Scattering. March 20, 2016

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Scarlett Blankenship
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1 Scattering March 0, 06 The scattering of waves of any kind, by a compact object, has applications on all scales, from the scattering of light from the early universe by intervening galaxies, to the scattering of protons into Higgs particles at the Large Hadron Collider. Between these extremes lie myriad phenomena from rainbows and the blue sky to geometric optics. The same basic principles apply to the description of all of these phenomena. Scattering of light depends on the size of the scatterer relative to the wavelength of the light. For wavelengths much smaller than the scattering object, geometric optics gives an adequate description. At larger wavelengths, corrections to geometric optics may be found. At the other extreme, it is possible to do a treatment in terms of lowest-order multipoles. Between these extremes, a full multipole treatment is required. Differential cross-section We will be especially interested in the differential cross-section: the probability of scattering into a given solid angle. In the radiation zone, the time-averaged radiated power per unit area is given by the Poynting vector, dp da ˆr S. Writing the time-averaged radiated power as S E H, gives the time-averaged radiated power per unit area dp da n E H For electric dipole or quadrupole fields in the radiation zone we found that H ε0 µ 0 n E, so this becomes dp da E n ε0 E µ 0 E Z 0 ε0 µ 0 n E Finally, we write the area element da r in terms of the solid angle, to get the time-averaged radiated power per unit solid angle, dp r E Z 0 If we desire the cross-section for a particular polarization, we consider only that component of the electric field, dp r ε E Z 0

2 In scattering experiments, a target is struck by many incoming waves, so we express the outgoing power as a probability for scattering in a given solid angle. To do this, we normalize by the incident power per unit area. The result is the differential cross-section dp scattered dp incident /da Z 0 r ε E scattered Z 0 ε E inc r ε E scattered ε E inc Because we divide by the incident power per unit area, the differential cross-section and the total cross-section have units of area. The total cross-section, σ, is found by integrating over all angles ˆ σ πˆπ 0 0 sin θdθdϕ The total cross-section may be thought of as the effective cross-sectional area of the target. For long wavelength or small scatterers, we may assume an incoming polarized plane wave, E sc ε 0 E 0 e ikn0 x H sc Z 0 n 0 E sc which induces electric and magnetic dipole moments, p, m, in the scatterer. Then the scattered radiation is the resulting dipole radiation, E sc k [n eikr p n c ] 4πɛ 0 r n m H sc Z 0 n E sc Now, substituting into the differential cross section, we have n, ε, n 0, ε 0 Z 0 r ε E scattered Z 0 ε 0 E inc r ε 4πɛ 0 k eikr r [ n p n c n m] E 0 [ 4πɛ 0 E 0 ε p n p n ] c ε n m [ 4πɛ 0 E 0 ε p n p ε n ] c m ε n 4πɛ 0 E 0 ε p + c m n ε We have therefore reduced the scattering problem to finding the induced polarization and magnetization. These induced properties generically involve the direction, n 0, and polarization, ε 0 of the incident light. Notice that the differential cross section depends on. This dependence, called Rayleigh s law, will be the case unless both dipole moments vanish - quadrupole radiation will depend on k 6, and so on.

