Joel A. Shapiro January 20, 2011

Transcription

1 Joel A. January 20, 2011

2 Course Information Instructor: Joel Serin X 3886, shapiro@physics Book: Jackson: Classical Electrodynamics (3rd Ed.) Web home page: contains general info, syllabus, lecture other notes, homework assignments, etc. Classes: ARC 207, Monday Thursday, 10:20 (sharp!) - 11:40 Homework: there will be one or two projects, homework assignments every week or so. Due dates to be discussed. Exams: a midterm a final. Office Hour: Tuesdays, 3:30 4:30, in Serin 325, or by arrangement.

3 Course Content Last term you covered Jackson, Chapters We will cover a large fraction of the rest of Jackson, but we will have to leave out quite a bit. Everything comes from Maxwell s Equations the Lorentz Force. We will discuss: EM fields in matter, at boundaries EM fields confined: waveguides, cavities, optical fibers Sources of fields: antennas their radiation, scattering diffraction Relativity, relativistic formalism for E&M Relativistic particles other gauge theories (maybe)

4 Maxwell s Equations At a fundamental level, electromagnetic fields obey Maxwell s equations with sources given by all charges currents: B 1 E c 2 t E + B t E = 1 ɛ 0 ρ all Gauss for E B = 0 Gauss for B = µ 0 Jall Ampère (+Max) = 0 Faraday charged particles experience the Lorentz force: ( ) F = q E + v B

5 Potentials As B has zero divergence 1 ( B = 0), there are vector fields A( r) for which B( r) = A( r). Then B t = A t, so Faraday s law tells us ( E + A ) = 0, t the term in parenthesis is 2 the gradient of some function, Φ. Then E = Φ A t. 1 B corresponds to a closed 2-form, which in Euclidean space means it is exact, B = da for some 1-form A. 2 So the 1-form E + A/ t is closed, exact, equal to d( Φ).

6 Gauge Invariance Because only the curl of A is determined by B, A is determined only up to a divergence 3. Though such an ambiguity will affect E, this can be compensated for in our choice of Φ. Thus the electromagnetic fields are unchanged by a gauge transformation: A A + Λ, Φ Φ Λ t. Because of gauge invariance, the physical equations (Maxwell) do not fully determine A Φ, but we may impose a gauge condition. Two popular choices are: A + 1 Φ c 2 t = 0 Lorenz gauge, A = 0 Coulomb gauge. 3 a closed or exact form

7 Solving the Wave Equation Either gauge choice is possible because a Λ to adjust A can be found by Green function for the wave or Laplace equation. With the Lorenz gauge, Maxwell s equations give 2 Φ 1 2 Φ c 2 t 2 = ρ ɛ 0 2 A 1 2 A c 2 t 2 = µ 0J Both of these are wave equations with specified source: 2 Ψ 1 c 2 2 Ψ t 2 = 4πf( x, t), with some form of boundary condition. If there are no boundaries, solution by Fourier transform is best.

8 Fourier transforms Let s review Fourier transforms. Given an integrable real or complex-valued function on the real line R, f : R C, we define its Fourier transform f : R C by f(ω) = This can be inverted, because 1 2π dω f(ω)e iωt = 1 2π = 1 2π = dt f(t)e iωt. dω dt f(t ) dt f(t )e iω(t t) dω e iω(t t) dt f(t )δ(t t) = f(t). so f(t) = 1 dω 2π f(ω)e iωt.

9 Fourier transforms convert derivatives to algebra If g(t) = d dt f(t), g(ω) = = = i dt e iωt d dt f(t) dt e iωt d ( 1 dt 2π dω ω 1 2π = iω f(ω). ) dω f(ω )e iω t dt e i(ω ω )t

10 Confusion Inconsistencies There are lots of arbitrary choices made here, not made consistently by different authors. Which is the fourier transform, f(t) f(ω) or the reverse, which is the reverse transform? Which one gets the minus sign in the exponential? In fact, everyone chooses exp(iωt) for f(t) f(ω) but exp( ikx) for f(x) f(k) for spatial functions. Which transform gets the 1/2π factor? f f does here, in 6.4 or do they each get a 1/2π, as on p. 69, or should we transform to frequency ν instead of angular frequency ω, with e 2πiνt, in which case we don t need any factors of 2π in front. That s what Wikipedia does, but no physics book I know of does.

