EFUW Summary

Andreas Biri, D-ITET 06.06.15 Pre-Maxwellian Electrodynamics Units 1J = 1 Nm = 1 Ws = 1 𝑉𝐶 , 1 𝐻 = 1 𝛺𝑠 Nm 1 W = 1 s = 1 VA , 1 𝛺 = 1 𝑉⁄𝐴 = 1 𝑊/𝐴2 𝑁𝑚 1 V = 1 W⁄A = 1 𝐽⁄𝐶 = 1 , 1 𝐶 = 1 𝐴𝑠 𝐶 1 𝑇 = 𝑉𝑠⁄𝑚2 = 1 𝑁⁄𝐴𝑚 , 1𝐹 = 1𝐶 ⁄𝑉 = 1 𝐴𝑠/𝑉 𝟏𝒆 = 𝟏. 𝟔𝟎𝟐 ∗ 𝟏𝟎−𝟏𝟗 𝑪, 𝟏 𝑪 = 𝟔. 𝟐𝟒 ∗ 𝟏𝟎𝟏𝟖 𝒆 𝑚𝑒 = 9.11 ∗ 10−31 𝑘𝑔 , 𝑔 = 9.81 𝑚 𝑠 −2 8 𝑐 = 2.99792458 ∗ 10 𝑚/𝑠 𝟏 𝜺𝟎 = 𝟖. 𝟖𝟓𝟒 ∗ 𝟏𝟎−𝟏𝟐 𝑨𝒔⁄𝑽𝒎 , 𝒄𝟐 = 𝜺𝟎 𝝁𝟎 −𝟕 −𝟐 −𝟐 𝛍𝟎 = 𝟒𝛑 ∗ 𝟏𝟎 𝐤𝐠 ∗ 𝐦 ∗ 𝐬 ∗ 𝐀 Basic Properties 𝑑𝑉 𝑑 ∫ 𝐸(𝑟, 𝑡) 𝑑𝑠 = − ∫𝐵(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = 𝑈𝑖𝑛𝑑 𝑑𝑡 𝐴 𝑑𝐴 Ampere’s law ∫ 𝐵(𝑟, 𝑡) 𝑑𝑠 = 𝜇0 ∫𝑗(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = 𝜇0 𝐼 𝑑𝐴 𝑑 𝜌(𝑟, 𝑡) 𝑑𝑡 Displacement current Faraday’s law / Induction law 𝑗𝑑𝑖𝑠𝑝 = 𝜀0 𝑑 𝐸 𝑑𝑡 Interaction of fields with matter Primary sources: 𝑐ℎ𝑎𝑟𝑔𝑒𝑠 𝜌 𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑗 Secondary sources: 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑐ℎ𝑎𝑟𝑔𝑒𝑠 𝑏𝑦 𝑓𝑖𝑒𝑙𝑑𝑠 𝐴 No magnetic monopoles, zero flux & electrostatics Polarisation P : ∫𝑑𝑉 𝑃(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = − ∫𝑉 𝜌𝑝𝑜𝑙 (𝑟, 𝑡) 𝑑𝑉 ∫ 𝐵(𝑟, 𝑡) 𝑛 𝑑𝑎 = 0, Electric displacement : 𝑑𝑉 ∫ 𝑗(𝑟, 𝑡) 𝑛 𝑑𝑎 = 0, 𝐴 ∫ 𝐸(𝑟, 𝑡) 𝑑𝑠 = 0 𝐷 = 𝜀0 𝐸 + 𝑃 = 𝜀0 𝜀𝑟 𝐸 𝑑𝐴 Polarisation current: due to bound charges Maxwell’s equations in integral form 𝑗𝑝𝑜𝑙 (𝑟, 𝑡) = 𝑑𝑉 [𝐵] = 𝑉 𝑠⁄𝑚2 , [𝐹] = 𝑁 = 𝑘𝑔 𝑚/𝑠 2 Magnetic field is relative counterpart of electric field ′ 2 𝑞 𝑛𝑟 ′ 𝑟 𝑑 𝑛𝑟 ′ 1 𝑑 [ ′2 + ( ′ 2 ) + 2 2 𝑛𝑟 ′ ] 4𝜋𝜀0 𝑟 𝑐 𝑑𝑡 𝑟 𝑐 𝑑𝑡 𝜌(𝑟) = ∑𝑛 𝑞𝑛 𝛿[ 𝑟 − 𝑟𝑛 ] Current density: 𝑗(𝑟) = ∑𝑛 𝑞𝑛 𝑟𝑛̇ 𝛿[ 𝑟 − 𝑟𝑛 ] ⃗ (𝑟, 𝑡) ] 𝑑𝑉 𝐹(𝑟, 𝑡) = ∫ [ 𝜌(𝑟, 𝑡) 𝐸⃗ (𝑟, 𝑡) + 𝑗(𝑟, 𝑡) 𝑥 𝐵 𝑑 𝑃(𝑟, 𝑡) 𝑑𝑡 [ 𝑃 ] = 𝐶/𝑚2 ∇ 𝐸(𝑥) = − ∇2 𝛷(𝑥) = 𝜌(𝑥) 𝜀0 𝑉 𝑑 ∫ 𝐸(𝑟, 𝑡) 𝑑𝑠 = − ∫𝐵(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 𝑑𝑡 𝐴 𝑑𝐴 𝑑 ∫ 𝐻(𝑟, 𝑡) 𝑑𝑠 = ∫ [ 𝑗(𝑟, 𝑡) + 𝐷(𝑟, 𝑡) ] ∗ 𝑛 𝑑𝑎 𝑑𝑡 𝑑𝐴 𝐴 Magnetisation M : ∫𝑑𝐴 𝑀(𝑟, 𝑡) 𝑑𝑠 = ∫𝐴 𝑗𝑚𝑎𝑔 (𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 Magnetic field : 𝐻= 1 𝜇0 𝐵−𝑀 [𝐴/𝑚] 𝐵 = 𝜇0 𝜇𝑟 𝐻 Magnetisation current: due to circular charges ∫ 𝐵(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = 0 𝑗𝑚𝑎𝑔 = ∇ 𝑥 𝑀 𝑑𝑉 Maxwell’s equation in differential form Conduction current: due to free charges ∇ ∗ 𝐷(𝑟, 𝑡) = 𝜌(𝑟, 𝑡) → ∇ ∗ 𝐸(𝑟, 𝑡) = 𝜌(𝑟, 𝑡)/𝜀0 ∇ 𝑥 𝐸(𝑟, 𝑡) = − 𝑉 Poisson: 1 𝑄 ∫ 𝜌(𝑟, 𝑡) 𝑑𝑉 = 𝜀0 𝑉 𝜀0 ∇ ∗ 𝑗(𝑟, 𝑡) = − 𝑑 ∫ 𝜌(𝑟, 𝑡) 𝑑𝑉 𝑑𝑡 𝑉 ∫ 𝐷(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = ∫ 𝜌(𝑟, 𝑡) 𝑑𝑉 𝐹(𝒓, 𝑡) = 𝑞 [ 𝐸(𝒓, 𝑡) + 𝑣(𝒓, 𝑡) ∗ 𝐵(𝒓, 𝑡) ] Charge density: 𝑑𝑉 ∫ 𝐸⃗ (𝑟, 𝑡) ∗ 𝑛⃗ 𝑑𝑎 = Natural constants 𝐸(𝑟0 , 𝑡) = ∫ 𝑗(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = − Gauss’ law 1. Introduction [𝐸] = 𝑉 ⁄𝑚 , Continuity equation / conservation of charges 2. Maxwell’s Equations EFUW Summary ∇ 𝑥 𝐻(𝑟, 𝑡) = 𝑑 𝐵(𝑟, 𝑡) 𝑑𝑡 𝑑 𝐷(𝑟, 𝑡) + 𝑗(𝑟, 𝑡) "𝑫𝒖𝒓𝒄𝒉𝒇𝒍𝒖𝒕𝒖𝒏𝒈" 𝑑𝑡 ∇ ∗ 𝐵(𝑟, 𝑡) = 0 𝑗𝑐𝑜𝑛𝑑 = 𝜎 𝐸 Total current 𝑗𝑡𝑜𝑡 = 𝑗𝑑𝑖𝑠𝑝 + 𝑗𝑐𝑜𝑛𝑑 + 𝑗𝑝𝑜𝑙 + 𝑗𝑚𝑎𝑔 𝑑 𝑑 = 𝜀0 𝐸 + 𝜎 𝐸 + 𝑃 + ∇ 𝑥 𝑀 𝑑𝑡 𝑑𝑡 1 Evanescent waves 3. The Wave Equation If ( 𝑘𝑥2 + 𝑘𝑦2 ) > 𝑘 2 , 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑒𝑣𝑎𝑛𝑒𝑠𝑐𝑒𝑛𝑡: Inhomogenous wave equations 1 𝑑2𝐸 𝑑 𝑑𝑃 = −𝜇0 (𝑗+ + ∇𝑥𝑀) 2 2 𝑐 𝑑𝑡 𝑑𝑡 𝑑𝑡 2 1𝑑 𝐻 𝑑𝑃 1 𝑑 2 𝑀 ∇𝑥∇𝑥𝐻+ 2 2 = ∇𝑥𝑗+ ∇𝑥 − 𝑐 𝑑𝑡 𝑑𝑡 𝑐 2 𝑑𝑡 2 2 𝑘𝑧 = √𝜔 ⁄𝑐 2 − ( 𝑘𝑥2 + 𝑘𝑦2 ) ∇𝑥∇𝑥𝐸+ Homogeneous solution in free space 1 𝑑2 No material or sources: ∇2 𝐸 − Else: ∇2 𝐸 − Interference of waves 𝑐 2 𝑑𝑡 2 Monochromatic: 𝐸(𝑟, 𝑡) = 𝑅𝑒 { 𝐸0 𝑒 ± 𝑖 (𝑘𝑥 𝑥+𝑘𝑦 𝑦)−𝑖𝜔𝑡 ∗ 𝑒 ∓ |𝑘𝑧 | 𝑧 } Decay exponentially in z-direction, only near sources Spectral representation −∞ ∞ 1 𝐸̂ (𝑟, 𝜔) = ∫ 𝐸(𝑟, 𝑡) 𝑒 𝑖𝜔𝑡 𝑑𝑡 2𝜋 −∞ Helmholtz equation 𝑘 = 𝜔/𝑐 𝐸̂ (𝑟, −𝜔) = 𝐸̂ ∗ (𝑟, 𝜔) , 𝛿(𝑥) = Dispersion relation ̂ (𝑟, 𝜔) = −𝑖𝜔𝐷 ̂ (𝑟, 𝜔) + 𝑗̂(𝑟, 𝜔) ∇𝑥𝐻 ∇ 𝑥 𝐸̂ (𝑟, 𝜔) = 𝑖𝜔𝐵̂ (𝑟, 𝜔) ∇ ∗ 