GreensFunctionsAndMaxwell

Green functions and Maxwell equations Ralf Vogelgesang Stuttgart January 28, 2004 Abstract These notes accompanied a course I gave at the Max Planck Institute for Festko¨rperforschung, Stuttgart in 2002. They are a somewhat “short-hand” overview of how to obtain Green functions from differential equations in general. I include a discussion of the notorious self-interaction term. Most interesting to me personally is the application of Green functions to optical scattering problems. Therefore I concentrate in the second part of the notes on optical scattering problems and how to solve them using Green functions. Starting “ab-initio” from the Maxwell equations, the complete way up to numerical evaluations in the discrete dipole approximation is delineated. Contents 1 Green function – one page summary 2 2 What does it mean? I 3 2.1 a single effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 What does it mean? II 3 3.1 a single cause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 Open questions 4 5 What happens at r = r′? 4 6 What’s wrong with the derivation? 4 6.1 Remedy I: “Separate Self-Cell” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 6.2 Remedy II: “Dimensional Regularization” . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 7 What’s the proper derivation? 5 7.1 What’s the proper derivation? I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 7.2 What’s the proper derivation? II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 8 How to get a Green function – some properties 6 8.1 How to get a Green function – I. guess! . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 8.2 How to get a Green function – II. directly . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 8.3 How to get a Green function – III. Fourier analysis . . . . . . . . . . . . . . . . . . . . . . 8 8.4 How to get a Green function – IV. Dyson’s Equation . . . . . . . . . . . . . . . . . . . . . 8 2 1 GREEN FUNCTION – ONE PAGE SUMMARY 8.5 What does Dyson’s equation mean? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 8.6 Significance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 How does the time figure in? 10 9.1 Scalar wave Green function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 10 Maxwell Equations 11 10.1 Other forms of the material equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 10.2 Macroscopic Polarization, etc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 11 Helmholtz Equation 12 11.1 Helmholtz Equation I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 11.2 Helmholtz Equation II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 11.3 Helmholtz Equation III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 12 Helmholtz and the background Green dyadic 13 12.1 Constant Dielectric Background Plus Variation . . . . . . . . . . . . . . . . . . . . . . . . 13 12.2 Obtaining Green Dyadic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 13 Review of important Green functions 14 14 Full Helmholtz Equation from background 14 15 Full Helmholtz Equation from complete 15 16 Discrete Dipole Approximation 15 17 Numerical Evaluation 15 1 Green function – one page summary 1. Differential equation to solve Lf(r) = F(r) . 2. Get (somehow!) the Green function associated with this equation, i.e., LG(r, r′) = 1δ(r, r′) . 3. Do the “trick” (hopelessly illegal! but it gets the right result...) Lf(r)− F(r) = 0 Lf(r)− ∫ δ(r, r′)F(r′)dr′ = 0 L { f(r)− ∫ G(r, r′)F(r′)dr′ } = 0 . 4. And therefore f(r) = ∫ G(r, r′)F(r′)dr′ . Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 3 2 What does it mean? I 2.1 a single effect f(r) = ∫ G(r, r′)F(r′)dr′ • What happens at r is influenced by everything that happens at all other places r′ where there are non-zero “drivers” F. • The individual influences are summed up (→ interference effects!). • The “strength” of the influence of r′ on r is determined by G(r, r′)—G “propagates” the action from r′ to r. tr � � � �> S S S So r′ � � � �� G(r, r′) 3 What does it mean? II 3.1 a single cause f(r) = ∫ G(r, r′)F(r′)dr′ Suppose, there is only one localized “driver” F(r) = F0δ(r, r0) ⇒ f(r) = F0G(r, r0) • The Green function is the response of space to a single localized source • In isotropic space, G(r, r0) = G(|r− r0|) and the effect are concentric spheres of equal strength tr0 ��� �> S S S So � � � �� G(r, r0)&% '$ c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 4 6 WHAT’S WRONG WITH THE DERIVATION? 4 Open questions • what happens at r′ = r? • what’s wrong with the derivation? what’s the proper derivation? • how does one obtain the propagator, i.e., Green function for a specific L? • in physical systems, how does the time figure in? 5 What happens at r = r′? LG(r, r′) = δ(r, r′) ⇒ f(r) = ∫ G(r, r′)F(r′)dr′ In order to have the action of a δ function, the Green function has to be very steep, must diverge for r′ → r. Exactly at r′ = r it is not definable as a normal function, but must be considered a distribution. 6 What’s wrong with the derivation? Looking at the above derivation more closely, we find that we exchanged integration and differentia- tion, like here L r f(r) = L r ∫ G(r, r′)F(r′)dr′ = ∫ L r G(r, r′)F(r′)dr′ = ∫ δ(r, r′)F(r′)dr′ = F(r) . Though it looks right, it is highly defective. Typically, the Green function diverges for r′ → r! The question then is “how badly?” Consider this example, ∇ ∫ x−1dx2 = ∇x = const Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 6.1 Remedy I: “Separate Self-Cell” 5 ?? = ∫ ∇x−1dx2 = ∫ x−2dx2 = lnx→ −∞ Whereas the left-hand side converges, the right-hand side does not. 6.1 Remedy I: “Separate Self-Cell” One way to perform the exchange (“commutation” of operators) ∂/∂x and ∫ is to exclude a small volume V ∗ around the pole and consider the limV ∗ → 0; ∫ V ∇ = lim V ∗→0 {∫ V−V ∗ ∇+ ∫ V ∗ ∇ } = lim V ∗→0 { ∇ ∫ V−V ∗ + ∫ V ∗ ∇ } = lim V ∗→0 { ∇ ∫ V−V ∗ + ∮ Σ∗ } = ∇P ∫ +L . 6.2 Remedy II: “Dimensional Regularization” Another way to perform the exchange is to temporarily change the dimension by �. In general, ∂n ∫ dxmx−k = ∫ dxm∂nx−k ∝ ∫ dxmx−k−n gives problems if m = k + n. Therefore take m→ m− �, perform the calculation, then take lim �→ 0. ∂n ∫ dxm+�x−k ∝ ∂nxm+�−k ∝ xm+�−k−n → x0∫ dxm+�∂nx−k ∝ ∫ dxm+�x−k−n ∝ x−k−n+m+� → x0 Note: fractional dimensions like m + � are called Hausdorff dimension, after the inventor of fractional calculus. 7 What’s the proper derivation? 7.1 What’s the proper derivation? I δ(r, r′) and G(r, r′) are distributions. That is, they are not functions and live in their own (vector-) space. c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 6 8 HOW TO GET A GREEN FUNCTION – SOME PROPERTIES A distribution D is always defined by its action on functions under integrals! D : f(x) 7→ ∫ D(x)f(x) = 〈D; f〉 . Similarly, the derivative of a distribution LD is defined by the action of D on the derivative of functions under integrals! LD : f(x) 7→ ∫ D(x)Lf(x) = 〈D;Lf〉 . 7.2 What’s the proper derivation? II The proper way in which to read the distribution equation that defines the Green “function”, LG(r, r′) = δ(r, r′) ; HOW TO READ THIS “EQUATION”? Like this:∫ G(r, r′)Lf(r)dr = ∫ δ(r, r′)f(r)dr . Using this view...apply ∫ dr′G(r, r′)� L′f(r′) = F(r′) G(r, r′)L′f(r′) = G(r, r′)F(r′)∫ G(r, r′)L′f(r′)dr′ = ∫ G(r, r′)F(r′)dr′∫ δ(r, r′)f(r′)dr′ = ∫ G(r, r′)F(r′)dr′ f(r) = ∫ G(r, r′)F(r′)dr′ 8 How to get a Green function – some properties 1. Dimensional analysis. If space is n-dimensional, the differential operator is r−m-dimensional, the Green function has di- mensions n−m (δ has dimensions r−n). Example: Poisson equation in 3d. 4Φ = 0 4G = 1δ ; The Green function’s dimension must be proportional to r−1. 