Causality. but that does not mean it is local in time, for = 1. Let us write ɛ(ω) = ɛ 0 [1 + χ e (ω)] in terms of the electric susceptibility.

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1 We have seen that the issue of how ɛ, µ n depend on ω raises questions about causality: Can signals travel faster than c, or even backwards in time? It is very often useful to assume that polarization is linear local in space, the polarizability is not time dependent, meaning D( x, ω) = ɛ(ω) E( x, ω), but that does not mean it is local in time, for D( x, t) = = = 1 2π 1 2π 1 2π dω D( x, ω)e iωt dω e iωt ɛ(ω) E( x, ω) dω e iωt ɛ(ω) dt E( x, t )e iωt Let us write ɛ(ω) = ɛ [1 + χ e (ω)] in terms of the electric susceptibility.

2 G(t t ) Let G(τ) be the fourier transform of χ e (ω), G(τ) = 1 2π dω e iωτ χ e (ω). Then we have the relation between D E given by { } D( x, t) = ɛ E( x, t) + dτ G(τ) E( x, t τ). Thus we see that D( x, t) depends linearly on the function E( x, t ) of time t, but not on the single value E( x, t). That is, the dependence is non-local in time. Of course if ɛ(ω) were constant, G(τ) δ(τ) we would have the local D( x, t) = ɛ E( x, t), but that is not the case generally. As D( x, t) E( x, t) are both real, D ( x, ω) = D( x, ω) similarly for E, so for real ω, ɛ (ω) = ɛ( ω). This means G(τ) is real (for real τ).

3 That the polarization at time t might depend on the electric field at some earlier time t is not surprising, but shouldn t it be blind to fields at later times? That is, shouldn t we insist G(τ) = for τ < )? Let s consider our oscillator strength model, with χ e (ω) = ω 2 P ω 2 ω2 iγω (or a sum of such contributions with different ω s γ s). Then G(τ) = ω2 P 2π = ω2 P 4πν e iωτ dω ω 2 ω2 iγω ( dω where ν = ω 2 γ2 /4. e iωτ ω + ν + iγ/2 e iωτ ). ω ν + iγ/2

4 Evaluating G(τ) for oscillators The integrals can be easily done by closing the contours in the complex plane. The integral dω e iωτ ω ± ν + iγ/2 can be done by closing the contour with a large semicircle in the upper or lower half plane, whichever gives zero contribution because of the exponential. For negative τ, e iωτ = e τ Imω, so the contribution of the green infinite-radius semicircle in the upper half plane vanishes, as the integr is analytic in the enclosed region, the integral is zero. Thus we do have G(τ) = for τ <.

5 For τ >, the contour is closed in the lower half plane, the integral is given by 2πi times the sum of the residues, which are e ±iν τ γτ/2. So the two terms for positive τ fill in the result: G(τ) = ω 2 P sin(ν τ) Θ(τ). ν Typical values for the lifetime of states, hence the line-widths of the photons emitted, give γ i from 1 7 /s to 1 9 /s, so the effective τ s are of the order of nanoseconds. While the response of D to E is not instantaneous, it is quick.

6 Model-independent ɛ s We will now find constraints on possible forms of ɛ(ω) without assumptions on the model of molecular behavior. We have seen that reality of the fields in the time domain requires ɛ (ω) = ɛ( ω) for real ω. We have only defined used ɛ(ω) for real ω, but if we would like to continue ɛ as a complex valued analytic function, we need to extend the constraint to ɛ (ω ) = ɛ( ω), χ e(ω ) = χ e ( ω). We will also insist that G(τ) is finite real for positive τ zero for negative τ. We might expect G τ which is true for dielectrics, but DC currents correspond to singular polarizability for DC conditions, with ɛ iσ/ω, as we saw in for zero-mode oscillators. This will come from G τ σ/ɛ. In any case, we will assume G does not blow up at infinity.