𝐵̂ (𝑟, 𝜔) = 0 2 𝑐=𝜆𝑓, 𝜔 𝑐2 𝑘 = 2𝜋⁄𝜆 𝐸 ⊥𝐻 ⊥𝑘‖𝑆 → 𝐸∗𝑘 =0=𝐻∗𝑘 =𝐸∗𝐻 1 (𝑘𝑥𝐸), 𝐻 = 𝜔 𝜇0 1 ( ∇ 𝑥 𝐸) , 𝐻= 𝑖 𝜔 𝜇0 Evanescent: 𝐼(𝑟) = 1 2 𝜀 0 √𝜇 |𝐸0 |2 ∗ 𝑒 −2 𝑘𝑧 𝑧 0 𝜀0 ⟨[ 𝐸1 + 𝐸2 ] ∗ [𝐸1 + 𝐸2 ] ⟩ = 𝐼1 + 𝐼2 + 2 ∗ 𝐼12 𝜇0 Coherent fields: monochromatic, same frequencies Maxwell’s equation in Fourier form ̂ (𝑟, 𝜔) = 𝜌̂(𝑟, 𝜔) ∇∗ 𝐷 𝜔 = 2𝜋𝑓 , 0 1 ∞ 𝑖𝑥𝑡 ∫ 𝑒 𝑑𝑡 2𝜋 −∞ + 𝑖𝑘 ∗ 𝑟 ∶ propagation in k-direction, outgoing waves − 𝑖𝑘 ∗ 𝑟 ∶ propagation against k-direction, incoming waves 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = 𝑘 2 = 𝑛2 ∗ 𝜀 0 √𝜇 |𝐸0 |2 |𝐸|2 = 𝐸 ∗ 𝐸 ∗ = < 𝐸, 𝐸 > 𝐼(𝑟) = √ ∞ 𝐸(𝑟, 𝑡) = ∫ 𝐸̂ (𝑟, 𝜔) 𝑒 −𝑖𝜔𝑡 𝑑𝜔 𝐸(𝑟, 𝑡) = 𝑅𝑒 { 𝐸0 𝑒 ± 𝑖𝑘∗𝑟 − 𝑖𝜔𝑡 } 1 2 Field pair 𝐸(𝑟, 𝑡) = 𝑅𝑒 { 𝐸(𝑟) ∗ 𝑒 −𝑖𝜔𝑡 } Plane / homogenous waves 𝐼(𝑟) = 1/𝑒 decay length for evanescent waves: 𝐿𝑧 = 1/(2 𝑘𝑧 ) 𝐸=0 Monochromatic waves time-harmonic: oscillate with one fixed frequency 𝜀0 ⟨ 𝐸(𝑟, 𝑡) ∗ 𝐸(𝑟, 𝑡) ⟩ 𝜇0 ∈𝑖ℝ 𝐸=0 𝑐 2 𝑑𝑡 2 𝑛2 𝑑 2 ∇2 𝐸(𝑟) + 𝑘 2 𝐸(𝑟) = 0 , 𝐼(𝑟) = √ 𝑓𝑜𝑟 𝑝𝑙𝑎𝑛𝑒 𝑤𝑎𝑣𝑒 𝑖𝑓 𝑛𝑜𝑡 𝑝𝑙𝑎𝑛𝑒 𝑤𝑎𝑣𝑒 𝜀 𝐼(𝑥) = 𝐼1 + 𝐼2 + √ 0⁄𝜇0 𝑅𝑒 { 𝐸1 ∗ 𝐸2∗ 𝑒 𝑖𝑘𝑥(sin 𝛼+sin 𝛽 ) } If they are real and polarized along the z-axis 𝐼(𝑥) = 𝐼1 + 𝐼2 + 2 √𝐼1 𝐼2 ∗ cos[ 𝑘𝑥 (𝑠𝑖𝑛 𝛼 + 𝑠𝑖𝑛 𝛽 ) ] Monochromatic waves 𝐸̂ (𝑟, 𝜔 ′) 1 = [ 𝐸(𝑟) 𝛿(𝜔′ − 𝜔) + 𝐸 ∗ (𝑟) 𝛿(𝜔′ + 𝜔) ] 2 For time-harmonic fields, the Maxwell equations simplify: ∇ ∗ 𝐷(𝑟) = 𝜌(𝑟) ∇ 𝑥 𝐸(𝑟) = 𝑖𝜔𝐵(𝑟) ∇ 𝑥 𝐻(𝑟) = −𝑖𝜔𝐷(𝑟) + 𝑗(𝑟) ∇ ∗ 𝐵(𝑟) = 0 Visibility / “Interferenzkontrast”: 𝜂= 𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛 , 𝐼𝑚𝑎𝑥 + 𝐼𝑚𝑖𝑛 𝜂 = 1 𝑓𝑜𝑟 𝐼1 = 𝐼2 𝐼𝑚𝑎𝑥 ∶ 𝑘𝑜𝑛𝑠𝑡𝑟𝑢𝑘𝑡𝑖𝑣 ↔ 𝐼𝑚𝑖𝑛 ∶ 𝑑𝑒𝑠𝑡𝑟𝑢𝑘𝑡𝑖𝑣 , ∆𝑟 = 𝜆/2 Period of interference: ∆ 𝑥 = 𝜆 / (sin 𝛼 + sin 𝛽 ) Incoherent fields The interference term vanishes due to ∆𝜔 = 𝜔1 − 𝜔2 𝐼(𝑥) = 𝐼1 + 𝐼2 2 4. Constitutive Relations Temporally dispersive: field depends on previous times Spatially dispersive: field depends on other locations 𝑘𝑖 = Piecewise homogenous media Inhomogeneities are entirely confined to the boundaries. ̂ (𝑘, 𝜔) = 𝜀0 𝜀(𝑘, 𝜔) 𝐸̂ (𝑘, 𝜔) 𝐷 ̂ (𝑘, 𝜔) 𝐵̂ (𝑘, 𝜔) = 𝜇0 𝜇(𝑘, 𝜔) 𝐻 𝜔 √𝜇𝑖 𝜀𝑖 = 𝑘0 ∗ 𝑛𝑖 𝑐 𝑘𝑧𝑖 = √𝑘𝑖2 − 𝑘∥2 = 𝑘𝑖 ∗ cos 𝜃𝑖 = 𝑘𝑖 √1 − sin 𝜃𝑖 2 Inhomogenous vector Helmholtz equations For time-harmonic fields ( ∇2 + 𝑘𝑖2 ) 𝐸𝑖 = −𝑖𝜔𝜇0 𝜇𝑖 𝑗𝑖 + 𝐷(𝑟) = 𝜀0 𝜀(𝜔) 𝐸(𝑟) 𝐵(𝑟) = 𝜇0 𝜇(𝜔) 𝐻(𝑟) ∇ 𝜌𝑖 𝜀0 𝜀 ( ∇2 + 𝑘𝑖2 ) 𝐻𝑖 = − ∇ 𝑥 𝑗𝑖 For time-dependent fields Can only be used in dispersion-free materials ( 𝜀(𝜔) = 𝜀 , 𝜇(𝜔) = 𝜇 ) , especially in vacuum. Electric & Magnetic Susceptibilities 𝜀 = ( 1 + 𝜒𝑒 ) 𝜇 = ( 1 + 𝜒𝑚 ) For all k-vectors, their components can be calculated: 5. Material Boundaries 𝑘∥ = √𝑘𝑥2 + 𝑘𝑦2 = √𝑘𝑖2 − 𝑘𝑧2𝑖 = 𝑘𝑖 ∗ sin 𝜃𝑖 Snell’s Law Mostly, no source currents and therefore homogenous. 𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2 Boundary conditions From these 6 equations, only 4 are linearly independent Polarized waves 𝑛 ∗ [ 𝐵𝑖 (𝑟) − 𝐵𝑗 (𝑟) ] = 0 𝑃(𝑟) = 𝜀0 𝜒𝑒 (𝜔) 𝐸(𝑟) 𝐵(𝑟) = 𝜒𝑚 (𝜔) 𝐻(𝑟) 𝑛 ∗ [ 𝐷𝑖 (𝑟) − 𝐷𝑗 (𝑟) ] = 𝜎(𝑟) 𝑛 𝑥 [ 𝐸⃗𝑖 (𝑟) − 𝐸⃗𝑗 (𝑟) ] = 0 ⃗ 𝑖 (𝑟) − 𝐻 ⃗ 𝑗 (𝑟) ] = 𝐾(𝑟) 𝑛𝑥[𝐻 Conductivity 𝐾 : surface current density, mostly zero 𝜎 : surface charge density, mostly zero 𝑗𝑐𝑜𝑛𝑑 (𝑟) = 𝜎(𝜔) 𝐸(𝑟) s – polarisation p-polarisation Electric permeability 𝜎 𝜀 =𝜀 +𝑖 𝜔𝜀0 ′ 𝑅𝑒{ 𝜀 } = 𝜀 ′ : 𝜎 𝐼𝑚{ 𝜀 } = : 𝜔𝜀0 energy storage 𝐸𝑖∥ = 𝐸𝑗∥ , 𝐻𝑖∥ = 𝐻𝑗∥ 𝐸1 (𝑟, 𝑡) = 𝑅𝑒{ 𝐸1 𝑒 𝑖( 𝑘𝑥 𝑥+𝑘𝑦 𝑦+𝑘𝑧1 𝑧−𝜔𝑡) } Helmholtz equation for isotropic materials ∇2 𝐸(𝑟) + 𝑘 2 𝐸(𝑟) = ∇2 𝐸(𝑟) + 𝑘02 𝑛2 𝐸(𝑟) = 0 Index of refraction: Skin depth: 𝐷𝑖⊥ = 𝐷𝑗⊥ Reflection & Refraction at plane interfaces energy dissipation 𝑛 = √𝜀 𝜇 𝜔 𝑘 = 𝑛(𝜔) ∗ 𝑘0 = √𝜀 𝜇 , 𝑐 𝐵𝑖⊥ = 𝐵𝑗⊥ , 𝐸1𝑟 (𝑟, 𝑡) = 𝑅𝑒{ 𝐸1𝑟 𝑒 𝑖( 𝑘𝑥 𝑥+𝑘𝑦 𝑦−𝑘𝑧1 𝑧−𝜔𝑡) } 𝑘 =𝜀𝜇 𝐷𝑆 = √2⁄𝜎𝜇0 𝜇𝜔 𝑘02 𝑘𝑥1 = 𝑘𝑥1𝑟 = 𝑘𝑥2 = 𝑘𝑥 , 𝑘𝑦1 = 𝑘𝑦1𝑟 = 𝑘𝑦2 = 𝑘𝑦 |𝑘1 |2 = |𝑘1𝑟 |2 = 𝑘12 = 𝑘02 𝑛12 ⟹ 𝑘𝑧1𝑟 = ±𝑘𝑧1 (𝑝) + 𝐸𝑖 s-polarized : perpendicular to the plane of incidence → parallel to the interface p-polarized : parallel to the plane of incidence Wave impedance 𝐸2 (𝑟, 𝑡) = 𝑅𝑒{ 𝐸2 𝑒 𝑖( 𝑘𝑥 𝑥+𝑘𝑦 𝑦+𝑘𝑧2 𝑧−𝜔𝑡) } Boundary conditions at = 0 : transverse comp. const. 