KevinYager at 17:17, 23 May 2022

2022-05-23T17:17:34Z

KevinYager: Created page with "==Derivation== ===Form Factor=== Assuming a particle is cylindrically-symmetric: :: $\rho(\mathbf{r}) = \rho(r,\phi,z) = \rho(r)$ We of course use [http://en.wikiped..."

2014-06-13T21:05:22Z

Created page with "==Derivation== ===Form Factor=== Assuming a particle is cylindrically-symmetric: ::<math>\rho(\mathbf{r}) = \rho(r,\phi,z) = \rho(r)</math> We of course use [http://en.wikiped..."

New page

==Derivation==
===Form Factor===
Assuming a particle is cylindrically-symmetric:
::<math>\rho(\mathbf{r}) = \rho(r,\phi,z) = \rho(r)</math>
We of course use [http://en.wikipedia.org/wiki/Cylindrical_coordinate_system cylindrical coordinates]:
::<math>
\mathbf{r}=(x,y,z)=(r \cos\phi, r \sin\phi, z)
</math>
We can take advantage of the cylindrical symmetry by rotating any candidate ''q''-vector into the <math>q_x,q_z</math> plane, eliminating the <math>q_y</math> component:
::<math> \mathbf{q}=(q_x,0,q_z) </math>
::<math>
\begin{alignat}{2}
\mathbf{q}\cdot\mathbf{r} & =q_x x + q_y y + q_z z \\
& = q_x r \cos\phi + q_z z \\
\end{alignat}
</math>

The form factor is (note that the integration limits in ''z'' define the particle size in that direction):
:<math>
\begin{alignat}{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^{i q_z z } \mathrm{d}z \right) \int\limits_{0}^{\infty} r \rho(r) \left ( \int\limits_{0}^{2 \pi} e^{i q_x r \cos \phi } \mathrm{d}\phi \right) \mathrm{d}r \\
& = \left( \left[ \frac{1}{i q_z}e^{i q_z z} \right]_{z=0}^{L} \right) \int\limits_{0}^{\infty} r \rho(r) \left ( \int\limits_{0}^{2 \pi} e^{i q_x r \cos \phi } \mathrm{d}\phi \right) \mathrm{d}r \\
& = \left( \frac{e^{i q_z L} - 1}{i q_z} \right) \int\limits_{0}^{\infty} r \rho(r) \left ( \int\limits_{0}^{2 \pi} e^{i q_x r \cos \phi } \mathrm{d}\phi \right) \mathrm{d}r \\

\end{alignat}
</math>

@@ Line 21: / Line 21: @@
 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_{z=0}^{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^{i q_z z }  \mathrm{d}z \right) \int\limits_{0}^{\infty}   r \rho(r) \left ( \int\limits_{0}^{2 \pi}   e^{i q_x r \cos \phi }  \mathrm{d}\phi  \right) \mathrm{d}r \\

Form Factor:Cylindrical symmetry - Revision history

KevinYager at 17:17, 23 May 2022

KevinYager: Created page with "==Derivation== ===Form Factor=== Assuming a particle is cylindrically-symmetric: ::\rho(\mathbf{r}) = \rho(r,\phi,z) = \rho(r) We of course use [http://en.wikiped..."

KevinYager: Created page with "==Derivation== ===Form Factor=== Assuming a particle is cylindrically-symmetric: :: $\rho(\mathbf{r}) = \rho(r,\phi,z) = \rho(r)$ We of course use [http://en.wikiped..."