Scattering I

R. Wordsworth

April 2, 2015

1 Rayleigh scattering from the resonant oscillator

From Lesson 7, we have

σ(ω) =8πe4

3m2c4

ω4

(ω20 − ω2)2 + α2ω2

. (1)

Equation for a forced classical (or equivalent quantum) oscillator, remember.This has several useful limits beyond heuristic derivation of the Lorentz line profile. For example,

for free electrons both ω0 and α are zero. Hence

σT =8πe4

3m2c4(2)

Scattering by a free electron is Thomson scattering. If we plug in the numbers we get σT =6.65× 10−25 cm2/molecule.

Now again assume α = 0 but do ω << ω0 to get

σR(ω) = σTω4

(ω20 − ω2)2

≈ σTω4

ω40

∝ λ−4. (3)

Rayleigh scattering. Tells us there is a 4th power λ dependence for a scatterer far from its resonantfrequency. Here the resonant frequencies are electronic (i.e. Lyman series in UV/EUV for hydro-gen). The formula says nothing about the scattering phase function, though, and the magnitudedoes not account for quantum effects.

2 Rayleigh scattering from Hertzian EM theory

A dipole moment is p = qx. The dipole induced by an EM field is p0 = αEext, where α is thepolarizability constant. α has units of volume and is a tensor in general. Here it is a scalar.

Hertz (1889) solution (can be derived from classical EM theory) for scattered E-field far fromthe dipole

E =1

c2

1

r

∂2p

∂t2sinθ0. (4)

θ0 is the angle between the scattered dipole moment and the observation direction.p is the scattered dipole moment in the E−field direction

p = p0e−ik(r−ct) = αEexte

−ik(r−ct). (5)

1

Hence∂2p

∂t2= −k2c2αEexte

−ik(r−ct) (6)

and

E = −k2

rαEexte

−ik(r−ct)sinθ0. (7)

where θ0 is the angle between the scattered dipole moment and the observer.We can define parallel (to solar beam) and perpendicular components

E⊥ = −k2

rαE⊥,exte

−ik(r−ct)sinθ1 (8)

E‖ = −k2

rαE‖,exte

−ik(r−ct)sinθ2. (9)

θ1 = π/2 always because the scattered dipole moment is normal to the scattering plane. Alsoθ2 = π/2−Θ. This means that

E⊥ = −k2

rαE⊥,exte

−ik(r−ct) (10)

E‖ = −k2

rαE‖,exte

−ik(r−ct)cosΘ (11)

Now remember

I =c|E|2

8π(12)

hence

I⊥ =k4

r2α2I⊥,ext (13)

I‖ =k4

r2α2I‖,extcos2Θ (14)

Now note for unpolarized light (sunlight) I⊥,ext = I‖,ext = Iext/2. Hence adding we get

I =Iextr2

α2k4 1 + cos2Θ

2(15)

IIext

=8π4

r2α2λ−4(1 + cos2Θ). (16)

This is the Rayleigh scattering formula.Or we can use defn for phase function with normalization to write

P(cosΘ) =3

4(1 + cos2Θ) (17)

Remember ∫ 2π

0

∫ π

0P(cosΘ)sinΘdΘdφ = 4π. (18)

and ∫ π

0(1 + cos2Θ)sinΘdΘ =

8

3(19)

2

henceIIext

=128π5

3r2α2λ−4P(cosΘ)

4π. (20)

The total scattered Rayleigh power [W] is

PR =

∫ΩI∆Ωr2dΩ (21)

where the integral is over solid angle. I∆Ω is a flux [W/m2]

PR =128π5α2

3λ4Iext∆Ω

∫Ω

P(cosΘ)

4πdΩ (22)

but ∫Ω

P(cosΘ)

4πdΩ ≡ 1 (23)

so

PR =128π5α2

3λ4Fext (24)

with Fext ≡ Iext∆Ω the incoming flux [W/m2]. Finally Rayleigh cross-section

σR ≡FRFext

=128π5α2

3λ4. (25)

Compare this to what we had before from the oscillator argument.Now α is given by the Lorentz-Lorenz formula

α =3

4πns

m2 − 1

m2 + 2(26)

See Liou or Feynman Lectures vol. 2 for a good derivation of this.ns is the number of scatterers (here, molecules) per unit volume. m is refractive index. In

practice for gases m ≈ 1, so

α ≈ 1

4πns(m2 − 1) (27)

so finally rewrite

σR =8π3(m2

r − 1)2

3n2sλ

4f(δ). (28)

and we’ve added an anisotropy factor f(δ).

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