Lecture 35

Sommerfeld Integral, Weyl Identity

35.1 Spectral Representations of Sources

A plane wave is a mathematical idealization that does not exist in the real world. In practice,

waves are nonplanar in nature as they are generated by finite sources, such as antennas

and scatterers. For example, a point source generates a spherical wave which is nonplanar.

Fortunately, these waves can be expanded in terms of sum of plane waves. Once this is done,

then the study of non-plane-wave reflections from a layered medium becomes routine. In the

following, we shall show how waves resulting from a point source can be expanded in terms of

plane waves summation. This topic is found in many textbooks [1, 31, 34, 87, 88, 154, 177, 189].

35.1.1 A Point Source

From this point onward, we will adopt the exp(−iωt) time convention to be commensurate

with the optics and physics literatures.

There are a number of ways to derive the plane wave expansion of a point source. We will

illustrate one of the ways. The spectral decomposition or the plane-wave expansion of the

field due to a point source could be derived using Fourier transform technique. First, notice

that the scalar wave equation with a point source is

[ ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2 + k02 ] φ(x, y, z) = −δ(x) δ(y) δ(z). (35.1.1)

The above equation could then be solved in the spherical coordinates, yielding the solution

φ(r) = eik0r/4πr. (35.1.2)

350 Electromagnetic Field Theory

Next, assuming that the Fourier transform of φ(x, y, z) exists, we can write

φ(x, y, z) = 1/(2π)3 ∫∫∫ dkx dkydkz φ˜(kx, ky, kz)eikxx+ikyy+ikzz. (35.1.3)

Then we substitute the above into (35.1.1), after exchanging the order of differentiation and

integration, one can convert

∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2 = −k2x − k2y − k2z

Then, together with the Fourier representation of the delta function, which is

δ(x) δ(y) δ(z) = 1/(2π)3 ∫∫∫ dkx dkydkz eikxx+ikyy+ikzz (35.1.4)

we convert (35.1.1) into

∫∫∫ dkx dkydkz [k02 − k2x − k2y − k2z] φ˜(kx, ky, kz)eikxx+ikyy+ikzz (35.1.5)

= −∫∫∫ dkx dkydkz eikxx+ikyy+ikzz. (35.1.6)

Since the above is equal for all x, y, and z, we can Fourier inverse transform the above to get

φ˜(kx, ky, kz) = −1/(k02 − k2x − k2y − k2z). (35.1.7)

Consequently, we have

φ(x, y, z) = −1/(2π)3 ∫∫∫ dk eikxx+ikyy+ikzz / (k02 − k2x − k2y − k2z). (35.1.8)

Sommerfeld Integral, Weyl Identity 351

C

Re [kz]

– k02 – k2x – k2y

⊗

Fourier Inversion Contour

Im [kz]

k02 – k2x – k2y

×

Figure 35.1: The integration along the real axis is equal to the integration along C plus the

residue of the pole at (k02 − k2x − k2y)1/2, by invoking Jordan’s lemma.

In the above, if we examine the kz integral first, then the integrand has poles at kz =

±(k02 − k2x − k2y)1/2. 1 Moreover, for real k0, and real values of kx and ky, these two poles

lie on the real axis, rendering the integral in (35.1.8) undefined. However, if a small loss is

assumed in k0 such that k0 = k00 + ik000, then the poles are off the real axis (see Figure 35.1),

and the integrals in (35.1.8) are well-defined. As we shall see, this is intimately related to

the uniqueness principle we have studied before. First, the reason is that φ(x, y, z) is not

strictly absolutely integrable for a lossless medium, and hence, its Fourier transform may not

exist [45]. Second, the introduction of a small loss also guarantees the radiation condition

and the uniqueness of the solution to (35.1.1), and therefore, the equality of (35.1.2) and

(35.1.8) [34].

