Thompson scattering

1. Introduction

The scattered radiation cross section varies with respect to the scattered angle. It shows where is probable to find a scattered radiation.

An electromagnetic wave propagates carrying two fields wich are its components. They are the electriv field, E and the magnetic fiel B. The vectors corresponding to these fields are perpendicular to each other and to the direction of the propagation (the poynting vector).

When an electromagnetic wave is incident on a charged particle, its electric and magnetic components exert a Lorentz force on the particle. The incident wave varies (is periodic) in time, so is the exerted force. Once the particle is accelerated, It emits, in its turn, a radiation.

In other words, the energy lost by the incident wave is gained (absorbed) by the particle target, setting it in a motion. This particle reemits this energy as electromagnetic radiation. This process is called scattering of the electromagnetic wave by the particle.

Two processes could occur: The exchange of energy is inelasctic (the emitted wave has not exactly the same energy as the incident wave, the difference is used by the electron to recoil). This is called Compton Scattering. If the exchange of energy is elasctic, the emitted wave has exactly the same energy (the same frequency) as the incident wave). The related process is called Thompson Scattering.

2. Thompson scattering

We consider the approximation of small velocities, so the magnetic component of the Lorentz force is neglected. The particle undergoes only the electric field effect, thus the equation of motion of the charged particle is approximately : f = q E. Also the particule is assumed to be free electron (that is its binding enery is less than the enegy hν of the incident photon).

We have two particular cases:

2.1. The wave is polarized:
Consider a linearly polarized, monochromatic, plane wave (E_l, B) incident on a particle (free electron). If the polarization is set over the z axis, the field E will remain parallel to z axis (vertical), so is the acceleration of the target electron (because F = q E). We know that the direction of the radiated electric field E is given by r x (r x a) ("r" is the position vector of the electric field from the charge). Hence the scattered wave will move from un element of scattering volume, in the plane (x,y), all over the z axis. The electric field of the scatterd wave will remains parallel to the incident field. The field E_l points towards a direction perpendicular to the palne (x,z). As it depends on the time, It will oscillate vertically over the an axis parallel to the z axis. The remitted (scattered) radiation wave goes in the "y" direction.

2.2. The general case: Now the incident radiation is not polarized (or randomly polarized). Then the acceleration of te particle target takes any direction in the space. So is its electric field E. For any direction of the incident field in space, we have the acceleration "a" that points in any direction. So is the corresponding electric field (E_a). For a given angle θ in the plane (y,z), the electric filed E will points in direction of its polarization vector P. The vextor P is in the (x,z) plane. The angle between r and P is φ (φ = π/2 - θ) that takes all the values from 0 to 2π around the x axis. Plus, for one value of φ we have to take all directions in the plane (x,z); that is all the possible values of the angle ψ, from 0 to 2π.

The calculation of the angular distribution of the scattering radiation cross section will have the following steps:

For each angle θ we consider all values fo the angle φ for all values of ψ.

We can write:
r projected over z axis gives h = r sin θ. Projected again over the polarization axis, gives d = r sin θ cos ψ. But this value is equal to r cox φ Thus:
cox φ = sin θ cos ψ (1)

3. Differential cross section of the Thompson scattering

Let's write:
E_int = E₀ sinωt , the expression of the incident electrical field. The vector electric field E is parallel to the z-axis; it becomes E cos θ in order to be perpendicular to the wave vector "r". For the related scattered electromagnetic wave, the electric field has the following expression:
E_out = (e/4πε₀r c²) a cos θ = (e/4πε₀r c²) a sin φ;
where "a" is the acceleration of the particle = ∂s/∂t, and s is any displacement of this particle from the origin, "e" is the charge of the particle (the electron), and m is its mass.

The equation of motion of the charged particle is: m a = e E_int = e E₀ sin ωt
Hence: a = (e/m) E₀ sin ωt (2)

The time average of each Poynting vector for the two electric fiels are:
S_int = c ε₀ <E_int²> = c ε₀ E₀² <sin²ωt>
S_out = c ε₀ <E_out²>
We have :
sin²x = (1-cos2x)/2
then :<sin²x>= ∫dx (1-cos2x)/2 /∫dx = 1/2 (3)
The integration is set from 0 to 2π
Thus:
S_int = c ε₀ E₀²/2 (4)
S_out = c ε₀ (e/4πε₀)² <a²> <sin²φ>/c⁴ r²

From the equations (2) and (3), we get <a²> = (1/2) (e E₀/m)²
<sin²φ> = ∫dψ sin²φ /∫dψ [ψ: from 0 to 2π] = (1/2π) ∫dψ (1 - cos²φ)
Using the formula (1), we have:
∫dψ cos²φ = ∫dψ cos²ψsin²θ = sin²θ ∫dψ cos²ψ

∫dψ cos²ψ = ∫dψ (1/2 + cos 2ψ) = π Hence:
<sin²φ> = (1/2π)(2π - π sin² θ) = (1 + cos²θ)/2 (5)
Then:
S_out = c ε₀ (e/4πε₀)²) .(1/2) (e E₀/m)²/ c⁴ r² . (1/2)(1 + cos²θ)
= (e² E₀) ² (1 + cos²θ) /4(4π)²ε₀ c³ r² m²

The differential scattering cross section per electron is given by:
dσ_Th/electron = S_out dA /S_inc
Where dA is the elementary surface = dV/dr = r² 2π sinθdθ = r² dΩ
Thus:
dσ_Th/electron = S_out r² dΩ /S_inc
dσ_Th/electron/ dΩ = r² S_out /S_inc

We have:
S_out /S_inc = (e²/4πε₀ m c²)² (1/r)² ( 1 + cos²θ)

Let: r_e = e²/4πε₀ m c² = 2.8 x 10^-15 m, the classical electron radius . Hence:
S_out /S_inc = (1/2) (r_e/r)² ( 1 + cos²θ)

Finally:
dσ/dΩ (Thomson-Electronic) = (r_e²/2)(1 + cos²θ) (6)
Which is the differential scattering cross section with respect to the solid angle: dΩ = 2πsin θdθ. This differential cross section depends only on the angle θ.

5. Total cross section