THE UNIVERSITY OF MICHIGAN 7322-2-T RADIATION DUE TO AN OSCILLATING DIPOLE OVER A LOSSLESS SEMI-INFINITE MOVING DIELECTRIC MEDIUM by Vittal P. Pyati Feb'ruary 1966 Report No. 7322-2-T on Grant NGR -2 3 -005-107 Prepared for NATIONAL AERONAUTICS AND SPACE ADMINISTRATION NASA-LANGLEY RESEARCH CENTER LANGLEY STATION HAMPTON, VIRGINIA

THE UNIVERSITY OF MICHIGAN 7322-2-T FOREWORD The material contained in this report was also used as a Dissertation for the partial fulfillment of the requirements for the degree of Doctor of Philosophy in the Department of Electrical Engineering, The University of Michigan.

THE UNIVERSITY OF MICHIGAN 7322-2-T A CKNOWLEDGEMENT The author is indebted to Professor Chen-To Tai for suggesting the problem and invaluable guidance and encouragement throughout the investigation. Thanks are also due to the members of the Thesis committee for their helpful suggestions. A portion of this investigation was conducted while the author was with the Radio Astronomy Observatory of The University of Michigan (NsG-572) and he is grateful to the Director, Professor Fred T. Haddock, for his support. The author wishes to thank Professor Ralph E. Hiatt, Head of the Radiation Laboratory for his encouragement. In addition, he wishes to express his sincere appreciation to Mrs. Claire F. White and Miss Madelyn L. Hudklns for typing the manuscript. iii

THE UNIVERSITY OF MICHIGAN 7322-2-T TABLE OF CONTENTS Page FOREWORD ii A CKNOWL.EDGEMENT iii LIST OF ILLUSTRATIONS v CHAPTER I INTRODUCTION AND STATEMENT OF THE PROBLEM 1 II ELECTRODYNAMICS OF MOVING MEDIA 3 2. 1 The Lorentz Transformation 3 2.2 Maxwell-.Minkowski Equations 4 2. 3 The Method of Potentials for Moving Media 6 m REFLECTION AND REFRACTION OF A PLANE ELECTROMAGNETIC WAVE AT THE BOUNDARY OF A MOVING DIELECTRIC MEDIUM 9 3. 1 Geometry of the Problem 9 3.2 Plane Waves in Moving Media 9 3. 3 The Modified Snell's Law 14 3. 4 Electric Field Perpendicular to the Plane of Incidence 23 3. 5 Electric Field Parallel to the Plane of Incidence 31 3. 6 Perpendicular Incidence 36 3. 7 Summary 39 IV OSCILLATING DIPOLE OVER A MOVING DIELECTRIC MEDIUM 40 4. 1 Introduction 40 4.2 Vertical Dipole 40 4.2. 1 Fourier Integral Method 40 4. 2. 2 Method of Weyl 49 4. 2. 3 Approximation of the Integrals; Asymptotic Forms 55 4. 2. 4 Numerical Results 77 4. 3 Horizontal Dipole in the Direction of the Velocity 84 4. 3. 1 Fourier Integral Method 84 4. 3. 2 Approximation of the Integrals; Asymptotic Forms 87 4. 3. 3 Numerical Results 93 V CONCLUSIONS 101 RE FERENCES 103 APPENDIX A: POINT CHARGE IN MOVING MEDIA; CERENKOV RADLATION 105 108 ABSTRA CT iv

THE UNIVERSITY OF MICHIGAN 7322-2-T LIST OF ILLUSTRATIONS Fig. No. Page 1. THE COORDINATE SYSTEMS 3 2. PLANE WAVE INCIDENCE ON A MOVING MEDIUM 10 3. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=2,.i=0 17 4. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n 2, i=450 18 5. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=2, i = 90~ 19 6. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=0.5, pi= 20 7. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=0.5, i=450 21 8. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=0.5,' i= 900 22 9. THE (x,y, z) and (xi,yl, z) COORDINATE SYSTEMS 24 10. BREWSTER'S ANGLE VS i FOR n = 2, i= 90~ 37 11. DIPOLE OVER A MOVING MEDIUM 41 12. PHYSICAL INTERPRETATION OF THE DIPOLE PROBLEM 51 13. PATH OF INTEGRATION IN THE a1l PLANE 52 14. DIPOLE SOURCE AND IMAGE 57 15. ILLUSTRATION OF SADDLE POINT METHOD 72 16. IN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n= 2, h=0 (values in dielectric have been scaled down by a factor of 2) 78 17. E JIN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE FORn=2, h=0.25X 79

THE UNIVERSITY OF MICHIGAN 7322-2-T LIST OF ILLUSTRATIONS (CONT'D) Fig. No. Page 18. jE IN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE FR n=2, h=0.5X 80 19. d IN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n = 0. 5, h=0 81 20. EI IN THE AIR IN THE XZ PLANE FOR A VERTICAL DIPOLE FR n=0. 5, h=0.25X 82 21. I |RIN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE FR n = 0.5, h=0.25X 83 22. IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n =2, h0 94 23. |E IN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FR n =2, h=0.25X 95 24. IE IN THE AIR IN THE YZ PLANE FOR A 4HORIZONTAL DIPOLE F6Rn=2, h=0.5X 96 25. JEIN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n =0. 5, 26. E0I IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FOR n 0.5, h 0 98 27. |,IIN THE AIR IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FbR n= 0. 5, h =0. 25X 99 28. kE IN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE 15Rk n=0. 5, h=0.25X 100 vi

THE UNIVERSITY OF MICHIGAN 7322-2-T CHAPTER I INTRODUCTION AND STATEMENT OF THE PROBLEM In the last decade, interest in the study of the electrodynamics of moving media has increased considerably. Based on Minkowski's theory, Nag and Sayiedl have presented an alternate derivation of Frank and Tamm's2 formula for Cerenkov radiation. Boundary value problems involving stationary charges and one or more moving dielectric media have been considered by Sayied3, Zelby4 and others. While Frank5 has analyzed the problem of an oscillating dipole in uniform motion, the complementary problem, in which the medium is in uniform motion and the source and the observer at rest, has been independently solved by Tai6 and Lee and Papas7. The present work is concerned with the following boundary value problem. 1. Radiation due to an oscillating (Hertzian) dipole over a lossless semi-infinite moving dielectric. medium. Here lossless means zero conductivity and the dipole source is assumed to be located in free space or vacuum which is stationary with respect to an observer in whose frame of reference all the fields will be determined. The problem may be regarded as an extension of Sommerfeld's8 dipole problem to moving media. The object of this study is, first to develop techniques of formulation of boundary value problems in moving media, and then to apply these techniques to the above problem to ascertain the extent to which the radiation patterns are modified due to the motion of the dielectric medium. It may be recalled that Weyl9 developed a method by which Sommerfeld's solution for a dipole over a flat earth could be interpreted as a btiiile of plane waves reflected and refracted by the earth at various angles of incidence. In order to give such a physical interpretation to the present problem, it is necessary that we extend Fresnl'ss results to moving media, namely: 2. Reflection and refraction of a plane electromagnetic wave at the boundary of a moving dielectric medium.

THE UNIVERSITY OF MICHIGAN 7322-2-T The outline of the present work will be as follows. In Chapter II, the electrodynamics of moving media largely following Sommerfeld'0 is presented. The method of potentials due to Tai6 is also introduced. The problem of reflection and refraction is treated in Chapter III and the dipole problem in Chapter IV. Asymptotic solutions using the method of saddle points are obtained. Finally, a discussion of the results along with some suggestions for future research appears in Chapter V.

THE UNIVERSITY OF MICHIGAN 7322-2-T CHAPTER II ELECTRODYNAMICS OF MOVING MEDIA 2. 1 The Lorentz Transformation Consider two coordinate systems as shown in Fig. 1, in which the y and y' axes coincide and the system S' is mrving with a uniform velocity yinthe y-direction with respect to S. For the case when the two origins O and 0' coincide at the z ZI S S' O o y O' yI X 1: TE FIG. 1: THE COORDINATE SYSTEMS instant t = t' = 0, the equations of transformation of the space-time coordinates from one system into another are given by Y' = Y (y-vt) y = (y' +vt') x' =x, z'= z x =x', z =z' (2.1) t'='y(t- f X) t= Y(t'+p x') The above is known as the Lorentz or the Lorentz-Minkowski transformation. The various constants appearing above are given by c = (oeo)-0) = velocity of light in free space or vacuum eo = permittivity of free space = (36rx109) 1 farads/m. po = permeability of free space = 4irx10-7 henries/m -./G

THE UNIVERSITY OF MICHIGAN 7322-2-T 2. 2 Maxwell-Minkowski Equations Consider an isotropic homogeneous lossless (zero conductivity) medium moving uniformly with a velocity v in some direction. Without loss of generality, we can choose this to be the y-direction and orient the axes as in Fig. 1. Now, according to the theory of relativity, Maxwell equations must have the same form in all inertial frames of reference, that is, they must be covariant under the Lorentz transformation (2. 1). Therefore in the unprimed or laboratory system, we have aB VxE= - (2.2) aD Vx H= J + a (2. 3) - at V- D= p (2. 4) V- B= O (2.5) V. J+ a=O (2.6) and by attaching primes, we get Maxwell equations in the primed system, for instance, (2. 2) becomes aB' V'x E'=- at' at, It may be noted that the divergence equations follow from the curl equations and the equation of continuity; hence do not yield any new information. To formulate a problem completely the constitutive relations must be known. These can be derived in the following manner. In the primed system where an observer is at rest with the medium, we have D' = eE' (2. 7) B' = tu H' (2. 8) where e and j are the permittivity and the permeability of the medium in the primed system. Now, according to Minkowski's theory, which is based upon the special theory of relativity, the fields in the two systems transform according to the following scheme.

THE UNIVERSITY OF MICHIGAN 7322-2-T E' = y. (E+ v x B) (2.9) B'= ~ (_B - x E) (2.10) H'= y. (H -vxD) (2.11) D' == (D+ - VxH) (2.12) where O O0 = 0 1 0 0 v=vy. Substituting (2. 7) and (2. 8) into the above, we obtain D+- VxH = (_E + xB) (2.13) e B - e laio x s = firH -iv x M i (2.14) These two relations were first derived by Minkowski in 1908. Solving for B and D in terms of E and H, we obtain the desired constitutive relations D= ca E + QxH (2.15) B = ae.H-Qx E (2.16) where (n2 -1)f A -= (1-n~) c Y n = A E = refractive index in the primed system a = 1-3 = to [a 0 O c =0 1, k2 = W2 A e 0 O a The system (2. 2) - (2. 5) can now be solved once the sources are specified by the method of potentials discussed in the next section. To illustrate the usefulness

THE UNIVERSITY OF MICHIGAN 7322-2-T of this type of formulation a modified version of Nag and Sayied's method for Cerenkov radiation is presented in Appendix A. 2. 3 The Method of Potentials for Moving Media The method developed here is due to Tai6. The Maxwell equations -ict assuming e variation are Vx E =iwB (2.17) Vx H =J -iwD (2.18) Making use of the constitutive (2. 15) and (2. 16) we obtain (V+iw2)xE =iwpL a- H (2.19) (V+icQ)xH ==-iw E a- E +J (2.20) Applying the transformation -I = e-iwY [-H1 (2.21) we get VxE1 = ioa' H1 (2.22) 1 —1 VxH1 = -i _a. El+eiwYJ (2.23) Now introduce the vector potential A1 such that,a a' Hl = VX Ca ~ A 1 -1 =-1 = -1 Substituting in (2. 22), we obtain El= iw&a- 1 Al- (2.25) where 01 is the scalar potential. Substituting for E1 and H1 in (2. 23), we obtain Vx 1i.LVx(al.1 Aje _WQYJ+k2 Al+iwLEc a. V. (2. 26)

THE UNIVERSITY OF MICHIGAN 7322-2-T The vector operator on the left hand side can be expanded in rectangular coordinates thus vx 1. [Vx(a l. Al] = [(VA1+aV)(V 1 (2. 27) where Aa A a Aa a ax aay az Now define, the gauge condition ik2a2 VA ika (2.28) -1 wi1 so that (2. 26) becomes (V X)Aj+ak2A = -,ae" yJ. (2.29) To integrate (2. 29) in an infinite domain we introduce the scalar Green's function G which satisfies the equation 2 + a2 +ak G=-6(rlr) (2.30) ay az j where r refers to source point. Two distinct cases depdnding upon the sign of a exist. Case 1: a>0 orv<c/n The solution can be obtained by dimensional scaling aGi eika LR1 GR = (2. 31) where R1, the modified distance is given by R1= [(x-x)2+a(y-yo)2+(z-zo) 2]1 Case2: a < 0 or v >c/n In this case, we have a two-dimensional Klein-Gordon equation and the

THE UNIVERSITY OF MICHIGAN 7322-2-T solution is given by J -acos(a2 kR2) if (yy) > G2 27rR2La -Yo (2.32) = 0 if a1!(y-yo) < p where a= lal R2= [a:-yO) 2p2 p = [(x-xo)2+(z-zo) ] If the medium is moving in the negative y-direction, replace (y-yo) by (yO-y). Once the Green's function is known, the solution for A1 in (2. 29) is given by A(r= a fei wQy O G ( )d (2. 33) and the electric and magnetic fields are given by _1k~a -iwQYa2k2 -1 w-a E=e-i Y[a2k2 -1'Ai+ W.A] (2. 34) w 1 H=e | aY[V( A1J (2. 35),~ H=e ~ ~ c~- ~ One final word is necessary. The vector and scalar potentials A1 and 01 introduced here do not form a four-vector in the Minkowski space. This is in contrast with the potentials employed by Lee and Papas7 which transform like four-vectors. The latter will not be used here.

