Set 5: Classical E&M and Plasma Processes

Transcription

1 Set 5: Classical E&M and Plasma Processes

2 Maxwell Equations Classical E&M defined by the Maxwell Equations (fields sourced by matter) and the Lorentz force (matter moved by fields) In cgs (gaussian) units D = 4πρ, B = 0, E = 1 B c t, H = 1 D c t + 4π c j. where ρ = charge density, j = current density, D = ɛe and B = µh with ɛ = dielectric constant µ = magnetic permeability In vacuum, ɛ = µ = 1 and the Maxwell equations simplify to E = 4πρ, B = 0, E = 1 c B t, B = 1 c E t + 4π c j,

3 Charge Conservation Maxwell equations are E and B symmetric aside from the lack of magnetic charges Divergence of Ampere s law charge conservation [ B = 1 E c t + 4π c j] 0 = 1 E + 4π c t c j 0 = 4π ρ c t + 4π c j 0 = ρ t + j

4 Source free propagation Wave Equation E = 0, B = 0, E = 1 B c t, B = 1 E c t, invariant under E B and B E, so work out equation for E Curl of Faraday s law wave equation [ E = 1 B c t ] ( E) 2 E = 1 B c t 2 E = 1 2 E c 2 t 2 2 E 1 2 E c 2 t = 0 2

5 Wave Equation Similarly for B [ 2 1c ] ( 2 E 2 t 2 B ) = 0 Wave solutions k = 2π/λ = 2πν/c [real part or superimposed k and k] ( E B ) = ( E0 e 1 B 0 e 2 ) e i(k x ωt) Wave solution in wave equation provides the dispersion relation k 2 1 c 2 ω2 = 0, ω = kc = 2πν

6 Wave Equation E and B fields related by Maxwell equations ik ê 1 E 0 = 0, ik ê 2 B 0 = 0, ik ê 1 E 0 = i ω c ê2b 0, ik ê 2 B 0 = i ω c ê1e 0,, so ê 1 k, ê 2 k, k ê 1 ê 2. So (ê 1, ê 2, ˆk) form a orthonormal right handed basis ik ê 1 E 0 = i ω c ê2b 0, ke 0 = ω c B 0, E 0 = B 0

7 Lorentz Force Lorentz force F = q [E + v c B ] where q is the charge For a distribution of charges in a volume, charge and current density j = ρv provide a force density f = ρe + 1 c j B Work and change in energy density W = F dx, du mech dt = ρe v = E j dw dt = F v = qe v

8 Field Energy Use energy conservation to define field and radiation energy 4π c j = H 1 D c t j = c ( H 1 ) D 4π c t = E j = c ( H 1 ) D E 4π c t [E ( H) = H ( E) (E H)] = c [ H ( E) (E H) 1 ] D 4π c t E [ E = 1 ] B, H = B/µ, D = ɛe c t = 1 [ 1 ] B 4π µ t B ɛ E E c (E H) t du mech dt du mech dt

9 Field Energy Rewrite field terms as a total derivative B (B B) = 2B t t, Equation forms a conservation law E (E E) = 2E t t with t (u mech + u field ) + S = 0 u field = 1 8π (ɛe 2 + 1µ B2 ) and the energy flux carried by the radiation, or Poynting vector S = c 4π E H

10 Field Energy For vacuum µ = ɛ = 1 and so field energy in a monocromatic wave, time averaged over the oscillation u field = 1 8π S = c 4π 1 2 (E2 0 + B0) 2 = E 0B 0 = c 8π E2 0 8π E2 0 which says that S / u field = c, (recall I ν /u ν = c) Energy Flux S = c 8π E2 0 = dν dω cos θi ν so the specific intensity of a monocromatic plane wave is a delta function in frequency and angle

11 Specific Intensity Actual processes will not be monocromatic but have some waveform associated with the acceleration of the emitter: superposition of plane waves E(t) = E(ω) = 1 2π E(ω)e iωt dω E(t)e iωt dt Total emission will be an incoherent superposition of these single particle sources E(t) <t>~t t

12 Specific Intensity Energy flux normal to propagation direction dw dtda = S = c 4π E2 (t) Total energy passing through da dw da = c 4π [ = c 4π [ 4π dw dadω = c E(ω) 2 E 2 (t)dt 0 E 2 (t)dt = 2π ] E(ω) 2 dω ] E(ω) 2 dω

13 Specific Intensity Now given a timescale T for single particle processes dw dadωdt = 1 T dw dadω = c T E(ω) 2 which given ω = 2πν is the specific flux. Given a resolved source divide by the source solid angle to get I ν plane wave distant source approx

