Spherical Waves. Daniel S. Weile. Department of Electrical and Computer Engineering University of Delaware. ELEG 648 Spherical Coordinates

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1 Spherical Waves Daniel S. Weile Department of Electrical and Computer Engineering University of Delaware ELEG 648 Spherical Coordinates

2 Outline Wave Functions 1 Wave Functions

3 Outline Wave Functions 1 Wave Functions 2 The Spherical Cavity Radial Waveguides

4 Outline Wave Functions 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

5 Outline Wave Functions 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

6 The Scalar Helmholtz Equation Maxwell s equations are complicated in spherical coordinates. We postpone examining them for a moment and begin with a scalar Helmholtz equation. The Scalar Helmholtz Equation ( 1 r 2 r 2 ψ ) + 1 ( r r r 2 sin θ ψ ) + sin θ θ θ As usual we substitute 1 r 2 sin 2 θ ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) 2 ψ φ 2 +k 2 ψ = 0

7 Multiplying the resulting equation by r 2 sin 2 θ, and dividing by ψ gives sin 2 θ R We now let ( d r 2 dr ) + sin θ dr dr Θ ( d sin θ dθ ) + 1 d 2 Φ dθ dθ Φ dφ 2 +k 2 r 2 sin 2 θ = 0 1 d 2 Φ Φ dφ 2 = m2, substitute, and divide by sin 2 θ. This gives 1 R ( d r 2 dr ) + 1 dr dr Θ sin θ ( d sin θ dθ ) m2 dθ dθ sin 2 θ + k 2 r 2 = 0

8 We may now separate the θ equation. The strange separation constant is n(n + 1) is chosen because the form of Θ(θ) depends on whether or not n Z. 1 Θ sin θ ( d sin θ dθ ) m2 dθ dθ sin 2 θ = n(n + 1) Substituting this into the previous equation, we find ( 1 d r 2 dr ) n(n + 1) + k 2 r 2 = 0 R dr dr

9 Thus, after separation, we are left with the Spherical Separated Equations 1 sin θ d dθ ( r 2 dr dr d dr ( sin θ dθ dθ ) [ ] + k 2 r 2 n(n + 1) R = 0 ) + [n(n + 1) m2 sin 2 θ ] Θ = 0 d 2 Φ dφ 2 + m2 Φ = 0 Note that there is no separation equation here since two of the independent variables refer to angles!

10 Outline Wave Functions 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

11 The Harmonic Equation The only familiar equation above is the harmonic equation d 2 Φ dφ 2 + m2 Φ = 0 As usual, its solutions are e ±jmφ, cos mφ, and sin mφ. Again, m must be an integer if Φ(φ) must be periodic.

12 The Spherical Bessel Equation The radial equation is of the form ( d r 2 dr ) [ ] + k 2 r 2 n(n + 1) R = 0 dr dr The solutions of this equation are called spherical bessel functions b n (kr). This term is general; there are Bessel functions j n (kr), Neumann functions y n (kr), and Hankel Functions h (1) n (kr) and h (2) n (kr). The spherical functions are related to the cylindrical functions by π b n (kr) = 2kr B n+ 1 (kr) 2

13 The Spherical Bessel Equation Each function has the same properties as the corresponding cylindrical function: j n is the only function regular at the origin. j n and y n represent standing waves. h (2) n is an outgoing wave, h (1) n is an incoming wave. Spherical wave functions are actually expressible in terms of more familiar functions: j 0 (kr) = sin kr kr h (1) 0 = ejkr jkr cos kr y 0 (kr) = kr h (2) 0 = e jkr jkr The higher order functions can be found from the recurrence formula.

14 Spherical Bessel Functions j 0 (x) j 1 (x) j 2 (x) j 3 (x) j 4 (x)

15 Spherical Neumann Functions y 0 (x) y 1 (x) y 2 (x) y 3 (x) y 4 (x)

16 The Associated Legendre Equation The final equation is of the form ( 1 d sin θ dθ ) ] + [n(n + 1) m2 sin θ dθ dθ sin 2 Θ = 0 θ Defining x = cos θ (and y = Θ) we find ] (1 x 2 ) d2 y dy 2x [n(n dx 2 dx + + 1) m2 1 x 2 y = 0. Since 0 θ π, 1 x 1. One set of solutions is regular for n Z, we call them associated Legendre functions of the first kind P m n (x). The other set, associated Legendre functions of the second kind, Q m n (x), are singular at x = 1.

