Russian Physics Journal, Vol. 51, No. 3, 2008

ELECTROMAGNETIC WAVES IN ISOTROPIC AND GYROTROPIC MEDIA Yu. G. Pavlenko and A. V. Seliverstov

UDC 535.012

Maxwell's equations are considered for waves in transparent inhomogeneous anisotropic and gyrotropic media with time dispersion. A uniaxial inhomogeneous crystal is taken for an anisotropic medium and an isotropic heterogeneous medium in a constant electric field for a gyrotropic medium. The solutions for flat-layered and cylindrical media are presented in terms of Hertz potentials.

INTRODUCTION In nonlinear optics, several particular solutions of Maxwell's equations have been obtained for an anisotropic uniaxial crystal with cubic nonlinearity [1, 2]. However, in the monographs on electrodynamics known to the authors, solutions for a linear anisotropic medium are not given. For this case, the solution of Maxwell's equations, like in the theory of waveguides, can be presented in the form of a superposition of transverse electric (TE) and transverse magnetic (TM) waves which are described by Hertz potentials. In optics, to them there correspond ordinary and extraordinary waves. In this paper, waves in anisotropic and gyrotropic flat-layered media are considered. The solutions are presented in the form of an expansion in the eigenfunctions of Maxwell's equations.

MAXWELL'S EQUATIONS FOR A MEDIUM WITH TIME DISPERSION For the case of an inhomogeneous nonferromagnetic medium with time dispersion, constitutive equations can be presented in the form B (t , x ) = μ 0 H (t , x ), +∞

Dm (t , x ) = ε0 ∫ dt ′ Gmn (t − t ′, x ) En (t ′, x ),

(1)

−∞

where Gmn (t − t ′, x ) is Green’s function which takes into account the linear response of the medium [1]. Then Eq. (1) and Maxwell's equations rot E = −∂B / ∂t ,

(2)

rot H = ∂D / ∂t ,

(3)

constitute an integro-differential system. We seek a solution to the system in the form of a wave packet:

M. V. Lomonosov Moscow State University, Moscow, Russia, tel.: +7 (495) 939-41-12. Translated from Izvestiya Vysshikh Uchebnykh Zavedenii, Fizika, No. 3, pp. 82–86, March, 2008. Original article submitted June 25, 2007. 314

1064-8887/08/5103-0314 ©2008 Springer Science+Business Media, Inc.

E (t , x ) = Re e (t , x ) exp(−iωt ), H (t , x ) = Re h(t , x ) exp(−iωt )

(4)

whose complex amplitudes vary adiabatically slowly: | ∂e / ∂t |<< | ωe |, | ∂h / ∂t |<< | ωh | . Using a new variable τ = t − t ′, in (1), we obtain ∞

Dm (t , x ) = ε0 Re exp(−iωt ) ∫ d τ Gmn (τ, x )en (t − τ, x ) exp(iωt ). −∞

Since the basic contribution to the integral is made by the values of ωτ << 1 , we have ∞

Dm (t , x ) = ε0 Re exp(−iωt ) ∫ d τ Gmn (τ, x )exp(iωτ)[en (t , x ) − τ∂en (t , x ) / ∂t + …] −∞

= ε0 Re [ε mn (ω, x )en (t , x ) + iε mn '(ω, x )∂en (t , x ) / ∂t + …]exp( −iωt ),

where ε mn '(ω, x ) ≡ ∂ε mn (ω, x ) / ∂ω , ε mn (ω, x ) is the permittivity tensor, ∞

ε m n(ω, x ) = ∫ Gmn (τ, x ) exp(iωτ). −∞

In the adiabatic approximation we have ∂Dm (t , x ) / ∂t = ε0 Re [−iωε mn (ω, x )en (t , x ) + (ω∂ε mn / ∂ω)∂en (t , x ) / ∂t + …]exp(−iωt ).

The amplitudes satisfy the equations rot e = iωμ 0 h − μ0 ∂h / ∂t ,

(5)

(rot h) m = −iωε0 ε mn en + ε0 (ωε mn )′∂en (t , x ) / ∂t.

(6)

In view of Eqs. (5) and (6), we obtain the mean electromagnetic energy density 1 u = [ε0 (ωε mn )′en* em + μ0 hn* hn ]. 4

(7)

THE SOLUTION OF FIELD EQUATIONS FOR A FLAT-LAYERED ANISOTROPIC MEDIUM Let us first find a solution to Eqs. (5), (6) for the case of a monochromatic wave, assuming that ∂e / ∂t = 0 , ∂h / ∂t = 0 . Let the z-axis be directed parallel to the optical axis of the crystal and the medium be inhomogeneous: ε mn (ω, x ) = ε mn (ω, z ) , ε11 = ε 22 = ε , ε33 = ε3 . The electric and magnetic field components e1 , e2 , h1 , and h2 are determined by the longitudinal components e3 and h3 , which play the role of Hertz potentials [3, 4]. The

electromagnetic field in the crystal is a superposition of two independent modes. 1. The transverse electric, TE, or the ordinary wave, is determined by the Hertz potential GO : e ( x, y, z ) = iμ0 ω rot GO , GO = gO ( z ) exp(ik1 x + ik2 y ), gO = {0, 0, g O }.

