Radiation Due to an Oscillating Dipole over a Lossless Semi-Infinite Moving Dielectric Medium, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Radiation Due to an Oscillating Dipole over a Lossless Semi-Infinite Moving Dielectric Medium and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! 7322-2-T COPY - THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL ENGINEERING Radiation L a bototory Rudiution Due to an Oscillating Dipole Over u Lossless Semi-lnfinite Moving Dielectric Medium b y VITTA1 P. PYATI , r- r[ . s February 1966 - Grant NGR-23-005-107 National Aeronautics and Space Administration Langley Research Center Langley Station Hampton, Virginia 23365 Contract M t h : Administered through: OFFICE O F RESEARCH ADMINISTRATION A N N A R B O R b r' THE U N I V E R S I T Y O F , ~ I C H I G A N 7322-2-T RADIATION DUE TO AN OSCILLATING DIPOLE OVER A LOSSLESS SEMI-INFINITE MOVING DIELECTRIC MEDIUM I * . by Vittal P. Pyati February 1966 Report No. 7322-2-T on Grant NGR -23-005-107 Prepared for NATIONAL AERONAUTICS AND SPACE ADMINISTRATION NASA-LANGLEY RESEARCH CENTER LANGLEY STATION HAMPTON, VIRGINIA T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T TABLE OF CONTENTS FOREWORD A CKNQWLEDGEMENT LIST OF ILLUSTRATmNS CHAPTER I I1 ELECTRODYNAMICS O F MOVING MEDIA INTRODUCTION AND STATEMENT OF THE PROBLEM 2.1 The Lorentz Transformation 2.2 Maxwell-Minkowski Equations 2. 3 The Method of Potentials for Moving Media III REFLECTION AND REFRACTION OF A PLANE ELECTRO- MAGNETIC WAVE AT THE BOUNDARY OF A MOVING DIELECTRIC MEDIUM 3.1 Geometry of the Problem 3.2 Plane Waves in Moving Media 3. 3 The Modified Snell's Law 3.4 Electric Field Perpendicular to the Plane of Incidence 3.5 Electric Field Parallel to the Plane of Incidence 3.6 Perpendicular Incidence 3.7 Summary OSCILLATING DIPOLE OVER A MOVING DIELECTRIC MEDIUM 4.1 Introduction 4.2 Vertical Dipole IV 4.2.1 Fourier Integral Method 4.2.2 Method of Weyl 4.2.3 Approximation of the Integrals; Asymptotic Forms 4.2.4 Numerical Results 4. 3.1 Fourier Integral Method 4.3.2 Approximation of the Integrals; Asymptotic Forms 4.3.3 Numerical Results 4.3 Horizontal Dipole in the Direction of the Velocity V CONCLUSIONS REFERENCES APPENDIX A: POINT CHARGE IN MOVING MEDIA; CERENKOV RADIATION ABSTRACT Page ii iii V 1 3 3 4 6 9 9 9 14 23 31 36 39 40 40 40 40 49 55 77 84 84 87 93 101 103 105 108 iv J . T H E U N I V E R S I T Y O F M I C H I G A N 7322-2 -T LIST OF ILLUSTRATIONS Fig. No. 1. THE COORDINATE SYSTEMS 2. PLANE WAVE INCIDENCE ON A MOVING MEDIUM 3. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n = 2 , @.=O ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n = 2 , @. =45O 1 4. 1 5. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n = 2 , 6. = 900 1 6. ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=O.5, p . = 0 ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=O.5, 8. = 45O ANGLE OF REFRACTION VS ANGLE OF INCIDENCE FOR n=O.5, @.= 90° THE k , y , z ) and kl,y,,z) COORDINATE SYSTEMS 1 7. 1 8. 1 9. io. BREWSTER'S ANGLE vs p FOR n = 2, p.= 90' 1 11. DIPOLE OVER A MOVING MEDIUM 12. PHYSICAL INTERPRETATION OF THE DIPOLE PROBLEM 13. 14. DIPOLE SOURCE AND IMAGE PATH OF INTEGRATION IN THE cy1 PLANE 15. ILLUSTRATION OF SADDLE POINT METHOD IN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n = 2, = 0 (values in dielectric have been scaled down by a factor of 2) IN THE AIR IN THE YZ PLANE FOR A VERTICAL DIPOLE R n = 2 , h=O.25X Page 3 10 17 18 19 20 21 22 24 37 41 51 52 57 72 78 79 V T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T LIST OF ILLUSTRATIONS (CONT'D) Fig. No. Page IN THE AIR IN THE Y Z PLANE FOR A VERTICAL DIPOLE R n = 2 , h=O.5X 80 19. pd IN THE XZ PLANE FOR A VERTICAL DIPOLE FOR n = 0.5, h = O 81 IN THE AIR IN THE XZ PLANE FOR A VERTICAL DIPOLE n = 0.5 , h = 0.25X 82 21. I I N THE AIR IN THE Y Z PLANE FOR A VERTICAL DIPOLE F$R n=O.5, h=O.25X 83 IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n = 2, 94 IN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE R n = 2 , h=O.25X 95 24. I IN THE AIR IN THE Y Z PLANE FOR A X~ORIZONTAL DIPOLE F $ R n = 2 , h=O.5X 96 25. E IN THE XZ PLANE FOR A HORIZONTAL DIPOLE FOR n =O. 5, b A0 97 26. b61 IN THE YZ PLANE FOR A HORIZONTAL DIPOLE FOR n =0.5, h = O 98 IN THE AIR IN THE XZ PLANE FOR A HORIZONTAL DIPOLE R n = 0.5, h=O.25X 99 IN THE AIR IN THE YZ PLANE FOR A HORIZONTAL DIPOLE n = 0.5 , h = 0.25X 100 vi 8 # i W' T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T CHAPTER II ELECTRODYNAMICS OF MOVING MEDIA 2 . 