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Maxwell’s Equations in Vacuum

Maxwell’s Equations in Vacuum. (1) . E = r / e o Poisson’s Equation (2) . B = 0 No magnetic monopoles (3)  x E = - ∂ B / ∂ t Faraday’s Law (4)  x B = m o j + m o e o ∂ E / ∂ t Maxwell’s Displacement Electric Field E Vm -1 Magnetic Induction B Tesla

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Maxwell’s Equations in Vacuum

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  1. Maxwell’s Equations in Vacuum (1) .E = r /eoPoisson’s Equation (2) .B = 0 No magnetic monopoles (3)  x E = -∂B/∂t Faraday’s Law (4)  x B = moj+ moeo∂E/∂t Maxwell’s Displacement Electric Field EVm-1 Magnetic Induction B Tesla Charge density r Cm-3 Current Density Cm-2s-1 Ohmic Conduction j = sEElectric Conductivity Siemens (Mho)

  2. Constitutive Relations (1) D= eoE + PElectric Displacement DCm-2 andPolarization PCm-2 (2) P = eoc EElectric Susceptibility c (3) e=1+cRelative Permittivity (dielectric function) e (4) D = eo e E (5) H = B / mo– M Magnetic Field HAm-1and Magnetization M Am-1 (6) M = cBB / moMagnetic Susceptibility cB (7) m=1 / (1-cB)Relative Permeability m (8) H = B / mmo m ~ 1 (non-magnetic materials), e ~ 1 - 50

  3. E P s+ s- E- E+ Electric Polarisation • Apply Gauss’ Law to right and left ends of polarised dielectric • EDep = ‘Depolarising field’ • Macroscopic electric field EMac= E + EDep= E - P/o s± surface charge density Cm-2 s± = P.n n outward normal E+2dA = s+dA/o Gauss’ Law E+ = s+/2o E- = s-/2o EDep= E+ + E- = (s++ s-)/2o EDep= -P/o P = s+ = s-

  4. Electric Polarisation Define dimensionless dielectric susceptibility c through P = ocEMac EMac = E– P/o oE = oEMac + P oE = oEMac + ocEMac= o (1 + c)EMac= oEMac Define dielectric constant (relative permittivity)  = 1 + c EMac = E/ E = eEMac Typical values for e: silicon 11.8, diamond 5.6, vacuum 1 Metal: e →∞ Insulator: e∞(electronic part) small, ~5, lattice part up to 20

  5. Electric Polarisation Rewrite EMac = E– P/o as oEMac + P = oE LHS contains only fields inside matter, RHS fields outside Displacement field, D D = oEMac + P = oEMac= oE Displacement field defined in terms of EMac (inside matter, relative permittivity e) and E (in vacuum, relative permittivity 1). Define D = oE where  is the relative permittivity and E is the electric field

  6. E + - + - P + - Gauss’ Law in Matter • Uniform polarisation → induced surface charges only • Non-uniform polarisation → induced bulk charges also Displacements of positive charges Accumulated charges

  7. Gauss’ Law in Matter Polarisation charge density Charge entering xz face at y = 0: Py=0DxDz Cm-2 m2 = C Charge leaving xz face at y = Dy: Py=DyDxDz = (Py=0 + ∂Py/∂yDy) DxDz Net charge entering cube via xz faces: (Py=0 -Py=Dy)DxDz = -∂Py/∂yDxDyDz z Charge entering cube via all faces: -(∂Px/∂x + ∂Py/∂y + ∂Pz/∂z)DxDyDz = Qpol rpol= lim (DxDyDz)→0Qpol/(DxDyDz) -. P = rpol Dz Py=0 Py=Dy y Dy Dx x

  8. Gauss’ Law in Matter Differentiate -.P = rpol wrt time .∂P/∂t + ∂rpol/∂t = 0 Compare to continuity equation .j + ∂r/∂t = 0 ∂P/∂t = jpol Rate of change of polarisation is the polarisation-current density Suppose that charges in matter can be divided into ‘bound’ or polarisation and ‘free’ or conduction charges rtot = rpol + rfree

  9. Gauss’ Law in Matter Inside matter .E = .Emac = rtot/o= (rpol+ rfree)/o Total (averaged) electric field is the macroscopic field -.P = rpol .(oE + P) = rfree .D = rfree Introduction of the displacement field, D, allows us to eliminate polarisation charges from any calculation. This is a form of Gauss’ Law suitable for application in matter.

