Past exam of the optics and electrodynamics course of the University of Cambridge / 2012

a

Solution

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance

R

E_{rad} (r, t) = \frac{q}{4 π ϵ _{0} c ^{2} R} n \times [n \times a (t - R / c)], B_{rad} = \frac{1}{c} n \times E_{rad} .

The acceleration is evaluated at retarded time. Radiation fields decay as

1/ R

, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as

1/ R^{2}

. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector gives

\frac{d P}{d Ω} = \frac{q ^{2} a ^{2} s i n ^{2} ϑ}{16 π ^{2} ϵ _{0} c ^{3}}, P = \frac{q ^{2} a ^{2}}{6 π ϵ _{0} c ^{3}} .

The integral uses

\int sin^{2} ϑ d Ω = 8 π /3

. This is the Larmor formula. A harmonically accelerated charge with peak acceleration

a_{0}

has mean power

q^{2} a_{0}^{2} / (12 π ϵ_{0} c^{3})

; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

Figure 1.
Original nonrelativistic radiation pattern, with zero power along the acceleration and maximum power transverse to it
. **Details**.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to

q^{2} γ^{6} [a^{2} - ∣ β \times a ∣^{2}] / (6 π ϵ_{0} c^{3})

, with

β = v / c

. The displayed simple

sin^{2} ϑ

pattern is restricted to the nonrelativistic limit.

b

Solution

Temporal coherence measures correlation between an optical field and a delayed copy. For a stationary scalar complex field with carrier convention

e^{- iω t}

Γ (τ) = ⟨ E^{*} (t) E (t + τ)⟩, g^{(1)} (τ) = \frac{Γ ( τ )}{Γ ( 0 )} .

The power spectrum

S (ω) \geq 0

and the field autocorrelation form a Fourier transform pair, the Wiener-Khinchin theorem:

g^{(1)} (τ) = \frac{\int S ( ω ) e ^{- iω τ} d ω}{\int S ( ω ) d ω} .

A Michelson interferometer supplies delay

τ = Δ L / c

. With equal arm intensities and matched polarization, its interferometric visibility is

∣ g^{(1)} (τ) ∣

; unequal intensities or mode mismatch lower visibility independently of temporal coherence. The relevant spectrum is the optical field spectrum, not the low-frequency spectrum of detector intensity fluctuations.

For a single Lorentzian optical line of angular-frequency full width at half maximum

Δ ω

, use the narrow-line detuning model

\frac{S ( ω )}{\int S d ω} = \frac{Δ ω / ( 2 π )}{( ω - ω _{0} ) ^{2} + ( Δ ω /2 ) ^{2}}, g^{(1)} (τ) = e^{- i ω_{0} τ} e^{- Δ ω ∣ τ ∣/2} .

(3)

Such a model describes a phase-diffusing monochromatic source. An

e^{- 1}

correlation time is

2/Δ ω

, and its coherence length is

2 c /Δ ω

. The Lorentzian extension over all detunings approximates a narrow optical line with

Δ ω ≪ ω_{0}

For two equal, mutually incoherent ideal monochromatic spectral lines,

S \propto δ (ω - ω_{1}) + δ (ω - ω_{2}), g^{(1)} (τ) = e^{- i \overset{ω}{ˉ} τ} cos (Δ ω_{12} τ /2) .

(4)

Here

\overset{ω}{ˉ} = (ω_{1} + ω_{2}) /2

and

Δ ω_{12} = ω_{2} - ω_{1}

. Visibility has zeros and revivals, so the overall spectral span alone does not determine a unique decay time. Giving each line a common finite Lorentzian linewidth multiplies this beating correlation by the corresponding exponential envelope.

Figure 1.
Original spectrum–coherence examples: a Lorentzian line and a doublet with visibility revivals
. **Details**.

These two examples distinguish irreversible envelope decay from cancellation and revival of discrete spectral components. The line profiles are explicit models, not claimed measurements of a particular source.

c

Solution

The electric potential and magnetic vector potential represent the fields through

B = \nabla \times A

and

E = - \nabla ϕ - \partial_{t} A

. They are not unique: a differentiable function

χ

gives

A \mapsto A + \nabla χ, ϕ \mapsto ϕ - \partial_{t} χ,

without changing either field. Gauge fixing restricts this freedom to suit the sources, boundaries or symmetry; it does not alter observables.

The Lorenz gauge

\nabla \cdot A + c^{- 2} \partial_{t} ϕ = 0

is Lorentz covariant. Maxwell's equations then reduce to sourced wave equations,

(\nabla^{2} - c^{- 2} \partial_{t}^{2}) ϕ = - ρ / ϵ_{0}, (\nabla^{2} - c^{- 2} \partial_{t}^{2}) A = - μ_{0} J .

