# Scattering by one electron


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}
\tag{3}
\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}
\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)
• J. Als-Nielsen and D. McMorrow, Elements of Modern X-ray Physics, 2nd ed., John Wiley & Sons, 2011.
[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}
}
###### References
[1] J. Als-Nielsen and D. McMorrow, Elements of Modern X-ray Physics, 2nd ed., John Wiley & Sons, 2011.
[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}
}