Physics 504, Physics 504, Spring 2010 Vector Spherical Harmonics Spring 2010 Electricity Electricity and and Last time, used scalar Green function on vector source. Magnetism This mixes spherical expansion with vectors in an Shapiro Shapiro awkward way, Multipole For example, ` = 1 mixed magnetic and electric Expansion I We will first finish up the ` = 1 term from the Green quadripole source contributions. Scattering Scattering function. This is giving us magnetic dipole and electric quadripole contributions. In doing spherical cavity, we expanded scalars, ~r E~ and · ~r H~ . Each satisfies Helmholtz away from sources, so for I We will briefly describe a more consistent way of · doing the angular expansion, using “vector spherical r>dthey are expanded in spherical bessels times harmonics”. This will more cleanly separate spherical harmonics, as we learned in lecture 5: magnetic multipoles and electric multipoles, and is (M) `(` +1) (M) consistent with what we did for the spherical cavity. ~r H~ = g`(kr)Y`m(θ, φ),~r E~ =0 · `m k · I We will skip most of the rest of chapter 9 and go on (E) `(` +1) (E) or ~r E~ = Z0 g`(kr)Y`m(θ, φ),~r H~ =0 to scattering of the electromagnetic waves. · `m − k · for magnetic multipole modes (M) or electric multipole modes (E).

In either case g` satisfies the spherical Bessel equation Physics 504, Physics 504, Spring 2010 Properties Spring 2010 2 Electricity Electricity ∂ 2 ∂ `(` +1) 2 and and Magnetism Magnetism 2 + 2 + k g`(kr)=0 ∂r r ∂r − r  1 Shapiro dΩX~ ∗ X~ `m = Shapiro `m · with solutions outside the source region proportional to Z `(` +1) `0(`0 +1) (1) Multipole p p Multipole h (kr) for outgoing waves. We found the transverse Expansion 1 Expansion ` dΩ L+∗ Y`∗m (L+Y`m) Scattering × Z 2 Scattering components are given by h !  1 ∗ ∗ i + L Y`m (L Y`m) ~ (M) ~ ~ (M) ~ ~ (M) 2 − − E`m = Z0g`(kr)LY`m, H`m = E`m !  −kZ0 ∇× +(L∗Y ∗ )(LzY`m) ~ (E) ~ ~ (E) Z0 ~ ~ (E) z `m or H`m = g`(kr)LY`m(θ, φ), E`m = i H`m i k ∇× 1 1 2 Y ∗ L L++ L+L +Lz Y`m = dΩ `m 2 − 2 − where L~ = i~r ~ . Z  `(` +1) ` (` +1) − × ∇ 0 0 p p 1 `(` +1) ~ ~ = dΩ Y ∗ Y`m For ` 1, define: X`m(θ, φ):= LY`m(θ, φ). p `m ≥ `(` +1) `0(`0 +1)Z p p = δ`` δmm These are called the vector spherical harmonics, and π 2π provide good basis functions for expanding our fields. where dΩ= 0 sin θdθ 0 dφ, and we have used R R R