Retarded solutions implement causal radiation conditions. This gauge is useful for radiation and relativistic calculations; residual transformations obey the homogeneous wave equation.^[ref]

The Coulomb gauge

\nabla \cdot A = 0

instead gives

\nabla^{2} ϕ = - ρ / ϵ_{0}

. Its electric potential is an instantaneous Poisson solution, but the accompanying magnetic vector potential makes the full physical electric field causal. It is useful in separating longitudinal electric fields from transverse radiation. In a source-free vacuum region, suitable radiation boundary conditions permit radiation gauge

ϕ = 0

\nabla \cdot A = 0

, leaving two transverse plane-wave polarizations. These extra conditions are not a gauge choice applicable unchanged to arbitrary nonzero charge distributions.

Different gauges redistribute information between potentials; the same boundary-value problem gives the same electric fields and magnetic fields.

3

Solution

Use

E (z, t) = Re [J e^{i (k z - ω t)}]

. A Jones vector

J = (J_{x}, J_{y})^{T}

records the two transverse complex amplitudes. The amplitude ratio and relative phase determine the polarization ellipse; a common nonzero scalar does not change the polarization. A fixed Jones vector describes fully polarized monochromatic light, not an unpolarized statistical mixture.

Linear birefringence gives orthogonal principal polarizations different refractive indices. Put

n_{y} - n_{x} = Δ n > 0

, so

x

is the fast axis. A plate of thickness

d

gives slow-minus-fast phase retardation

δ (ω) = \frac{ω d Δ n ( ω )}{c} .

The thinnest quarter-wave plate has

d = \frac{π c}{2 ω _{0} Δ n ( ω _{0} )} = \frac{λ _{0}}{4Δ n ( ω _{0} )} .

Here

λ_{0} = 2 π c / ω_{0}

is the vacuum wavelength. For linear input at angle

θ

to the fast axis, the output at design frequency is

(cos θ, i sin θ)^{T}

up to common phase. Axis-aligned input remains linear; equal nonzero components at

4 5^{\circ}

modulo

9 0^{\circ}

give circular polarization; other input angles give elliptical polarization. There is no need to spatially separate the two components.^[ref]

For waveplate retardance detuning, define

ϵ = \frac{ω d Δ n ( ω )}{c} - \frac{π}{2} ≃ \frac{d}{c} [Δ n_{0} + ω_{0} Δ n_{0}^{'}] (ω - ω_{0}) .

(3)

Neglecting material dispersion reduces this to

ϵ = (π /2) (ω - ω_{0}) / ω_{0}

. Remove the common fast-axis phase. Then

Q = diag (1, i e^{i ϵ})

. A similar half-wave plate of twice the thickness has twice the retardation, giving

H = diag (1, - e^{2 i ϵ})

. The shared error parameter presumes the same material and matched thickness ratio.

For the half-wave and quarter-wave achromatic circularizer, the first fast-axis angle is

α = 1 5^{\circ}

and the second is

γ = α + β = 7 5^{\circ}

. With the stated passive rotation convention,

J_{out} = R (- γ) QR (γ) R (- α) H R (α) (10) .

(4)

An explicit expression valid for the given shared error is obtained by putting

z = e^{i ϵ}

and defining

v_{x} = cos^{2} α - z^{2} sin^{2} α, v_{y} = sin α cos α (1 + z^{2}),

(5)

J_{out} = ((cos^{2} γ + i z sin^{2} γ) v_{x} + (1 - i z) sin γ cos γ v_{y} (1 - i z) sin γ cos γ v_{x} + (sin^{2} γ + i z cos^{2} γ) v_{y}) .

(6)

ϵ = 0

, the half-wave plate rotates the linear input to

3 0^{\circ}

, which is

- 4 5^{\circ}

relative to the second fast axis. Thus

J_{0} = \frac{e ^{i 7 5^{\circ}}}{2} (1 - i) .

(7)

This is circular polarization. With our time convention its fixed-point electric vector is proportional, after a time-origin shift, to

(cos ω t, - sin ω t)

; this specifies the rotation without relying on conflicting handedness conventions.

Expanding the exact product to first order gives

J_{out} = (0.183 + 0.683 i 0.683 - 0.183 i) + ϵ (- 0.433 + 0.116 i 0.116 + 0.433 i) + O (ϵ^{2}) .

(8)

These coefficients are rounded to three significant figures. Before rounding, the derivative is exactly

J_{0}^{'} = iκ J_{0}

, with

κ = (3 - 3) /2

. The Jones-component ratio sensitivity therefore vanishes:

\frac{J _{y}}{J _{x}} = \frac{J _{0 y} ( 1 + iκ ϵ )}{J _{0 x} ( 1 + iκ ϵ )} + O (ϵ^{2}) = - i + O (ϵ^{2}) .

(9)

Only common phase changes at first order. This proves compensation for the prescribed input near the design frequency, not exact achromatic performance throughout the visible spectrum for arbitrary input polarization.