2 (𝑠) 𝐸𝑖 = 𝐸𝑖 𝑍𝑖 = √ 𝜇0 𝜇 𝜀0 𝜀 s-polarized wave (𝑠) 𝐸𝑖 = 𝐸𝑖 𝐻𝑖 = 𝑒𝑦 𝑘𝑧 (𝑠) 1 𝑘𝑥 (𝑠) ( − 𝑖 𝐸𝑖 𝑒𝑥 + 𝐸 𝑒𝑧 ) 𝑍𝑖 𝑘𝑖 𝑘𝑖 𝑖 3 Fresnel Coefficients 𝜇2 𝑘𝑧1 − 𝜇1 𝑘𝑧2 𝑟𝑠 = 𝜇2 𝑘𝑧1 + 𝜇1 𝑘𝑧2 𝜀2 𝑘𝑧1 − 𝜀1 𝑘𝑧2 𝑟𝑝 = 𝜀2 𝑘𝑧1 + 𝜀1 𝑘𝑧2 2 𝜇2 𝑘𝑧1 𝑡𝑠 = 𝜇2 𝑘𝑧1 + 𝜇1 𝑘𝑧2 2 𝜀2 𝑘𝑧1 𝜇2 𝜀1 𝑡𝑝 = √ 𝜀2 𝑘𝑧1 + 𝜀1 𝑘𝑧2 𝜇1 𝜀2 6. Energy and Momentum Maxwell stress tensor Field momentum Poynting’s Theorem 𝐺𝑓𝑖𝑒𝑙𝑑 = Explains the relation between electromagnetic fields and their energy and therefore extends Maxwell’s equations. 1 ∫ [ 𝐸 𝑥 𝐻 ] 𝑑𝑉 𝑐2 𝑉 Time-averaged mechanical force Poynting vector: energy flux density, parallel to k-vector They solve the two following equations: (𝑠) (𝑠) (𝑠) 𝑘𝑧 (𝑠) 𝑘𝑧 (𝑠) 𝑘𝑧 1 1 (𝑠) ( − 1 𝐸1 + 1 𝐸1𝑟 ) = ( − 2 ) 𝐸2 𝑍1 𝑘1 𝑘1 𝑍2 𝑘2 Time average: ⟨ 𝑆(𝑟) ⟩ = 1 2 Total internal reflection: transmitted field evanescent Far-field: ⟨ 𝑆(𝑟) ⟩ = 1 1 2 𝑍𝑖 Total transmission Intensity 𝑛2 p-pol: Brewster angle : tan 𝜃1 = s-pol: not possible, always partly reflected (𝑝) 𝐸2 = [ 𝑘𝑧1 = 𝑘1 √1 − sin2 𝜃1 , Index of refraction: 𝑊= 𝑛̃ = ]𝑒 𝑖 ( 𝑘𝑥 𝑥+𝑘𝑧2 𝑧 ) |𝐸(𝑟)|2 ∗ 𝑛𝑟 Maxwell’s stress tensor ⃡ = [ 𝜀0 𝜀𝑬𝑬 + 𝜇0 𝜇 𝑯𝑯 − 𝑻 1 ⃡] ( 𝜀 𝜀𝐸 2 + 𝜇0 𝜇𝐻2 )𝑰 2 0 Monochromatic plane wave, normal to interface Part of the field is reflected at the material, superposed: Density of electromagnetic energy 𝑡 𝑝 𝑘𝑧2 / 𝑘2 (𝑠) 𝐸1 𝑡 𝑠 (𝑝) 𝐸1 𝑡 𝑝 𝑘𝑥 / 𝑅𝑒{ 𝐸(𝑟) 𝑥 𝐻 ∗ (𝑟) } 𝐼(𝑟) = |⟨ 𝑆(𝑟) ⟩| Evanescent fields − 𝐸1 𝑑𝑉 Radiation pressure 𝜀2 𝑘𝑧1 = 𝜀1 𝑘𝑧2 (𝑟 𝑝 = 0) 𝑛1 ⃡ (𝑟, 𝑡) ⟩ ∗ 𝑛(𝑟) 𝑑𝑎 ⟨𝐹⟩ = ∫ ⟨𝑇 𝑆=𝐸𝑥𝐻 𝐸1 + 𝐸1𝑟 = 𝐸2 1 [𝐷∗𝐸 +𝐵∗𝐻] 2 1 𝐸(𝑟, 𝑡) = 𝐸𝑜 𝑅𝑒{ [ 𝑒 𝑖𝑘𝑧 + 𝑟 ∗ 𝑒 −𝑖𝑘𝑧 ] ∗ 𝑒 −𝑖𝜔𝑡 } ∗ 𝑛𝑥 𝐻(𝑟, 𝑡) = √ 𝑊 = 4 [ 𝐷(𝑟) 𝐸 ∗ (𝑟) + 𝐵(𝑟) 𝐻 ∗ (𝑟) ] Time-harmonic: 𝑘2 𝜀0 𝐸 𝑅𝑒{ [ 𝑒 𝑖𝑘𝑧 − 𝑟 ∗ 𝑒 −𝑖𝑘𝑧 ] ∗ 𝑒 −𝑖𝜔𝑡 } ∗ 𝑛𝑦 𝜇0 0 Radiation pressure 𝑘𝑧2 = 𝑘2 √1 − 𝑛̃2 sin2 𝜃1 √𝜀1 𝜇1 √𝜀2 𝜇2 Critical angle (afterwards evanescent/TIR): = Total power : generated or dissipated inside the surface 𝑃 𝑛𝑧 = 𝑃̅ = ∫ ⟨ 𝑆(𝑟) ⟩ ∗ 𝑛 𝑑𝑎 = ∫ 𝐼(𝑟) 𝑑𝑎 𝑛1 𝑑𝑉 𝑛2 𝜃𝑐 = arcsin 1⁄𝑛̃ Frustrated total internal reflection 𝑛2 < 𝑛3 < 𝑛1 𝑛2 𝑎) 𝜃1 < 𝑎𝑟𝑠𝑖𝑛 ( ) ∶ 𝑏𝑜𝑡ℎ 𝑝𝑟𝑜𝑝. 𝑛1 𝑏) 𝐹𝑇𝐼𝑅: 𝑛2 𝑒𝑣𝑎𝑛. , 𝑛3 𝑎𝑔𝑎𝑖𝑛 𝑝𝑟𝑜𝑝. 𝑛3 𝑐) 𝜃1 > 𝑎𝑟𝑠𝑖𝑛 ( ) ∶ 𝑏𝑜𝑡ℎ 𝑒𝑣𝑎𝑛. 𝑛1 𝑑𝑉 First two terms of stress tensor have no contribution, yield Energy transport by evanescent waves 𝑃= Dielectric interface; irradiated by plane wave under TIR ⟨𝑆⟩𝑧 = ⃗ 1 𝜀0 𝜀 𝑘 𝑅𝑒{ 𝐸𝑥 𝐻𝑦∗ − 𝐸𝑦 𝐻𝑥∗ } = 0, 𝐻 = √ [ 𝑥𝐸] 2 𝜇0 𝜇 𝑘 1 ⟨𝑆⟩𝑥 = 𝑅𝑒{ 𝐸𝑦 𝐻𝑧∗ − 𝐸𝑧 𝐻𝑦∗ } ≠ 0 2 = 1 𝜀2 𝜇2 (𝑝) 2 (𝑠) 2 sin 𝜃1 ( |𝑡 𝑠 |2 |𝐸1 | + |𝑡 𝑝 |2 |𝐸1 | ) ∗ 𝑒 −2𝛾𝑧 √ 2 𝜀1 𝜇1 𝐹 1 ⃡ (𝑟, 𝑡)⟩ ∗ 𝑛𝑧 𝑑𝑎 𝑛𝑧 = ∫ ⟨𝑇 𝐴 𝐴 𝐴 Reflectivity: 𝐼0 [ 1 + 𝑅] , 𝑐 𝐼0 = 𝜀0 𝑐 𝐸02 2 𝑅 = |𝑟|2 = 1 − 𝑇 𝑅 = 1: 𝑝𝑒𝑟𝑓𝑒𝑐𝑡𝑙𝑦 𝑟𝑒𝑓𝑙𝑒𝑐𝑡; 𝑅 = 0 ∶ 𝑛𝑜𝑡 Perfectly reflecting material has twice the radiation pressure of a non-reflecting material 4 Scalar Green function 7. Radiation [ ∇2 + 𝑘 2 ] 𝐺0 (𝑟, 𝑟 ′ ) = 𝛿( 𝑟 − 𝑟 ′ ) Only accelerated charge can give rise to radiation. The smallest radiating unit is a dipole, an electromagnetic point source, and is used with the superposition principle. Dyad: tensor of order two ( rank one ) Non-vanishing field components in spherical coordinates ′ 𝐺0 (𝑟, 𝑟 ′ ) = 𝑒 𝑖𝑘|𝑟−𝑟 | 4𝜋 |𝑟 − 𝑟 ′ | Electric and magnetic dipole fields Scalar and Vector potentials 𝐸(𝑟) = 𝑑 𝐸(𝑟, 𝑡) = − 𝐴(𝑟, 𝑡) − ∇ 𝛷(𝑟, 𝑡) 𝑑𝑡 𝜔2 𝜇0 𝜇 ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 ′ ) 𝑝 1 𝐻(𝑟) = −𝑖𝜔 [ ∇ 𝑥 ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 ′ ) ] 𝑝 = (∇𝑥𝐸) 𝑖𝜔𝜇0 𝑝 cos 𝜗 