THE UNIVERSITY OF MICHIGAN 7322-2-T CHAPTER III REFLECTION AND REFRACTION OF A PLANE ELECTROMAGNETIC WAVE AT THE BOUNDARY OF A MOVING DIELECTRIC MEDIUM 3. 1 Geometry of the Problem As shown in Fig. 2, the region z < 0 is filled by a medium, with a permeability p, and a permittivity e, moving uniformly in the y-direction with a velocity v. The region z > 0 is free space (,o, co) and is stationary. A plane electromagnetic wave traveling in free space in an arbitrary direction is incident upon the interface; as a result there will be a reflected wave and a transmitted wave. Let the orientation of the three waves be as in Fig. 2, the azimuthal angles being measured from the x-axis. 3. 2 Plane Waves in Moving Media In order to represent the transmitted field, we need plane wave solutions in the moving medium. The Maxwell equations in the absence of sources and -iwt for e variations are given by VxE= iwB (3.1) VxH = -iwD (3.2) V B=0,'7 D=0 ~ (3.3) Making use of the constitutive relations (2. 15) and (2. 16) we obtain (V+iw Q)xE = iuu ac' H (3.4) (V7+iwQ)x H = -iwe a E (3.5) The plane wave solutions which we are seeking can be represented thus E=E e (3.6) _O HH eK (3.7) - O where the first factor denotes complex amplitude, K the propagation constant and

THE UNIVERSITY OF MICHIGAN 7322-2-T Incident z Reflected 10 I I- (A, E) o, Transmitted FIG. 2: PLANE WAVE INCIDENCE ON A MOVING MEDIUM 10

THE UNIVERSITY OF MICHIGAN 7322-2-T A is given by D = x sin et cos t +y sin t sin t -z cos t. (3.8) Here, the subscript t refers to the transmitted wave, et is the angle between the negative z-axis and the direction of propagation as shown in Fig. 2 and Ot is the azimuthal angle. For our purposes, it is sufficient to treat the complex amplitudes as being independent of the coordinates x, y and z. In component form (3. 6) can be written as E={(Eo E )e i, (3.9).ox oy oz To solve (3. 4) and (3. 5), we first let He -i EIy Et (3.10) e It and find that E' and H' satisfy Vx E' = iw c a H' (3.11) VxH' =-iwea c- E' (3.12) We make one more substitution H' = - 1' - = )2la H" (3.13) to get Vx -1 E_" = iW IH" (3.14) Vx -1. H" = -iwE E" (3.15) It may be noted that the above equations imply V H" =V E" =0. (3.16) +The primes used here should not be confused with the similar notation used in the moving reference frame. The latter will not be used in this chapter. 11

THE UNIVERSITY OF MICHIGAN 7322-2-T Eliminating one variable at a time between (3. 14) and (3. 15), we find that E" and H" satisfy the modified vector wave equation Vx [-1 (Vkx.t = k ) (3.17) Expanding the left hand side according to (2. 27) and making use of (3. 16), we find E" and H" satisfy (Va * V)E" +ak2 E_" = 0 (3.18) (Va. V) H" + ak2 H" = 0 (3.19) These equations separate into three scalar equations in rectangular coordinates of the form 2 1 2 2 [i-2+3 -2-~+ +ak (3.20) ax2 a ay az2 where b stands for one of the components. Plane wave solutions of the above are given by i(klx +k2y -k3z) e (3.21) provided k1, k2, k3 satisfy the characteristic equation k2 2k + 2 + k2 -ak2 =. (3.22) 1 a 3 Now, we are ready to construct plane wave solutions in the moving medium starting with either the electric or the magnetic field. Thus setting i(klx+k2Y-k3z) E"= (E" E",E" )e (3.23) - ox' oy oz we get E=e-it cr-1* E" 1El' 1 E i [k1x-(k2-w)y-k3] =( a E" E" - E" )e (3. 24) a ox' oy' a oz 12

THE UNIVERSITY OF MICHIGAN 7322-2-T The amplitudes in (3. 23) must satisfy the relation (3. 16), so that kl E'x+k2El" -k E" = 0 (3. 25) 1 ox oy 3 oz Comparing (3. 24) with (3. 9), we get from the phase functions K sin Otcos it = k1 (3. 26a) K sintsint = (k2-w2Q) (3. 26b) K COS et = k3 (3. 26c) substituting for kl, k2 k3 in (3. 22) we get the following dispersion relation for K 2i.2 2o+ 1(+2 2 2 2=a 2 K sin 2fcos2t - (KsinOtsinft+wQ)2 + cos Ot-ak =o (3.27) Solving for K /ko, the refractive index of the moving medium, we get K = 1 [-2,y 2(n2_1)cos2 a]-1 2, (1-n2)cos a +[1+ 2(n2-1)(1-f2cos2 (3.28) +[i+ 22)( (3.28) where a is the angle between the direction of propagation and the velocity of the medium, (cos a = sin St sinft).. This expression checks with that of Papas1l (see page 231, Eq. 61 ). We also note that the amplitudes in (3. 9) will have to satisfy the following relation sintOSEox+ - (K sinotsint+ow Q)Ey -IccosOE = O (3. 29) Ksinetcos6tE+-ox a - oy t oz The magnetic field H can be obtained from (3. 4). Making use of (3. 27) and (3. 29), one can show that the H field thus obtained satisfies (3. 5). From this it follows that the divergence relations in (3. 3) are also satisfied. We conclude this section by summarizing the method of construction of plane wave solutions in moving media. 13

THE UNIVERSITY OF MICHIGAN 7322-2-T Summary: To construct plane wave solutions in the moving medium, we set setE = (Eox, Ey Eoz )ei (3. 9) subjecting the amplitudes to the condition (3. 29), K satisfying the dispersion relation (3. 27) and { given by (3. 8). The magnetic field is given by H = ~ x1 *V+iwQ)xE - Uw/ = - Alternatively, one can start with the magnetic vector by setting H = (Hox, H H )e K (3.30) - ox oy oz and obtain the electric field from E= - ~ (V+ iwQ)xH everything else remaining the same except replacing E by H in (3. 29). 3. 3 The Modified Snells Law -iot The incident and the reflected waves satisfy for e variation Vx E = iB_ V B = 0 (3.31) VxH=-iwD V D= 0 (3.32) and the constitutive relations being B_ = oH D_ = o E (3.33) The plane wave solutions are well known and the phase functions are of the form iko iko e, e where 2 2 k =W c, E = x sin Oicos ~i+y sin Oisin 0i-z cos. (3. 34) I7 = xsin rcos d+y sin Orco sr+y sin rin rr+Z cos r35) 14

THE UNIVERSITY OF MICHIGAN 7322-2-T Here subscripts i and r stand for the incident and reflected waves respectively and the angles measured as in Fig. 2. Now, in order to match the boundary conditions at the interface z= 0, it is clear that the phase functions of the incident, reflected and the transmitted waves must be identical when z = 0. This is possible only if a) i - r = -t b) 0i = 0r (3. 36) c) kosin0i= K sint )t Physically this means that the three waves are coplanar (the plane =i will be called the plane of incidence), and the angle of reflection is equal to the angle of incidence. These two results am no different from the case in which the lower medium is not moving. The Snell's law, which relates the angle of refraction to the angle of incidence is, however, modified according to c) in (3. 36). Making use of these relations in (3. 27), substituting for a and Q2, we get after some simplification sint=sii 12- (-sinOisini) (3. 37) which is the modified Snell's law. We note that except when Oi=0, 7r, the formula is affected by a change of sign of 1. The angle of total reflection can be obtained by setting sinOt=l and solving for Oi. For the case of Oi = 0, we get sin = ( ). (3.38) Excluding the non-moving case (1=0), from the inequality n2-i2 > 2 i < n X if n2 > 1 1_~2 <, 15

THE UNIVERSITY OF MICHIGAN 7322-2-T we conclude that the phenomena of total reflection occurs only if n2 < 1 and the value of this angle is less than the corresponding value in the non-moving case. The next sections are devoted to the boundary value problem of reflection and refraction. To facilitate analysis, different cases based on the polarization of the incident wave are treated separately. The modifications introduced in the Snell's law due to the movement of the dielectric medium are depicted in Figs. 3-8 for n=2 and n= 0.5. 16

THE UNIVERSITY OF MICHIGAN 7322-2-T 30 /-0.30 25 0 I ~~~~~~~~~,: 0.75,200I0 5 1 5 30 45 60 75 90 8i (degrees) FIG 3: ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n= 2, i= 0 17

8t (degrees) c~n O ~0 0. 0 0 (0' 0 -z. t'r ~~~~0~~~~~~~~~~~ -0.. 0 0 U0~~~~~~~~~~~ to 0 01 FG4ANEORFAINVAG z (~~~~~0 "1 0 9~~~~~.~ 0 FIG. 4: ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n = 2, ~i 450 (D~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

THE UNIVERSITY OF MICHIGAN 7322-2-T -0.75 55 50/0 45 400 40~/ =0.30 0 S -30 I /X 2519 20 5 LO 15 30 45 60 75 90 19

8t (degrees) -I NE CP 01 ) 00 (DH o 0 0 0. 0 ) 0 0 01 Z m CL 6+6:AGOFRFATNVANLOINIECFRn.5 o" a~~~~~~~~~~~ iftj~~~~~~~" -C NP I 01 O~~~~~~~~~~~~~~~~~ FIG. 6: ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR. n=0. 5, ~.= 0.

THE UNIVERSITY OF MICHIGAN 7322-2-T o LO 90 80 =-.31 P=o0 /3=0.3 80 - 0 0) Z 20 0 5 10 15 20 25 30 35 0 I0&, (degrees) 21

THE UNIVERSITY OF MICHIGAN 7322-2-T 90 80 - / =-.3 =0o /.3 70 #60.~ 50403020 I0 5 10 15 20 25 30 35 40 45 &i (degrees) FIG. 8: ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=0.5, pi= 900 22

THE UNIVERSITY OF MICHIGAN 7322-2-T 3. 4 Electric Field Perpendicular to the Plane of Incidence In the coordinate system (xl, Y1, z ) resulting from a rotation of x, y axes as shown in Fig. 9, the incident electric field is given by ik (YlsinO.-zcosOi) E.=E e i (3. 39) — 1 o0 The two coordinate systems are connected by the following relations X sin~~i cosoi =L: [sin.i cosf7] [xi (3. 40) Lyj tLcosoi sin0i LY It is easier to work with the (x, y, z) system instead of (x1, Y1, Z) in spite of the fact that the incident field has a simpler form in the latter system; the reason being the analysis of the transmitted wave would then be greatly complicated. In the (x, y, z) system the incident field becomes ik e Ei=(I1, I2, 0)e (3. 41) where I1=Eosin6i, I2=-E cosoi and 5 given by (3. 34). The magnetic field is given by H.= Vx E. 0 so that in component form (suppressing phase factors) k 0 H. =- I cos0. ix W0o 2 1 H. = - I1 coso (3.42) liy L0/1 1 1o HiZ =-o Sin0i (I2cos i-Ilsini ) 23

THE UNIVERSITY OF MICHIGAN 7322-2-T Ylz-plane is the plane of incidence 9. y i Y1 X1 x FIG. 9: THE (x, y, z) AND (xl, Yi' z) COORDINATE SYSTEMS. 24

THE UNIVERSITY OF MICHIGAN 7322-2-T The reflected field can be represented by E = (R1, R2, R3)eikol (3. 43) where rj is given by (3. 35) and the amplitudes satisfy the condition V E = 0 which yields R sinO cos +RsinO sinr in+R3 cose = 0 (3. 44) The magnetic field is given by -i H =-VxE r (E1O -r so that in component form we have (suppressing phase factors) H w (R3in sinrSi-R2cos ) rx W/12 3 rr2 r ko ry w=- (R3 sinOr cos r- RlCOSr ) (3. 45) ko Hrz w sinOr(R2cos r-Rl sinr). The transmitted wave can be represented by E5; = (T1, T2, T )e (3. 46) -t T' T2' T3)e where ~ is given in (3. 8) and the amplitudes satisfy K sineOtcostT,+I (K Sinotsinotf wQ)T2-K cos tT3 0 (3. 47) The magnetic field is given by -i -1 H = - a-1 [(V+iw Q2)xEj 25