3 Example : Scattering by a small dielectric sphere. The outgoing electric field For a dielectric sphere much smaller than the wavelength, we may treat the electric field as momentarily constant across the sphere. Recall the solution for a dielectric sphere in a constant field. We start with a pair of series solutions for the potential inside and out, Φ in A l r l P l cos θ l0 Φ out E 0 r cos θ + B l r l+ P l cos θ where the first term in Φ out gives the constant field at large distances from the sphere. The remaining terms incorporate the boundary conditions at the origin and infinity. Then, equating the tangential E and normal D fields we equate like coefficients to find the solution l0 with the field outside being Φ in 3ɛ 0 E 0 r cos θ Φ out E 0 r cos θ + ɛ ɛ 0 E 0 a 3 r cos θ E out Φ out where we used ˆn r Φ out ˆθ r ˆn r θ Φ out E 0 r cos θ + ɛ ɛ 0 E 0 a 3 r E 0ˆk + ɛ ɛ 0 E 0 a 3 r 3 ˆn cos θ + cos θ ˆθ r ɛ ɛ0 E 0 a 3 r 3 E 0ˆk + ɛ ɛ 0 E 0 a 3 r 3 ˆn cos θ + ˆθ sin θ E 0ˆk + ɛ ɛ 0 E 0 a 3 r 3 3ˆn cos θ E 0ˆk + ɛ ɛ 0 E 0 a 3 r 3 3ˆn cos θ ˆk E 0 r cos θ + ɛ ɛ 0 E 0 a 3 θ r cos θ ˆθ sin θ ˆn cos θ ˆθ sin θ ˆθ ˆn sin θ ˆk cos θ ˆk ˆn cos θ ˆθ sin θ Notice that the potential inside is just proportional to z r cos θ, so the induced electric field inside is parallel to the applied field, but changed generally reduced in magnitude by We know that a dipole p pˆk at the origin produces an electric field E 3ˆn ˆn p p 4πɛ 0 r 3 p 4πɛ 0 r 3 3ˆn cos θ ˆk 3 3ɛ0 ɛ+ɛ 0.

4 Comparing this dipole field to the non-constant part of the exterior field, ɛ ɛ 0 E 0 a 3 r 3 3ˆn cos θ ˆk p 4πɛ 0 r 3 3ˆn cos θ ˆk we see that we can identify the dipole strength as ɛ p 4πɛ 0 E 0 a 3 ɛ0 ˆk. Differential cross section Now return to the differential cross-section. Since there is no magnetic dipole moment, we set m 0, leaving n, ε, n 0, ε 0 4πɛ 0 E 0 ε p From here there are several cases, depending on the polarizations of the incoming and outgoing waves. We have computed the dipole moment assuming the electric field is in the z-direction, but to find the angular distribution it is easier to let the incoming wave propagate in the z-direction. Rotating the coordinate system so that the incoming wave moves in the ˆk-direction with polarization in the ˆε 0 -direction, we have E E 0ˆε 0 e ikz iωt ɛ p 4πɛ 0 E 0 a 3 ɛ0 ˆε 0 where the polarization vector, ˆε 0, may be in any combination of the x and y directions. Notice that the induced polarization is parallel to the polarization of the incident wave. It makes no difference which direction we choose for the incident polarization, since changing it is simply a change in the origin of the ϕ coordinate. We may always choose the x-axis to be the polarization direction, ε 0 î For the outgoing wave, the possible linear polarizations are in the directions orthogonal to the outward radial unit vector, n. It is convenient to write out the unit vectors for spherical coordinates: n î sin θ cos ϕ + ĵ sin θ sin ϕ + ˆk cos θ ˆθ î cos θ cos ϕ + ĵ cos θ sin ϕ ˆk sin θ ˆϕ î sin ϕ + ĵ cos ϕ It is easy to check that these are orthonormal. The outward moving, scattered wave may have polarization in any combination of the ˆθ and the ˆϕ directions. We can now compute the differential cross-section. If the measured polarization is in the ˆθ direction, n, ε ˆθ, n 0 ˆk, ε 0 î 4πɛ 0 E 0 ε p ɛ 4πɛ 0 E 0 ˆθ 4πɛ 0 E 0 a 3 ɛ0 î 4πɛ 0 E 0 4πɛ0 E 0 a 3 ɛ ɛ0 ˆθ î 4πɛ 0 E 0 4πɛ0 E 0 a 3 ɛ ɛ0 cos θ cos ϕ ɛ ɛ0 a 6 cos θ cos ϕ 4

