FFresnel’s equations in statics and quasistatics
Johannes Skaar Department of Technology Systems, University of Oslo, Box 70, NO-2027 Kjeller, Norway ∗ Fresnel’s equations describe reflection and transmission of electromagnetic waves at an interfacebetween two media. It turns out that these equations can be used in quasistatics and even statics,for example to straightforwardly calculate magnetic forces between a permanent magnet and a bulkmedium. This leads to a generalization of the classical image method.
I. INTRODUCTION
Can electrostatic or magnetostatic phenomena be de-scribed by Fresnel’s equations for reflection and transmis-sion at an interface? For example, if a permanent mag-net is located in the vicinity of a semi-infinite mediumwith permeability µ , can we calculate the attraction us-ing Fresnel’s equations? The answer is not obvious, asthe eigenmodes for frequency ω > ω = 0 the eigenmodes arelongitudinal. Moreover, a crucial step in the textbookderivation [1, 2] of the Fresnel equations is to relate theelectric and magnetic fields. Derivations limited to prop-agating, plane waves use the constant ratio between theelectric and magnetic field amplitudes, equal to the waveimpedance. Clearly these derivations do not apply instatics, where the electric and magnetic fields are de-coupled. Alternatively, one may use possibly complexwavevectors, and express the magnetic field H from theelectric field E using Faraday’s law, ωµ H = k × E , (1)for a plane wave with wavevector k . Here, we have thecomplications that in statics, ω = 0, and also, the electricfield is longitudinal, k × E = 0.Despite these challenges, we will prove that the Fresnelequations apply even in statics (Sec. III). For the specialcase with a point charge in the vicinity of a conductorplane or dielectric half-space, the Fresnel equations leadto the classical image method from electrostatics. Sim-ilarly, the Fresnel equations give the image method forcalculating the fields when a magnetic source is locatedin the vicinity of a magnetic medium.In quasistatics the Fresnel equations turn out to be use-ful to calculate the interaction between a time-varyingmagnetic source and a conducting medium (Sec. IV).Also here, the Fresnel equations lead to an image method,but with the reflection coefficient as a spatial low-pass fil-ter, acting on the field from the image. This gives valu-able information about the strength of the interaction asa function of spatial frequency or characteristic size ofthe source.We also consider the cases where a static electric ormagnetic field source is located in the vicinity of a mov-ing medium. Also here we obtain relatively simple image ∗ [email protected] methods for calculating the electromagnetic field, evenfor relativistic velocities. Dependent on the orientationof the source, certain interesting effects arise. For ex-ample, when describing electric reflection from magneticsources (or vice versa), the image is Hilbert transformed.With the filtered image model, we can also confirm therecently reported nonreciprocity associated with movingmedia [3].The structure of the paper is as follows. In Sec. II theFresnel equations of electrodynamics, and their conven-tional proof, are reviewed. In the last part of the section,another proof is given, which will turn out to be valid instatics as well. In Sec. III we consider a Fourier decom-position of the static fields, and prove that the Fresnelequations lead to the image methods from electrostat-ics and magnetostatics. In Sec. IV we use the Fresnelequations in quasistatics, to describe interaction betweene.g. a time-varying magnetic source and a conductingmedium. Finally, we consider a moving medium in Sec.V. II. FRESNEL’S EQUATIONS INELECTRODYNAMICS