14 Stokes Parameters Specific intensity is related to quadratic combinations of the field. Define the intensity matrix (time averaged over oscillations) E E Hermitian matrix can be decomposed into Pauli matrices P = E E = 1 2 (Iσ 0 + Q σ 3 + U σ 1 V σ 2 ), where ( 1 0 ) ( 0 1 ) ( 0 i ) ( 1 0 ) σ 0 = 0 1, σ 1 = 1 0, σ 2 = i 0, σ 3 = 0 1 Stokes parameters recovered as Tr(σ i P)

15 Stokes Parameters Consider a general plane wave solution E(t, z) = E 1 (t, z)ê 1 + E 2 (t, z)ê 2 E 1 (t, z) = A 1 e iφ 1 e i(kz ωt) E 2 (t, z) = A 2 e iφ 2 e i(kz ωt) Explicitly: I = E 1 E 1 + E 2 E 2 = A A 2 2 Q = E 1 E 1 E 2 E 2 = A 2 1 A 2 2 U = E 1 E 2 + E 2 E 1 = 2A 1 A 2 cos(φ 2 φ 1 ) V = i E 1 E 2 E 2 E 1 = 2A 1 A 2 sin(φ 2 φ 1 ) so that the Stokes parameters define the state up to an unobservable overall phase of the wave

16 Polarization Radiation field involves a directed quantity, the electric field vector, which defines the polarization Consider a general plane wave solution E(t, z) = E 1 (t, z)ê 1 + E 2 (t, z)ê 2 E 1 (t, z) = ReA 1 e iφ 1 e i(kz ωt) E 2 (t, z) = ReA 2 e iφ 2 e i(kz ωt) or at z = 0 the field vector traces out an ellipse E(t, 0) = A 1 cos(ωt φ 1 )ê 1 + A 2 cos(ωt φ 2 )ê 2 with principal axes defined by E(t, 0) = A 1 cos(ωt)ê 1 A 2 sin(ωt)ê 2 so as to trace out a clockwise rotation for A 1, A 2 > 0

17 Polarization. Define polarization angle e' 2 e 2 e' 1 Match ê 1 = cos χê 1 + sin χê 2 ê 2 = sin χê 1 + cos χê 2 E(t) χ e 1 E(t, 0) = A 1 cos ωt[cos χê 1 + sin χê 2 ] A 2 cos ωt[ sin χê 1 + cos χê 2 ] = A 1 [cos φ 1 cos ωt + sin φ 1 sin ωt]ê 1 + A 2 [cos φ 2 cos ωt + sin φ 2 sin ωt]ê 2

18 Polarization Define relative strength of two principal states A 1 = E 0 cos β A 2 = E 0 sin β Characterize the polarization by two angles A 1 cos φ 1 = E 0 cos β cos χ, A 1 sin φ 1 = E 0 sin β sin χ, A 2 cos φ 2 = E 0 cos β sin χ, Or Stokes parameters by I = E 2 0, U = E 2 0 cos 2β sin 2χ, A 2 sin φ 2 = E 0 sin β cos χ Q = E 2 0 cos 2β cos 2χ V = E 2 0 sin 2β So I 2 = Q 2 + U 2 + V 2, double angles reflect the spin 2 field or headless vector nature of polarization

19 Special cases Polarization If β = 0, π/2, π then only one principal axis, ellipse collapses to a line and V = 0 linear polarization oriented at angle χ If χ = 0, π/2, π then I = ±Q and U = 0 If χ = π/4, 3π/4... then I = ±U and Q = 0 - so U is Q in a frame rotated by 45 degrees If β = π/4, 3π/4, then principal components have equal strength and E field rotates on a circle: I = ±V and Q = U = 0 circular polarization U/Q = tan 2χ defines angle of linear polarization and V/I = sin 2β defines degree of circular polarization

20 Natural Light A monochromatic plane wave is completely polarized I 2 = Q 2 + U 2 + V 2 Polarization matrix is like a density matrix in quantum mechanics and allows for pure (coherent) states and mixed states Suppose the total E tot field is composed of different (frequency) components E tot = i E i Then components decorrelate in time average E tot E tot = ij E i E j = i E i E i

21 Natural Light So Stokes parameters of incoherent contributions add I = i I i Q = i Q i U = i U i V = i V i and since individual Q, U and V can have either sign: I 2 Q 2 + U 2 + V 2, all 4 Stokes parameters needed

22 Polarized Radiative Transfer Define a specific intensity vector : I ν = (I ν1, I ν2, U, V ) where I = I ν1 + I ν2, Q = I ν1 I ν2 di ν ds = α ν(s ν I ν ). Source vector in practice can be complicated Thomson collision based on differential cross section e' 1 e' 2 Θ e- Thomson Scattering dσ T dω = 3 8π Ê Ê 2 σ T, e 2 e 1

