## 25 November 2012

### Power laws in small-angle scattering - part II

In the first part I showed that the SAXS intensity scattered by a platelet system goes like $$I(q) \sim q^{-2}$$, at least in some intermediate (but as yet unspecified) q range. Here I will show that for thin rods this dependence becomes $$q^{-1}$$, I will then derive the terminal (Porod) behaviour $$q^{-4}$$ and briefly consider the transition between these two regimes.

### Rods: α = 1

For a rod, the electron density profile:
$\rho (\mathbf{r}) = \delta(x) \delta(y) \mathrm{Cst}(z)$ with the same notations as in part I. Its Fourier transform $$\tilde{\rho} (\mathbf{q}) = \mathrm{Cst}(q_x) \mathrm{Cst}(q_y) \delta(q_z)$$ and, as in part I, I will assume the same form for the intensity $$I(\mathbf{q})$$. The "dual" object of a rod under Fourier transform is a platelet:
When spreading this intensity over reciprocal space we must keep in mind that the intersection of the plane with a sphere of constant $$q$$ is a great circle (shown in red in the figure above). Thus, the total contribution $$2 \pi q I_0$$ increases with $$q$$, but it must also be divided by the surface area of the sphere, yielding for the two spheres: $I(q_0) = \frac{2 \pi q_0 I_0}{4 \pi q_0^2} = \frac{I_0}{2 q_0} \quad \mathrm{and} \quad I(2 q_0) = \frac{4 \pi q_0 I_0}{4 \pi (2 q_0)^2} = \frac{I_0}{4 q_0}$ so that $$I(2 q_0) = I(q_0) /2$$ and $$\alpha = 1$$.

### Interfaces: α = 4 (Porod)

The density profile of an interface is invariant under x and y translations and is a Heaviside (step) function along z: $\rho (\mathbf{r}) = \mathrm{Cst}(x) \mathrm{Cst}(y) H(z)$ The Fourier transform of $$H(z)$$ is $$1/q_z$$, as one can see either by direct evaluation or by noting that the derivative of $$H(z)$$ is the Dirac delta etc. Finally, $\left | \tilde{\rho}(\mathbf{q}) \right |^2 = \delta(q_x) \delta(q_y) {q_z}^{-2}$ As for the platelet in part I, the scattering is confined along a rod perpendicular to the interface, but its intensity, instead of being constant, decreases along its length. Further spreading this signal over the sphere adds an additional $${q}^{-2}$$ factor, for a final $${q}^{-4}$$ dependence.

### Crossover

Although both are infinitely extended in the plane, a single interface (Porod) and two interfaces very close together (thin plate) exhibit very different scattering laws. However, at high enough scattering vector, all objects reach the Porod regime. We will discuss this crossover for the case of a plate with finite thickness. The Fourier transform of this object along its normal is easily shown to be a cardinal sinus, so that: $$\left | \tilde{\rho}(q_z) \right |^2 = \left [ \frac{\sin(q_z a)}{q_z a} \right ] ^2$$. Close to the origin, ie. for $$q_z a \ll 1$$, this function is flat, hence the approximation $$\sim \mathrm{Cst}(q_z)$$ we used for the platelet in part I. At high $$q$$, on the other hand, it behaves as $$q^{-2}$$, yielding the Porod law. To put it differently, when the typical scale $$L = 2\pi/q$$ over which we observe the object is much larger than its thickness, we are in the "platelet" regime and do not resolve the two interfaces. When $$L \ll 2a$$, on the other hand, we only observe "one interface at a time", justifying the treatment above and thus the $$q^{-4}$$ law. This regime is "terminal" insofar there is no smaller typical length scale in the system, until that of the composing atoms or molecules.