~ 2 1 1 2 ~ ~ L = 2 L L+ + 2 L+L + Lz, dΩ(LΦ)∗Ψ= dΩΦ∗LΨ, Physics 504, Physics 504, − − Spring 2010 Spring 2010 and dΩY ∗ Y`m = δ`` δmm . R R `m Electricity The justification for claiming ~r E~ and ~r H~ satisfy the Electricity So theR vector spherical harmonics are an orthonormal set: and · · and Magnetism required them to be divergenceless, Magnetism Shapiro which ~r E~ is not in the presence of sources. Shapiro dΩX~ ∗ X~ `m = δ`` δmm . · ~ ~ ~ ~ `m · The trick is to evaluate E 0 := E + iJ/ω0,so~r E 0 and Z Multipole · Multipole Expansion ~r H~ do satisfy inhomogeneous Helmholtz equations with Expansion · Scattering sources given by ρ and J~, with the latter supplemented by Scattering We also have any intrinsic magnetization. This is somewhat messy, given in section 9.10, but we will dΩX~ ∗ ~r X~ `m `m not elaborate. Z ·  ×  1 = dΩ L~ ∗Y`∗m (~r L~ )Y`m A major use has the sources given by quantum `(` +1) ` (` +1)Z · × 0 0   mechanical operators for atomic or nuclear structure, and p 1 p ∗ ~ ~ the vector potential is then a for outgoing dΩY`m L (~r L)Y`m `(` +1) `0(`0 +1)Z · × photons, giving a decay probabilities rather than p p1 radiation power flux. But we will skip this as well, and ∗ ~ ~ dΩY`m~r (L L)Y`m =0. `(` +1) `0(`0 +1)Z · × proceed to discuss scattering of electromagnetic waves. p p because L~ L~ = iL~ ,so~r (L~ L~ )=~r L~ =0. × · × · Physics 504, Physics 504, Scattering of waves Spring 2010 Cross section Spring 2010 Electricity Electricity Currents create fields, but fields affect the motion of and Scattering is measured by cross sections. and Magnetism dσ Magnetism charges too. Mutual reaction. For classical particle dynamics, is the area of the Scattering by small scatterers: Shapiro dΩ Shapiro 1 incident beam which gets scattered into the solid angle Incident wave in directionn ˆi: Multipole Multipole Expansion Expansion iknˆi ~x dΩ. E~ = ~iEie · , H~ =ˆni E~ /Z0, dσ power scattered into dΩ inc inc × inc Scattering Scattering iωt For wave, = . with e− understood, k = ω/c. dΩ incident flux If scatterer is small, its radiation is dominated by dipole 1 ~ ~ 1 ~ ~ Flux rˆ Esc Hsc∗ = rˆ Esc rˆ Esc∗ = terms, electric dipole moment p~ and magnetic dipole 2 ·  ×  2Z0 ·  ×  ×  moment m~ . 1 ~ ~ ~ Esc Esc∗ ,asˆr Esc = 0. But the outgoing wave Far from scatterer, r λ, 2Z0 · ·  consists of two polarizations, and we can ask what the 2 ikr k e cross section is for a given polarization, ~, for an incident E~sc = [(ˆr p~) rˆ rˆ m/c~ ] . 4π0r × × − × wave with polarization ~i.So H~ =ˆr E~ /Z . sc × sc 0 2 ~ ∗ E~sc dσ 2 · Wave radiates in all directions. (ˆr,~;ˆni,~i)=r , dΩ 2 1 ~ E~ Notation changes from Jackson: ~0 ~i, and i generally for i∗ inc → · incident wave. His ~n rˆ. →

Physics 504, Physics 504, Spring 2010 Dielectric Spring 2010 Electricity Electricity and Last term you found (Jackson 4.56) that a non-magnetic and Magnetism dielectric sphere of radius a has a static induced dipole Magnetism Shapiro moment Shapiro r 1 3 ~ 4 Multipole p~ =4π0 − a Einc, Multipole dσ k 2 Expansion r +2 Expansion (ˆr,~;ˆni,~i)= 2 ~ ∗ p~ +(ˆr ~ ∗) m/c~ , dΩ (4π0Ei) | · × · | Scattering and of course m~ =0.So Scattering 2 ~ dσ 4 6 r 1 2 where we need to know the p~ and m~ induced by Einc. (ˆr,~;ˆn ,~ )=k a − ~ ∗ ~ . i i | · i| dΩ r +2 If scatterers are small ( λ), polarization response should  The scattered wave has the electric field in the plane of dσ 4 4 be quasi-static, independent of ω,so k ω . This the incident polarization andr ˆ;if~ ∗ ~i the amplitude is dΩ ∝ ∝ ⊥ is known as Rayleigh’s law. zero. If the incident wave is unpolarized, say coming in the z direction, we may take the average over polarization in φ, ~i = (cos φ, sin φ, 0). If we are looking at an angle θ,say withr ˆ = (sin θ, 0, cos θ), the two polarization vectors are ~ = (cos θ, 0, sin θ) in the scattering plane and k ~ =(0, 1, 0) perpendicular to it. ⊥

Physics 504, Physics 504, Spring 2010 Unpolarized and Total Cross Sections Spring 2010 2 Electricity Electricity 2 2 2 2 and and Then ~ ∗ ~i = cos θ cos φ and ~ ∗ ~i = sin φ, with Magnetism Magnetism k · | ⊥ · | 1 2 1 If we don’t measure the polarization of the scattered light, average values (over φ of cos θ and respectively. So Shapiro Shapiro 2 2 the unpolarized cross section is the sum of the two, 4 6 2 Multipole Multipole dσ k a r 1 2 Expansion 4 6 2 Expansion k = − cos θ, dσ k a r 1 2 dΩ 2 r +2 Scattering = − 1 + cos θ Scattering dΩ 2 r +2 4 6 2 !  dσ k a r 1 ⊥ = − . dΩ 2  +2 and the total cross section is the integral of this over r π 2π dΩ= sin θdθ dφ, 0 0 The polarization is defined by the difference over the sum, R R 2 1 r 1 dσ dσ σ = πk4a6 − d(cos θ) 1 + cos2 θ ⊥ k 2 r +2 Z 1 dΩ − dΩ sin θ − !  Π(θ):= = 2 for dielectric sphere 2 dσ dσ  1 + cos θ  8π 4 6 r 1 k = k a − . ⊥ + dΩ dΩ 3 r +2

We will skip subsection C.