1 The Imentz Transformation Consider two coordinate systems as shown in Fig. 1, in which the y and y1 axes coincide and the system S' is mbving tKith a uniform vdodty v in the y-direction with respect to S. For the case when the two origins 0 and 0' coincide at the Z' S' V Y L! Y' 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 I y' = y(y-vt) y = y(y'+vt') x' = x, z' = z x = x', z =z ' t' = y(t- gx) t = y(t '+Ex') The above is known as the Lorentz or the Lorentz-Minkowski transformation. The various constants appearing above are given by c = (Po€o) -' = velocity of light in free space or vacuum eo = permittivity of free space = ( 3 6 a ~ l O ~ ) - ~ farads/m. p, = permeability of free space = 4 ~ x 1 0 - ~ henries/m p = v/c . 3 T H E U N I V E R S I T Y O F M I C H I G A N 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 - at V X E = -- a s at V x H = J + - - - (2 .3) (2.6) and by attaching primes, we get Maxwell equations in the primed system, for instance, (2.2) becomes 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 ' = p€J' (2.8) where E: and p 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. . . r T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T where -. v=vy" . E+? v x g = € ( _ E + I X _ B ) Substituting (2. 7) and (2.8) into the above, we obtain 1 (2.9) (2.10) (2.11) (2.12) (2.13) - B - 2- V x_H = p(_H --x g (2.14) These two relations were first derived by Minkowski in 1908. Solving for _B and _D in terms of _E and g , we obtain the desired constitutive relations n = ( - E )* = refractive index in the primed system pOE0 1-82 a = 1-n2 6 2 ' % = w2c10E, (2.15) (2.16) The system (2.2) - (2.5) can now be solved once the sources are specified by the method of potentials discueeed in the next section. To illustrate the usefulness 5 T H E U N I V E R S I T Y O F M I C H I G A N 7 322 -2 -T solution is given by where . u (2.32) If the medium is moving in the negative y-direction, replace (y-yo) by (y,-y). Once the Green's function is known, the solution for A in (2.29) is given -1 bY (2.33) and the electric and magnetic Cields are given by (2.34) (2.35) One final word is necessary. The vector and scalar potentialsL1 and f1 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. 8 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T CHAPTER III REFLECTION AND REFRACTION OF A PLANE ELECTROMAGNETIC WAVE AT THE BOUNDARY OF A MOVING DIELECTRIC MEDIUM 3 . 1 Geometrv of the Problem ~~ ~ A s shown in Fig. 2, the region z < 0 is filled by a medium, with a per- meability p, and a permittivity E, moving uniformly in the y-direction with a velocity v. The region z > 0 is free space bo, eo) and is stationary. A plane electromagnetic wave traveling in free space in an arbitrary direc- tion is incident upon the interface; as a result there will be a reflected wave and a transmitted wave. azimuthal angles being measured from the x-axis. Let the orientation of the three waves be as in Fig. 2 , the 3 . 2 Plane Waves in Moving Media In order to represent the transmitted field, we need plane wave solutions in the moving medium. for e-iwt variations are given by The Maxwell equations in the absence of sources and VxE= - i w g (3 .1) VxH_ = -iwD_ (3.2) V * B = O , - V * D = O (3 .3) Making use of the constitutive relations (2.15) and (2.16) we obtain The plane wave solutions which we a re seeking can be represented thus i K K i K K E = E e H = H e -0 - -0 - (3.4) (3.5) (3 .6) (3.7) where the first factor denotes complex amplitude, K the propagation constant and 9 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T Reflected Transmitted FIG. 2: PLANE WAVE INCIDENCE ON A MOVING MEDIUM 10 b - Y c T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T The amplitudes in (3.23) must satisfy the relation (3.16), so that klE" +$E'' -k3E" = 0 . ox OY 02 Comparing (3.24) with (3.9), we get from the phase functions (3.25) K sin Otcos f t = kl K sin Opingt = (k2-ws2) K COS et = kg (3.26a) (3.26b) (3.264 substituting for kl, %, k3 in (3.22) we get the following dispersion relation for K K 2 2 sin et 1 (K sinet s ingt+ua) 2 2 2 + K cos et-ak2 = o (3.27) , a Solving for K /ko , the refractive index of the moving medium, we get (3.28) where CY is the angle between the direction of propagation and the velocity of the medium, (cos a = sin 0 sinp ) This expression checks with that of Papas" (see page 231, Eq. 61 ). We also note that the amplitudes in (3.9) will have to satis- fy the following relation t t -- (3.29) 1 K sinOtcOS$tE +- (KshOtsingt+wQ)E -KCOSOtE = O OX a OY 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 T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T Summary: To construct plane wave solutions in the moving medium, we 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 Alternatively, one can start with the magnetic vector by setting and obtain the electric field from everything else remaining the same except replacing E by H in (3.29). 