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Coherent Backscattering from Multiple Scattering Systems - KOPS ...

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2 Theory<br />

Figure 2.8: Radiative transfer. Radiative transfer theory describes the photon flux<br />

<strong>from</strong> volume element dV through area element dS without any intermediate scattering.<br />

The result are the two series of image points sketched in fig. 2.7, which we can simplify by<br />

using an effective sample thickness L ′ = L + 2z 0 :<br />

ρ(A → B, t) 2 surfaces =<br />

∞<br />

∑<br />

= ρ(A → B, t) − ρ(A → B ′ ∞<br />

k , t) − ∑<br />

k=1 k=1<br />

ρ(A → B ′′<br />

k , t)<br />

= e− τ<br />

t ∞<br />

√ 3 ∑ e ( − ⃗r ⊥,B −⃗r ⊥,A) 2 +(2mL ′ +z B −z A ) 2<br />

4Dt − e ( − ⃗r ⊥,B −⃗r ⊥,A) 2 +(2mL ′ −z B −z A ) 2<br />

4Dt (2.10)<br />

4πDt m=−∞<br />

2.4.2 Radiative transfer theory<br />

The image point method gives an intuitive picture of the photon density distribution, but fails<br />

to explain the influence of internal reflections at the sample boundaries and to give a value<br />

for the average penetration depth z 0 . This can be accomplished by radiative transfer theory,<br />

where we follow the approach of [58].<br />

Let us consider the flux of photons through a small area dS inside the sample, which for<br />

simplicity we place at the origin and perpendicular to the z-axis, as sketched in fig. 2.8. The<br />

number of photons scattered <strong>from</strong> a volume element dV directly through dS is given by<br />

dS cos θ<br />

the product of the number of photons ρ(⃗r)dV in dV, the fractional solid angle that<br />

represents the cross section of dS for the photons coming <strong>from</strong> dV, the speed of photon<br />

transport v, and the loss e − r<br />

l ∗ due to scattering between dV and dS. The total flux in the<br />

negative z-direction j − dS can then be obtained by integrating over the upper half space with<br />

z > 0, the flux in positive direction j + dS by integration over the lower half space with z < 0:<br />

4πr 2<br />

∫<br />

j ∓ dS =<br />

z > < 0 ρ(⃗r) · v · e − r<br />

l ∗ ·<br />

dS cos θ<br />

4πr 2<br />

dV<br />

14

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