the time rate of change of the excitation. The description of such a mechanismis best performed in the frequency domain, where ε(ω) and µ(ω) play the role of transfer functions.The fact that they depend on frequency is called dispersion : hence free space is non dispersive. Ingeneral ε(ω) and µ(ω) are complex:

ε = ε − jε µ = µ − jµ

It can be shown that the volume density of dissipated power in a medium is related to theirimaginary part

P diss = 1

2ε|E|2 + 1

2µ|H|2

Notice that ε, µ are positive in a passive medium because of the phasor time convention

exp( jω0t). Some authors use the convention exp(− jω0t): in this case passive media have neg-ative ε, µ. Clearly with the time convention we adopt, negative ε, µ characterize activemedia, such as laser media.

When the dielectric contains free charges, the presence of an electric field E(r,ω) gives rise toa conduction current density Jc(r,ω):

Jc(r,ω) = γ (ω) E(r,ω)

where γ (ω) is the conductivity of the dielectric, measured in S/m. The previous equation is themicroscopic form of Ohm’s law of circuit theory.

The conduction current enters into the second Maxwell’s equation (1.13), which becomes

∇ ×H = jωεE + γ E + J

where the term J is the (independent) source. It is customary to incorporate the conductioncurrent into the polarization current by means of an equivalent permittivity. Indeed we can write

jωεE + γ E = jω

ε − j γ ω

E = jωεeqE

with εeq = ε − j(ε + γ/ω). In practice the subscript eq is always dropped: the imaginary part of

ε takes into account all loss mechanisms, both those due to molecular and atomic transitions andthose due to Joule effect.


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In the case of low loss dielectrics one often introduces the loss tangent

tan δ = ε

ε

so that we can writeε = ε(1 − j tan δ )

Values of tan δ 10−3 : 10−4 characterize low loss dielectrics.It is interesting to note that for fundamental physical reasons, there is relationship between

the real and the imaginary part of the dielectric permittivity and of the magnetic permeability.Indeed, in the case of the permittivity, just as a consequence of causality, ε(ω) − ε0 and ε(ω) areHilbert transforms of each other, that is

ε(ω) − ε0 = 1π

P

∞−∞

ε(α)

α − ω dα

ε(ω) = 1

πP

∞−∞

ε(α) − ε0α − ω dα

These equations are called Kramers-Krönig relations. The symbol P denotes the Cauchy principalvalue of the integral, that is, for f (0) = 0

P

∞−∞

f (x)

x dx = lim

a→0

−a−∞

f (x)

x dx +

∞a

f (x)

x dx

The constitutive equations (1.15) imply that the inductions are parallel to the applied fields, whichis true in the case of isotropic media but not in the case of crystals. These media are said to beanisotropic and are characterized by a regular periodical arrangement of their atoms. In this casethe permittivity constitutive equations must be written in matrix form:

DxDyDz

=

εxx εxy εxzεyx εyy εyz

εzx εzy εzz

E xE y

E z

In the case of an isotropic dielectric, the matrix is a multiple of the identity: ε = εI.

1.4 Boundary conditions

Maxwell’s equations (1.13) are partial differential equations (PDE), valid in every point of a givendomain, which can be finite or infinite. If it is finite, we must supply information about the nature

of the material that forms the boundary. In mathematical terms, we must specify the boundaryconditions, i.e. the values of the state variables on the boundary.

Often the boundary is a perfect electric conductor (PEC), that is a material with infiniteconductivity. Note that copper is such a good conductor that up to microwave frequencies it canbe modeled as a PEC. If the conductivity is infinite, the electric field must vanish everywhere inthe volume of a PEC, otherwise the conduction current would be infinite. The first Maxwell’sequation shows that also the magnetic field is identically zero, provided the frequency is not zero.Since we are essentially interested in time-varying fields, we conclude that in a PEC both fieldsvanish identically. At the surface, since the conduction current cannot have a normal component,


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only the tangential component of the electric field must be zero. If ν̂ is the unit normal at theboundary, this condition can be written

ν̂ × E = 0 on the boundary (1.16)Indeed, ν̂ × E is tangential to the boundary, as shown in Fig. 1.5.

ˆν

tgE

ˆν

Figure 1.5. Boundary condition at the surface of a perfect electric conductor. Relationship betweenthe tangential component of the electric field Etg and ν̂ ×E

Sometimes the permittivity or the permeability change abruptly crossing a surface in the do-main. By applying some integral theorems to Maxwell’s equations, it can be shown that thefollowing continuity conditions hold

ν̂ × (H(rΣ+) − H(rΣ−)) = 0 ν̂ × (E(rΣ+) − E(rΣ−)) = 0 (1.17)ν̂

·(B(rΣ+)

−B(rΣ−)) = 0 ν̂

·(D(rΣ+)

−D(rΣ−)) = 0 (1.18)

where ν̂ is the normal to the surface and rΣ± are infinitely close points, lying on opposite sides of the surface, as shown in Fig. 1.6. It can also be proved that if a surface current Js or Ms exist,then the fields are discontinuous

ˆν

+Σr

−Σr

Σ1 1

ε µ

2 2ε µ

Figure 1.6. Boundary conditions at the surface of separation between different dielectric media

ν̂ × (H(rΣ+) − H(rΣ−)) = Js ν̂ × (E(rΣ+) − E(rΣ−)) = MsSince it can be proved that when a PEC is present in a magnetic field, the induced current flowsonly on the surface of it and the magnetic field is identically zero in the PEC, we can write

ν̂ × H(rΣ+) = Js (1.19)


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Notice that this is an equation that does not force a condition on H but allows the computation of Js. Hence the boundary condition to be enforced at the surface of a PEC is only (1.16). Notice alsothat the units of a surface electric current are A/m and those of a surface magnetic current V/m.

This is necessary for the validity of the previous equations, but it is also obvious for geometricalreasons, as Fig. 1.7 shows.

sJ

γ

ˆν

Figure 1.7. Surface electric current induced on a perfect electric conductor (PEC). γ is a curvelying on the PEC surface. Js is a current density per unit length measured along γ (A/m)

Sometimes an approximate boundary condition of impedance type is used: it is a linear relationbetween the tangential electric and magnetic fields, called also a Leontovich boundary condition,that can be written

ν̂ × E(rΣ+) = Z s (ν̂ × H(rΣ+)) × ν̂ (1.20)in terms of a suitable surface impedance. This condition is typically applied when the boundary isa real metal and one desires a more accurate model than just a PEC. The double vector producton the right hand side makes the tangential electric and magnetic field orthogonal on the surface.If the surface is not smooth but has a sub-wavelength wire structure, the surface impedance is nota scalar but a tensor (matrix).

If the domain is infinite and sources are all at finite distance from the origin, the only necessaryboundary condition is that the field is only outgoing at large distance from the origin.

