Form Factor:Cylindrical symmetry

Derivation

Form Factor

Assuming a particle is cylindrically-symmetric:

$\rho (\mathbf {r} )=\rho (r,\phi ,z)=\rho (r)$

We of course use cylindrical coordinates:

$\mathbf {r} =(x,y,z)=(r\cos \phi ,r\sin \phi ,z)$

We can take advantage of the cylindrical symmetry by rotating any candidate q-vector into the $q_{x},q_{z}$ plane, eliminating the $q_{y}$ component:

$\mathbf {q} =(q_{x},0,q_{z})$

${\begin{alignedat}{2}\mathbf {q} \cdot \mathbf {r} &=q_{x}x+q_{y}y+q_{z}z\\&=q_{x}r\cos \phi +q_{z}z\\\end{alignedat}}$

The form factor is (note that the integration limits in z define the particle size in that direction):

${\begin{alignedat}{2}F(\mathbf {q} )&=\int \rho (\mathbf {r} )e^{i\mathbf {q} \cdot \mathbf {r} }\mathrm {d} V\\&=\int \limits _{z=}^{L}\int \limits _{\phi =0}^{2\pi }\int \limits _{r=0}^{\infty }\rho (r)e^{i\mathbf {q} \cdot \mathbf {r} }r\mathrm {d} r\mathrm {d} \phi \mathrm {d} z\\&=\int \limits _{0}^{L}\int \limits _{0}^{2\pi }\int \limits _{0}^{\infty }\rho (r)e^{i(q_{x}r\cos \phi +q_{z}z)}r\mathrm {d} r\mathrm {d} \phi \mathrm {d} z\\&=\left(\int \limits _{0}^{L}e^{iq_{z}z}\mathrm {d} z\right)\int \limits _{0}^{\infty }r\rho (r)\left(\int \limits _{0}^{2\pi }e^{iq_{x}r\cos \phi }\mathrm {d} \phi \right)\mathrm {d} r\\&=\left(\left[{\frac {1}{iq_{z}}}e^{iq_{z}z}\right]_{z=0}^{L}\right)\int \limits _{0}^{\infty }r\rho (r)\left(\int \limits _{0}^{2\pi }e^{iq_{x}r\cos \phi }\mathrm {d} \phi \right)\mathrm {d} r\\&=\left({\frac {e^{iq_{z}L}-1}{iq_{z}}}\right)\int \limits _{0}^{\infty }r\rho (r)\left(\int \limits _{0}^{2\pi }e^{iq_{x}r\cos \phi }\mathrm {d} \phi \right)\mathrm {d} r\\\end{alignedat}}$