Before going to statics, we review the Fresnel equationsin electrodynamics [1]. We consider a setup as depictedin Fig. 1. For z < (cid:15) and permeability µ , while for z > (cid:15) and µ . The media can be dispersive, but are assumed tobe linear, isotropic, homogeneous, time-shift invariant,and passive . Moreover, the media are assumed to bespatially nondispersive, in the sense that the permittiv-ity and permeability are local and describe all inducedcharges and currents. A source produces an electromag-netic field which can be expanded into monochromatic,plane waves. These waves will have their electric fieldperpendicular (TE) or parallel (TM) to the plane of in-cidence. Considering an incident TE wave, the electricfield can be written E = (cid:40) ˆ y (cid:0) E i e ik x x + ik z z + E r e ik x x − ik z z (cid:1) , z < , ˆ y E t e ik x x + ik z z , z ≥ . (2) For gain media, there is a complication to identify the correctsign of the longitudinal wavevector based on causality [4–7]. a r X i v : . [ phy s i c s . c l a ss - ph ] F e b ǫ , µ zz = − d yxǫ , µ k k source FIG. 1. The Fresnel setup. A source is located at least d to theleft of the boundary. The source produces a field which canbe decomposed into plane waves. One of them has incidentwavevector k . Here E i , E r , and E t are the incident, reflected, and trans-mitted electric field amplitudes, respectively. We haveoriented the coordinate system such that the electric fieldpoints in the ˆ y -direction. The wavevectors can then beexpressed as k = ( k x , , k z ) (incident), ( k x , , − k z ) (re-flected), and k = ( k x , , k z ) (transmitted), with thedispersion relations k x + k z = (cid:15) µ ω , (3a) k x + k z = (cid:15) µ ω . (3b)The Fresnel equations can be expressed in several dif-ferent forms. We will use a form which remains validfor lossy media and/or evanescent modes. The stan-dard way to derive the equations is as follows. From(2) and continuity of the tangential electric field, we ob-tain E i + E r = E t . Assuming ω (cid:54) = 0 and applying (1), wecan find the associated magnetic fields for the incident,reflected and transmitted waves. Continuity of the tan-gential magnetic field gives ( E i − E r ) k z /µ = E t k z /µ .Combining the two continuity equations we find r TE E ≡ E r E i = µ k z − µ k z µ k z + µ k z , (4a) t TE E ≡ E t E i = 2 µ k z µ k z + µ k z . (4b)These two coefficients express the reflected and trans-mitted electric field amplitudes relative to the incident electric field.To express the corresponding ratio between magnetic field amplitudes (but still TE polarization), we considerthe ˆ x - and ˆ z -components separately. Using Faraday’slaw (1), the incident magnetic field ( H i x , , H i z ) can be expressed H i x = − k z ωµ E i , (5a) H i z = k x ωµ E i . (5b)Similarly, we can express the reflected and transmittedmagnetic fields from the electric fields. For example, thisgives us a transmission coefficient for the ˆ x -componentof the magnetic field t TE H x = − k z /ωµ − k z /ωµ · E t E i = k z µ k z µ t TE E . (6)We end up with two sets of Fresnel equations: r TE H x = − µ k z − µ k z µ k z + µ k z , (7a) r TE H z = µ k z − µ k z µ k z + µ k z , (7b) t TE H x = 2 µ k z µ k z + µ k z , (7c) t TE H z = 2 µ k z µ k z + µ k z . (7d)As evident from the symmetry of the two Maxwellcurl equations, the analogous expressions for TM (mag-netic field perpendicular to the plane of incidence) canbe found by interchanging (cid:15) and µ : r TM H = (cid:15) k z − (cid:15) k z (cid:15) k z + (cid:15) k z , (8a) t TM H = 2 (cid:15) k z (cid:15) k z + (cid:15) k z , (8b)and r TM E x = − (cid:15) k z − (cid:15) k z (cid:15) k z + (cid:15) k z , (9a) r TM E z = (cid:15) k z − (cid:15) k z (cid:15) k z + (cid:15) k z , (9b) t TM E x = 2 (cid:15) k z (cid:15) k z + (cid:15) k z , (9c) t TM E z = 2 (cid:15) k z (cid:15) k z + (cid:15) k z . (9d)When deriving the Fresnel equations (4), we used thetransverse electromagnetic modes of the system, mean-ing that ω was assumed to be nonzero. We also made theexplicit assumption ω (cid:54) = 0 when applying (1). An alter-native way of proving the Fresnel equations, which turnsout to survive the static limit, is the following: FromGauss’ law, we have k · E = 0 (10)in the homogeneous media to the left or right of theboundary, away from the source. Eq. (10) is valid forthe incident, reflected, and transmitted wave, separately.Note that the Fourier decomposition of the source fieldat the plane z = − d leads to real k x and k y (while k z asresulting from the dispersion relation may be complex).Orienting coordinates such that k is perpendicular to ˆ y , k x E x + k z E z = 0 . (11)Consider the TM case, where E is in the plane of inci-dence, E = E x ˆ x + E z ˆ z . (12)Requiring the tangential electric field to be continuous,we get E