Sometimes the geometry of dielectric and metal bodies possesses sharp edges or sharp vertices(e.g. plates, cubes, cones), as shown in Fig. 1.8. In this case some field components can becomeinfinite at the geometric singularity: however the fields must always be locally square integrable.In physical terms this condition means that the electromagnetic energy contained in a finite neigh-borhood of the singularity must always be finite. Apart from these cases of true singularities of the geometry, fields are always regular, i.e. differentiable. This is to be noted in particular whenthe geometry singularity is only apparent because it is related to the particular coordinate system.For example if we use cylindrical coordinates in a homogeneous medium the fields must be regularin the origin even if the coordinates have a singularity there.

Figure 1.8.


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Chapter 2

Waves in homogeneous media

At this point we have finished the preliminaries. We have decided to use the electric and magneticfield as state variables and we have the relevant equations:

∇ × E = − jωµH − M∇ × H = jωεE + J+ boundary conditions

(2.1)

where for brevity we have dropped the dependence of all variables on r, ω, but it is understood.The line “boundary conditions” contains the form of the domain where we want to compute thefields created by the given sources J,M and information on the nature of the material that formsthe boundary. In general permittivity and permeability are functions of r and provide information

on the shape of the bodies in the domain and on their nature. In this way the problem is well posedand it can be shown it has a unique solution, provided there is at least a region in the domainwhere energy is dissipated. Needless to say the problem (2.1) can be very complicated and can besolved only by approximate numerical techniques. For this reason it is useful to proceed by smallsteps, by analyzing first a very idealized problem that is so simple to allow an analytical solution.

2.1 Plane waves

Let us start by assuming that the domain of interest is infinite and the medium is homogeneousand lossless, so that ε, µ are real constants. The only boundary condition to enforce is that thefield is regular everywhere, in particular at infinity.

Moreover we assume that sources are identically zero. This is, at first sight, a strange assump-tion since it would seem to imply that also the fields must be identically zero! However, if thereare no sources, Maxwell’s equations become a homogeneous system of differential equations: it iswell known that homogeneous differential equations have nontrivial solutions. Let us review someexamples:

• d

dxf (x) = 0, x ∈ R ⇒ f (x) = const

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• Harmonic oscillator

d2

dt2 + ω20 f (t) = 0, t ∈ R ⇒ f (t) = A cos ω0t + B sin ω0t

• Transmission line

− ddz

V = jω L I

− ddz

I = jω C V , z ∈ R ⇒

V (z) = V +0 e−jkz + V −0 e

+jkz

I (z) = Y ∞V +0 e

−jkz − Y ∞V −0 e+jkz

These are actually generalizations of the simple case of a linear system of algebraic equations

A x = 0

which has nontrivial solutions if the matrix A is non invertible.

So the problem we want to solve is ∇ × E = − jωµH∇ × H = jωεE

(2.2)

Equations (2.2) are written in a coordinate-free language. However, in order to solve them itis necessary to select a coordinate system. Several choices are at our disposal, the more commonbeing cartesian, cylindrical and spherical coordinates. The corresponding solutions of (2.2) willbe plane, cylindrical and spherical waves, respectively. The simplest case is the first and we startwith that.

Recalling the expression of the ∇ operator in cartesian coordinates

∇ = x̂ ∂ ∂x

+ ŷ ∂

∂y + ẑ

∂

∂z

and that the medium is homogeneous, it is clear that (2.2) is a linear system of constant coefficientequations. On the basis of the experience with ordinary differential equations, we can expect thatthe solution is of exponential type, hence we assume tentatively

E(r) = E0 exp(− jkxx) exp(− jkyy) exp(− jkzz) (2.3)and likewise for the magnetic field, where E0 and kx,ky ,kz are constants to be found. The constantskx,ky,kz have dimensions rad/m and are wavenumbers along the three coordinate axes. It isconvenient to work with a vector formalism, even if the coordinate system is fixed. By recallingthat r = xx̂ + yŷ + zẑ and defining the wavevector k = kxx̂ + kyŷ + kz ẑ, the assumed form of the

solution isE(r) = E0 exp(− jk · r) H(r) = H0 exp(− jk · r) (2.4)

Before substituting it into (2.2) it is useful to compute

∇ exp(− jk · r) =

x̂ ∂

∂x + ŷ

∂

∂y + ẑ

∂

∂z

exp(− jkxx) exp(− jkyy) exp(− jkzz)

= (− jkxx̂ − jkyŷ − jkz ẑ)exp(− jkxx) exp(− jkyy) exp(− jkzz)= − jk exp(− jk · r)


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Moreover we recall the identity of vector calculus

∇ ×(A(r)φ(r)) = φ(r)

∇ ×A(r) +

∇φ(r)

×A(r)

so that the substitution of (2.4) into (2.2) yields

− jk × E0 exp(− jk · r) = − jωµH0 exp(− jk · r)

− jk × H0 exp(− jk · r) = jωεE0 exp(− jk · r)Canceling common factors

− jk × E0 = − jωµH0

− jk × H0 = jωεE0Note that the fact that these equations do not contain the variable r any longer confirms the

correctness of the assumption (2.4).Recalling the properties of vector products we learn that E0, H0, k form a righthanded triple

of mutually orthogonal vectors. Next, to proceed, we eliminate H0 between the two equations. Tothis end, we solve the first equation with respect to H0:

H0 = k × E0

ωµ (2.5)

and substitute into the second one

k × (k × E0) + ω2µεE0 = 0The double vector product can be expanded

k (k · E0) − (k · k) E0 + ω2µεE0 = 0The first term is zero because of the orthogonality of k and E0, hence

k · k − ω2µεE0 = 0 (2.6)We are interested in nontrivial solutions of this equation, so that the following condition must hold

k · k = ω2µε (2.7)A relationship between frequency and wavenumbers is called in general a dispersion relation . Wecan read it from right to left or vice versa: in the first instance it tells us what the frequencymust be so that the field distribution (2.4) with a specific wavevector k is a solution of Maxwell’s

equations. From this point of view, the value of ω can be considered to be a resonance frequency of the structure. Notice that the requirement that the solution be regular everywhere (in particularat infinity) forces the wavector to be real. Apart from this condition there is no constrain on thepossible values of the wavenumbers, hence the resonance frequencies of the system are infinite innumber and even distributed continuously. We start seeing a property that characterizes all fieldproblems. Whereas lumped element circuits have a finite number of resonances, distributed systemsalways have an infinite number of them. Moreover, if the structure has finite size its resonancesare denumerably infinite: this means that they can be labeled with integers ω1, . . . , ωn, . . .. If thestructure, as in this case, has infinite size, then the resonances should be labeled with a continuous


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variable. To simplify the notation, we omit this labeling variable and remember that ω can takeevery real value.

The dispersion equation can also be read from left to right: in this case the frequency isconsidered fixed and we look for the wavevectors that satisfy eq.(2.7). It is convenient to introducea unit vector ŝ, called direction of propagation, directed along k, so that k = kŝ; then the dispersionrelation becomes

k = ω√

εµ (2.8)

Clearly the direction ŝ can be whatever, only the wavenumber k is specified. In other words forany given frequency there are an infinite number of waves with arbitrary directions of propagation.

