Lecture 10 February 25, 2010

Transcription

1 Lecture 10 February 5, 010 Last time we discussed a small scatterer at origin. Interesting effects come from many small scatterers occupying a region of size d large compared to λ. The scatterer j at position x j has an E inc with an extra factor of e ikˆn i x j, in the scattered wave, r needs to be replaced by r x j. Assuming we are observing from far away, r d, the variations of the r in the denominator or the ˆr s are r r x j not important, but the effect in x j the oscillating exponential is, r x j we should approximate n i e ik r x j e ikr e ikˆr x j So the amplitude for the scattered wave due to j has an extra factor of e ikˆn i x j ikˆr x j = e i q x j, with q = k(ˆn i ˆr).

2 The amplitudes for all the scatterers need to be added before squaring to find the flux, so we have dσ dω = k 4 (4πɛ 0 E i ) [ ɛ p j + (ˆr ɛ ) m j /c] e i q x j j If all the scatterers react the same way, p j m j can be factored out of the sum, we appear to have a single scatterer with a structure factor F( q) = e i q x j = e i q ( xj x j ). j j j. The nature of F( q) depends on how the scatterers are distributed.

3 Structure Factor Large number of romly positioned scatterers: phases rom superposition incoherent. Only the terms with i = j contribute, F( q) = N, except for q = 0. Coherent scattering N, so incoherent scattering is very faint. Crystaline structure: with a regular array we can get even less scattering. Consider a one dimensional array of N scatterers each displaced by a from the previous. N 1 F( q) = j=0 e ij q a = 1 e in q a 1 e i q a =N sin (N q a/) (N sin( q a/)). For lattice spacings a λ but total extent Na λ, the fraction is (sin x/x) for x = N q a/. x 1 (sin x/x) 1 unless q a is comparable or smaller than 1/N.

4 So except for forward scattering, we have destructive interference. In three dimensions, the same thing happens unless the Bragg condition holds for some pair of scatterers, q d = nπ for some d the separation between two scatterers, not too far apart. In that case there will be some fraction of N interfering constructively, the structure factor will be proportional to N. But if the lattice spacing is much less than λ, this will happen only for forward scattering. So a perfect crystal with a λ is uniform material with permittivity ɛ permeability µ, without scattering. But suppose small fluctuations, ɛ = ɛ + δɛ( x), µ = µ + δµ( x).

5 Applying Maxwell Maxwell in medium but without sources applies: As D = 0, D = ( ) = D D = ( ) D = ( ) ( D ɛe ɛ ( ) E. } {{ } B t last term: ɛ B t So altogether, = ɛ t ( ) B µ H + ɛ µ t } {{ H } D ɛ µ D ( ) t = ( D ɛe + ɛ t ( ) B µ H. D t (1)

6 This equation is exact. Good approximations: δɛ, δµ small, treat to first order, as sources. Can treat full field D as harmonic, e iωt so D satisfies inhomogeneous Helmholtz equation with k := µ ɛω, all fields perturbations on an incident plane wave D inc ( x) = D i e ik n i x µ B inc ( x) = ɛ n i D inc ( x), the fields in the source term, to first order in the variations, will be D ɛe = δɛ( x) D ɛ inc ( x) B µh = δµ( x) B µ inc ( x)

7 the correction will then be the scattered wave given by the Green s function D D inc = 1 d 3 x eik x x 4π x x { 1 ɛ ( δɛ( x ) D ) inc ( x ) + i ɛω µ ( δµ( x ) B ) } inc ( x ) Integration by parts: Note 1 V A = S n A 0 if A vanishes sufficiently at infinity, therefore V d3 x f( x ) A( x ) ( ) V d3 x f( x ) A( x ). For the B inc term, f( x ) is the Green function, eik x x x x = R eikr R 3 [ikr 1], with R = x x 1 See lecture notes

8 For the D inc term, we also need d 3 x f( x ) A( x ) V ( [ = d 3 x f( x ) A( x ] ) ) A V ( d 3 x f( x )) A( x ) V d 3 x A( x ) f( x ) V + d 3 x A( x ) ( ) f( x ) V d 3 x A( x ) f( x ). V Again f( x ) = e ik x x / x x is the Green s function for + k, so for the second term, outside the region of scattering (where we can ignore the δ( x x ) term) we have k V d3 x A( x )e ik x x / x x.

