Theoretische Physik 2: Elektrodynamik (Prof. A-S. Smith) Tutorial 14

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1 WiSe Prof. Dr. A-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Theoretische Physik : Elektrodynamik Prof. A-S. Smith) Tutorial 4 Problem 4. Reection of an electromagnetic wave at a conducting mirror A plane polarized electromagnetic wave of frequency ω in free space is incident normally on the at surface of a nonpermeable medium of conductivity σ 0 and a constant background susceptibility χ m > 0. a) First consider the medium. Show that for harmonically time-varying elds, Et) Re E ω e iωt etc., the polarization P χ m E and the current density j σ E in Ampère's equation can be eliminated in favor of a complex dielectric permittivity, H ω iω c εω) E ω with εω) ε m + 4πiσ ω, ε m + 4πχ m. b) The incident wave is partially reected and absorbed by the medium. Choosing the z-axis perpendicularly to the at surface, a suitable Ansatz for the electric eld is given by { e ikz + r e ikz for z < 0 empty space), E ω z) E i t e iqz e κz for z > 0 medium). Determine the wave numbers q, k as well as the decay rate κ by solving the corresponding wave equations. c) Formulate appropriate matching conditions for the electromagnetic elds at the interface z 0) and determine the reection amplitude r and the transmission amplitude t. Calculate the re- ection coecient R r and the transmission coecient T R. d) Evaluate the time averaged Poynting vector in both half spaces and interpret your result. S c Re E 4π ω H ) ω. e) Specialize your results for the case of good conductor σ ωε m, i.e., ε m can be neglected, and discuss the decay rate κ and the reection coecient R. Argue that the displacement current is small compared the current density j in this case and show that the electromagnetic elds in the medium fulll diusion equations rather than wave equations.

2 f) In the opposite limit of a poor conductor, σ ωε m, the decay rate becomes large to the wavelength of the incident wave. Determine the absorption length κ and the reection coecient R in this case. Hints: The parts ad can be solved independently of each other. The calculation of b) may be done for general complex εω); the results of c) should be expressed in terms of q, k and κ, the Poynting vector in d) in terms of R and κ. Problem 4. Reflection and transmission of electromagnetic waves A plane electromagnetic wave meets the planar interface of two dielectric media, with permeabilities and permittivities, ε and µ, ε, respectively, at an arbitrary angle of incidence α and arbitrary polarisation defined by an angle ϑ between the electric field vector and the plane of incidence. Find the reflection and transmission coefficients for the wave. Hint: any electric field vector can be written as a linear combination of two perpendicular vectors: one parallel and one perpendicular to the plane of incidence.) Problem 4.3 Paraxial beams Consider a monochromatic beam of angular frequency ω kc propagating essentially along the positive z-direction. a) Argue that the components of the electric eld allow for a representation as E x, z; t) e iωt d k π) a k ) expi k x + ik z), where k k k )/ is to be eliminated in favor of k. b) The complex amplitudes a k ) are assumed to contribute only for k k. Expanding the square root, k k k /k to leading order in k /k, show that the eld assumes the following form E x, z; t) e ikz iωt E x, z), where the envelope function E is slowly varying along z on the scale of a wavelength, z E ke. Relate the envelope to the amplitudes a k ). Show that the eld equation for the envelope is: i z E x, z) k E x, z). In particular, the eld equation is rst order in the z-direction. c) Evaluate the electric eld E x, z; t) for a Gaussian amplitude function a k ) exp ) 4 w 0k, w 0 > 0, and show that the intensity I E exhibits a Gaussian prole in the perpendicular direction x and a width that depends on z. Where is the width minimal? Due date:

3 WiSe Prof. Dr. A-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Theoretische Physik : Elektrodynamik Prof. A-S. Smith) Solutions to Tutorial 4 Solution of Problem 4. Reection of an electromagnetic wave at a conducting mirror a) Maxwell's equations for a harmonic time varying eld Et) Re E ω e iωt read Ampère: Faraday: Hω 4π c j ω + iω c Eω iω c B ω Coulomb: Dω 0 Gauÿ: Bω 0 Using the constitutive equations D E + 4π P + 4πχ m ) E and B H as well as Ohm's law, j σ E, Ampère's law takes the form inside the medium. D ω H ω iω c εω) E ω with εω) ε m + 4πiσ ω b) The wave equation is as usual ) { εω)ω c + Eω B ω } 0 and ε m + 4πχ m In each region, the solutions are plane waves polarized parallel to the mirror plane due to the normal incidence), { e ikz + re ikz for z < 0, E ω z) E i te iqz e κz for z > 0. The amplitude of the reected wave E r re i is determined by the reection amplitude r. Similarly, the transmission amplitude t determines the transmitted electric eld E t te i. Plugging in the Ansatz into the wave equation yields outside the medium ω c E i e ikz + r e ikz) + E i k e ikz k r e ikz) 0 k ω c

4 and inside the medium [ ] εω)ω c + iq κ) te i e iq κ)z 0 iq κ) εω) ω c real part: Substituting κ in the rst equation, yields a quadratic equation in q, with the solution q κ ε ω) ω c, imaginary part: κq ε ω) ω c. [ ε q ω) ω ] q c ε ω) ω c, q 4 ε ω) ω c q ε ω) ω 4 4 c 4 0, q ω c [ ε ω) + ε ω) + ε ω) ], and the wave numbers follow, q k ε ω) + ε ω) + ε ω), kε ω) κ ε ω) +. ε ω) + ε ω) c) The wave is incident normally, hence both the electric and magnetic eld are parallel to the interface. The transverse component of the electric eld is continuous at the boundary, + r)e i te i + r t. Similarly the transverse component of the magnetic induction H is continuous, thus in our case the magnetic eld B is continuous. From Faraday's law we calculate the magnetic eld, { B ω z) c iω e ikz re ikz for z < 0, ze ω z) E i tk q + iκ)e iqz e κz for z > 0. Matching both solutions at the boundary z 0, r tk q + iκ) k r) + r)q + iκ), yields the reection and transmission amplitudes, r k q + iκ) k + q + iκ) and t + r k k + q + iκ). The reection and transmission coecients R and T are obtained as R r rr k q) + κ k + q) + κ 4kq k + q) + κ q k t T.

5 d) Remember that the Poynting vector for harmonically varying elds is given by St) c 4π Et) Ht) c 4π E ω e iωt + E ωe iωt After time averaging only two terms survive In front of the mirror z < 0) Inside the mirror z > 0) H ω e iωt + H ωe iωt S c E ω H ω + E ω H ω 4π 4 Re c 4π E ω H ω S [ c Re 4π E i e ikz + re ikz) E i ). e ikz r e ikz)] c E i 8π Re r r e ikz + re ikz) S c E i R) 8π S Re c 4π E ite iqz e κz E i c E i t 8π q k e κz S c E i R)e κz 8π t q iκ ) e iqz e κz k The Poynting vector is continuous at z 0 by construction. Energy is dissipated due to the nite conductivity of the mirror. e) Good conductor means ε ω) ε m 4πσ/ω ε ω). In this limit the wavelength in the conductor equals the decay length q κ k ε ω)/ k, since alway ε m >. The reection coecient is obtained by neglecting k in comparison to q κ, R 4kq k + q) + κ k κ kδ. The magnetic eld inside a good conductor is much larger than the electric eld. There is a phase shift of π/4 between the oscillations of E and H; the ratio of the envelope is given by κ/k. The characteristic decay length κ for the case of a material characterized by a conductivity σ is known as skin penetration depth δ c/ πσω. As expected, it decreases with increasing conductivity, but more importantly it depends on frequency. For optical frequencies and silver the Skin penetration depth is much smaller than the wavelength of visible light, kδ, and lies in the range of a few nanometers! Hence on the scale of the penetration depth the incident wave appears as spatially homogeneous. The response of the metal is phase shifted and damped with respect to the excitation. 3