4𝜋𝜀0 𝜀 𝑝 sin 𝜗 𝐸𝜗 = 4𝜋𝜀0 𝜀 𝐸𝑟 = 𝐻𝜑 = 𝑒 𝑖𝑘𝑟 2 2 2𝑖 𝑘 [ 2 2− ] 𝑟 𝑘 𝑟 𝑘𝑟 𝑒 𝑖𝑘𝑟 2 1 𝑖 𝑘 [ 2 2− −1] 𝑟 𝑘 𝑟 𝑘𝑟 𝑝 sin 𝜗 𝑒 𝑖𝑘𝑟 2 𝑖 𝜀0 𝜀 𝑘 [ − − 1] √ 4𝜋𝜀0 𝜀 𝑟 𝑘𝑟 𝜇0 𝜇 1. Far-field component is purely transverse 2. The near-field is dominated by the electric field Radiation patterns and power dissipation 𝐵(𝑟, 𝑡) = ∇ 𝑥 𝐴(𝑟, 𝑡) Dyadic Green function ( tensor ) 𝐴 ∶ 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 , 𝛷 ∶ 𝑠𝑐𝑎𝑙𝑎𝑟 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 Dipole radiation ′) 1 = [⃡ 𝐼 + 2 ∇ ∇ ] 𝐺0 (𝑟, 𝑟 ′ ) 𝑘 Radial component of the Poynting vector ⟨𝑆𝑟 ⟩ = In Cartesian coordinates: 𝑅 = |𝑹| = |𝑟 − 𝑟 ′ | 𝑝(𝑡) = 𝑞(𝑡) 𝑑𝑠 𝑑 𝑑 𝑞(𝑡) 𝑝(𝑡) = [ 𝑛𝑠 ] 𝑑𝑠 = [ 𝑗0 𝑑𝑎] 𝑑𝑠 = 𝑗0 𝑑𝑉 𝑑𝑡 𝑑𝑡 Vector potential of a time-harmonic dipole 𝛷(𝑟, 𝑡) = 𝑅𝑒{ 𝛷(𝑟) 𝑒 −𝑖𝜔𝑡 } [ ∇2 + 𝑘 2 ] 𝐴(𝑟) = −𝜇0 𝜇 𝑗0 (𝑟) 1 [ ∇2 + 𝑘 2 ] 𝛷(𝑟) = − 𝜌 (𝑟) 𝜀0 𝜀 0 Vector potential 𝑨 𝒂𝒕 𝒓 𝒅𝒖𝒆 𝒕𝒐 𝒂 𝒅𝒊𝒑𝒐𝒍𝒆 𝒑 𝒂𝒕 𝒓′ ′ 𝑒 𝑖𝑘|𝑟−𝑟 | 𝐴(𝑟) = −𝑖𝜔𝜇0 𝜇 𝑝 4𝜋 |𝑟 − 𝑟 ′ | = 𝜇0 𝜇 ∫ 𝐺0 (𝑟, 𝑟 𝑗0 𝑉 (𝑟 ′ ) With it, the resulting radiated power can be found by 2𝜋 𝑃̅ = ∫ 0 𝜋 ∫ | ⟨𝑆𝑟 ⟩ | 𝑟 2 sin 𝜗 𝑑𝜗 𝑑𝜑 0 Radiated power of a dipole 𝑑 𝑗(𝑟, 𝑡) = 𝑝(𝑡) 𝛿( 𝑟 − 𝑟0 ) 𝑑𝑡 ′) 𝑒 𝑖𝑘𝑅 𝑖𝑘𝑅 − 1 3 − 3𝑖𝑘𝑅 − 𝑘 2 𝑅2 𝑹 𝑹 [(1+ 2 2 ) ⃡ 𝑰 + ] 4𝜋𝑅 𝑘 𝑅 𝑘 2 𝑅2 𝑅2 Near- , intermediate- and far-field Current density of the elemental dipole 𝐴(𝑟, 𝑡) = 𝑅𝑒{ 𝐴(𝑟) 𝑒 −𝑖𝜔𝑡 }, ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 ′ ) = 1 𝑅𝑒 { 𝐸𝜗 𝐻𝜑∗ } 2 𝑑𝑉′ 1. 𝑅 ≪ 𝑘 → 𝑜𝑛𝑙𝑦 𝑡𝑒𝑟𝑚𝑠 𝑤𝑖𝑡ℎ ( 𝑘𝑅 )−3 𝑠𝑢𝑟𝑣𝑖𝑣𝑒 ⃡⃗ 𝑁𝐹 = 𝑮 𝑒 𝑖𝑘𝑅 1 𝑹𝑹 [ −⃡ 𝑰+3 2 ] 4𝜋𝑅 𝑘 2 𝑅2 𝑅 2. 𝑅 ≈ 𝑘 → 𝑜𝑛𝑙𝑦 𝑡𝑒𝑟𝑚𝑠 𝑤𝑖𝑡ℎ ( 𝑘𝑅 )−2 𝑠𝑢𝑟𝑣𝑖𝑣𝑒 ⃡𝑮⃗ 𝐼𝐹 = 𝑒 𝑖𝑘𝑅 𝑖 𝑹𝑹 [ ⃡ 𝑰− 3 2 ] 4𝜋𝑅 𝑘 𝑅 𝑅 3. 𝑅 ≫ 𝑘 → 𝑜𝑛𝑙𝑦 𝑡𝑒𝑟𝑚𝑠 𝑤𝑖𝑡ℎ ( 𝑘𝑅 )−1 𝑠𝑢𝑟𝑣𝑖𝑣𝑒 ⃡𝑮⃗ 𝐹𝐹 = 𝑒 𝑖𝑘𝑅 4𝜋𝑅 [ ⃡ 𝑰− 𝑹𝑹 ] 𝑅2 Intermediate-field is 90° 𝑜𝑢𝑡 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒 in respect to the near- and far-fied. 𝑃̅ = |𝑝|2 𝑛3 𝜔4 |𝑝|2 𝜔 𝑘 3 = 4𝜋𝜀0 𝜀 3 𝑐 3 12 𝜋 𝜀0 𝜀 In order to describe the radiation characteristic, we calculate the power radiated at an infinitesimal unit solid angle 𝑑𝛺 = sin 𝜗 𝑑𝜗 𝑑𝜑 and normalize it: ⟨𝑆𝑟 ⟩ 𝑟 2 𝑃̅ (𝜗, 𝜑) 3 = 2𝜋 𝜋 = sin2 𝜗 2 8𝜋 𝑃̅ ∫0 ∫0 ⟨𝑆𝑟 ⟩ 𝑟 sin 𝜗 𝑑𝜗 𝑑𝜑 Near-field: in direction of p Far-field: perpendicular to p 5 Dipole Radiation in arbitrary environments Sources with arbitrary time-dependence 8. Angular Spectrum Energy dissipation of dipole is influenced by surroundings. Utilize Fourier transform: superposition of time harmonic 𝑧 𝐸(𝑧 = 0, 𝑡) = 𝑓(𝑡) → 𝐸(𝑧, 𝑡) = 𝑓 (𝑡 − ) 𝑐 Series expansion of an arbitrary field in terms of plane 𝑃̅ = ∞ 𝜇0 𝐴(𝑟, 𝑡) = 𝜇̃(𝑡) ∗ ∫ 𝐺0 (𝑟, 𝑟 ′ , 𝑡) ∗ 𝑗0 (𝑟 ′ , 𝑡) 𝑑𝑉 ′ 4𝜋 −∞ 𝑑𝑊 1 = − ∫ 𝑅𝑒 { 𝑗 ∗ ∗ 𝐸 } 𝑑𝑉 𝑑𝑡 2 𝑉 By using the dipole’s current density ( 𝑟0 : 𝑑𝑖𝑝𝑜𝑙𝑒 𝑜𝑟𝑖𝑔𝑖𝑛 ) = ∞ 1 ′ ∫ 𝑒 𝑖𝜔 [ 𝑡−𝑛(𝜔)|𝑟−𝑟 |⁄𝑐 ] 𝑑𝜔 4𝜋|𝑟 − 𝑟 ′ | −∞ Dispersion-free materials: 𝜔 𝑃̅ = 𝐼𝑚 { 𝑝∗ ∗ 𝐸(𝑟0 ) } 2 3 𝐺0 (𝑟, 𝑟 ′ , 𝑡) = |𝑝|2 𝜔 ⃡ (𝑟0 , 𝑟0 ) } ∗ 𝑛𝑝 ] [ 𝑛𝑝 ∗ 𝐼𝑚{ 𝐺 2 𝑐 2 𝜀0 𝜀 In an inhomogeneous surrounding, the field is a superposition of the primary and scattered field: 𝑛(𝜔) = 𝑛 , 𝜇(𝜔) = 𝜇 𝑗0 (𝑟 ′ , 𝑡 − |𝑟 − 𝑟 ′ | 𝑛⁄𝑐 ) ∫ 𝑑𝑉 ′ |𝑟 − 𝑟 ′ | 𝑉 𝜇0 𝜇 𝐴(𝑟, 𝑡) = 4𝜋 1 𝜌0 (𝑟 ′ , 𝑡 − |𝑟 − 𝑟 ′ | 𝑛⁄𝑐 ) 𝜙(𝑟, 𝑡) = ∫ 𝑑𝑉 ′ |𝑟 − 𝑟 ′ | 4𝜋𝜀0 𝜀 𝑉 Dipole fields in time domain Assume dipole in vacuum ( 𝑘 = 𝜔⁄𝑐 , 𝜀 = 1 ) 𝐸(𝑟0 ) = 𝐸0 (𝑟0 ) + 𝐸𝑆 (𝑟0 ) |𝑝| ̅̅̅ 𝑃0 = Contribution of 𝐸0 : 2 𝜔 12𝜋 𝜀0 𝜀 𝑘3 𝑃̅ 6𝜋𝜀0 𝜀 1 = 1+ 𝐼𝑚{ 𝑝∗ ∗ 𝐸𝑆 (𝑟0 ) } ̅̅̅ |𝑝|2 𝑘 3 𝑃0 cos 𝜗 2 2 𝑑 [ 3 + 2 ] 𝑝(𝜏) | 𝜏 = 𝑡 − 𝑟 4𝜋𝜀0 𝑟 𝑐𝑟 𝑑𝜏 𝑐 𝐸𝜗 (𝑡) = − sin 𝜗 1 1 𝑑 1 𝑑2 [ 3+ 2 + 2 ] 𝑝(𝜏) | 𝜏 = 𝑡 − 𝑟 4𝜋𝜀0 𝑟 𝑐𝑟 𝑑𝜏 𝑐 𝑟 𝑑𝜏 2 𝑐 sin 𝜗 𝜀0 1 𝑑 1 𝑑2 𝐻𝜑 (𝑡) = − + 2 ] 𝑝(𝜏) | 𝜏 = 𝑡 − 𝑟 √ [ 2 4𝜋𝜀0 𝜇0 𝑐𝑟 𝑑𝜏 𝑐 𝑟 𝑑𝜏 2 𝑐 Fields emitted by arbitrary sources 𝐸(𝑟) = 𝑖 𝜔 𝜇0 𝜇 ∫ ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 ′ ) 𝑗(𝑟 ′ ) 𝑑𝑉 ′ 𝑉 Far-field generated by acceleration of the dipole charges, intermediate-field by velocity and near-field by position. Lorentzian power spectrum 𝐻(𝑟) = ∫ [ ∇ 𝑥 ⃡⃗⃗⃗ 𝐺0 (𝑟, 𝑟 ] 𝑗(𝑟′) 𝑑𝑉 ′) 𝑉 𝐸𝑟 (𝑡) = ′ 𝑑𝑊 1 |𝑝|2 sin2 𝜗 𝜔02 𝛾02 ⁄4 = [ ] (𝜔 − 𝜔0 )2 + 𝛾02 ⁄4 𝑑𝛺𝑑𝜔 4𝜋𝜀0 4𝜋 2 𝑐 3 𝛾02 (and evanescent) waves with variable amplitudes and propagation directions. For this, we we draw an arbitrary axis 𝑧 and consider the field E in a plane 𝑧 = 𝑐𝑜𝑛𝑠𝑡. 