i x + r TM E x E i x = t TM E x E i x , (13)where the reflection and transmission coefficients so farare unknown. Rather than expressing continuity of themagnetic field, we require the normal displacement fieldto be continuous: (cid:15) E i z + (cid:15) r TM E z E i z = (cid:15) t TM E z E i z . (14)From (11), the connection between the Fresnel coeffi-cients for the z and x components are r TM E z = E r z E i z = k x k z E r x − k x k z E i x = − r TM E x , (15a) t TM E z = E t z E i z = − k x k z E t x − k x k z E i x = k z k z t TM E x . (15b)Here subscripts “r” and “t” stand for reflected and trans-mitted, respectively. Canceling out the incident fieldsfrom (13) and (14), and using (15), we obtain (9). III. FRESNEL’S EQUATIONS IN STATICS
We now assume static conditions, i.e., ω = 0. First weconsider an electrostatic source, which is a static chargedistribution located at least d to the left of the boundary,see Fig. 1. A central point to enable the use of Fresnel’sequations, is a plane wave expansion of the field. Asdetailed in the Appendix, a plane wave expansion is pos-sible in statics as well; however the field is longitudinal.The electrostatic field for − d < z < E ( k x , k y ) = ( k x , k y , k z ) U e ik x x + ik y y + ik z z + ( k x , k y , − k z ) V e ik x x + ik y y − ik z z , (16)where k x and k y are real, and k z = i (cid:113) k x + k y . (17)For z >
0, the field can be written as a superposition of E ( k x , k y ) = ( k x , k y , k z ) W e ik x x + ik y y + ik z z . (18) Here the amplitudes U , V , and W are arbitrary functionsof k x and k y . Morover, k x and k y are the Fourier vari-ables corresponding to x and y , respectively (i.e., theyare the transversal “spatial frequencies” of the source).The quantity k z describes the z -dependence of eachFourier component, and is given by (17) as resulting fromMaxwell’s equations. Eq. (17) can be seen as the disper-sion relation in the static limit.To apply the proof of Fresnel’s equations from the pre-vious section, we identify the first and second term in(16) as the “incident wave” E i and “reflected wave” E r ,respectively, and (18) as the “transmitted wave” E t . Wenote that k · E i = 0, ( k x , k y , − k z ) · E r = 0, and k · E t = 0,since (17) implies k = 0. Furthermore, the “TM” condi-tion (12) is satisfied, if coordinates are rotated such that k = k x ˆ x + k z ˆ z . Apparently, in statics the label “TM”refers to electric sources, while, as we shall see later, “TE”refers to magnetic sources.We conclude that (9) applies to the case with an elec-trostatic source. Since (17) is valid in both media, theequations can be simplified to r E ≡ VU = E r x E i x = E r y E i y = − E r z E i z = (cid:15) − (cid:15) (cid:15) + (cid:15) , (19a) t E ≡ WU = E t x E i x = E t y E i y = E t z E i z = 2 (cid:15) (cid:15) + (cid:15) . (19b)Note that in statics, there is only a single polarization, asthe field is longitudinal. Thus it makes sense to write (19)for both x and y directions, not requiring the coordinatesystem to be rotated such that k y = 0.Remarkably, the reflection and transmission coeffi-cients are independent of k x and k y . Therefore, whenconsidering an actual finite-size source, the coefficientscan be moved outside the two-dimensional inverse Fouriertransform with respect to k x and k y . Thus the fieldfor z > t E (Fig.2b). The field for z < z = 0,and multiplied by the factor r E (Fig. 2a).Note that the image charge interpretation above is ageneralization of the classical result for a point chargeabove a dielectric half-space, see e.g. Jackson [8], Chapt.4.4. In addition it coincides with the image chargemethod for a point charge above a conducting plane; thenwe find r E = − For k x = k y = 0 the proof does not apply. However, this 1Dspecial case is trivial in electrostatics; the electric displacementfield (in the ˆ z -direction) ends up uniform. ǫ , µ ǫ , µ source zǫ , µ ǫ , µ zt · source r · mirroredsource(a) Observation point z < :(b) Observation point z > : FIG. 2. (a) The field for z < z = 0 and multiplied by r . (b) The field for z > t . In theelectrostatic case r = r E and t = t E . In the magnetostaticcase r = ± r H and t = t H . The upper sign is used if the sourceis characterized by a magnetization density, while the lowersign is used when the source is described by a current density. field in the area z <