2. Free space. Vacuum is isotropic; neither location of origin nor direction can matter. Therefore, the vacuum Green function must be a function of the distance r = |r− r′| only. G(r, r′) = f(r) = G(r′, r) Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 8.1 How to get a Green function – I. guess! 7 8.1 How to get a Green function – I. guess! For a given differential operator L that does not explicitly depend on location one can often guess the form of G. For example, for the Poisson equation in 3d, L = 14 G(r, r′) ∝ 1|r′ − r| . Testing LG, ∫ 1 r 4f(r)dr3 = {∫ ∇(1 r ∇f(r))dr3 = 0 } − ∫ (∇1 r )∇f(r)dr3 = − {∫ ∇((∇1 r )f(r))dr3 = 0 } + ∫ (41 r )f(r)dr3 = f(0) ∫ ∇∇1 r dr3 = f(0) lim R→0 ∮ S ∇1 r dS = f(0) lim R→0 −n R2 4piR2n = −4pif(0) Conclusion: it works, if we set G(r, r′) = −1 4pi|r′ − r| . Is it unique? Not at all! Any homogeneous solution f(r), for which Lf = 0 may be added. Poisson’s Equation: e.g., any constant G(r, r′) = −1 4pi|r′ − r| + V0 . (Electric reference potential.) 8.2 How to get a Green function – II. directly Sometimes one can obtain G by direct integration. For example, ∂2 ∂x2 G(x) = δ(x)∫ ∂2 ∂x2 G(x)dx = ∫ δ(x)dx ∂ ∂x G(x) = θ(x) + C1∫ ∂ ∂x G(x)dx = ∫ θ(x) + C1dx G(x) = θ(x)x+ C1x+ C2 . The integration constants may be set by imposing boundary conditions. c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 8 8 HOW TO GET A GREEN FUNCTION – SOME PROPERTIES 8.3 How to get a Green function – III. Fourier analysis The 4-potential field A = (a, φ) for Maxwell’s equations (Lorentz gauge); the wave equation ∂2A = J , where ∂2 = ∂igij∂j = µBεB∂2/∂t2 − ∇·2, and gij = (−,−,−,+) the appropriate metric. (J is the 4-current.) The scalar Green function for this differential equation is defined by ∂2gB(x;x′) = δ(x, x′) . (1) First go to the reciprocal domain by �(x) = 1 (2pi)4 ∫ ∞ −∞ �(k)e−ikxdk4 �(k) = ∫ ∞ −∞ �(x)e+ikxdx4 , and note in passing that δ(x′ − x) = ∫ e−ik(x−x ′) (2pi)4 dk4 . Thus ∂2 ∫ gB(k, x′)e−ikx (2pi)4 dk4 = ∫ e−ik(x−x ′) (2pi)4 dk4 ⇒ −k2gB(k, x′) = e+ik(x′) Back-transform... ⇒ gB(x, x′) = −1(2pi)4 ∫ e−ik(x−x ′) k2 dk4 . In conventional notation, this reads gB(r, t; r′, t′) = ∫ ∫ e−ik(r−r ′)+iω(t−t′) k2 − ω2µBεB dkdω (2pi)4 . Upon realizing an explicit Fourier transformation of the time-dependent part (take τ = t − t′), becomes gB(r, r′, ω˜) = ∫ e−ik(r−r ′) k2 − ω˜2µBεB dk (2pi)3 , 8.4 How to get a Green function – IV. Dyson’s Equation Suppose we have solved LBGB = δ. We want to find the Green function for L = LB − δε(r). LG(r′′, r′) = 1δ(r′′, r′){ LB − δε(r′′) } G(r′′, r′) = 1δ(r′′, r′) Apply ∫ dr′′GB(r, r′′)� Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 8.5 What does Dyson’s equation mean? 9 G(r, r′)− ∫ dr′′GB(r, r′′)δε(r′′)G(r′′, r′) = GB(r, r′) The necessary relation is Dyson’s Equation, given in a somewhat abbreviated operator notation, G = GB +GBδεG . 