Considering again eq.(2.6), we see that if the dispersion relation is satisfied, the vector E0can be arbitrarily chosen, provided it is orthogonal to ŝ. For a given ŝ, there are two linearlyindependent waves, but they are degenerate, because they have the same value of the wavenumberk. The corresponding magnetic field is obtained by the impedance relation (2.5):

H0 = ω√ εµŝ × E0

ωµ =

ε

µŝ × E0 = Y ŝ × E0 (2.9)

where the wave admittance has been introduced. Its inverse is the wave impedance Z = 1/Y . Infree space it has the value

Z 0 =

µ0ε0

≈ 120π ≈ 377 Ω

In conclusion a solution of the problem (2.2) is

Eŝ(r) = E0 exp(− jkŝ · r)Hŝ(r) = H0 exp(− jk ŝ · r)

(2.10)

where ŝ is the direction of propagation, k = ω√ εµ, H0 = Y ŝ × E0 and E0, H0, ŝ are mutuallyorthogonal: electromagnetic waves are transverse. Since the problem is linear, the general solutionof (2.2) can be written as a linear combination of waves of the type (2.10) with all possible directionsŝ.

Wavefronts are defined to be the surfaces on which the phase Φ(r) of the wave is constant. Inthis case

Φ(r) = −kŝ · r = constantis the equation of a family of planes perpendicular to ŝ: hence the fields (2.10) are called plane waves because the wavefronts are planes.

Let us now study the polarization of plane waves. This requires going back to time domain viaeq.(1.8)

E (r,t) = R {E0 exp(− jk ŝ · r)exp( jω0t)}= E0 cos (ω0t − kŝ · r) − E0 sin (ω0t − kŝ · r)

(2.11)

We see clearly from this equation that the type of polarization in every point of space is specifiedby E0, which is the electric field in the origin. What changes from point to point is the timeevolution, due to the propagation delay. The constant phase surfaces of the time varying field areω0t − kŝ · r = const, from which we find

ŝ · r = ω0k

t − const


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E

H

r

ŝ

û

Figure 2.1. Wavefronts of a plane wave with direction of propagation ŝ. They move at the phasevelocity along ŝ. The vector û denotes an arbitrary direction

This means that the wavefronts are not fixed but are moving. Indeed, consider a specific wavefront,

say the zero-phase one, i.e. the one for which const=0, as shown in Fig.2.1; the vector r that denotesits points is such that its projection on ŝ increases linearly with time. In other words, the planemoves as a whole with speed

v ph = ω0

k =

ω0ω0

√ εµ

= 1√

εµ =

c√ εrµr

(2.12)

This velocity is called phase velocity of the wave, because it has been defined by means of theconstant phase surfaces. Now consider an arbitrary straight line with direction û, an let P be itsintersection with the zero-phase wavefront. The velocity of P is

v ph(û) = 1

û · ŝc√

εrµr

Clearly, for all directions û = ŝ this velocity is larger than c/

√ εrµr. Notice, however, that even

when this velocity is larger than the speed of light in empty space, the theory of relativity is notviolated. Indeed no matter or energy is moving in the direction of û, but only a mathematicalpoint, e.g. a maximum or a node of the oscillation. This concept is at the basis of the fact thatthe phase velocity in a waveguide is always greater that the speed of light in vacuum.

Fig.2.2 shows the fields of a linearly polarized plane wave propagating in the ŝ = ẑ direction:remember that the electric and magnetic field are orthogonal. For clarity, the field vectors havebeen drawn only for a number of points on the z axis, even if they are defined in every point of space.

Wave propagation is always associated to energy flow. The Poynting vector has the meaningof p ower density (per unit surface) associated to the wave. Let us compute the Poynting vector Sin the case of a plane wave:

S = 12

(E × H∗) = 12

(E0 exp(− jk ŝ · r) × H0∗ exp( jk ŝ · r))

= 1

2 (E0 × H0∗) = 1

2 (E0 × (̂s × E0∗)Y ∗)

= 1

2

|E0|2Z

ŝ

(2.13)

where we used the impedance relation (2.9) and the property

E0 × (ŝ× E0∗) = |E0|2ŝ− (E0 · ŝ) E∗0 = |E0|2ŝ (2.14)


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,

z t E( )

, z t H( )

Figure 2.2. Snapshot of a linearly polarized plane wave propagating in the ŝ = ẑ direction

because of the orthogonality between E0 and ŝ. Note that the Poynting vector is real, henceno reactive power is associated to plane waves in a lossless medium. We see that the activepower density (magnitude of S) associated to a plane wave is constant: this implies that the totalpower, obtained by integration over the whole wavefront, is infinite. Hence, a single plane waveis not a physically realizable field. This property, however, does not destroy the usefulness of theconcept. Indeed, since Maxwell’s equations are linear, the superposition principle holds and linearcombinations of plane waves are also solutions. It turns out that a continuous sum (integral) of plane waves not necessarily has infinite power: indeed all physically realizable fields can always berepresented as integrals of plane waves.

We recall that to each plane wave not only a power flow is associated, but also a flow of linear momentum and of angular momentum. In particular the linear momentum flow, which hasdirection ŝ, is responsible for the radiation pressure , that explains, for instance, the shape of comettails and has been considered as a possible “engine” for interplanetary travels.

The properties of plane waves do not change much if the dielectric is lossy. In this case thepermittivity is complex and the dispersion relation (2.8) becomes

k = ω

(ε − jε)µ = β − jα, α ≥ 0where β is the true phase constant, measured in rad/m and α is the attenuation constant, measuredin Nepers/m. The electric field, for instance, obeys the propagation law

E(r) = E0 exp(− jβ ̂s · r) exp(−αŝ · r) (2.15)Clearly the magnitude of the field is no longer constant in space and the wavefronts are also surfacesof constant field magnitude. Obviously, it must be remarked that the value of the phase velocitycannot be computed by (2.12), but it is given by

v ph = ω

β

and the wavelength is

λ = 2π

β =

v phf


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Indeed, both the phase velocity and the wavelength are defined on the basis of the phase of thewave and β is exactly the phase rate-of-change, measured, as said above, in rad/m.

Let us compute the power flow.

S = 1

2 (E × H∗) = 1

2 (E0 exp(− jk ŝ · r) × H0∗ exp( jk∗ŝ · r))

= 1

2 (E0 × (ŝ× E0∗)Y ∗ exp(−2αŝ · r)) = 1

2Y ∗|E0|2 exp(−2αŝ · r)ŝ

(2.16)

where we have used again the property (2.14). In this case the Poynting vector is complex. Theactive power density of the wave decreases during propagation because part of it is transferred tothe dielectric in the form of heat.

All the plane waves considered up to now are called uniform because their propagation direc-tion ŝ is real. If we go through all the steps of the derivation, we realize that even if ŝ is complex(whatever this means!) the expression (2.10) is a valid solution of Maxwell’s equations, although

only in a halfspace. Such a generalization leading to non uniform plane waves is required whensolving a scattering problem where a plane wave is incident on the interface separating two di-electrics. It is to be remarked that also the plane wave (2.15) is a valid solution only in a halfspace.Indeed, if ŝ · r → −∞, the field diverges, which is not physically acceptable.