3. 3 The Modified Snells Law The incident and the reflected waves satisfy for e-iwt variation Vx E - = iwB_ Vx H - = -iwD - and the constitutive relations being V * B = O - V * D = O - B _ = p H E=EoE_ . O--’ (3.30) (3.31) (3.32) (3.33) The plane wave solutions are well hown and the phase functions are of the form where k 2 = w p 2 E 0 0 0 5 = x sin 8 cos g.+y sin eisin gi-z COS e. i 1 1 rl = x sin e,cos 8 +y sin Orsin gr+z cos e, . r (3. 34) (3.35) 14 . r a T H E U N I V E R S I T Y O F M I C H I G A N 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 wave8 must be identical when z = 0. This is possible only if (3.36) I I Physically this means that the three waves are coplanar (the plane g=gi will be called the plane of incidence), and the angle of reflection is equal to the angle of incidence. "heee two results am no different from the case in which the lower medium is not moving. The &ell'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 R, we get after some simplification (3.37) which is the modified Snell's law. We note that except when fi=O, T, the formula is affected by a change of sign of . The angle of total reflection ca.n be obtained by setting sin0 t=l and solving for Oi. For the case of Bi = 0, we get (3.38) Excluding the non-moving case (PO), from the inequality 15 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2 -T 3: 3c 2: n cn a8 a# CI, L - a8 Y 2c 5 I O 5 p.0.75 // Bi (degrees) 18 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T 55 50 45 40 35 3c 22 20 I, e; IC 5 15 30 45 60 75 90 8. (degrees) I 19 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2 -T 90 80 70 IC p =t0.3 I 5 IO 15 20 25 30 . ei (degrees) 20 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T c 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 c t The two coordinate systems are connected by the following relations (3.39) It is easier to work with the (x, y, z) system instead of (xl, yl, 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 5 0 E.=(I I , 0)e 1 1’ 2 (3.41) where I1=Eosinvi, $=-Eocosgi and -E given by (3. 34). The magnetic field is given bY so that in component form (suppressing phase factors) 23 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2 -T 1 X J X FIG. 9: THE (x, y, z) ANI’ (xl, yl, z) COORDINATE SYSTEMS. I 24 I THE U N I V E R S I T Y O F M I C H I G A N 7322-2 -T The reflected field can be represented by where t) is given by (3.35) and the amplitudes satisfy the condition 0- E = 0 which yields -r (3.43) R sine cos9 r 2 +R sin6 r singr+R3cos8r= 0 . (3.44) 1 r The magnetic field is given by so that in component form we have (suppressing phase factors) The transmitted wave can be represented by i K S = (T T T )e Et 1’ 2’ 3 where < is given in (3.8) and the amplitudes satisfy 1 K sinetcosVt T1+ a ( Ksin8tsin$t+w S2)T2- K cos 8 t T 3- 0 The magnetic field is given by (3.46) (3.47) 25 T H E U N I V E R S I T Y O F M I C H I G A N 7 322 -2 -T continuity of tangential E _ (E) and Snell’s law ensure the continuity of normal E@)). The solution for the six h o w n s related by Equations (3.44)’ (3.47) and (3.50) - (3.53) can be conveniently carried out as follows: R =T -I 1 1 1 R =T -I 2 2 2 (3.59) (3. 60) (3.61) T T T satisfying 1’ 2’ 3 (3.62) T2 T sin ne c0sg.t sine ne sinV.+wa)-T Kcose =O 1 t i a t 1 3 t 2 k 0 K kO k tane.sine.sing.cos V.+T (--case + - c o s e i + s taneisineisin gi) 1 1 1 1 2 a l - C tCco PO (3. 63) k K 2 k , +T2 tanOisineisingicosgi+T3 -sinetcosfi = - Ilcosei . (3. 64) PO P PO The solution of the above equations for the general case is quite tedious though not impossible. We will consider only two special cases; when Vi= 0, 90 and also set I-( = po. 0 Case 1: p = po vi = o (3.65a) 28 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T The fields are given by iEo(xsinei- z cos e,) -ut E.=(O, -Eo, 0)e 1 7 - E =(R -r 1’ E =(T t 1’ . The amplitudes of the transmitted and reflected waves are given by cos B.(sec O.+b cos 8 ) 3 1 1 t T = - - - 2Eo’ (n2-1) R =T 1 1 R =T +E 2 2 0 R ~ = -T tan8 1 i where (3. 