Thomson scattering

Thomson Scattering PhL 10.20.08 Summary ( all in Jackson units) 1 History: 2 1. Dipole Radiation Formulas 2 2. Application to Thomson scattering: radiation hits a free charge 3 3. Concerning the ε ε' factor. 4 4. The Thomson Cross section averaged over incoming polarization. 5 5. The total cross section (averaged over polarization) 6 6. Doing the polarization sums to get BD page 131 equation F. 7 7. Range of applicability 8 8. Comparison with Rayleigh Scattering. 8 Summary ( all in Jackson units) The general sinusoidal-time electric-dipole power radiation formula is this: dP/dΩ = (c/8π) k4 p2 sin2ψ where sinψ ≡ x cosψ ≡ where radiating dipole is p, and observer is a distant location r. If the observer is located at spherical coordinates r,θ,φ (large r), and if the incoming radiation has = , the above can be written as dP/dΩ = (c/8π) k4 p2 [ 1 - sin2θ cos2(φ-φp) ] where [ 1 - sin2θ cos2(φ-φp) ] = sin2ψ where φp indicates the direction of incoming polarization in the x-y plane. If a free electric charge is irradiated with an E&M plane wave, the charge acquires an electric dipole moment of this magnitude p = Eo e2/(mω2) = Eo e2/(mk2c2) and when this is inserted into the formula above, we get dP/dΩ = (c/8π) Eo2 [ e2/mc2]2 sin2ψ and we see no frequency dependence in the scattering. If we define a cross section according to dP = Sodσ where So = (c/8π) |Eo|2 , we get the following differential cross section for Thomson scattering dσ/dΩ = [ e2/mc2]2 sin2ψ = ro2 sin2ψ e2/mc2 = ro = "classical electron radius" If the initial polarization is ε = , the final polarization is ε' ={ - cosψ }/sinψ (both unit vectors) and ε ε' = sinψ. If we average over the continuum of initial polarization directions , the cross section becomes. d/dΩ = ro2/2 (1 + cos2θ) = ro2/2 (2 - sin2θ) If we then integrate over dΩ to get the total cross section, the result is = (8π/3)ro2 = (8/3) (π ro2) which is very small because ro is very small. History: "J.J." Thomson 1856-1940 came of with the theory of Thomson scattering in 1906, and this theory applies any place you have free charges -- plasmas. From Google books, There are many avenues leading to these results, so today let's pick a simple one. 1. Dipole Radiation Formulas On page 178 of Chapter 6 Jackson shows Maxwell's equations. On page 179 he shows that you can in effect replace the fields with the potential Aμ and you end up with some ugly coupled equations for the potential. He notes that the fields are unchanged by a gauge transformation Aμ → Aμ + ∂μΛ and this allows us to select a Lorentz Gauge in which ∂μAμ = 0 as in 6.36. In this gauge the messy equations decouple and become ( using H-L units just for the moment) Aμ = + Jμ using the BD metric gμv (∂a ∂a) = This allows the Green's function method to be used to find Aμ as an integral of Jμ. Starting on page 183, Jackson computes this Green's function for the equation ψ = f just to be general. The Green's is the usual δ(t - r/c)/r as shown in 6.64 page 185 and this leads to the solution ψ as a d4x integral over f times the Green's function as shown in 6.65. If you do the time integral, you get 6.66 which is then a d3x integral of 1/r times the "retarded" f. We then jump ahead to Jackson Chapter 9 page 269 where we see our solution of Aμ = + Jμ written in this manner 9.3 (now in Jackson units) A(x) = 1/c ∫d3x' Jret(x') / | x - x'| This is a very general result which we shall apply to our very simple Thomson scattering problem. It is best at this point to express A as a d4x' integral as shown in 9.2. Then when we install a sinusoidal time function for the current J(t') as in 9.1, we get an integral of δ(t' + r/c - t)dt' exp(-iωt') = exp(-iωt) exp(+ikr) since k = ω/c. We then define A(x,t) = A(x)e-iωt and then we end up with 9.3. It says this: A(x) =1/c ∫d3x' J(x') eikr/r and you see where that famous eikr/r factor comes from: (1) the 1/r is present in the Green's function, (2) the eikr comes from the retardation effect so we wind up phase as we move away. Another thing we notice is that A points in the same direction as J ! Now if we go far away from a localized current distribution, we can make the Born approximation and pull the eikr/r out of the integral to get A(x) = eikr/rc ∫d3x' J(x') as shown page 271 9.13. Jackson does a parts integration on the above integral, uses the continuity equation (which picks us up a factor of iω = ikc from the time derivative of ρ, and parts