5 Radiation (Chapte 11) 5.1 Electic dipole adiation Oscillating dipole system Suppose we have two small sphees sepaated by a distance s. The chage on one sphee changes with time and is descibed by q(t) =q 0 cos ωt and on the othe one q(t) = q 0 cos ωt This can be achieved by sepaating chages fom one sphee to the othe though a wie. The sum of chages on two sphees is zeo. We shall calculate the adiated fields. We shall conside long wavelength adiation at a lage distance fom the dipole. The assumptions ae: I. s fa field II. s c/ω =1/k =2πλ long wavelength III. À 2πλ fa field O s λ This chage system is actually an oscillating dipole: p = q 0 s cos ωt = p 0 cos ωt whee p 0 = q 0 s 1/note16
Scala potential Fo simplicity, we shall use the exponential fieldpesentationinthespheicalcoodinate system. φ(, θ,t)= 1 " q0 exp(iωt + ) q # 0 exp(iωt ), 4πε 0 R + R whee t ± = t R ± /c, and R + and R aedistancefom~ to the chages q + (t) andq (t), espectively. q R + = 2 +(s/2) 2 s cos θ q = 1+(s/2) 2 (s/)cosθ µ ' 1 s 2 cos θ assumption I : s/ 1 Similaly µ R = 1+ s 2 cos θ So ωt ± = ω t µ 1 s c 2 cos θ = (ωt k) ± ks whee we used ω = k/c. The scala potential is then φ(, θ,t) = q 0 exp[i(ωt k)] exp ³ i ks cos θ 2 4πε 0 R + Note also So 1 R ± = φ(, θ,t) = q 0 exp[i(ωt k)] 1 exp ³ i ks 2 R ³ 1 s 2 cos θ ' 1 cos θ " µ 1+ s 2 cos θ µ 1 s 2 cos θ exp µ 1 ± s 2 cos θ Ã Ã exp i ks! i ks!# 2/note16
= q "à à 0 exp[i(ωt k)] exp i ks + s à à 2 cos θ exp i ks +exp = q " Ã! 0 exp[i(ωt k)] ks 2i sin ' q 0 exp[i(ωt k)] iks cos θ + s cos θ! à exp i ks!!! à i ks!!# whee we used the assumption I,sk 1, in the last appoximation. Finally, we obtain φ(, θ,t)= q 0s cos θ exp[i(ωt k)] (1 + ik) 2 The actual potential is the eal pat of the above + s cos θ cos à ks φ(, θ,t) = Re φ = p 0 cos θ Re{[cos(k ωt) i sin(k ωt)][1 + ik]} 2 = p 0 cos θ [cos(k ωt)+k sin(k ωt)], 2 whee p 0 = q 0 s. Thee ae two pats in the scala potential epesenting adiation in two diffeent egions: (A) Fo the static case, ω 0andk = ω/c = 0 and the potential in both zones educes to the familia dipole potential φ = p 0 cos θ = ~p 0 ~ 2 3 (B) Radiation zone (assumption I), k À 1(o À λ),!# note (C) Nea field zone, k 1 note φ ' p 0 cos θ k sin(k ωt) φ 1 adiation zone φ ' p 0 cos θ cos(k ωt) 2 φ 1 2 nea field zone 3/note16
Vecto potential While the chage is diven back and foth between the sphees, thee is a cuent flowing though the connecting wie dq I (t) = dt ~e z = ωq 0 sin ωt~e z in the egion s/2 <z<s/2. The etaded vecto potential is in the ~e z diection A z = µ Z s/2 0 I(t ) 4π s/2 R dz = µ Z s/2 µ 0 1 4π s/2 R ωq 0 sin ω t R dz c Since the integal contains the limits ±s/2 whichisthefist ode in s, ther in the integand needs to be expanded to keep only zeo-th ode of s. Since z s/2 So ~A = µ Z s/2 0 1 4π s/2 = µ 0ωq 0 s R = 2 + z 2 2z cos θ ' ωq 0 sin ω sin µ ω µ t c t c dz ~e z ~e z = µ 0ωp 0 sin(k ωt) ~e z In ode to calculate ~ E and ~ B in the spheical coodinate system, the unit vecto ~e z should be decomposed ~e z =cosθ~e sin θ~e θ The components of the vecto potential ae 4/note16