65b) (3. 66) (3.67) (3.68) (3. 69) (3.70) (3. 71) (3.72) (3.73) (3. 74) (sec ei+b cos et) b ’2(2 - 1) M=b( - COS e +COS ei)(b+cos etsec ei)+ a t ( 1 -B2 (3.75) 29 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T . Making use of (3.65) to eliminate et' we get the following convenient expressions for R1 and % 0 2 2E R 1 N =- p(n2-l)sin eicos ei (3. 72a) A significant feature of the above results is that the reflected and transmitted waves have components not originally present in the incident wave. 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 Because of this, the An exception, however, occurs when vi= 90' which is discussed next. Case 2: p = p g. = 900 + 0' 1 sinei (3. 76) and K can be determined from the relation K sin Bt = ko sin ei . + This case was considered by Tai before. Actually, this Chapter is an extension of Tai's work. 30 - T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T - The transmitted field can be represented by %= (T1, T2, T3)eiK' where 5 is given by (3.8) and the amplitudes satisfy 1 K sinetcosfl T + - ( K s i n O sing +wQT -Kcose T =O t l a t t 2 t 3 The electric field is given by so that in component form (stppressing phase factors) 1 E = - [T sine C O S ~ +T cos et t y W € 3 t t 1 (3.88) (3.47) (3.89) The boundary conditions (3.49) lead exactly to the same set of equations as before except that we replace 1.1, by co and p by E . Thus R =T -I (3.90) R =T -I (3.91) 1 1 1 2 2 2 R 3 =-tane.(T 1 1 COS gi+T2 sin 92 (3.92) T , T , T satisfying 1 2 3 (3.93) T2 T K sine cosf.+ - (Ksine sinf.+wQ)-T Kcose =O 1 t i a t 1 3 t 1 Eo 1 a€ T k0 -tan B.sin Bisingicosfi+T2(L cos 8 + kO - cos e.+ 5 taneisineisin 2 Vi) €0 €0 T3 2ko €0 + - ( K S h e sinf.+wO)= - I ~ C O S ~ ~ a€ t 1 (3.94) 33 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T T (-we. k0 sineicos 2 f.+ kO - cosei+ - K coset) E €0 1 €0 0 K 2k0 k - tanOisinOisingicosgi+T - sine cosf. = - I cosOi 3 E t 1 E 1 0 (3.95) As in the cme of perpendicular polarization,we will consider only two cases of incidence; when f . = 0, 90’ . 1 Case1: p = p g i = 0 0’ The expressions for K and 8, are given by (3.65) and the fields are given by L i [ko(x s i n q z cosOi)-Wt] H.=(O, -Ho, 0)e T 1, 2’ 3 1 i (x sine.+z cose.)-d] H =(R R R )e 0 1 1 The amplitudes of the transmitted and reflected waves are given by T = - - 2HoP (2-1) cose.(n 2 secei+b coset) 3 (1-8) 1 R =T 1 1 R =T +H 2 2 0 3 1 i R = -T tme where b = ( (3. 96) (3.97) (3.98) (3.99) (3.100) (3.101) (3.102) (3.103) (3.104) . T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T b (secei+ - cos 6 ) p2 (2- 1) (1-Pl2 n 2 t + (3. 105) Eliminating 6 we get the following convenient expressions for the reflected field. t’ 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 gi=900, which is discussed next. Case 2: ,u = p 0’ 1 The expression for et is given by (3. 76) and K can be found from the modified g.=9o0. Snell’s law. The fields are given by - - i L o ‘ k ~sine.-zcosei)-wtJ 1 A 7 0 H.=xH e (3. 106) Eliminating K and Bt we get for the reflection and transmission zoefficients (3. 108) 35 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T A -i(KZ+Wt) E =xE T e -t 0 1 k# - W o R = 1 k p + K p 0 0 Case 2: H. in the x-direction. 1 H . = h o e -i(k++wt) 1 A ko - i (k, z+&) E.=y - 1 WEo Hoe i(k0z-d) H k i(k0z-d) H =?H R e E =-?&onl e -r w Eo -r 0 1 2k E where 0 T = 1 koE+ K k E - K E 0 0 0 0 R = 1 k E + K E (3. 114) (3. 115) (3. 116) (3. 117) (3. 118) (3. 119) (3. 120) (3.121) (3.122) (3. 123) (3. 124) (3. 1'5) (3. 126) (3. 127) T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T The situations are similar to the non-moving case. Furthermore, in Case 1 and in Case 2: (3. 128) (3.129) These two ratios are not equal contrary to the non-moving case; the deviation is, however, of the order 3 since (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. 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 90' (or 180°), 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). 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. The resuiting wave Finally, a word of 39 T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T CHAPTER IV OSCILLATING DIPOLE OVER A MOVING DIELECTRIC MEDIUM 4.1 Introduction The geometry of the problem is shown in Fig. 11. This is similar to that at a height of Fig. 2 except that there is an oscillating dipole of moment 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 a re 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 9 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 III. 