Editing Thomson scattering (section)

==Description of the phenomenon== 
Thomson scattering describes the classical limit of electromagnetic radiation scattering from a free particle. An incident plane wave accelerates a charged particle which consequently emits radiation of the same frequency. The net effect is to scatter the incident radiation.<ref>{{Cite book |last=Jackson |first=John David |title=Classical electrodynamics |date=1975 |publisher=J. Wiley |isbn=978-0-471-43132-9 |edition=2nd |location=New York Chichester (GB) Brisbane (Australia) [etc.]}}</ref>{{rp|679}}

Thomson scattering is an important phenomenon in [[plasma physics]] and was first explained by the physicist [[J. J. Thomson]]. As long as the motion of the particle is non-[[special relativity|relativistic]] (i.e. its speed is much less than the speed of light), the main cause of the acceleration of the particle will be due to the electric field component of the incident wave. In a first approximation, the influence of the magnetic field can be neglected.<ref name=Froula/>{{rp|15}} The particle will move in the direction of the oscillating electric field, resulting in [[Dipole#Dipole radiation|electromagnetic dipole radiation]]. The moving particle radiates most strongly in a direction perpendicular to its acceleration and that radiation will be [[Polarization (waves)|polarized]] along the direction of its motion. Therefore, depending on where an observer is located, the light scattered from a small volume element may appear to be more or less polarized.

[[Image:Thomson scattering geometry.png|thumb|right|alt=Thomson scattering geometry|Incident photon comes from the left, with its electric field perpendicular to its path.  It hits the scattering electron, which absorbs it and vibrates, matching the incident field, generating the outgoing field.  The outgoing photon field matches the electron's motion, absorbs some of the energy, and exits to the bottom.]]

In the diagram, everything happens in the plane of the diagram.  Electric fields of the incoming and outgoing wave can be divided up into perpendicular components. Those perpendicular to the plane are "tangential" and are not affected.  Those components lying in the plane are referred to as "radial".  The incoming and outgoing wave directions are also in the plane, and perpendicular to the electric components, as usual.   (It is difficult to make these terms seem natural, but it is standard terminology.)

It can be shown that the amplitude of the outgoing wave will be proportional to the cosine of <math>\chi</math>, the angle between the incident and scattered outgoing waves. The intensity, which is the square of the amplitude, will then be diminished by a factor of cos<sup>2</sup>(<math>\chi</math>). It can be seen that the tangential components (perpendicular to the plane of the diagram) will not be affected in this way.

The scattering is best described by an [[emission coefficient]] which is defined as ''ε'' where ''ε'' ''dt'' ''dV'' ''d''Ω ''dλ'' is the energy scattered by a volume element <math>dV </math> in time ''dt'' into solid angle ''d''Ω between wavelengths ''λ'' and ''λ''+''dλ''. From the point of view of an observer, there are two emission coefficients, ''ε''<sub>''r''</sub> corresponding to radially polarized light and ''ε''<sub>''t''</sub> corresponding to tangentially polarized light. For unpolarized incident light, these are given by:
<math display="block">\begin{align}
\varepsilon_t &= \frac{3}{16\pi} \sigma_t In \\[1ex]
\varepsilon_r &= \frac{3}{16\pi}\sigma_t In \cos^2\chi
\end{align}</math>

where <math>n</math> is the density of charged particles at the scattering point, <math>I</math> is incident flux (i.e. energy/time/area/wavelength), <math>\chi</math> is the angle between the incident and scattered photons (see figure above) and <math>\sigma_t</math> is the Thomson [[Cross section (physics)|cross section]] for the charged particle, defined below. The total energy radiated by a volume element <math>dV </math> in time ''dt'' between wavelengths ''λ'' and ''λ''+''dλ'' is found by integrating the sum of the emission coefficients over all directions (solid angle):
<math display="block">
\int\varepsilon \, d\Omega = \int_0^{2\pi} d\varphi \int_0^\pi d\chi (\varepsilon_t + \varepsilon_r) \sin \chi = I \frac{3 \sigma_t}{16\pi} n  2 \pi (2 + 2/3) = \sigma_t I n.
</math>

The Thomson differential cross section, related to the sum of the emissivity coefficients, is given by
<math display="block">
\frac{d\sigma_t}{d\Omega} = \left(\frac{q^2}{4\pi\varepsilon_0 mc^2}\right)^2 \frac{1+\cos^2\chi} 2
</math>
expressed in [[SI]] units; q is the charge per particle, m the mass of particle, and <math>\varepsilon_0</math> a constant, the [[permittivity]] of free space. (To obtain an expression in [[Centimeter gram second system of units|cgs units]], drop the factor of 4{{pi}}''ε''<sub>0</sub>.) Integrating over the solid angle, we obtain the Thomson cross section
<math display="block">
\sigma_t = \frac{8\pi} 3 \left(\frac{q^2}{4\pi\varepsilon_0 mc^2}\right)^2
</math>
in SI units.

The important feature is that the cross section is independent of light frequency. The cross section is proportional by a simple numerical factor to the square of the [[classical electron radius|classical radius]] of a [[point particle]] of mass ''m'' and charge ''q'', namely<ref name=Froula/>{{rp|17}}
<math display="block">\sigma_t = \frac{8\pi} 3 r_e^2</math>

Alternatively, this can be expressed in terms of <math>\lambda_c</math>, the [[Compton wavelength]], and the [[Coupling constant|fine structure constant]]:
<math display="block">
\sigma_t = \frac{8 \pi} 3 \left(\frac{\alpha \lambda_c}{2\pi}\right)^2
</math>

For an electron, the Thomson cross-section is numerically given by:<ref name="NIST">{{cite web | url = http://physics.nist.gov/cgi-bin/cuu/Value?sigmae | title = National Institute of Standards and Technology | access-date = 3 February 2015}}</ref>
<math display="block">
\sigma_t =\frac{8 \pi} 3 \left(\frac{\alpha \hbar c}{m c^2}\right)^2 = 6.652 458 7321(60)\times 10^{-29} \text{ m}^2 \approx 66.5 \text{ fm}^2 = 0.665 \text{ b}
</math>