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.

### Platelets: α = 2

For a platelet, this simplification yields for the electron density profile:

\[ \rho (\mathbf{r}) = \mathrm{Cst}(x) \mathrm{Cst}(y) \delta(z)\]

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 \(\rho (\mathbf{r})\), \(\tilde{\rho} (\mathbf{q})\). Since the Fourier transform of a constant is a Dirac delta and viceversa, we simply have:

\[ \tilde{\rho} (\mathbf{q}) = \delta(q_x) \delta(q_y) \mathrm{Cst}(q_z)\]

The intensity scattered at a given wave vector \( I(\mathbf{q}) = | \tilde{\rho} (\mathbf{q})|^2\), and we'll admit that the latter function has the same form as \( \tilde{\rho} (\mathbf{q}) \). This is plausible: if a distribution is perfectly localized, we expect its square to share this property. Rigorously speaking, however, the square of a Dirac delta makes no sense (within classical distribution theory) so a proper derivation involves considering a finite size for the object, taking the modulus square of its Fourier transform and only letting the size go to infinity in the last step. Finally, 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.

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...

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