Electrodynamics II: Lecture 9 Multipole radiation

Amol Dighe

Sep 14, 2011 Outline

1

2 Electric dipole radiation

3 Magnetic dipole and electric radiation Outline

1 Multipole expansion

2 Electric dipole radiation

3 Magnetic dipole and electric quadrupole radiation ~ rad Potential Aω 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 ω. The results for a general time dependence can be obtained by integrating over all frequencies (inverse Fourier transform), of course. 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 a case, we know that

~ ~0 rad µ Z eik|x−x | A~ (~x) = 0 ~J (~x0) d 3x 0 (1) ω 4π ω |~x − ~x0|

~ Given Aω, 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. ~ rad ~ rad Bω and Eω for monochromatic sources

The radiative part of the magnetic field is then

~ ~ ~ Bω = ∇ × Aω = ikˆr × Aω (2)

Note that here ~r = ~x, to be consistent with standard convention. The radiative part of the electric field can be obtained in this ~ ~ monochromatic case by using ∇ × Bω = µ00(−iω)Eω (note that there is no current at large r):

ic2 E~ = ∇ × B~ = cB~ × ˆr (3) ω ω ω ω

~ ~ Thus, Eω and Bω fields are orthogonal to ˆr, orthogonal to each other, and their magnitudes differ simply by a factor of c. Long-distance approximation

Since the sources are confined to a finite region, there will be some distance d such that |~x0| < d. We shall work in the approximation d  (1/k)  r, where r = |x 0|. In this approximation, we will be able to expand the radiation fields in a suitable form. Since |~x0|  |~x|, one can approximate |~x − ~x0| = r − ˆr · ~x0 (4)

This allows us to expand ` 1 1 1 X x 0  = = P (cos θ0) (5) |~x − ~x0| r − ˆr · ~x0 r r ` ` where θ0 is the angle between ~r and ~x0. Keeping only the leading term, the vector potential becomes ikr µ e Z ~ 0 A~ = 0 ~J (~x0)e−ik·~x d 3x 0 (6) ω 4π r ω Long-distance approximation continued

0 Since |~k · ~x0|  1, we can expand the e−i~k·~x term:

ikr n Z µ0 e X (−ik) A~ = ~J (~x0)(ˆr · ~x0)nd 3x 0 (7) ω 4π r n! ω

Note that ~k = kˆr. This is the “multipole expansion”. Note that the subleading terms in 1/|~x − ~x0| are not included here, which is fine as long as d/r << kd, i.e. the expansion parameter in 1/|~x − ~x0| is much 0 smaller than the expansion parameter in eik|~x−~x |. At sufficiently large distances, this will always be true. However for practical situations, this needs to be checked. There is a general expression, valid even for intermediate distances, which we’ll give on the next slide. For the purposes of this lecture, the approximation given above will suffice. Radiation potential at intermediate distances

0 An expansion for eik|~r−~x |/|~r − ~x0| exists in terms of legendre polynomials, spherical Bessel functions and Hankel functions, which we give here without proof: 0 eik|~r−~x | X = ik (2n + 1)P (cos θ0)j (k|~x0|)h (kr) (8) |~r − ~x0| n n n At k|~x0| << 1, we have 2nn! j (k|~x0|) = (k|~x0|)n (9) n (2n + 1)! For kr >> 1, we have eikr h (kr) = (−i)n+1 (10) n kr Using these two, the long-distance approximation gives 0 eik|~r−~x | eikr X 2nn! = (−ik)n |~x0|nP (cos θ0) (11) |~r − ~x0| r (2n)! n which should match our expansion (not checked explicitly yet). Outline

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 (12) ω 4π r ω

The integral may be written in a more familiar form by integrating R (J(~x0) · 1)d 3x 0 by parts, and then using the continuity equation ∇0 · J(~x0) = −∂ρ(~x0)/∂t = −iωρ(~x0): Z Z ~ 0 3 0 0 ~ 0 0 3 Jω(~x )d x = − ∇ · J(~x )~x d x (13) Z = −iω ~x0ρ(~x0)d 3~x0 = −iω~p (14)

where ~p is the . The n = 0 term thus represents the electric dipole radiation: µ eikr AED = 0 (−iω)~p (15) ω 4π r ~ ~ Electric dipole radiation: Eω, Bω and radiated power

