Electronic Science and Technology Rabindra Kishore Mishra Lecture 5- Hertzian Dipole EST-E-335: Radiation System 3 Credits Electronic Science and Technology Rabindra Kishore Mishra Outline ➢Recap ➢Time Harmonic Functions ➢Time Harmonic Lorenz Condition ➢Time Harmonic Helmholtz Equation ➢Helmholtz Equation: Solution ➢Green’s Function ➢Time Harmonic Fields ➢Dielectric Polarization Revisited ➢Hertzian Dipole ➢Field distribution • Comes from Maxell’s curl equations Electronic Science and Technology Rabindra Kishore Mishra Recap 1 ✓∇ × 𝑯 = 𝒋 + 𝜕𝑡𝑫 ✓∇ × 𝑬 = −𝜕𝑡𝑩 ➢∇ ∙ 𝑩 = 0 • 𝑩 = 𝜇𝑯 = 𝜵 × 𝑨 ➢∇ × 𝑬 = −𝜕𝑡𝑩 • 𝑬 + 𝜕𝑡𝑨 = −𝜵𝜑 ➢Potentials A and ϕ are not unique ➢For any scalar function f(r,t), the following gauge transformation leaves E and B invariant: • 𝜑 ′ = 𝜑 − 𝜕 𝑡𝑓 • 𝑨′ = 𝑨 + 𝜵𝑓 ➢𝑬′ = − 𝜕𝑡 𝑨′ − 𝜵𝜑’ ⇒ 𝑬′ = −𝜕𝑡 𝑨 + 𝜵𝑓 − 𝜵(𝜑 − 𝜕𝑡𝑓)=E ➢ −𝜇𝒋 𝜵2𝑨 1 2 − 2 𝜕𝑡 𝑨 𝑐 −𝜵 𝜵. 𝑨 + 1 𝜕𝜑 𝑐2 𝑡 Lorenz Gauge condition = =0 Electronic Science and Technology Rabindra Kishore Mishra Recap 2 ➢If A, ϕ did not satisfy the constraint 1 𝜵. 𝑨 + 2 𝜕𝑡𝜑 =0 •f be solution of the inhomogeneous wave equation 1 1 2 𝜵. 𝑨 + 2 𝜕𝑡𝜑 = ( 2 𝜕𝑡 𝑓 − 𝜵2𝑓) 𝑐 𝑐 𝑐 • Use Gauge transformation, ✓𝜑 ′ = 𝜑 − 𝜕𝑡𝑓; 𝑨′ = 𝑨 + 𝜵𝑓 1 𝜵. 𝑨 + 2 𝜕𝑡𝜑 ′ 𝑐 1 1 2 = 𝜵. 𝑨 + 2 𝜕𝑡𝜑 − ( 2 𝜕𝑡 𝑓 𝑐 𝑐 − 𝜵2𝑓) ′ 1 ➢ 𝜌 𝜀 = 𝜵. 𝑬 = 1 1 1 2 2 𝜕𝑡 𝜑 𝑐 − 𝜵2𝜑 1 ➢ 𝜵2𝜑 − 𝑐2 𝜕𝑡2 𝜑 = − 𝜀 𝜌; 𝜵2𝑨 − 𝑐2 𝜕𝑡2 𝑨 = −𝜇 𝒋 • 𝜵. 𝒋+ 𝜕𝑡𝜌=0 Continuity Equation ➢Fields (H and E): 𝑯 = 𝜇1 𝜵 × 𝑨; 𝑬 = − 𝜵𝜑 − 𝜕𝑡𝑨 = −𝜕𝑡𝑨 + 𝑐2 𝛻 𝛻 . 𝑨 𝑑𝑡 f(t)=[A(r,t), 𝜑(r,t)] Field Point Electronic Science and Technology Rabindra Kishore Mishra Recap 3 Volume V Source Point u(r’,t’)=[J(r’,t’) ,𝜌(r’,t’)] d3r’ ➢𝑨 𝒓, 𝑡 = 𝑅 ′ 𝒓 ,𝑡− 𝑐 𝜇𝒋 1 4𝜋 𝑅 ′ 3 𝑑𝒓 ➢𝝋 𝒓, 𝑡 = 𝜌 1 4𝜋𝜀 𝑅 ′ 𝒓 ,𝑡− 𝑐 𝑅 𝑑3𝒓′ ➢Function W(r, t) is harmonic in time • It contains only one frequency: W(r, t)= W(r)ejwt