2.2 Cylindrical waves

To solve eqs.(2.2) it is also possible to use a cylindrical coordinate system instead of a cartesianone. The mathematics is considerably more complicated in this case. The reason is that the unitvectors of the cylindrical coordinate system are not constant but change from point to point. Asa consequence, the expression of the differential operators is no longer with constant coefficients

and the solutions are no longer of exponential type, but are expressed in terms of Bessel functions.These special functions of mathematical physics were actually introduced, along with many others,in order to solve the wave equation.

If a cylindrical or spherical coordinate system is used, Maxwell’s equations (2.2) are not attackeddirectly but are first transformed into a single second order equation. We write them again forconvenience ∇ × E = − jωµH

∇ × H = jωεESince it is a system of equations, we can eliminate one of the two unknowns. We solve the firstequation with respect to H and substitute in the second

H = ∇ ×E

− jωµ

∇ × (∇ × E) = ω2εµEAs expected, the second equation contains only the electric field: the price to pay for it is that itis second order in the space derivatives; it is called the curl-curl equation . However its form canbe simplified recalling the identity

∇ × (∇ × E) = ∇(∇ · E) − ∇2E = −∇2E


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where we have used the fact that (2.2) do not have sources, hence also the charge density ρ(r) iszero and the electric field has zero divergence, ∇ · E = ρ = 0. We obtain in this way the vector Helmholtz equation

∇2E + ω2εµE = 0 (2.17)Even if it is written in coordinate-free language, its meaning is easily understood in cartesiancoordinates only, where

E(r) = E x(r)x̂ + E y(r)ŷ + E z(r)

Since the unit vectors are not function of r, each cartesian component of the electric field satisfiesHelmholtz equation, which then becomes scalar:

∇2ψ + ω2εµψ = 0

where ψ(r) denotes any component of E . It is interesting to note that even if we are using cartesian components to represent the electric field, we are not forced to use necessarily cartesian coordinates

to specify the observation point, i.e. the components of r. By using the classical method of theseparation of variables in cylindrical coordinates, we can find

ψ(ρ,ϕ,z) = ψ0H (2)m (kρρ)e

−jkzze−jmϕ (2.18)

where m = 0,±1,±2, . . ., kρ ∈ [0,∞) and kz identify the various outgoing cylindrical waves. Thesethree parameters play the role of kx, ky, kz in the case of plane waves. The function H

(2)m (kρρ) is

a Hankel function of second kind and order m. Its asymptotic expansion is

H (2)m (kρρ) ∼

2

πkρρ exp

− j(kρρ − m π

2 − π

4)

The dispersion relation is

kz =

ω2εµ − k2ρNotice that all three components have this form, but the values of ψ0 for each must be interrelatedso that the resulting vectors E and H satisfy Maxwell’s equations.

We are not going to describe in detail the properties of these waves. To explain the name, itis enough to say that the wavefronts are cylinders having ẑ as axis, at least in the case m = 0 andkρ = ω

√ εµ.

2.3 Spherical waves

The case of spherical waves is similar, from a certain point view, to that of cylindrical waves. Againthe mathematics is fairly complicated and new special functions are introduced. In this case thescalar Helmholtz equation is solved in spherical coordinates and the result is

ψ(r,ϑ,ϕ) = ψ0h(2)l (kr)P

ml (cos ϑ)e

−jmϕ

where l = 0,1,2, . . . and −l ≤ m ≤ l identify the various outgoing spherical waves. The functionsh(2)l (kρ) are spherical Hankel functions of second kind and order l, whereas P

ml (cos ϑ) are associated

Legendre polynomials of degree l and order m. Again, the various solutions for the three cartesian


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1

1

ε

µ

2

2

ε

µ

3

3

ε

µ

4

4

ε

µ

t

r

i

Figure 2.3. Scattering from a stratified dielectric: i, incident wave; r reflected wave; t,transmitted wave. For clarity, the couple of plane waves existing in each of the internallayers has not been indicated

components must be related so that the resulting E and H satisfy Maxwell’s equations. Theasymptotic expansion of the spherical Hankel functions is

h(2)l (kr) ∼ 1kr exp− j(kr − m π

2 − π

4)

Hence the wavefronts are spheres with center in the origin and this justifies the name.

2.4 Waves in non homogeneous media

The case that has been considered, namely a homogeneous medium filling the whole space is highlyidealized. In a realistic situation, ε(r), µ(r) are not constant and obviously the plane waves (2.10)are not solution of Maxwell’s equations (2.2). In order to consider a simple case, let us assumethat the medium is piecewise homogeneous and that the interfaces between the different materials

are planar: the structure is called a stratified dielectric . In the left half space an incident planewave is assumed. In each layer, plane waves are solutions of (2.2), but the continuity conditions(1.17) must be obeyed. It can be proved that in each one of the internal layers two plane wavesare present, one forward (incident on the following interface) the other backward (reflected fromthe following interface); in the right half space only one, because the medium extends to infinityand there is no other interface. All these plane waves have the same transverse (to z) componentof the wavevector and their amplitudes can be easily determined so that the continuity conditionsare satisfied. The single wave in the fourth medium is the transmitted field, the second one in thefirst medium is the reflected field, as sketched in Fig. 2.3.

If the interfaces are not planar, the problem becomes much more difficult. Consider, for ex-ample, the case of Fig. 2.4, where a plane wave is incident on a cylinder with parameters ε2, µ2,embedded in a homogeneous medium with parameters ε1, µ1. It can be shown that the continuity

conditions require that an infinite number of plane waves are excited, each one with the right am-plitude. Collectively. these are called scattered waves . Hence the difficulty of the problem stemsfrom the necessity to solve a linear system with an infinite number of unknowns.

If the medium is not even piecewise homogeneous but arbitrarily inhomogeneous, no analyticalsolution is at our disposal. It is, however, to be mentioned that when the variations of ε(r), µ(r)are small on the wavelength scale, a well known approximate method can be used, i.e. GeometricalOptics. Whereas the plane waves discussed up to now can be defined global plane waves since eachone is defined over the whole space, the elementary geometrical optics field is a local plane wave .For instance, a spherical wave in free space can be approximated by a collection of local plane


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i

s

ss

s

1

1

ε

µ

2

2

ε

µ

Figure 2.4. Scattering from a non planar interface: i, incident wave; s, scattered waves. For clarity,the plane waves existing inside the cylinder have not been indicated

waves because its wavefront (a sphere) can be approximated locally by the relevant tangent plane.

The k vectors of these local plane waves define a vector field, whose field lines are the geometrical optics rays . It turns out that rays are also the field lines of the Poynting vector field: hence a plotof the rays provides information about the power flow.

Geometrical optics is a very powerful technique, but sometimes yields definitely wrong results.This happens when rays cross in a point or along a line, because in this case it predicts a field of infinite intensity. These singularities are called caustics and the focus of a converging lens is anexample: in such a point the electromagnetic field can be very large but is certainly finite. Hencegeometrical optics can be safely used only away from caustics.