6 The relation R kδ is known as Hagen-Rubens law. Thus a good conductor reects almost all of the incoming intensity, i.e., serves as a mirror. Supplement: Metals usually become transparent at even higher frequencies, the reason being that the inertia of the electrons becomes important. Thus the mass of the electrons prevents currents to instantaneously respond to the eld, or in other words the conductivity becomes itself frequency-dependent. An appropriate model for this eect is encoded in the Drude conductivity σω) ne τ/m iωτ where n is the density of conduction) electrons, and τ refers to the collision time. Indeed for small frequencies the conductivity becomes real and frequency independent σω τ ) σ 0 ne τ/m. For high frequencies the conductivity decreases and becomes purely imaginary, σω τ) ine /mω, characteristic for a collisionless plasma. The displacement current can be neglected for a good conductor since ωd ω ωε ω)e ω 4πj ω ωε ω)e ω, H 4πσ c This is actually Ampère's law in its original form, Maxwell's only) contribution is irrelevant here. Taking the curl and employing Faraday's law, yields with Gauÿ's law and B H E. H) 4πσ c E H) H 4πσ c tb, t H D H, D c 4πσ ; of course, the electric eld obeys the same diusion equation. Conclusion: Dissipative systems can sustain waves albeit damped ones. A classic example is a viscous incompressible) uid excited by a periodic shear force at the surface with the so-called kinematic viscosity ν η/ρ. t v ν v f) A poor conductor is a weakly absorbing medium, q k ε m kn, κ ε ω)k q ε ω)k n πσ nc k The index of refraction of a corresponding transparent medium is given by n ε m. The wavelength in the medium equals the one of a non-absorbing one. The absorption is usually quantied in terms of the decay of intensity Iz) exp αz), Beer's law. Comparison yields, α 4πσ nc. Correspondingly, the reection coecient reduces to the one of a transparent medium R ) k q k + q ) n. + n 4

7 Solution of Problem 4. Reflection and transmission of electromagnetic waves Let us rst dene the coordinate system and symbols: k ε, x ε, µ k k β α ˆn γ z z 0 We will take that wave incomes through the medium with ε,. All quantities related to incident, refracted and reected wave are noted with pure, primed and doubleprimed letters, respectively. So, k, k and k are the wave vectors of these three waves, respectively; ˆn is a unit normal vector directed from medium ε, into medium ε, µ. The indices of refraction are n µ µ 0 ε ε 0 and n µ µ 0 ε ε 0. ) Now electric and magnetic eld vectors of the three waves are: E e ik r iωt µ ε, B k E), k E E 0e ik r iωt, B µ ε k k E ), ) 0e ik r iωt, B µ ε k k ), where the wave numbers have the magnitudes: k k k ω ε ; k k ω µ ε. 3) The spatial variation of all elds must be the same at z 0 because of boundary conditions: This relation has 3 obvious consequences:. all three wave vectors must lie in a plane;. in the notation of Fig., 4) reads k r) z0 k r) z0 k r) z0 4) k sin α k sin β k sin γ, 5) so: a) the angle of incidence equals the angle of reection since k k and b) sin α sin γ k k ω µ ε ω n. 6) ε n In terms of elds ), the boundary conditions at z 0 normal components of D and B and tangential components of E and H continuous) are: + 0) ε 0) n 0 7) k E0 + k 0 k E 0) n 0 8) E0 + 0 E 0) n 0 9) k E0 + k ) 0 k E ) ) 0 n 0. 0) µ 5

8 The electric eld vector in the plane perpendicular to the initial wave is shematically shown in Fig.. below. E In applying these boundary conditions it is convenient to consider two separate situations, one in which the incident plane wave is linearly polarized with its polarization vector perpendicular to the plane of incidence the plane dened by k and ˆn), and the other in which the polarization vector is parallel to the plane of incidence. The general case can be obtained by appropriate linear combinations of the two results. Now we consider the two cases separately.. E perpendicular to plane of incidence. x ϑ E E z B ε, x ε, µ k E E k B β γ E k B α z 0 ˆn z Since the electric elds are perpendicular to ˆn, the rst condition 7) yields nothing, and the third gives + 0 E 0 0. ) Since k k B 0 B 0 k ) ˆn k cos π α) k sin α ) and analogous for k and k ), second condidtion 8) gives which, using 5), reproduces ). In analogy with ), so the relation 0) gives k sin α + k 0 sin β k E 0 sin γ 0, 3) k ) ˆn k sin π α) k cos α, 4) ε 0 ) cos α cos γ 0, 5) µ where we used 3) for respective wave vectors. If we take get E 0 cos α + cos α ε µ cos γ 6) from ) and put it into 5), we cos α µ ε cos α + ε µ n n sin α 6