Electronic Science and Technology Rabindra Kishore Mishra Time Harmonic Functions ➢𝑨 𝒓, 𝑡 = 𝑅 ′ 𝒓 ,𝑡− 𝑐 𝜇𝒋 1 4𝜋 𝑅 𝑑3𝒓′ 𝑅 𝑗𝜔(𝑡− ) 𝑐 𝒓′ 𝑒 1 𝜇𝒋 𝑨 𝒓, 𝑡 = 𝑨 𝒓 = න 𝑑3𝒓′ 4𝜋 𝑅 𝑗𝜔𝑡 ′ 𝑒 𝑗𝑘𝑅 𝑒 𝜇 𝒋 𝒓 𝜔 𝑗𝜔𝑡 ′ 3 𝑨 𝒓 𝑒 = න 𝑑 𝒓 ,𝑘 = 4𝜋 𝑅 𝑐 𝑒 𝑗𝜔𝑡 1 𝜇 𝒋 𝒓′ 𝑒 𝑗𝑘𝑅 3 ′ 𝑨 𝒓 = න 𝑑𝒓 4𝜋 𝑅 (Fourier Transform of j ?) ➢𝜵. 𝑨 𝒓, 𝑡 + Electronic Science and Technology Rabindra Kishore Mishra Time Harmonic Lorenz Condition • 𝜵. 𝑨 𝒓 𝑒 • 𝜵. 𝑨 + 𝑗𝜔𝑡 1 2 𝜕𝑡𝜑 𝑐 + 𝑗𝜔 2 𝜑 𝑐 𝒓, 𝑡 =0 1 2 𝜕𝑡 𝜑 𝑐 𝒓 𝑒 𝑗𝜔𝑡 = 0 = 0 ⇒ 𝜵. 𝑨 = −𝑗𝜔𝜀𝜇𝜑 ➢𝜵2𝑨 Electronic Science and Technology Rabindra Kishore Mishra Time Harmonic Helmholtz Equation − ➢𝜵2𝑨 𝒓 1 2 2 𝜕𝑡 𝑨 = −𝜇 𝒋 𝑐 𝜔2 + 2 𝑨 𝒓 = −𝜇 𝑐 𝒋 𝒓 ⟹ 𝜵2𝑨 𝒓 + 𝑘 2 𝑨 𝒓 = −𝜇 𝒋 𝒓 ➢𝜵2𝑨 𝒓 + 𝑘 2 𝑨 𝒓 = −𝜇 𝒋 𝒓 • Inhomogeneous wave equation Electronic Science and Technology Rabindra Kishore Mishra Helmholtz Equation: Solution ➢Solution Technique 1) Obtain Green’s function G 2) Convolve j with the G ➢Finding Green’s Function 1) Replace A with G in the L.H.S. 2) Replace the R.H.S. with delta function 3) Find solution of the corresponding differential equation ➢𝜵2𝐺 Electronic Science and Technology Rabindra Kishore Mishra Green’s Function 2 𝒓 + 𝑘 𝐺 𝒓 = −𝛿 3 𝒓 • It is homogeneous everywhere except at r=0 ✓𝜵2𝐺 𝒓 + 𝑘 2 𝐺 𝒓 = 0 • General form of the solution ✓𝐺 𝒓 = 𝑒𝑥 𝑝 −𝑗𝑘𝒓 4𝜋𝑟 − Is equivalent to the response of the potential to an impulse excitation ➢Solution for the Potential A • 𝑨 𝒓 = 𝒓 𝒋 𝜇 ′ 𝐺(𝒓 − 𝒓′)𝑑3𝒓′ ➢𝑯 = Electronic Science and Technology Rabindra Kishore Mishra Time Harmonic Fields 1 𝜵 𝜇 𝜵. 𝑨 = −𝑗𝜔𝜀𝜇𝜑 −𝜵. 𝑨 ⇒𝜑= 𝑗𝜔𝜀𝜇 ×𝑨 ➢𝑬 = −𝜵𝜑 − 𝜕𝑡𝑨 ➢𝑬 = ➢𝑬 = 𝜵(𝜵.𝑨) − 𝑗𝜔𝜀𝜇 1 (𝜵 𝑗𝜔𝜀𝜇 𝑗𝜔𝑨 𝜵. 