2D Fourier transform ( 𝑘𝑥 , 𝑘𝑦 are the spatial frequencies) ∞ 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) = 1 ∬ 𝐸(𝑥, 𝑦, 𝑧) 𝑒 −𝑖(𝑘𝑥 𝑥+𝑘𝑦 𝑦) 𝑑𝑥 𝑑𝑦 4𝜋 2 −∞ ∞ 𝐸(𝑥, 𝑦, 𝑧) = ∬ 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) 𝑒 𝑖 (𝑘𝑥 𝑥+𝑘𝑦 𝑦) 𝑑𝑘𝑥 𝑑𝑘𝑦 −∞ 𝑘𝑧 = √( 𝑘 2 − 𝑘𝑥2 − 𝑘𝑦2 ) 𝐼𝑚(𝑘𝑧 ) ≥ 0 , 𝑘 = √𝜇𝜀 𝜔 𝑐 Evolution of the Fourier spectrum 𝐸̂ along the 𝑧-axis 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) = 𝐸̂ ( 𝑘𝑥 , 𝑘𝑦 ; 0) ∗ 𝑒 ± 𝑖 𝑘𝑧 𝑧 + : wave propagation in the half space 𝑧 > 0 - : wave propagation in the half space 𝑧 < 0 Angular Spectrum Representation ∞ 𝐸(𝑥, 𝑦, 𝑧) = ∬ 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 0) ∗ 𝑒 𝑖 (𝑘𝑥 𝑥+𝑘𝑦 𝑦 ± 𝑘𝑧 𝑧) 𝑑𝑘𝑥 𝑑𝑘𝑦 −∞ ∞ ∆ 𝜔 = 𝛾0 , 𝛾0 ∶ 𝑑𝑎𝑚𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 −∞ Total radiated energy 𝑊= ̂ (𝑘𝑥 , 𝑘𝑦 ; 0) ∗ 𝑒 𝑖 (𝑘𝑥 𝑥+𝑘𝑦 𝑦 ± 𝑘𝑧 𝑧) 𝑑𝑘𝑥 𝑑𝑘𝑦 𝐻(𝑥, 𝑦, 𝑧) = ∬ 𝐻 |𝑝|2 𝜔04 4𝜋𝜀0 3 𝑐 3 𝛾0 Express wave at any point by Fourier transform in 𝑧 = 0 6 Using Maxwell: 𝐻 = ̂𝑥 = 1⁄ 𝐻 𝑍 𝜇𝜀 1 𝑖𝜔𝜇𝜇0 𝜇 𝜇 ( ∇ 𝑥 𝐸 ) , 𝑍𝜇𝜀 = √ 0 𝜀 𝜀 0 Wavevector 𝑘 almost parallel to the 𝑧-axis ( 𝑘𝑥 , 𝑘𝑦 ≪ 𝑘) 𝑠 = ( 𝑠𝑥 , 𝑠𝑦 , 𝑠𝑧 ) = ( 𝑘𝑧 = 𝑘 √1 − (𝑘𝑥2 + 𝑘𝑦2 )/𝑘 2 ≈ 𝑘 − [ (𝑘𝑧 ⁄𝑘) 𝐸̂𝑥 − (𝑘𝑥 ⁄𝑘 ) 𝐸̂𝑧 ] ̂𝑧 = 1⁄ 𝐻 𝑍 [ (𝑘𝑥 ⁄𝑘 ) 𝐸̂𝑦 − (𝑘𝑦 ⁄𝑘) 𝐸̂𝑥 ] 𝜇𝜀 Far-field Approximation [ (𝑘𝑦 ⁄𝑘 ) 𝐸̂𝑧 − (𝑘𝑧 ⁄𝑘) 𝐸̂𝑦 ] ̂𝑦 = 1⁄ 𝐻 𝑍 𝜇𝜀 Paraxial Approximation ( 𝑘𝑥2 + 𝑘𝑦2 ) 2𝑘 𝐸∞ (𝑠𝑥 , 𝑠𝑦 ) = −2𝜋𝑖 𝑘 𝑠𝑧 𝐸̂ ( 𝑘𝑠𝑥 , 𝑘𝑠𝑦 ; 0) ̂=0 𝑘 ∗ 𝐸̂ = 𝑘 ∗ 𝐻 with ̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) = 𝑒 ± 𝑖 𝑘𝑧 𝑧 𝐻 Optical transfer function (OTF): ̂ ( 𝑘𝑥 , 𝑘𝑦 ; 𝑧) 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 0) 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) = 𝐻 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 0) ̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) Filter function : 𝐻 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 𝑧) ̂ acts as a low-pass filter (only 𝑘𝑥2 + 𝑘𝑦2 < 𝑘 2 can pass), 𝐻 𝑘 𝜔02 ∆𝑥 ≈ Maximal resolution: 1 𝑘 = 𝜆 2𝜋𝑛 Calculation of the fields 𝐸(𝑥, 𝑦, 𝑧) = 𝐸(𝑥, 𝑦 ; 0) ∗ 𝐻(𝑥, 𝑦 ; 𝑧) 𝐸(𝑧 = 𝑐𝑜𝑛𝑠𝑡. ) = 𝐶𝑜𝑛𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ( 𝐸(𝑧 = 0), 𝐻 ) ∞ 𝐻(𝑥, 𝑦, 𝑧) = ∬ 𝑒 𝑖 [ 𝑘𝑥 𝑥+𝑘𝑦 𝑦 ± 𝑘𝑧 𝑧] 𝑑𝑘𝑥 𝑑𝑘𝑦 −∞ spectrum at 𝑧 = 0 contributes to the far-field in 𝑠 direction. The rest gets cancelled by destructive interference. Therefore, the far-field behaves as a collection of rays 2 where each ray is characterized by a particular plane wave of the original angular spectrum representation. 1 2 2 Beam radius: 𝜔(𝑧) = 𝜔0 ( 1 + 𝑧 2 ⁄𝑧0 ) Wavefront radius: 𝑅(𝑧) = 𝑧 (1 + 𝑧02 ⁄𝑧 2 ) Phase correction: 𝜂(𝑧) = arctan 𝑧/𝑧0 Transverse size: 𝜌 , 𝑠𝑜 𝑡ℎ𝑎𝑡 ∶ as evanescent waves are omitted. There is always a loss of information from the near field to the far field. 