17 Legendre Polynomials If m = 0, and n Z the solutions become orthogonal polynomials; Legendre Polynomials of degree n. These are given by the well-known Rodrigues Formula d n P n (x) = 1 2 n n! dx n (x 2 1) n These polynomials are not orthonormal; they are normalized so P n (1) = 1. The first few are: P 0 (x) = 1, P 1 (x) = x, P 2 (x) = (3x 2 1)/2... How can I find the rest?

18 Associated Legendre Facts If ν / Z, the functions P ν (x) and P ν ( x) are independent. Therefore, if ν Z, the Q must be used. The functions P ν (x) (and, for that matter, P m ν (x)) for ν / Z are not regular at x = 1 either. When m 0, they are called Associated Legendre Functions. Of course, in practice, we use P m n (cos θ) and Q m n (cos θ). (We call these L m n (cos θ).) Finally, it is important to note that L m n (x) = 0 for m > n.

19 Outline Wave Functions 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

20 The Problem Wave Functions Our normal approach uses the scalar Helmholtz equation to solve for a vector potential component. This generally requires choosing a Cartesian component. In spherical coordinates, there is no Cartesian component! One approach is to set fields to be, say, TM z anyway. Then A = u z µψ = u r µψ cos θ u θ µψ sin θ, where ψ is a solution to the Helmholtz equation in spherical coordinates. Why is this approach unpopular?

21 A Better Approach Wave Functions Our approach will look for fields TM r and TE r by letting A = u r µa r and F = u r ɛf r. Our problem is that 2 A r ( 2 A) r so that A r is not a solution of the Helmholtz equation. To find the equation the vector potential solves, we go back to the beginning.

22 Spherical Vector Potential Because B = 0 and B = µh, we define A through H = 1 µ A From Faraday s Law, E = jωµh = jω A This implies or (E + jωa) = 0 E + jωa = Φ

23 Spherical Vector Potential Plugging this into the source-free Ampère-Maxwell law, we find ( ) 1 µ A = jωɛ ( Φ jωa) or Now assume Then A k 2 A = jωµɛ Φ. A = A r u r. 1 A r A = u θ r sin θ φ u φ 1 r A r θ

24 Spherical Vector Potential Now ( A) r = 1 r 2 sin θ ( A) θ = 1 2 A r r r θ 1 2 A r ( A) φ = r sin θ r φ [ ( sin θ A ) r ] A r θ θ sin θ 2 φ and [ Φ jωµɛ Φ = jωµɛ u r r + u 1 Φ θ r θ + u φ 1 r sin θ ] Φ φ

25 Spherical Vector Potential We can equate the u θ and u φ components immediately by choosing jωµɛφ = A r r. Plugging this into the u r equation gives 2 A r r r 2 sin θ 1 2 A r r r r 2 sin θ 2 A r r 2 r + 1 r 2 sin θ [ θ [ θ [ θ ( sin θ A r θ ( sin θ A r θ r ( sin θ A r θ r ) A r sin θ 2 φ ) + 1 sin θ ) + 1 sin θ 2 A r 2 φ r 2 A r 2 φ r ] + k 2 A r = 0 ] + k 2 A r r ] + k 2 A r r = 0 = 0

26 Spherical Vector Potential This equation is recognized as the Scalar Helmholtz Equation (for TM r waves) [ 2 + k 2] ( ) A r = 0 r It is not to hard to imagine the Scalar Helmholtz Equation (for TE r waves) [ 2 + k 2] ( ) F r = 0 r with the gauge condition F r r = jωɛµφ m.

27 Spherical Vector Potential Thus, in short, if ψ a and ψ f solve the Helmholtz equation, we have A = µψ a r F = ɛψ f r where r = ru r. In terms of these we have The Fields E = rψ f + 1 jωµ rψa H = rψ a + 1 jωɛ rψf

28 Spherical Vector Potential The components may be written ɛe r = 1 ( ) 2 jωµ r 2 + k 2 A r ɛe θ = 1 F r r sin θ φ A r jωµr r θ ɛe φ = 1 F r r θ A r jωµr sin θ r φ µh r = 1 ( ) 2 jωɛ r 2 + k 2 F r µh θ = µh φ = 1 r 1 A r r sin θ φ F r jωɛr r θ A r θ F r jωɛr sin θ r φ

29 The Schelkunoff Bessel Functions A r and F r do not solve the Helmholtz equation. If ψ solves the Helmholtz equation, the vector potential may be rψ. It is therefore helpful to define new radial eignefunctions. The most common (and obvious) definition is that due to S. A. Schelkunoff Schelkunoff Bessel Functions πkr ˆB n (kr) krb n (kr) = 2 B n+ 1 (kr) 2