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 pat- terns 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. T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T Green’s function for the Helmholtz equation which has the following integral representation due to Sommerfeld where R1= [Tx2+y2+(z-h)g @ . The secondary excitation can be represented by the integral (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 dis- cussed in Chapter 11 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 where Al=[p;+;-akq 4 Ib , ReA120 . We a r e now ready to formulate the problem. It m y be pointed out that in the non-moving case considered by Sommerfeld, the z-component of A - 43 T H E U N I V E R S I T Y O F M I C H I G A N 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 (4.9) ? Ao(z-h)+Fze-"l e i(PlX+P2Y) dPldP2 (4.10) 2 where C = -iwp0m/87r . The sign convention in the primary excitation should be chosen s o as to ensure the convergence of the integrals. + sign for 0 < z < h (Region 2) - sign for h < z < 00 (Region 1) (4.11) b) Lower Half Space (4.12) (4.13) The problem can now be solved, in principle at least, since all the four unlmown functions can be determined from the following boundary conditions. a) continuity of tangential E - and H_ b) continuity of normal D - and B - (3.49) In Chapter III, where w e considered the problem of reflection and refraction of a plane electromagnetic wave at the boundary of a semi-infinite moving medium, 44 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-'I' l - 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 Fourierintegral 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. (4.14) (4.15) (4.16) (4.19) 45 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T We note that these two relations can be obtained from (4.29) - (4.32) thus 2 (4.33) = ip2(4. 29)-tp1(4. 30) (4.34) = p2(4. 31)+(4. 32) Setting p = poJ solving the system (4.29)- (4. 3 a J we get for the unknown functions (4.35) 0 h (4.36) (4.37) (4. 38) where iwpo m 87r c = - 2 and (4.39) 48 T H E U N I V E R S I T Y O F M I C H I G A N 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 alter- nate 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 inte- gration in (4.14) - (4.25) to polar coordinates but first a few remarks are necessary. In Chapter In, 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 inci- dence tbugh 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 . T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T induced in the dielectric. Referring to Fig. 12, one may also interpret t?iese two contributions as a direct field and an indirect o r reflected field reaching an obser- ver at a point P. With a view to express these fields as integrals of elementary plane waves, consider the integral representation (4.6) Introducing polar coordinates defined by p1 = k sin a COS/^^ 0 1 1 p = k sincr s in0 2 0 1 where a is complex and 1 (4.40) p, real, varying from 0 - 2n, the above relation becomes m ib p ? ko(h-z)cos QJ (4.41) ikoR1= 5 1; -ioo - dR1 e - 2T ' e R1 0 where f : O < z < h -: h < z < m dR = s i n a d a dp 1 1 1 1 p = ko(x sin a cos - 1 1 +y sin crl sin p,) (4.42) and the path of integration in the complex a plane is as indicated in Fig. 13. The integrand in (4.41) is easily recognized as a plane wave in the directions al, 0,; in fact (4.41) represents the spherical wave function as a superposition of plane waves with real directions for which 0 < cy < - and complex directions for 1 7r 1 2 50 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T . where 5 = x s i n a cos /3 +ysina s in0 -z cos a 1 1 1 1 1 I = - - G s i n a s inp k 0 1 c 2 ko sinal cos p1 . (4.44) 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 71 ikor) (4.45) dR1 T =rrm (R1> R2> R3)e 0 0 where r ) = x sin CY cos Pl+y s in (Pisin Pl+z cos al 1 R = - i 3 kocos al(sinalsin/31F,-cos alF Po Y R2= - - i 3 k sinalcos ctlcos PIF, Po O i 3 R3= - k sinalcos a cos p F . I-10 O 1 1 Y (4.46a) (4.46b) (4 .46~) We also note that R sin altos P1+Rpin Q sin P1+R3 cos al= 0 . 