Last Latexed: February 28, 2010 at 21:04 1 504: Lecture 11 Last Latexed: February 28, 2010 at 21:04 2

Physics 504, Lecture 11 downfield from the scatterer, “behind” the scatterer, we must find that the ~ March 1, 2010 scatterer has reduced that power by the appropriate amount. If we take ki in the z direction, and measure the power crossing a plane of constant large Copyright c 2009 by Joel A. Shapiro z, the power flux is given by

1 ~ ~ ~ ~ P = ρdρdφ Re Ei + Es Bi∗ + Bs∗ . 1 The Optical Theorem 2µ0 × z Z h   i The optical theorem relates the scattering amplitude in the forward direc- The power that would have arrived without the scatterer is the term without tion to the total scattering cross section. It may be familiar from quantum the scattered fields. As they fall off as 1/r at large r, we can ignore the mechanics, where it is related to conservation of probability, but it first arose term with two scattered fields, and thus the power removed from the beam in optics, hence its name. is ∆P ,where To follow Jackson’s presentation in 10.11 requires following sections in − which he gives details of diffraction and§ scattering that we are skipping. I 1 ∆P = ρdρdφ Re E~i B~ ∗ + E~s B~ ∗ s i z will present an alternate argument which doesn’t require such details. 2µ0 Z × × 2 h i Consider the scattering by a scatterer of finite size of an incident plane E0 ρ = | | dρ dφ wave 2ωµ0 Z r ~ ~ ~ ~ ~ ~ ikr+iki ~x ~ iki ~x iωt Re ~i k f ∗(k, ki) e− · Ei = E0 ~i e · − × ×    1 1 ~ ~ ~ ~ ~ ~ iki ~x iωt ikr iki ~x Bi = ki Ei = ki ~i E0 e · − +e − · f~(~k,~ki) ~ki ~i ∗ ω × ω × × × z   iωt The scattered fields will also have a e− dependence, assuming the scatterer The contribution for very large values of z will come from ρ √z so the responds linearly in a time-invariant fashion, so we will drop all such factors. 2 2 ∼ 2 angle goes to zero, ~k = ~ki, kr ~ki ~x = k(r z)=k(√z + ρ z) kρ /2z, The scattered wave is proportional to the incident wave and is given at so − · − − ≈ ~ ~ ~ 1 large distances by the scattering amplitude f(k, ki)as ikr i~ki ~x e − · 2π ∞ ikρ2/2z 2π ∞ iku/z 2π eikr ρdρdφ ρdρe = du e = i . ~ ~ ~ ~ √ 2 + 2 z 0 z 0 k Es(~x)= f(k, ki)E0 Z z ρ ≈ Z Z r 1 eikr Thus B~ s(~x)= ~k E~s = E0 ~k f~(~k,~ki). ω × ωr × 2 π E0 ∆P = | | Re i~ f~∗(~k ,~k )~k + i~ ~kf~∗(~k ,~k ) ωµ k i i i i i i The total fields are E~ = E~i + E~s and B~ = B~ i + B~ s. 0 − · · The total power removed from the incident beam must be the incident i~i ∗ f~(~ki,~ki)~ki i(~ki f~(~ki,~ki))~i ∗ power flux times the total cross section, which includes both the absorption · − · z 2 cross section and the scattering cross section. If we measure the power far π E0 ~ ~ ~ ~ ~ ~ = | | Re i~i f ∗(k, ki)+0+i~i ∗ f(ki, ki) 0 ωµ0 − · · − 1Here ~ is in the direction of , and we are looking only at “elastic scattering” with   k ~r 2 2 k = k = k . It would probably be better to define f as a tensor multiplying ~ E , but π E0 ~ ~ ~ | | | i| i 0 = | | Im ~i ∗ f(ki, ki) . we will not need that generality. Though we haven’t made it explicit, f~ depends on ~ . − ωµ0 · i   504: Lecture 11 Last Latexed: February 28, 2010 at 21:04 3 504: Lecture 11 Last Latexed: February 28, 2010 at 21:04 4

The power flux in the incident beam is where we integrated by parts for the last expression. The last term is

