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Module I: Electromagnetic Waves Lectures 10-11: Multipole Radiation

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Module I: Electromagnetic Waves Lectures 10-11: Multipole Radiation Module I: Electromagnetic waves Lectures 10-11: Multipole radiation Amol Dighe TIFR, Mumbai Outline 1 Multipole expansion 2 Electric dipole radiation 3 Magnetic dipole and electric quadrupole radiation Coming up... 1 Multipole expansion 2 Electric dipole radiation 3 Magnetic dipole and electric quadrupole radiation ~ Vector potential A0 for monochromatic sources We are interested in calculating the radiative components of EM fields and related quantities (like radiated power) for a charge / current distribution that is oscillating with a frequency !. The results for a general time dependence can be obtained by integrating over all frequencies (inverse Fourier transform). We have already seen that it is enough to know about the current distribution (we are interested only in radiative parts), since the charge distribution is related to it by continuity. In such ~ 0 ~ 0 −i!t a case, with J(~x ; t) = J0(~x )e , we get 0 µ Z eikj~x−~x j A~ (~x; t) = 0 e−i!t ~J (~x0) d 3x 0 (1) 4π 0 j~x − ~x0j ~ ~ −i!t Given A(~x; t) = A0(~x)e , the rest of the quantities can be easily calculated in terms of it. We shall omit the “rad” label in this lecture, it is assumed to be everywhere except when specified. B~ rad and E~ rad for monochromatic sources At large distances, 0 µ Z e−i~k·~x A~ (~x; t) = 0 ei(kjx|−!t) ~J (~x0) d 3x 0 (2) 4π 0 j~x − ~x0j Taking ~x = r ^r and neglecting terms that go as (1=r 2), the radiative part of the magnetic field is ~ ~ ~ −i!t B(~x; t)= r × A(~x; t) = ik^r × A0(~x)e (3) The radiative part of the electric field can be obtained in this monochromatic case by using (1=c2)@E~ (~x; t)=@t = r × B~ (~x; t) (there is no current at large j~xj): c2 E~ (~x; t)= r × B~ (~x; t) = cB~ (~x) × ^re−i!t (4) (−i!) 0 Thus, E~ (~x; t) and B~ (~x; t) fields are orthogonal to ^r, orthogonal to each other, and their magnitudes differ simply by a factor of c. Long-distance, long-wavelength approximation For wavelength large compared to the source size, j~k · ~x0j 1, 0 and we can expand the e−i~k·~x term (note that ~k = k^r): n Z 0 n µ0 X (−ik) (^r · ~x ) A~ (~x) = eikr ~J (~x0) d 3x 0 (5) 0 4π n! 0 j~x − ~x0j If we approximate j~x − ~x0j ≈ r, i.e. neglect the corrections proportional to (d=r) where d is the source size, we get a simpler form ikr n Z µ0 e X (−ik) A~ (~x) = ~J (~x0)(^r · ~x0)nd 3x 0 (6) 0 4π r n! 0 This is the approximate form of the “multipole expansion”, and works for a few lower-order multipoles. Note that this approximation is fine as long as the expansion parameter in j~x − ~x0j is much smaller than the expansion parameter in 0 eikj~x−~x j, i.e. d=r << kd, or r λ. The complete expression for multipole expansion, valid even for intermediate distances, is given on the next page. Radiation potential at intermediate distances 0 An expansion for eikj~x−~x j=j~x − ~x0j exists in terms of Legendre polynomials, spherical Bessel functions and Hankel functions, which we give here without proof: 0 eikj~x−~x j X = ik (2n + 1)P (cos θ0)j (kj~x0j)h (kr) (7) j~x − ~x0j n n n At kj~x0j << 1, we have 2nn! j (kj~x0j) = (kj~x0j)n (8) n (2n + 1)! For kr >> 1, we have eikr h (kr) = (−i)n+1 (9) n kr Using these two, the long-distance approximation gives 0 eikj~x−~x j eikr X 2nn! = (−ik)n j~x0jnP (cos θ0) (10) j~x − ~x0j r (2n)! n which matches our expansion to the two leading orders (check). Coming up... 1 Multipole expansion 2 Electric dipole radiation 3 Magnetic dipole and electric quadrupole radiation The n = 0 term in the multipole expansion The leading (n = 0) term in the multipole expansion is µ eikr Z A~ (0) = 0 ~J (~x0)d 3x 0 (11) 0 4π r 0 The integral may be written in a more familiar form through the steps Z Z ~ 0 3 0 0 0 ~ 0 3 0 J0(~x )d x = − ~x [r · J0(~x )]d x (12) 0 ~ 0 0 and using the continuity equation r · J0(~x ) == −i!