1 1 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 -iw Ry waves is slightly involved. By bringing the factor e in accordance with the translation property (4.26), the phase function will assume under the integral sign the form i(plx+p2y)+~~ z e where i h (4.47) 5 3 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T In order that the above may represent a plane wave in the direction %, p2 in the moving medium as shown in Fig. 12, we make use of the modified Snell's law "2 k sin a = K sin 0 1 P, = P2 and since K satisfies (3.27) with et = %, f t = p2, we get .II A''. = -i K cos CY 1 2 the negative square root being chosen to ensure the convergence of integrals. The phase function (4.47) now becomes i~ [xsincr, cos p 2 +y sin % 2 s i n p -z cos 4 e (4.48) which has the desired form. Using the above relations in (4.23) - (4.25), we get rr f T k -ia> i u 5 (4.49) where < = x sin %cos p +y sin a sin ,!I -z cos CY 2 2 2 2 { ( K s h %Sin p2 -tW R)G"'+ K cos %G'" Y ik2cos a 0 T = 1 a/J Z 2 ik cos& T = - K sina, cos p G::: 2 w 2 2 z 2 Y 2 K s i n q ~ c o s G::: . 1 ik cosa 0 T = 3 aC1 We also note that T2 T K sin@ cos@ + - ( K s i n a s inp +wR)-T C O S Q = 0 . 1 2 2 a 2 2 3 2 (4.50a) (4.50b) (4 .50~) The integrand in (4.49) may now be identified wit11 the transmitted field (3.88). 5 4 T H E U N I V E R S I T Y O F M I C H I G A N 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 compli- cated 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 Bremmer and others . The results are 12 in which A i s the Hessian determinant 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 T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T where We will now use the result given by (4.51) to obtain an asymptotic expansion of I. Introducing polar coordinates defined by (4. 56) ( z + ~ ) = R ~ c o s e2 J where R2 is the distance from the image point as shown in Fig. 14, the integral becomes R2f I= lJ A(P1JP2)e dPldP2 where f = i(p sin 8 cos pkp2sin e2sin @-(p2-ki)pcos 6 4 c 1 2 Setting the partial derivatives equal to zero, we get = E sin e2cos f- *l pFose21 = O ' G- =[isin, s in9 - ap2 @q Saddle point s occurs at p =k sin8 cos 9 p =k sine s ing 1 0 2 2 0 2 provided we A = - S take (P2-kzp= -ik 0 cos 8 2 ' A direct calculation shows that 1 2 2 k cos O2 0 58 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T and n i J A , = - 2 k cos8 0 The reason for choosing negative square root will be given shortly. Thus, the asymptotic expansion of (4.55) becomes dPldP2 ikoR2 -00 e N 2 k cos 8 A(k sin 8 cos fJ k sin 02sinf) as % t a0 - (4.57) 0 R2 i o 2 0 2 As an example, consider the first term in (4.14) which corresponds to the primary field. For z > h, we get mkz ikoRl E = -sin e COS e COS 9 e Xp 1 1 4a Eo% which is the same as that given by (4.53). Hessian determinant has been chosen to yield consistent results. The contribution due to the saddle point yields for the reflected waves Thus the negative square root in the ikoR2 2 2 (4.58) i w o m e 2 R2 Ds - cos e sin e Q c sin e2+sin 9 [ ( c z ~ ) +an (l-ad 2 2 A = y r 47r 2 2 an -sin 02) -iwp m e *oR2 0 A = zr 47r R2 *cos e2 an -sin2e2cos 2 1 9- - ( ~ c b i n e2sin@ c 2 a -a(l-a)n 2 s i n 2 ~ ~ s i n 2 f - n c sin e2sin g(l+sin 2 e2+m sin e2siii 9) 1) (4.59) 59 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T where D = (an -sin e )+an 2 [a-sin 2 e2(acos 2 #+sin 2 S I 2 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: $! = 0 (upper sign) or 9 = 180° (lower sign) Primary Field y p = O E d2 ikoRl E = sin el- - 2 o e ZP O R 1 60 (4. 62) (4.63) (4.64) (4.60) The fields can be obtained either by applying the saddle point method directly to the integral representations given by (4.14) to (4.19) or by substituting (4.58) and (4.59) into (4.2) and (4.3) provided the differentiations are replaced according to the following scheme. 1 - - a - ik s ine cos# ax 0 2 a - = ik s in8 s in9 a Y 0 2 az o - - a - ik COS e2 . (4. 61) H = O ZP mk2 ikoRl - O e 5 E = + s i n e c o s e Xp 1 1 47Mo T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T E = ; sine2c0s e2 1 - Y r c 2F 14 e i%R2 R2 z r (n2cos 02+F) 471. f0 where (4.79) (4.80) (4.81) Once again, we note that the above results, apart from constant factors, are the same as those in Chapter m, Section 5, for the case gi = 90°. Using the approximations (4. 