2 2 1 E E k 1 z0 z d ~ ~ 0 ~ 0 ∞ ikR ~ ( ; ˆ ) Re (Ei Bi∗)z = | | Re ~i k ~i = | | e dR −2 f k, θ, φ kez 2µ × 2ωµ × × z 2ωµ ik z0 z R d cos θ 0 0    0 Z| − | so the total cross section must be which, provided the indicated derivative is not singular, falls off like 1/R. Dropping that term, we have 2∆Pωµ0 4π σ = − = Im ~ ∗ f~(~k ,~k ) . Tot 2 i i i d 2π E0 k k · NE0 ikz ik(z0 z) | |   E~s = i dze dφe − f~ (k, 0,φ; keˆz) k Z0 Z0 This is the optical theorem. NE d =2πi 0 eikz0 f~(k, 0, 0; keˆ ) k z 1.1 Index of Refraction Thus the total electric field at points far beyond the slab is I mentioned that the forward scattering is related to the index of refraction. To see k i 2πiNd E~ (~x)=E eikz ~ + f~(k, 0) , this, consider a thin slab of scattering ma- θ 0 i k x k R ! terial, of number density N, thickness d, ρ i~ki ~x with an incident wave E~i = E0~ie · ,with which is a plane wave with a shifted phase, amplitude and polarization, but ~ = ˆ in the direction, normal to the z is still an exact solution of the free space wave equation. Therefore this ki kez z 0 d z0 surfaces of the slab. Let us observe the expression should hold right up to the edge of the slab. If we project on the 3 ~ field at a large distance z0.Eachd x in original polarization, we see the effect of a thickness dz of material on ~i ∗ E 3 wavefront 1 ~ · the slab has Nd x scatterers contributing is to multiply it by 1 + 2πik− N~i ∗ f(k, 0)dz, so integrating this effect for · a scattered wave a larger distance would give

ikR 1 ~ e ~ 2πik− N~i ∗ f(k,0)z ikz iki ~x 3 ~ ∗ E~ (~x)=e · E e , dE~ = f~(k, θ, φ;~k )E e · Nd x i 0 s R i 0 · d 2π ikR ikz ∞ e 1 z0 z that is, we have the vacuum value k replaced by nk, where the index of E~ = NE dze dφ ρdρ f~ k, cos− ,φ; keˆ s 0 − z refraction n is given by Z0 Z0 Z0 R   R   2 2 2 As R = ρ +(z0 z) , ρdρ= RdR,so 2πN~ ∗ f~(k, 0) − =1+ i n ·2 . ikR k ∞ e ~ 1 z0 z ρdρ f k, cos− − ,φ; keˆz Thus we see that the index of refraction is given by the forward scattering Z0 R   R   ∞ ikR 1 z0 z amplitude. = dR e f~ k, cos− − ,φ; keˆz z0 z R Jackson makes some comments about improving the assumptions, in par- Z| − |     1 ticular that the field experienced by each scatterer is unaffected by the others. ikR ~ 1 z0 z ∞ = e f k, cos− − ,φ; keˆz He tells us the principal effect of fixing this is to evaluate the scattering cross ik R R= z0 z     | − | 1 section at the wavenumber suitable for the dielectric rather than for free ∞ ikR d ~ 1 z0 z e dR f k, cos− − ,φ; keˆz space. −ik z0 z dR R Z| − |     504: Lecture 11 Last Latexed: February 28, 2010 at 21:04 5

One effect we can certainly handle is the absorption — as the forward scattering has an imaginary part which gives the total scattering, and the optical theorem relates that to the absorption from the beam, and hence the i imaginary part of the k = nki =Rek + 2 αki,so

4πN ~ ~ ~ α = Nσtot = Im ~i ∗ f(k, k) . k ·   Jackson also warns us that the optical theorem requires the full, correct scattering amplitude f, and that various approximations for f may give an imaginary part inconsistent with the total cross section which depends on f 2 . For a small loss-less dielectric sphere we found a real scattering | | amplitude because r is real, which would give zero for σtot by the optical theorem, which is clearly wrong. But in section 7.10D you discussed the Kramers-Kronig relation, in which analyticity based§ on causality gave an integral relationship giving the real part of r as an integral over frequency of its imaginary part, so a purely real  1 is inconsistent. r −