The Poynting vector is therefore given by: S= e2 16π20c|n∧(n∧β

Electrodynamique S´erie 13: Thomson Scattering 2015

We discuss here the Thomson scattering of a point particle by an incoming electro- magnetic monochromatic wave. The radiation fields are given by:

E= e 4π₀c

1 R

n∧(n∧β)˙

, B= e

4π₀c² 1 R

h n∧

(n∧β)˙ ∧n i

, (1)

where R is the distance between the observer and the particle. The Poynting vector is therefore given by:

S= e²

16π²0c|n∧(n∧β)|˙ ² n

R², (2)

so the emitted power per unit solid angle is:

dP

dΩ = e²

16π²0c|n∧(n∧β)|˙ ². (3)

The incoming electric field of the incident wave can be written as:

E= E0e^{i(k·x−ωt)}+ c.c. =E0e^{i(k·x−ωt)}+ c.c. =

2(Re)E₀cos(k·x−ωt)−2(Im)E₀sin(k·x−ωt), (4) where E₀ is now a real constant and satisfies the conditions:

^∗·= 1, ·k= 0. (5) The equation of motion for the free particle is therefore:

mv˙ =e(E0e^{i(k·x−ωt)}+ c.c.), (6)

from which one can derive ˙β. The incoming energy flux can be derived from the Poynting vector, Sin= 20c|E₀|² = 20cE₀², from which we find the cross section as:

dσ dΩ = 1

Sin

dP

dΩ = e² 16π²0c

1 20cE₀²

e mc

2

|n∧(n∧E)|². (7)

The quantity |n∧(n∧E)|² must be averaged over time and over the initial polarization of the electromagnetic wave. Averaging over time, we get:

|n∧(n∧E)|² = |n∧h

n∧(E0e^{i(k·x−ωt)}+^∗e^{−i(k·x−ωt)} i

|² =

2 (n∧(n∧E0))·(n∧(n∧^∗E0)) = 2E₀²|n∧(n∧)|². (8) We can rewrite the last result using

[n∧(n∧)]_i=_ijkn_jklmn_lm = (δ_ilδ_jm−δ_imδ_jl)n_jn_lm= (n·)n_i−_i = (δ_ij−n_in_j)_j. (9) Averaging over polarizations, we then find:

<|n∧(n∧)|²>=<(δ_ij−n_in_j)_j(δ_ik−n_in_k)^∗_k >= (δ_ij−n_in_j)(δ_ik−n_in_k)< _j^∗_k> . (10)

Now, lives in the plane orthogonal to the wave vector of the electromagnetic wave;

assuming the wave is propagating in the z direction, z = (0,0,1), must be a vector in the (x, y) plane. If the incoming wave is not polarized, then< _j^∗_k > must be the most symmetric tensor in the (x, y) plane, so we can write:

< _j^∗_k >=a(δ_jk−z_jz_k), (11)

where ais a constant that can be obtained by noticing thatδ_iji^∗_j = 1. So we have:

1 =δij < i^∗_j >=a(3−1) = 2a, (12)

so that a= 1/2. We have finally that the average over polarizations can be written as:

< _j^∗_k >= 1

2(δ_jk−z_jz_k). (13)

We can now compute the average over time and polarization of the quantity in Eq. (9):

<|n∧(n∧)|² > = (δij−ninj)(δ_ik−nin_k)1

2(δ_jk−zjz_k) = 1

2(δ_jk−njn_k)(δ_jk−zjz_k)

= 1

2(3−1−1 + (n·z)²) = 1

2(1 + cos²θ).

(14) Putting together the results so far obtained in Eq. (7), we find:

dσ

dΩ = e⁴ 16π²c⁴(0m)²

1

2(1 + cos²θ). (15)

The total cross section is easily found from the previous expression:

σ_tot= Z

dΩdσ dΩ =

1 4π₀

2 e² mc²

2

8π

3 . (16)

In the case of the electron, the quantity r0= e²

4π0mc² = 2.82·10⁻¹³cm (17)

is called classical electron radius.

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