**Problem 1:**

The z = 0 plane is perfectly conducting. A point electric dipole
oscillating in time as **p**(t) = p_{0}(**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) = p_{0}(z/z)exp(-iωt)
at position (0,0 -h). | |

Details of the calculation: (a) For a dipole p(t) = p_{0}(z/z)exp(-iωt)
at the origin the radiation field at position r is
E _{R}(r,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}
exp(-iω(t - r/c))sinθ (θ/θ)
= -(1/(4πε _{0}c^{2}r))ω^{2}p_{0}
exp(i(kr - ωt))sinθ (θ/θ).B_{R}(r,t) = -(1/(4πε_{0}c^{3}r))ω^{2}p_{0}
exp(i(kr-ωt))sinθ (φ/φ).For the dipoles at (0,0,h) and (0,0 -h) we therefore have B_{R}(r,t)
= -(1/(4πε_{0}c^{3}))ω^{2}p_{0}
exp(-iωt)) (φ/φ) [sinθ_{1}
exp(k∙(r
- hz/z))/|r - hz/z| + sinθ_{2}
exp(k∙(r
+ hz/z))/|r + hz/z|].At large distances (r >> h) we have 1/| r
± hz/z|
≈ 1/r, sinθ_{1}
= sinθ_{2} = sinθ.We only have to be careful with the phase. B_{R}(r,t) = -(1/(4πε_{0}c^{3}r))ω^{2}p_{0}
sinθ exp(i(kr - ωt)) (φ/φ) [exp(ik∙hz/z))
+ exp(-ikhz/z))].B_{R}(r,t) = -(1/(2πε_{0}c^{3}r))ω^{2}p_{0}
sinθ exp(i(kr - ωt)) (φ/φ) cos(khcosθ)
= B (φ/φ).E_{R}(r,t) = cB (θ/θ).S = (1/μ_{0})E×B
, <S>
= (r/r)(1/(8π^{2}ε_{0}c^{3}r^{2}))ω^{4}p_{0}^{2}
sin^{2}θ cos^{2}(khcosθ).< S> is proportional to sin^{2}θ
cos^{2}(khcosθ).In the long wavelength limit kh = (2π/λ)h << 1 and < S>
= (r/r)(1/(8π^{2}ε_{0}c^{3}r^{2}))ω^{4}p_{0}^{2}
sin^{2}θ.For a single dipole p(t) = p_{0}(z/z)exp(-iωt)
at the origin we have < S>
= (r/r)(1/(32π^{2}ε_{0}c^{3}r^{2}))ω^{4}p_{0}^{2}
sin^{2}θ.The total power radiated by a single dipole is <P> = [ω ^{4}p_{0}^{2}/(12πc^{3}ε_{0})]
.Here < S>
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.<P> = <∫ S∙dA>
=
ω^{4}p_{0}^{2}/(6πε_{0}c^{3})
proportional to ω^{4}. |

**Problem 2:**

A non-relativistic positron of charge q_{e} and velocity **v**_{1}
(v_{1 }<< c) impinges head-on on a fixed nucleus of charge Zq_{e}. 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 **v**_{2}. Taking radiation loss into account (but
assuming it is small), find v_{2} as a function of v_{1} 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 E _{2}
and v_{2}. This works if the energy radiated is much less than
the characteristic energy of the system. | |

Details of the calculation: E = ½mv _{1}^{2} = ½mv^{2} + Ze^{2}/r.
v(r) = (v_{1}^{2} - 2Ze^{2}/(mr))^{1/2}, with
e^{2} = q_{e}^{2}/(4π^{2}ε_{0}).dv/dt = (dv/dr)v = Ze ^{2}/(mr^{2}).The Larmor formula gives the power radiated by an accelerating charge. P = 2e ^{2}a^{2}/(3c^{3}) = [2e^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}
= -dE/dt = -(dE/dr)v.dt = dr/v. ΔE = -∫Pdt = -∫[2e ^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}dt
= -2∫_{rmin}^{∞}[2e^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}dr/v= -[4e ^{6}Z^{2}/(3c^{3}m^{2})]∫_{rmin}^{∞}r^{-4}dr/v(r).v(r) = v _{1}(1 - 2Ze^{2}/(mrv_{1}^{2}))^{1/2}.
v(r_{min}) = 0.r _{min} = 2Ze^{2}/(mv_{1}^{2}).ΔE = -[4e ^{6}Z^{2}/(3c^{3}m^{2}v_{1})]∫_{rmin}^{∞}r^{-4}dr/(1
- r_{min}/r)^{1/2}.Let x = r _{min}/r, dx = -r_{min}dr/r^{2}.ΔE = -[4e ^{6}Z^{2}/(3c^{3}m^{2}v_{1}r_{min}^{3})]∫_{0}^{1}x^{2}dx/(1
- x)^{1/2}.From an integral table: ∫ _{0}^{1}x^{2}dx/(1
- x)^{1/2} = [2(8 + 4x + 3x^{2})/15](1 - x)^{1/2}|_{0}^{1}
= -16/15.ΔE = [4e ^{6}Z^{2}/(3c^{3}m^{2}v_{1}r_{min}^{3})]16/15
= (8/45)mv_{1}^{5}/(Zc^{3}).For a nonrelativistic v _{1} we have ΔE <<
E.ΔE/2 is the radiation loss during the approach. The radiation loss on the way out is also ΔE/2. ½mv _{2}^{2} = ½mv_{1}^{2} -
ΔE, v_{2}^{2} = v_{1}^{2}
- (16/45)v_{1}^{5}/(Zc^{3}).v _{2} ≈ v_{1}(1 - (8/45)v_{1}^{3}/(Zc^{3})). |