23 Polarized Radiative Transfer Ê and Ê denote the incoming and outgoing directions of the electric field or polarization vector. Thomson scattering into e 1 : I ν2 I ν2 but I ν1 does not scatter More generally if Θ is the scattering angle then referenced to the plane of the scattering α ν = n e σ T and cos 2 Θ S ν = 3 dω I ν 8π 0 0 cos Θ cos Θ But to calculate Stokes parameters in a fixed coordinate system must rotate into the scattering basis, scatter and rotate back out to the fixed coordinate system

24 Plasma Effects Astrophysical media are typically ionized so that radiation does not propagate in a vacuum but through an ionized plasma. However the plasma is typically so rarified that only the very lowest frequency radiation is affected Maxwell equations for plane wave radiation exp[i(k r ωt)] with sources ik E = 4πρ, ik B = 0, ik E = i ω c B, ik B = iω c E + 4π c j, Medium is globally neutral but electric field of the radiation cause a high frequency electron drift current charge via continuity

25 Plasma Sources Lorentz force m v = ee v = ee iωm Current density carried by electrons of number density n Charge conservation j = nev = ne2 E iωm σe σ = ine2 ωm conductivity iωρ + ik j = 0 ρ = k j ω = σ ω k E

26 Maxwell equation with ρ Plasma Frequency i ik E = 4π σ ω k E ( 1 4πσ ) k E = 0 ωi ik ɛe = 0 with the dielectric constant ɛ = 1 4πσ ωi = 1 ω2 p ω 2 = 1 4πne2 mω [ω 2 ] 2p = 4πne2 m

27 Plasma Frequency Likewise the Maxwell equation with j ik B = 4π c j iω c E ( ) 4π = c σ iω c ik B = i ω c ɛe E So that the Maxwell equations become source free equations ik ɛe = 0, ik B = 0, ik E = i ω c B, ik B = iω c ɛe,

28 Wave Equation Wave equation becomes (similarly for B E) i[k (k E)] = ik 2 E = i ω c k B = iω2 c 2 ɛe Modified dispersion relation k 2 = ω2 c ɛ = ω2 2 c 2 ( 1 ω2 p ω 2 ) ω 2 ω 2 p k = 1 c If ω < ω p then k is imaginary and the wave function has an exponential suppression - waves don t propagate below the plasma frequency ν p = ω ( p n ) 1/2 2π = 0.01 MHz 1cm 3

29 Plasma Cutoff & Refraction For the ionosphere n 10 4 cm 3 and radio waves at < 1MHz cannot propagate For ISM n < 1 cm 3 and the cut off is a much smaller < 0.01 MHz The phase velocity defines the index of refaction v p = ω k c n r n r ɛ = 1 ω2 p ω 2 Radio waves can be refracted according to Snell s law and change their direction of propagation along the path s dn rˆk ds = n r

30 Dispersion Measure For wave packet propagation the relevant quantity is the group velocity defined by demanding that the phase remain stationary for constructive interference φ(k) = kz ω(k)t φ k = 0 = z ω k t = z v gt v g = ω k = c c( ω2 p ω 2 c ω 2 p ω 2 ) [ω ω p] Photons effectively gain a mass leading to a delay in arrival times

31 Dispersion Measure For a pulse of radiation from a pulsar t p = d 0 ds d v g c + 1 2cω 2 d 0 ω 2 pds t p = d c + 2πe2 mcω 2 [ d 0 ] nds D t p ω = 4πe2 mcω 3 D Change in arrival time with frequency dispersion measure distance given a mean n

32 Faraday Rotation In an external magnetic field B 0 = B 0 e 3 the electron responds to the magnetic field as well as the electric field of the radiation m dv dt = ee + e c v B 0 Examine the propagation of circularly polarized states Take a trial solution E ± (t) = E ± e iωt [e 1 ± ie 2 ] v ± (t) = v ± e iωt [e 1 ± ie 2 ] imωv ± [e 1 ± ie 2 ] = ee ± [e 1 ± ie 2 ] + e c v ±B 0 [ e 2 ± ie 1 ] = [ee ± ± i e c B 0v ± ][e 1 ± ie 2 ]

33 Faraday Rotation v ± = Conductivity iee ± m(ω ± ω B ) i(ωm ± e c B 0)v ± = ee ± [ ω B = eb ] 0 mc σ = j ± E ± = env ± E ± = ɛ ± = 1 4πσ ωi = 1 4πne 2 m(ω ± ω B )ω ie 2 n m(ω ± ω B ) = 1 ω 2 p ω(ω ± ω B )

34 Faraday Rotation. Right and left polarizations travel at different velocities: disperson relation for ω ω B and ω ω p k ± = ω c ɛ± ω c [ ωp 2 ( 1 ± ω ) ] p ω 2 ω t 1 t 1 t 2 + t 2 x2 equal speed t 1 t 1 + linear polar. t 2 t 2 x2 unequal speed Considering linear polarization as a superposition of right and left circular polarization, the difference in propagation speeds will lead to a Faraday rotation of the linear polarization rotated

