Power laws in small-angle scattering - part I

The small-angle X-ray scattering (SAXS) spectrum of particles with a well-defined shape (such as rods or platelets) is often characterized by a power-law dependence: \( I(q) \sim q^{-\alpha}\), where the exponent \( \alpha \) is directly related to the particle geometry. For "compact" particles, the large-\( q \) intensity scales as \( q^{-4}\) (Porod regime). Below, I'll give the most compact and yet -hopefully- understandable derivation I can think of for these power laws.

To simplify the derivation, we'll consider these objects as infinitely thin and infinitely large, meaning that we'll be looking at them on length scales much larger than their thickness and much smaller than their lateral extension. The approximation is legitimate, since it is in this range of length (or, conversely, scattering vector) that the power-law regimes are encountered.
As discussed above, the Patterson function is similar to the density and thus we will apply the same approximation to \(P(\mathbf{r})\), which is the natural descriptor of the system, due to its intimate relation with the intensity \(I(\mathbf{q}) = \left | \tilde{\rho}(\mathbf{q}) \right |^2\). 

Platelets: α = 2

For a platelet, this simplification yields for the Patterson function:
\[ P (\mathbf{r}) = \mathrm{Cst}(x) \mathrm{Cst}(y) \delta(z)\]
where "Cst" (constant) means that the density does not vary as a function of \(x\) and \(y\). Of course, a constant does not need an argument, but we will specify it in order to keep track of the space dimensions.

We now need the Fourier transform of \(P (\mathbf{r})\), \(\mathcal{F}^{-1} [P (\mathbf{r})] = \tilde{P} (\mathbf{q})\). Since the Fourier transform of a constant is a Dirac delta and viceversa, we simply have:
\[ \tilde{P} (\mathbf{q}) = \delta(q_x) \delta(q_y) \mathrm{Cst}(q_z)\] From the Wiener-Khinchine theorem, the intensity scattered at a given wave vector is precisely \( I(\mathbf{q}) = \tilde{P} (\mathbf{q})\). Thus, the intensity scattered by a platelet perpendicular to the \(z\) axis is a thin rod parallel to \( q_z\), as shown in the Figure below.

Fourier transform of a platelet.

In solution, colloidal particles assume all possible orientations, so that this intensity is spread evenly over the sphere of constant \( q\). Consider two such spheres, with radii \( q_0\) and \( 2 q_0\). The rod contributes to each sphere the same amount, namely twice its (constant) intensity \( 2 I_0\), shown as red dots at the poles. This signal must however be divided by the surface area of the sphere, an area that increases as the square of the radius: \[ I(q_0) = \frac{2 I_0}{4 \pi q_0^2} \quad \mathrm{and} \quad I(2 q_0) = \frac{2 I_0}{4 \pi (2 q_0)^2}\] so that \( I(2 q_0) = I(q_0) /4\) and, finally, \( \alpha = 2\).

More power laws coming in part II of this post...