**Problem 3:**

Let **E**_{0} = E_{0} **k**. The Abraham-Lorentz force equation for a damped, charged,
oscillator driven by an electric field **E**_{0}exp(-iωt) in the dipole approximation is

d^{2}**r**'/dt^{2} + Γ d**r**'/dt - τ d^{3}**r**'/dt^{3}
+ ω_{0}^{2} **r**' = (q/m)**E**_{0}exp(-iωt),

where Γ, τ, and ω_{0} are
constants, q is the charge and m is the mass of the oscillator.

Using this
and the expression for the radiation electric field, **E**_{rad}(**r**,t)
= -(4πε_{0})^{-1}[(q/(c^{2}r'')]**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 = (e^{2}/(mc^{2}))^{2 }(**k**∙**n**)^{2}[ω^{4}/((ω_{0}^{2}
- ω^{2})^{2} + ω^{2}Γ_{t}^{2})],

where e^{2} = q^{2}/(4πε_{0}) and Γ_{t} = Γ +
ω^{2}τ.

Solution:

Concepts: The scattering cross section | |

Reasoning: Power radiated into the solid angle dΩ = <dP> = <incoming intensity> * dσ. | |

Details of the calculation: Solve the equation of motion. d ^{2}r'/dt^{2} + Γ dr'/dt - τ d^{3}r'/dt^{3}
+ ω_{0}^{2} r' = (q/m)E_{0}exp(-iωt) k.Let r' = r'k = r_{0}'exp(-iωt)k. Thendr'/dt = -iωr _{0}'exp(-iωt), d^{2}r'/dt^{2} =
-ω^{2}r_{0}'exp(-iωt), d^{3}r'/dt^{3}
= iω^{3}r_{0}'exp(-iωt).(-ω ^{2} - iωΓ - iω^{3}τ + ω_{0}^{2})r_{0}'
= (q/m)E_{0}. r' = (q/m)E _{0}exp(-iωt/(-ω^{2} - iωΓ_{t} + ω_{0}^{2}),
where Γ_{t} = Γ + ω^{2}τ.a = -ω^{2}(q/m)E_{0}k exp(-iωt)/(-ω^{2}
- iωΓ_{t} + ω_{0}^{2}).Calculate S.For convenience, place the oscillator at the origin. Let r =
r(θ, φ), then r'' = r.E_{rad}(r,t) = -(4πε_{0})^{-1}[(q/(c^{2})]a_{⊥}(t
- r/c)= (4πε _{0})^{-1}[(q/(c^{2}r)]ω^{2}(q/m)E_{0}
sinθ
(θ/θ) exp(-iωt_{r})/(-ω^{2} - iωΓ_{t} + ω_{0}^{2})= (e ^{2}/(mc^{2})) (sinθ/r) (θ/θ) ω^{2}E_{0}
exp(-iωt_{r})/(ω_{0}^{2} - ω^{2} - iωΓ_{t}),with t _{r} = t - r/c.S((θ, φ) = (r/r)(Re(E_{rad}))^{2}/(μ_{0}c).ω ^{2} exp(-iωt_{r})/(ω_{0}^{2} - ω^{2}
- iωΓ_{t}) = |ω^{2}/(ω_{0}^{2} - ω^{2}
- iωΓ_{t})|exp(-iωt_{r})exp(iξ)= [ω ^{4}/((ω_{0}^{2} - ω^{2})^{2} -
ω^{2}Γ_{t}^{2})]^{½}exp(-i(ωt_{r} -
ξ))S(θ, φ) = (r/r )cos^{2}(ωt_{r} - ξ) (e^{4}/(m^{2}c^{4}))
(sin^{2}θ/r^{2}) (μ_{0}c)^{-1}E_{0}^{2}ω^{4}/((ω_{0}^{2}
- ω^{2})^{2} - ω^{2}Γ_{t}^{2}).< S(θ, φ)> = (r/r )½(e^{4}/(m^{2}c^{4}))
(sin^{2}θ/r^{2}) (μ_{0}c)^{-1}E_{0}^{2}ω^{4}/((ω_{0}^{2}
- ω^{2})^{2} - ω^{2}Γ_{t}^{2}).<dP> = <S(θ, φ)>r ^{2} dΩ = power radiated into the solid angle dΩ.<dP> = <incoming intensity> * dσ. dσ/dΩ = <S(θ, φ)>r ^{2}/( ½(μ_{0}c)^{-1}E_{0}^{2})= ½(e ^{4}/(m^{2}c^{4})) sin^{2}θ ω^{4}/((ω_{0}^{2}
- ω^{2})^{2} - ω^{2}Γ_{t}^{2}).k∙n = k∙(θ/θ) = -sinθ, (k∙n)^{2}
= sin^{2}θ. |