2.5 Propagation in good conductors

Apart from the case of optical fibers, guided wave propagation is possible in structures containingmetal conductors. Examples are coaxial cables, parallel wire transmission lines, microstrip lines,waveguides with any cross section. Since the metals used in the applications (such as copper) arecharacterized by a very large conductivity, in a first approximation they can be considered to beperfect conductors (PEC), an assumption that greatly simplifies the study. However, in order tobuild more accurate numerical models of real devices, it is necessary to take into account the finiteconductivity of real metals. In this section we consider the propagation of plane waves in goodconductors, in order to draw some conclusions pertaining to transmission systems.

Metals are characterized by so a large conductivity that the displacement currents can be safelyneglected with respect to the conduction currents, so that some simplifications in the generalformulas of Section 2.1 are possible. Starting with the wavenumber,

km = ω

(ε − jγ /ω)µ ≈ ω

(− jγ /ω)µ

if γ

ωε 1 (good conductor)

Recalling that − j = ±1 − j√

2


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and that Imk ≤ 0 for a passive medium, we find

km = 1 − j

√ 2√

ωγµ = 1 − j

δ (2.19)

where we have introduced the so called skin depth

δ =

2

ωµγ (2.20)

which, of course, should not be confused with the loss angle, introduced in Section 1.3, indicatedwith the same symbol. This relation can also be written

f δ =

1

πµγ = const

where f is the frequency and the constant depends only on the material. For instance, in the caseof copper, γ = 5.8 · 107 S/m and µ = µ0 = 4π · 10−7 H/m, hence f δ = 0.0661

√ Hzm (2.21)

The reason for the name will be explained below.

The wave impedance is computed with the same approximation:

Z m =

µ

ε − jγ /ω ≈

µ

− jγ /ω =

jµω

γ =

1 + j√ 2

ωµ

γ

that isZ m = Rs(1 + j) (2.22)

where we have introduced the surface resistance

Rs =

ωµ

2γ =

1

γδ

for which we can writeRs√

f =

πµ

γ = const

iE

r E

t E

x

z

t Hr H

iH

Figure 2.5. Good conductor in a plane wave field. In the free space region both an incident anda reflected wave exist, in the metal only the transmitted one. The wavevector of the transmittedwave is drawn dashed, to indicate that it is complex.


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Again, in the case of copper,Rs√

f = 2.6090 · 10−7 Ω/

√ Hz

Consider now a (highly idealized) conductor in the form of a half space, which faces free space,with a linearly plane wave incident normally on it, as shown in Fig. 2.5. The tangential electricand magnetic fields are continuous at the interface, then the ratio of their magnitudes is the samein z = 0− and in z = 0+. But in z = 0+ this ratio is Z m by definition, so we can easily understandthat the expressions of the electric fields are

Ei = E 0 e−jk0zx̂

Er = ΓE 0 ejk0zx̂

Et = (1 + Γ)E 0 e−jkmzx̂

where the reflection coefficient is

Γ =

Z m−

Z 0Z m + Z 0

Since |Z m| Z 0, Γ is close to −1. Indeed,

1 + Γ = 2Z mZ m + Z 0

≈ 2Z mZ 0

= 2(1 + j)

πε0f

γ

from which

Γ ≈ −1 + 2(1 + j)

πε0f

γ

In the case of copper,Γ

≈ −1 + 2(1 + j)

·6.9252

·10−10 f

(frequency in Hz). We can also say that the metal enforces an impedance type boundary condition(see (1.20)) with Z m as surface impedance.The magnetic field is

Hi = Y 0E 0 e−jk0zŷ

Hr = −Y 0ΓE 0 ejk0zŷHt = Y 0(1 − Γ)E 0 e−jkmzx̂ ≈ 2Y 0E 0 e−jkmzŷ

(2.23)

Note that the total magnetic field at the interface is approximately twice the incident one becauseΓ is very close to −1.The electric field in the metal produces a conduction current in the x̂ direction

Jc = γ Et = 2(1 + j)

πε0f γE 0 e

−jkmzx̂

In the case of copper, this becomes

Jc = 2(1 + j) · 0.402

f E 0 e−jz/δ e−z/δ x̂

(frequency in Hz) where we have used (2.19). The magnitude of this current density is maximumat the interface and then decays exponentially in the metal. At a depth z = δ , it has reduced by afactor 1/e = 0.368. Eq.(2.21) allows a simple computation of δ for various frequencies, reported in


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Table 2.1. Skin depth for copper at various frequencies

Frequency Skin depth

50 Hz 9.3 mm1 kHz 2.1 mm1 MHz 66.1 µm1 GHz 2.1 µm

Table 2.1. We see clearly that as the frequency increases, the current density remains appreciableonly in a very thin layer close to the metal surface, which can be considered as its “skin”. Even if this analysis strictly refers to a metal half space, we can use it to draw qualitative conclusions inthe case of finite thickness conductors or even round conductors, provided the thickness is muchlarger than the skin depth. At the power frequency of 50Hz, the skin depth is so large that thecurrent has a uniform distribution in ordinary wires. At the frequency of 1MHz, instead, most of the conductor copper is not used. At microwave frequencies, a few microns of copper deposited onan insulator perform as an excellent conductor.The consequence of the skin depth change with frequency is that the resistance of a conductor isan increasing function of frequency: indeed, the “effective” cross-section of the conductor decreasesas the frequency increases. This phenomenon is generally called skin effect .

Let us compute the impedance of the structure of Fig. 2.5, viewed as a current carrying con-ductor. Since the fields and the current density does not depend on y , we consider a strip of unitlength in this direction. We compute first the total current I , flowing in the x̂ direction, per unitlength along y :

I = ∞

0

Jc(x,z) · x̂dz = ∞

0

J c0e−j(1−j)z/δdz = J

c0δ 1 + j

(2.24)

Notice that the dimensions of I are correctly A/m, since J c0 is a surface current density with value

J c0 = 2(1 + j)

πε0f γE 0 (2.25)

Next, consider a unit length in the x̂ direction of this conductor and compute the potential differ-ence along this length by integrating the electric field E x along the x axis (y = 0, z = 0). Notethat E x does not depend on x, hence E x itself coincides numerically with this potential difference.Finally, the impedance per unit width in the y direction and unit length in the x direction is givenby

Z pul = E x(0,0)

I

= J c0/γ

J c0δ/(1 + j)

= 1 + j

γδ

= Z m

where we used (2.22). In conclusion we have this remarkable result: the impedance seen by acurrent flowing through a square of unit sides coincides with the wave impedance in the metal.Notice that, apart from the similarity in the symbols,

Z m = E xH y

hence it is a completely different concept. Moreover, since the conductor we are considering hasunit width in the y direction, unit length in the x direction (and infinite thickness in the z direction)


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the previous analysis shows why often the value of the surface resistance Rs is expressed in Ω/(read Ohm per square).

As a final remark, we note that the material becomes a perfect conductor when γ → ∞. Inthese conditions, the skin depth vanishes and the value of the current density at the interface tendsto infinity, according to (2.25). Nevertheless, we see from (2.24) that the total current is finite andits value, independent of γ is

I = J c0δ

1 + j = 2

ε0µ0

E 0

where (2.20) and (2.25) have been used. This means that in a perfect conductor the current densitycan be written

J c(x,z) = 2

ε0µ0

E 0δ (z) = J σδ (z)

On the other hand, from (2.23) we see that in the limit γ → ∞ the magnitude of the total magneticfield at the z = 0− interface coincides with J σ. Taking the directions of the vectors into account,

we conclude that if a perfect conductor is immersed in an electromagnetic field, on its surface acurrent density Jσ (A/m) appears, such that

Jσ = ν̂ × Hwhere ν̂ is the normal to the PEC surface, pointing toward free space. In practice, this is the proof of Eq.(1.19).