7 For large ω, χ e is determined by G(τ) near τ =. Indeed, if G(t) = t n d n G n! dt n, n χ e (ω) = 1 d n G n! dt n t n e iωt n = d n G dt n ( iω) (n+1) = i G() n ω G () ω Note G() = by continuity from negative τ s, so the leading term is 1/ω 2. With G thus well-behaved, χ e (ω) = G(τ)e iωτ dτ is a well defined integral for all ω with Im ω, except for the possible pole at ω =. Thus χ e (ω) is an analytic function in the upper half plane.

8 Therefore, by Cauchy s theorem, for z in the upper half plane, χ e (z) = 1 χ e (ω ) 2πi C ω z dω with C the contour consisting of the black real axis the green semicircle, going over under z if it lies on the real axis. Because χ e (ω) goes to zero at infinity, we can discard the green semicircle. If z = ω + iδ, with δ > χ e (z) = 1 2πi dω χ e (ω ) ω ω iδ. We are interested in z approaching the real axis from above, δ, so we may use ( ) 1 1 ω ω iδ = P ω + iπδ(ω ω), ω where the principal part P means P dω 1 ω ω f(ω ) := lim ɛ ( ω ɛ ) dω + ω+ɛ ω ω f(ω ).

9 The δ(ω ω) term just cancels half the left h side, so doubling it, for real ω, we have χ e (ω) = 1 iπ P = 1 iπ P = 1 iπ P dω χ e(ω ) ω ω ( dω χe (ω ) ω ω χ e( ω ) ω + ω ) ( dω χe (ω ) ω ω χ e(ω ) ω + ω Taking real imaginary parts separately, Re χ e (ω) = 1 ( 1 π P dω Im χ e (ω ) ω ω + 1 ) ω + ω = 1 π P dω Im χ e (ω 2ω ) ω 2 ω 2 Im χ e (ω) = 1 ( 1 π P dω Re χ e (ω ) ω ω 1 ) ω + ω = 1 π P dω Re χ e (ω 2ω ) ω 2 ω 2 )

10 Maxwell s Equations in Linear Media Recall the basis equations: H D t E + B t D = 1 ɛ ρ Gauss for D B = Gauss for B = µ J Ampère (+Max) = Faraday plus the Lorentz force: F = q( E + v B) the constitutive relations (in frequency space) D = ɛe, B = µ H.

11 Interface between conductor non-conductor Consider the interface between a dielectric an good conductor c. Conductor c: if perfect, no E. Surface charge Σ eddy currents can prevent fields from penetrating, so no H inside conductor. Γ Across the interface: n ξ Faraday on loop Γ E continuous conductor Gauss on pillbox S B continuous c S Thus just outside the conductor E =, B =.

12 Good (not perfect) conductor For good (not perfect) conductor, take J = σ E with large conductivity σ. Assume time-dependence e iωt Let ξ be distance inside conductor. H varies rapidly with ξ. H c = J + D t σ E, E c = B t = iωµ ch c Rapid variation with depth ξ dominates, = ˆn ξ, E c = 1 J σ = 1 σ ˆn H c ξ, Hc = i ˆn E c ωµ c ξ

13 E c = 1 J σ = 1 σ ˆn H c ξ, Hc = i ˆn E c ωµ c ξ so ˆn H c = ˆn H c = ( i ˆn ˆn E c ωµ c ξ ( [ = i σωµ c ˆn = ˆn ) i 2 ( σωµ c ξ 2 ˆn H ) c. ]) ˆn 2 Hc ξ 2 Simple DEQ, exponential solution, with δ = H c = H e ξ/δ e iξ/δ, 2 µ c ωσ, H is tangential field outside surface of conductor.

14 E inside conductor at boundary From H c = H e ξ/δ e iξ/δ, E c = 1 σ ˆn H c ξ = µ c ω 2σ (1 i)ˆn H e ξ/δ e iξ/δ, which means, by continuity, that just outside the conductor µ c ω E = 2σ (1 i)ˆn H.