**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 = I_{0}cos(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 180 ^{o} 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) = p_{0}cos(ωt) isE_{R}(r,t) = -(1/(4πε_{0}c^{2}r))(d^{2}p_{⊥}(t
- r/c)/dt^{2}). (SI units)Let p_{0} = p_{0}(z/z), then E_{R}(r,t)
= -(1/(4πε_{0}c^{2}r))ω^{2}p_{0} cos(ω(t -
r/c)) sinθ (θ/θ).B_{R}(r,t) = (r/(rc))×E_{R}(r,t)
= -[1/(4πε_{0}c^{3}r)]ω^{2}p_{0} cos(ω(t -
r/c)) sinθ (φ/φ).S(r,t) = [1/(4πε_{0})^{2}][1/(c^{5}r^{2}μ_{0})] ω^{4}p_{0}^{2}
cos^{2}(ω(t - r/c)) sin^{2}θ (r/r).
= [1/(16π ^{2}ε_{0}r^{2}c^{3})]ω^{4}p_{0}^{2}
cos^{2}(ω(t - r/c)) sin^{2}θ (r/r). The intensity varies as sin ^{2}θ.<P> = <∫ S∙dA> = ω^{4}p_{0}^{2}/(12πε_{0}c^{3}),
proportional to ω^{4}.(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/(4πε_{0}))(3(p∙r)r/r^{5}
- p/r^{3}),B(r) = (μ_{0}/(4π))(3(m∙r)r/r^{5}
- m/r^{3}).Maxwell's equations in the regions outside the sources are ∇∙E = 0, ∇×E = -∂B/∂t, ∇∙B
= 0, ∇×B = (1/c^{2})∂E/∂t.Let E' = -c^{2}B, 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/c^{2})∂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 E_{R}(r,t) = -(1/(4πε_{0}c^{2}r))(d^{2}p_{⊥}(t
- r/c)/dt^{2}), B_{R}(r,t) = r×E_{R}(r,t)/(rc).E'_{R}(r,t) has the same form, except we replace
p/(4πε_{0}) by -c^{2}mμ_{0}/(4π) = -m/(4πε_{0}).Therefore E'_{R}(r,t) = (1/(4πε_{0}c^{2}r))(d^{2}m_{⊥}(t
- r/c)/dt^{2}), B'_{R}(r,t) = r×E'_{R}(r,t)/(rc).The far fields of an oscillating magnetic dipole therefore are B_{R}(r,t) = (1/(4πε_{0}c^{4}r))(d^{2}m_{⊥}(t
- r''/c)/dt^{2}), E_{R}(r,t) = -cr×B_{R}(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'))/(4πε_{0}rc).

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)[∂(rA_{θ})/∂r - ∂A_{r}/∂θ] (φ/φ)
= B (φ/φ).B = -[ikp/(4πε _{0}rc)] [-iksinθ + sinθ/r] exp(i(kr - ωt')).Taking the real part we have B = -[k ^{2}p sinθ/(4πε_{0}rc)] [cos(kr - ωt')
- sin(kr - ωt')/(kr)].(b) ∇×B = (1/c^{2})∂E/∂t = -(iω/c^{2})E,
since E is proportional to exp(-iωt).∇×B = (r/r) (1/(rsinθ))∂(Bsinθ)/∂θ - (θ/θ) (1/r))∂(rB)/∂r= -( r/r) [2k^{4}p cosθ/(4πε_{0}c)] [1/(kr)^{2}
+ i/(kr)^{3}]exp(i(kr - ωt'))+ ( θ/θ) [k^{4}p sinθ/(4πε_{0}c)]
[[1/(kr)^{2} + i/(kr)^{3} - i/(kr)]exp(i(kr - ωt'))].Taking the real part we have E _{r} = [2ck^{4}p cosθ/(4πε_{0}ω)][sin(kr - ωt')/(kr)^{2}
+ cos(kr - ωt')/(kr)^{3}],E _{θ} = [ck^{4}p sinθ/(4πε_{0}ω)][sin(kr - ωt')/(kr)^{2}
+ cos(kr - ωt')(1/(kr)^{3} - 1/(kr)].(c) radiation zone: E and B are proportional to 1/r. E_{R} = E_{R} (θ/θ),E _{R}
= -[ck^{4}p sinθ/(4πε_{0}ω)] cos(kr - ωt')/(kr)
= -[ω^{2}p sinθ/(4πε_{0}c^{2})] cos(kr - ωt')/r,B_{R} = B_{R} (φ/φ),B _{R}
= -[k^{2}p sinθ/(4πε_{0}rc)] cos(kr - ωt') =
-[ω^{2}p sinθ/(4πε_{0}c^{3})] cos(kr - ωt')/r.B_{R}(r,t) = r×E_{R}(r,t)/(rc). |