Thomson scattering, Rayleigh scattering and all that First consider scattering of (classical) electromagnetic waves by free electrons, called Thomson scattering. The electron experiences a force due to the incident electric ¯eld ~ i~k¢~r ¡ i!t E =² ^ E0 e where² ^ represents the polarization direction and² ^ ¢ ~k = 0. The equation of motion for the charged particle is m~rÄ = ¡eE:~ We recall that in the dipole approximation (valid since the wavelength of light is large compared to the scatterer) the power emitted per unit solid angle is 2 dP e 2 2 = 2 3 ha i sin θ d­ 16¼ ²0c where ha2i represents the time-averaged squared acceleration. The equation of motion gives the acceleration directly and the time averaged squared acceleration is e2 h a2 i = jE j2 : 2m2 0 This yields, upon rearranging, à ! 2 2 2 dP e ²0cjE0j 2 = 2 sin θ : d­ 4¼²0 mc 2 The term in the brackets is known as the classical electron radius: 2 e ¡15 2 = 2:8 £ 10 m : (1.1) 4¼²0 mc Now recall that the time-averaged Poynting vector which is the incident energy per 2 1 unit area per unit time is I ´ ²0cjE0j =2. The di®erential scattering cross section is de¯ned by dσ dP = d­ (1.2) d­ I 1From our previous work we have 1 c 1 S~ = E~ £ B~ =n ^ B~ ¢ B~ =n ^ E~ ¢ E~ ¹0 ¹0 ¹0c given that E~ = c B~ £ n^. The time-averaged Poynting vector is given by 1 c h S~ i =n ^ E~ ¢ E~ ¤ =n ^ B~ ¢ B~ ¤ : 2¹0c 2¹0 2 In the case of plane waves the power incident per unit area per unit time is hS~i ¢ n^ = ²0cjE0j =2. 1 and is the area of the wavefront which delivers the same power as is scattered into a given solid angle d­. So we obtain, dσ = r2 sin2 θ : (1.3) d­ c The total cross section is obtained by integrating over d­: Z ¼ 2 2 8¼ 2 σ = 2¼ rc dθ sin θ sin θ = rc : (1.4) 0 3 This is known as the classical Thomson cross section. When the incident energy of the photonh! ¹ becomes comparable to the rest mass of the electron quantum mechanical e®ects become important. This is the case of Compton scattering. 2 Rayleigh scattering Now consider scattering from a harmonically bound electron. In the simplest ap- proximation we have seen before it is described by the equation of motion given below: e ~rÄ + γ ~r_ + !2~r = ¡ E:~ 0 m We have solved this equation before and found ¡e=m ~ ~r = 2 2 E: !0 ¡ iγ! ¡ ! The time-averaged squared acceleration is à ! 2 4 Ä Ä e ! h ~r ¢ ~r i = 2 2 2 2 2 2 : 2m (! ¡ !0) + γ ! Thus the scattering cross-section known as the Rayleigh scattering cross section2 is given by à ! !4 σR = σT 2 2 2 2 2 (1.5) (! ¡ !0) + γ ! where σT is the Thomson scattering cross section. For strong binding (i.e., !0 >> !) we have µ ! ¶4 σR ¼ σT : (1.6) !0 So the total scattering cross section is given by the inverse fourth power of the wave- length and this was derived by Rayleigh. This formula is valid for scattering of light by air molecules (N2, O2) and shows that air molecules scatter the shorter wavelengths(blue) preferentially thus leading to blueness of the sky. Explain redness of the sunset by scattering of the blue light away from the line of sight. The wavelength of red light is 650nm while that of blue light is 450nm and their ratio is 1:44. Thus blue light is scattered 1:444 ¼ 4:3 times as e®ectively as red light. 2In the ¯rst paper, Lord Rayleigh, Phil. Mag. 12, 81-101 (1881) he used the concept of luminiferous ether and an elastic solid theory of the ether; he gave a derivation based on Maxwell's theory in Phil. Mag. 47, 375-394 (1899). 3.