# Difference between revisions of "Thomson Scattering"

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## Basics

Electric fields exert a force on charged particles causing them to accelerate. As seen in the derivation of the Larmor formula, accelerated charged particles emit radiation. So overall, incoming radiation might be off-scattered by charged particles. Here, we will consider incoming radiation being off-scattered by an electron, the so called Thomson scattering. Note that this requires a low energy, low intensity ${\displaystyle {\vec {E}}}$-field and a low energy electron. Other limits will be discussed later in the course, see Compton Scattering.

An incoming electromagnetic wave with an angular frequency ${\displaystyle \omega _{0}}$ passes an electron of mass ${\displaystyle m_{e}}$. Model this wave in its easiest form by

${\displaystyle E=E_{0}\sin(\omega _{0}t){\hat {z}}.\,\!}$

The electron is accelerated by this external electric field, so that its motion is described by

${\displaystyle m{\ddot {z}}=eE_{0}\sin(\omega _{0}t).\,\!}$

The root-mean-square average acceleration, which will be important for determining the radiated power is

${\displaystyle a_{\text{rms}}={\sqrt {\langle a^{2}\rangle }}={\frac {eE_{0}}{2m_{e}}}.\,\!}$

In order to oscillate with the same frequency as the incoming radiation, the photon energy needs to be significantly smaller than the rest energy ${\displaystyle m_{e}c^{2}}$ of the electron. This statement is equivalent saying that the wavelength of the incoming wave needs to be larger than the Compton wavelength ${\displaystyle h/m_{e}c}$, a characteristic scale at which quantum mechanics becomes important. At the same time, we assumed the electron to be at low velocity and the amplitude of the incoming wave to be small, otherwise a relativistic description is needed.

We now try to determine the cross-section of this interaction. A trick to be used here is to expand the differential power scattered into a solid angle with ${\displaystyle d\sigma }$.

${\displaystyle {\frac {dP}{d\Omega }}={\frac {dP}{d\sigma }}{\frac {d\sigma }{d\Omega }}=\langle S\rangle {\frac {d\sigma }{d\Omega }}={\frac {cE_{0}^{2}}{8\pi }}{\frac {d\sigma }{d\Omega }},\,\!}$

where ${\displaystyle \langle S\rangle ={\frac {dP}{d\sigma }}}$ is the flux passing the cross-section of the electron, which is related to the amplitude of the incoming field via the time-averaged Poynting flux.

Integrating the above equation and recalling the Larmor formula for ${\displaystyle P}$, we obtain:

{\displaystyle {\begin{aligned}\sigma _{T}&={\frac {8\pi }{cE_{0}^{2}}}P\\&={\frac {8\pi }{cE_{0}^{2}}}{\frac {e^{2}a_{\text{rms}}^{2}}{3c^{3}}}\\&={\frac {8\pi }{3}}{\frac {e^{4}}{m_{e}^{2}c^{4}}}={\frac {8\pi }{3}}r_{0}^{2},\end{aligned}}\,\!}

where ${\displaystyle r_{0}}$ is the classical electron radius, which can be easily re-derived by requiring that the potential Coulomb energy makes up all of the electron’s rest energy by bringing in charge within a radius ${\displaystyle r_{0}}$ that suffices this condition.

Concluding, the cross-section is off from a pure, classical cross-section for this interaction by a factor of ${\displaystyle 8/3}$.

${\displaystyle \sigma _{T}={\frac {8\pi }{3}}r_{0}^{2}=6.65\cdot 10^{-25}{\text{cm}}^{2}\,\!}$

## Examples

The Power spectrum of BAO over angular scales.

Thomson scattering is important on very different scales in Astronomy. For example:

• At high temperatures, Thomson scattering can dominate the photon diffusion in a star and thus determines the energy flux making its way out of the star.
• At the epoch of recombination (${\displaystyle z\approx 1100}$), Thomson scattering caused the high opacity of the plasma. The effects of Thomson scattering are, for example, visible in the diffusion damping of Baryon Acoustic Oscillations. For high l-moments (${\displaystyle l\geq 1000}$), Thomson scattering set the length over which photons could diffuse. This directly determines the damping of small scales (high moments) as inhomogeneities can be leveled out by photons.