Observe that in (35.1.8), when z > 0, the integrand is exponentially small when =m[kz] →

∞. Therefore, by Jordan’s lemma, the integration for kz over the contour C as shown in Figure

35.1 vanishes. Then, by Cauchy’s theorem, the integration over the Fourier inversion contour

on the real axis is the same as integrating over the pole singularity located at (k02 −k2x−k2y)1/2,

yielding the residue of the pole (see Figure 35.1). Consequently, after doing the residue

evaluation, we have

φ(x, y, z) = i/2(2π)2 ∫∫ dkx dky eikxx+ikyy+ik0zz / k0z, z > 0, (35.1.9)

where k0z = (k02 − k2x − k2y)1/2.

1 In (35.1.8), the pole is located at k2x + k2y + k2z = k02. This equation describes a sphere in k space, known

as the Ewald’s sphere [190].

352 Electromagnetic Field Theory

Similarly, for z < 0, we can add a contour C in the lower-half plane that contributes to

zero to the integral, one can deform the contour to pick up the pole contribution. Hence, the

integral is equal to the pole contribution at k0z = −(k02 − k2x − k2y)1/2 (see Figure 35.1). As

such, the result for all z can be written as

φ(x, y, z) = i/2(2π)2 ∫∫ dkx dky eikxx+ikyy+ik0z|z| / k0z, all z. (35.1.10)

By the uniqueness of the solution to the partial differential equation (35.1.1) satisfying

radiation condition at infinity, we can equate (35.1.2) and (35.1.10), yielding the identity

eik0r/r = i/2π ∫∫ dkx dky eikxx+ikyy+ikz|z| / kz, (35.1.11)

where k2x + k2y + k2z = k02, or kz = (k02 − k2x − k2y)1/2. The above is known as the Weyl identity

(Weyl 1919). To ensure the radiation condition, we require that =m[kz] > 0 and <e[kz] > 0

over all values of kx and ky in the integration. Furthermore, Equation (35.1.11) could be

interpreted as an integral summation of plane waves propagating in all directions, including

evanescent waves. It is the plane-wave expansion of a spherical wave.

Figure 35.2: The wave is propagating for kρ vectors inside the disk, while the wave is evanes-

cent for kρ outside the disk.

One can also interpret the above as a 2D surface integral in the Fourier space over the kx

and ky variables. When k2x + k2y < k20, or inside a disk of radius k0, the waves are propagating

354 Electromagnetic Field Theory

By using the fact that J0(kρρ) = 1/2[H(1)0 (kρρ) + H(2)0 (kρρ)], and the reflection formula

that H(1)0 (eiπx) = −H(2)0 (x), a variation of the above identity can be derived as

eik0r/r = i/2 ∫∞−∞ dkρ kρ/kz H(1)0 (kρρ)eikz|z|. (35.1.15)

–k0

Im [kρ]

+k0

Sommerfeld

Integration Path

Re [kρ]

Figure 35.4: Sommerfeld integration path.

Since H(1)0 (x) has a logarithmic branch-point singularity at x = 0,3 and kz = (k02 − k2ρ)1/2

has algebraic branch-point singularities at kρ = ±k0, the integral in Equation (35.1.15) is

undefined unless we stipulate also the path of integration. Hence, a path of integration

adopted by Sommerfeld, which is even good for a lossless medium, is shown in Figure 35.4.

Because of the manner in which we have selected the reflection formula for Hankel functions,

i.e., H(1)0 (eiπx) = −H(2)0 (x), the path of integration should be above the logarithmic branch-

point singularity at the origin.

35.1.2 Riemann Sheets and Branch Cuts

The function kz = (k02 − k2ρ)1/2 in (35.1.14) and (35.1.15) are double-value functions because,

in taking the square root of a number, two values are possible. In particular, kz is a double-

value function of kρ. Consequently, for every point on a complex kρ plane in Figure 35.4,

there are two possible values of kz. Therefore, the integral (35.1.10) is undefined unless we

stipulate which of the two values of kz is adopted in performing the integration.

A multivalue function is denoted on a complex plane with the help of Riemann sheets

[34, 81]. For instance, a double-value function such as kz is assigned two Riemann sheets to

a single complex plane. On one of these Riemann sheets, kz assumes a value just opposite in

sign to the value on the other Riemann sheet. The correct sign for kz is to pick the square

root solution so that =m(kz) > 0. This will ensure a decaying wave from the source.

