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:

\begin{equation}

\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

\end{equation}

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:

\begin{equation}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

\end{equation}

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

\begin{equation}

\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})

\end{equation}

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 \(I(q)\):

\begin{equation}\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})

\end{equation}

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:\begin{equation}

\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})

\end{equation}

#### 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:

\begin{equation}

\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})}

\end{equation}

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:

\begin{equation}

\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)).

\end{equation}

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:

\begin{equation}

\label{eq:avgtheta2}

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

\end{equation}

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.↩}

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