Another example that we consider now is that of sea water: because of the salt contained in it,the conductivity is γ = 5 S/m, whereas the relative permittivity, up to the microwave region, doesnot change very much and will be taken to be εr = 80. We compute the complex wavenumber andthe attenuation constant by the general equation

km = ω (ε − jγ /ω)µThe results are the following:

• At f = 100Hz, γ /(2πf ε0εr) = 1.1234 · 107, so sea water behaves as a good conductor;k = (4.4429 − j4.4429) · 10−2m−1 α = 0.3859dB/m

• At f = 10kHz, γ /(2πf ε0εr) = 1.1234 · 105, so sea water behaves as a good conductor;k = (0.4429 − j0.4429)m−1 α = 3.8590dB/m

• At f = 1GHz, γ /(2πf ε0εr) = 1.1234, so the displacement currents cannot be neglected;

k = (209.7536 − j94.1066)m−1

α = 817.3998dB/m

• At f = 10GHz, γ /(2πf ε0εr) = 0.1123, so the displacement currents cannot be neglected;k = (1877.5266 − j105.1341)m−1 α = 913.1833dB/m

Obviously, at microwave frequency, the attenuation of sea water precludes the possibility of commu-nicating with submarines during subsurface navigation. This becomes possible at low frequencies,where, however, the available bandwidth is very narrow.


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Chapter 3

Radiation in free space

The fundamental problem in electromagnetics is computing the fields created by a specified set of sources in a given region of space. This means that the functions ε(r), µ(r) are assigned, as wellas the form of the region boundary and the material of which it consists. Then the sources arespecified in terms of electric and magnetic current densities J(r), M(r).

In order to understand the basic mechanism of radiation, it is convenient to consider first ahighly idealized problem, wherein the sources radiate in an infinite homogeneous medium. Laterwe will see how to apply the results of this chapter to the real antenna problem.

3.1 Green’s functions

The radiation problem is mathematically formulated as ∇ × E = − jωµ0H − M∇ × H = jωε0E + J

(3.1)

in an infinite homogeneous domain that we assume to be free space. These equations are linear withconstant coefficients and the independent variable is r. We can interpret them as the equations of a Linear Space Invariant system (LSI), where the source currents play the role of input and theradiated fields that of output, see Fig. 3.1. The box represents a system with two inputs and twooutputs.

E(r)

H(r)

J(r)

M(r)

EJG

HJG

HMG

EMG

Figure 3.1. Linear system view of the radiation phenomenon

28


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LSI systems are clearly a multidimensional generalization of Linear Time Invariant (LTI) sys-tems. Let us review briefly the properties of the latter. LTI systems, as shown in Fig. 3.2 arecompletely characterized in time domain by their impulse response h(t), that is the output that is

obtained when the input is a Dirac delta function δ (t). An arbitrary (continuous) input x(t) canbe represented as a linear combination of pulses thanks to the sifting property of the delta function

x(t) =

∞−∞

x(t)δ (t − t)dt

Because of linearity, the response y (t) to x(t) can be found by convolution

y(t) = h(t) ∗ x(t) = ∞−∞

h(t − t)x(t)dt

Alternatively, an LTI system can be characterized by its transfer function : when the input is x(t) =exp( jωt), the output is proportional to it and the constant of proportionality is, by definition, H (ω),so that y(t) = H (ω)exp( jωt). It can be proved that the impulse response and the transfer functionof a system are related by a Fourier transform

H (ω) =

∞−∞

h(t) exp(− jωt)dt

( ) x t ( )h t ( ) y t ( ) X ω ( ) H ω ( )Y ω

Figure 3.2. Time domain and frequency domain description of an LTI system

As a preparation for (3.1), let us consider the simpler case of an infinite, uniform transmissionline excited by a distribution of voltage and current generators, vs(z) and is(z), as shown in Fig. 3.3.Since these generators are distributed continuously, vs(z) and is(z) are densities per unit length of generators described, as usual, in terms of their open circuit voltage (V/m) and short circuitcurrent (A/m), respectively. The differential equations of the system are

−dV

dz = j ωLI + vs

−dI dz

= j ωCV + is(3.2)

++ ++

( )s

v z

( )s

i z

Figure 3.3. Infinite uniform transmission line with distributed voltage and current generators.


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( )V z

I z( )( )s

v z

( )s

i z sVi

G

s IiG

s Iv

G

sVvG

Figure 3.4. Linear system view of the transmission line with distributed generators.

Generally, transmission lines are excited at one end by a generator that acts as a transmitter.The model shown in Fig. 3.3 refers to a situation of electromagnetic compatibility, where a lineis excited by an electromagnetic wave that couples to the line along a certain segment of it. It

is easy to recognize that this is the one dimensional analogue of Maxwell’s equations (3.1). Theproblem is again LSI and can be schematized as in Fig. 3.4. Hence, as suggested by this picture,the solution can be expressed as

V (z) = Z ∞

∞−∞

GV is(z − z)is(z)dz + ∞−∞

GV vs(z − z)vs(z)dz

I (z) =

∞−∞

GIis(z − z)is(z)dz + Y ∞ ∞−∞

GIvs(z − z)vs(z)dz

The system here has two inputs and two outputs: each output depends on both inputs, so thatin practice there are four Green’s functions, each one a pure number. They can be obtainedby applying the spatial Fourier transform to the system equations (3.2). However, by the verydefinition of Green’s function

• GV is(z) is the voltage wave V (z)/Z ∞ created by a unit amplitude current generator locatedin z = 0

• GV vs(z) is the voltage wave V (z) created by a unit amplitude voltage generator located inz = 0

• GIis(z) is the current wave I (z) created by a unit amplitude current generator located inz = 0

• GIvs(z) is the current wave I (z)Y ∞ created by a unit amplitude voltage generator located inz = 0

so that they can be found by simple circuit theory, just recalling that the input impedance of an

infinitely long line is Z ∞. The resulting expressions are

GV is(z) = GIvs(z) = −1

2e−jk|z|

GV vs(z) = GIis(z) = −1

2sgn(z)e−jk|z|

where sgn(z) is the sign function

sgn(z) =

1 if z > 0−1 if z


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As another preparatory example before tackling (3.1), let us consider the case of sound waves.It can be shown that the excess pressure p(r) with respect to the background pressure satisfies thescalar Helmholtz equation