30 The Schelkunoff Bessel Functions The Schelkunoff Bessel Functions solve the equation [ ] d 2 dr 2 + k 2 n(n + 1) ˆB r 2 n = 0 The solutions for either vector potential can now be written as C mn ˆBn (kr)l m n (cos θ)h(mφ). m n

31 Outline Wave Functions The Spherical Cavity Radial Waveguides 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

32 Transverse Electric Modes The Spherical Cavity Radial Waveguides Consider a spherical resonator of radius a. The field must be finite at the origin. The field must be finite at the poles (i.e. θ = 0 and θ = π. The field must be 2π-periodic in azimuth. Thus we choose the Modal F r = ɛĵn(kr)p m n (cos θ) { cos mφ sin mφ (In the future we ignore the sin mφ for simplicity.) }

33 Transverse Electric Modes The Spherical Cavity Radial Waveguides Now E θ and E φ are not proportional to derivatives with respect to r, so we need Ĵ n (ka) = 0 This will be satisfied if where, as might be expected, ka = u np Ĵ n (u np ) = 0, in particular, u np is the p th root of Ĵn

34 Transverse Magnetic Modes The Spherical Cavity Radial Waveguides Transverse magnetic modes are similar. We start with Modal A r = µĵn(kr)p m n (cos θ) { cos mφ sin mφ } In this case, E θ and E φ are proportional to derivatives with respect to r, so we need Ĵ n(kr) = 0 Therefore k = u np a

35 Resonant Frequencies The Spherical Cavity Radial Waveguides Given the values for k, it is easy to find the Resonant Frequencies (f r ) TE mnp = (f r ) TM mnp = u np 2πa µɛ u np 2πa µɛ It is important to note that many of these modes are degenerate (i.e. have the same cutoff frequency.)

36 Degeneracies Wave Functions The Spherical Cavity Radial Waveguides Cutoff frequencies in spherical waveguides do not depend on m or on whether they are even or odd. The lowest order TE modes have n = p = 1, u np = The following modes have the same cutoff: ( (F r ) 0,1,1 = Ĵ r ) cos θ ( a (F r ) even 1,1,1 = Ĵ r ) cos θ cos φ ( a (F r ) odd 1,1,1 = Ĵ r ) cos θ sin φ a In this case, the reason for the degeneracy is clear; these are all the same mode, rotated in space. How many modes have n = 2, p = 1?

37 Outline Wave Functions The Spherical Cavity Radial Waveguides 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

38 Conical Waveguide Wave Functions The Spherical Cavity Radial Waveguides Consider the cone θ = θ 1. Let s find the field: The field should be outwardly travelling. The field should be 2π-periodic in azimuth. The field should be finite at the poles: If θ 1 > π/2, the field must be finite at θ = 0. If θ 1 < π/2, the field must be finite at θ = π. Either way, if we choose TM r functions and assume outward propagation and cos mφ dependence, (A r ) mν = Pν m (cos θ) cos(mφ)ĥ(2) ν (kr)

39 Conical Waveguide Wave Functions The Spherical Cavity Radial Waveguides To satisfy the boundary condition, we need E r = E φ = 0. Neither E r nor E φ is proportional to a derivative of A r with respect to θ. To make the field vanish at θ = θ 1 we need This is an equation in ν! P m ν (cos θ 1 ) = 0.

40 Conical Waveguide Wave Functions The Spherical Cavity Radial Waveguides To satisfy the boundary condition, we need E r = E φ = 0. Neither E r nor E φ is proportional to a derivative of A r with respect to θ. To make the field vanish at θ = θ 1 we need P m ν (cos θ 1 ) = 0. This is an equation in ν! For TE modes, we have (F r ) mν = Pν m (cos θ) cos(mφ)ĥ(2) ν (kr) subject to [ ] d dθ Pm ν (cos θ) = 0 θ=θ 1

41 Biconical Waveguide The Spherical Cavity Radial Waveguides Consider now the waves in the region between θ = θ 1 and θ = θ 2. Here, neither θ = 0 or θ = π is involved, so we can (and must) use either Q n (x) or P n ( x). We still have azimuthal 2π-periodicity, so m Z. We assume outward travel, and cos mφ dependence. Then we may write (A r ) mν = [P m ν (cos θ)p m ν ( cos θ 1 ) P m ν ( cos θ)p m ν (cos θ 1 )] cos(mφ)ĥν(kr) Note that this function vanishes already at θ = θ 1 by design.