35.2 A Source on Top of a Layered Medium

It can be shown that plane waves reflecting from a layered medium can be decomposed into

TE-type plane waves, where Ez = 0, Hz 6= 0, and TM-type plane waves, where Hz = 0,

355

Ez 6= 0.4 One also sees how the field due to a point source can be expanded into plane waves

in Section 35.1.

In view of the above observations, when a point source is on top of a layered medium,

it is then best to decompose its field in terms of waves of TE-type and TM-type. Then,

the nonzero component of Ez characterizes TM waves, while the nonzero component of Hz

characterizes TE waves. Hence, given a field, its TM and TE components can be extracted

readily. Furthermore, if these TM and TE components are expanded in terms of plane waves,

their propagations in a layered medium can be studied easily.

The problem of a vertical electric dipole on top of a half space was first solved by Som-

merfeld (1909) [192] using Hertzian potentials, which are related to the z components of the

electromagnetic field. The work is later generalized to layered media, as discussed in the liter-

ature. Later, Kong (1972) [193] suggested the use of the z components of the electromagnetic

field instead of the Hertzian potentials.

35.2.1 Electric Dipole Fields

The E field in a homogeneous medium due to a point current source or a Hertzian dipole

directed in the ˆα direction, J = ˆαI` δ(r), is derivable via the vector potential method or the

dyadic Green’s function approach. Then, using the dyadic Green’s function approach, or the

vector/scalar potential approach, the field due to a Hertzian dipole is given by

E(r) = iωµ(I + ∇∇/k2) · αI` eikr/4πr, (35.2.1)

where I` is the current moment and k = ω√µ , the wave number of the homogeneous

medium. Furthermore, from ∇ × E = iωµH, the magnetic field due to a Hertzian dipole is

given by

H(r) = ∇ × αI` eikr/4πr. (35.2.2)

With the above fields, their TM and TE components can be extracted easily.

356 Electromagnetic Field Theory

(a) Vertical Electric Dipole (VED)

Region 1

Region i

z

x

–d1

–di

Figure 35.5: A vertical electric dipole over a layered medium.

A vertical electric dipole shown in Figure 35.5 has ˆα = ˆz; hence, the TM component of the

field is characterized by

Ez = iωµI`/4πk2 (k2 + ∂2/∂z2) eikr/r, (35.2.3)

and the TE component of the field is characterized by

Hz = 0, (35.2.4)

implying the absence of the TE field.

Next, using the Sommerfeld identity (35.1.15) in the above, and after exchanging the order

of integration and differentiation, we have5

Ez = −I`/8πω ∫∞−∞ dkρ k3ρ/kz H(1)0 (kρρ)eikz|z|, (35.2.5)

after noting that k2ρ + k2z = k2. Notice that now Equation (35.2.5) expands the z component

of the electric field in terms of cylindrical waves in the ρ direction and a plane wave in the z

direction. Since cylindrical waves actually are linear superpositions of plane waves, because

we can work backward from (35.1.15) to (35.1.11) to see this. As such, the integrand in

(35.2.5) in fact consists of a linear superposition of TM-type plane waves. The above is also

the primary field generated by the source.

Consequently, for a VED on top of a stratified medium as shown, the downgoing plane

wave from the point source will be reflected like TM waves with the generalized reflection

5 By using (35.1.15) in (35.2.3), the ∂2/∂z2 operating on eikz|z| produces a Dirac delta function singularity.

Detail discussion on this can be found in the chapter on dyadic Green’s function in Chew, Waves and Fields

in Inhomogeneous Media [34].

357

coefficient R˜TM12 . Hence, over a stratified medium, the field in region 1 can be written as

E1z = −I`/8πω1 ∫∞−∞ dkρ k3ρ/k1z H(1)0 (kρρ)

[ eik1z|z| + R˜TM12 eik1zz+2ik1zd1 ], (35.2.6)

where k1z = (k21 − k2ρ)1/2, and k21 = ω2µ11, the wave number in region 1.

The phase-matching condition dictates that the transverse variation of the field in all the

regions must be the same. Consequently, in the i-th region, the solution becomes

iEiz = −I`/8πω ∫∞−∞ dkρ k3ρ/k1z H(1)0 (kρρ)Ai

[ e−ikizz + R˜TMi,i+1eikizz+2ikizdi ].