∇2 + ω2V 2s

p(r) = −S (r) (3.3)

where V s is the sound velocity and S (r) is a source term. This equation corresponds to the pictureof Fig. 3.5, where S (r) is the input and p(r) the output. In this case the system has only oneinput and one output but it is multidimensional, since both depend on the three independentvariables x, y, z. In perfect analogy with LTI systems, LSI systems are completely characterizedin space domain by their “impulse response” G(r), which is traditionally called Green’s function .This is the output of the system when the input is a point source located at the origin of thecoordinate system, which can be represented mathematically by a three-dimensional Dirac deltafunction S (r) = δ (r) = δ (x)δ (y)δ (z). The fundamental property of this multidimensional Dirac δ function is

δ (r)dr =

∞−∞

∞−∞

∞−∞ δ (x)δ (y)δ (z)dxdydz = 1

When the input is an arbitrary function, the output is found by (three-dimensional) convolution

p(r) =

G(r − r)S (r)dr

Alternatively, an LSI system can be characterized in the spectral domain. When the input is aharmonic function of x, y , z , that is S (r) = exp(− j(kxx + kyy + kzz)) = exp(− jk · r), the outputis proportional to it and the coefficient of proportionality is, by definition, the transfer functionH (k). Again, transfer function and Green’s function of the same system are related by a Fouriertransform: however, in this case, it is triple, since it operates on the three variables x, y, z. Thecouple of inverse and direct 3-D Fourier transform is given by

G(r) = 1

(2π)3

H (k)exp(− jk · r)dk

H (k) =

G(r) exp( jk · r)dr

(3.4)

where dk = dkx dky dkz. It can be shown that in the case of free space, the transfer function is

H (k) = 1

k2 − ω2/V 2s(3.5)

and the corresponding Green’s function is

G(r) = exp(− jk0r)4πr

(3.6)

( )S r ( )G r ( ) p r ( )S k ( ) H k ( ) p k

Figure 3.5. Space domain and spatial frequency domain description of the sound radiation phenomenon


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with k0 = ω /V s denoting the wavenumber. This expression describes a diverging spherical wave.Indeed, the constant phase surfaces are obviously r = const, a series of concentric spheres withcenter in the origin. Moreover, assuming that the source is harmonic with frequency ω0, the

expression of the Green’s function in time domain is

g(r,t) = R

exp(− jk0r)

4πr exp( jω0t)

=

cos(ω0t − k0r)4πr

from which it is evident that the phase velocity is

V ph = ω0/k0 = V s > 0

As another well known example of LSI system, let the frequency ω go to zero in (3.3), so thatthe Helmholtz equation becomes Poisson equation. This, for example, relates the electric potentialV (r) to a charge distribution ρ(r), which acts as its source:

∇2V (r) = −ρ(r)εThe transfer function associated to this equation is (from (3.5))

H (k) = 1

εk2

and the corresponding Green’s function (from (3.6))

G(r) = 1

4πεr

We recognize immediately this expression as the potential generated by a point charge q = 1 C ina dielectric with permittivity ε.

We are ready now for Maxwell’s equations (3.1), which are still more complicated because inaddition to being multidimensional and multiple input/output, they are vector equations: thismeans that the output is a vector that is not necessarily parallel to the input. This implies thateach of the four Green’s function is not a scalar but a linear operator (a tensor ), which, in a basis,is represented by a 3 × 3 matrix. This means that, differently from the case of sound waves, theGreen’s function is not directly the field radiated by a point source. The source is really a pointbut is also a vector, which can have all possible orientations. From a certain point of view, we cansay that the Green’s tensor yields the field radiated in a given point by a point source in the originwith all possible orientations. This concept will be better clarified in Section 3.2.

In coordinate-free language

E(r) =

− jω µ0 GEJ(r − r

)

·J(r)dr

− GEM(r − r)

·M(r)dr

H(r) =

GHJ(r − r) · J(r)dr − jω ε0

GHM(r − r) · M(r)dr

(3.7)

To check the dimensions of the various Green’s functions, it is useful to note that

ωµ0 = k0Z 0 ωε0 = k0Y 0 = k0/Z 0

Hence we recognize that GEJ and GHM are measured in m−1, GEM and GHJ in m

−2. Theexplicit expressions of the various dyadic Green’s functions can be obtained by applying the Fourier


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transform technique to (3.1). The most appropriate coordinate system that can be used to showthe result is the spherical one, because the source is a point. It can be shown that the matricesrepresenting the Green’s functions are:

GEJ(r,ϑ,ϕ) = GHM(r,ϑ,ϕ) ↔ A 0 00 B 0

0 0 B

exp(− jk0r)

4πr

GEM(r,ϑ,ϕ) = GHJ(r,ϑ,ϕ) ↔ − jk0

0 0 00 0 −C

0 C 0

exp(− jk0r)

4πr

(3.8)

where the wavenumber is k0 = ω√

ε0µ0. Note that the row and column indices are r̂, ϑ̂, ϕ̂,respectively. In dyadic form

GEJ(r,ϑ,ϕ) = GHM(r,ϑ,ϕ) = Ar̂r̂ + Bϑ̂ϑ̂ + Bϕ̂ϕ̂ exp(− jk0r)

4πr

GEM(r,ϑ,ϕ) = GHJ(r,ϑ,ϕ) = − jk0C ϕ̂ϑ̂− ϑ̂ϕ̂

exp(− jk0r)4πr

where

A = 2

j

1

k0r +

1

(k0r)2

B = 1 − 12

A = 1 − j 1k0r

− 1(k0r)2

C = 1 − j 1k0r

Consistently with the fact that the source is a point, the Green’s function does not depend on theangular variables.

The behavior of the three coefficients at large distance from the source, k0r → ∞ (i.e. r λ)is

A = O

1

k0r

B → 1C → 1

In this region, usually called far field region , the expressions of the Green’s functions simplify andbecome

GEJ(r,ϑ,ϕ) = GHM(r,ϑ,ϕ) ∼ Itr exp(− jk0r)4πr

GEM(r,ϑ,ϕ) = GHJ(r,ϑ,ϕ) ∼ − jk0r̂ × Itr exp(− jk0r)4πr

(3.9)

where Itr is the transverse to r identity dyadic, see Appendix. This operator, when applied to anarbitrary vector, produces as a result the projection of the vector on the plane perpendicular to r.The operator r̂ × Itr adds a further 90◦ counterclockwise turn around r.


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Conversely, in the near field region , k0r → 0 (i.e. r λ), the coefficients become

A∼

2

(k0r)2

B ∼ − 1(k0r)2

C ∼ − j 1k0r

3.2 Elementary dipoles

As discussed previously, the Green’s function is the basic tool for the computation of the fieldradiated by any source by means of eq. (3.7). However, it is convenient to start with the simplestone, i.e. a point source, and this will help in understanding the properties of the Green’s functions.