42 Biconical Waveguide The Spherical Cavity Radial Waveguides The vector potential must also vanish at θ = θ 2. This implies P m ν (cos θ 2 )P m ν ( cos θ 1 ) P m ν ( cos θ 2 )P m ν (cos θ 1 ) = 0 This is an (awful, transcendental, obscure) equation for ν. Similar considerations apply to the TE modes.

43 The TEM Mode Wave Functions The Spherical Cavity Radial Waveguides The bicone is a two-conductor waveguide, and therefore supports a TEM mode. The easiest way to study it is to consider a special formulation of the TM 00 mode. Thus, let (A r ) 00 = jµq 0 (cos θ)ĥ(2) 0 (kr) = ln cot θ 2 e jkr. (The extra factor of j is included just to make the expression real.)

44 The TEM Mode Wave Functions The Spherical Cavity Radial Waveguides Now, we first note that d dφ cot θ 2 = tan θ 2 csc2 θ 2 ( 1 ) = sin θ 2 cos θ 2 = csc θ This implies Similarly, E θ = 1 2 A r jωɛµr r θ = H φ = 1 r ke jkr ωɛr sin θ A r θ = e jkr r sin θ

45 The TEM Mode Voltage The Spherical Cavity Radial Waveguides Notice E θ H φ = η. This is to be expected from TEM waves. We can also define Voltage V (r) = θ 2 θ 1 E θ rdθ = η ln cot θ 1 2 cot θ 2 2 e jkr

46 The TEM Mode Current The Spherical Cavity Radial Waveguides Similarly we have Current 2π I(r) = H φ r sin θdφ = 2πe jkr 0 We can also compute the Transmission Line Characteristic Impedance Z = V I = η 2π ln cot θ1 2 cot θ 2 2

47 Outline Wave Functions Wave Transformations 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

48 Plane Wave Expansion Wave Transformations A wave traveling in the z-direction must have the expansion e jz jr cos θ = e a n j n (r)p n (cos θ) n=0 Multiplying both sides by sin θ and integrating from 0 to π gives π 0 e jr cos θ P n (cos θ) sin θdθ = 2a n 2n + 1 j n(r) after invoking orthogonality. Computing the integral yields a n = j n (2n + 1)

49 Plane Wave Expansion Wave Transformations We thus have the Spherical Expansion of a Plane Wave e jr cos θ = j n (2n + 1)j n (r)p n (cos θ) n=0 Similar theorems can translate cylindrical Bessel functions into spherical harmonics, etc.

50 Addition Theorem Wave Functions Wave Transformations We know that the solution to is 2 ψ + k 2 ψ = δ(r r ) ψ(r) = e jk r r 4π r r = j 4π h(2) 0 (kr) We can write this in terms of origin centered spherical harmonics in two ways. The difference centers on the special nature of the z-axis. The solution must have rotational symmetry with respect to the r axis. Let the angle with r be ξ. It can be shown geometrically that cos ξ = cos θ cos θ + sin θ sin θ cos(φ φ )

51 Addition Theorem Wave Functions Wave Transformations We also know from Our study of spherical coordinate special functions, and Symmetry, that the solution can be written as ψ = j 4π n=0 c n j n (r < )h (2) n (r > )P n (cos ξ) By interpreting ξ as θ and, say, µψ as A z we find The Addition Theorem h (2) 0 ( r r ) = (2n + 1)j n (r < )h (2) n (r > )P n (cos ξ) n=0

52 The Other Addition Theorem Wave Transformations If we do not invoke the polar symmetry we just used, we can solve the problem using associated Legendre functions in their full glory. This leads to The Legendre Function Addition Theorem P n (cos ξ) = n m=1 ɛ m (n m)! (n + m)! Pm n (cos θ)p m n (cos θ ) cos m(φ φ )

53 Outline Wave Functions Wave Transformations 1 Wave Functions 2 The Spherical Cavity Radial Waveguides 3 Wave Transformations

54 The Set Up Wave Functions Wave Transformations Consider a sphere of radius a, illuminated by the Incident Wave Ex i = E 0 e jkz = jkr cos θ E 0 e Hy i = E 0 η e jkz = E 0 jkr cos θ η e Our wave is not TE r or TM r. We need to expand it as a combination of these types of modes. To do this, we need the r components of E and H.