THE UNIVERSITY OF MICHIGAN 7322-2-T so that in component form we have (suppressing the phase factors) H = 1 T3(K sinOtsinyt+ w Q)+T2 K COS o tx aL 3 t = a TsinO coso+T cosO (3.48) ty-tl Wm 3t t t1 H=- I TKsinO cOSO -T (K sinO sint+w ~)I tz aw T2 t t 12 tt The boundary conditions at the interface z = 0 are a) continuity of tangential components of E and H (3. 49) b) continuity of normal components of D and B (3. 49) These yi eld six equations and along with (3. 44) and (3. 47) we have eight equations between the six unknowns. This need not disturb us, in fact, b) above is automatically satisfied as will be shown shortly. The continuity of the tangential components yields I1+R T1 (3.50) I2+R2=T2 (3.51) k 0 I2cosOi+R3sinOrsinrR2cos Or] = 1- T3(K sinOtSinot+w Q)+T2 K cosOt] (3. 52) k LI cosO. -R3sinO rcos~r+R~~osOr] -K tTrsinltcos~t —TicosO1 ~ E 3O r r 1 wt tp 3 t t 1 4 (3.53) Before solving this system, we will show that (3. 49b) is automatically satisfied. Apart from the phase factor, at z = 0+, 26

THE UNIVERSITY OF MICHIGAN 7322-2-T k ko BZ= w sinOi(I2cosoi-Ilsin0i)+ w sinOr(R2cosr-_Rlsin r) Substituting for R1 and R2 from (3. 50) and (3. 51) and noting that Oi=Or and ei=0r, this becomes ko w sinOi(T2cos.i-TlSinoi ) and at z = 0B = aHt +2Etx = -- sin 0t(T2 cos 0i-T1 sin i) (3. 55) which is the same as (3. 54) because of Snell's law (3. 36). Similarly, at z = 0+ D = e R (3. 56) z o3 and at z = 0D = ea E -QHtx. (3. 57) z tz tx Instead of directly substituting for Etz, we express in terms of Ht thus tz' -t Et= jl La-1. (V+ i w Q)xH = a [(CK sinstsinot+oQ )Ht KsintcostHt so that the right hand side of (3. 57) becomes K sinOt(sintHtx-cosotHty) Because of the continuity of tangential H_, we can substitute the left-hand sides of (3. 52) and (3. 53) for Htx and Hty respectively and after some simplification, we get K sinOe k t o (3.58) 2 sine. 3 L0 I 1 o which is the same as (3. 56) because of Snell's law. Thus we conclude that the 27

THE UNIVERSITY OF MICHIGAN 7322-2-T continuity of tangential E (H) and Snell's law ensure the continuity of normal B(D). The solution for the six unknowns related by Equations (3. 44), (3. 47) and (3. 50) - (3. 53) can be conveniently carried out as follows: R1=T -I1 (3.59) R2=T2-I2 (3.60) R3= -tanei(T 1cos i+T2sini) (3. 61) T1' T2, T3 satisfying T1IKsinOtcosT2 -(K sinOtsinfi+wQ )-T 3KcosOt=0 (3.62) + (KsinO sino+oWQ)= -I2cos Oi (3. 63) k k T1(- taneisinOicos+. i+ costoi T MO anO sino l0 1 scos t k K 2ko +T -3tanOisinOisinicosoj+T3 sinOtcosi =l I coso + 2 (Ks =-t I03os O i(3.63) (3. 64) The solution of the above equations for the general case is quite tedious though not impossible. We will consider only two special cases; when Ai= 0, 900 and also set Au =,L. Case 1: ALp -i 0 ci=~ sinOt{2j lsinO. (3. 65a) 28

THE UNIVERSITY OF MICHIGAN 7322-2-T K ~ = [ k 1_ 2 (3. 65b) The fields are given by i[k (xsin i-zcos 0i)- t] E. =(0, -Eo 0) e (3. 66) -i 0' i [k (xsinoi+zcosoi)-di E =(R1, R2, R3)e (3. 67) iK (xsinO -zcosOt) -ct] Et=(T1, T2, T3)e (3. 68) The amplitudes of the transmitted and reflected waves are given by 2Eo] (n 2_1) T sin0. cos. (3. 69) 1 M (12) 1 1 2E T- M b cosO.(b+cosOtsecO.) (3. 70) 2 V-1 t i 2E 1 (n2_ T = (n 1) cos 0i(sec 0 +b cos 0 ) (3. 71) RI=T1 (3.72) R 2=T2+Eo (3.73) R3= -T 1 tan 0i (3. 74) where Mb =) (s 12 t 1 t 1 (1-_2)2 (3.75) 29

THE UNIVERSITY OF MICHIGAN 7322-2-T Making use of (3. 65) to eliminate Ot, we get the following convenient expressions for R1 and R2 2E R1= N 1(n2-1)sin0icos20i (3. 72a) f2 2- 2 2 n2-f32 i2b] R2=E E 1- 0Cos O~i [-/)cosOi+(1) ( _-sin20i) (3. 73a) where N=(n2 -2)cos2 i-(1-n2 32)sin2+n2(1_2) +(1-32)(j+n2)cosi_ sin2o (3. 75a) A significant feature of the above results is that the reflected and transmitted waves have components not originally present in thee incident wave. Because of this, the phenomena of the reflection and refraction by a moving medium cannot be completely described by merely specifying the reflection and transmission coefficients R and T defined by R=, TF E. An exception, however, occurs when.i= 900 which is discussed next. Case 2: o =, i = 900 + sinot = [+ 2(1-3 sin Oi)2 sin0i (3. 76) and K can be determined from the relation K sint = k sin Oi +This case was considered by Tai before. Actually, this Chapter is an extension of Tai's work. 30

THE UNIVERSITY OF MICHIGAN 7322-2-T The fields are given by i [ko(y sin 0.-z cos 0i)-]E. =E e (3.77) — 1 O A=E isin(0t-0i) i [k(y sin 0.+z cos 0e)-wct] E=XE e (3.78) o sin(0t+0i) 2 sinOtcos 0.sin0t i[K(ysin t-z cosot) -cA: _Et o sin(0+0 e The reflection coefficient R and and transmission coefficient T are given by sin(0t-0i) duced via (3. 80) 2 sin0 cos 0. t 1 T = ( - (3.781) s in(tn+0i) ti1 The proceduresults are is sdenticalar to the stapionaryevious case modified by velocity terms intro- start with the duced via (3field. Eliminating 09, we get for the reflection coefficient sin(t-ik) sin os cosin 0 2 ==1- =1-(3.78a) in(n the sin(ste+0) this becomes where F= Eos20i~ n2-1 (1-~Isin0i)2] (3.82) 3. 5 Electric Field Parallel to the Plane of Incidence magnetic field. Referring to Fig. 9, we have ik (y sin O.-z cos 0. ) H.=H e (3.83) -1 O In the (x, y, z) system this becomes 31

THE UNIVERSITY OF MICHIGAN 7322-2-T iko H.=(I1, I2, 0)e (3.84) where ~ is given by (3. 34) and Ii=H sin i, I2=-Hocos i. The electric field is given by E.=- Vx H. i WE -i1 0 so that in component form (suppressing phase factors) k E. = cos 0 k E. iy I cos. (3.85) iy WE1 i1 k iz WE sin0i(I2cos Oi-Ilsin Oi) 0 The reflected field can be represented by H =(R1, R2, R3)eiko (3.86) where r1 is given by (3. 35) and the amplitudes satisfy the condition V H =0 which yields Rlsin 0rcos Or+R2sin 0rsin Or+R3cos =0. (3.44) The electric field is given by i E =- VxH - WE - 0 so that in component form (suppressing phase factors) ko rx w Eo(R3sin ersin Or-R2cos Or) ko Er= W (R3sin ercs0 r-Rlcos r)0 (3.87) ko E=- — sinOr(R2cos6r.-R Sinr ) 32

THE UNIVERSITY OF MICHIGAN 7322-2-T The transmitted field can be represented by -t (1' 2' 388) where { is given by (3. 8) and the amplitudes satisfy K sin tcosotTl+ I (K sinOtsint+w Q)T2- KcosOtT3=0 (3. 47) The electric field is given by = (V + i )xHt] so that in component form (sUppressing phase factors) E _ T3( K sinOtSint+fwQ)+T2 K cs~ t] tx awo E E = -- T3sinOtcost+Tlcos O (3.89) Etz=- awe [T2 KsinOtcos t-Tl(K sinOtsinot+w )1 The boundary conditions (3. 49) lead exactly to the same set of equations as before except that we replace po by E and p by E. Thus R,=T -I1 (3.90) R2=T2 -I2 (3.91) R3 =-tan 0i(T1 cos 0i+T2 sin 0i) (3. 92) T1 T2' T3 satisfying TlC sintcOS.+- (sin sin+W)-TKcosO= (3.93) 1 t 1 i a t 3iisn- ) -T3 sisii =( T1 eo tan 0sin 0sin0icosi+T2(K — cos O+ — cos 0i+ - tanOisinOisin2oi) T3 2ko +a (Ksinotsini+CW 2)=, I2cosOi (3. 94) 33

THE UNIVERSITY OF MICHIGAN 7322-2-T ko 2 ko K T(eotan.0. sinOicosi + coseOi+ cosOt) 1 o o k 2k +T2 o tan.isisinsinficOSoi+T3 e sitC~S~i E IlCsoi 0o (3. 95) As in the case of perpendicular polarization,we will consider only two cases of incidence; when Vi = 0, 900~. Casel: P =o i= 0 The expressions for K and 0t are given by (3. 65) and the fields are given by i Cko(x sinOi-z cos i)-w ] H. =(0 -H, O)e (3. 96) i [ (x sinoi+z cosoi) —t H =(R1 R2, R)e (3. 97) i (x sin0t-z cos0t)-wl H =(T T T )e (3. 98) H=(T 1' T2' T3)e -t 1'1 2' 3(3.98) The amplitudes of the transmitted and reflected waves are given by 2Ho (2_1) T1= o ( 2) sin0.cosO. (3. 99) 2H b T2 = M cos 0i(b-n2cos0tsecoi) (3. 100) 2Ho ( 2 3= M T3= ~ M (i32) cosO0(n2secOitbcos0t) (3.101) R,=T1 (3.102) R=2T2+Ho (3.103) R 3=-T tan 0. i (3. 104) 3 1 1 where b=( )2 1-f 34

THE UNIVERSITY OF MICHIGAN 7322-2-T b 2 M=b( -cos 0+cos 0.)(b +n 2cosO sec.i) an2 t 1 t 1 p2(n21)2 b + 2 ( - (secOi+ 2 cos 0t ) (3. 105) Eliminating Ot, we get the following convenient expressions for the reflected field. 2H R N 3(n2-l)sin0.cos2. (3. 102a) Ri N 1 R2=H i- ~ coso0i[(n -2)cos0i+n2(1-32) ( n32 -sin20i)'] (3. 103a) where N is given by (3. 75a). Besides the remarks already made in connection with perpendicular polari - zation following (3. 75a), an additional feature is that for no angle of incidence does the reflected wave vanish. Hence the Brewster angle phenomenon has no parallel in the present case. An exception, however, occurs when Oi=900, which is discussed next. Case 2: p = po' oi=90. The expression for 0t is given by (3. 76) and K can be found from the modified Snell's law. The fields are given by i ks iyinozcoso.)-t] H.= =H e (3. 106) i o Fk 2cos0 K i [(ysino.+zcos ot)-w] H =xH KE (3. 107) kcosO i —k ocos t A 2kcos0i =H ] e (3. 108) kcosO + - k cos Eliminating K and 8t, we get for the reflection and transmission ooefficients

THE UNIVERSITY OF MICHIGAN 7322-2-T 2F R=1- 2 (3. 109) n cosO.+ F 1 2n2cos ei T= 2n005 (3.110) n2cos0i+ F where F is given by (3. 82) F = [cos 0i+ (1- sinO i)2 (3. 82) The angle of incidence 00 for which the reflected wave vanishes is found by setting (3. 109) equal to zero. This is the modified Brewster's angle and is given by sin0 = 3[+n(1 2) ( +n2)1/2 (3.111) n2+1-n2J2 In Fig. 10, 00 is plotted as a function of [ for n = 2. 3.6. Perpendicular Incidence Finally, we have to discuss the case of perpendicular incidence (i =0) which is rather trivial compared to the previous ones. Though the distinction between the two kinds of polarization disappears, we have to still consider separately the two cases in which the incident electric or magnetic field is in the x-direction. First, we note that 0i=0r =t=0 Case 1: E. in the x-direction. E.=XE e-i(koz+t) (3. 112) -1 e (3.113 36

THE UNIVERSITY OF MICHIGAN 7322-2-T 90 80 70 60 o 4020' n0 o o > z 30 < I! I l l.2.4.6.8 1.0 37