  10. Ampère’s Law in Matter Ampère’s Law Problem! Continuity equation A steady current implies constant charge density,so Ampère’s law is consistent with the continuity equation for steady currents Ampère’s law is inconsistent with the continuity equation (conservation of charge) when the charge density is time dependent

  11. Ampère’s Law in Matter Add term to LHS such that taking Div makes LHS also identically equal to zero: The extra term is in the bracket extended Ampère’s Law Displacement current (vacuum)

  12. k M= sin(ay) k j i jM= curl M = a cos(ay) i Ampère’s Law in Matter Total current • Polarisation current density from oscillation of charges in electric dipoles • Magnetisation current density variation in magnitude of magnetic dipoles

  13. Ampère’s Law in Matter ∂D/∂t is the displacement current postulated by Maxwell (1862) In vacuum D= eoE∂D/∂t = eo∂E/∂t In matter D = eoe E∂D/∂t = eoe ∂E/∂t Displacement current exists throughout space in a changing electric field

  14. Maxwell’s Equations in vacuum in matter .E = r /eo.D = rfreePoisson’s Equation .B = 0 .B = 0 No magnetic monopoles  x E = -∂B/∂t  x E = -∂B/∂t Faraday’s Law  x B = moj+ moeo∂E/∂t  x H = jfree+ ∂D/∂t Maxwell’s Displacement D = eoeE = eo(1+ c)E Constitutive relation for D H = B/(mom)= (1- cB)B/moConstitutive relation for H

  15. .AdV A.ndA Divergence Theorem 2-D 3-D • From Green’s Theorem • In words - Integral of A.ndAover surface contour equals integral of div Aover surface area • In 3-D • Integral of A.ndAover bounding surface S equals integral of div VdVwithin volume enclosed by surface S • The area element ndA is conveniently written as dS

  16. .V dv V.dS Differential form of Gauss’ Law • Integral form • Divergence theorem applied to field V, volume v bounded by surface S • Divergence theorem applied to electric field E Differential form of Gauss’ Law (Poisson’s Equation)

  17. local value of x A n outward normal (x a) .ndA dA dr A.dr local value of A Stokes’ Theorem 3-D • In words - Integral of (x A).n dAover surface S equals integral of A.drover bounding contour C • It doesn’t matter which surface (blue or hatched). Direction of dr determined by right hand rule. A C

  18. Faraday’s Law

  19. B j dℓ S Differential form of Ampère’sLaw Integral form of law: enclosed current is integral dS of current density j Apply Stokes’ theorem Integration surface is arbitrary Must be true point wise

  20. Maxwell’s Equations in Vacuum Take curl of both sides of 3’ (3)  x ( x E) = -∂ ( x B)/∂t = -∂ (moeo∂E/∂t)/∂t = -moeo∂2E/∂t2  x ( x E) = (.E) - 2E vector identity - 2E = -moeo∂2E/∂t2 (.E = 0) 2E -moeo∂2E/∂t2 = 0 Vector wave equation

  21. Maxwell’s Equations in Vacuum Plane wave solution to wave equation E(r, t) = Re {Eoei(wt - k.r)}Eoconstant vector 2E =(∂2/∂x2 + ∂2/∂y2 +∂2/∂z2)E = -k2E .E = ∂Ex/∂x + ∂Ey/∂y +∂Ez/∂z = -ik.E= -ik.Eoei(wt - k.r) If Eo || k then .E≠ 0 and  x E = 0 If Eo┴k then .E= 0 and  x E≠0 For light Eo┴k and E(r, t) is a transverse wave

  22. l r|| r k r Consecutive wave fronts Maxwell’s Equations in Vacuum Eo Plane waves travel parallel to wave vector k Plane waves have wavelength 2p /k

  23. Maxwell’s Equations in Vacuum • Plane wave solution to wave equation • E(r, t) = Eoei(wt - k.r)Eoconstant vector • moeo∂2E/∂t2 = - moeow2Emoeow2 = k2 w =±k/(moeo)1/2 = ±ckw/k = c = (moeo)-1/2 phase velocity w(k) w= ±ck Linear dispersion relationship k

  24. Maxwell’s Equations in Vacuum Magnetic component of the electromagnetic wave in vacuum From Faraday’s law  x ( x B) = moeo∂( x E)/∂t = moeo∂(-∂B/∂t)/∂t = -moeo∂2B/∂t2  x ( x B) = (.B) - 2B - 2B = -moeo∂2B/∂t2 (.B = 0) 2B -moeo∂2B/∂t2 = 0 Same vector wave equation as for E

  25. Maxwell’s Equations in Vacuum If E(r, t) = Eoex ei(wt- k.r)and k || ezand E|| ex(ex, ey, ezunit vectors)  x E = -ikEoeyei(wt- k.r)= -∂B/∂t From Faraday’s Law ∂B/∂t = ikEoeyei(wt- k.r) B = (k/w) Eoeyei(wt- k.r)= (1/c) Eoeyei(wt- k.r) For this wave E|| ex, B|| ey, k || ez, cBo = Eo More generally -∂B/∂t = -iwB =  x E  x E = -ik x E -iwB = -ik x E B = ekx E / c