ρ0(~x ): Z Z ~ 0 3 0 0 0 3 0 J0(~x )d x = −i! ~x ρ0(~x )d ~x = −i!~p (13) where ~p is the electric dipole moment. The n = 0 term thus represents the electric dipole radiation: µ eikr AED(~x) = 0 (−i!)~p (14) 0 4π r Electric dipole: B~ , E~ and radiated power The magnetic and electric fields can immediately be written as µ eikr B~ ED(~x) = ik^r × A~ ED(~x) = 0 (ck 2)^r × ~p (15) 0 0 4π r µ eikr E~ ED(~x) = cB~ ED(~x) × ^r = 0 (c2k 2)(^r × ~p) × ^r (16) 0 0 4π r The Poynting vector N~ (~x; t) = E~ (~x; t) × H~ (~x; t) is normal to both, (^r × ~p) and [(^r × p) × ^r], i.e. along ^r, as expected. 1 µ 1 hN~ (~x)i = 0 k 4c3j^r × ~pj2^r (17) 2 (4π)2 r 2 µ = 0 k 4c3j~pj2 sin2 θ ^r (18) 32π2r 2 The average power radiated per solid angle is then dP µ = hN~ i · r 2^r = 0 k 4c3j~pj2 sin2 θ (19) dΩ 32π2 Electric dipole radiation: salient features The radiated power is proportional to the fourth power of frequency. This results in the blue colour of the sky: the sunlight induces dipoles in the air molecules, which then radiate, giving out more light at high frequencies, i.e. near the blue end of the spectrum. The angular dependance is sin2 θ, i.e. there is no radiation in the direction of the dipole, most of the radiation is in the equatorial plane. At large wavelengths (λ L), antennas (discussed in the last lecture) also emit dipole radiation. Coming up... 1 Multipole expansion 2 Electric dipole radiation 3 Magnetic dipole and electric quadrupole radiation n = 1 term in the multipole expansion The n = 1 term in the expansion is µ eikr Z A(1)(~x) = 0 (−ik) ~J (~x0)(^r · ~x0)d 3x 0 (20) 0 4π r 0 Using (~x0 × ~J) × ^r = (^r · ~x0)~J − (^r · ~J)~x0, the integral may be separated into two parts: Z ~ 0 0 3 0 J0(~x )(^r · ~x )d x = IMD + IEQ (21) where Z 1 I = [~x0 × ~J (~x0)] × ^r d 3x 0 (22) MD 2 0 Z 1 I = [(^r · ~x0)~J (~x0) + (^r · ~J (~x0))~x0]d 3x 0 (23) EQ 2 0 0 These two terms correspond to the magnetic dipole and the electric quadrupole components, respectively, as we shall see. Magnetic dipole radiation Since the magnetic dipole moment is defined as Z 1 m~ = [~x0 × ~J (~x0)]d 3x 0 (24) 2 0 ~ the component of A0 corresponding to IMD becomes µ eikr A~ MD(~x) = 0 (−ik)(m~ × ^r) (25) 0 4π r This immediately leads to µ eikr B~ MD(~x) = (ik)^r × A~ MD(~x) = 0 k 2 ^r × (m~ × ^r) (26) 0 0 4π r µ eikr E~ MD(~x) = cB~ MD(~x) × ^r = 0 k 2 [^r × (m~ × ^r)] × ^r (27) 0 0 4π r And the average power radiated per unit area is 1 µ 1 hN~ (~x)i = 0 k 4c3jm~ j2 sin2 θ ^r (28) 2 (4π)2 r 2 where θ is the angle between m~ and ^r. Electric quadrupole radiation (1) The remaining component of A0 is the electric quadrupole part (as will be clear soon): µ eikr Z 1 AEQ = 0 (−ik) [(^r · ~x0)~J (~x0) + (^r · ~J (~x0))~x0]d 3x(29)0 0 4π r 2 0 0 µ eikr (−k 2c) Z = 0 ~x0(^r · ~x0)ρ (~x0)d 3x 0 (30) 4π r 2 0 µ eikr (−k 2c) 1 Z = 0 Q~ (^r) + ^r r 02ρ (~x0)d 3x 0 (31) 4π r 2 3 0 Here, Q~ (^r) is the component of the electric quadrupole moment along ^r, i.e. ~ X Qα = Qαβ^rβ ; (32) with Z 0 0 02 0 3 0 Qαβ ≡ (3xαxβ − r δαβ)ρ0(~x )d x ; (33) the electric quadrupole moment. Electric quadrupole: B~ , E~ and power radiated ~ ~ Now we can calculate B0(~x) and E0(~x): µ eikr −ik 3c BEQ(~x) = 0 ^r × Q~ (^r) (34) 0 4π r 6 µ eikr −ik 3c2 E EQ(~x) = 0 (^r × Q~ (^r)) × ^r (35) 0 4π r 6 The average Poynting vector is 1 µ 1 k 6c3 hN~ (~x)i = 0 j^r × Q~ (^r)j2 ^r (36) 2 (4π)2 r 2 36 The average power radiated per unit solid angle is dP µ k 6c3 = 0 j^r × Q~ (^r)j2 (37) dΩ 4π 288 Comment on Electric quadrupole radiation If the charge distribution is azimuthally symmetric, and has a reflection symmetry about z axis (spheroidal distribution is a special case of this), then Qxy = Qyz = Qxz = 0 ; Qxx = Qyy = Q0 Qzz = −2Q0 (38) In such a case, it can be shown that the power radiated is dP µ k 6c3 = 0 jQ j2 sin2 θ cos2 θ (39) dΩ 4π 32 0 where θ is the angle between ^r and Q~ (^r).
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