THE UNIVERSITY OF MICHIGAN 7322-2-T Er = XEoRle 1 (3.114) =y e i(k z-wt) (3. 115) -r lOp O 1 E-t E T -i(z+ot) (3. 116) k ET1 r 11 h Hqt= - ~ —-- I(n2 -2) /2(1-2) l/2z3(n2-1z ) (3. 117) u/p (1-j2) where T- 2kot (3.118) 1 kot - K / R k 0 K J(3. 119) 1 kou+Kco 0 0 Case 2: H. in the x-direction. -1 -i(koz+tt) H. =xH e-ioz+) (3.120) _ O E i=y He( (3.121) 1 WE0 o H H = R ei(lZ-wt) (3.122) -r o 1 H =NH T e-i(KZ+wt) (3. 124) -t o l kH Ek=o o1 1/ 2-i(KZ+Wt) E(1- 0 0 1( ny(n2 ) 2 ( - 1)+(n21) ei t) (3. 125) -~t ] c~l - where 2k 2k E k0E+ K(3.126) k c-K E 0 0 (3. 127) 1 k E+IK3 O O 38

THE UNIVERSITY OF MICHIGAN 7322-2-T The situations are similar to the non-moving case. Furthermore, in Case 1 k =rI | kpl- K~ H(3. 128) and in Case 2: IEr K -k e K k -k 2P (3.129) El ei - +k 0 Kk,+k2 (3. 129) These two ratios are not equal contrary to the non-moving case; the deviation is, however, of the order 12 since Kk k[1+. (n2-1)] (3.130) This completes our study of the problem of reflection and refraction at a moving boundary. We close this chapter by presenting a summary. 3.7 Summary The problem of reflection and refraction of a plane electromagnetic wave traveling in free space and striking a moving dielectric boundary has been solved in this chapter. The solution proceeded in a logical fashion by first determining plane wave solutions in an unbounded moving medium. The resulting wave number, hence the refractive index, of the moving medium was found to be a function of the velocity, the direction of propagation, and n, the refractive index in a rest frame, as given by (3. 28). The rest of the analysis was carried out in a straightforward manner. Snell's law was modified according to (3. 37). Except when the azimuthal angle of the incident wave was 900 (or 1800), the results were found to be quite complicated, the reflected and refracted waves having components notoriginally present in the incident wave and Brewster's angle being absent. The results in the exceptional case were found to be quite similar to the non-moving case and the modified Brewster angle given by (3. 111). Finally, a word of caution is necessary. Though the results of this chapter are valid whatever the value of n (real), some modifications are necessary if total reflection occurs. 39

THE UNIVERSITY OF MICHIGAN 7322-2-T CHAPTER IV OSCILLATING DIPOLE OVER A MOVING DIELECTRIC MEDIUM 4. 1 Introduction The geome try of the problem is shown in Fig. 11. This is similar to that of Fig. 2 except that there is an oscillating dipole of moment me at a height h above the interface. Because of the asymmetry introduced by the motion of the dielectric, in order to take care of the general case corresponding to an arbitrarily oriented dipole, it is necessary that we consider the three cases in which the dipole is oriented along each of the three axes, whereas in the non-moving case considered by Sommerfeld two orientations only were sufficient. The case of the vertical dipole is considered first. Two methods of solution are presented. In one, the problem is formulated in terms of Fourier integral representations of the vector and scalar potentials appropriate in each of the regions shown in Fig. 11. In the other method, all the fields are expressed as integrals of plane waves over all possible directions. The latter method, originally due to Weyl, has the advantage of providing a physical interpretation to the dipole problem by reducing it to the reflection and refraction problem considered in Chapter m. Next, the case of the y-directed dipole (parallel to the velocity) is treated and that of the x-directed dipole (perpendicular to the velocity) being omitted since the method of solution is no different from the previous cases. Electric field patterns in the two principal planes (xz and yz ) are included. 4. 2 Vertical Dipole 4.2. 1 Fourier Integral Method. First let us define a two-dimensional Fourier integral. x1 ~~II5r ri(P lx+P2Y) r -i(p u+p2v) f(x, y)= 2 I dpdPldP2e 1 dudv e f(u,v) ~ (4.1) -Go -as 40

THE UNIVERSITY OF MICHIGAN 7322-2-T z Region 1 (h < z < oo ) iwt Region2 (0 < z < h) k-=-(p c h Cu, 0CO) _ 97 v Region 3 (-oo < z < O) 4, ) x FIG. 11: DIPOLE OVER A MOVING MEDIUM 41

THE UNIVERSITY OF MICHIGAN 7322-2-T Henceforth, we will assume that the functions are well behaved so that the above representation is valid. Referring to Fig. 11, the fields in the upper half space are given bythe well known relations ik2 -- E= k2 A+ VV A (4.2) H = VxA (4.3) O/ - where A, the vector potential, satisfies V2 A_+ko A = i m 6(x) 6 (y) 6 (z-h). (4. 4) In component form, the vector equation (4. 4) separates into the following three scalar equations. V2A +k 2A = 0 (4. 5a) x O X V2A +k2A =0 (4. 5b) y o y V 2A +k2 A = i wp, m 6(x) 6 (y)6(z-h) (4. 5c) z 0 Z' The solution of the third equation above consists of two parts; a primary excitation due to the dipole source, and a secondary excitation due to currents induced in the moving medium while the solution of the first two (4. 5a, 4. 5b) is accounted for by secondary excitation alone. Speaking mathematically, the primary excitation is regular everywhere in the upper half space except at (O, O, h) and the secondary excitation is regular everywhere in this half space. Since the boundary over which the fields are to be matched is of infinite extent in the x and y directions, it is clear that each of these excitations should be expressed as a double Fourier integral in these two directions. This will serve as the necessary groundwork to formulate the problem. Now, the primary excitation in Az, apart from a constant, is the 42

THE UNIVERSITY OF MICHIGAN 7322-2-T Green's function for the Helmholtz equation which has the following integral representation due to Sommerfeld iik0R1 _ 1 (Plx+P2y)-XoIz-h R e 2j X e dpldp2 (4. 6) 1 I 1o where Xo= (p2-k2) /2 Re 2o>0 2 2 2 p2= p2 + P2 R1= [2+y2+(z-h)2]l/ The secondary excitation can be represented by the integral ((Plx+P2Y)-Xoz] F(p1, P2)e dPldP2 (4. 7) where F is an amplitude function and the only requirement being that the integral be regular throughout the region z > 0. The potentials appropriate in the lower half space have already been discussed in Chapter II and in the present problem, there is no primary excitation. Each component of the vector potential, which must therefore be regular and satisfy (3. 20), can be represented by the integral Od (Plx+p2y)+Xl] G(p1 p2 )e (4.8) where p2 - ak2, Re 1 0 P2 -2 ReX)>0 We are now ready to formulate the problem. It may be pointed out that in the non-moving case considered by Sommerfeld, the z-component of A 43

THE UNIVERSITY OF MICHIGAN 7322-2-T alone was sufficient to match the boundary conditions. In the present problem, because of the motion of the dielectric in the y-direction, it is reasonable to expect that the y-component will also be needed. a) Upper Half Space A =0 x pe x+P2Y)- Pldp2 (4. 9) Ay = Fy(Pl, P2)e 1dP2 (4-9) C + z-h) -oZ e i(plx+P2Y) A = +F ze e dPldP2 (4.10) eX1 X+ dd where C = -iw/,om/8r2. The sign convention in the primary excitation should be chosen so as to ensure the convergence of the integrals. + sign for 0 < z <h (Region 2) (4.11) - sign for h < z < oo (Region 1) J b) Lower Half Space Ax - 0 A O D(PlX+P2Y)+X1l = GZ(p1,p2)e dpldp2. (4.13) The problem can now be solved, in principle at least, since all the four unknown functions can be determined from the following boundary conditions. a) continuity of tangential E and H (3.49) b) continuity of normal D and B_ In Chapter II, where we considered the problem of reflection and refraction of a plane electromagnetic wave at the boundary of a semi-infinite moving medium, 44

THE UNIVERSITY OF MICHIGAN 7322-2-T it was shown that b) above follows from a) and Snell's law. This is also true in the present problem even though we do not make use of Snell's law in an explicit manner; because of the Fourier integral representation, Snell's law in fact enters implicitly. Now, let us compute the electric and magnetic fields. a) Upper Half Space. The fields are obtained by substituting (4. 9) and (4. 10) into (4. 2) and (4. 3) and differentiating under the integral sign. ik2 i(p1x+p2y) ( + (z-h) ik2 i(p1x p2y)+ o (z-h) — E =y dp1dp2e 1 _ip2 Ce E 2= -kz Y dPldP2e ip2e -2 2 o' +[p 2 -iP2Ao~z e J 0(4. 15) dpdpe pC (h +ei(PlXeo +p2 y (4.17) -2 o X 0 5:3i(p X+P Y) ip1C (z-h) -z o y +ip2e 2F+XoF e (4.17) H =- dPldP2e + i e 1 (4.18) o. y 12_0 e |o H z dp 1dp\ e ip py e (4. 19) -Oa 45

THE UNIVERSITY OF MICHIGAN 7322-2-T The sign convention is given by (4. 11). b) Lower Half Space. Substituting (4. 12) and (4. 13) into (2. 34) and (2. 35) gives ik2a2 iy (plx+p2y)+lXz] _ E =e dp dp e 1(P2Gy-iGzY) (4.20) [222 y "p2AGW (4. 21) ik2a2 -iW yaL_ - y) Eika =e dpldp2e w0~ ~ ~ ~ - Y P 2-2) ik a -iW y2 -z =- e dPldP2 e 12 L2+ 1 p2)G XApG (4.22) i Qi(Plxa P2Y z 122 xH = -[y e dy ddp dp2e [l -G (4.24) 00 1 i Q (PdPld X+P2Y)+e (4 -iu a2y The disturbing factor e wY appearing in (4.20) - (4.25) can be brought under the integral sign by invoking the translation property of Fourier transforms given below. e-i~ot g(t)= G(w+o)e iwtdo, (4.26) -00 46

THE UNIVERSITY OF MICHIGAN 7322-2-T where g(t) and G(o) are the Fourier transform pair Go -iot G() = g(t)e dt (4.27) 2 g g(t) =) G(w) e itdo (4.28) -00 Applying the boundary conditions (3. 49a), we obtain the following set of four equations arranged in the order of continuity of H, H, E and E at z = 0. The asterisk * indicates that the argument has been changed from P2 to (p2+w Q) for instance x =F 2 1 etc. 1 li +a (p2L+w %e) 1 1 a 2 0 - P2 1 F P2 0P2 +w o 0Q)W 2. 2 p2 p2o p2 2 ua+w )X 1 ip2X O -- -ka (4. 29) - (4. 32) Before solving the above system of equations, we will verify that the boundary conditions (3. 49b) are automatically satisfied. The continuity of D and B at z z z = 0 yields 47

THE UNIVERSITY OF MICHIGAN 7322-2-T ip 1 2 2 2 -X0h -F -F + 2 G+-I p -( +w (P2+) G= (4.33) o a y ka ka Z a 2 We note that these two relations can be obtained from (4. 29) - (4. 32) thus (4. 33) = ip2(4. 29)-+p (4. 30) (4. 34) = P2(4. 31)+(4. 32) Setting I = p,0. solving the system (4.29)- (4.32), we get for the unknown functions F= 2Ceh wp2+p 2[( )2+ak (1a (4.35) F CeX | 1+ (a- +p2-ak2)+a(-a)k2p2+pwQ(p2w Q+kO) (4.36) yG D " p2Cae + P i Q- 2+k22(1-a)+wi (4. 37) GI' =a [Ce +Fz (4. 38) where iopOo m 8 - 2 and D =k2p2- 2)+ak2a22_2 l 2 2 22 iao k'L 1J'2 oJ'~o1 +2k p2L Q-(wQ,(pw-k (4. 39) 48

THE UNIVERSITY OF MICHIGAN 7322-2-T Thus our formulation in terms of the y- and z-components of the vector potential does indeed lead to a solution. Evaluation of the infinite integrals is all that remains to be done. It turns out that this is indeed a formidable problem in itself. Even in the non-moving case, where a single integral is involved, closed form solutions are not possible and the situation is much worse in the present case. Before we take up the evaluation of the integrals, it is in order to present an alternate formulation of the problem. 4.2.. 2 Method of Weyl. Weyl developed a method by which Sommerfeld's solution for a dipole over flat earth could be interpreted as a bundle of plane waves reflected and refracted by the earth at various angles of incidence. The alternate formulation to be presented here would not only extend a similar concept to the present problem but serve as an independent check on the results obtained in the previous section. This is easily accomplished by changing the variables of integration in (4. 14) - (4.25) to polar coordinates but first a few remarks are necessary. In Chapter III, where the problem of reflection and refraction of a plane electromagnetic wave was considered, in order to facilitate analysis, we distinguished between two kinds of polarization depending upon whether the incident electric field was perpendicular or parallel to the plane of incidence. In the present problem, since the dipole is vertical, lines of H in the upper half space are circles, hence perpendicular to the meridian planes and every such plane is a plane of incidence as shown in Fig. 9. The electric field is, therefore, parallel to the plane of incidence though not perpendicular to the direction of propagation. Our aim would then be to show that the results of the vertical dipole problem are the same as those of the reflection and refraction problem in which the incident electric field is parallel to the plane of incidence. In the last section, prior to the formulation of the problem, attention was drawn to the fact that the fields in the upper half space are caused by a primary excitation due to the dipole itself and a secondary excitation due to currents 49

