\(

\newcommand{\diff}{\mathrm{d}}

\renewcommand{\deg}{^{\circ}}

\newcommand{\e}{\mathrm{e}}

\newcommand{\A}{\unicode{x212B}}

\)

The ability of an electron to scatter an x-ray is described in terms of *differential scattering length* defined as follows:

\begin{align}

\label{Eq:dif_scattering_length}

\tag{1}

\left( \frac{\diff \sigma}{\diff \Omega}\right) \equiv \frac{I_{\rm{SC}}}{\Phi_0 \Delta \Omega},

\end{align}

where \(\Phi_0\) is the strength of the incident beam (the number of photons passing through a unit area per second), \(I_{\rm{SC}}\) is the number of scattered photons recorded per second in a detector positioned at a distance \(R\) away from the object, \(\Delta \Omega\) is a solid angle of the detector.

The values of the right-hand side in Eq. \(\eqref{Eq:dif_scattering_length} \)is also described by the electric fields of incoming and scattered x-ray with \(\Phi_0 = c \left| \vec{E}_{\rm{in}}\right|^2 /\hbar \omega\) and \(I_{\rm{SC}} = cR^2\Delta \Omega \left| \vec{E}_{\rm{rad}}\right|^2 /\hbar \omega\)

as follows1:

\begin{align}

\label{Eq:dif_scattering_length_electric_field}

\tag{2}

\left( \frac{\diff \sigma}{\diff \Omega}\right) = \frac{\left| \vec{E}_{\rm{rad}}\right|^2 R^2}{\left| \vec{E}_{\rm{in}}\right|^2}.

\end{align}

In a classical model of the elastic scattering of the x-ray, the scattered x-ray is generated by the electron forcedly vibrated by the electric field of the incoming x-ray 2. The radiated field is proportional to the charge of the electron \(-e\), and to the acceleration, \(a_X(t’)\), cause by the electric filed of the incident x-ray, which is a linearly-polarized in \(x-z\) plane, evaluated at a time \(t’\) earlier than the observation time \(t\) since the speed of light is finite value of \(c\). Thus, the electron field of the radiated x-ray is written as

\begin{align}

\label{Eq:rad_electric_field}

\tag{3}

E_{\rm{rad}}(R,t) \propto \frac{-e}{R}a_{X}(t’)\sin \Phi,

\end{align}

where \(t’ = t – R/c\). The full acceleration from the force on the electron is evaluated with Newton’s equation of motion as

\begin{align}

\label{Eq:acceleration}

\tag{4}

a_{X}(t’) =\frac{-e E_{0}\e^{-i\omega t’}}{m}

= \frac{-e}{m}E_{\rm{in}}\e^{i\omega (R/c)}

= \frac{-e}{m}E_{\rm{in}}\e^{ikR},

\end{align}

where \(E_{\mathrm{in}} = E_{0}\e^{-i\omega t}\) is the electric field of the incoming x-ray. Therefore, Eq. \(\eqref{Eq:rad_electric_field}\) can be rearranged to be

\begin{align}

\label{Eq:ratio_of_electric_field}

\tag{5}

\frac{E_{\mathrm{rad}}(R,t)}{E_{\mathrm{in}}} \propto \left( \frac{e^2}{m} \right)\frac{\e^{ikR}}{R} \sin \Psi.

\end{align}

In order to complete the derivation of the differential cross section of the electron, it is necessary to check the dimension of both members of Eq. \(\eqref{Eq:ratio_of_electric_field}\). First, the left-hand side is definitely dimensionless. On the other hand, the dimension of \(\e^{ikR}/R\) is the inverse of the length. Therefore, the proportionality coefficient of Eq. \(\eqref{Eq:ratio_of_electric_field}\) must have units of length. By noting that in SI units the Coulomb energy at distance \(r\) from a point charge \(-e\) is \(e^2/(4\pi \epsilon_{0}r)\) while the dimensionally the energy is also described as the form of \(mc^2\), the proportionality coefficient \(r_{0}\) is then written as

\begin{align}

\label{Eq:classic_radius_of_electron}

\tag{6}

r_{0} = \left( \frac{e^2}{4\pi \epsilon_{0} m c^{2} } \right) = 2.82 \times 10^{-5} \ \A.

\end{align}

This value is known as the Thomson scattering length or classical radius of the electron. By generalizing the relationship of the electric fields of incident and radiated x-rays, the ratio of the radiated electric field to the incident electric field described in Eq. \(\eqref{Eq:ratio_of_electric_field}\) is reduced to

\begin{align}

\label{Eq:complete_ratio_of_electric_field}

\tag{7}

\frac{E_{\mathrm{rad}}(R,t)}{E_{\mathrm{in}}} = -r_{0} \frac{\e^{ikR}}{R} \left| \hat{\vec{\epsilon}} \cdot \hat{\vec{\epsilon}}’\right|,

\end{align}

where “\(-\)” indicates the radiated x-ray has a different phase from that of the incident x-ray by 180\(\deg\) because the charge of the electron is negative, \(\hat{\vec{\epsilon}}, \hat{\vec{\epsilon}}’\) are unit vectors for the electric field of the incident and radiated x-rays.

Therefore, the differential cross-section becomes

\begin{align}

\label{Eq:derived_cross_section}

\tag{8}

\left( \frac{\diff\sigma}{\diff\Omega} \right) = r_{0}^{2}\left| \hat{\vec{\epsilon}} \cdot \hat{\vec{\epsilon}}’\right|^{2}.

\end{align}

The factor of \(\left| \hat{\vec{\epsilon}} \cdot \hat{\vec{\epsilon}}’\right|^{2}\) is called the Polarization factor and its value is dependent on the polarization of the incoming x-ray and experimental geometry:

\begin{align}

\label{Eq:polarization_factor}

\tag{9}

P = \left| \hat{\vec{\epsilon}} \cdot \hat{\vec{\epsilon}}’\right|^{2} =

\begin{cases}

1 & \text{horizontal linear polarization: vertical scattering plane} \\

\cos^{2} \Psi & \text{horizontal linear polarization: horizontal scattering plane}\\

\left( 1 + \cos^{2} \Psi \right)/2 & \text{unpolarized source: x-ray tube}

\end{cases}

\end{align}

The resultant equation predicts that the scattering intensity becomes very weak if \(\Psi\) is around 90\(\deg\) when a detector is scanned in the horizontal plane with the light source of horizontal linear polarization, which is definitely unfavorable to the usual scattering measurement3. This is the reason why a detector is vertically scanned in a synchrotron radiation facility, whose polarization is generally linear in the horizontal plane due to the electron orbit in a storage ring.

###### For further study…(This article was written based on the following book)

###### References

[Bibtex]

```
@book{Als-Nielsen2011,
author = {Als-Nielsen, J and McMorrow, D},
edition = {2nd},
publisher = {John Wiley \& Sons},
title = {{Elements of Modern X-ray Physics}},
year = {2011}
}
```