## Assignment 4, solutions

Problem 1:

The z = 0 plane is perfectly conducting.  A point electric dipole oscillating in time as p(t) = p0(z/z)exp(-iωt) is placed at a height h above the conducting plane at the position (0,0,h).
(a)  Calculate the angular distribution of the radiated power at large distances.
(b)  Calculate the total power radiated in the long wave length limit.

Solution:

 Concepts: The method of images, dipole radiation Reasoning: The near field is the instantaneous Coulomb field.  We can therefore model the field due to the surface charge distribution in the region z > 0 as identical to that of an image dipole p(t) = p0(z/z)exp(-iωt) at position (0,0 -h). Details of the calculation: (a)  For a dipole p(t) = p0(z/z)exp(-iωt) at the origin the radiation field at  position r is  ER(r,t) = -(1/(4πε0c2r))ω2p0 exp(-iω(t - r/c))sinθ (θ/θ) = -(1/(4πε0c2r))ω2p0 exp(i(kr - ωt))sinθ (θ/θ). BR(r,t) = -(1/(4πε0c3r))ω2p0 exp(i(kr-ωt))sinθ (φ/φ). For the dipoles at (0,0,h) and (0,0 -h) we therefore have BR(r,t) = -(1/(4πε0c3))ω2p0 exp(-iωt)) (φ/φ) [sinθ1 exp(k∙(r - hz/z))/|r - hz/z| + sinθ2 exp(k∙(r + hz/z))/|r + hz/z|]. At large distances (r >> h) we have 1/|r ± hz/z| ≈ 1/r, sinθ1 = sinθ2 = sinθ. We only have to be careful with the phase.BR(r,t) = -(1/(4πε0c3r))ω2p0 sinθ exp(i(kr - ωt)) (φ/φ) [exp(ik∙hz/z)) + exp(-ikhz/z))].BR(r,t) = -(1/(2πε0c3r))ω2p0 sinθ exp(i(kr - ωt)) (φ/φ) cos(khcosθ) = B (φ/φ).ER(r,t) = cB (θ/θ). S = (1/μ0)E×B , = (r/r)(1/(8π2ε0c3r2))ω4p02 sin2θ cos2(khcosθ). is proportional to sin2θ cos2(khcosθ).In the long wavelength limit kh = (2π/λ)h << 1 and = (r/r)(1/(8π2ε0c3r2))ω4p02 sin2θ.For a single dipole p(t) = p0(z/z)exp(-iωt) at the origin we have = (r/r)(1/(32π2ε0c3r2))ω4p02 sin2θ.The total power radiated by a single dipole is

= [ω4p02/(12πc3ε0)] .Here increases by a factor of 4, but we only integrating in the z > 0 plane.  We have 2 times the power radiate by a single dipole.

= <∫S∙dA> = ω4p02/(6πε0c3) proportional to ω4.

Problem 2:

A non-relativistic positron of charge qe and velocity v1 (v1 << c) impinges head-on on a fixed nucleus of charge Zqe.  The positron which is coming from far away (∞), is decelerated until it comes to rest and then accelerated again in the opposite direction until it reaches a terminal velocity v2.  Taking radiation loss into account (but assuming it is small), find v2 as a function of v1 and Z.

Solution:

 Concepts: Motion in a central field, energy conservation, the Larmor formula Reasoning: The positron moves in a central field.  It has no angular momentum about the position of the nucleus (at the origin).   If we neglect radiation losses, we can find its velocity and acceleration as a function of distance from energy conservation.  Using the Larmor formula, we then can find the energy lost in a small time interval dt or a small distance dr.  Integrating these losses over the path, we find E2 and v2.  This works if the energy radiated is much less than the characteristic energy of the system. Details of the calculation: E = ½mv12 = ½mv2 + Ze2/r.  v(r) = (v12 - 2Ze2/(mr))1/2, with e2 = qe2/(4π2ε0). dv/dt = (dv/dr)v = Ze2/(mr2). The Larmor formula gives the power radiated by an accelerating charge. P = 2e2a2/(3c3) = [2e2/(3c3)](Ze2/(mr2))2 = -dE/dt = -(dE/dr)v. dt = dr/v. ΔE = -∫Pdt = -∫[2e2/(3c3)](Ze2/(mr2))2dt = -2∫rmin∞[2e2/(3c3)](Ze2/(mr2))2dr/v = -[4e6Z2/(3c3m2)]∫rmin∞r-4dr/v(r).v(r) = v1(1 - 2Ze2/(mrv12))1/2.  v(rmin) = 0.rmin = 2Ze2/(mv12).ΔE = -[4e6Z2/(3c3m2v1)]∫rmin∞r-4dr/(1 - rmin/r)1/2.Let x = rmin/r,  dx = -rmindr/r2.ΔE = -[4e6Z2/(3c3m2v1rmin3)]∫01x2dx/(1 - x)1/2.From an integral table:∫01x2dx/(1 - x)1/2 = [2(8 + 4x + 3x2)/15](1 - x)1/2|01 = -16/15.ΔE = [4e6Z2/(3c3m2v1rmin3)]16/15 = (8/45)mv15/(Zc3).For a nonrelativistic v1 we have ΔE << E.ΔE/2 is the radiation loss during the approach.  The radiation loss on the way out is also ΔE/2.½mv22 = ½mv12 - ΔE,  v22 = v12 - (16/45)v15/(Zc3).v2 ≈ v1(1 - (8/45)v13/(Zc3)).