55 The Incident Field Wave Functions Wave Transformations Let us work with E i r and find the TM part. E i r = cos φ sin θe i x = E 0 cos φ jkr Plugging in the plane wave expansion we find E i r = E 0 cos φ jkr θ (e jkr cos θ ) j n j n (kr) θ P n(cos θ) n=0 This can be simplified using the Schelkunoff functions and derivative formula for Legendre functions: E i r = je 0 cos φ (kr) 2 j n (2n + 1)Ĵn(kr)Pn(cos 1 θ) n=1

56 The Incident Field Wave Functions Wave Transformations We want to derive this radial field from A r. We thus write A i r = E 0 ω cos φ a n Ĵ n (kr)pn(cos 1 θ) n=1 By deriving E i r from this potential and setting it equal to E i r from the previous slide, we can find the a n. We find E i r = je 0 cos φ (kr) 2 a n n(n + 1)Ĵn(kr)Pn(cos 1 θ) n=1 Setting this equal to the previous E i r we find a n = j n (2n + 1) n(n + 1)

57 The Incident Fields Wave Functions Wave Transformations Using duality, we can thus write the Incident Wave A i r = E 0 ω cos φ a n Ĵ n (kr)pn(cos 1 θ) F i r = ɛe 0 k n=1 sin φ a n Ĵ n (kr)pn(cos 1 θ) n=1 Each of these incident waves generates a scattered wave independently.

58 The Scattered Fields Wave Transformations We can also write expressions for the Scattered Wave A s r = E 0 ω cos φ b n Ĥ (2) n (kr)pn(cos 1 θ) F s r = ɛe 0 k n=1 sin φ n=1 c n Ĥ (2) n (kr)p 1 n(cos θ) Remember: the a n are known. The two sets of unknowns (b n and c n ) are from the independent TM r and TE r waves.

59 The Solution Wave Functions Wave Transformations Computing the E θ and E φ components from the vector potentials given and forcing them to vanish gives the coefficients: The Solution b n = a n Ĵ n(ka) Ĥ (2) n (ka) c n = a n Ĵ n (ka) Ĥ (2) n (ka) Now we can easily find the scattered field. Indeed, we can find the far field from Ĥ (2) n (kr) kr j n+1 e jkr

60 The Far Scattered Field Wave Transformations Plugging this in and retaining only the fields that decay as r 1, we find The Far Field E s θ = je 0 kr e jkr cos φ E s φ = je 0 kr e jkr sin φ [ j n b n sin θp 1 P 1 ] n (cos θ) c n(cos θ) n sin θ n=1 [ j n P 1 ] b n(cos θ) n c n sin θpn 1 (cos θ) sin θ n=1 We are most interested in the backscattered field E s θ (θ = π, φ = π)

61 An Aside: Radar Cross Section Wave Transformations We can get an idea of the size of something by how much energy it scatters. This idea is formalized in the radar cross section (RCS) or echo area A e. This is, or course, a far field quantity. Definition The RCS is that area which intercepts an amount of power, which, when reradiated isotropically, produces a power at the receiver equal to that actually observed.

62 An Aside: Radar Cross Section Wave Transformations If the power in the incident wave is S i, the power that would be intercepted is A e S i. If this power, reradiated isotropically, creates the power received, then S r = A es i 4πr 2 We thus have the Radar Cross Section ) A e = lim (4πr 2 Sr r S i = lim r (4πr 2 E s 2 ) E i 2

63 The RCS Wave Functions Wave Transformations Plugging our scattered fields into the RCS formula, and using the wonderfully obscure formulas P 1 n(cos θ) sin θ sin θp 1 n (cos θ) θ π θ π ( 1)n n(n + 1) 2 ( 1)n n(n + 1) 2 and using the Wronskian of Schelkunoff Bessel functions, we find The Radar Cross Section of a Metal Sphere A e = λ2 ( 1) 2 (2n + 1) 4π (ka)ĥ(2) (ka) n=1 Ĥ (2) n n 2

64 Wave Transformations RCS (m 2 ) of a 1m Radius Sphere vs. Frequency (Hz) x 10 8

65 Wave Transformations RCS (m 2 ) of a 1m Radius Sphere vs. Frequency (Hz) x 10 9

66 Observations Wave Functions Wave Transformations At low frequencies A e ka 0 9λ2 4π (ka)6 In other words, for small spheres A e λ 4 This law was first invoked by Lord Rayleigh to explain the blueness of the sky. What happens for high frequencies?

67 Observations Wave Functions Wave Transformations At low frequencies A e ka 0 9λ2 4π (ka)6 In other words, for small spheres A e λ 4 This law was first invoked by Lord Rayleigh to explain the blueness of the sky. What happens for high frequencies? A e ka πa 2

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