0, while we have t E = 0 which meansthat the field vanishes for z > ∇ × H = 0. Moreover, ∇ · B = 0 everywhere. For theelectrostatic situation treated above, we had ∇ × E = 0and ∇ · D = 0 away from the source. Thus the analogy E → H and D → B makes it possible to express themagnetic field as a superposition of fields of the form H ( k x , k y ) = ( k x , k y , k z ) S e ik x x + ik y y + ik z z + ( k x , k y , − k z ) T e ik x x + ik y y − ik z z , (20)for z <
0, and H ( k x , k y ) = ( k x , k y , k z ) G e ik x x + ik y y + ik z z (21)for z >
0. Furthermore, we can use the Fresnel equations in the same way as above, now using the “TE” coefficients r H ≡ TS = H r x H i x = H r y H i y = − H r z H i z = µ − µ µ + µ , (22a) t H ≡ GS = H t x H i x = H t y H i y = H t z H i z = 2 µ µ + µ . (22b)Again the reflection and transmission coefficients areindependent of k x and k y . Eqs. (22) therefore lead tothe following image source interpretation: The magneticfield for z < z = 0, and multiplied by r H ,according to (22a). Thus, if the source is a permanentmagnet with magnetization M , the magnetization of theimage source is found by reflecting the source’s magne-tization about z = 0 and multiplying by r H . However,if the source is a current distribution, the current mustbe multiplied by − r H after reflection. This sign shift canbe viewed as a consequence of the current being a polarvector, while H and M are axial. The image source in-terpretation is a generalization of the well known resultfrom magnetostatics [8].As an example, consider a bar magnet with magne-tization in the ˆ z -direction, located a distance d from amagnetic medium with µ = ∞ . Assuming µ = µ (per-meability in vacuum), we have r H = −
1. An imaged barmagnet would have magnetization in the − ˆ z -direction,but after multiplication with r H the magnetization pointsin the ˆ z -direction. We therefore get a magnetic attrac-tion equal to the attraction between two identical barmagnets at 2 d distance. IV. FRESNEL’S EQUATIONS INQUASISTATICS
In quasistatics, provided k x (cid:29) ω (cid:112) | (cid:15) µ | , (23a) k x (cid:29) ω (cid:112) | (cid:15) µ | , (23b)we have k z ≈ k z ≈ ik x . Therefore, the electrostaticresult (with a charge distribution as source) and mag-netostatic result (with a current distribution/permanentmagnet as source) will also be relevant to the quasistaticsituation. An electrostatic source influences a dielec-tric/conductor medium, while a magnetostatic source in-fluences a magnetic medium. However, in quasistaticsthe two other combinations are relevant as well; a time-dependent current distribution/magnet as a source anddielectric/conductor media, or a time-dependent chargedistribution and magnetic media. We consider the formercombination here.By moving or changing the magnetic source, althoughslowly, there will be induced eddy currents in the ǫ , µ zz = − d yxǫ , µ M FIG. 3. The source is a 2D “permanent” magnet with time-dependent magnetization in the ˆ z -direction. In practice, thetime dependency can be realized by moving the magnet inthe ˆ x or ˆ z -direction, or by inducing magnetization in a fer-romagnetic material using currents in the ˆ y -direction. Themagnet is infinite and uniform along the y -axis. Medium 1 isvacuum, with (cid:15) and µ equal to the free space values (cid:15) and µ , respectively. Medium 2 is a nonmagnetic conductor withcomplex permittivity (cid:15) . medium. We now analyze this magnetic response withFresnel’s equations. Let medium 1 be vacuum ( (cid:15) = (cid:15) ),and medium 2 be a nonmagnetic medium with conduc-tivity σ described in the form of a complex, relative per-mittivity (cid:15) /(cid:15) = 1 + iσ/(cid:15) ω ≈ iσ/(cid:15) ω . The skin depthis δ = (cid:112) /ωµ σ. (24)We consider a source which is uniform (and infinite)in the y -direction, with bound or free currents in the ± ˆ y -direction. For example, the source may be a time-varying “permanent” magnet with magnetization in theˆ z -direction (see Fig. 3). Then the magnetic field is in-dependent of y , and has only ˆ x and ˆ z components. Thisleads to an electric field pointing in the ˆ y -direction (TE).The quantities we need in Fresnel’s equations are k z = (cid:112) ω /c − k x , (25a) k z = (cid:112) iωµ σ − k x = (cid:112) i /δ − k x . (25b)Since the medium is passive, the sign of the square rootis taken such that Im k z >