(8)

From Eq. (5) we find h = rot rot GO . Hence, the amplitudes of the fields are determined as 315

e = eO exp(ik1 x + ik2 y ), h = hO exp(ik1 x + ik2 y ), eO1 = −μ 0 ωk2 gO , eO2 = μ 0 ωk1 gO , eO3 = 0, hO1 = ik1∂ 3 gO , hO2 = ik2 ∂ 3 g O , hO3 =

(9)

k⊥2 g O ,

where k⊥2 = k12 + k22 . Substituting e and h in (6), we obtain a second-order equation: d 2 gO dz 2

⎡ ⎛ ω ⎞2 ⎤ + ⎢ ε ⎜ ⎟ − k⊥2 ⎥ g O = 0. ⎣⎢ ⎝ c ⎠ ⎦⎥

(10)

2. The transverse magnetic, TМ, or the extraordinary wave, is determined by the Hertz potential GE : h( x, y, z ) = −iε0 ωrot GE , GE = g E ( z ) exp(ik1 x + ik2 y ), g E ( z ) = (0, 0, g E ).

(11)

For this case, we have hE1 = ε0 ωk2 g E , hE2 = −ε0 ωk1 g E , hE3 = 0.

Substituting h in (6), we obtain the amplitudes of the electric field as eE1 = i (k1 / ε)∂ 3 g E , eE2 = i (k2 / ε)∂ 3 g E , eE3 = (k⊥2 / ε3 ) g E .

(12)

2 d 1 dg E ⎡⎛ ω ⎞ k⊥2 ⎤ + ⎢⎜ ⎟ − ⎥ g E = 0. ε3 ⎦⎥ dz ε dz ⎣⎢⎝ c ⎠

(13)

Now Eq. (7) yields

A FLAT MONOCHROMATIC WAVE IN A HOMOGENEOUS UNIAXIAL CRYSTAL

For the case of a homogeneous crystal, the solutions of Eqs. (10), (13) have the form gO = a exp(ikO3 z ), g E = b exp(ikE3 z ),

where kO3 and kE3 are the roots of the equations ε (ω / c ) 2 = k 2 ,

(14)

k⊥2 k32 ⎛ ω ⎞ 2 + =⎜ ⎟ , ε3 ε ⎝c⎠

(15)

which describe the surfaces of the wave vectors kO = (k1 , k2 , kO3 ) and kE = (k1 , k2 , kE3 ). Putting k = (ω / c)n, we obtain from (9) and (12) the eigenvectors of the electric field strength for the cases of an ordinary and an extraordinary wave [5]:

316

⎛ −n1nE3 / ε ⎞ ⎛ −n2 ⎞ ⎟ 1 ⎜ 1 ⎜ ⎟ n1 ⎟ , eE = eO = ⎜ −n2 nE3 / ε ⎟ . n⊥ ⎜⎜ n ⊥ ⎜ ⎟ ⎟ 2 ⎝ 0 ⎠ ⎝ n⊥ / ε3 ⎠

(16)

The eigenvectors satisfy the equation n2 em − nm (ne ) − ε mn en = 0

(17)

with the eigenvalues nO3 and nE3 . The eigenvectors of the magnetic field strength are given by hλ = k × eλ / ωμ 0 , λ = O, E. Note that the Fourier components of the electric induction vectors d O = ε0 (εeO1 , εeO2 , 0) and d E = ε0 (εeE1 , εeE2 , ε3eE3 ) satisfy the common property d O d E = 0 . In view of the relation μ0 h 2 = ε0 ε mn em en , we obtain from (7) the mean energy density u=

1 ε0 (∂ω2 ε mn / ∂ω)em en . 4ω

(18)

The general solution of Eqs. (2) and (3) has the form ⎡ C ⎤ C E (t , x ) = Re ⎢ 1 eO exp(−iωt + ikO x ) + 2 eE exp(−iωt + ikE x ) ⎥ . Ní ⎣ Nî ⎦

(19)

The normalizing coefficients N O = ε0 (∂ω2 ε / ∂ω) / 4ω, N E = ε0 [(∂ω2 ε / ∂ω)(nE3 / ε) 2 + (∂ω2 ε3 / ∂ω)(n⊥ / ε3 ) 2 ]4ω2

are chosen so that the electromagnetic energy density be u = ω | C1 |2 +ω | C2 |2 . If we neglect the dispersion, we have N O = ε0 ε / 2ω and N E = ε0 / 2ω .