Figure 1.
Original comparison of single-plate and composite polarization errors and the composite electric-vector trajectory
. **Details**.

a

Solution

A transverse static electric field can induce linear birefringence through the electro-optic Kerr effect. Define its coefficient by

Δ n = λ_{0} K E_{dc}^{2}

, with the principal axes parallel and perpendicular to the applied field. A length

d_{K}

then gives

δ_{K} = 2 π K d_{K} E_{dc}^{2}

For input polarized at

4 5^{\circ}

to those axes, switching from zero retardation to

∣ δ_{K} ∣ = π /2

switches linear to circular polarization. Other input angles generally become elliptical, and axis-aligned input stays linear. The required field magnitude for the lowest-order setting is

∣ E_{dc} ∣ = \frac{1}{4∣ K ∣ d _{K}} .

A Kerr cell is suitable as a switchable quarter-wave plate. The field can be changed electronically without mechanically rotating the axes; actual switching time depends on the material response and driving circuit. This is induced linear birefringence, not polarization-selective absorption.^[ref]

b

Solution

The longitudinal Faraday effect produces circular birefringence: right- and left-circular components have different phase velocities. Linear input contains equal amplitudes of both circular eigenpolarizations. A relative phase

δ_{F}

rotates its linear polarization through

δ_{F} /2

, preserving linearity in an ideal transparent sample. Conventionally the signed rotation is

ϑ_{F} = V B_{∥} d_{F}

, with Verdet constant

V

A pure Faraday rotator cannot serve as the required switchable quarter-wave plate. It rotates a line into another line rather than producing the required relative phase between fixed orthogonal linear components. Applying a field that gives a nominal quarter-cycle circular-eigenmode retardation simply rotates linear polarization by

4 5^{\circ}

. Circular dichroism or a separate linear retarder would change the device, and is not the stated pure Faraday mechanism.

4

Solution

Take

x^{μ} = (c t, x, y, z)

and metric signature

(+, -, -, -)

. The four-current, four-potential and four-gradient are

J^{μ} = (c ρ, j), A^{μ} = (ϕ / c, A), \partial_{μ} = (c^{- 1} \partial_{t}, \nabla), \partial^{μ} = (c^{- 1} \partial_{t}, - \nabla) .

The first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:

\partial_{μ} J^{μ} = \partial_{t} ρ + \nabla \cdot j = 0.

Indeed,

J^{' μ} = Λ^{μ}_{ν} J^{ν}

and

\partial_{μ}^{'} = (Λ^{- 1})^{α}_{μ} \partial_{α}

, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

a

Solution

Let

r

point transversely away from the line and define

φ = x \times r

. Cylindrical Gauss's law in the line's rest frame gives

E^{'} = \frac{λ ^{'}}{2 π ϵ _{0} r ^{'}} r, B^{'} = 0.

The transverse distance is unchanged by the axial boost:

r = r^{'}

. Applying the inverse of the given field transformation to a boosted uniform line charge,

E = \frac{γ λ ^{'}}{2 π ϵ _{0} r} r, B = \frac{u \times E}{c ^{2}} = \frac{μ _{0} γ λ ^{'} u}{2 π r} φ .

Here

γ = (1 - u^{2} / c^{2})^{- 1/2}

and all longitudinal field components vanish. Substituting back verifies

B^{'} = γ (B - u \times E / c^{2}) = 0

and

E^{'} = γ (1 - u^{2} / c^{2}) E = E / γ

. The sign of

λ^{'}

determines the radial and circulation directions.

b

Solution

A material segment of the line with rest length

d x^{'}

has laboratory length

d x = d x^{'} / γ

, while its charge

d Q

is unchanged. Thus

λ = γ λ^{'}, I = λ u = γ λ^{'} u .

Equivalently the line-integrated four-current transforms from

(c λ^{'}, 0)

(γ c λ^{'}, γ u λ^{'})

. Gauss's and Ampère's laws in

S

now give

E_{r} = λ / (2 π ϵ_{0} r)

and

B_{φ} = μ_{0} I / (2 π r)

, exactly the fields found in part (a). Charge density is frame-dependent even though the charge of the same set of particles is invariant.

c

Solution

S

, adding the stationary opposite line cancels charge density and electric field but contributes no current. Hence

λ_{tot} = 0, I_{tot} = λ u, E_{tot} = 0, B_{tot} = \frac{μ _{0} λ u}{2 π r} φ .

The neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:

λ_{tot}^{'} = - \frac{γ u I _{tot}}{c ^{2}} = - γλ \frac{u ^{2}}{c ^{2}}, I_{tot}^{'} = γ I_{tot} = γλ u .

The same result follows by adding the original resting line

(λ^{'}, 0)

to the new negative line, whose primed density and current are

(- γλ, + γλ u)

. In particular

λ^{'} - γλ = - γλ u^{2} / c^{2}

because

λ = γ λ^{'}

Transforming the total fields gives

E_{tot}^{'} = - \frac{γλ u ^{2}}{2 π ϵ _{0} c ^{2} r} r, B_{tot}^{'} = \frac{μ _{0} γλ u}{2 π r} φ .

(3)

For positive

λ

u \times B

points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from

λ_{tot}^{'}

and

I_{tot}^{'}

reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.

d

Solution