Electromagnetic Radiation Flashcards
Maxwell’s eqs (+ free space)
∮ₛ E.dS = Qₑₙ꜀/∈₀
∮ₛ B.dS = 0
∮꜀ E.dL = -d/dt ∫ₛB.dS
∮꜀ B.dL = µ₀Iₑₙ꜀ = µ₀∫ₛ(j+∈₀ ∂E/∂t).dA
———————————-
∇.E = ρ/ϵ₀
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀(J+ϵ₀ ∂E/∂t)
———–Free Space————————–
∇.E = 0
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀ϵ₀ ∂E/∂t
Gauss, no magnetic monopoles, Faraday, Ampere
Stokes’ theorem
∫ₛ ∇ ⨯ v dS = ∮ₗ v ⋅ dl
Divergence theorem
∫ᵥ∇ ⋅ v dV = ∫ₛv ⋅ dS
E-field for a point charge
E(r)= 1/4π∈₀ q/r²
Coulomb’s Law
F = q₁q₂/4π∈₀r²
Lenz’s law
An induced electric field and consequent current will generate magnetic fields to oppose the initially varying magnetic field.
Get the continuity equation from Amperes law
Ampere’s law
∇ × B = µ₀(J+ϵ₀ ∂E/∂t)
Divergence of both sides
∇ · (∇ × B) = µ₀∇ · J + µ₀ϵ₀∇ · ∂E/∂t
Vector identity ∇ · (∇ × B) = 0, divide by µ₀
0 = ∇ · J + ϵ₀∇ · ∂E/∂t = ∇ · J + ϵ₀∂/∂t(∇ · E) = ∇ · J + ϵ₀ ∂/∂t (ρ/ϵ₀)
Continuity equation
0 = ∇ · J + ∂ρ/∂t
Poisson’s equation
∇²φ = −ρ/ϵ₀
What is a conservative force
A force with the property that the work done in moving a particle between two points is independent of the path taken.
Electric potential of a point charge
Φ = −∫E = q/4πε₀R²
Electric potential of a charge distribution
Φ(r) = 1//4πε₀ ∫ᵥ ρ(r’)/|r-r’| dV’
Electric field from potential
E = −∇φ
Magnetic vector potential with Coulomb gauge (∇ · A = 0)
From ∇ x (∇ · A) = 0
and ∇ · B) = 0
get B = ∇ × A,
leads to
∇²A = −µ₀J
Genral form of the Biot-Savart law
B(r) = µ₀/4π ∫ᵥ J(r’)×(r − r’)/|r − r’|³ dV’
Electro- and magnetostatics equations
∇ × A = B
∇²A = −µ₀J
∇ · A = 0
E = −∇ · φ
∇²φ = −ρ/ε₀
∇ · E = ρ/ε₀
Electro- and magnetodynamics equations
Magneto-
B = ∇ × A
∇ × E = −∂B/∂t
E + ∂A/∂t = −∇ · Φ
∇ × B = µ₀J + µ₀ε₀ ∂E/∂t
∇²A − µ₀ε₀ ∂²A/∂t² − ∇(∇ · A + µ₀ε₀ ∂Φ/∂t) = −µ₀J
////////////////////////////////////////////////////////////
Electro-
∇ · E = ρ/ε₀
∇ · (−∇Φ − ∂A/∂t ) = ρ/ε₀
∇²Φ + ∂/∂t (∇ · A) = −ρ/ε₀
////////////////////////////////////////////////////////////
Reduces Maxwell into two coupled equations
Lorenz gauge
Choosing a potential that satisfies
∇ · A = −µ₀ε₀ ∂Φ/∂t = 0
allows uncoupling of dynamic equations
∇²A − µ₀ε₀ ∂²A/∂t² = −µ₀J
∇²Φ − µ₀ε₀ ∂²Φ/∂t² = −ρ/ε₀
Lorentz Force
Single charge
F = q(E + v × B)
Charge distribution
F = ∫ᵥ ρ(E + v × B) dV
Relativistic formulae and Lorentz transformation to a field
γ = E/E₀
where E is total energy of a moving particle, and E₀ = m₀c² is the rest energy given the rest mass m₀.