THE RAY VELOCITIES OF WAVES

The energy transferred by an electromagnetic wave is determined by the Poynting vector S = E × H . Substituting (19) in this formula, in view of (6), we obtain the mean value s = 〈 S 〉 = sO + sE , sO =

1 1 ε0 cnO eO2 C12 , sE = ε0 c[nE eE2 − eE (nE eE )]C22 . 2 NO 2NE

(20)

The magnitude of the vector s determines the wave intensity. Substituting (16) in (20), we find sO =

1 1 ε0 c(n1 , n2 , nO3 )C12 , sE = ε0 c(n1 / ε3 , n2 / ε3 , nE3 / ε)C22 . 2 NO 2 NE

From (16) and (20) we find sλ eλ = 0 and sλ hλ = 0, λ = O, E. The mean electromagnetic energy density is u = uO + uE . The ray velocities of an ordinary and an extraordinary wave are given by

317

sO n , vO = c , ε uO

(21)

sE ⎛n n n ⎞ , vE = c ⎜ 1 , 2 , E3 ⎟ . uE ⎝ ε 3 ε3 ε ⎠

(22)

vO =

vE =

Note that for an extraordinary wave the equation of the surface of the wave vectors (15) can be written in a form not related to a certain coordinate system [1]. Let l be a unit vector directed along the symmetry axis of a crystal. Then the equation of the surface (15) can be presented in the form f = (nl ) 2 / ε + [n 2 − (nl ) 2 ] / ε3 − 1 = 0 . Since the ray vector s is perpendicular to the surface, we have s ~ ∂f / ∂n s = n / ε3 + (nl )l (1/ ε − 1/ ε3 ) . Propagation of a quasiplane wave If λ | d ε / dz |<< 1 and λ | d ε3 / dz |<< 1 , one can seek a solution to Eqs. (12), (13) in a ray approximation: gO = a exp[iψ O ( z )] , g E = b exp[iψ E ( z )] . Then from (12) and (13) we find ψ O ( z ) = ∫dzkO3 ( z ), kO = [εω2 / c 2 − k⊥2 ]1/2 , ψ E ( z ) = ∫dzkE3 ( z ), kE = [εω2 / c 2 − εk⊥2 / ε3 ]1/2 .

Let us consider a wave beam whose transverse dimensions are much less than the wavelength. We seek a solution to Eqs. (5), (6) in the form of a quasiplane wave [1, 2, 6]: e (t , x ) = b(t , x ) exp[ik1 x + ik2 y + iψ λ ( z )], b(t , x ) = Aλ (t , x )e ( λ ) ( z ).

Equations (5), (6) are equivalent to the equation −(rot rot e ) m + (ω / c) 2 ε mn en + iε0 μ0 (ω2 ε mn )′∂en / ∂t = 0.

After substitution, in a ray approximation, we obtain −(rot[k ⋅ b]) m − [k ⋅ rot b]m + ε0 μ 0 (ω2 ε mn )′∂bn / ∂t = 0.

Compose a scalar product of both parts with the vector bm and take into account the relations d E eO = d O eE = 0 , ε0 (ω2 ε mn )′bm ∂bn / ∂t = 4ωuλ Aλ ∂Aλ / ∂t , and [b ⋅ [k ⋅ b]] = 2ωμ 0 sλ Aλ2 , div sλ = 0 .

As a result, we obtain the equation sλ ∇Aλ2 + uλ ∂Aλ2 / ∂t = 0

that describes the process of energy transfer along the direction of the ray velocity.

SOLUTION OF A FIELD EQUATION FOR A MEDIUM CONFINED BY A CYLINDRICAL SURFACE

Let a homogeneous uniaxial crystal is confined by a cylindrical surface. In this case, to solve the boundary problem on the propagation of a wave incident on the surface of a uniaxial crystal, it is necessary to use cylindrical coordinates ρ , ϕ , z . Let the z-axis be directed parallel to the optical axis of the crystal. Assume that ε mn (ω, x ) = ε mn (ω, ρ) , ε11 = ε 22 = ε , and ε33 = ε3 in the region 0 ≤ ρ ≤ R . Then the particular solution of Eqs. (5), (6)

can be presented in the form 318

e ( x ) = [eρ (ρ), eϕ (ρ), ez (ρ)]exp(ik z z + imϕ), h( x ) = [hρ (ρ), hϕ (ρ), hz (ρ)]exp(ik z z + imϕ),

where m = 0, ±1, ±2,… . The components of the field strengths depend on the parameters m and k3 . In the cylindrical coordinates, the Fourier components of the electrical induction vector are written as d = (εeρ , εeϕ , ε3ez ) . Substituting d ( x ) and h( x ) in (5), (6), we obtain the system of equations iωμ0 hρ =