THE UNIVERSITY OF MICHIGAN 7322-2-T induced in the dielectric. Referring to Fig. 12, one may also interpret these two contributions as a direct field and an indirect or reflected field reaching an observer at a point P. With a view to express these fields as integrals of elementary plane waves, consider the integral representation (4. 6) 1 ikR1 _ 1 _ (px+p2y)-Xo Iz-hII - e - e dpdp2 (4.6) -00 Introducing polar coordinates defined by P1 = kosin a1 cos 1) (4. 40) P2= k sin1 sin31 j where al is complex and 31 real, varying from 0 - 27r, the above relation becomes 27T 21* P ko(h-z)cs all 1ie ik= e (h)c (4.41) where +: 0 <z<h -: h<z <oo d21= sin aldaldl) k_- _p = ko(x sin alicos l+y sin a1 sin (4.42) p = (x2+y2)V1 and the path of integration in the complex a1 plane is as indicated in Fig. 13. The integrand in (4.41) is easily recognized as a plane wave in the directions a,',1; in fact (4. 41) represents the spherical wave function as a superposition of plane waves with real directions for which 0 < a1 < - and complex directions for 50

THE UNIVERSITY OF MICHIGAN 7322-2-T P Dire( 1 2 T! ~SP= eflected h OP=R _ _ S__IP = R V -a' Transmitted S FIG. 12: PHYSICAL INTERPRETATION OF THE DIPOLE PROBLEM 51

THE UNIVERSITY OF, MICHIGAN 7322-2-T Imag Ca1 plane 0 7r-2 real FIG. 13: PATH OF INTEGRATION IN THE a1 PLANE which a1 is situated between r /2 and F -i Co. The latter correspond to positive imaginary values of cos ca and therefore are exponentially attenuated in the zdirection (evanescent waves). Moreover, in Region 1, ca is measured from the positive z-axis and in Region 2, from the negative z-axis which is thus also the angle of incidence at which an elementary plane wave meets the dielectric surface as shown in Fig. 12. Changing the variables of integration in (4.17) - (4.19) to polar coordinates defined by (4.40), we get for the primary magnetic field 2w vp7 -ico iko(~+h cos a1) H= J (I1, I2 0)e d21 (4.43) 0 o 52

THE UNIVERSITY OF MICHIGAN 7322-2-T where = x sin c1 cos I1+y sin a1 sin l-Z cos 1a i1=- HC k2sin a, sin /1 0 C I2 s c c P (4.44) i2- C k2 sineacs l *'1 The integrand in (4. 43) may now be identified with the incident field (3. 84). Similarly, for the reflected field in the half space z > 0, we get 27 r — io o r r2 -Diko0 -r (1' R2' RR3)e ddR1 (4.45) where r = x sin cacos /31+y sin acsin 31+z cos c1 i 3 R1=o k cos 1(sin a sin 1Fz-cos aF (4. 46a) R2= -k k3 sin alc os n F4os. 46b) R3= i kO sinalcos alcos 1F. (4. 46c) We also note that Rlsin calcos 1l+R2sin alsin ll+R3 cos a1= 0 The integrand in (4.45) may now be identified with the reflected field (3. 86). Theprocess of expressing the fields in the lower half space in terms of plane waves is slightly involved. By bringing the factor e sunder the integral sign in accordance with the translation property (4.26), the phase function will assume the form i(plx+p2y)+X z47) e (4.47) where where 2 2 12 =Pl+ - (P2+ Q)2 -ak2]2 53

THE UNIVERSITY OF MICHIGAN 7322-2-T In order that the above may represent a plane wave in the direction a2, f32 in the moving medium as shown in Fig. 12, we make use of the modified Snell's law k sin a K sin A1= g2 and since K satisfies (3. 27) with t = a2' Ot = /32' we get X = -i K COS 2 the negative square root being chosen to ensure the convergence of integrals. The phase function (4. 47) now becomes iK x sin cz2 cos g2+y sin e2sin 2-z cos (4.48) e (4.48) which has the desired form. Using the above relations in (4.23) - (4.25), we get rT "-io i K -tH } 22 (T1, T2, T3)e dal (4.49) 00 where = x sin a2cos 82+y sin a2sin 2-z cos e2 T k~so = 1 t$ ~(Ksin a2sin 2+WQ)G'+KCoS 02G; l (4.50a) ik2cos a T2 = -_ K sina CoS cGG2 (4. 50b) ik2 cos a T K sin COS 32G (4. 50c) We also note that T1K sinacos K Sina sin2c 2+ a ( sin2sin2+ )-T3c 2 The integrand in (4.49) may now be identified with the transmitted field (3. 88). 54

THE UNIVERSITY OF MICHIGAN 7322-2-T The electric field can be similarly expressed as integrals of plane waves and the resulting set of equations for the unknown functions would be identifical to (3. 90) - (3. 95). This shows that the problem of the dipole reduces to the problem of reflection and refraction. 4.2. 3. Approximation of the Integrals; Asymptotic Forms. We are now faced with the task of evaluating a series of double Fourier integrals such as those in (4. 9) and (4.12). The integrands involved in each case are too complicated to permit even one integration exactly. However, in the present problem, it is sufficient to obtain an asymptotic expansion because the first term in such an expansion corresponds to the far zone field which is of major interest in a radiation problem. One of the most important methods of obtaining asymptotic expansions is the method of saddle points. The two-dimensional case has been discussed by 12 Bremmer and others. The results are rf(p 1,p2) ___ PP21 2 dp dp T- A as r- oo (4. 51) A(p2)e 1 2 rs rs in which A is the Hessian determinant 2 2 af af ap2 A= a2f a2f ap2ap1 ap2 and the subscript s denotes that quantities are to be evaluated at the saddle point which is found by simultaneously equating to zero all the partial derivatives of f. In the above integral A and f are assumed to be sufficiently regular and its approximation is derived by replacing f by its Taylor series about the saddle point and cutting off terms beyond the second order. 55

THE UNIVERSITY OF MICHIGAN 7322-2-T a) Fields in the Upper Half Space. The primary vector potential is given by -ihop m ikoR1 A =A =0, A= e (4. 52) xp yp zp 4.5R where R1 is the distance between the dipole source and the point of observation P as shown in Fig. 14. The far zone (radiation) fields are found by substituting (4. 52) into (4. 2) and (4.3) and retaining l/R1 terms only. mk2 ikoR1 E: -sinB 0cos lCos o e R xp 1 I- -47r (mk2 ikoR1, (4.53) E =-sin 0cos 1sin 4r e R1 yp I I44T o i mko2 ikoR1 E =sin2o 0 e zp 1 4ir c R1 muko eikoR1 H =sin, sin 4_- e R1 mwko eikoRg H =-sin 01cos 4r - H =0 zp where 01 is measured from the positive z-axis as shown in Fig. 14. Now, in the integral representations (4. 9) and (4. 10), Fy and Fz contribute to the reflected waves. Examination of (4. 35) and (4. 36) reveals that integrals of the type given below are involved in each case. Dd [(Plx+2y)-Xo(z+h)] I= dPldp2A(pl, P2)e (4.55) -00 56

THE UNIVERSITY OF MICHIGAN 7322-2-T 1 /1 OP=R 2 h OP=R SP=R2 3 CO, E) i 3~oA/ /6 FIG. 14: DIPOLE SOURCE AND IMAGE 57

THE UNIVERSITY OF MICHIGAN 7322-2-T where xo= _(p2-ko) We will now use the result given by (4. 51) to obtain an asymptotic expansion of I. Introducing polar coordinates defined by x=R2sin O2cos (4. 56) y=R2sin 02sin (z+h)=R2cos 02 where R2 is the distance from the image point as shown in Fig. 14, the integral becomes.GQ R2f I= J A(pl, p2)e dpldP2 where f = [(sin 02cos psi sin V) (p2-k2)l/co 02] Setting the partial derivatives equal to zero, we get af = i sin o2 2sin -(- 1P = O Saddle point s occurs at pl =kosin 02cos 0 p2=kosin 02sin y provided we take (p2-k2>/2= -ik cos 0. A direct calculation shows that s 2 2 o2 2 kcos 02 58

THE UNIVERSITY OF MICHIGAN 7322-2-T and Js k cos 02 o 2 The reason for choosing negative square root will be given shortly. Thus, the asymptotic expansion of (4.55) becomes, C i(PlX+P2y)-o(z+h)z I= JA(p1 P2 l2 dpdp)e -00 ikoR2 2 2 -2 kcos 02A(kosin 2cosi, kosin 2sin 5) 7) As an example, consider the first term in (4.14) which corresponds to the primary field. For z > h, we get k2 ikR E = -sinO lcosO 4I Cos e which is the same as that given by (4. 53). Thus the negative square root in the Hessian determinant has been chosen to yield consistent results. The contribution due to the saddle point yields for the reflected waves ik R iw em 2 in ci si Fc2 2 A = e - os sin in +si n [(c) +an (1 yr 4ar R2 D 2s 2 2a (4.58) ik R A_ e o22 n2 2 A -sin 02) zr 4r R2D +a cos02 Fan2-sin2e2cos2- a1 (2c Fsin 2sino)2] -a(l-a)n2 sin2s2sin2 i-nc sin 2sin (l+sin02+ (4. 59) 59

THE UNIVERSITY OF MICHIGAN 7322-2-T where D =t (an 2-sin 02)+an [a-sin 022(a cos2 +sin2 +a(l+n2)cos 2 [an2-sin2O2cos20- - (c +sin 02sin)2J 12 -20 c sin 02sin_-(Qc )2(1-sin2O2cos20) } (4. 60) The fields can be obtained either by applying the saddle point method directly to the integral representations given by (4.14) to (4.19) or by substituting (4. 58) and (4. 59) into (4. 2) and (4. 3) provided the differentiations are replaced according to the following scheme. ax = iko sin 02 cos a =ik sinO2sin i (4.61) ay o 2 = ik cos 2 az o 2 The resulting expressions are too lengthy for the general case. We will consider only two principal planes, namely the xz and yz planes in detail. Case 1: ~ = 0 (upper sign) or ~ = 1800 (lower sign) Primary Field H =0 xp xmkok eikoR1 H = + sin 01 (4. 62) yp+ 1 4 R1 H =0 mk2 ikoR1 _k o e E = sinOCosO e (4.63) xp =sinc 1 4vE R mk2 ikoR1 E =sin2 oe (4. 64) zp sin01 47r E R1 60

THE UNIVERSITY OF MICHIGAN 7322-2-T Reflected Field m2 ikoR2 H -=2 N(n2-l)sin2 02cos202 4 (4.65) xrN 2 47 R2 H =+sin 02 2 1 2- [-(ln22)n22)sin220+(1-l2)cos 02 In22 i2 ] _sin 2 o e R2 (4. 66) zr xr 2 ko E =cos 0 H - (4.68) xr 2 yr 2 eik mk2 ikoR2 E 2 (n2-1)sin202cos 02 4 e yr N 2 2 4.IE R2 (4.69) -kO E =+sin 0 H - (4. 70) zr 2 yr we where 0 N=n2(1'2 )-(1-n232)sin2 02+(n2-32)cos 202+(1-/2)(l+n2)cos 2:2- -sin2 021" (4. 71) which is the same as (3. 75a) with 0i=02. Moreover the above results, apart from constant factors, are the same as those in Chapter Ill, Section 5 for the case i=0. For numerical calculations the above results can be put in more convenient forms. Making use of the following approximations for points of observation remote from the dipole source (See Fig. 14). R R-hcos 0 (4.72) R2 R+hcosO 61

THE UNIVERSITY OF MICHIGAN 7322-2-T the total electric field in the upper half space in spherical coordinates (R, 0, O) is given by ER= 0 R-i r-ik hcosO ik hcos0 O 0 o 2 L(1n2.sin2 2 n2-2 22 22] mk 2 ik~R H = +n s R (.75sin g0o) p l47r R H mrk ik0oR E n 2sin ~cos0s e (4.76) H =sin 0 (4. 77) xp I 4=r r Reflected Field: H = +isin0 I oi1 mkOw eik~R (4. 78) H =H =0 yr zP E =0 xr 62 mk2 ikoR1 E =c-sin elcos e1 (4o76) mk2 ikoR1 E =sin 20 o e (4.77) zp 1 47r co R Reflected Field: =sin 2F.mk.w eko xr -sin 4 R2 (4. 78) (n2cosO2, F) 4mk oRei~2 H =H =0 yr zr E =0 xr 62

THE UNIVERSITY OF MICHIGAN 7322-2-T 2 ikoR2 2F 0mko E =+sin22coCoss o [ 1 2 enk R2 (4.79) yr 2 2 2 47r R2 2 ikoR2 E = sin 0 1- _ 2F2 mko e (4.80) zr (n2cos;2+F) 2 (4.80) where F =[cos202+ n (1 + /]sinO2) (4.81) Once again, we note that the above results, apart from constant factors, are the same as those in Chapter III, Section 5, for the case i = 900. Using the approximations (4. 72), we have for the total electric field ikhcosO ik hcos 1 cos+F( i ik E 0 b) Fields in the Lower Half Space. The unknown functions G and G in Y z (4. 12) and (4. 13) are related to G and G" given by (4. 37) and (4. 38) as follows. y z G (Pi I P2 G"' "' 1.~ ink2 y(1 2y 12 |(4.83) It, A convenient set of polar coordinates in this half space are x = R sin cos c+ ) y = R sino sinpo (4. 84) z = -Rcos where F is the angle shown in Fig. 14 and R, G have their usual meaning. Due to the presence of the factor elth in G and Gz, integrals of the following type 63