Particle kinetic energy is
T = E − E₀
β factor is
β = v/c = √1 − 1/γ²’
In vector form, v = βc
Fields transform as
E’ =γ(E + cβ × B) − γ²/γ+1 β(β · E)
B’ =γ(B − 1/c β × E) − γ²/γ+1 β(β · B)
Energy and energy density in an electromagnetic field
U = UE + UB
= ε₀/2 ∫ᵥ E² dV + 1/2µ₀ ∫ᵥ B² dV
Energy density is the integrand.
* Electric field
ε₀E²/2
* Static magnetic field
B²/(2µ₀)
Field energy from energy density around a stationary point charge
Electric field
E(r) = 1/4πε₀ q/r² r̂
Energy density
1/2 ε₀E² = q²/32π²ε₀r⁴
Integrate for energy in the field.
* Field does not depend on θ or φ, so integration over those limits gives 4π.
* Integrate over all r
U = ∫∞ᵣ₀ q²/32π²ε₀r⁴ 4πr² dr
= 1/2 q²/4πε₀ ∫∞ᵣ₀ dr/r²
= 1/2 q²/4πε₀ 1/r₀
Becomes infinite if r₀ → 0, meaning point charge has infinite field energy, clearly not
physical.
Instead, let r₀ equal the classical electron radius
r₀ = 1/4πε₀ e²/mₑc² ≡ rₑ,
Now get
U = 1/2 mₑc²
Power exerted by the electromagnetic field upon a single charge
Work done over small distance dl is W = F · dl, power exerted by the field upon the charge is
P = dW/dt
= d/dt (F · dl)
= F · v
= q(E + v × B) · v
= qE · v
B term goes to zero as magnetic forces do no work.
If +ve charge moves in same direction as E, the charge gains energy and E does work on the charge. If the charge moves opposite to E, the charge does work on the field.
Power exerted by the electromagnetic field upon a current density
P = ∫ᵥ ρE · v dV
= ∫ᵥ E · J dV
Derive the Poynting vector
Using Faraday’s law
B · ∂B/∂t = −B · (∇ × E)
Using Ampere’s law
µ₀ε₀E · ∂E/∂t = E · (∇ × B) − µ₀E · J
Combine
B · ∂B/∂t + µ₀ε₀E · ∂E/∂t = −µ₀E · J − (B · ∇ × E − E · ∇ × B)
Use vector calculus identity
∇ · (E × B) = B · (∇ × E) − E · (∇ × B)
And rewrite terms with time derivatives as
E · ∂E/∂t = 1/2 ∂/∂t (E · E)
B · ∂B/∂t = 1/2 ∂/∂t (B · B)
Can now write
E · J = −1/2 ∂/∂t (ε₀E² + 1/µ₀ B²) − 1/µ₀ ∇ · (E × B)
Define the Poynting vector
S = 1/µ₀ (E × B)
What is the Poynting vector
There is power present in an electromagnetic field. Its energy flux and direction is given by the Poyting vector
Poynting’s theorem
The theorem is a differential form of energy conservation
−∂U/∂t = ∇ · S + J · E
where U is the energy density.
In the case that no work is done, J · E = 0
We get the continuity equation
∂U/∂t = −∇ · S
Derive plane wave equations in a vacuum from Maxwell’s equations
Electric
Faraday’s law
∇ × E = −∂B/∂t
Take curl
∇ × (∇ × E) = −∂/∂t (∇×B)
Vector identity and Ampere’s law
∇(∇ · E) − ∇²E = −∂/∂t (µ₀ε₀ ∂E/∂t)
Gauss’ law in vacuum/ with no charge,
∇ · E = 0
Wave equation
∇²E − µ₀ε₀ ∂²E/∂t² = 0
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Magnetic
Ampere’s law
∇ × B = µ₀ε₀ ∂E/∂t
Take curl
∇ × (∇ × B) = µ₀ε₀ ∂/∂t (∇×E)
Vector identity and Faraday’s law
∇(∇ · B) − ∇²B = −µ₀ε₀ ∂/∂t (∂E/∂t)
No magnetic monopoles
∇ · B = 0
Wave equation
∇²B − µ₀ε₀ ∂²B/∂t² = 0
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Two wave equations, waves travelling with speed
c = 1/√µ₀ε₀’
Fields in a linear, isotropic, non-conducting, dielectric medium
Modified permittivity
ε₀ → ε = εᵣε₀
Modified permeability
µ₀ → µ = µᵣµ₀
D = εE
H = 1/µ B
D-field is the electric displacement field and H-field is the magnetic field
Waves in a linear, isotropic, non-conducting, dielectric medium
Wave equation
∇²E − µε ∂²E/∂t² = 0
Speed of wave
v = 1/√µε’
= 1/√µᵣµ₀εᵣε₀’
= c/√µᵣεᵣ’
= c/n
Refractive index n of the dielectric is given by
n = √µᵣεᵣ’
Speed of light is
c = 1/√µ₀ε₀’
Can also write wave equation as
∇²E − 1/c² ∂²E/∂t² = 0
Plane wave solution for a wave moving along the z-axis
Solutions are of form
E = E₀exp(i(ωt±kz))
where
E₀ =
(Eₓ₀
Eᵧ₀)
and
Eₓ = f(z − ct) + g(z + ct)
f and g are arbitrary functions
Same for B-field
Components of form (ωt − kz) describe disturbances in the +z direction, and components of form (ωt + kz) describe disturbances in the −z direction.