5 Notice that a 6 has units of area as expected. If we measure the polarization in the ˆϕ direction, we have n, ε ˆϕ, n 0 ˆk, ε 0 î 4πɛ 0 E 0 ε p ɛ 4πɛ 0 E 0 ˆϕ 4πɛ 0E 0 a 3 ɛ0 î ɛ ɛ0 a 6 ˆϕ î ɛ ɛ0 a 6 sin ϕ If we want the initial polarization to be in the y-direction, it is simply a matter of replacing ϕ by ϕ π in the results above. For outgoing polarization in the ˆθ direction this gives, n, ε ˆθ, n 0 ˆk, ε 0 ĵ ɛ ɛ0 a 6 cos θ cos ϕ π ɛ ɛ0 a 6 cos θ sin ϕ while for the polarization in the ˆϕ direction, we have n, ε ˆϕ, n 0 ˆk, ε 0 ĵ ɛ ɛ0 a 6 sin ϕ π ɛ ɛ0 a 6 cos ϕ.3 Unpolarized incoming wave When the incoming light is unpolarized, we average over the possible incoming polarizations. For outgoing polarization in the ˆθ direction, this gives n, ε ˆθ, n 0 ˆk [ n, ε ˆθ, n 0 ˆk, ε 0 î + n, ε ˆθ, n 0 ˆk, ] ε 0 ĵ [ ɛ ɛ0 ɛ a 6 cos θ cos ɛ0 ϕ + a 6 cos θ sin ϕ] ɛ ɛ0 a 6 cos θ which depends only on θ, while for the outgoing polarization in the ˆϕ direction, n, ε ˆϕ, n 0 ˆk [ n, ε ˆϕ, n 0 ˆk, ε 0 î + n, ε ˆϕ, n 0 ˆk, ] ε 0 ĵ [ ɛ ɛ0 ɛ a 6 sin ɛ0 ϕ + a 6 cos ϕ] k4 a 6 ɛ ɛ0 which has no angular dependence. There is no preferred ϕ dependence in either case. 5

6 .4 Final polarization not measured As a final possibility, suppose we have unpolarized light coming in, and we do not measure the outgoing polarization. Then the result is the sum of the results for the outgoing radiation, n, n 0 ˆk.5 Total cross-section n, ε ˆθ, n 0 ˆk k4 a 6 ɛ ɛ0 + + cos θ n, ε ˆϕ, n 0 ˆk The total light scattered gives an estimate of the size of the scatterer. In the present case, for unpolarized light, we integrate over all angles, ˆ σ ˆπ 0 ˆ π 0 k4 a 6 ɛ ɛ0 + cos θ The integral is so the total cross section is + cos θ 6π 3 σ 8 ɛ ɛ0 3 πa a 4 which in this case is the cross-sectional area, πa, times a dimensionless factor depending on the ratio a λ. 3 Example : Scattering by a small, perfectly conducting sphere For a perfectly conducting sphere, the boundary conditions change. In problem you are asked to work out this case, so here we simply state the results for the electric and magnetic dipole strengths: p 4πɛ 0 a 3 E inc m πa 3 H inc For linear polarization, E inc and H inc are orthogonal and orthogonal to the direction of propagation, E inc E incˆε 0 e in0 x iωt H inc µ 0 c n 0 E inc so the dipole strengths are perpendicular as well. We can immediately write the differential cross-section, n, ε, n 0, ε 0 4πɛ 0 E inc ε p + c m n ε 4πɛ 0 E inc ε 4πɛ 0 a 3 πa 3 E inc H inc n ε c 6