THE UNIVERSITY OF MICHIGAN 7322-2-T are involved in the determination of A and A y z Rf(pl, P2) 1I= A(P1, P2)e dp1dp2 (4.85) -ao where f= (Plsino cos O+p2sini sin 0)-hcos - h Xo*:P1 2 2] and 2 2 P2 a.2 k1= i1+ -akIn order to obtain an asymptotic expansion of (4.85), we first determine the saddle point of f. Setting the partial derivatives equal to zero, we get af P1cosV hpI -- = i sino cos - - = 0 ap- R X0, a f p2COS? h(p2- Q2) = i sino/sin asin X RX =0 The solution of above when h / 0 is quite difficult and will not be considered here. When h = 0, we get P =a/ k -) (4. 86a) R(a P2 a 3/2 k(Y) (4. 86b) provided we take X i a; k ( R-) (4. 866) Ra where R =(x2+ay2+z2),2=R [in2V(cos2+a sin20)+cos2l'A (4.87) 64

THE UNIVERSIT'Y OF MICHIGAN 7322-2-T At the saddle point Ra 2 s aRkz 1/2 Ra f = ia k( R) s R For consistent results, we take = - k [sin2c(cos2asin20)+cos2] s ak cos 0 Using (4. 51), the asymptotic expansion of (4. 85) becomes 2r a k cos A(alakx/R, a' ky/R )e Ra In. a a as R co (4.88) i sin2 (cos2+a sin2R)+cos2] R Since, as in the case of upper half space we are going to consider the fields in the two principal planes only, the asymptotic expressions of A and A for the y z general case will not be given here. It may be noted that when using (2. 34) and (2. 35) to determine the far zone fields, the differentiations are to be replaced according to the following scheme. a 1/'x = ia k () Ra = ia3/2 k( ) (4.89) ay Ra az R Case 1): h=O, 0=0 (upper sign) or -= 1800(lower sign). ial/2nkoR 3/22 iom e 2n(n) cos 2 22 A Cos ~ a cos c -(Qc) an sin 2j( ~y ~ 47r R D (1-n212) (4.90) 65

THE UNIVERSITY OF MICHIGAN 7322-2-T A iWoPm eiankoR 2a52n2 sin A =-cs ~nco +ae 2a 2 2 A =- R D coso ncoso+' [-(Qc)-ansin z 4ir R D (4.91) where D = n Os2+a-(Qc 2Cos2 -a2n2sin2, +a2 (n2+1)coso -(Q c)-2 an2sin2j (4.92) The magnetic field is given by ink H= a-2 cos i A (4.93) x y a o inko H = i +2 sin 0 A (4.94) a o inko IZt= + si n ~ A. (4 95) a /~ A point worth noting is that the above results cannot be obtained by substituting 0ti= / in (3. 99) to (3. 101); Ott has drawn attention to it for the non-moving case. The electric field is given by E = + sin l cos / A (4.96) x a z E = iw A (4.97) Y Y E = sin2 A (4.98) z a z or in spherical coordinates ER= 0 (4.99) R iL E= - sin O A (4. 100) 0 a z E- + i WA (4.101) Y 66

THE UNIVERSITY OF MICHIGAN 7322-2-T Case 2): h=0, 0=90~(upper sign)or p=270~(lower sign) iwI, m iaVynkoRj 3,k _ 3 2 o e 2a3n c Cos - n2 47 R 2 Dcos + sin) y DR +n a c + a (l-a)sin ~ (4. 102) m ia Lnko RI ( 2 o e 2a ncos n = c os Co z 47r R R 2 D..(QCna3 sino,)2W t~n n+ --- sin;;l } (4.103) where D=n -— sin2 n -(e- s in a +n2(QC si)2o D=n -2 sin 1'2 (4.104) [os 2 + a sin2a 1/2 (4.105) The fields are given by ink H [os / A +sinA e (4.106) H =H =0 y z E =0 x E =iw Cs [cosbA +sinoA e 1iwQy (4.107) Y 2 Y E = iW s in 0 A +_ cos 0Aj e (4.108) or in spherical coordinates ER=0 R 67

THE UNIVERSITY OF MICHIGAN 7322-2-T E0= sinAz+cosA e (4.109) E = c) Fields in the Free Space Side of the Interface for Low Velocities. The asymptotic forms obtained thus far have certain limitations. First of all, since the expansions are only up to the first order term, they are of no avail should the coefficient of this term vanish. This is precisely what happens when the point of observation P moves very close to the interface as shown in Fig. 14 and is far removed from the dipole. In such a case 0 lt0T A 1 2 2~(4. 110) R =R 12 and substituting the above in (4. 52), (4. 58) and (4. 59),. we note that both the components of the vector potential, hence the fields, vanish. Next consider the expression under the square root sign in (4. 60). Substituting for a and Q2, rearranging, we get for this expression 1+ I-2 (1-Isin 0sin 6)2-sin20 = [1+n (1-,sin0sin )2]E-sin20] (4.111) which becomes negative if total reflection occurs. This would give rise to complications which go much deeper than just making the amplitude of the reflected waves complex. To get an idea of the nature of these complications, it is imperative that we examine the method of saddle points in greater detail. While distorting the given path of integration into the path of steepest descents through the saddle point, one might sweep across the singularities of the integrand. In such a case, the path of integration must be deformed to avoid the singularities and in the final result their contributions included. These difficulties also occur in Sommerfeld's original problem and have been thoroughly discussed by Ott 68

THE UNIVERSITY OF MICHIGAN 7322-2-T In the present problem, it is almost impossible either to obtain higher order terms or to examine the singularities because the integrands are unwieldy, and a double integral instead of a single integral is involved. The situation eases considerably if one integration can be carried out exactly. We will, therefore, make some reasonable approximations to achieve this. First, a and f2 are expanded in Taylor series about 1 = 0 n21-i )13- =(n21) (1+n232+...) l-n2132 C C (4.112) a = = l+(n2-1)12+ < (4.112) _1-n2 n22 For low velocities, it is sufficient to retain only the first term, so that Q 2(n2-1) c f> (4. 113) a~l1 Making use of these approximations in (4. 35) and discarding higher order terms in 3, we get 2 2iCwQ p e F 2 2p (4.114) Y ko(A+n X )(A+Xo) where A=(X2+2 w Qo P2) Qo =(n2-l) l (4. 115) X=(p2 k2. Because of the troublesome factor 2 w Qop2 occurring in the denominator, the above still cannot be integrated over one of the variables. Expanding the denominator in Taylor series and retaining only the first order term in 1, we get 69

THE UNIVERSITY OF MICHIGAN 7322-2-T -X h 2iC 2op e~ F 2 (4.116) y 2 2 k (X+n2Xo)(X+X) o 0 which is the desired low velocity approximation. Similarly from (4. 36), we get c-Xoh 2x xp%(X+nX) 1- 2 W 2 (4. 117) z 2 2 2 2 z O X+nX Lk OX(X+X o)(X+n2X o ( Substituting for Fy and Fz in (4. 9) and (4.10), introducing polar coordinates defined by P1=P cos v, 2=p sinv ( x= pcos, y psin J118) where p is the cylindrical distance and making use of the relations 2r JZ eiZ c~s (V-)dv (4.119) inT 27risin~J!(Z) sinve (v o)dv where J stands for the Bessel function of the first kind, we get o o wmB m3 X -x -X (z+h) A = J (pp)e p3dp (4.120) y 2r k X+n2X 0 o o A iqum FikoRl ejkoR2 2 r Jo(P ) -X (z+h) ~~A "i Ro2 ) A 4a R1 R2 J(+n2Xo) 1 A0 mg OD 0 o+n X)(X -X) - (z+h) 4 -~ k~ sing X(Xon2) Jl(pp)e pdp T rnoigt canb converted fm o +(4. 121) The range of integration can be converted from - to + using7 tle relations

THE UNIVERSITY OF MICHIGAN 7322-2-T n+lP 1 H(l\(pP). ~.p dp (4. 122) 0 ~ -02 where * denotes any arbitrary function of p. It may be noted if j = 0, A =0 y and only first three terms in Az remain which checks with known results. Let us consider the first integral in A z OD j i (p) (z~h) OD H(1)(pp) -A.(z+h) 2Jepp) -d( z) pdp e epdp (4.123) 0 (X+nXo) (X+n X) Approximation of the above integral has occupied the attention of several investigators beginning with Sommerfeld. Besides, Ott', Nomura has given a thorough treatment and we will draw freely from their work. Similar approximations can be carried out on (4. 12) and (4. 13) which pertain to the lower half space. There is no need to give the complete expressions here but it is enough to note that the exponent will be of the form (p2- k2)2 z Xz e~p2-k2 = e ~ (4. 124) The integral I1 will now be transferred to a complex p plane defined by p=kosingp, p = 91+L2 (4.125) 7r I For the path of integration L running from- +i o to - -ico through the origin as 2 2 shown in Fig. 15, it is necessary that we choose X = -ik cos p = -ik (n2sin-sin2p)12 Im(n2-sin2p)2 } 0 0 so that (4. 124) is bounded as z - -oo. The integral now becomes ik~ 1l iko(z+h)cosp I1= 2 A()H (k psinp)e sinrp dp (4.126) L 71

THE UNIVERSITY OF MICHIGAN 7322-2-T /~~~ // ~. v15 ILUTRTO OFSCDDErIS'MT 2~~7' 2 ~ -'' 00,~~~~~~~~~~~~~~~~~~~~~~~/ 2~ ~ 2 ranch cu f n is real / /~ FIG. 15' ILLUSTRA~ON OF SADDLE POINT METHOD~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 72~~~~~~~~~~~~~~~~~~~~~~~~~~0

THE UNIVERSITY OF MICHIGAN 7322-2-T where A os w = (n2-sin2,) 1 (n cos + w) Let us now examine the singularities of A(u) in the strip -ir< A1 (71. For convenience, we will consider n as being complex. However, this does not imply that the results obtained thus far can be extended to moving conducting media because certain points in their electrodynamics have not yet been fully resolved. The singular points of A are: a) Branch points of w defined by sin u = + n; since sin u=sin(7r -u), there exist four in number vl, v2, v3 and v4 of which v2 and v4 can be considered as the reflection of vl and v3 about the origin. For real values of n the branch points lie on the real axis if n(1 and on the vertical lines A + 2- if n>l1. Corresponding to the two combinations of signs of w, the integrand is double valued and its Reimann surface has two sheets. These sheets are connected with one another by the branch cuts along the lines Imw=O running from the branch points to co as shown in Fig. 15 for the case In <1. If n is real and is less than 1, the branch cut emanating from v2 degenerates into a portion of the real axis from -sin-in and the origin and the negative imaginary axis. Similar remarks apply to the remaining branch cuts. The upper (lower) sheet is specified in which Imw is greater (lesser) than zero. The path of integration L lies on the upper sheet. b) Four poles of first order obtained by setting the denominator equal to zero. A simple calculation shows that the poles are given by sinp= + n (4.127) (l+n2 Whether they lie on the upper or lower sheet can be ascertained by examining the 73

THE UNIVERSITY OF MICHIGAN 7322-2-T relation w cos - P (4. 128) P 2 n Let ia 7 n = In e a, 0 t< 2i/ (1 +n2)1+= n2 1 e2i 34 then w = +Iwpl ei(2-) p Since Im wp> O in the upper sheet, we have to choose the positive sign, so that w 2|2 I- ee in the upper sheet. n n Substituting in (4. 128), we get cos =- Wp e-iB 0<< Since cosp=cos(j1+ it2)=cos 1 coshCi2-i sinl1sinhp2 the position of the pole in the upper sheet is given by T 2hu w1e' i r24~ Thus when n is real the pole lies on the real axis between 7r /2 and t, coinciding with fr + iO when n = 0, moving left as n increases and approaching 7r /2 + iO as n -- oo. The inverse point is also pole lying on the upper sheet. This completes the discussion on the singularities of the function A in (4. 126) In order to proceed with the saddle point method, the Hankel function in (4. 126) is replaced by its asymptotic value 74