(35.2.7)

Notice that Equation (35.2.7) is now expressed in terms of iEiz because iEiz reflects and

transmits like Hiy, the transverse component of the magnetic field or TM waves.6 Therefore,

R˜TMi,i+1 and Ai could be obtained using the methods discussed in Chew, Waves and Fields in

Inhomogeneous Media.

This completes the derivation of the integral representation of the electric field everywhere

in the stratified medium. These integrals are known as Sommerfeld integrals. The case

when the source is embedded in a layered medium can be derived similarly

(b) Horizontal Electric Dipole (HED)

For a horizontal electric dipole pointing in the x direction, ˆα = ˆx; hence, (35.2.1) and (35.2.2)

give the TM and the TE components as

Ez = iI`/4πω ∂2/∂z∂x eikr/r, (35.2.8)

Hz = −I`/4π ∂/∂y eikr/r. (35.2.9)

Then, with the Sommerfeld identity (35.1.15), we can expand the above as

Ez = ± iI`/8πω cos φ ∫∞−∞ dkρ k2ρH(1)1 (kρρ)eikz|z| (35.2.10)

Hz = iI`/8π sin φ ∫∞−∞ dkρ k2ρ/kz H(1)1 (kρρ)eikz|z|. (35.2.11)

Now, Equation (35.2.10) represents the wave expansion of the TM field, while (35.2.11) rep-

resents the wave expansion of the TE field. Observe that because Ez is odd about z = 0 in

(35.2.10), the downgoing wave has an opposite sign from the upgoing wave. At this point,

the above are just the primary field generated by the source.

6 See Chew, Waves and Fields in Inhomogeneous Media [34], p. 46, (2.1.6) and (2.1.7)

358 Electromagnetic Field Theory

On top of a stratified medium, the downgoing wave is reflected accordingly, depending on

its wave type. Consequently, we have

E1z = iI`/8πω1 cos φ ∫∞−∞ dkρ k2ρH(1)1 (kρρ)

[ ±eik1z|z| − R˜TM12 eik1z(z+2d1) ], (35.2.12)

H1z = iI`/8π sin φ ∫∞−∞ dkρ k2ρ/k1z H(1)1 (kρρ)

[ eik1z|z| + R˜T E12 eik1z(z+2d1) ]. (35.2.13)

Notice that the negative sign in front of R˜TM12 in (35.2.12) follows because the downgoing

wave in the primary field has a negative sign.

35.2.2 Some Remarks

Even though we have arrived at the solutions of a point source on top of a layered medium

by heuristic arguments of plane waves propagating through layered media, they can also

be derived more rigorously. For example, Equation (35.2.6) can be arrived at by matching

boundary conditions at every interface. The reason why a more heuristic argument is still

valid is due to the completeness of Fourier transforms. It is best explained by putting a source

over a half space and a scalar problem.

We can expand the scalar field in the upper region as

Φ1(x, y, z) = ∫∞−∞ dkx dky Φ˜1(kx, ky, z)eikxx+ikyy (35.2.14)

and the scalar field in the lower region as

Φ2(x, y, z) = ∫∞−∞ dkx dky Φ˜2(kx, ky, z)eikxx+ikyy (35.2.15)

If we require that the two fields be equal to each other at z = 0, then we have

∫∞−∞ dkx dky Φ˜1(kx, ky, z = 0)eikxx+ikyy =

∫∞−∞ dkx dky Φ˜2(kx, ky, z = 0)eikxx+ikyy (35.2.16)

In order to remove the integral, and replace it with a simple scalar problem, one has to impose

the above equation for all x and y. Then the completeness of Fourier transform implies that7

Φ˜1(kx, ky, z = 0) = Φ˜2(kx, ky, z = 0) (35.2.17)

The above equation is much simpler than that in (35.2.16). In other words, due to the

completeness of Fourier transform, one can match a boundary condition spectral-component

by spectral-component. If the boundary condition is matched for all spectral components,

than (35.2.16) is also true.

7 Or that we can perform a Fourier inversion on the above integrals.

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