Assignment 4, solutions

Problem 1:

Two equal magnitude, opposite sign charges are located at either end of a molecule of mass M and length l.  The molecule rotates end over end (a nonrelativistic tumbling motion) with an initial rotational period T  (cT >> l).  How long will it take the molecule to lose 1/10 of its rotational energy by electromagnetic radiation?

Solution:

• Concepts:
The radiation field of an accelerating dipole, the Larmor formula
• Reasoning:
The energy lost per unit time given by the Larmor formula
P = -dE/dt = <(d2p/dt2)2>/(6πε0c3).  (SI units)
The two point charges move in a circle and produce a rotating dipole.  Let the dipole rotate about the x-axis.
p(t) = p0cos(ωt) k - p0sin(ωt) j.
d2p/dt2 = -ω2p,  (d2p/dt2)2 = ω4p02.
• Details of the calculation:
P = -dE/dt = ω4p2/(6πε0c3) is the radiated power.
The rotational kinetic energy is E = ½Iω2, with I = Ml2/12.
(I assume that the mass M is distributed uniformly along the length l.  you could also read the problem as implying that a mass M/2 is located at the position of each charge and the rod is massless.  The I = Ml2/4.)
Approximate solution:  -∆E = -(dE/dt)∆t = -E/10.
∆t = Iω26πε0c3/(20ω4p2) = Ml2πε0c3/(40ω2p2) = MT2ε0c3/(160 πq2).
Exact solution:
-dE/dt = ω4p2/(6πε0c3) = 4E2p2/(6 I2πε0c3).
-dE/E2 = ω4p2/(6πε0c3) = 4p2dt/(6 I2πε0c3).
-(6 I2πε0c3/(4p2))∫E00.9E0dE/E2 = ∫dt = ∆t.
∆t = [MT2ε0c3/(16 πq2)] [1/0.9 - 1].

Problem 2:

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,
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 = <dP> = <incoming intensity> * 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.
= (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.
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).
<S(θ, φ)> = (r/r )½(e4/(m2c4)) (sin2θ/r2) (μ0c)-1E02ω4/((ω02 - ω2)2 - ω2Γt2).
<dP> = <S(θ, φ)>r2 dΩ = power radiated into the solid angle dΩ.
<dP> = <incoming intensity> * dσ.
dσ/dΩ =  <S(θ, φ)>r2/( ½(μ0c)-1E02)
= ½(e4/(m2c4)) sin2θ ω4/((ω02 - ω2)2 - ω2Γt2).
kn = k∙(θ/θ) = -sinθ,  (kn)2 = sin2θ.

Problem 3:

For a dipole antenna at the origin with complex electric dipole moment p(t) = p0exp(-iωt) k the electric and magnetic fields in the far (radiation) zone are
E
(r,t) = Eθ(θ/θ),  Eθ = -[1/(4πε0)](p0/k2)(sinθ/r)exp(i(kr - ωt)),  k = ω/c,
B(r,t) = Bφ(φ/φ),  Bφ = Eθ/c.
(a)  If the real dipole moment is p(t) = p0cos(ωt) k  with p0 real, what is the instantaneous power radiated per unit solid angle, dP(r,θ,φ,t)/dΩ?   What is the time average?
(b)  If there is a second dipole at -d0 with p(t) = p0cos(ωt + α) k, with kd0 << 1, what is the instantaneous power radiated by the pair of dipoles in the x-z plane,  dP(r,θ,φ,t)/dΩ?
What is the time average?  What are the minimum and maximum values of the time average with respect to variations of the phase α?
(c)  A large number, N, of dipoles, as above, is located near the origin with Nkd0 << 1 and with phases α0 randomly distributed. What is the time averaged dP(r,θ,φ,t)/dΩ for the system?

Solution:

• Concepts:
• Reasoning:
We are asked to characterize the power radiated by an oscillating dipoles.
• Details of the calculation:
(a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p0cos(ωt) k is
E(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t - r/c))sinθ (θ/θ).  B(r,t) = (E/c)(φ/φ).
S(r,t) = (1/μ0)(E × B) = [1/(4αε0)2][1/(c5r2μ0)]ω4p02 cos2(ω(t - r/c))sin2θ (r/r).
dP = S∙(r/r) r2dΩ = [1/(4πε0)] [ω4/(4πc3)]p02 cos2(ω(t - r/c))sin2θ dΩ
= instantaneous power radiated into dΩ.
<dP> = <S>∙(r/r) r2dΩ = [1/(4πε0)] [ω4/(8πc3)]p02 sin2θ dΩ
= average power radiated into dΩ.