THE UNIVERSITY OF MICHIGAN 7322-2-T 2n+1 H (Z)M i e (4. 129) and introducing polar coordinates defined by (4. 56) we get ike 7/4 ( ik R2cos(-02) I =7 j Fp)e dp (4. 130) (2k7r kR2sine2) L where cos 1p (sin p)l/2 F(p) = - 1/2 n2cosp+(n2 -sin2p The exponent ikR2 cos (P-2)=koR2 sin(l -e2)sinhlu2+i cos( 1-0 )coshp] has negative real part in the hatched area of Fig. 15 (-r+02p1<02 above and 02<1!<7r+ 02 below the real axis) in which the above integral converges. A simple calculations shows that the saddle point is given by i s=02 and the path of steepest descents is given by Re cos(-02) = 1. e. cos 2)os2 = 1 and is denoted by L in Fig. 15. As the angle 02 varies from 0 - 2, L just shifts parallel to itself. Let us now find out what part the singularities of A in (4. 126) play in the process of distorting the given path of integration L into the path of steepest descents Ls. If n is real and less than one, a finite portion of Ls would lie on the lower sheet (indicated by broken lines in Fig. 15) for 02(sin-ln. Since the condition Im(n2-sin2,p)l2 > 0 is not needed in (4. 126), this is of no consequence. However, if 02> sin-1n, the path of integration will have to go around the branch cut from v1 (details about which can be found in Ott'sl3 work). The branch cut integrations are not important in the present problem; hence will not be included. Similar 75

THE UNIVERSITY OF MICHIGAN 7322-2-T remarks apply when n > 1. Regarding poles, from previous discussions it is clear that none is swept across unless 02 -' r /2 and n >> 1 when the pole and the saddle point come arbitrarily close to each other. ott14 and van der Waerden 16 have presented a modified saddle point method to take care of such a situation. Their results are not needed here since the dielectric medium we have in mind is an ionized gas whose index of refraction will be less than unity. Thus, for the present, the singularities of the integrand in (4. 126) play no significant part. To get an idea of the fields in the free space close to the dielectric, it is necessary to carry out the saddle point method of integration up to second order terms. Following Ottl4, we make the following substitutions - 02 = t cos t = 1 + is2 and taking s = +I eiT/4 sin t we get C ikR2 cos(_-02) ik R 2-i 7r O 2 Fj^t~e ~ 2; 2)d," g o2 4 -kos F;u) ds r~IA)e d 2= e e L cost S -00 2 One now expands the integrand in power series in s and integrates term by term to obtain ik R2- 4 1 72 2 e F(02)+ 2ik + O ( k 2 oR2 R2 where G(W) F= i'1 cos 2 ~-02) Substituting in (4. 130), we get 76

THE UNIVERSITY OF MICHIGAN 7322-2-T ikoR2 G"(02) zIl~H- e F(sz~f ~ ikR )+ j (4. 131) R 2(sine2);2 (2+ 2 and for 02 7/ 2 2 ik p -in e 1 k (1-n2) 2 (4. 0 where p is the cylindrical distance. The remaining integrals in (4. 120) and (4. 121) can be approximated in a similar fashion. The final results are given by A=r o0 n2+1 eik~ A 27r n+2 p2 (4. 133) = (4. 134) A = k (12) Ln4 (2n4_n2+1) Bsine (4.134) The fields can be determined from (4. 2) and (4. 3) bearing in mind that ax =iko cos a = ik sino (4. 135) ay 0 -a =0 az This completes the solution of the problem of the vertical dipole over a moving medium. 4.2.4 Numerical Results. For polar plots ED8 and IEe | are given by IE~ i EPE0 (bar denotes conjugate). These are depicted in Figs. 16 - 21 for n = 2 and n = 0. 5. In the course of 77

600 300 00 300 600 H z nil Air 0f3.O45 z.0 1.5 1.0 r0 0001. 1.5 2.0 ~ Dielectric ddle000 0 ~~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ H 60~~~~~Q FIG. 16: IE0IIN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n=2., h=O (values in dielectric have been s caled down by a factor of 2)

30P 0~ 300 /3 =0.4 z rri 60 6 ~~~~~~~~~~~~~~~~~~~~~~0~ z FIG. 17: lEO) IN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE FOR n2, h0.25X ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I o ~~~~~~~~~~~~~~~~~~~~~~~~~-...I1 1.0.75.5.25.25 Y.5.75 I OP 1TTFL2 FZG. 17- I~e ZN THE Am mTHE YZ PLAN FOR A VERTZAL DmOLE FORn-2, h:0.25

30~ 00 30 -z =0.45.Jo o" 60 H0 / 1.0.75.5.25.25 Y.5.75 1.0 FIG. 18: 1E01 IN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE FOR n=2, h=0. 5X

300 00 300 60 606 300 06 300 600 60IG 30 0 30h 600 FIG. 19: IEIIN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n0. 5, h0

,-1 300 0 300 =30 = /0.45 Z 6( 0081, 1 ~~~~~600 ~~~" co03~~~ I \e I P\3 ~~~~Cx0 r~~~~~~\3,~~ \ C 2.0 1.5 1.0.5.5 1.0 1.5 2.0 FIG. 20: 1EO IN THE AIR IN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n 0.5, h: 0.25X

300 0Q 30 z /3=04 0 0 6 0 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ H e~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l ~60 6so"~0* ~~~~~~~~ I 1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3 CC~~~~~~~~~~~~~~~~~~~~-3II " FIG. 21 1E0 IN TH AI NTEY LN O ETIA IOEFRn.,h. \ /~~L t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C3 2.0 1.5 1.0.5.5 1.0 1.5 2.0 Z: FIG. 21' JE0/I H I H ZPAEFR VRIA TOEFRn05 ~.5

THE UNIVERSITY OF MICHIGAN 7322-2-T numerical calculations uncertainty regarding the sign of the square root has been resolved in the following manner. Take for example the expression for E0 given by (4. 73). The quantity under the square root also happens to be the wave number K of the transmitted wave in the problem of reflection and refraction discussed in Chapter III. In order that fields may not become infinite it is clear that a positive square root must be chosen, i. e. (2 - sin 0)2=i (sin20- ). (4. 137) 13s2 1_2 The symbol X in all figures stands for wavelength in free space and should not be confused with the same notation used elsewhere. 4. 3 Horizontal Dipole in the Direction of the Velocity 4. 3. 1 Fourier Integral Method. In the non-moving case considered by Sommerfeld both y and z-components of the vector potential were needed to satisfy the boundary conditions. The same is true in the present case and the method of solution is similar to the vertical dipole problem. a) Upper Half Space: A =O x A y C+F eo i 2 dpdp (4. 138) JA F e 2dpdp2 (4. 139) -OD where 84

THE UNIVERSITY OF MICHIGAN 7322-2-T C=87r (p2-k2 Re Xo> 0 (4. 140) +: O<z (h -: h(z(oo b) Lower Half Space: Ax O O r E1PX+p Y)+xld A = G e 1 dpldp2 (4. 141) -00 OD i(PlX+P2Y)+A1 A = J G e dp dp (4. 142) -aO where 2 1 p2_ak 12 Re/2 0 (12 + a P2 ) The continuity of H, H. E and E at z = 0, yields the following set of x y x y equations. 85

THE UNIVERSITY OF MICHIGAN 7322-2-T X iP21 -i(p2_ ) Po go pa, a2 Y 0 -1 0 -1 F Mo Ma z -P2 iko P2 +Q -iXl 2 k22 22 k2a2 (12) 2 ka- 2 2 0 0 0 Ceoh O = x P2 (4. 143)-(4. 146) k2 2 (1 —) We note that the system matrix is the same as in the case of the vertical dipole. Setting M = uo, solving for the unknowns, we get 86

THE UNIVERSITY OF MICHIGAN 7322-2-T F =Ce -h A2i )k2o(an2rOi+p2-ak2 )+a(1a)k2p2 o w2(p2+p k2O Y XO D o 2+P2 o+Pj5 (4. 147) Feoh 2 2 20 F2CeD kp -(p +W2Q)(k +p2WQ) 1 (4.148) 2iCae ~ 2 -', 2 G~r D k(X1 +an Xo)tP2 1 (4.149) G': =aF (4. 150) z z where D is given by (4. 39) 4. 3. 2 Approximation of the Integrals - Asymptotic Forms a) Fields in the Upper Half Space: The primary'vector potential is given by =0A-i~m ik~R1 A =0, A =-i..e,A 0 (4.151) xp yp 4 R1 zp and using the asymptotic formula (4.57), we get for the reflected vector potential A =0 xr Ayr 4r R (1- (an -sin 02) yr /2 R D 2 L 2 2 +an2cos 02 an-sin202 cos2 - 1 ( Qc +sin 02 s in)2 -a(l-a) n2 s2in2 02 sin2 sin ino(l+sin202+Q * sin02in i m ik R2 0 e 222 _ _A' 4- -R * cos 02(a n -l)sinO2sinp -Qc E+sin 02sin a +Qc sine2sin (4.153) where Ds is given by (4. 60). 87

THE UNIVERSITY OF MICHIGAN 7322-2-T The fields in the two principal planes are given by: Case 1): 0=0 (upper sign) or p=180~(lower sign). Primary Field: E =0 X 2 ik R mnk o 1 O e E o (4. 154) yp 4i c R1 E =0 zp mk ikR1 H =-cosO o e xp 1 47r R H =0 ikR (4.155) mwk o 1 H =+ sinO e zp 1 47 R1 Reflected Field: 2 ik R mk o 2 E = +-N (n2-1)sinO2cos2O24 (4.156) xr N R o 2 E = + E tanO (4.158) k zr~1-p (1-x2)-(1-n2 xr)2n2e2+(1- 2)n2 Hxr = -cos 0 - up E yr (4.159) xr 2cqo yr ik0R2 -2 m2o e H yr- + N (n-l)sin2cose (4. 160) yr N 2 2 47r R8

THE UNIVERSITY OF MICHIGAN 7322-2-T k H =+sinO E (4. 161) zr 2 0oW yr where N is given by (4. 71). Making use of the approximations (4. 72), we get for the total electric field in the upper half space ER=0 2 ik (R+hcose) nk o E_=- 2 B(n -l)sinOcosO 4 e (4 162) 0 N(0) 4r E R r-ik hcos ik hcosoe o o0 0 2 2 (lg2 2 2 2 E + ie 0+e 22hCS L1 2O Ln2(1 l-n2)-(l-n2 2)sin 0 +(l- )n coso( sin (4.0163) Case 2):-=90~(upper sign) or 0=270~(lower sign). Primary Field: mwk ikoR H =-coso o e xp 1 47r R (4.164) H =H= 0 yp zp E =0 XIp ink2 ik R1 ik2 ikoR1 2 o e Ep= + sin 0 47rcos R zp 1 I4 44rc0 RI89 H =H =O~~~~8

THE UNIVERSITY OF MICHIGAN 7322-2-T Reflected Field: m 2k ikoR2 HE- 2F jcosO o e H =2 - nU ncos 02+F 2 47 2(4. 166) H =H =0 yr zr E =0 xr o 2 mk2 ik R2 =+ 2F 7 e z cose2+F 20 R Ezr= +L1- XsinO~cos02 cos o R2 (4. 168) where F is given by (4. 81). Making use of the approximations (4. 72), we get for the total electric field in the upper half space E =0 ~R i(-ik hcosO ik hcosO 2 ikR + 0g Ee0 2 oF(0) 4ie R ( E - cos _ __-e (4. 169) 0 O~~~~~~~n cos+F() 4 Ep= 0. b) Fields in the Lower Half Space. As in the case of the vertical dipole, the problem of obtaining asymptotic forms for the fields when h =/ 0 is extremely difficult. The results when h = 0 can be obtained by using the formula (4 88). Case 1): h=0, =0O(upper sign) or V=1800(lower sign). 90

THE UNIVERSITY OF MICHIGAN 7322-2-T iwJt m ial/2nkoR 2a3/2n cos Ob )2] 447r R D where D is given by (4. 92). The fields can be obtained by substituting the above into (4. 93) - (4. 101). Case 2): h=0, #=900(upper sign) or 1=2700 (lower sign). z 45. R D2 cos + a2L -(C + ___sink] ipr/om eial/ nkoRn 2atsdn cos Ca2n A= i s a"n sin- (/ z 47r R X2D 3 2c A Co { X + -i.sei.a(n2-b) (4.173) where D and i are given by (4. 104) and (4. 105). The fields can be obtained by substituting above into (4. 106) - (4. 109). c) Fields in the Free Space Side of the Interface for Low Velocities. The procedure is identical to that used in the case of the vertical dipole. For small velocities; 2 ~Ce-'oh 2,- v x 2 c - 0a / P2 Y XO X+X0 k 2x (X+X ) (>,-an2 1 10.0

THE UNIVERSITY OF MICHIGAN 7322-2-T -Xoh X -X) k2 W 2 2iCe (x0x) o o z 2 p2 2 0 k (X+n2X) ( X+X o)X+n2X ) -P2 7X)oX lQ } (4.175) -2Lxo(X+n2Xo) X(Xko)(+n2x) Substituting in (4. 138) and (4. 139), introducing polar coordinates defined by (4. 118) and making use of the relation 2~ sin2v eiZc~s(v-)d= o E(Z)+cos 2J2(Z)] (4. 176) 0 in addition to those in (4. 119), we get i4 m ikoRl ikoR2 o(PP) -)(z+h) y 47T L R1 R2 (X+X0) d 0 0d - ~3 sin 0 J1(pp)e p 4d 2ok 0 X(Xk ~0 (4.177) iwi (mA -o(z+h) p2f Com (41X ) A sino j Q pd(pp)e O - 0d - * JOPP)e ~(z+h)PdP _ ~7 L ( 0 4)Z Z2k (X+n) +o)2 -(+no k3, Bsn (l(pp)+cos20J (P~) e- ) p3dp (4.178) ~p92 d 2 92