Plane wave solution, as it is uniform in directions perpendicular to propagation.
Dispersion relation
c = ω/|k|
= 1/√µ₀ε₀’
where k is the wavenumber (or wave-vector) where the wavelength is λ = 2π/k
Relate E- and B-fields from their plane wave solutions
Faraday’s law
∇ × E = −∂B/∂t
Substitute plane wave solutions
|x̂…………………………………….ŷ……………………….ẑ|
|∂/∂x……………………………..∂/∂y……………….∂/∂z|
|Eₓ₀exp(i(ωt−kz)…..Eᵧ₀exp(i(ωt−kz)……………0|
= −iω(Bₓ₀x̂ + Bᵧ₀ŷ) exp(i(ωt−kz)
= (+x̂(ikEᵧ₀) − ŷ(ikEₓ₀)) exp(i(ωt−kz))
Match terms in x̂ and ŷ
Bₓ₀ = −k/ω Eᵧ₀
Bᵧ₀ = k/ω Eₓ₀
AKA
B₀ = k/ω ẑ × E₀
Taking dot product
E · B = −(k/ω Eᵧ₀) Eₓ₀ + (k/ω Eₓ₀) Eᵧ₀ = 0.
E and B are perpendicular to one another and
B₀ = k/ω E₀ = E₀/c
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For an arbitrary direction k̂:
E(r, t) = E₀exp(i(ωt−k·r)
B(r, t) = 1/c k̂ × E = 1/ω K x E
Using
c = ω/|k|
Different types of polarization
General wave propagating in +z direction
E = (Eₓ, Eᵧ, 0)
with individual components
Eₓ = E₁exp(i(ωt−kz+φ₁))
Eᵧ = E₂exp(i(ωt−kz+φ₂))
Polarisation is determined by ratio E₁/E₂ and phase difference φ₁ − φ₂, can choose φ₁ = 0 and φ₂ = φ.
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- Linear Polarisation
φ = 0
E = (E₁, E₂, 0) cos(ωt − kz)
E oscillates along a fixed plane - Left-Hand (LH) Circular polarisation
φ = π/2
E₁ = E₂ - Right-Hand (RH) Circular polarisation
φ = −π/2
E₁ = E₂ - Elliptical Polarisation
E₁ ≠ E₂ - Note that handedness is w.r.t. z coming out of page
B direction and magntitude compared to E
|B| = |E|/c
B direction is π/2 compared
to E
Energy density for a plane wave in vacuum
Assuming a linearly polarised electric field
E = E₀x̂ cos(ωt−kz)
B = B₀ŷ cos(ωt−kz)
where B₀ = E₀/c as expected.
Volumetric energy density at given z
UE = ε₀/2 E₀² cos²(ωt−kz)
UB = 1/2µ₀ B₀² cos²(ωt−kz)
In a plane electromagnetic wave, there is
equal energy density in the E field and B field, UE = UB
Can combine for total energy
U = UE + UB = ε₀E₀² cos²(ωt−kz)
Time average of total electromagnetic wave energy
U = UE + UB = ε₀E₀² cos²(ωt−kz)
〈U〉= ε₀E₀²〈cos²(ωt−kz)〉
= 1/2 ε₀E₀²
= ε₀E²ᵣₘₛ
where Eᵣₘₛ = E₀/√2’
Time average of Poynting vector for a plane wave in vacuum
E = E₀x̂ cos(ωt−kz)
B = B₀ŷ cos(ωt−kz)
S = 1/µ₀ E×B
〈S〉= 1/µ₀ (E₀B₀) (1/2)
where factor 1/2 comes from time average of the cos² term.