The magnetic and electric fields can immediately be written as µ eikr B~ ED = ikˆr × A~ ED = 0 (ck 2)ˆr × ~p (16) ω ω 4π r µ eikr E~ ED = cB~ ED × ˆr = 0 (c2k 2)(ˆr × ~p) × ˆr (17) ω ω 4π r ~ ~ ∗ ~ ω The Nω = Eω × H is normal to both, (ˆr × ~p) and [(ˆr × p) × ˆr], i.e. along ˆr, as expected. The average rate of energy radiated is 1 µ 1 hN~ i = 0 k 4c3|ˆr × ~p|2ˆr (18) 2 (4π)2 r 2 µ = 0 k 4c3|~p|2 sin2 θ ˆr (19) 32π2r 2 The average power radiated per solid angle is then dP µ = hN~ i · r 2ˆr = 0 k 4c3|~p|2 sin2 θ (20) 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 class) also emit dipole radiation. Outline

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) = 0 (−ik) ~J (~x0)(ˆr · ~x0)d 3x 0 (21) ω 4π r ω

Using (~x0 × ~J) × ˆr = (ˆr · ~x0)~J − (ˆr · ~J)~x0, the integral may be separated into two parts: Z ~ 0 0 3 0 Jω(~x )(ˆr · ~x )d x = IMD + IEQ (22)

where Z 1 I = [~x0 × J(~x0)] × ˆr d 3x 0 (23) MD 2 Z 1 I = [(ˆr · ~x0)~J(~x0) + (ˆr · ~J(~x0))~x0]d 3x 0 (24) EQ 2

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 (25) 2 ~ the component of Aω corresponding to IMD becomes µ eikr A~ MD = 0 (−ik)(m~ × ˆr) (26) ω 4π r This immediately leads to µ eikr B~ MD = (ik)ˆr × A~ MD = 0 k 2 ˆr × (m~ × ˆr) (27) ω ω 4π r µ eikr E~ MD = cB~ MD × ˆr = 0 k 2 [ˆr × (m~ × ˆr)] × ˆr (28) ω ω 4π r And the average power radiated per unit area is 1 µ 1 hN~ i = 0 k 4c3|m~ |2 sin2 θ ˆr (29) 2 (4π)2 r 2 where θ is the angle between m~ and ˆr. Electric quadrupole radiation

(1) The remaining component of Aω is the electric quadrupole part (as will be clear soon):

µ eikr Z 1 AEQ = 0 (−ik) [(ˆr · ~x0)~J(~x0) + (ˆr · ~J(~x0))~x0]d 3x 0 (30) ω 4π r 2 µ eikr (−k 2c) Z = 0 ~x0(ˆr · ~x0)ρ(~x0)d 3x 0 (31) 4π r 2 µ eikr (−k 2c) 1 = 0 Q~ (ˆr) (32) 4π r 2 3

Here, Q~ (ˆr) is the component of the electric quadrupole moment along ˆr, i.e. ~ X Qα = Qαβrβ , (33)

with Z 0 0 02 0 3 0 Qαβ ≡ (3xαxβ − r δαβ)ρ(~x )d x , (34)

the electric quadrupole moment. ~ ~ Electric quadrupole: Bω, Eω and power radiated

~ ~ Now we can calculate Bω, Eω:

µ eikr −ik 3c BEQ = 0 ˆr × Q~ (ˆr) (35) ω 4π r 6 µ eikr −ik 3c2 E EQ = 0 (ˆr × Q~ (ˆr)) × ˆr (36) ω 4π r 6

The average Poynting vector is

1 µ 1 k 6c3 hN~ i = 0 |ˆr × Q~ (ˆr)|2 ˆr (37) 2 (4π)2 r 2 36

The average power radiated per unit solid angle is

dP µ k 6c3 = 0 |ˆr × Q~ (ˆr)|2 (38) 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 (39)

In such a case, it can be shown that the power radiated is

dP µ k 6c3 = 0 |Q |2 sin2 θ cos2 θ (40) dΩ 4π 32 0

where θ is the angle between ˆr and Q~ (ˆr). The gravitational radiation has a similar form to the electric quadrupole radiation, except one has to deal with time-dependent mass distribution rather than time-dependent charge distribution. Recap of topics covered in this lecture

~ ~ ~ Calculating Bω and Eω from Aω (for their radiative components) Multipole expansion when |~x0| < λ < |~x| Electric dipole radiation as the leading term in multipole expansion Separating magnetic dipole moment and electric quadrupole moment contributions from the subleading term ~ ~ Eω, Bω, Poynting vector, average rate of radiated power, and the angular distribution of radiated power