(b)  E = E1 + E2.
E1(r,t) = -(1/(4πε0c2r))ω2p0 cos(ωt - kr) sinθ (θ/θ).
E2(r,t) = -(1/(4πε0c2r'))ω2p0 cos(ωt - kr' + α) sinθ' (θ'/θ').
r' = r + d0i.  r' = |r + d0i| ≈ r(1 +  d0ir/r2) = r + d0cos(π/2 - θ) = r + d0sin(θ).
kr' ≈ kr + kd0sin(θ).
Since kd0 << 1 and r >> d0 (radiation zone), we can write
E2(r,t) ≈ -(1/(4πε0c2r))ω2p0 cos(ωt - kr + α) sinθ (θ/θ).
B1(r,t) = (E1/c)(φ/φ), B2(r,t) = (E2/c)(φ/φ).
S(r,t) = (1/μ0)(E1 + E2) × (B1 + B2.)
= [1/(4πε0)2][1/(c5r2μ0)]ω4p02sin2θ (r/r) *
[cos2(ωt - kr) + cos2(ωt - kr + α) + 2 cos(ωt - kr) cos(ωt - kr + α)].
dP = S∙(r/r) r2dΩ.
<dP> = <S>∙(r/r) r2dΩ = [1/(4πε0)] [ω4/(8πc3)]p02 sin2θ [½ + ½ + 2cos(α)]dΩ
= [1/(4πε0)] [ω4/(8πc3)]p02 sin2θ [2 cos2(α/2)]dΩ
= average power radiated into dΩ.
Minimum:  α = nπ, n = odd, cos2(α/2) = 0, <dP>/dΩ = 0.
Maximum:  α = nπ, n = even, cos2(α/2) = 1, <dP>/dΩ = [1/(4πε0)] [ω4/(2πc3)]p02 sin2θ.

(c)  For a large number N of dipoles with randomly distributed phases we get
<dP>/dΩ = N * <dP>/dΩ from part (a). Interference effects average out.

Problem 4:

A thin linear antenna of length d is excited in such a way that the sinusoidal current makes a full wavelength of oscillation as shown in the figure.
(a)  Calculate the power radiated per unit solid angle and plot the angular distribution of the radiation.
(b)  Determine the total power radiated and find a numerical value for the radiation resistance R by setting <P> = <I2>R.  You can leave your answer in integral form.

Solution:

• Concepts:
• Reasoning:
We can treat the antenna as made up of small sections, each section emitting dipole radiation.
• Details of the calculation:
(a)  Let the antenna lie on the z-axis, from z = -λ/2 to z = λ/2.
Consider a small section of the antenna.  Treat it like a dipole.
Let p = q dz (z/z), q = q0(z) cosωt.
Then I(z) = dq(z)/dt = -ωq0(z) sinωt = I0(z) sin(ωt),  I0(z) = -ωq0(z).
p0(z) = q0(z) dz = -(I0(z)/ω)dz.
For this antenna I0(z) = I0sin(kz).
A small section of the antenna of length dz' located at z' a distance R = r - (z/z)z' from the observation point therefore produces
dER(r,t) = (1/(4πε0c2R))ω2(I0(z)/ω) cos(ωt - kR) sinθ' dz' (θ'/θ')
= (1/(4πε0c2R))ω2(I0/ω) cos(ωt - kR) sinθ' sin(kz') dz' (θ'/θ').
We have r >> λ,  r >> z', Therefore θ' ≈ θ,  1/(kR) ≈ 1/(kr), since kR ≈ kr - kz' cosθ.
For the phase we use cos(ωt - kR) ≈ cos(ωt - kr + kz' cosθ)
dER(r,t) = [1/(4πε0c3kr)]ω2I0 sinθ cos(ωt - kr + kz' cosθ)  sin(kz') dz' (θ/θ)
We now can find ER(r,t) due to the antenna.
ER(r,t) = (θ/θ)[1/(4πε0c3kr)]ω2I0 sinθ ∫-λ/2λ/2cos(ωt - kr + kz' cosθ)  sin(kz') dz'.

-λ/2λ/2cos(ωt - kr + kz' cosθ)  sin(kz') dz'
= (1/k)cos(ωt - kr)∫πcos(kz' cosθ)  sin(kz') dkz'
- (1/k)sins(ωt - kr)∫πsin(kz' cosθ)  sin(kz') dkz'.
The first term is an odd function of kz' and therefore zero.
πsin(ax) sin(x') dx = [½sin (a-1)x/(a-1) - ½sin (a+1)x/(a+1)]|π = 2sin(aπ)/(1 - a2)
ER(r,t) = (θ/θ) [1/(4πε0c3k2r)]ω2I0 sinθ sin(ωt - kr) 2sin(πcosθ)/sin2θ.
S = (r/r) (1/(μ0c))|ER(r,t)|2 = (r/r)[1/(4π2ε0r2c sin2θ)]I02 sin2(ωt - kr) sin2(πcosθ).
<dP> = <S>∙(r/r) r2dΩ = [1/(8π2ε0c sin2θ)]I02 sin2(πcosθ)dΩ.
The plot below compares the angular radiation pattern of a dipole (sin2θ) and a full-wave antenna (sin2(πcosθ)/sin2θ).

(b)  <P> = 2π∫0πsinθdθ <dP>/dΩ = (1/(4π2ε0c)]I020πdθ sin2(πcosθ)/sinθ.
<P> = <I2>R.  R = [1/(2πε0c)] ∫0πdθ sin2(πcosθ)/sinθ.
Numerical integration:  R = 93 ohm.

Problem 5:

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)]∫rminr-4dr/v(r).
v(r) = v1(1 - 2Ze2/(mrv12))1/2.  v(rmin) = 0.
rmin = 2Ze2/(mv12).
ΔE = -[4e6Z2/(3c3m2v1)]∫rminr-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)).