9 For large r, we have So altogether where e ik x x 1 x x ( A ) ( f) A sc = k 4π = e ikr e ikˆr x, 1/r, f = ik r ˆreikr e ikˆr x, = k r ˆr Aˆre ikr e ikˆr x. D = D inc + eikr r A sc, { d 3 x e ikˆr x δɛ( x ) ( ˆr D ɛ ) inc ( x ) ɛω δµ( x } ) ˆr B k µ inc ( x ). ˆr

10 The differential cross section for light with polarization ɛ is ɛ A sc dσ dω = = D inc [ k 4π d 3 x e i q x { ɛ ɛ δɛ( x ) δµ( x ) ɛ µ } ] ( ɛ ˆr) (ˆn i ɛ i ), with q = k(ˆn i ˆr).

11 Our first application is to consider molecules in a dilute gas as a fluctuation in ɛ from the vacuum at a point. With an induced dipole moment p j = ɛ 0 γ mol E( xj ) we have δɛ = ɛ 0 γ mol δ( x x j ) j we assume no magnetic moments, so δµ = 0. Then dσ dω = k4 16π γ mol ɛ ɛ F( q) where for a dilute gas we have an incoherent sum F( q) is the number of scattering molecules, except for q = 0, the forward direction. For the dilute gas as a whole the dielectric constant ɛ r = ɛ/ɛ 0 = 1 + Nγ mol, where N is the number density of molecules.

12 The total scattering cross section per molecule is then k 4 σ = 16π N ɛ r 1 dω ɛ ɛ i ɛ The polarization factor is ɛ ( ɛ i ɛ ) ( ɛ ɛ i ) = 1 ˆr ɛ i, as ɛ ɛ j ɛ k + ˆr j ˆr k = δ jk. Consider light incident in the z direction with ɛ i = ˆx, so ˆr ɛ = sin θ cos φ, the integral dω ɛ ɛ i = σ = π 0 k4 6πN ɛ r 1 = π sin θdθ dφ(1 sin θ cos φ) = 8π/3, 0 k4 6πN n 1 k4 n 1 3πN where n = ɛ r is assumed to deviate only slightly from 1.

13 The intensity of the beam I(z) = I(0)e αz falls exponentially with distance with the attenuation coefficient α due to the scattering. In a slice of width dz, there are Ndz scatterers per unit area, each scattering an area σ of the beam, so there is a fractional loss of Nσdz in distance dz, α = Nσ k4 3πN n 1. This is Rayleigh scattering. Note that it is a method of determining the number of molecules, so an approach which was used historically to determine Avagadro s number.

14 In the previous discussion we assumed no corrolation in the positions of the scatterers. This is not a good approximation in denser fluids. A better approximation is to consider ɛ to be the mean permittivity of the fluid but take into account density fluctuations. From the Clausius-Mossotti relation (J4.70) we have ɛ r = 3 + Nγ mol 3 Nγ mol = dɛ r dn = 9γ mol (3 Nγ mol ) = (ɛ r 1)(ɛ r +), 3N so the variation of ɛ in a region of fluid with varying density is δɛ = (ɛ r 1)(ɛ r + ) δn. ɛ 0 3N How do we evaluate δn?

15 In a fluid in equilibrium with a reservoir at constant pressure temperature, the probability that a given piece of fluid occupies a volume V is exp G(V )/k B T, where G is the Gibbs free energy k B is Boltzmann s constant. In terms of the isothermal compressibility β T = 1 V ( ) V p T = ( ) 1 V G V, the mean square deviation of ( V ) = k B T V β T, ( N) = k B T N /V β T. See Reif, p300

16 So the total (for all the particles in the volume) differential cross section is dσ NV = k4 dω 16π ɛ ɛ d 3 i q x δɛ( x) xe ɛ = k4 16π ɛ ɛ ɛ r 1)(ɛ r + ) 3Nɛ r d 3 x d 3 x e i q ( x x ) δn( x)δn( x ). If we assume the correlation length for density fluctuations is much less than the wavelength, we may take e i q ( x x ) 1 the integrals give V (δn) = N k B T β T.

17 As for the blue sky, the attenuation coefficient is just α = Nσ the angular integral is dω ɛ ɛ ɛ = 8π/3, so α = k4 6πN = ω 4 6πNc 4 (ɛ r 1)(ɛ r + ) 3ɛ r Nk B T β T (ɛ r 1)(ɛ r + ) 3 Nk B T β T. The most important feature of this is that at the critical point the compressibility β T blows up, so the fluid becomes opalescent. I am going to skip the sections on diffraction. This has been or is covered in our optics courses.