7 4πɛ 0 E inc ε 4πɛ 0 a 3 E inc πa 3 µ 0 c n 0 E inc n ε ε 4πɛ 0 4πɛ 0 a 3ˆε 0 πa 3 ɛ 0 n 0 ˆε 0 n ε a 6 ε ˆε 0 n 0 ˆε 0 n ε Now compute the differential cross-section in the case of unpolarized incident light. average over incident polarizations, ˆε 0 and ˆε 0, where This means we to find ˆε 0 ˆε 0 n, ε, n 0, ε 0 sin θ n 0 n sin θ n 0 n n 0 a 6 ε ˆε 0 n 0 ˆε 0 n ε ˆε 0,ˆε 0 For the outgoing polarization perpendicular to the plane of incidence we still have ε sin θ n 0 n but for the outgoing polarization perpendicular to the plane of incidence we need Notice that ε sin θ n 0 n n ˆε 0 ε ˆε 0 ε 0 ˆε 0 ε 0 ˆε 0 ε cos θ Outgoing polarization parallel to the plane of incidence vector products, we will need For the parallel case, working through the ε ˆε 0 cos θ ε ˆε 0 0 n 0 ˆε 0 n ε n 0 ˆε 0 n ε n0 ˆε 0 n ε 0 n0 ˆε 0 n ε n0 ˆε 0 n ε 7

8 Therefore, a 6 ε ˆε 0 n 0 ˆε 0 ˆε 0,ˆε 0 k4 a 6 ε ˆε 0 k4 a 6 cos θ n ε n0 ˆε 0 n ε + k4 a 6 ε ˆε 0 n 0 ˆε 0 n ε Outgoing polarization perpendicular to the plane of incidence the vector products For the perpendicular case, we need ε ˆε 0 0 ε ˆε 0 n 0 ˆε 0 n ε n 0 ˆε 0 n ε cos θ n0 ˆε 0 n ε n0 ˆε 0 n ε 0 so the perpendicular contribution to the differential cross-section is k4 a 6 ε ˆε 0 n0 ˆε 0 n ε k4 a 6 cos θ k4 a 6 cos θ If the final polarization is unmeasured, we sum these, giving + k4 a 6 ε ˆε 0 n 0 ˆε 0 n ε + k4 a 6 cos θ + k4 a 6 cos θ [ k4 a 6 cos θ cos θ + + cos θ + ] 4 4 cos θ [ 5 k4 a 6 4 cos θ + 5 ] 4 cos θ [ 5 a 6 8 cos θ + 5 ] 8 cos θ We define the polarization of the scattered radiation to be the difference between the parallel and perpendicular cross-sections, normalized by the total differential cross-section, Π [ ] 8

9 a [ ] cos θ cos θ [ 5 8 cos θ cos θ cos θ cos θ sin θ 5 cos θ 8 cos θ + 5 [ k4 a 6 cos θ k4 a 6 cos θ cos θ cos θ + 4 ] cos θ + 4 cos θ [ cos θ ] ] 4 Collections of scatterers When light travels through a medium, it encounters many scatterers, so the scattering is a superposition of the results of many scatterings. Suppose there are scatterers located at positions, x i. Then since the fields vary as e ik x there will be factors E x i, B x i e ikn0 xi associated with the corresponding induced dipole moments, p i, m i e ikn0 xi Recalling that the total differential cross-section depends on E B, we will have a sum over conjugate pairs of phase factors: [ 4πɛ 0 E 0 ε p i + ] c m i n ε e ikn0 xi e ikn xi Define and assume all the scatterers are identical, so that i,j ikq x i ik n 0 n x i p i p m i m Then the sum applies only to the phase factor, giving an overall factor of F q e ikq xi i e ikq xi e ikq xj i j i,j e ikq xi xj There are two important limiting cases of this. When the scatterers are randomly distributed, as in a gas, the different phases tend to cancel, so only the diagonal terms contribute, F q ij ij N e ikq xi xj 9

10 where N is the total number of scatterers. The second limiting case is when the scatterers form some sort of regular lattice. For a perfect lattice, the same thing happens, with the effect of a scatterer at x i cancelling the effect of another scatterer at x i. The wave progresses only in the forward direction picture a clear crystal of pure quartz, for example. Scatterings do occur as a result of thermal vibrations which make the lattice imperfect. Jackson gives and explicit example of an exact result in eq

Lecture 10 February 25, 2010

Lecture 10 February 25, 2010 Lecture 10 February 5, 010 Last time we discussed a small scatterer at origin. Interesting effects come from many small scatterers occupying a region of size d large compared to λ. The scatterer j at position