𝑨 + 𝑘 2 𝑨) ➢Homework: Prove 𝑬 = 𝟏 𝑗𝜔𝜀𝜇 𝜵 × 𝜵 × 𝑨 in source-free region ➢Assumptions Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole 1) 2) 3) 4) 5) Point dipole Located at origin (i.e. r’=0) Oriented along z Time harmonic in nature (i.e. exp(jwt) time variation) Dipole moment p ✓Corresponding Polarization (i.e. dipole moment per unit volume) P(r)=p𝛿 3 𝒓 ➢polarization current and charge densities • 𝒋𝑝𝑜𝑙 = Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole 𝜕𝑷 𝜕𝑡 = 𝑷 𝑒𝑥𝑝(𝑗𝜔𝑡), 𝜌𝑝𝑜𝑙 = −𝜵. 𝑷 𝑒𝑥𝑝(𝑗𝜔𝑡) ✓𝒋(r)=𝑗𝜔𝒑𝛿 3 𝒓 , 𝜌(r)= − p.∇𝛿 3 𝒓 ➢Potentials • 𝑨 𝒓 = 𝜇 0 𝒋 𝒓′ 𝐺(𝒓 − 𝒓′)𝑑3𝒓′ = 𝜇0 න 𝑗𝜔𝒑𝛿 3 𝒓′ 𝐺 𝒓 − 𝒓′ 𝑑3𝒓′ = 𝑗𝜔𝜇0 𝒑 𝐺 𝒓 •𝜑 𝒓 = 1 𝜀0 =− 𝜌 𝒓′ 𝐺(𝒓 − 𝒓′)𝑑3𝒓′ 1 p.∇𝛿 3 𝜀0 1 𝜵.𝑨 𝒓′ 𝐺(𝒓 − 𝒓′)𝑑3𝒓′ = − p.∇ 𝐺 𝒓 = 𝑗𝜔𝜇 𝜀 𝜀0 0 0 ➢Fields (using 𝑘 = Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole 𝑯 𝜔 𝑐 1 = 𝜵×𝑨 𝜇0 1 = 𝜵 × 𝑗𝜔𝜇0 𝒑 𝜇0 = 𝜔 𝜇0 𝜀0 ) 𝐺 𝒓 = −𝑗𝜔𝒑 × 𝜵𝐺 𝒓 1 𝑬= 𝜵 𝒑 ∙ 𝜵 + 𝑘2𝒑 𝐺 𝒓 𝜀0 ➢𝑬 = Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole 1 𝑗𝜔𝜀0 1 𝑗𝜔(𝜵 𝑗𝜔𝜀0 𝜵×𝑯 = × 𝒑 × 𝜵𝐺 𝒓 ) 1 1 = (𝒑 𝜵 ∙ −𝑟Ƹ 𝑗𝑘 + 𝐺 𝒓 𝜀0 𝑟 1 −(𝒑 ∙ 𝜵)[−𝑟Ƹ 𝑗𝑘 + 𝐺 𝒓 ]) 𝑟 ➢𝑯 = −𝑗𝜔𝒑 × 𝜵𝐺 𝒓 Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole = 𝑗𝜔 𝑗𝑘 + ➢ 1 𝑬= 𝑗𝜔𝜀0 1 = 𝑗𝜔𝜀0 1 𝑟 (𝒑 × 𝑟)𝐺 Ƹ 𝒓 𝜵×𝑯 𝜵 × 𝑗𝜔 𝑗𝑘 + 1 𝑟 (𝒑 × 𝑟)𝐺 Ƹ 𝒓 Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole 1 ➢𝑬 = 𝜀0 1 = 𝜀0 1 𝑟𝜕 Ƹ 𝑟 × 𝑗𝑘 + (𝒑 × 𝑟 Ƹ −𝒑 1 3𝑟Ƹ 𝑟∙𝒑 𝑗𝑘 + 𝐺 𝑟 𝑟 𝑘2 + 𝑟)𝐺 Ƹ 𝒓 𝒓 𝑟Ƹ × (𝒑 × 𝑟)𝐺 Ƹ 𝒓 𝜀0 ➢Change of dipole location to r0 • P(r)=p𝛿 3 𝒓 − r0 ; 𝑹 = 𝒓 − r0 • 𝑟Ƹ → 𝑹 •𝐺 𝒓 →𝐺 𝑹 } Fields unaffected ➢k=0 • Static fields Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole ✓No radiation ➢Radiated fields • Terms decreasing as 1/r ➢Near-fields • non-radiating • drop faster than 1/r • important in new area of near-field scanning/ probing ➢𝑯𝑟𝑎𝑑 𝑟 = 𝑗𝜔𝑗𝑘 𝒑 × 𝑟Ƹ 𝐺 𝒓 Electronic Science and Technology Rabindra Kishore Mishra Hertzian Dipole ➢𝑬𝑟𝑎𝑑 𝑟 𝑘2 = 𝑟Ƹ 𝜂0 𝜀0 𝑘2 = 𝑟Ƹ × 𝜀0 𝑘2 = 𝑟Ƹ × 𝜀0 ×𝒑 𝑒𝑥 𝑝 −𝑗𝑘𝒓 4𝜋𝑟 𝒑 × 𝑟Ƹ 𝐺 𝒓 𝒑 × 𝑟Ƹ ➢𝜂0 𝑯𝑟𝑎𝑑 = 𝑟Ƹ × 𝑬𝑟𝑎𝑑 𝑒𝑥 𝑝 −𝑗𝑘𝒓 4𝜋𝑟