Problem 3:

Let E0 = E0 k.  The Abraham-Lorentz force equation for a damped, charged, oscillator driven by an electric field E0exp(-iωt) in the dipole approximation is
d2r'/dt2 + Γ dr'/dt - τ d3r'/dt3 + ω02 r' = (q/m)E0exp(-iωt),
where Γ, τ, and ω0 are constants, q is the charge and m is the mass of the oscillator.
Using this and the expression for the radiation electric field, Erad(r,t) = -(4πε0)-1[(q/(c2r'')]a(t - r''/c),  where r'' = r - r'(t - |r - r'|/c), show that the differential cross section for scattering of radiation of frequency ω and polarization n = (θ/θ)  is
dσ/dΩ = (e2/(mc2))2 (kn)24/((ω02 - ω2)2 + ω2Γt2)],
where e2 = q2/(4πε0) and Γt = Γ + ω2τ.

Solution:

 Concepts: The scattering cross section Reasoning: Power radiated into the solid angle dΩ = = * dσ. Details of the calculation: Solve the equation of motion. d2r'/dt2 + Γ dr'/dt - τ d3r'/dt3 + ω02 r' = (q/m)E0exp(-iωt) k. Let r' = r'k = r0'exp(-iωt)k.  Then dr'/dt = -iωr0'exp(-iωt),  d2r'/dt2 = -ω2r0'exp(-iωt),  d3r'/dt3 = iω3r0'exp(-iωt). (-ω2 - iωΓ - iω3τ + ω02)r0' = (q/m)E0.  r' = (q/m)E0exp(-iωt/(-ω2 - iωΓt + ω02), where Γt = Γ + ω2τ. a = -ω2(q/m)E0k exp(-iωt)/(-ω2 - iωΓt + ω02). Calculate S. For convenience, place the oscillator at the origin.  Let r = r(θ, φ), then r'' = r. Erad(r,t) = -(4πε0)-1[(q/(c2)]a⊥(t - r/c) = (4πε0)-1[(q/(c2r)]ω2(q/m)E0 sinθ (θ/θ) exp(-iωtr)/(-ω2 - iωΓt + ω02) = (e2/(mc2)) (sinθ/r) (θ/θ) ω2E0 exp(-iωtr)/(ω02 - ω2 - iωΓt), with tr = t - r/c. S((θ, φ) = (r/r)(Re(Erad))2/(μ0c). ω2 exp(-iωtr)/(ω02 - ω2 - iωΓt) = |ω2/(ω02 - ω2 - iωΓt)|exp(-iωtr)exp(iξ) = [ω4/((ω02 - ω2)2 - ω2Γt2)]½exp(-i(ωtr - ξ)) S(θ, φ) = (r/r )cos2(ωtr - ξ) (e4/(m2c4)) (sin2θ/r2) (μ0c)-1E02ω4/((ω02 - ω2)2 - ω2Γt2). = (r/r )½(e4/(m2c4)) (sin2θ/r2) (μ0c)-1E02ω4/((ω02 - ω2)2 - ω2Γt2). = r2 dΩ = power radiated into the solid angle dΩ. = * dσ. dσ/dΩ =  r2/( ½(μ0c)-1E02) = ½(e4/(m2c4)) sin2θ ω4/((ω02 - ω2)2 - ω2Γt2). k∙n = k∙(θ/θ) = -sinθ,  (k∙n)2 = sin2θ.

Problem 4:

(a)  Consider the radiation from an oscillating electric dipole.  Find E, B and the Poynting vector at a large distance from the dipole and integrate your result to find the total radiation.
(b)  From the symmetry of Maxwell’s equations and the form of the electric and magnetic field of an oscillating electric dipole, deduce the field of an oscillating magnetic dipole.
The near field must resemble the field of a dipole formed by a small current loop of radius a (a << c/ω), and current I = I0cos(ωt).