another minus) , and we end up with what we have been looking for here, our dipole radiation formula. The Born approximation just made above is thus equivalent to keeping only the dipole radiation part of the multipole radiation series. Our final dipole result then is this A(x) = -ikp eikr/r p = ∫d3x' x' ρ(x') = electric dipole moment of charge distributions which is 9.16. Again, A points in the same direction as p. Using this for A, we can then compute the fields in the radiation zone to get B = k2 eikr/r x p = p k2 eikr/r x = (p k2 eikr/r) sinψ E = B x = (p k2 eikr/r ) { - cosψ } // see below sinψ ≡ x cosψ ≡ as shown 9.19. The Poynting vector on page 205 is S = (c/8π)E x B* (with energy density U = (1/8π)E2). We have an outgoing plane wave so S = (c/8π) |B|2 and S dA = (c/8π) |B|2dΩ since dA = r2dΩ . The power flow into solid angle dΩ is then dP/dΩ = r2 (c/8π) |B|2 = r2 (c/8π) 2 k4 p2 /r2 * sin2ψ = (c/8π) k4 p2 sin2ψ // Jackson 9.23 p272 and this is the usual dipole antenna radiation result which is a donut pattern which peaks to the sides of the dipole antenna, but we express things in terms of p instead of current J. 2. Application to Thomson scattering: radiation hits a free charge Suppose we zap a free charged particle (m, e in Jackson units) with an E field plane wave. We have F = eEo = ma so a = (e/m)Eo . Assuming e-iωt time dependence for E we get v = a/(-iω) and then r = a/(-iω)2 = -a/ω2= -(e/mω2) Eo. The dipole moment is p = er ( in general it is Σqiri) so we get p = -(e2/mω2) Eo. So now we have a dipole moment to throw into our formulas above, with ω = kc. We have p = Eo e2/(mω2) = Eo e2/(mk2c2) = electric dipole magnitude Then the above radiation formula is dP/dΩ = (c/8π) k4(Eo2 e4/(m2k4c4))sin2ψ = (c/8π) Eo2 [ e2/mc2]2 sin2ψ where again cosψ = and we can think of = = the polarization of the incoming plane wave scattering beam. The incoming beam had So = (c/8π) |Eo|2 so set dP = Sodσ to define a cross section, then we get dσ/dΩ = [ e2/mc2]2 sin2ψ sinψ ≡ x cosψ ≡ = ro2 sin2ψ e2/mc2 = ro = "classical electron radius" = 3 x 10-13 cm You could hardly imagine a simpler formula. The distance dimension for the cross section has to come from the three constant shown, and the resulting ro gets the name shown. The number ro gives the general distance scale of E&M scattering from an electron. Notice that the scattering is independent of the frequency of the incoming light! Our general dipole radiation formula has k4 p2, but then our specific p ~ 1/k2 and the k's cancel. 3. Concerning the ε ε' factor. We know what our radiation zone fields look like for dipole radiation B = k2 eikr/r x = (p k2 eikr/r ) x E = B x = (p k2 eikr/r ) ( x ) x = (p k2 eikr/r ) { - ( ) } = (p k2 eikr/r ) { - cosψ } Thus, we know the direction of the final polarization unit vector ε' ε' = { - cosψ }/sqrt(1+cos2ψ - 2cos2ψ) = { - cosψ }/sqrt(1-cos2ψ) = { - cosψ }/sinψ and therefore ε ε' = { - cosψ }/sinψ = ( 1 - cos2ψ)/sinψ = sinψ which is the result I was looking for, ε ε' = sinψ so we can then say dσ/dΩ = = ro2 sin2ψ = ro2 (ε ε')2 // Thomson scattering where ε = and ε' = the polarization of the scattered beam going out in direction . Notice that electron spin does not enter our all-classical calculation at all. The above result agrees with the non-rel limit given in BD1 page 131 equations D and E. There is a distinction to notice however between the above formula and the BD one. In our formula here, ε = is the assumed initial polarization, and given that fact, we get exactly ε' as the polarization of the photon after scattering. We don't sum over final polarizations because our ε' is preselected to be in the exact direction of the final polarization. We could sum in the direction perp to this and that would add 0. Below we will average over the continuum of all possible positions of ε = , whereas in BD we will average by taking ½ the sum of initial polarizations. We expect results to be the same. 4. The Thomson Cross section averaged over incoming polarization. Suppose you have two unit vectors and in spherical coordinates. Then you can express each in Cartesian coordinates as = (sinθcosφ, sinθsinφ, cosθ) = (sinθpcosφp, sinθpsinφp, cosθp) then = sinθcosφ sinθpcosφp + sinθsinφsinθpsinφp + cosθcosθp Now suppose is tilted down to θp = 90 degrees and has no z component. Then we have = sinθcosφ cosφp + sinθsinφsinφp = sinθ (cosφ cosφp + sinφsinφp) = sinθ cos(φ-φp) = cosψ We have in mind a plane wave