72), we have for the total electric field E =O R mk2 ikoR -ihhcose %hcoseE- 2F(B) j\ -- 47r o eo e R 2 (4.82) +e n cose+F(e E G O * f b) Fields in the Lower Half Space. The unknown functions G and G in (4.12) and (4.13) are related to G ‘ I . and G‘” given by (4.37) and (4.38) as follows. J, Y Z Y Z A convenient set of polar coordinates in th is half space a re x = R sin$cosg y = R sin@sing (4.83) (4.84) .e, z = -RCOS$ where @ is the angle shown in Fig. 14 and R, 9 have their usual meaning. Due to the presence of the factor e-’oh in G::: and G% , integrals of the following type Y 6 3 T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T are involved in the determination of A and A . Y Z Rf(P1, P,) 00 I = dPldP2 where h f = E(plsin$ cos sin$ sin g)-\cos - E x~:J 2 (4.85) In 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 The solution of above when h # 0 is quite difficult and will not be considered here. When h = 0, we get provided we take X1= i a q z k ( - ) Ra where (4.86a) (4.86b) (4.864 (4.87) T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T At the saddle point 0 R f = i a lbk ( 3 ) S For consistent results, we take Using (4.511, the asymptotic expansion of (4.85) becomes ia@ k Ra (4.88) a i Ein2+(cos2f+a si n2f)+cos2@] R 2 r a k c o s ~ A ( a @ kx/Ra, a 3F ky/R )e I N asR-,a> 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 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. Y 2 ax - - a - ia3b k( Y) aY Ra (4.89) Case 1): h=O, q = O (upper sign) or $= 180°(lower sign). \ (4.90) 65 T H E U N I V E R S I T Y O F M I C H I G A N 7322 -2 -T (4.109) 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 a re 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 \ and el% e2z e 5- (4.110) R 1 2 =R "i 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 S2, rearranging, we get for this expression which becomes negative if total reflection occurs. This would give rise to com- plications 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 im- perative that we examine the method of saddle points in greater detail. While distorting the given path of integration into the path of steepest descents throug!: 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. in Sommerfeld's original problem and have been thoroughly discussed by Ott These difficulties also occur 13,14 . 68 T H E U N I V E R S I T Y O F M I C H I G A N 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 con- siderably if one integration can be carried out exactly. We will, therefore, make some reasonable approximations to achieve this. First, a and 0 are expanded in Taylor series about B = 0 1 . s2- J n2 For low velocities, it is sufficient to retain only the first term, so that \ C (4.112) (4.113) I a% 1 Making use of these approximations in (4.35) and discarding higher order terms in P, we get 2 - b h 2iCWR-p e U F G k (A+ZA )(A+A ) 0 0 0 where I Because of the troublesome factor 2 w flop2 (4.114) (4.115) 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 P, we get 69 T H E U N I V E R S I T Y O F M I C H I G A N 7 322-2 -T -A h 2 0 2iCwQop e k2(A+r? Ao)(A+h0) F W 0 (4.116) which is the desired low velocity approximation. Similarly from (4. 361, we get Substituting for F and FZ in (4.9) and (4. lo), introducing polar coordinates defined by Y p =pcos v , 1 x = pcospl, p =ps inv y =ps ing 2 where p is the cylindrical distance and making use of the relations 7 J iZcos(v-@dv where J stands for the Bessel function of the first kind, we get -A (z+h) A = 0 P3dP 0 (4.118) (4.119) (4.120) 70 I T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T where 2 * , cos p 2 , w = (n -sin p ) A = (n cosp+w) Let us now examine the singularities of Ab) in the strip -a< p ~a . 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: For 1 a) Branch points of w defined by sin p = n; since sin p=sin(r -p), there exist four in number v , v , v and v of which v and v can be considered as the 1 2 3 4 2 4 reflection of v and v about the origin. For real values of n the branch points lie on the real axis if n < 1 and on the vertical lines p = ? '1 if n > 1. ponding to the two combinations of signs of w, the integrand is double valued and its Reimann surface has two sheets. These sheets a re connected with one another by the branch cuts along the lines Iw=O running from the branch points to OD as shown in Fig. 15 for the case Inl<l. If n is real and is less than 1, the branch cut emanating from v degenerates into a portion of the real axis from -sin-ln and the origin and the negative imaginary axis. Similar remarks apply to the remaining branch cuts. The upper (lower) sheet is specified in which Imwis greater (lesser) than zero. The path of integration L lies on the upper sheet. 1 3 Corres- 1 2 2 b) Four poles of first order obtained by setting the denominator equal to zero. A simple calculation shows that the poles are given by (4.127) Whether they lie on the upper o r lower sheet can be ascertained by examining the 73 T H E U N I V E R S I T Y O F M I C H I G A N 7 322 -2 -T relation W P 2 c o s p = - - - P n (4. 