35 Faraday Rotation Phase φ ± = φ 2 = 1 2 d 0 d k ± ds 0 (k + k )ds = 1 2c d φ/2 gives the rotation of linear polarization θ = φ 2 = 1 4πe 2 2cω 2 m = 2πe3 m 2 c 2 ω 2 B 0 nds e mc 0 d 0 ω 2 p ω 2 ω Bds B 0 nds More generally B 0 B the line of sight component Given an average n measure B e.g. magnetic field of ISM, cluster

36 E & M Potentials Introduce the vector and scalar potential to simplify source calculation B = 0 implies that B = A, where A is the vector potential So Faraday s law becomes E = 1 c [E + 1 c t A] = 0 implying a scalar potential φ ( A) t E + 1 c A t = φ

37 Gauge Potentials (φ, A) allow for gauge freedom. Given a change in the potentials through an arbitrary field ψ A = A + ψ φ = φ 1 ψ c t the observable E and B fields invariant E = 1 A c t φ = 1 A c t 1 c = 1 A c t φ B = A = A t ψ φ + 1 c t ψ

38 Lorentz Gauge Gauge freedom allows one to choose a convenient gauge to simplify equations Choose a gauge where the relationship between the potentials is A + 1 c Maxwell equations simplify to φ t = 0 2 φ 1 c 2 2 φ t 2 2 A 1 c 2 2 A t 2 = 4πρ = 4π c j

39 Retarded Potentials Green function solutions (propagate a δ function disturbance; superimpose to get arbitrary source. [See Jackson] Looks like electrostatics but accounts for the finite propagation time of light φ(r, t) = [ρ]d 3 r r r [j]d 3 r A(r, t) = 1 c r r where the [] denotes evaluation at the retarded time [f](r, t) = f(r, t 1 c r r )

40 Lienard-Wiechart Potential. Consider a single charge on a trajectory r 0 (t) with velocity u = ṙ 0 (t) q r 0 (t') ρ(r, t) = qδ(r r 0 (t)) j(r, t) = quδ(r r 0 (t)) Scalar potential r observer φ(r, t) = d 3 r dt ρ(r, t ) r r δ(t t + 1 c r r ) = d 3 r dt qδ(r r 0 (t )) δ(t t + 1 r r c r r ) = q dt δ(t t + R(t ) 1 ) R(t) = r r c R(t 0 (t) ) R(t') n r'

41 Lienard-Wiechart Potential Change variables so that explicit integration possible t = t t + R(t ) c dt = dt + 1 c Ṙ(t )dt = [1 + 1 c Ṙ(t )]dt Ṙ = ṙ 0 = u, ˆn = R R, Ṙ ˆn = u ˆn 2RṘ = 2Ṙ R Ṙ = Ṙ ˆn = u ˆn κ(t ) c Ṙ(t ) = 1 1 c u ˆn which is the origin of relativistic beaming effects

42 Lienard-Wiechart Potential Thus since t = 0 t = t R(t )/c Similarly φ(r, t) = q dt δ(t 1 ) κ(t )R(t ) = A(r, t) = [ qu ] cκr q = κr t =0 [ q ] κr For non-relativistic velocities κ = 1 u/c ˆn 1 and potential are just retarded versions of electrostatic potentials For u c then enhanced radiation along u n from relativistic beaming

43 E&B Field Plug and chug with E = 1 A c t φ, E = E vel + E rad B = A B = ˆn E Velocity field falls off as 1/R 2, β = u/c as a generalization of Coulomb s law [ ] (ˆn β)(1 β 2 ) E vel = q κ 3 R 2 The radiation field depends on the acceleration and falls off as 1/R so that there is a flux E 2 1/R 2 that propagates to infinity E rad = q [ ] ˆn ((ˆn β) β) c κ 3 R

44 Larmor Formula. Larmor formula: non relativistic case β 1 u n x u n E rad = q ˆn (ˆn β) c R = q ˆn (ˆn u) c 2 R Let u/ u ˆn = cos Θ Θ E rad So flux E rad = q u Rc 2 sin Θ S = c c E2 4π radˆn = 4π q 2 u 2 R 2 c 4 sin2 Θ = 1 4π q 2 u 2 R 2 c 3 sin2 Θ

45 Larmor Formula Power per unit angle da = R 2 dω dw dtdω = q2 u 2 4πc 3 sin2 Θ P = dw dt = q2 u 2 4πc 3 dω sin 2 Θ = 2q2 u 2 So dipole pattern of radiation is perpendicular to acceleration and polarization is in plane spanned by u and ˆn Larmor formula can be used more generally in that one can transform to a frame where the particles are non-relativistic via Lorentz transformation 3c 3