in the direction, where we know that polarization would then have to be in the x-y plane so the above would apply. Then we have cosψ = = sinθ cos(φ-φp) sin2ψ = 1 - sin2θ cos2(φ-φp) Now, we know our dσ/dΩ ~ sin2ψ from above, and we know that φp gives the direction of the incident polarization in the x-y plane. If we want a cross section which is averaged over incoming polarization, we do this: < (ε ε')2> = <sin2ψ > = 1 - sin2θ <cos2(φ-φp)> = 1 - ½ sin2θ = ½ (1 + cos2θ) since [ and see Portis page 539 for verification ] <cos2(φ-φp)> = (1/2π) dφp cos2(φ-φp) = (1/2π) dφp (1/2) [ 1 + cos(2[φ-φp]) ] = ½ so we then have d/dΩ = ro2 (ε ε')2 = ro2 <sin2ψ > = ro2/2 (1 + cos2θ) = ro2/2 (2 - sin2θ) If we take the non-rel limit of BD1 page 131 F, we set k=k' and the above formula is replicated! Recall that θ is the spherical angle coordinate of observation point r where incoming photons have = . In the above computation, for each initial polarization we get a "donut" radiation pattern with axis along (left plot showing sin2ψ = 1 - sin2θ cos2(φ-φp) ). That is, no radiation in the ± direction. When we average these donuts around azimuth, everything is additive on the top and bottom so we get peaks in the directions θ=0 and θ = π, namely ± where sinθ = 0. But things are reduced on the sides due to the effect of the donut hole nulls, so on the right is a plot of r = (2 - sin2θ) with 2 on the top and bottom, but only 1 on the sides. The photon is coming up from the bottom, so scattering is most in the forward and backward directions. 5. The total cross section (averaged over polarization) Now we can integrate the result of the previous section over all solid angle to get = (ro2/2) ∫ dΩ (1 + cos2θ) = π ro2 dx (1+x2) = 2π ro2dx (1+x2) = 2π ro2 (1 + 1/3) = (8π/3)ro2 = (8/3) π ro2 in agreement with Portis page 540 (20) and BD1 page 132 A. 6. Doing the polarization sums to get BD page 131 equation F. Assuming incoming along so we have these incoming polarizations (1) = (2) = The outgoing photon has these possible states: '(1) = '(2) = Then I can put ½ in front of KN, then just manually sum on these polarizations: Σpol (ε ε')2 = ((1) '(1))2 + ( (1) '(2) )2+ ((2) '(1) )2 + ((2) '(2) )2 = ( )2 + ()2 + ( )2 + ()2 = sin2φ + cos2θcos2φ + cos2φ + cos2θ sin2φ = = 1 + cos2θ Of course constant terms in the NK formula will do this under the same summation Σpol a = 4a So, if the K-N formula were this: KN = K * (a + 4(ε ε')2 - 2 ) Then our result would be KN = K/2 * (4a + 4 + 4cos2θ - 8) = 2K * (a + 1 + cos2θ - 2) = 2K * (a + cos2θ - 1) = 2K ( a - sin2θ). This then explains how we get from BD 7.74 to BD F. 7. Range of applicability We used F = ma to derive the electrons motion and dipole moment. This means the electron velocity is v << c . Looking at item 2 above, we have a = (e/m)Eo and v = a/(-iω) so this suggests that we need (e/m)Eo/ω << c or Eo << ω ( mc/e) = ω ( mc/e) Another overlooked factor is quantum mechanics and the recoil of the electron, so we have been assuming ω << mc2 . Combining these we get Eo (e/mc) << ω << mc2 We noted that the electron is supposed to be free, as in a plasma, but atomic electrons can be considered "free" and to do Thomson scattering as long as the incoming photon energy is not near any bound state transition energies. In general, x-ray photons do Thomson scattering from the electrons in atoms, but you have to modify the formulas somewhat introducing a "form factor". 8. Comparison with Rayleigh Scattering. In the Thomson case, we have p = er = -e/mω2 Eo . This 1/ω2 dependence cancels the k4 factor present in the general dipole radiation formula, so Thomson scattering is independent of wavelength. It is also pretty weak because the classical electron radius is very small. In the Rayleigh case, we have P = χeE or some such formula indicating we have an induced dipole moment in an atom or molecule or larger object. In this case, the susceptibility χe is often only weakly dependent on ω, so when we apply the general dipole radiation formula, we get dσ/dΩ ~ k4 ~ ω4 so Rayleigh scattering has a very strong 4th power frequency dependence and this is why the sky is blue and all that good stuff. Here we see the same (1 + cos2θ) angular factor we had above in the Thomson case, ro2/2 (1 + cos2θ), which arises from doing the polarization average. In Rayleigh, when the scattering object gets larger than the photon wavelength, the problem gets harder and it is called Mie scattering.