11 transforms to ( 2 + k 2) Ψ( x, ω) = 4π f( x, ω), Back to solving the wave equation So if we fourier transform in time, our wave equation ( 2 1c 2 ) 2 t 2 Ψ = 4πf( x, t) with k := ω/c. For a fixed ω this is now just a three-dimensional differential equation, depending on x. It is the inhomogeneous Helmholtz equation. It can be solved by finding the suitable Green function: G k ( x, x ) ( 2 + k 2) G k ( x, x ) = 4πδ 3 ( x x ). In general we might be interested in the Green function with some specified spatial boundary conditions. Let us first consider, however, an unrestricted region, for which the Green function will be translationally invariant, G k ( x, x ) = G k ( x x ), rotationally symmetric, G k ( x, x ) = G k ( x x ).

12 Free space Helmholtz Green function In that case, with R = x x, we have 4 1 R dr 2 (RG k(r)) + k 2 G k (R) = 0 for R > 0. d 2 Thus G k (R) = AG (+) (R) + BG ( ) (R), with G (±) k (R) = e±ikr R. As G k ( r) = k 2 G k ( r) 4πδ 3 ( r), if we integrate inside a small sphere of radius ɛ use Gauss law: G k ( r) = ê r G k ( r). r ɛ r =ɛ 4 In spherical coordinates, 2 Ψ is given by Jackson s equation 3.1, which on a function with no angular dependence reduces to r 1 d 2 (rφ)/dr 2. We will derive 3.1 generalizations of other coordinate systems in a few weeks.

13 On the small sphere, the radial component of the gradient of G k ( r) is ê r G k ( r) = 1 + ikɛ ɛ 2 Ae ikɛ 1 ikɛ + ɛ 2 Be ikɛ A + B ɛ 0 ɛ 2 so the surface integral is 4π(A + B). On the other h, when integrating k 2 G k ( r) 4πδ 3 ( r), over the small sphere, the first term vanishes as the volume integral ( r 3 ) dominates the Green function ( 1/r), the integral of δ 3 ( r) = 1. Thus A + B = 1. Again, G (±) k ( r ) = e±ik r. AG (+) + (1 A)G ( ) is a r Green function for the Helmholtz equation.

14 Time Dependent Green Function Green functions depend on boundary conditions. We left ambiguity in behavior for t ±, so have arbitrariness in A, B. The Green function for wave equation in space-time: ( 2 x 1c 2 ) 2 t 2 G (±) ( x, t, x, t ) = 4πδ 3 ( x x )δ(t t ) can be found by fourier transforming (t ω) the source into 4πδ 3 ( x x )e iωt, so the fourier transform of the Green function is G (±) k ( x x )e iωt = G (±) k (R)e iωt, where R = x x. Taking the inverse transform (ω t) we have G (±) ( x, t, x, t ) = 1 2π where τ = t t. = 1 2π dω G (±) ω/c (R) e iω(t t ) dω e±iωr/c R e iωτ

15 Note, as we might expect, that it depends only on the difference of the two points in spacetime (on x x t t ). Also note the integral over ω gives a simple delta function, G (±) ( x, t, x, t ) = G (±) (R, τ) = 1 ( R δ τ R ) c = δ ( t [ t x x /c ]) x x. The delta function requires t = t R/c to contribute, R/c is always nonnegative, so for G (+) only t t contibutes, or sources only affect the wave function after they act. Thus G (+) is called a retarded Green function, as the affects are retarded (after) their causes. On the other h, G ( ) is the advanced Green function, giving effects which preceed their causes.

16 In Out Fields If the sources f( x, t) are only nonzero in a finite time interval [t i, t f ], any solution of the wave equation Ψ( x, t) must obey the homogeneous wave equation for t < t i for t > t f. As that equation is deterministic, if we define Ψ in ( x, t) Ψ out ( x, t) to be solutions, for all t, of the homogeneous equation which agree with Ψ( x, t) for t < t i for t > t f respectively, we must have Ψ( x, t) = Ψ in ( x, t)+ d 3 x dt G (+) ( x x, t t )f( x, t ) = Ψ out ( x, t)+ d 3 x dt G ( ) ( x x, t t )f( x, t ).