Solution:

 Concepts: The radiation field of a point charge moving non-relativistically, the Larmor formula Reasoning: A dipole can be viewed as a positive and a negative point charge oscillating 180o out of phase.  Oscillating charges are accelerating charges. Details of the calculation: (a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p0cos(ωt) is ER(r,t) = -(1/(4πε0c2r))(d2p⊥(t - r/c)/dt2).  (SI units) Let p0 = p0(z/z), then ER(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t - r/c)) sinθ (θ/θ). BR(r,t) = (r/(rc))×ER(r,t) = -[1/(4πε0c3r)]ω2p0 cos(ω(t - r/c)) sinθ (φ/φ). S(r,t) = [1/(4πε0)2][1/(c5r2μ0)] ω4p02 cos2(ω(t - r/c)) sin2θ (r/r).  = [1/(16π2ε0r2c3)]ω4p02 cos2(ω(t - r/c)) sin2θ (r/r).  The intensity varies as sin2θ.

= <∫S∙dA> = ω4p02/(12πε0c3), proportional to ω4. (b)  Let p at the origin be the source of E, m at the origin be the source of B. For the near fields we have E(r) = (1/(4πε0))(3(p∙r)r/r5 - p/r3), B(r) = (μ0/(4π))(3(m∙r)r/r5 - m/r3). Maxwell's equations in the regions outside the sources are ∇∙E = 0,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = (1/c2)∂E/∂t. Let E' = -c2B,  B' = E,  i.e. E' is just a scaled B, and B' is just a scaled E. Then ∇∙E' = ∇∙B' = 0, ∇×E' = -∂B'/∂t, ∇×B' = (1/c2)∂E'/∂t. E' and B' satisfy the same equations as E and B, so with the same boundary conditions they have the same solutions. We know the far fields of an oscillating electric dipole. For a dipole at the origin ER(r,t) = -(1/(4πε0c2r))(d2p⊥(t - r/c)/dt2),  BR(r,t) = r×ER(r,t)/(rc). E'R(r,t) has the same form, except we replace p/(4πε0) by -c2mμ0/(4π) = -m/(4πε0). Therefore E'R(r,t) = (1/(4πε0c2r))(d2m⊥(t - r/c)/dt2),  B'R(r,t) = r×E'R(r,t)/(rc). The far fields of an oscillating magnetic dipole therefore are BR(r,t) = (1/(4πε0c4r))(d2m⊥(t - r''/c)/dt2),  ER(r,t) = -cr×BR(r,t)/r.

Problem 5:

The vector potential A(r,t) of an oscillating dipole p at the origin is A(r,t) = -ik p exp(i(kr - ωt'))/(4πε0rc).
Note:  complex notation, the real part matters.
(a)  Let p = p k.  Use this vector potential to calculate the magnetic field.
(b)  Use Maxwell's equations to calculate the accompanying electric field.
(c)  Find the fields in the radiation zone.

Solution:

 Concepts: Maxwell's equations, radiation fields Reasoning: The radiation field decreases as 1/r. Details of the calculation: (a)  B = ∇×A.  A = A k = A[(r/r) cosθ  - (θ/θ) sinθ]. B = (1/r)[∂(rAθ)/∂r - ∂Ar/∂θ] (φ/φ) = B (φ/φ). B = -[ikp/(4πε0rc)] [-iksinθ + sinθ/r] exp(i(kr - ωt')). Taking the real part we have B = -[k2p sinθ/(4πε0rc)] [cos(kr - ωt') - sin(kr - ωt')/(kr)]. (b)  ∇×B = (1/c2)∂E/∂t = -(iω/c2)E,  since E is proportional to exp(-iωt). ∇×B = (r/r) (1/(rsinθ))∂(Bsinθ)/∂θ - (θ/θ) (1/r))∂(rB)/∂r = -(r/r) [2k4p cosθ/(4πε0c)] [1/(kr)2 + i/(kr)3]exp(i(kr - ωt')) + (θ/θ) [k4p sinθ/(4πε0c)] [[1/(kr)2 + i/(kr)3 - i/(kr)]exp(i(kr - ωt'))]. Taking the real part we have Er = [2ck4p cosθ/(4πε0ω)][sin(kr - ωt')/(kr)2 + cos(kr - ωt')/(kr)3], Eθ = [ck4p sinθ/(4πε0ω)][sin(kr - ωt')/(kr)2 + cos(kr - ωt')(1/(kr)3 - 1/(kr)]. (c)  radiation zone:  E and B are proportional to 1/r. ER = ER (θ/θ), ER = -[ck4p sinθ/(4πε0ω)] cos(kr - ωt')/(kr) = -[ω2p sinθ/(4πε0c2)] cos(kr - ωt')/r, BR = BR (φ/φ), BR = -[k2p sinθ/(4πε0rc)] cos(kr - ωt') =  -[ω2p sinθ/(4πε0c3)] cos(kr - ωt')/r. BR(r,t) = r×ER(r,t)/(rc).