A = µ 0ωp 0 cos θ sin(k ωt) A θ = µ 0ωp 0 sin θ sin(k ωt) A φ = 0 Summay of the potentials in the adiation zone (k À 1): φ ' p 0 cos θ k sin(k ωt) Electic and magnetic fields The fields can be calculated based on ~A ' µ 0ωp 0 sin(k ωt)(cos θ~e sin θ~e θ ) ~E = φ ~ A t ~B = ~ A 5/note16
Let s now calculate deivatives of the scala and potential vectos: ~ A t = µ 0ω 2 p 0 cos(k ωt)(cos θ~e sin θ~e θ ) = ω2 p 0 1 c cos(k ωt)(cos θ~e 2 sin θ~e θ ), whee c =1/ µ 0 ε 0 has been used. φ = φ ~e + 1 φ θ ~e θ " = p # 0 cos θ k sin(k ωt)+p 0 cos θ 2 k2 cos(k ωt) ~e = p 0 sin θ k sin(k ωt)~e 2 θ kp 0 {[ cos θ sin(k ωt)+k cos θ cos(k ωt)]~e 2 sin θ sin(k ωt)~e θ } ' k2 p 0 cos θ cos(k ωt)~e We have used the appoximation k À 1 fo the adiation zone. Since c = ω/k So φ = ω2 p 0 1 c cos θ cos(k ωt)~e 2 ~E = φ A ~ t = ω2 p 0 1 c cos θ cos(k ωt)~e 2 + ω2 p 0 1 c cos(k ωt)(cos θ~e 2 sin θ~e θ ) = ω2 p 0 1 c sin θ cos(k ωt)~e 2 θ The magnetic field ~B = A ~ = ~e 2 sin θ ~e θ sin θ θ ~e φ φ A A θ sin θa φ {z } =0 6/note16
" 1 (Aθ ) = ~e φ " = ~e φ µ 0 ωp 0 = ~e φ µ 0 ωp 0 = ~e φ µ 0 ωp 0 2 A # θ µ 1 cos θ sin(k ωt) # ( sin θ sin(k ωt)) θ k sin θ cos(k ωt)+ 1 sin θ sin(k ωt) [ k sin θ cos(k ωt)+sinθsin(k ωt)] ' µ 0ωp 0 k sin θ cos(k ωt)~e φ (k À 1) = µ 0ω 2 p 0 4πc sin θ cos(k ωt)~e φ As we can see, ~ E and ~ B ae in phase. The vectos ~ E, ~ B and ~ k = k~e ae pependicula to each othe. ~ k is the wave popagation diection. Radiation powe The Poynting vecto ~S = 1 E ~ B ~ µ 0 = 1 " # ω2 p 0 1 µ 0 c sin θ cos(k ωt)~e 2 θ 7/note16
" µ 0ω 2 # p 0 4πc sin θ cos(k ωt)~e φ = 1 " ω 2 2 p 0 sin θ cos(k ωt)# ~e ε 0 c 3 4π (1) The Poynting vecto (o powe popagation diection) is adially outwad in the ~e diection, i.e., ~ k diection. The time aveaged Poynting vecto magnitude is <S>= ω4 p 2 0 sin 2 θ 32π 2 ε 0 c 3 2 The aveage powe though a spheical suface can be integated <P > = = = Z <S>2πsin θdθ ω 4 p 2 Z π 0 sin 2 θ 2π 2 sin θdθ 32π 2 ε 0 c 3 0 2 ω 4 p 2 Z π 0 sin 3 θdθ 16πε 0 c 3 0 8/note16
The integal So Z π 0 sin 3 θdθ = = Discussions Z π Z π sin 2 θd cos θ = (cos 2 θ 1)d cos θ 0 0 π 1 3 cos3 θ cos θ = 4 3 <P >= 1 ω 4 p 2 0 4πε 0 3c 3 0 1) The fields ae popotional to 1/ 2) The Poynting vecto is popotional to 1/ 2 The powe going though any spheical suface is a constant Z Z <P >= <S>2πsin θdθ = <S> 2 dω = constant, consistent with the law of enegy consevation. 3) The angula powe distibution is given by f(θ) = d< P > dω = ω4 p 2 0 32π 2 ε 0 c 3 sin2 θ The maximum adiation is at θ = π/2 and minimum at θ =0, π 9/note16
Why is the sky blue duing the day and ed duing sunset? These questions can be answeed by the electic dipole adiation. Please ead Example 11.1 on Page 449. Phys 463, E & M III, C. Xiao 10 / note16