## 19 January 2016

### Scattering from a bunch of parallel wires

I got interested in this problem by trying to understand a result in [1], and also because it may be useful for some stuff I'm currently working on. Consider a collection of $$N$$ very (infinitely) long objects, parallel to the $$z$$-axis and whose centers have positions $$\mathbf{R}_{j}$$ in the $$(x,y)$$ plane. We are interested in the orientationally averaged scattering signal.
Left: diagram of the system showing the wires parallel to the $$z$$-axis. Right: In the $$(x,y)$$ plane, object $$j$$ is centered in $$\mathbf{R}_{j}$$.

The density function $$\rho (\mathbf{r})$$ is:

\label{eq:rhor}
\rho (\mathbf{r}) = C(z) \left\lbrace \left[ \sum_{j=1}^{N} \delta (\mathbf{r}_{\bot} - \mathbf{R}_{j}) \right] * F(\mathbf{r}_{\bot})\right\rbrace

where $$\mathbf{r} = (x,y,z) = (\mathbf{r}_{\bot}, z)$$ and $$F(\mathbf{r}_{\bot}) = F(\left| \mathbf{r}_{\bot} \right|)$$ is the (isotropic) in-plane density profile of the objects. $$*$$ is the convolution operator. $$C(z)$$ is the $$z$$ modulation of the density; for infinitely long objects it should be a constant, but for now we describe it by a gate function of length $$L$$: $$C(z) = \Pi_L(z)$$ and we'll take the $$L \to \infty$$ limit later.
The Fourier transform of $$\rho (\mathbf{r})$$ is given by:

\label{eq:rhoq}
\rho (\mathbf{q}) = \int \mathrm{d}^3 \mathbf{r} \, \mathrm{e}^{-i \mathbf{q} \mathbf{r}} \rho (\mathbf{r}) = C(q_z) \left\lbrace \left[ \sum_{j=1}^{N} \mathrm{e}^{-i \mathbf{q}_{\bot} \mathbf{R}_{j}} \right] \cdot F(\mathbf{q}_{\bot})\right\rbrace

and the intensity $$I (\mathbf{q}) \sim \rho^* (\mathbf{q}) \rho (\mathbf{q})$$ (with $$^*$$ the complex conjugation) writes:

\label{eq:Iq}
I (\mathbf{q}) = \left| C(q_z)\right|^2 \left| F(q_{\bot})\right|^2 \left\lbrace \left[ \sum_{j,k=1}^{N} \mathrm{e}^{-i \mathbf{q}_{\bot} (\mathbf{R}_{j} - \mathbf{R}_{k})} \right]\right\rbrace = \left| C(q_z)\right|^2 \left| F(q_{\bot})\right|^2 S(\mathbf{q}_{\bot})

In polar coordinates, $$\mathbf{q}_{\bot} = (q_{\bot}, \phi)= q_{\bot}(\cos(\phi), \sin(\phi))$$ and the form factor only depends on its modulus because it is isotropic, like its real-space counterpart. The distance between objects $$j$$ and $$k$$ is $$\mathbf{R}_{jk} = \mathbf{R}_{j} - \mathbf{R}_{k}$$ and it makes an angle $$\phi_{jk}$$ with the $$x$$ axis. The last factor in \eqref{eq:Iq} is the in-plane structure factor $$S(\mathbf{q}_{\bot})$$.

We denote the polar angle (between $$\mathbf{q}$$ and the $$z$$ axis) by $$\theta$$, so that $$q_z = q \cos (\theta)$$ and $$q_{\bot} = q \sin (\theta)$$. We are interested in the orientationally averaged intensity $$I(q)$$:

\label{eq:Iqavg}
I(q) = \left\langle I (\mathbf{q}) \right\rangle _{\theta, \phi}= \frac{1}{4 \pi} \int_{0}^{\pi} \sin (\theta) \mathrm{d} \theta \int_{0}^{2 \pi} \mathrm{d} \phi \, I (\mathbf{q})

It is convenient to begin by evaluating the integral over $$\phi$$, since the first two factors in \eqref{eq:Iq} do not depend on this angle.
Two cases appear:

#### In-plane liquid structure

$$S(\mathbf{q}_{\bot})$$ is isotropic (due to an additional ensemble average). We then have simply:

\label{eq:avgiso}
I(q,\theta) = \left\langle I (\mathbf{q}) \right\rangle _{\phi}= \left| C(q_z)\right|^2 \left| F(q_{\bot})\right|^2 S(q_{\bot})

#### Solid structure

$$S(\mathbf{q}_{\bot})$$ is not isotropic. The $$\phi$$ average consists in $$N(N+1)/2$$ terms (grouping together the $$jk$$ and $$kj$$ contributions) of the form:
$\frac{1}{2 \pi} \int_{0}^{2 \pi} \mathrm{d} \phi \cos \left[ q_{\bot} R_{jk} \cos(\phi - \phi_{jk}))\right] = J_0 (q_{\bot} R_{jk})$
yielding:

\label{eq:avgphi}
I(q,\theta) = \left\langle I (\mathbf{q}) \right\rangle _{\phi}= \left| C(q_z)\right|^2 \left| F(q_{\bot})\right|^2 \overbrace{\left[ \sum_{j , k=1}^{N} J_0 (q_{\bot} R_{jk}) \right]}^{S(q_{\bot})}

where, in the third factor, we retrieve once again an isotropic $$S(q_{\bot})$$.