𝑠𝑧 = √1 − ( 𝑠𝑥2 + 𝑠𝑦2 ) = 𝑧⁄𝑟 Only one plane wave with wave vector 𝑘 of the angular 2 𝜔0 − 𝜌2 (𝑧) 𝑖 [ 𝑘∗𝑧 − 𝜂(𝑧) + 𝑘𝜌2 ⁄2𝑅(𝑧) ] 𝐸(𝜌, 𝑧) = 𝐸0 𝑒 𝜔 𝑒 𝜔(𝑧) Where 𝜌 = √𝑥 2 + 𝑦 2 , 𝑧0 = As linear response theory 𝑒 𝑖𝑘𝑟 𝑟 Far-field entirely defined by Fourier spectrum at 𝑧 = 0. Propagation and Focusing of Fields Output : Evanescent waves vanish. Now only ( 𝑘𝑥2 + 𝑘𝑦2 ) < 𝑘 2 Gaussian Beams (does NOT fulfil Maxwell) Divergence-free : Input : 𝑥 𝑦 𝑧 𝑘𝑥 𝑘𝑦 𝑘𝑧 , , )=( , , ) 𝑟 𝑟 𝑟 𝑘 𝑘 𝑘 |𝐸(𝑥,𝑦,𝑧)| |𝐸(0,0,𝑧)| 𝐸̂ (𝑘𝑥 , 𝑘𝑦 ; 0) = = 1 𝑒 𝐸(𝑥, 𝑦, 𝑧) = Spaghetti formula 𝑧0 = 𝑘 𝜔02 , 2 Numerical aperture (NA): 𝜃= 𝑖 𝑟 𝑒 − 𝑖𝑘𝑟 𝑘𝑥 𝑘𝑦 𝐸∞ ( , ) 2𝜋 𝑘𝑧 𝑘 𝑘 Rayleigh range 𝒛𝟎 : distance form the beam waist to where the beam radius has increased by a factor of √2 Beam stays roughly focused over a distance of 𝟐 𝒛𝟎 . Gouy phase shift: 180° shift from 𝑧 → −∞ 𝑡𝑜 𝑧 → ∞ ∬ 2 ) ≤ 𝑘2 (𝑘𝑥2 + 𝑘𝑦 𝐸∞ ( 𝑘𝑥 𝑘 … 𝑒 𝑖 [ 𝑘𝑥 𝑥 + 𝑘𝑦 𝑦 ± 𝑘𝑧 𝑧 ] 2 𝑘 𝜔0 𝑁𝐴 ≈ 2𝑛/𝑘𝜔0 𝑖 𝑟 𝑒 − 𝑖𝑘𝑟 2𝜋 , 𝑘𝑦 𝑘 )… 1 𝑑𝑘 𝑑𝑘 𝑘𝑧 𝑥 𝑦 𝐸 𝑎𝑛𝑑 𝐸∞ 𝑓𝑜𝑟𝑚 𝑎 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑝𝑎𝑖𝑟 𝑎𝑡 𝑧 = 0 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑘𝑧 ≈ 𝑘, 𝑡ℎ𝑖𝑠 𝑝𝑎𝑖𝑟 𝑖𝑠 𝑝𝑒𝑟𝑓𝑒𝑐𝑡. Object plane: 𝑧=0 Image plane: 𝑧 = 𝑧0 (𝐹𝑎𝑟 𝑓𝑖𝑒𝑙𝑑: 𝑧0 → ∞ ) Fourier optics: 𝑘𝑧 ≈ 𝑘 , R only dependent on z 7 Fresnel & Fraunhofer diffraction The Point-Spread function 9. Waveguides & Resonators Measure of the resolving power of an imaging system: the narrower the function, the better the resolution. Resonators confine electromagnetic energy Waveguides guide this electromagnetic energy 9.1 Resonators Consider a rectangular box with sides 𝐿𝑥 , 𝐿𝑦 , 𝐿𝑧 We now search solutions for Helmholtz: [ ∇2 + 𝑘 2 ] 𝐸 = 0 (𝑥) 1. Ansatz: 𝐸𝑥 (𝑥, 𝑦, 𝑧) = 𝐸0 𝑋(𝑥) 𝑌(𝑦) 𝑍(𝑧) ; 𝐸𝑦 = … 𝑟 2 = (𝑥 − 𝑥 ′ )2 + (𝑦 − 𝑦 ′ )2 + 𝑧 2 2(𝑥𝑥 ′ − 𝑦𝑦 ′ ) 𝑥 ′2 + 𝑦 ′2 = 𝑅2 [ 1 − + ] 𝑅2 𝑅2 2. Separation of variables 1 𝑑2𝑋 1 𝑑2𝑌 1 𝑑2𝑍 + + + 𝑘2 = 0 𝑋 𝑑 𝑥 2 𝑌 𝑑 𝑦2 𝑍 𝑑 𝑧2 Determine the field at the observation point using Huygens 3. Set constants to − 𝑘𝑥2 , −𝑘𝑦2 , −𝑘𝑧2 , which implies principle of “summing up” elementary spherical waves: ′ ′ ∫ 𝐴(𝑥 , 𝑦 𝑧=0 ′ We can set 𝑟(𝑥 , 𝑦 ′) ′) 𝑒 − 𝑖 𝑘 𝑟(𝑥 ,𝑦 𝑟(𝑥 ′ , 𝑦 ′ ) ′) ′ 𝑑𝑥 𝑑𝑦 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = 𝑘 2 = ′ ≈ 𝑅 in the denominator due to the large distance between source and observer. However, we Due to the loss of evanescent waves (with their high spatial frequencies) and the finite angular collection, the point appears as a function with finite width. cannot neglect interference effects in the exponent. 𝜔2 2 𝑛 (𝜔) 𝑐2 4. We obtain tree separate equations 𝑑2𝑋 𝑑2𝑌 𝑑2𝑍 + 𝑘𝑥2 𝑋 = + 𝑘𝑦2 𝑌 = 2 + 𝑘𝑧2 𝑍 = 0 2 2 𝑑𝑥 𝑑𝑦 𝑑𝑧 5. → 𝐸𝑥 (𝑥, 𝑦, 𝑧) = 𝐸0(𝑥) [𝑐1,𝑥 𝑒 − 𝑖𝑘𝑥𝑥 + 𝑐2,𝑥 𝑒 𝑖𝑘𝑥𝑥 ] … With the paraxial approximation, we get: … [𝑐3,𝑥 𝑒 − 𝑖𝑘𝑦𝑦 + 𝑐4,𝑥 𝑒 𝑖𝑘𝑦𝑦 ][𝑐5,𝑥 𝑒 − 𝑖𝑘𝑧 𝑧 + 𝑐6,𝑥 𝑒 𝑖𝑘𝑧 𝑧 ] 6.Boundary conditions: 𝐸𝑥 (𝑦 = 0) = 𝐸𝑥 (𝑦 = 𝐿𝑦 ) = 0 = 𝐸𝑥 (𝑧 = 0) = 𝐸𝑥 (𝑧 = 𝐿𝑧 ) 7. Use ∇ ∗ 𝐸 = 0 , ( ∇ ∗ 𝐻 = 0, ∇ 𝑥 𝐸 = 0 ) Maximum extend of the source at 𝑧 = 0 𝐷 ⁄2 = 𝑀𝑎𝑥 { √𝑥 ′ 2 + 𝑦 ′ 2 } For 𝑧0 ≫ 𝐷 , we can use Fraunhofer. Else, we use Fresnel. Magnification: 𝑀= 𝑓 𝑛1 ⁄𝑛2 ∗ 2⁄𝑓 1 Numerical aperture: 𝑁𝐴 = 𝑛1 sin(𝑀𝑎𝑥[𝜃1 ]) 𝑓 = 2⁄𝑓 sin(𝑀𝑎𝑥[𝜃2 ]) 1 Airy disk radius: ∆𝑥 = 0.6098 The transition between Fraunhofer and Fresnel happens around the Rayleigh range 𝑧0 𝑧0 = 1⁄8 𝑘 𝐷2 , 𝜔0 = 𝐷⁄2 𝑀𝜆 𝑁𝐴 𝑥 𝑦 𝑧 ] sin [ 𝑚 𝜋 ] sin [ 𝑙 𝜋 ] 𝐿𝑥 𝐿𝑦 𝐿𝑧 𝑥 𝑦 𝑧 (𝑦) 𝐸𝑦 (𝑥, 𝑦, 𝑧) = 𝐸0 sin [𝑛 𝜋 ] cos [ 𝑚 𝜋 ] sin [ 𝑙 𝜋 ] 𝐿𝑥 𝐿𝑦 𝐿𝑧 𝑥 𝑦 𝑧 (𝑧) 𝐸𝑧 (𝑥, 𝑦, 𝑧) = 𝐸0 sin [𝑛 𝜋 ] sin [ 𝑚 𝜋 ] cos [ 𝑙 𝜋 ] 𝐿𝑥 𝐿𝑦 𝐿𝑧 (𝑥) 𝐸𝑥 (𝑥, 𝑦, 𝑧) = 𝐸0 cos [𝑛 𝜋 8 𝑛 (𝑥) 𝑚 (𝑦) 𝑙 (𝑧) 𝐸 + 𝐸 + 𝐸 =0 𝐿𝑥 0 𝐿𝑦 0 𝐿𝑧 0 Quality factor Due to losses such as absorption and radiation, the discrete frequencies broaden to a finit line width ∆𝜔 = 2𝛾 Dispersion relation / mode structure of the resonator 2 𝑛2 𝑚2 𝑙 2 𝜔𝑛𝑚𝑙 + 2 + 2 ] = 2 𝑛2 (𝜔𝑛𝑚𝑙 ) , 2 𝐿𝑥 𝐿𝑦 𝐿𝑧 𝑐 𝑄 = 𝜔0 /𝛾 Density of States (DOS) Measure for how long energy can be stored in a resonator Due to the losses, the electric field diminishes: 𝜔0 𝐸(𝑟, 𝑡) = 𝑅𝑒{ 𝐸0 (𝑟) exp [ (𝑖𝜔0 − ) 𝑡] 2𝑄 Finite-size box with equal length: 𝐿 = 𝐿𝑥 = 𝐿𝑦 = 𝐿𝑧 Where 𝜔0 is one of the resonant frequencies 𝜔𝑛𝑚𝑙 . 