0. The magnetic field for z < H ( x, z ) = ˆ x (cid:90) H i x ( k x ) (cid:2) e ik z z + r H e − ik z z (cid:3) e ik x x d k x + ˆ z (cid:90) H i z ( k x ) (cid:2) e ik z z − r H e − ik z z (cid:3) e ik x x d k x , (26) x (m) | r H | f=1 Hzf=10 Hzf=100 Hz FIG. 4. The reflection coefficient (27) as a function of char-acteristic size Λ x = 2 π/k x of the source for three differentfrequencies. The vertical lines indicate ten times the value ofthe skin depth, 10 δ . The medium is Cu. for some input spectra H i x ( k x ) and H i z ( k x ), and reflec-tion coefficient r H ≡ r TE H x = k z − k z k z + k z . (27)In other words, we can still use an image source to cre-ate the field for z <
0; however its field is filtered inthe wavenumber domain using r H = r H ( k x ). This filterresponse is plotted in Fig. 4 for Cu ( σ ≈ · S/m)for three different frequencies. We observe that the finedetails (small Λ x ) are filtered away, such that the imagegets smoother than the source. As seen from the figure,the cutoff size is around 10 δ .For Λ x (cid:28) δ (cid:28) λ (where λ is the free-space wave-length), we obtain r H = − i k x δ . (28)Using the permeability of Cu, we find that for Λ x = 1 cmthe magnetic response due to eddy currents is larger thanthe magnetic response due to the permeability, as longas f > .
01 Hz. Thus, the permeability of Cu ceases todescribe the magnetic response in this setup already atthe frequency 0.01 Hz.Eq. (26) expresses the magnetic field due to the source(first term in the brackets) and due to the resulting, in-duced currents in the medium (second term in the brack-ets). Thus (26)-(27) describe the interaction between thesource and the medium. The reflection coefficient (27)quantifies the strength of the interaction (Fig. 4). Although not important here, these spectra are related by k x H i x + k z H i z = 0. Similarly to (26), the magnetic field for z > H ( x, z ) = ˆ x (cid:90) H i x ( k x ) t H e ik z z + ik x x d k x + ˆ z (cid:90) H i z ( k x ) t H k z k z e ik z z + ik x x d k x , (29)for the transmission coefficient t H ≡ t TE H x = 2 k z k z + k z . (30) V. MOVING MEDIUM
We finally consider the case where medium 2 moveswith velocity v in the x -direction, as seen from the labsystem. The coordinates of this lab system are shown inFig. 1. We assume that medium 1 is vacuum ( (cid:15) = (cid:15) and µ = µ ), and medium 2 is characterized by (cid:15) = (cid:15) r (cid:15) and µ = µ r µ , where (cid:15) r and µ r are the relative permit-tivity and permeability, respectively. The source maybe electrostatic (fixed charges) or magnetostatic (time-independent currents or permanent magnet), and is fixedin the lab system. We are interested in the reflectioncoefficients and coupling between electric and magneticfields due to the moving medium 2. For the case with amagnetic source and a moving conducting medium, thisresponse occurs as a result of eddy currents being inducedin the conducting medium. We then obtain a connectionto the analysis in Sec. IV. In certain cases we will findthat the response can be significant even for low (nonrel-ativistic) velocities.We consider each spatial Fourier component of thesource separately (with a certain k x and k y ), and writethe static electric and magnetic fields for − d < z < E = ( k x , k y , k z ) U e ik x x + ik y y + ik z z + ( k x , k y , − k z ) V e ik x x + ik y y − ik z z , (31a) H = ( k x , k y , k z ) S e ik x x + ik y y + ik z z + ( k x , k y , − k z ) T e ik x x + ik y y − ik z z , (31b)where k z = i (cid:113) k x + k y . For an electrostatic source weare given U and S = 0, and want to find r E = V /U (reflection coefficient) and c EH = T /U (coupling). For amagnetostatic source we are given S and U = 0, and willfind r H = T /S (reflection coefficient) and c HE = V /S (coupling). The strategy will be to transform (31) tothe rest system of medium 2 using the usual relativistic transformation formulas [9] E (cid:48) x = E x , (32a) E (cid:48) y = γE y − γvB z , (32b) E (cid:48) z = γE z + γvB y , (32c) B (cid:48) x = B x , (32d) B (cid:48) y = γB y + γvc E z , (32e) B (cid:48) z = γB z − γvc E y , (32f)where γ = 1 / (cid:112) − v /c . We then use the electromag-netic boundary conditions to match the fields (32) withthose in medium 2.In general we have that ( ω/c, k ) is a four-vector, suchthat [9] ω (cid:48) c = γ (cid:16) ωc − vc k x (cid:17) , (33a) k (cid:48) x = γ (cid:16) k x − vc ωc (cid:17) . (33b)This follows from the fact that phase is scalar:e i k · r − iωt = e i k (cid:48) · r (cid:48) − iω (cid:48) t (cid:48) . (34)Here t (cid:48) and r (cid:48) = ( x (cid:48) , y, z ) are the time and space coor-dinates in the rest system of medium 2, and ω (cid:48) is thefrequency in this system. Furthermore, k = ( k x , k y , k z )and k (cid:48) = ( k (cid:48) x , k y , k z ) are the wavevectors in the lab sys-tem and the rest system of medium 2, respectively. Inour case we have static conditions ( ω = 0) in the labsystem, which means that ω (cid:48) = − γvk x , (35a) k (cid:48) x = γk x , (35b)e ik x x = e ik (cid:48) x x (cid:48) − iω (cid:48) t (cid:48) . (35c)For simplicity, we will now consider two special caseswith a two-dimensional source (which is homogeneousalong the third direction). A. Case k x = 0 and k y (cid:54) = 0 Consider first the case where the source is homoge-neous in the x -direction. In this case the source pro-duces only components with k x = 0. Then ω (cid:48) = ω = 0and k (cid:48) = k = (0 , k y , k z ). The fields are therefore staticand longitudinal, even in the rest system of medium 2.For z > E (cid:48) = (0 , k y , k z ) F e ik y y + ik z z , (36a) H (cid:48) = (0 , k y , k z ) G e ik y y + ik z z , (36b)where k z = i | k y | , and F and G are arbitrary functionsof k y . Matching fields (32) and (36) according to theMaxwell boundary conditions, we obtain the reflectionand coupling coefficient r E ≡ VU = γ (cid:18) − (cid:15) r (cid:15) r + v c − µ r µ r (cid:19) , (37a) c EH ≡ TU = i sgn( k y ) γ vc η (cid:18) − (cid:15) r (cid:15) r + 1 − µ r µ r (cid:19) , (37b)for the electrostatic source ( U (cid:54) = 0, S = 0). For themagnetostatic source ( U = 0, S (cid:54) = 0), r H ≡ TS = γ (cid:18) v c − (cid:15) r (cid:15) r + 1 − µ r µ r (cid:19) , (38a) c HE ≡ VS = − i sgn( k y ) γ vc η (cid:18) − (cid:15) r (cid:15) r + 1 − µ r µ r (cid:19) . (38b)We have defined η = (cid:112) µ /(cid:15) as the vacuum waveimpedance. In (37)-(38) the relative permittivity andpermeability must be evaluated at zero frequency.The reflections (37a) and (38a) can be interpretedas being produced by electrostatic and magnetic imagesources, respectively. This interpretation corresponds ex-actly to that in Sec. III except that the reflection coef-ficients r E and r H now are given by (37a) and (38a).The differences compared to the situation with v = 0are of order v /c and therefore negligible unless v is arelativistic velocity.However, for an electrostatic source, we also get a “re-flected” magnetic field, and for a magnetostatic source,we also get a “reflected” electric field. For the case witha magnetostatic source, the resulting electric field needsnot be negligible in the non-relativistic case. Assumingthe parenthesis in (38b) is ∼
1, the electric field becomes vη/c times the source magnetic field. This correspondsto an electric field strength E ∼ vB , where B is the fluxdensity from the source. For a neodymium magnet with B ∼ v = 1 m/s, we get E ∼ − i sgn( k y )function. This filtering operation amounts to convolvingby 1 /πy , which results in a Hilbert transform of the im-age with respect to y . B. Case k x (cid:54) = 0 and k y = 0 Consider next the case where the source is homoge-neous in the y -direction, i.e., k y = 0. For a nonzero k x ,we observe from (35) that ω (cid:48) (cid:54) = 0, despite the source be-ing static in the lab system. Thus the electromagneticfield for z > E (cid:48) = (cid:20) (0 , , F + ( k (cid:48) z , , − k (cid:48) x ) G(cid:15) r (cid:15) ω (cid:48) (cid:21) e i k (cid:48) · r (cid:48) − iω (cid:48) t (cid:48) , (39a) H (cid:48) = (cid:20) (0 , , G + ( − k (cid:48) z , , k (cid:48) x ) Fµ r µ ω (cid:48) (cid:21) e i k (cid:48) · r (cid:48) − iω (cid:48) t (cid:48) . (39b)Here F and G are arbitrary functions of k x . From thedispersion relation we have k (cid:48) z = (cid:112) (cid:15) r µ r ω (cid:48) /c − k (cid:48) x = γi | k x | (cid:112) − (cid:15) r µ r v /c , (40)where we have used (35). The sign of the square rootis taken such that Im k (cid:48) z >