THE UNIVERSITY OF MICHIGAN 7322-2-T Carrying out the asymptotic expansions up to the second term, we get for 02 7r/2 mUqP eikoP A =- Ll+3(+n )sinvj (4.179) y 27rk0(l-n2) P A =m + (ln-2)sin20;_ esin (4. 180) z 27rk 7j271 (n2_1)+ (n2_1) [(n p This completes the solution. of the problem of y-directed horizontal dipole over a moving medium. 4. 3. 3 Numerical Results Polar plots ofl E0 and [E | are depicted in Figs. 22 - 28 for n=2 and n=O. 5. In naming the figures, the words "y-directed horizontal dipole" have been abbreviated to'horizontal dipole". 93

THE UNIVERSITY OF MICHIGAN 7322-2-T 300 00 30~ 60o 60* 60 0 60 300 0~ 30~ FIG, 22: \E IN TIHE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n -2, h= 0 94

300 00 300 c~ 0 ___=0.45 __. / 60 / 600 ool~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~, L -~~~~~~~~~~y0 i ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I 2.0 1.5 1.0.5.5 1.0 1.5 2.0 FIG. 23: IE0\ IN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FOR n:2, hO0.25X

THE UNIVERSITY OF MICHIGAN 7322-2-T 300 0~ 300' 0 /.75 DYOLE FOR n=2, h=O. 5 96

THE UNIVERSITY OF MICHIGAN 7322-2-T 0 0~ 30 03 — /3=0.45 -~~~~~~~0 ~22 60 600 60~ 300 0~ 30~ 600 FIG. 25: 1E01 IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n=0. 5, h=O 97

300 00 300 60 -om-0 6dN 0.0,~~~~~~~~~~~~~~~304 ~ 0 0~~~~~~~~~~~~~H 60 CD~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C 00~~~~~~~~~~~~~~~~~~~ 2.01.5.05. 1. 1. 2. Dielectric 600 300 00 300 600 FIG. 26: j IE IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FOR n=O.5, h=O "I 0

30~ 00 30~ =0~~~~~~~~c 43:Q0.45 Z C<: C) 60 2.0 1.5 1.0.5.5 2.05 FIG. 27: IE6 IN THE AIR IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n0. 5, h0.5 / /I~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ I~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ — I~~~~~~ 2.0 1.5 1.0.5.5 1.0 1.5 2.0 FIo. 27. IE~lI T A I H PA F HRZNA IO F n. h. 5

300 00 300 Ir P=0-45 — r 6oh;:I 6o~ 1o -,.3 ~ 20 1.5 1.0.5 3 5 1.0 1.5Z 0I.8: II ALOR 20~ y 0 A I D0 2.0 1.5 1.0.5.15 1.0 1.5 z. FIG. 28: jE~jIN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FOR n=O.5, h0O. 25X

THE UNIVERSITY OF MICHIGAN 7322-2-T CHAPTER V CONCLUSIONS From the studSr undertaken in this work, the following conclusions can be drawn. 5. 1. When a plane electromagnetic wave, traveling in free space, strikes a uniformly moving semi-infinite dielectric medium, the incident, the reflected and the transmitted waves are coplanar and the angle of reflection is equal to the angle of incidence. However, Snell's law, hence the angle of refraction, is modified. The reflected and transmitted waves possess components not present in the incident wave and furthermore when the incident wave is polarized with its electric field parallel to the plane of incidence, there is no angle of incidence (Brewster's angle) for which the reflected wave vanishes. An exception to these results occurs when the plane of incidence is parallel to the velocity. In this case, there is a strong resemblance to the non-moving case. 5. 2. The problem of an oscillating dipole over a moving medium can be formulated in two ways. In one, Fourier integral representation of the vector and scalar potentials are employed and in the other, the electric and magnetic fields are expressed as integrals of elementary plane waves. The latter formulation has the merit of emphasizing the connection between the dipole and reflection-refraction problems. The solution by either formulation is in the form of integrals which cannot be evaluated in closed form. However, using the saddle point method, asymptotic expansions can be obtained. The first term in these expansions corresponds to the radiation field and from the numerical calculations, it is observed that in order to produce any perceptible change in the radiation patterns, the velocity must be comparable with that of light. 101

THE UNIVERSITY OF MICHIGAN 7322-2-T Finally, we would like to suggest some allied problems for future research. The case of the magnetic dipole over moving medium is a straightforward extension and the complementary problem where the sources are located in the moving medium should not prove to be too difficult. Extension of any of the present results to lossy dielectrics is altogether quite a different matter since some delicate questior pertaining to the electrodynamics of conducting media in motion need to be settled first. 102

THE UNIVERSITY OF MICHIGAN 7322-2-T REFERENCES 1. Nag, B. D. and A. M. Sayied (1956), "Electrodynamics of Moving Media and the Theory of Cerenkov Effect," Proc. Roy. Soc. (London), 235, Series A, 544-551. 2. Tamm, Ig. (1939), "Radiation Emitted by Uniformly Moving Electrons, " J. of Phys. (USSR), Nos. 5-6, 439-454 3. Sayied, A. M. (1958), "The Cerenkov Effect in Composite Media, " Proc. Phys. Soc. (London), 71, 399-404. 4. Zelby, L. W. (1962), "The Theory of Cerenkov Effect Based on Lorentz Transformation, " J. Appl. Phys., 33, 2995-2998. 5. Frank, I. M. (1943), "Doppler Effect in a Refractive Medium, " J. of Phys., (USSR), VII No.2, 52-67. 6. Tai, C. T. (June 1965), "Radiation in Moving Media, " Lecture Notes for Summer Antenna Course, The Ohio State University. 7. Lee, K. S. H. and C. H. Papas (1964), "Electromagrnetic Radiation in the Presence of Moving Simple Media, " J. Math. Phys., 5 1668-1672. 8. Sommerfeld, A. (1964), Partial Differential Equations in Physics, Academic Press. 9. Weyl, H. (1919), "Ausbereitung elektromagnetischer Wellen iiber einen ebenen Leiter," Ann. Physik, 60 481-500. (An account of'Weyl's article may be found in; Stratton, J. A. (1941), Electromagnetic Theory, McGrawHill Book Company, New York. ) 10. Sommerfeld, A. (1964), Electrodynamics, Academic Press. 11. Papas, C. H. (1965), Theory of Electromagnetic Wave Propagation, McGrawHill Book Company, New York. 12. Bremmer, H. (1949), Terrestrial Radio Waves, Elsevier Publishing Company. 13. Ott, H. (1942), "Reflexion und Brechung von Kugelwellen; Effekte 2. Ordnung, Ann. Physik, 41 443-466. 14. Ott, H. (1943), "Die Sattelpunktsmethode in der Umgebung eines Pols mit Anwendungen auf die Wellenoptik und Akustik, " Ann. Phys., 43 393-403. 103

THE UNIVERSITY OF MICHIGAN 7322-2-T 15. Nomura, Y (March 1953), "On the Theory of Propagation of Electric Waves over a Plane Surface of Homogeneous Earth (on Sommerfeld's Surface Waves), " Tohuku University Research Institute Scientific Report, B - (Elect. Comm.), 5, No. 3-4, 203-214. 16. van der Waerden, B. L. (1951), "On the Method of Saddle Points, " Appl. Sci. Res, B2, No., 33-45. 17. Cohen, M. H. (1961), "Radiation in a Plasma-I: Cerenkov Effect," Phys. Rev., 123, 711-721. 104

THE UNIVERSITY OF MICHIGAN 7322-2-T APPENDIX A POINT CHARGE IN MOVING MEDIA: CERENKOV RADIATION The problem of a point charge in uniform motion can be successfully treated in two ways. In one, the frame of reference is chosen to be at rest with respect to the medium whereas in the other, it is chosen to be at rest with respect to the charge. Nag and Sayiedl used the latter approach to derive Frank and Tamm's formula for Cerenkov radiation. Using the same approach, we will now show that the fields can be derived in a more direct and simpler fashion. In the case of the moving medium considered in Chapter II, for an observer in the unprimed system, the fields due to a point charge q located at the origin satisfy Vx E= 0 (A. 1) VxH= 0 (A.2) V. D = q 6(x) 6(y) 6(z) (A. 3) V B= 0. (A. 4) Introducing vector and scalar potentials defined by B = Vx A (A. 5) E =- V. (A. 6) Nag and Sayied have shown that aA+1 +2 - 6 6(x) 6(y) 6(z) (A.7) ax2 a ay2 2 a A =o, A =-Q2, A =0 (A. 8) where a _2 v l-n2 32 = (n2-1) n = ( c )l/2 (1-n212) c bo Co 105

THE UNIVERSITY OF MICHIGAN 7322-2-T The solution of (A. 7) when a > 0 yields the usual Lienard-Wiechert potentials. The case a-< 0, i. e. n22 > 1, corresponds to the Cerenkov effect and a formal solution is still possible if one recognizes that the form of the equation now resembles a two-dimensional wave equation the Green's function of which is known. Following Cohen17, we have q 2 2 f- a+ 1/2 >(x2+z2)1/2 =O ~if+ - yllay (x2+z2)l/2?47Tac ~2(x~2+Z.~' (A. 9) =0 if+ 1/2 a (22)1/2 where a = lal and + sign gives retarded, and - sign advanced potentials and the physics of the situation helps us pick the correct one. Since the Cerenkov cone trails behind the particle, it is clear that the retarded potential is appropriate when the velocity is in the positive y-direction and advanced potential for the negative y-direction. Since (A, io/c;) transforms like a 4-vector, we have in the primed system = (-vA)= e2q [(y+vtt)2(n2_ 2 l)(x,2+z,2)] 1/2 y 4i E if (y'+vt')>(n22-_1))12(x'2+z'2)1/2 (A. 10) =0 if (y1+vt')<(n2p2-l)l/2(x'2+z'2 )1/2 and A, =:(A -;~ 2 n v' ^,(A. 11) To the primed observer the charge appears to move in the negative y' direction and the potentials are given by (A. 10) and (A. 11) when the velocity exceeds the critical value c/n. These formulas check with those of Frank and Tamm. Further, it may be noted that the potentials in the primed system satisfy the 106

THE UNIVERSITY OF MICHIGAN 7322-2-T gauge condition V'. A'+n a =0 (A. 12) but the same is not true in the unprimed system since the gauge condition is not invariant unless n = 1. 107

THE UNIVERSITY OF MICHIGAN 7322-2-T ABSTRACT Two boundary value problems in the electrodynamics of moving media are solved in this dissertation. In both problems, there is a lossless dielectric filling one half space and moving parallel to its surface with a uniform velocity; the remaining half space is vacuum. The primary problem involves the determination of the radiation field due to an oscillating dipole located in vacuum. A secondary problem, namely reflection and refraction of a plane wave striking the moving dielectric, is solved as a preliminary to the more difficult problem above. The motivation for the present study is to introduce techniques for formulating boundary value problems in the electrodynamics of moving media and to ascertain if any corrections are warranted in practical problems of similar nature where the velocities are quite small compared to that of light. Following the well known work of Sommerfeld, Minkowski's theory of the electrodynamics of moving media is developed. A modified set of vector and scalar potentials appropriate in moving media is introduced. These potentials are found to have closed form solutions. Starting from the Maxwell-Minkowski equations, plane wave solutions in moving media are determined. Once this is accomplished, the solution of the reflection-refraction problem is found to be quite straightforward. Certain interesting features are revealed. First, Snell's law is modified, and the extent of this modification is indicated by a set of graphs depicting the angle of incidence versus the angle of refraction for different velocities and indices of refraction of the dielectric. Secondly, the reflected and transmitted waves possess components not present in the incident wave and furthermore, when the incident wave is polarized with its electric field parallel to the plane of incidence, there is no angle of incidence (Brewster's angle) for which the reflected wave vanishes. An exception to these results occurs when the plane of incidence is parallel to the velocity. In this case, there is a strong resemblance to the non-moving case. Exact expressions for the fields when the plane of incidence coincides with the two principal planes (perpendicular or parallel to the velocity) are given. 108

THE UNIVERSITY OF MICHIGAN 7322-2-T The dipole problem is considered next. The two cases of a vertical and a horizontal dipole over a moving medium are treated in detail. In each case, the problem is formulated in terms of double Fourier integral representations of the potentials appropriate in each region and a formal solution is obtained. An alternate formulation (the method of Weyl) is presented for the case of the vertical dipole in which all fields are expressed as integrals of the plane waves. The purpose of this is to emphasize the connection between the reflectionrefraction and dipole problems. Using the saddle point method, asymptotic forms for the fields are obtained. Electric field patterns in the two principal planes are included. It is observed that in order to produce any perceptible change in the radiation patterns, the velocity must be comparable with that of light. Some allied problems for future research are suggested. 109

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