𝜋2 [ → 𝑛, 𝑚, 𝑙 ∈ ℤ0 2 𝐿 𝜔𝑛𝑚𝑙 𝑛2 + 𝑚2 + 𝑙 2 = [ ∗ 𝑛(𝜔𝑛𝑚𝑙 )] 𝜋 𝑐 If 𝑛, 𝑚, 𝑙 ∈ ℝ ∶ 𝑟0 = [ 𝜔𝑛𝑚𝑙 𝐿 𝑛(𝜔𝑛𝑚𝑙 )⁄(𝜋𝑐)] Spectrum of the stored energy density 𝑊𝜔 (𝜔) = 𝜔02 𝑊𝜔 (𝜔0 ) 4𝑄2 (𝜔 − 𝜔0 )2 + (𝜔0 ⁄2𝑄 )2 Cavity Perturbance (Disturbance theory of a resonator) Particle absorption or a change of the index of refraction can lead to a shift of the resonance frequency The number of different modes in interval [ 𝜔 … 𝜔 + ∆𝜔] 𝑑𝑁(𝜔) 𝜔2 𝑛3 (𝜔) ∆𝜔 = 𝑉 ∆𝜔 𝑑𝜔 𝜋2 𝑐3 States that there are more modes for higher frequencies. Density of States (DOS) 𝜌(𝜔) = 𝜔2 𝑛3 (𝜔) 𝜋2 𝑐3 DOS: number of modes per unit volume V and unit frequency ∆𝜔 Number of modes 𝜔2 𝑁(𝜔) = ∫ ∫ 𝜌(𝜔) 𝑑𝜔 𝑑𝑉 𝑉 𝜔1 Example: Power emitted by a dipole 𝜋 𝜔2 |𝑝|2 𝜌(𝜔) 𝑃̅ = 12 𝜀0 𝜀 Unperturbed system (𝜔0 ∶ 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦) ∇ 𝑥 𝐸0 = 𝑖𝜔0 𝜇0 𝜇(𝑟) 𝐻0 , Waveguides Used to carry electromagnetic energy from A to B Parallel-Plate waveguides Material with 𝑛(𝜔) sandwiched between two conductors TE-Mode: TM-Mode: TE-Modes: electric field parallel to surfaces of the plates Ansatz: plane wave propag. at angle 𝜃 to surface normal 𝐸1 (𝑥, 𝑦, 𝑧) = 𝐸0 𝑛𝑦 𝑒 [ −𝑖𝑘𝑥 cos 𝜃+𝑖𝑘𝑧 sin 𝜃 ] Coming from the upper plate, it gets reflected at the bottom: 𝐸2 (𝑥, 𝑦, 𝑧) = − 𝐸0 𝑛𝑦 𝑒 [ 𝑖𝑘𝑥 cos 𝜃+𝑖𝑘𝑧 sin 𝜃 ] ∇ 𝑥 𝐻0 = − 𝑖𝜔0 𝜀0 𝜀(𝑟) 𝐸0 Perturbed system ( ∆𝜀, ∆𝜇 ∶ 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑝𝑎𝑟𝑎𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒) ∇𝑥𝐸 = 𝑖𝜔𝜇0 [ 𝜇(𝑟) 𝐻 + ∆𝜇(𝑟) 𝐻 ] ∇ 𝑥 𝐻 = − 𝑖𝜔𝜀0 [ 𝜀(𝑟) 𝐸 + ∆𝜀(𝑟) 𝐸 ] Bethe-Schwinger cavity perturbation formula Superposition of the fields: 𝐸(𝑥, 𝑦, 𝑧) = 𝐸1 + 𝐸2 = −2𝑖 𝐸0 𝑛𝑦 𝑒 𝑖 𝑘 𝑧 𝑠𝑖𝑛𝜃 sin(𝑘 𝑥 cos 𝜃) Quantisation of the normal wavenumber: field must fulfil boundary condition at the upper plate: 𝐸(𝑥, 𝑦, 𝑑) = 0 ∫ [ 𝐸0∗ 𝜀0 ∆𝜀(𝑟) 𝐸 + 𝐻0∗ 𝜇0 ∆𝜇(𝑟) 𝐻 ] 𝑑𝑉 𝜔 − 𝜔0 = − ∆𝑉 𝜔 ∫𝑉[ 𝜀0 𝜀(𝑟) 𝐸0∗ 𝐸 + 𝜇0 𝜇(𝑟) 𝐻0∗ 𝐻 ] 𝑑𝑉 Assuming a small effect of the perturbation on the cavity: 𝐸 = 𝐸0 , 𝐻 = 𝐻0 ∫ [ 𝐸0∗ 𝜀0 ∆𝜀(𝑟) 𝐸0 + 𝐻0∗ 𝜇0 ∆𝜇(𝑟) 𝐻0 ] 𝑑𝑉 𝜔 − 𝜔0 = − ∆𝑉 𝜔 ∫𝑉[ 𝜀0 𝜀(𝑟) 𝐸0∗ 𝐸0 + 𝜇0 𝜇(𝑟) 𝐻0∗ 𝐻0 ] 𝑑𝑉 For a weakly-dispersive medium: 𝜔 − 𝜔0 ∆𝑊 𝑊0 = − ⇔ 𝜔 = 𝜔0 [ ] 𝜔 𝑊0 𝑊0 − ∆𝑊 no electric field in propagation direction no magnetic field in propagation direction sin[ 𝑘𝑑 cos 𝜃 ] = 0 → 𝑘𝑑 cos 𝜃 = 𝑛 𝜋 𝑘𝑥 = 𝑘 cos 𝜃 → 𝑘𝑥𝑛 = 𝑛 𝜋⁄𝑑 , 𝑛 ∈ {1,2, … } As 𝑘 2 = 𝑘𝑥2 + 𝑘𝑧2 , we can find the propagation constant 𝑘𝑧𝑛 = √𝑘 2 − 𝑘𝑥2𝑛 = √𝑘 2 − 𝑛2 [𝜋⁄𝑑 ]2 , 𝑛 ∈ { 1,2, … } 𝑛 = 0 : zero-field (trivial solution for TE-modes) 𝑛𝜋 𝑑 > 𝑘 : exponential decay just like evanescent waves → High-pass filter 9 Cut-off frequency 𝜔𝑐 = Watch out: 𝑇𝐸00 does not exist! Therefore, 𝑛 = 0 = 𝑚 is not a valid solution; the lowest frequency modes are hence 𝑇𝐸01 𝑎𝑛𝑑 𝑇𝐸10 . Hollow Metal Waveguides 𝑛𝜋𝑐 , 𝑑 𝑛(𝜔𝑐 ) 𝑛 ∈ { 1, 2, … } Propagation constant / longitudinal wavenumber Below the cut-off frequency, waves cannot propagate 𝜔 > 𝜔𝑐 ∶ 2 𝜔𝑛𝑚 𝑛2 𝜋 2 𝑚2 𝜋 2 𝑘𝑧 = √𝑘 2 − 𝑘𝑡2 = √ 2 𝑛2 (𝜔𝑛𝑚 ) − [ 2 + 2 ] 𝑐 𝐿𝑥 𝐿𝑦 𝑃ℎ𝑎𝑠𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣𝑝ℎ = 𝜔⁄𝑘𝑧𝑛 𝐺𝑟𝑜𝑢𝑝 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣𝑔 = 𝑑𝜔 ⁄𝑑𝑘𝑧𝑛 TM-Modes: magnetic field parallel to surfaces of plates Ansatz: plane wave propag. at angle 𝜃 to surface normal 𝐻1 (𝑥, 𝑦, 𝑧) = 𝐻0 𝑛𝑦 𝑒 [ −𝑖𝑘𝑥 cos 𝜃+𝑖𝑘𝑧 sin 𝜃 ] Coming from the upper plate, it gets reflected at the bottom: 𝐻2 (𝑥, 𝑦, 𝑧) = 𝐻0 𝑛𝑦 𝑒 [ 𝑖𝑘𝑥 cos 𝜃+𝑖𝑘𝑧 sin 𝜃 ] As there is no zero mode, there is always a cut-off. Ansatz: 𝐸(𝑥, 𝑦, 𝑧) = 𝐸 𝑥𝑦 (𝑥, 𝑦) ∗ 𝑒 𝑖𝑘𝑧 𝑧 𝐸 = 𝐸𝑡𝑟𝑎𝑛𝑠𝑣 + 𝐸𝑙𝑜𝑛𝑔 = 𝐸 𝑥 𝑛𝑧 + ( 𝐸 ∗ 𝑛𝑧 ) 𝑛𝑧 The transverse field components can be calculated using the longitudinal field components: 𝐻(𝑥, 𝑦, 𝑧) = 𝐻1 + 𝐻2 = 2 𝐻0 𝑛𝑦 