0. Note that for dispersivemedia, we must evaluate the relative permittivity andpermeability at the frequency ω (cid:48) = − γvk x . When thisfrequency is negative, the symmetry relations (cid:15) r ( − ω (cid:48) ) = (cid:15) ∗ r ( ω (cid:48) ) and µ r ( − ω (cid:48) ) = µ ∗ r ( ω (cid:48) ) are used [1].By matching the tangential electric and magnetic fieldsof (32) and (39) at the boundary z = 0, and using (35),we obtain the reflection coefficients r E ≡ VU = γ (cid:112) − (cid:15) r µ r v /c − (cid:15) r γ (cid:112) − (cid:15) r µ r v /c + (cid:15) r , (41a) r H ≡ TS = γ (cid:112) − (cid:15) r µ r v /c − µ r γ (cid:112) − (cid:15) r µ r v /c + µ r , (41b)where the relative permittivity (cid:15) r and permeability µ r must be evaluated at the frequency ω (cid:48) = − γvk x . Thecoupling coefficients between electric and magnetic fieldsin the lab system vanish in this case.For a nondispersive medium 2, the reflection coeffi-cients (41) are independent of k x . Thus the field for z < v/c (cid:28)
1. Then the reflection coefficientsreduce to the v = 0 results in Sec. III, unless thepermittivity and/or permeability are extremely large for ω (cid:48) = − γvk x . However, this is exactly what happens fora conductor, for which (cid:15) r ( ω (cid:48) ) ≈ iσ/(cid:15) ω (cid:48) and µ r = 1. Inthis case, (41b) can be written r H = (cid:112) iµ σv/k x − (cid:112) iµ σv/k x + 1 . (42)Thus we obtain a considerable reflection when the dimen-sionless, magnetic Reynold’s number µ σv/k x (cid:38)
1. ForCu and v = 1 m / s, this corresponds to spatial variationsof the source which is 2 π/k x (cid:38) . x (m) | r H | v=0.1 m/sv=1 m/sv=10 m/s -0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 x (m) | F - r H | ( m - ) v=0.1 m/sv=1 m/sv=10 m/s FIG. 5. The reflection coefficient r H = r H ( k x ) (42) as afunction of characteristic size of the source Λ x = 2 π/k x (up-per plot), and the inverse Fourier transform of r H ( k x ) (lowerplot). be found by considering the source as time-dependent(or quasistatic), as a result of a velocity − v in the ˆ x -direction. Then we can use the result in Sec. IV to-gether with ω = − vk x to deduce (42) in the nonrela-tivistic limit. Despite giving the correct answer, sucha method is somewhat incomplete unless the relativis-tic transformation formulas (32) are taken properly intoaccount.In the special case when the source is large and variesslowly with x , such that µ σv/k x (cid:29) r H = 1. Then the imagemethod from Sec. III applies straightforwardly. In gen-eral, however, we must filter the image of the source with r H = r H ( k x ). This filter function is plotted with respectto characteristic size Λ x = 2 π/k x in Fig. 5. Also theinverse Fourier transformed filter function (point sourceimage) is plotted with respect to x . We observe the in-teresting result that the point source image is one-sided(i.e., vanishes for x > k x , since the branch point is in thelower half-plane. By closing the integration path witha large semi-circle in the upper half-plane, the inverseFourier transform can therefore be proved to vanish forpositive x . We finally note that since the image is notan even function, the system does not obey Lorentz reci-procity [3]. VI. CONCLUSION
By formulating the proof of Fresnel’s equations withonly electric fields, or with only magnetic fields, it turnsout that the relations are valid in statics. Since staticfields are longitudinal, “TM” applies to the case with anelectrostatic source, while “TE” applies to the case with amagnetostatic source. The results are independent of thespatial frequency of the source, which gives a connectionto the classical image charge or image current methods.In quasistatics the Fresnel equations may give resultsdependent on spatial frequency; in this case the imagemethod must be combined with a filter. For the casewhere a static source is located in the vicinity of a mov-ing medium, we also obtain image methods to describethe reflection and coupling between electric and magneticfields. Dependent on the orientation of the source relativeto the velocity of the medium, certain interesting effectsarise, including Hilbert transform of image, and Lorentznonreciprocity.