  26. Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for m= 1, rfree= 0,jfree= 0 Maxwell’s equations become  x E = -∂B/∂t  x H = ∂D/∂t H = B /moD = eoeE  x B = moeoe∂E/∂t  x ∂B/∂t = moeoe∂2E/∂t2  x (- x E) =  x ∂B/∂t = moeoe∂2E/∂t2 -(.E) + 2E = moeoe∂2E/∂t2 . e E= e . E = 0 since rfree= 0 2E - moeoe∂2E/∂t2 = 0

  27. Maxwell’s Equations in Matter • 2E - moeoe∂2E/∂t2 = 0 E(r, t) = Eoex Re{ei(wt - k.r)} • 2E = -k2E moeoe∂2E/∂t2 = - moeoe w2E • (-k2+moeoe w2)E = 0 • w2 = k2/(moeoe)moeoew2 = k2 k = ± w√(moeoe) k = ± √ew/c • Let e = e1-ie2be the real and imaginary parts of e and e = (n-ik)2 • We need √e = n-ik e = (n-ik)2 = n2- k2 - i 2nke1= n2- k2 e2= 2nk • E(r, t) = Eoex Re{ ei(wt- k.r) } = Eoex Re{ei(wt- kz)} k|| ez • = Eoex Re{ei(wt - (n - ik)wz/c)} = Eoex Re{ei(wt - nwz/c)e- kwz/c)} Attenuated wave with phase velocity vp = c/n

  28. Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for m= 1, rfree= 0,jfree= s(w)E Maxwell’s equations become  x E = -∂B/∂t  x H = jfree + ∂D/∂t H = B /moD = eoeE  x B = mojfree+ moeoe∂E/∂t  x ∂B/∂t = mos∂E/∂t + moeoe∂2E/∂t2  x (- x E) =  x ∂B/∂t = mos∂E/∂t+moeoe∂2E/∂t2 -(.E) + 2E = mos∂E/∂t+moeoe∂2E/∂t2 . e E= e . E = 0 since rfree= 0 2E - mos∂E/∂t - moeoe∂2E/∂t2 = 0

  29. Maxwell’s Equations in Matter 2E - mos∂E/∂t - moeoe∂2E/∂t2 = 0 E(r, t) = Eoex Re{ei(wt - k.r)} k|| ez 2E = -k2E mos∂E/∂t= mosiwEmoeoe∂2E/∂t2 = - moeoe w2E (-k2-mosiw+moeoew2 )E = 0 s >> eoe w for a good conductor E(r, t) = Eoex Re{ ei(wt - √(wsmo/2)z)e-√(wsmo/2)z} NB wave travels in +z direction and is attenuated The skin depth d = √(2/wsmo) is the thickness over which incident radiation is attenuated. For example, Cu metal DC conductivity is 5.7 x 107 (Wm)-1 At 50 Hz d = 9 mm and at 10 kHz d = 0.7 mm

  30. Mion k melectron kMion Si ion Bound electron pair Bound and Free Charges Bound charges All valence electrons in insulators (materials with a ‘band gap’) Bound valence electrons in metals or semiconductors (band gap absent/small ) Free charges Conduction electrons in metals or semiconductors Resonance frequency wo ~ (k/M)1/2 or ~ (k/m)1/2 Ions: heavy, resonance in infra-red ~1013Hz Bound electrons: light, resonance in visible ~1015Hz Free electrons: no restoring force, no resonance

  31. Mion k melectron kMion Bound and Free Charges Bound charges Resonance model for uncoupled electron pairs

  32. Mion k melectron kMion Bound and Free Charges Bound charges In and out of phase components of x(t) relative to Eo cos(wt) in phase out of phase

  33. Bound and Free Charges Bound charges Connection to c and e e(w) Im{e(w)} w = wo w/wo Re{e(w)}

  34. s(w) Re{e(w)} wo = 0 Drude ‘tail’ w Im{s(w)} Bound and Free Charges Free charges Let wo→ 0 in c and e jpol = ∂P/∂t

  35. Energy in Electromagnetic Waves Rate of doing work on a moving charge W = d/dt(F.dr) F = qE + qv x B = d/dt{(q E+ q vx B) . dr} = d/dt(q E . dr) = qv . E = j(r) = q vd(r - r’) W = fields do work on currents in integration volume Eliminate j using modified Ampère’s law  x H = jfree+ ∂D/∂t W =

  36. Energy in Electromagnetic Waves Vector identity .(A x B) = B . () - A. () W = becomes W = = = - W = = - Local energy density Poynting vector

  37. Energy in Electromagnetic Waves Energy density in plane electromagnetic waves in vacuum

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