9 If we take E 0 cos α + cos α ε n n ε µ n n sin α from ), from 5) we get 0 cos α cos α + ε cos γ µ ε cos γ µ 6) cos α. 6) cos α + n µ n sin α cos α µ n n sin α. 7) cos α + n µ n sin α. E parallel to plane of incidence. ε, x ε, µ k B B k E β γ α ˆn B Ek z z 0 Since the magnetic elds are perpendicular to ˆn, the second condidtion 8) yields nothing. Since and analogous for E 0 ˆn sin π α) cos α 8) and E 0 ), the third relation 9) reads 0 ) cos α E 0 cos γ 0, 9) and since k is perpendicular to ˆn, the fourth condition 0) gives + ε 0 ) 0. 0) µ Using 7) and Snell's law 5) reproduces previous result. The relative amplitudes from 9) and 0) are: E 0 ε µ cos α + cos α cos γ cos α µ ε ε µ µ µ cos α + µ cos α cos γ ε µ cos α + cos γ ε cos α µ ε cos α + sin γ ε µ n n sin α cos α µ n n cos α + µ n n µ n n cos α + n sin α n n n sin α cos α n n sin α ). ) 7

10 These are well known Frasnel relations. If we now dene in perpendicular E case we have we note E E ) a cos γ cos α ; b µ n n, 3) E 0 + a and and in the parallel E case we have we note E E ) 0 ab + ab 4) E 0 a + b and 0 b a b + a. 5) Note that b and a are given in terms of known data α,, ε, µ, ε ). Now the general eld E is the linear combination of perpendicular and parallel elds: From here, But so Analogously, But so +, sin ϑ, cos ϑ, E 0 E 0 + E 0, E 0 E 0 sin ϑ, E 0 E 0 cos ϑ, 6) E 0, E 0 E 0 sin ϑ, 0 E 0 cos ϑ. E 0 E 0 sin ϑ sin ϑ cot ϑ E 0 E 0 + a sin ϑ sin ϑ sin ϑ cot ϑ E E 0 sin ϑ sin ϑ sin ϑ + cot ϑ + a. a + b + a ) + a cot ϑ a + b + a a + b cot ϑ, E 0 sin ϑ sin ϑ sin ϑ ab + cot ϑ + ab. b a b + a ab + ab ) ab sin ϑ + + ab sin ϑ + a) + cos ϑ a + b). 7) b a) + ab) b + a) ab) cot ϑ, The intensity of radiation is generally given by the Poynting vector: BE E µεe I S ˆk µ E B ˆk µ µ ε µ ) b a cos b + a ϑ. 8) ε µ E E 0 cos k r ωt) ε µ. 9) 8

11 Here we are talking about the intensity falling on the interface, and the interface is at an angle α or β, or γ) to the wave front. Thus the intensities are: I ε E 0 cos α I cos γ 30) µ I cos β, and reection and transmission coecients are: ) R I E ) I 0 ab ) b a sin ϑ + cos ϑ + ab b + a ) T I I ε E 0 cos γ sin ) µ ε cos α 4ab ϑ + a) + cos ϑ a + b). Solution of Problem 4.3 Paraxial beams a) The wave equation in three dimensions reads as usual ) c t E x, z; t) 0 for each component of the electric and magnetic eld. Fourier transform, 0! ) c t E x, t) d 3 k π) 3 k ω The general solution is obtained by c ) E k)e i k x ωt), with the same ω for all wave vectors in case of a monochromatic wave. Thus we have to impose k k + k ω /c, and the k -integration can be carried out, E x, z; t) e iωt d k π) a k ) expi k x + ik z), k ω /c k. There is of course a second branch propagating in the negative z-direction E x, z; t) e iωt which we ignore in the following. b) Expanding the square root k ω/c) for k k, yields d k π) b k ) expi k x ik z), k ω /c k k k k /k k k k E x, z; t) e ikz ωt) E x, z), E x, z) d k π) a k ) expi k x iz k /k). 9