Appendix A: Plane wave expansion of static fields
We consider a general 3D charge distribution as thesource, located at least a distance d from the interface.In electrostatics the magnetic field is zero or constant,and we express E = −∇ φ , where ∇ φ = 0 away fromthe source and the interface. For each z , φ viewed asa function of x and y can be expressed as a 2D Fourierintegral: φ ( x, y, z ) = (cid:90) (cid:90) Φ( k x , k y , z )e ik x x + ik y y d k x d k y , (A1)where Φ( k x , k y , z ) is some function. Using ∇ φ = 0 weobtain (cid:18) − k x − k y + d d z (cid:19) Φ( k x , k y , z ) = 0 , (A2)with general solutionΦ( k x , k y , z ) = iU ( k x , k y )e ik z z + iV ( k x , k y )e − ik z z . (A3)Here U and V are arbitrary functions of k x and k y . Wehave factored out an i for later convenience, and defined k z = i (cid:113) k x + k y . (A4)Backsubstitution into (A1) we have φ = i (cid:90) (cid:90) (cid:2) U ( k x , k y )e ik z z + V ( k x , k y )e − ik z z (cid:3) · e ik x x + ik y y d k x d k y . (A5)The associated electric field E = −∇ φ becomes E = (cid:90) (cid:90) ( k x , k y , k z ) U ( k x , k y )e ik x x + ik y y + ik z z d k x d k y + (cid:90) (cid:90) ( k x , k y , − k z ) V ( k x , k y )e ik x x + ik y y − ik z z d k x d k y . (A6)The result (A6) is in particular valid in the region − d 0. For the unbounded region z > z . The result is φ = i (cid:90) (cid:90) W ( k x , k y )e ik x x + ik y y + ik z z d k x d k y (A7) and E = (cid:90) (cid:90) ( k x , k y , k z ) W ( k x , k y )e ik x x + ik y y + ik z z d k x d k y (A8)for some function W ( k x , k y ). Apparently there are notransverse fields in electrostatics.If the source is a static current distribution or perma-nent magnet, we have ∇ × H = 0 and ∇ · B = 0 awayfrom the source. Thus we can express H = −∇ ψ , where ∇ ψ = 0, completely analogous to the electrostatic situ-ation. [1] L. D. Landau, E. M. Lifshitz, and L. P. Pitaevskii, Electro-dynamics of continuous media (Pergamon Press, Oxford,1984).[2] B. E. Saleh and M. C. Teich, Fundamentals of photonics,2nd ed. (John Wiley & Sons, Inc., 2007).[3] J. Prat-Camps, P. Maurer, G. Kirchmair, and O. Romero-Isart, Phys. Rev. Lett. , 213903 (2018).[4] J. Skaar, Phys. Rev. E , 026605 (2006). [5] B. Nistad and J. Skaar, Phys. Rev. E , 036603 (2008).[6] A. A. Kolokolov, Phys. Usp. , 931 (1999).[7] H. O. H˚agenvik, M. E. Malema, and J. Skaar, Phys. Rev.A , 043826 (2015).[8] J. Jackson, Classical Electrodynamics (Wiley, 1999).[9] L. D. Landau and E. M. Lifshitz,