−iωε0 εeρ =

∂e im 1 ∂ρeϕ im ez − ik z eϕ (23a), iωμ0 hϕ = ik z eρ − z (23b), iωμ0 hz = − eρ , ∂ρ ρ ρ ∂ρ ρ

∂h im 1 ∂ρhϕ im hz − ik z hϕ (24a), −iωε0 εeϕ = ik z hρ − z (24b), −iωε0 ε3ez = − hρ . ∂ρ ρ ρ ∂ρ ρ

(23c);

(24c)

From Eqs. (23a), (24b) we obtain the components νeϕ = −iωμ0

∂hz m ∂h m − k z ez , νhρ = ik z z + ωε0 ε ez , ∂ρ ρ ∂ρ ρ

(25)

where ν = ε(ω / c) 2 − k z2 . From Eqs. (23b), (24a) we obtain the components νeρ = ik z

∂ez m ∂e m − ωμ 0 hz , νhϕ = iωε0 ε z − k z hz . ∂ρ ρ ∂ρ ρ

(26)

Substitution of eρ and eϕ in (23c) yields the second-order equation ⎡ ⎛ ω ⎞2 ⎤ 1 ∂ ∂hz m 2 ρ − 2 hz + ⎢ ε ⎜ ⎟ − k z2 ⎥ hz = 0. ρ ∂ρ ∂ρ ρ ⎣⎢ ⎝ c ⎠ ⎦⎥

After substitution of hρ and hϕ in (24c), we obtain the equation ⎤ ε ⎡ ⎛ ω ⎞2 ∂e m 2 1 ∂ ερ z − 2 ez + 3 ⎢ε ⎜ ⎟ − k z2 ⎥ ez = 0. ερ ∂ρ ∂ρ ρ ε ⎢⎣ ⎝ c ⎠ ⎥⎦

The electromagnetic field is a superposition of two independent modes. The transverse electric, TE ( ez = 0 ), or the ordinary wave, is determined by the Hertz potential hz . The transverse magnetic, TМ ( hz = 0 ), or the extraordinary wave, is determined by the Hertz potential ez .

ELECTROMAGNETIC WAVES IN AN INHOMOGENEOUS GYROTROPIC MEDIUM

A gyrotropic medium can be created in an isotropic body placed in a constant magnetic field. In this case, the permittivity tensor is hermitian. Then the Fourier components of the electrical induction vector can be presented in the form d ( z ) = ε 0 ( εe + i e × g ) ,

319

where g is the gyration vector and ε( z ) is the permittivity [1]. Let the z-axis be directed in line with the vector g . Assume that a plane wave is incident on a plane-parallel layer of the heterogeneous medium in the direction of the z-axis. From (5), (6) we obtain the equations iωμ 0 h1 = −∂ 3e2 , −∂ 3 h2 = −iωε0 (εe1 + ige2 ),

iωμ 0 h2 = ∂ 3e1 ,

(27)

∂ 3 h1 = −iωε0 (εe2 − ige1 ),

(28)

which yield the system of equations d 2 e1 / dz 2 + (ω / c) 2 (εe1 + ige2 ) = 0, d 2 e2 / dz 2 + (ω / c) 2 (εe2 − ige1 ) = 0.

(29)

To these equations there correspond two eigenvectors: ⎛ e1(1) ⎞ 1 ⎛ 1 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ q1 , ⎝ e2(1) ⎠ 2 ⎝ −i ⎠

⎛ e1(2) ⎞ 1 ⎛1⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ q2 , ⎝ e2(2) ⎠ 2 ⎝ i ⎠

where the functions q1 and q2 are solutions of the equations d 2 q1,2 / dz 2 + k32± ( z )q1,2 = 0, k32± ( z ) = (ω / c) 2 (ε ± g ).

The general solution e1 = (q1 + q2 ) / 2 , e2 = (q1 − q2 ) / 2i corresponds to the propagation of circular polarization waves. The authors are grateful to V. C. Zhukovsky for the discussion of this work.

REFERENCES

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L. D. Landau and E. M. Lifshits, Theoretical Physics. Electrodynamics of Continuous Media [in Russian], Fizmatlit, Moscow (2001). M. B. Vinogradova, O. V. Rudenko, and A. P. Suhorukov, Theory of Waves [in Russian], Nauka, Moscow (1979). L. D. Goldshtein and N. V. Zernov, Electromagnetic Waves [in Russian], Sov. Radio, Moscow (1956). L. A. Vainshtein, Electromagnetic Waves [in Russian], Sov. Radio, Moscow (1957). Yu. G. Pavlenko, Hamiltonian Methods in Electrodynamics and Quantum Mechanics [in Russian], Moscow State University Publ., Moscow (1985). Yu. A. Kravtsov and Yu. I. Orlov, Geometrical Optics of Heterogeneous Media [in Russian], Nauka, Moscow (1980).