For either \eqref{eq:avgiso} or \eqref{eq:avgphi}, we still need to average over $$\theta$$. Now is a good moment to introduce the simplifying assumption of infinite extent along $$z$$, yielding $$\lim\limits_{L \to \infty} \frac{\left| C(q_z)\right|^2}{L} = 2 \pi \delta (q_z)$$ (see below for the details). The integrated intensity diverges linearly with $$L$$ (while the intensity at the origin $$I(q_z = 0)$$ diverges quadratically with $$L$$), so we'll stop the limiting process when this parameter reaches $$H$$, the actual length of the objects (or beam size, whichever is shorter), but we'll assume that $$H$$ is much larger than all other length scales involved and use the Dirac delta anyway:

\label{eq:avgtheta}
I(q) = \left\langle I (q,\theta) \right\rangle _{\theta} = \frac{H}{2} \int_{0}^{\pi} \sin (\theta) \mathrm{d} \theta \, 2 \pi \delta (q \cos(\theta)) \left| F(q \sin(\theta))\right|^2 S(q \sin(\theta)).

Using $$\displaystyle \delta \left[ q \cos (\theta) \right] = \frac{1}{\left| q \sin (\theta) \right| } \delta (\theta - \pi/2)$$ ([2], Theorem 7.45), we finally get:

\label{eq:avgtheta2}
I(q) = \frac{2 \pi H}{2 q} \left| F(q)\right|^2 S(q)

where, to be perfectly clear, $$I$$ is a function of $$q = \left| \mathbf{q} \right|$$ and $$F$$ and $$S$$ are functions of the in-plane scattering vector $$q_{\bot}$$, to be evaluated in $$q_{\bot} = q$$, since $$q_{z} = 0$$. The $$q$$ in the denominator appears formally from the derivative of the Dirac delta, but it can be understood intuitively as a result of the scattering signal, initially confined to the equator of a sphere (with circumference $$2 \pi q$$) being spread by the averaging over the entire surface (of area $$4 \pi q ^2$$), resulting in "dilution" by a factor of $$2q$$.

Inserting in \eqref{eq:avgtheta2} the $$S(q_{\bot})$$ obtained in \eqref{eq:avgphi} yields Equation (7) in reference [1].

#### Bringing in the $$\delta$$ function

Let us prove the $$\left| C(q_z)\right|^2 =2 \pi L \delta (q_z)$$ relation used in \eqref{eq:avgtheta}. We have
$C(z) = \Pi_L(z) = \left\lbrace \begin{array}{ll} 1 & \left| z\right| \leq L/2 \\ 0 & \text{otherwise} \end{array} \right.$
and $$C(q_z) = L \, \mathrm{sinc} \left( q_z \frac{L}{2}\right)$$. It can then be shown ([2], Example 8.19) that $$\displaystyle \frac{\left|C(q_z)_L \right| ^2}{L} = L \, \mathrm{sinc} ^2 \left( q_z \frac{L}{2}\right)$$ is a Dirac sequence and it converges towards $$2 \pi \delta (q_z)$$. I will not prove it here, but it is easy to see that $$\left|C(q_z)_L \right| ^2$$ becomes narrower as $$L$$ increases. As to the prefactor, note that, with our particular convention for the Fourier transform used in \eqref{eq:rhoq}, the one-dimensional Parseval-Plancherel relation is written ([2], p. 612, line 3):
$\int \mathrm{d}q_z f^*(q_z) g(q_z) = 2 \pi \int \mathrm{d}z f^*(z) g(z)$

Thus, the trivial integral $$\displaystyle \int \mathrm{d}z \left| C(z) \right| ^2 = L$$ yields $$\displaystyle \int \mathrm{d}q_z \left| C(q_z) \right| ^2 = 2 \pi L$$, a result more difficult to obtain by direct integration.

[1] S. Rols, R. Almairac, L. Henrard, E. Anglaret and J.-L. Sauvajol, Diffraction by finite-size crystalline bundles of single wall nanotubes, Eur. Phys. J. B 10, 263-270 (1999).
[2] Walter Appel, Mathematics for Physics and Physicists, Princeton University Press,
2007.