12 With slowly varying I mean that the envelope does not change appreciably at the scale of a wavelength λ z E E Indeed π k ze π k k k ) E E since by assumption k k where the amplitude a k ) is important. Direct substitution yields i z E x, z) k E x, z). An alternative derivation is by substituting the Ansatz E E x, z)e ikz ωt) in the wave equation, ) ) ] c t z E e [k ikz iωt ω c E E ik ze z E 0, and requiring that the envelope is slowly varying, z E k z E. Then by k ω /c, the rst order eld equation follows. Compare with the Schrödinger equation of quantum mechanics, i t Ψ x, t) m Ψ x, t). Thus here z plays the role of time, and the Schrödinger equation is only two-dimensional. c) For a Gaussian amplitude a k ), calculating the envelope of the electric eld reduces to solving a Gaussian integral, d k E x, z) π) exp ) 4 w 0k exp i k x iz ) k k d [ k π) exp w 0 + iz ) ] k k + i x k, w introducing the complex beam parameter qz) ik 0 + iz ) and σ qz) k ik, E x, z) exp d k π) exp ) σ k + i x k d k [ π) exp σ k i x ) ] σ x σ ) x d k σ π) exp ) σ k ; Substituting u σ k, the remaining integral yields /πσ. Thus, the nal result for the electric eld is E x, z; t) e ikz ωt) x ) + iz/kw0 exp w0 + iz/k, ) where we have omitted a constant prefactor /w0 π. Introducing the Rayleigh length z R πw 0 λ kw 0, 0

13 the intensity follows as Iz) Ez) + z/z R ) exp x w 0 [ + z/z R) ] The eld mode of a laser described by Eq. ) is called the fundamental Gaussian mode because its Gaussian intensity prole perpendicularly to the propagation axis. It minimizes the product of divergence and transversal extension. The radius of the Gaussian beam is dened as the distance to the z-axis where the intensity has fallen o to /e of its maximum value, wz) w 0 + z/zr ) The minimal width w0 is attained for z 0, the beam waist. Far away from the Rayleigh range, z z R, the width grows linearly with z; the angle between this straight line and the beam axis is called divergence of the beam, ϑ. kw 0 If the coordinate z is interpreted as time, the divergence of the beam is analog to the spreading of a propagating wave packet with dispersion. Further discussion of the result: The prefactor of the electric eld may be rewritten, ). + iz/kw 0 + iζ w 0 wζ) w 0 wζ) iζ with ζ : z/z R ) + ζ [ ] cos arctanζ)) i sin arctanζ)) w 0 exp i arctanζ)). wζ) Hence, there is a longitudinal phase delay or Gouy phase of the beam, ψz) arctanz/z R ); it describes the rapid phase change of the beam when traversing the beam waist at z 0. Splitting the Gaussian factor in real and imaginary parts, x ) [ exp w0 + iz/k exp x wz) + ix z/z ] R wz), reveals that surfaces of constant phase are not planes perpendicular to the propagation axis; rather, they are curved. Neglecting the Gouy phase for z z R, the condition of constant phase, ζ kz + w0 + ζ ) x const, denes parabolas in the z-x plane with apex curvature radius Rz) kw 0 + ζ ζ z [ + z/z R ) ]. Then the nal form of the electric eld of the fundamental Gaussian mode reads E x, z; t) w [ ] { [ ] } 0 wz) exp x wz) exp ik z + x iωt iψz). Rz)

14 Fig. : Intensity gray scale), width red line) and wave fronts blue lines) of a Gaussian beam. The gures have size 0z R 0z R, the upper panel is for parameters w and z R. In the lower panel w 0 and z R ), the two blue circles close to the center reect the rapid phase drop due to the Gouy phase.