8.5 What does Dyson’s equation mean? G(r, r′) = GB(r, r′) + ∫ dr′′GB(r, r′′)δε(r′′)G(r′′, r′) • Propagation from r to r′ is done by the background propagator plus contributions from all scat- terers (δε(r′′)) • Again, interference effects occur because of the ∫ dr′′δε(r′′)� • Effect of scatterer δε(r′′) is to take a disturbance propagated from r by the “old” propagator, and send it to r′ by the “new” tr t r′ t δε(r′′) @ @ @ @ @ @ @ @I @ @ @ @ @ @ @ @I � � � �� � � � �� PP PP PP PP PP PPi GB GB G 8.6 Significance Interchangeable view if F(r) = δε(r)f(r)! Especially useful if we have f = finc+fsca—an “incident” and a “scattered” part, for which LBfinc = 0. Either think of F(r) as external driving forces LBf(r) = F(r) ⇒ fsca(r) = ∫ GB(r, r′)F(r′)dr′ = ∫ GB(r, r′)δε {finc + fsca} dr′ or as internal scattering centers, L = { LB − δε } Lf(r) = 0 Lfsca(r) = −Lfinc(r) = δεfinc(r) ⇒ fsca(r) = ∫ G(r, r′)δεfincdr′ c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 10 9 HOW DOES THE TIME FIGURE IN? 9 How does the time figure in? 9.1 Scalar wave Green function In isotropic, three dimensional space of permeability µB and dielectric constant εB , we have to find the Green function for ( ∇·2 − ∂ 2 ∂ct2 ) A = −f(r, t) , that is ( ∇·2 − ∂ 2 ∂ct2 ) gB(r, t; r′, t′) = −δ(r, r′)δ(t, t′) . (2) Go to the time-Fourier space, by applying∫ �(t) exp(iωt)dt, leading to(∇·2 + k2B) gB(r, ω; r′, t′) = −δ(r, r′) exp(iωt′) , where we use the convenient abbreviation k2B = µBεBω 2 = ω2/c2. Guess gB(r, r′) = gB(|r− r′|) (isotropic free space). Re-write ∇·2 in spherical coordinates; neglect any φ, θ-dependence, 1 R ∂2 ∂R2 [R · gB(R)] + k2BgB(R) = −δ(R) ∂2 ∂R2 [R · gB(R)] + k2B [R · gB(R)] = −δ(R) . Homogeneous solution (except R = 0): R · gB(R) = A exp(+ikBR) +B exp(−ikBR) . Normalize the constants A,B by requiring, in the limit δV, δr → 0∫ (∇·2 + k2B)gB(R)dR = − ∫ δ(R)dR → A+B = 1 4pi , Thus, gB(R) = A exp(+ikBR) 4piR + (1−A)exp(−ikBR) 4piR . Consider g±B(r, ω, r ′, t′) = exp(±ikB |R|+ iωt′) 4pi|R| , Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 11 which, after the inverse time-Fourier transform (1/2pi) ∫ �(ω) exp(−iωt)dω, reads g±B(|R|, t− t′) = δ (±√µBεB |R|+ t′ − t) 8pi2|R| . “An influence at (r, t) has to originate from an event at (r′, t′) that happened at the time t′ = t∓ |R|/c.” Only the “−” gives the conventional causal connection, i.e., an event is always caused by something in the past. By convention, we call the “retarded” Green function g+B(|R|, t − t′) the Green function, with its spatial part gB(r, r′) := exp(ikB |R|) 4pi|R| . (3) 10 Maxwell Equations ∇·D = ρ (4) ∇·B = 0 (5) ∇×E = −B˙ (6) ∇×H = J+ D˙ (7) together with the material equations D = εE (8) H = µ−1B (9) J = σE (10) form the Maxwell equations (in SI units). Notice that we consider ε, µ, and σ as second-rank tensors. 10.1 Other forms of the material equations In strong fields, ε, µ, and σ have to take non-linearities into account, that is, ε = f(E,B), etc. For instance, in second-harmonic generation problems: Di = ∑ jk εijkEjEk In optical activity problems: D ∝ k×E 10.2 Macroscopic Polarization, etc. Instead of using ε, one may separate the “vacuum part” ε01, giving (neglecting higher terms like quadrupoles, etc.) D = ε0E+ (ε− ε01)E = ε0E+P c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 12 11 HELMHOLTZ EQUATION where the macroscopic polarization P is P = ε0χE = ( ε− ε01 ) E χ = (ε− ε01) . The electric susceptibility χ contains all the information about matter—i.e., scatterers. 11 Helmholtz Equation 11.1 Helmholtz Equation I Apply ∇× to (∇×E = −B˙), ∇×∇×E = −∇×B˙ . (11) Express ∇×B with ∇×H = J+ D˙ ∇×(µ−1B) = σE+ ˙(εE) , where we also made use of Eqs. (8,9,10). 11.2 Helmholtz Equation II Assume explicit pure single-frequency dependence for the fields (“Fourier-decomposition”), E = E(r) exp (−iωt) B = B(r) exp (−iωt) . The material constants, however, we take as time-independent (i.e., ε˙ = µ˙ = σ˙ = 0) ε = ε(r) µ = µ01 σ = σ(r) =: −iωε σ (r) . ε σ is introduced merely for later convenience; ε∗ = ε+ ε σ . It allows us to absorb the effects of electrical conductivity in one, complex-valued dielectric constant. 11.3 Helmholtz Equation III ∇×(µ−1B) = σE+ ˙(εE) becomes ∇×B = −µ0iωε∗E , and allows us to express the vectorial wave equation ∇×∇×E = −∇×B˙{∇×∇×− ω2µ0ε∗}E = 0 . Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 13 12 Helmholtz and the background Green dyadic 12.1 Constant Dielectric Background Plus Variation Assume ε∗(r) = ε01 + δε(r) . ε0 is the “background” dielectric of free space (i.e.vacuum)—assumed isotropic. δε(r) are the scattering centers distributed in space. Now the Helmholtz equation becomes ∇×∇×E− ω2ε0µ0E = ω2µ0δεE . (12) 12.2 Obtaining Green Dyadic For the background Green function, LBGB(r, r′) = 1δ(r, r′) , (13) where LB = {∇×∇×− ω2ε0µ01} . The Green function is actually a 3× 3 tensor (a dyadic) of Green functions. Note that ∇×∇× ≡ ∇(∇·)−4⇒ great resemblance to the Poisson (or Laplace-) equation, for which L = {4− ω2ε0µ0} 1 “Probably”, the Green function can be expressed in terms of the scalar Green function gB(r, r′) = exp(ikB |R|) 4pi|R| . And indeed, GB(r, r′)ij := { 1 ij + ∇·i∇·j k2B } gB(r, r′) Explicit expression GB(r, r′)ij = 1ijgB(r, r ′) + 1 ij ikB |R| − 1 k2B |R|2 gB(r, r′) + RiRj 3− 3ikB |R| − k2B |R|2 k2B |R|4 gB(r, r′) . where we use the convenient abbreviations kB = √ ε0µ0ω and R = r− r′. Contributions ∝ |R|−1 give rise to far-field, those ∝ |R|−2,−3 are the near-field. c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 14 14 FULL HELMHOLTZ EQUATION FROM BACKGROUND 13 Review of important Green functions Operator Green function 4 14pi|R| 4− ∂2∂ct2 δ(|R|/c+t′−t) 8pi2|R| 4+ k2B exp(ikB |R|)4pi|R| ∇×∇×− k2B1 = ∇·i∇·j −41ij − k2B1ij { 1 ij + ∇·i∇·j k2B } exp(ikB |R|) 4pi|R| Note, in all cases we assume 3-dimensional space (i, j = x, y, z). (Note also, I did not care too much about the signs of the Green functions ...) 14 Full Helmholtz Equation using the background Green func- tion For the general case, L = {∇×∇×− ω2µ0ε(r′)} = LB − ω2µ0δε(r′) , we can now immediately write down the general solution in terms of the background dyadic (remember Dyson); Esca(r) = ∫ V GB(r, r′)ω2µ0δε(Esca +Einc)dr′ . The volume V only has to include all space where δε(r) 6= 0. This is an implicit equation in Esca(r ∈ V ). All Esca(r ∈/ V ) are completely determined by the Esca(r ∈ V ). Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 15 15 Full Helmholtz Equation using the complete Green function Alternatively, we may aspire to obtain first the general Green dyadic. Once we have accomplished this, and. . . . . . if we can distinguish an incident and scattered radiation field, E = Einc+Esca, where LBEinc = 0, we have Esca(r) = ∫ V G(r, r′)ω2µ0δε(r′)Einc(r′)dr′ . (14) This is an explicit equation in Esca. G is known and different setups for Einc are readily studied by direct evaluation. Again, all E(r 6∈ V ) are completely determined by the E(r ∈ V ). 16 Discrete Dipole Approximation To turn this into a numerically useful equation, we first consider V divided into small mesh volumes V = ⋃ ∆Vn, which allows us to read ∫ V → ∑ n ∫ ∆Vn ; the integrals may—for sufficiently small ∆Vn—be approximated by∫ ∆Vn I(r′)dr′ ≈ I(rn)∆Vn , that is, we assume the entire effect of ∆Vn may be collapsed into that of one discrete dipole at rn. 17 Numerical Evaluation Using the discrete dipole approximation, the two options for solving the complete Helmholtz equation turn into Esca(r) = ∫ V GB(r, r′)ω2µ0δε(Esca +Einc)dr′ ⇒ Escan = ∑ m GBnmω2µ0δεm(Escam +Eincm)dVm ∼ Mnm {Escam +Eincm} c© Ralf Vogelgesang, January 28, 2004 work in progress, do not quote! Green functions and Maxwell equations 16 17 NUMERICAL EVALUATION Esca(r) = ∫ V G(r, r′)ω2µ0δε(r′)Einc(r′)dr′ ⇒ Escan = ∑ m Gnmω2µ0δεmEincmdVm ∼ MnmEincm Green functions and Maxwell equations work in progress, do not quote! c© Ralf Vogelgesang, January 28, 2004 Green function -- one page summary Wh