17 The solution Ψ( x, t) is not determined by the sources alone, because we could have an arbitrary free wave coming from t =, but we may view the difference Ψ out Ψ in as the field produced by the sources after they have acted. If the sources are considered fixed, independent of the field, this will not depend on the incident wave, but if the sources are there because of the incident wave, we may view this difference as the scattered wave. We will skip Jackson 6.5

18 In Ponderable Media Up to now, we have talked fundamental fields interacting with all charges. Restricting our focus to distances large compared to m 10 fm, matter consists of point charge nuclei Ze electrons with charge e. On an atomic scale we would have wildly fluctuating fields dependent on the detailed positions on innumerable atoms. For many purposes we are not interested in the detailed fields, which we will now call e b, but in their averages over regions large compared to an atom, say R = 10 8 m. With a smearing function such as f( x 1 ) = π 3/2 R 3 e x 2 /R 2, which gives f( x)d 3 x = 1, we define macroscopic fields E( x, t) = e( x, t) := d 3 x f( x ) e( x x, t), B( x, t) = b( x, t) := d 3 x f( x ) b( x x, t).

19 These averaged fields are known as the electric field magnetic induction. The latter is an unfortunate historical necessity, as the magnetic field will be defined differently in a moment. The homogeneous Maxwell equations are linear, so the averaging passes right through, we have B = 0, E B + t = 0. The averaging of the charges, however, is more interesting. We divide the charges into some we consider free, with a charge density η free, some we consider to be there in response to the fields, with a charge density we call η bound, because it is usually identified as charges bound to the molecules. The free charges can be described by their positions x j, with η free ( x) = j q jδ 3 ( x x j ). The bound charges are better described in terms of the center of mass of the molecule x n the charge s displacement from it, x jn with x j = x n + x jn.

20 The bound density η n ( x, t) for one molecule n is a sum over its bound charges j(n), so η bound ( x, t) = n η n( x, t) = n j(n) q jδ( x x n x jn ), the smeared density for molecule n is η n ( x, t) = q j d 3 x f( x )δ( x x x n x jn ) j(n) = j(n) q j f( x x n x jn ) j(n) ( q j f( x x n ) x jn α f( x x n ) x α α ) x jn α x jn β f( x x n ) x α x β αβ The displacements of the bound charges x jn from their molecule s center is small compared to the averaging scale R, so we may exp f in a power series, which separates out the x jn dependence from the x x n.

21 Multipole Moments We define the multipole moments for the molecule: q n = j(n) q j, p n = j(n) q j x jn, Q n αβ = 3 j(n) q j x jn α x jn β are the monopole (or charge), dipole quadripole moments of the molecule about the center of mass. Aside: Jackson uses Q, presumably because this includes a monopole monopole moment 3 P q jx 2 jn in addition to the true quadripole part.

22 The averaged charge density is then equivalent to a density of charges, dipoles quadripoles given by the density of molecules. Let us define a macroscopic charge density ρ( x, t) = η free + n q n δ( x x n ), where the n is over the molecules, considers the net charge of any molecule n as located at its center of mass. Performing the smearing for the bound charges, ρ(x, t) = η free + n q n f( x x n ). This gives the first term in η n ( x, t).

23 Polarization Define the macroscopic polarization P ( x, t) = p n δ( x x n ) = n d 3 x f( x x ) n p n δ( x x n ) = n p n f( x x n ) = n,j q j x jn f( x x n ) which is the smeared out dipole moment of the molecules. Then P ( x, t) = n,j,α q j x jn α This gives the second term in η n ( x, t). x α f( x x n ).

24 between Q Q n Quadripole Moment Next define the macroscopic quadripole density 5 Q αβ ( x, t) = 1 Q n αβ δ( x x n ) 6 n = 1 d 3 x f( x x ) Q n αβ δ( x x n ) 6 n = 1 Q n αβ f( x x n ) = 1 q j x jn α x jn β f( x x n ) 6 2 n Then Q αβ ( x, t) x α x β αβ = 1 2 q j x jn α x jn β f( x x n ). 2 x α x β n,j n,j This gives the third term in η n ( x, t). 5 Caution: note the apparent inconsistency by a factor of 6

25 Maxwell s Gauss Law We see comparing these to our first expansion that η( x, t) = ρ( x, t) P ( x, t) + αβ x α x β Q( x, t). Then upon smearing Gauss law for the microscopic electric field e( x, t), ɛ 0 E( x, t) = η( x, t), gives us ρ( x, t) = ɛ 0 E( x, t) + P ( x, t) αβ 2 x α x β Q αβ. Let so that Dα ( x, t) = ɛ 0 Eα ( x, t) + P α ( x, t) β D( x, t) = ρ( x, t). x β Q αβ, This is the macroscopic Gauss Law for D.