Consider first an elementary source of electric type located at the origin of the coordinate system,modeled by the current distribution

J(r) = Meδ (r)

The vector Me is called the electric dipole moment of the current distribution and is measuredin Am (recall that the dimensions of the three dimensional δ function are m−3). An arbitrarycurrent distribution can be characterized by its moments . This concept is used also in the theoryof probability: if ξ is a random variable with density function W ξ(x), moments of all orders canbe defined by

mn = E {ξ n} = ∞−∞

xnW ξ(x)dx

where E { } denotes the expectation value. In the case of the current distribution, the role of W ξ(x) is played by J(r), but the situation is more complicated because its vector nature implies

that the moments beyond the first are tensors. The first moment (dipole moment) is a vector,defined by

Me =

J(r)dr (3.10)

In the case of the point source introduced above, thanks to the properties of the delta function, theprevious equation becomes an identity and we understand the reason for the name of the coefficient.From a practical point of view, we can imagine to obtain this source by a limiting process, startingfrom a rectilinear current I , whose length l is progressively reduced without changing the aspectratio (diameter/length of the wire), while, at the same time, the current is increased, so that thevalue of the integral (the dipole moment I l) remains constant.

Introducing the dipole current into (3.7), we find that the radiated fields are given by

E(r) = − jωµ0

GEJ(r − r) · Meδ (r)dr = − jωµ0GEJ(r) · Me

H(r) =

GHJ(r − r) · Meδ (r)dr = GHJ(r) · Me

(3.11)

Since we know the expressions of the matrices representing the Green’s functions in the sphericalbasis, it is necessary to express the vector Me in the same basis. Let us assume that the polaraxis of the coordinate system is in the direction of Me, i.e. assume Me = M eẑ. Note thatthis step is allowed because the Green’s function does not depend on the angular variables, as a


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consequence of the isotropy of free space. It is to be remarked, as a general rule, that the Green’sfunction depends only on the structure and, hence, shares its symmetries. This choice guaranteesthe simplest description of the radiated field. Since the radiated field must have the direction of

Me as a symmetry axis, orienting the polar axis of the coordinate system in this direction allowsthe expressions to be independent on ϕ.

Me = (Me · r̂)̂r + (Me · ϑ̂)ϑ̂ + (Me · ϕ̂)ϕ̂= M e(ẑ · r̂)̂r + M e(ẑ · ϑ̂)ϑ̂ + M e(ẑ · ϕ̂)ϕ̂= M e cos ϑr̂ − M e sin ϑϑ̂

(3.12)

where we have exploited (A.12). Now recalling the expression of the Green’s function (3.8), weobtain

E(r) = − jωµ0 exp(− jk0r)4πr

A 0 00 B 0

0 0 B

cos ϑ− sin ϑ

0

M e

= − j Z 0M e exp(− jk0r)2rλ

A cos ϑr̂ − B sin ϑϑ̂

(3.13)where use has been made of

ωµ0 = k0Z 0 = 2π

λ Z 0

and Z 0 is the wave impedance. Concerning the meaning of (3.12), it is to be remarked that thematrix (3.8) represents the Green’s function in the spherical basis consisting of the unit vectorsr̂, ϑ̂, ϕ̂ defined in the observation point r. Hence, even if the source is located in the origin, itscomponents are evaluated in the basis associated to the point r.We can proceed similarly for the magnetic field:

H(r) = − jk0 exp(− jk0r)4πr

0 0 00 0 −C 0 C 0

cos ϑ− sin ϑ0

= j

M e exp(− jk0r)2rλ

C sin ϑϕ̂

(3.14)

In conclusion, the electromagnetic field radiated by an electric dipole has the following expression

E(r) = − j Z 0M e exp(− jk0r)2rλ

2

j

1

k0r +

1

(k0r)2

cos ϑr̂ −

1 − j 1

k0r − 1

(k0r)2

sin ϑϑ̂

H(r) = jM e exp(− jk0r)

2rλ 1 − j 1

k0r sin ϑϕ̂(3.15)

This wave has two components of electric field and only one of magnetic field. Imagine a geo-graphical system of coordinates such that the direction of the dipole moment defines the directionof the earth axis. The angle ϑ is the colatitude (= 90◦−latitude), the angle ϕ is the longitude.Then the electric field is contained in the meridian planes and the magnetic field is tangent to theparallels. This type of wave is called TM (Transverse Magnetic) since the magnetic field has noradial component. We recognize also that the radial component of the electric field is dominantclose to the source, but negligible with respect to the others at large distance. Here the wave isessentially TEM, since neither field has a (significant) radial component.


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Let us compute the energy budget by means of the Poynting theorem. The Poynting vector is

S = E×

H∗ = Z 0M

2e

4r2λ2BC ∗ sin2 ϑr̂ + AC ∗ sin ϑ cos ϑϑ̂

Compute the components

BC ∗ =

1 − j 1

k0r − 1

(k0r)2

1 + j

1

k0r

= 1 − j 1

k0r − 1

(k0r)2 + j

1

k0r +

1

(k0r)2 − j 1

(k0r)3

= 1 − j 1(k0r)3

AC ∗ =

j

2

k0r +

2

(k0r)2

1 + j

1

k0r

= j

2

k0r +

2

(k0r)2 − 2

(k0r)2 + j

2

(k0r)3

= j

2

k0r +

2

(k0r)3

Substituting in the previous equation we get

S = Z 0M

2e

4r2λ2

1 − j 1

(k0r)3

sin2 ϑr̂ + j

2

k0r +

2

(k0r)3

sin ϑ cos ϑϑ̂

According to Poynting theorem, the surface density of active power flow is

dP

dΣ =

1

2R{S · ν̂ } (3.16)

where ν̂ is the normal to the surface element. In order to compute the total radiated active power,we have to evaluate the flux of the Poynting vector across a closed surface surrounding the source.For maximum simplicity we choose a sphere of radius r :

P rad = 1

2R

S · ν̂ dΣ = 1

2

2π

0

π

0

Z 0M 2e4r2λ2

sin2 ϑ r2 sin ϑ dθdϕ

= 1

2

Z 0M 2e

4λ2 2π

π0

sin3 ϑ dθ

= 1

2

Z 0M 2e 2π

3λ2

(3.17)

Here we have used the following facts

• the normal to the spherical surface is ν̂ = r̂• the area element in spherical coordinates is dΣ = r2 sin ϑ dθdϕ• the integrand does not depend on ϕ, so the ϕ integration yields the 2π factor• the ϑ integration yields π

0

sin3 ϑ dθ = 4

3

The factor 1/2 has been left explicit to make it clear that M e is a peak value, that is the timedomain dipole moment is Me(t) = M e cos(ω0t). If, on the contrary M e is an effective value, thefactor 1/2 has to be dropped.


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We notice that the total radiated power does not depend on the radius of the sphere chosento compute it. Algebraically this is the result of the cancellation between the r2 factor in thedenominator of the Poynting vector and the one in the area element dΣ. To get a more physical

explanation, consider the fluxes through two concentric spheres of different radii: if they weredifferent, power would be lost or generated in the shell, which is impossible by conservation of energy in a lossless medium.

In (3.17) we took the real part of the integral. The imaginary part is the reactive power

Q = 1

2I

S · ν̂ dΣ = −1

2

2π0

π0

Z 0M 2e

4r2λ21

(k0r)3 sin2 ϑ r2 sin ϑ dθdϕ

= −12

Z 0M 2e

4r2λ21

(k0r)3 2π

π0

sin3 ϑ dθ

= −12

Z 0M 2e 2π

3λ21

(k0r)3

It is reas

Lecture Notes Em Fields Nov 2012 - Renato Orta

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