Lecture 11 March 1, 2010

Transcription

1 Physics 54, Spring 21 Lecture 11 March 1, 21 Last time we mentioned scattering removes power from beam. Today we treat this more generally, to find the optical theorem: the relationship of the index of refraction the forward scattering amplitude.

2 The Optical Theorem The optical theorem relates the total cross section to the forward scattering amplitude. In Quantum Mechanics, this is conservation of probability. Here conservation of energy. We do this differently from Jackson. Consider scatterer of finite size in an incident plane wave: Physics 54, Spring 21 E i = E ɛ i e i k i x iωt B i = 1 ω k i E i = 1 ω k i ɛ i E e i k i x iωt Assume scattering is linear, time-invariant physics, so everything e iωt, make implicit. scattering amplitude f( k, k i, so E s ( x = eikr r f( k, k i E B s ( x = 1 ω k E s = eikr ωr E k f( k, k i.

3 Linearity assures frequency unchanged k = k = k i, elastic scattering. The total fields are E = E i + E s B = B i + B s. Consider the power flowing past planes k i, one way in front of the scatterer, one way in back. The total power removed from the incident beam = incident power flux times σ tot, which includes absorption scattering cross sections. Take k i ẑ. Total power across z downstream is P = 1 [( ρ dρ dφe Ei + 2µ ( E s B i + ] Bs The E i B i part of this is what would have been the power without any scattering. Each of E s B s falls off as 1/r, so the product falls off as 1/r 2 is negligible for large r. Thus if P is the change in the power of the beam ( δp is the power lost, P = 1 [ ρ dρ dφe Ei B 2µ s + E s ] Bi. z z. Physics 54, Spring 21

4 P = 1 [ ρ dρ dφe Ei B 2µ s + E s ] Bi 1 E 2 = ρ dρ dφ 2ωµ r [ ( e ɛ i k f ( k, k i e ikr+i k i x ( +e ikr i k i x f( k, ] ki ki ɛ i z z Physics 54, Spring 21 For large z, ρ z so the angle goes to zero, k = k i, kr k i x = k(r z = k( z 2 + ρ 2 z kρ 2 /2z. ρ dρ dφ eikr i k i x z 2 + ρ 2 2π z = 2π z ρ dρ e ikρ2 /2z du e iku/z = i 2π k.

5 Thus P = π E 2 ( ωµ k e i ɛ i f ( k i, k i k + i ɛ i k f ( k i, k i i ɛ i f( k i, k i k i i( k i f( k i, k i ɛ i z = π E 2 ( e i ɛ i ωµ f ( k, k i + + i ɛ i f( k i, k i = 2π E 2 ( Im ɛ i ωµ f( k i, k i. Physics 54, Spring 21 The power flux in the incident beam is 1 e ( E 2µ i B i z = E 2 ( ( e ɛ i k ɛi 2ωµ = E 2 k z 2ωµ so the total cross section must be σ Tot = 2 P ωµ E 2 = 4π ( ɛ k k Im i f( k i, k i. This is the optical theorem.

6 Index of efraction Consider thin slab, thickness d Number density N incident wave E i = E ɛ i e i k i x, ki = kê z, observe from far downstream, z. Each d 3 x in slab has Nd 3 x scatterers, so wavefront ki θ x k ρ z d z Physics 54, Spring 21 de s = eik f(k, θ, φ; k i E e i k i x Nd 3 x d 2π E s = NE dze ikz dφ ρ dρ e ik As 2 = ρ 2 + (z z 2, ρ dρ = d, so f ( ( k, cos 1 z z, φ; kê z

7 = ρ dρ eik f z z = 1 ik eik f 1 ik ( ( k, cos 1 z z ( ( d e ik f k, cos 1 z z z z ( ( k, cos 1 z z e ik d d d f, φ; kê z, φ; kê z, φ; kê z ( k, cos 1 ( z z = z z, φ; kê z Physics 54, Spring 21 where we integrated by parts for the last expression. The last term is 1 ik z z e ik d z z 2 d d cos θ f (k, θ, φ; kê z which, provided the indicated derivative is not singular, falls off like 1/.

8 Dropping that term, we have E s = i NE k d 2π dze ikz dφe ik(z z f (k,, φ; kêz = 2πi NE d e ikz f (k,, ; kêz k Thus the total electric field at points far beyond the slab is ( E( x = E e ikz ɛ i + 2πiNd f(k,, k This is a plane wave, exact solution of the free space wave equation, though with shifted phase, amplitude, polarization. Thus it holds right up to back edge of the slab. What was the effect of the slab? Project on original polarization initial ɛ i E has been multiplied by Physics 54, Spring πik 1 N ɛ i f(k, dz.

9 Integrating for finite thickness, Physics 54, Spring 21 ɛ i E( x = e 2πik 1 N ɛ i f(k,z E e ikz, That is, our wave has k replaced by nk, with n = 1 + 2πN ɛ i f(k, k 2, which is the index of refraction. Conclusion: The index of refraction is given by the forward scattering amplitude.

10 Caveats We assumed each scatterer feels only the incident field. Better treatment says evaluate f( k at the wavenumber in the medium, not vacuum. Physics 54, Spring 21 Absorption will give imaginary part to k = nk i = e k + i 2 αk i with α = Nσ tot = 4πN ( k Im ɛ i f( k, k. Only full f will satisfy the optical theorem. Approximations suitable for Im f in the forward direction may not work for σ tot f 2. Example: small lossless dielectric sphere, f is real! So there is scattering but zero total cross section can t be. In 7.1D Kramers-Kronig said e ɛ r 1 is given by dω of Im ɛ r (ω, so a purely real ɛ r can only be 1.