128) Let ia I 2 OSCY<- n = 1.. , (l+n )= ll+n2 I e2Q , /3< a 2 then Since Tm w > 0 i n the upper sheet, we have to choose the positive sign, so that P -iP in the upper sheet. 2 n Substituting in (4.128), we get Wp -i/3 c o s p = - I s l e , o , ( p < ; * P Since cosp=cos(p t ip2)=cosp coshp -i sinp sinh 1 1 2 1 % the position of the pole in the upper sheet is given by e Thus when n is real the pole lies on the real axis between ?r /2 and T , coinciding with 7r + io when n = 0, moving left as n increases and approaching 7r/2 + io as n -00. the discussion on the singularities of the function A in (4.126) The inverse point is also pole lying on the upper sheet. This completes In order to proceed with the saddle point method, the Hankel function in (4.126) is replaced by its asymptotic value 74 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T and introducing polar coordinates defined by (4.56) we get where The exponent ik R cosb-8 )=k R s i n b -8 )si* t icosh -8 )coshl 0 2 2 0 2 L 1 2 2 1 2 (4.129) (4. 130) has negative real part in the hatched area of Fig. 15 (++e (cc <8 above and 8 <p <a+e2 below the real axis) in which the above integral converges. A simple calculations shows that the saddle point is given by p =€I and the path of steepest descents is given by 2 1 2 2 1 s 2 i.e. cos (p -e )cosp2 = 1 1 2 Re cos(,,-8 ) = 1 , 2 I and is denoted by L in Fig. 15. A s the angle02 varies from 0 - - , 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. S 2 S If n is real and less than one, a finite portion of L, would lie on the lower sheet (indicated by broken lines in Fig. 15) for €I2(sin-'n. Since the condition Im(n2-sin 2 @ p) > O is not needed in (4.126), this is of no consequence. However, if 02> sin-In, the path of integration will have to go around the branch cut from v1 (details about which can be found in 0 tt's13 work). The branch cut integrations a r e not important in the present problem; hence will not be included . Similar 75 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T 0 0 u) 0 0 FC) 0 0 0 0 rr) 0 0 (D L .- 0 0 tD / , c 78 T H E E . ~ U N I V E R S I T Y O F M I C H I G A N 7322-2-T "0 CD d c\1 0 c IZ 0 Frc d t-c 0: W 79 T H E U N I V E R S I T Y O F M I C H I G A N 80 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T l - ~ , - 83 T H E U N I V E R S I T Y O F M I C H I G A N 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 Ee 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. - 1 (4. 137) 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. Jn the non-moving case considered by Sommerfeld both y and z-components of the vector potential were needed to satisfy the boundary conditions. similar to the vertical dipole problem. The same is true in the present case and the method of solution is a) Upper Half Space: A = I F e Z Z dPldP2 (4. 138) (4.139) where 84 I l - T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T A 0 =(p2-<?, ReAo>,O b) Lower Half Space: - - -OD -OD where 5 = (pl 2 + ,p2-ak2)’, 1 2 Re \) /O . The continuity of H H E and E at z = 0, yields the following set of equations. x’ y’ x Y (4.140) (4.141) (4.142) 85 T H E U N I V E R S I T Y O F M I C H I G A N 7322-2-T The fields in the two principal planes are given by: Case 1): f=O (upper sign) or $!=180°(lower sign). Primarv Field: E =O E = O ZP I ikoRl o e mwk xp 1 4%- H =-case - - R1 ik R H =O 0 1 YP + H = - sine - - o e mwk ZP 1 4a R1 Reflected Field: E zr =;Exrtan02 k, H = -COS e - " E xr 2 w n Y r v ikoR2 - 2 2 *o e H = + - /3(n -l)si118~cosO~ 7 - R2 Y r N 88 (4.154) (4.155) (4.156) (4.157) (4.158) (4.159) (4.160) THE U N I V E R S I T Y O F M I C H I G A N 7322-2-T E H = _ sin8 -0 k + z r 2 UPo Y r (4.161) 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 E =O R (4.162) Case 2):f=90~(upper sign) or 8=270~(lower sign) Primary Field: *ORl H =-case - - R1 o e 11u3k Xp 1 47r H =H = O YP ZP E =O Xp mk2 ikoRl 2 o e E =COS e - - YP 1 4 % s mk2* ikoRl - o e E = + s i n e cos0 - - ZP 1 147r€0 5 (4.164) (4.165) 89 T H E U N I V E R S I T Y O F M I C H I G A N '7322-2-'I' Reflected Field: 2F o e n cos6 +F xr 2 (4.166) H =H =O yr zr E =O xr 2F o e d 2 ikoR2 ]sine case - o e - E =+[I- 22' 2 2 4 n ~ g R2 n cos0 +F zr 2 (4.167) (4.168) where F is given by (4.81). Making use of the approximations (4. 72), we get for the total electric ficld in the upper half space ik R 0 (4. 169) -ik hcose ik hcose -e 0 0 b) Fields in the Lower Half Space. A s 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=O, f=O(upper sign) or f=180°(10wer sign). 90