𝑒 𝑖 𝑘 𝑧 𝑠𝑖𝑛𝜃 cos(𝑘 𝑥 cos 𝜃) 𝑥𝑦 𝐸𝑧 (𝑥, 𝑦) = 𝐸0𝑧 sin [ 𝑛𝜋 𝑚𝜋 𝑥 ] sin [ 𝑦 ], 𝐿𝑥 𝐿𝑦 𝑛, 𝑚 ∈ { 1, … } Optical Waveguides For very high / optical frequencies (200-800 THz) , metal waveguides become lossy. Boundary condition at top interface 𝑧 = 𝑑 leads to 𝑛 ∈ { 0, 1, 2, … } 𝑘𝑧𝑛 = √𝑘 2 − 𝑛2 [𝜋⁄𝑑 ]2 , 𝑛 ∈ { 0, 1, 2, … } TEM / 𝑇𝑀00 – Mode In contrast to the TE-modes, there exists a mode for 𝑛 = 0. This mode does NOT have a cut-off frequency like all other TM- and TE-modes. For 𝑇𝑀00 ∶ 𝑥𝑦 𝐻𝑧 = 0 𝑛 = 0 𝑜𝑟 𝑚 = 0 lead to zero-field solutions and are forbidden. The lowest frequency mode is 𝑇𝑀11 . Superposition of the fields: 𝑘𝑑 cos 𝜃 = 𝑛𝜋 , TM-Modes : 𝑘𝑧 = 𝑘 This is a transverse electric field: neither the electric nor the magnetic field show in the direction of propagation. TE-Modes: 𝑥𝑦 𝐻𝑧 (𝑐, 𝑥) 𝑥𝑦 𝐸𝑧 = 0 = 𝐻0𝑧 cos [ 𝑛𝜋 𝐿𝑥 𝑥] cos [ For Total Internal Reflection (TIR), we require: 𝑚𝜋 𝐿𝑦 𝑦 ] , 𝑛, 𝑚 ∈ {0,1, … } Transverse wavenumber: 𝑛2 𝑚2 𝑘𝑡2 = [ 𝑘𝑥2 + 𝑘𝑦2 ] = 𝜋 2 [ 2 + 2 ] , 𝐿𝑥 𝐿𝑦 𝑛, 𝑚 ∈ {0, 1, … } Frequency of the 𝑇𝐸𝑛𝑚 modes: 𝜔𝑛𝑚 = 𝜋𝑐 𝑛2 𝑚2 √ 2+ 2 , 𝑛(𝜔𝑛𝑚 ) 𝐿𝑥 𝐿𝑦 𝑛, 𝑚 ∈ {0,1, … } 1. 𝑛1 > 𝑛2 ( core optically denser material) 2. Angle to surface normal 𝜃 > 𝜃𝑐 = arctan [ 𝑛2 𝑛1 ] In contrary to metal waveguides, we have evanescent fields that stretch out into the surrounding medium with the lower index of refraction 𝑛2 . These fields ensure that no energy is radiated away from the waveguide. 10 Work per time unit 10. Various 𝑑𝑊 = 𝐹 ∗ 𝑣 = [ 𝑞 𝐸 + 𝑞 (𝑣 𝑥 𝐵) ] ∗ 𝑣 = 𝑞 𝐸 𝑣 𝑑𝑡 𝑃= Gradient operator ( grad ) Only the electric field can work, magnetic fields doesn’t TM-Modes Fraunhofer approximation 𝑘1 = [ 𝑘𝑥1 , 0, 𝑘𝑧 ] 𝑘2 = [ 𝑘𝑥2 , 0, 𝑘𝑧 ] Medium 1: Medium 2: 𝑒 𝑖𝑘|𝒓−𝒓′ | ≈𝑒 𝑖𝑘(𝑟 − 𝒓∗𝒓′ ) 𝑟 1 1 = |𝒓 − 𝒓′ | 𝑟 , Divergence operator ( div ) Small angle approximation sin(𝜃) ≈ 𝜃 , cos(𝜃) ≈ 1 , tan(𝜃) ≈ 𝜃 Force of an impulse Rotation operator ( rot ) 𝑃𝑘𝑟𝑎𝑓𝑡 = 𝐹 ∗ 𝑣 , 𝐹 =𝑚∗ 𝑑𝑣 𝑑𝑝 = , 𝑑𝑡 𝑑𝑡 𝑝 =𝑚∗𝑣 Energies 𝐸𝑝𝑜𝑡 = 𝑞 ∗ 𝛷(𝑥) 1 𝐸𝑘𝑖𝑛 = 𝑚 𝑣 2 2 𝐸𝑘𝑖𝑛 + 𝐸𝑝𝑜𝑡 = 𝑐𝑜𝑛𝑠𝑡. Laplacian operator ( ∆ ) Taylor In order to be evanescent outside the waveguide, we require 𝑘𝑧 > 𝑘2 . However, to propagate inside, we need 𝑘2 < 𝑘𝑧 < 𝑘1 𝑓(𝑥) ≈ 𝑓(𝑥0 ) + Rules 𝑑𝑓(𝑥0 ) 1 𝑑2 𝑓(𝑥0 ) (𝑥 − 𝑥0 ) + (𝑥 − 𝑥0 )2 + ⋯ 𝑑𝑥 2! 𝑑𝑥 2 Series ∞ If we solve those fields, we receive 𝑝 𝑥 𝑒 = ∑ 𝑖=0 𝑝 1 + 𝑟𝑎𝑏 (𝑘𝑧 ) 𝑟𝑏𝑐 (𝑘𝑧 ) 𝑒 2𝑖𝑘𝑥1 𝑑 = 0 𝑝 𝑝 Here, 𝑟𝑎𝑏 𝑎𝑛𝑑 𝑟𝑏𝑐 are the Fresnel coefficients for p-pol. 𝑥𝑘 𝑥2 𝑥3 = 1+𝑥+ + +⋯ 𝑘! 𝑥 6 , Integrals Gauss theorem: Area ↔ Volume ∞ 𝑥 sin(2𝑥) ∫ sin 𝑥 𝑑𝑥 = − , 2 4 2 ∫ 𝐹(𝑟, 𝑡) ∗ 𝑛 𝑑𝑎 = ∫ ∇ ∗ 𝐹(𝑟, 𝑡) 𝑑𝑉 TE-Modes 𝑑𝑉 Here, we receive a similar equation: 𝑠 𝑠 (𝑘𝑧 ) 𝑟𝑏𝑐 (𝑘𝑧 ) 𝑒 2𝑖𝑘𝑥1 𝑑 = 0 1 + 𝑟𝑎𝑏 −∞ 𝑉 Stokes theorem: Line ↔ Area ∫ 𝐹(𝑟, 𝑡) 𝑑𝑠 = ∫[ ∇ 𝑥 𝐹(𝑟, 𝑡) ] ∗ 𝑛 𝑑𝑎 𝑑𝐴 ∫ 𝜋 ∫ sin3 𝑥 𝑑𝑥 = 0 4 , 3 sin(𝑥) 𝑑𝑥 = 𝜋 𝑥 𝜋 ∫ sin4 𝑥 𝑑𝑥 = 0 3𝜋 8 𝐴 11 Additionstheoreme 11. Tabellen Reihenentwicklungen 𝜋 𝑖 = √1 = 𝑒 𝑖 2 tan′ 𝑥 = 1 + tan2 𝑥 sin2 𝑥 + cos 2 𝑥 = 1 cosh2 𝑥 − sinh2 𝑥 = 1 Doppelter und halber Winkel cos(𝑧) = cos(𝑥) cosh(𝑦) − 𝑖 sin(𝑥) sinh(𝑦) sin(𝑧) = sin(𝑥) cosh(𝑦) + 𝑖 cos(𝑥) sinh(𝑦) Umformung einer Summe in ein Produkt Summe der ersten n-Zahlen Umformung eines Produkts in eine Summe Geometrische Reihe 12 Fourier-Korrespondenzen Fourier-Tabelle Fourier-Funktionen Eigenschaften der Fourier-Transformation Partialbruchzerlegung (PBZ) Reelle Nullstellen n-ter Ordnung: 𝐴1 𝐴2 𝐴𝑛 + + …+ (𝑥 − 𝑎𝑘 ) (𝑥 − 𝑎𝑘 )2 (𝑥 − 𝑎𝑘 )𝑛 Paar komplexer Nullstellen n-ter Ordnung: 𝐵1 𝑥 + 𝐶1 𝐵𝑛 𝑥 + 𝐶𝑛 + …+ + [(𝑥 − 𝑎𝑘 )(𝑥 − 𝑎𝑘 )]𝑛 (𝑥 − 𝑎𝑘 )(𝑥 − 𝑎𝑘 ) (𝑥 − 𝑎𝑘 )(𝑥 − 𝑎𝑘 ) = (𝑥 − 𝑅𝑒)2 + 𝐼𝑚2 13 Ableitungen Stammfunktionen Standard-Substitutionen 14 15 16

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