〈S〉= 1/2µ₀ √µ₀ε₀’ E₀²
= 1/√µ₀ε₀’ 1/2 ε₀E₀²
= c〈U〉
EM wave with energy density 〈U〉 transfers energy to another location at velocity c.
Derive radiation pressure
For any particle
E² = p²c² + m₀²c⁴
For photons m₀ = 0, therefore E = pc.
p = E/c
Consider a volume of space containing an electromagnetic wave (that has an energy density), and define a momentum density, Pₐ.
Magnitude of momentum density:
|Pₐ| =〈U〉/ c
=〈S〉/ c²
Define radiation pressure from
Pressure = Force/Area
Volume of electromagnetic field that passes through area A in time ∆t is
V = Ac∆t
Impulse is
pₜₒₜ = PₐAc∆t = F ∆t
F = PₐAc
Radiation pressure
Pᵣ = F/A = Pₐc =〈U〉
Radiation pressure is equal to the energy density
Summarise the relationship between radiation pressure Pᵣ and the electromagnetic field quantities
Pᵣ =〈U〉
=|〈S〉|/c
= 1/µ₀c |〈E × B〉|
Why does a stationary charge not radiate
There is no B-field component (B=0)
Why does a charge with constant velocity not radiate
Lorentz transform the field
Direction of motion, field lines compressed by factor of 1/γ
E(r) = q/4πε₀ 1−β²/(1−β²sin²θ)³’² r̂/|r|²
where θ = 0 is along the direction of motion.
Eectric flux through dA varies as 1/r² through any θ, just as in the stationary case. Therefore no radiation, as it requires E and B to both vary as 1/r.
Can also transform to the rest frame of the charge, where it is stationary and therefore not emitting photons.
Equations of motion for scalar and vector potentials
∇²Φ − 1/c² ∂²Φ/∂t² = −ρ/ε₀
∇²A − 1/c² ∂²A/∂t² = −μ₀J
These are wave equations with a source term involving ρ and J
Static solutions for scalar and vector potentials
Time variation is zero (∂/∂t = 0)
∇²Φ = −ρ/ε₀
Poisson’s equation with general solution
Φ(r) = 1/4πε₀ ∫ᵥ ρ(r′)/|r−r′| dV′
Similarly, for magnetic vector potential A
A(r) = μ₀/4π ∫ᵥ J(r′)/|r−r′| dV′
What is delayed time
Time factoring in the time it takes information to travel.
τ = t − |r−r′|/c
Second term is time taken for information to travel from r′ to r.
Potentials in the time dependent case
Delayed potentials
Φ(r, t) = 1/4πε₀ ∫ᵥ ρ(r′,τ)/|r−r′|dV′
A(r, t) = μ₀/4π ∫ᵥ J(r′,τ)/|r−r′|dV′
They satisfy the wave equations and the Lorenz Gauge.
Field from an accelerated point charge
Point charge initially at rest, subjected to a short period of acceleration ∆t, after which time is moving at a constant velocity u ≪ c.
u = a∆t
Will now have three regions.
* Close to the charge, observer sees new location and speed, field lines emanate radially outwards.
* Far from the charge, observer at distance r still sees still sees field at the original stationary location; a time r/c has not yet elapsed.
* In between there is a boundary which moves away from the charge’s location at c.
In free space, ∇ · E = 0, so there can be no discontinuities in the field lines. Moving boundary must appear as a kink in the electric field lines. (Fig 2.3, page 31)
u ≪ c, can say electric field lines are approximately parallel inside and outside the kink.
Observer looking at time t at angle θ to the final motion of the charge.
Charge has final location ut, observer at θ sees components u⊥t and u‖t. (Fig 2.4 pg 32)
Related to kink,
E⊥/E‖ = u⊥t/c∆t
Additionally,
u⊥ = a⊥∆t
Can state
E⊥ = E‖ a⊥t/c = E‖ a⊥r/c²
since r = ct
Already know that
E‖ = 1/4πε₀ q/r²
same field as if charge were stationary.
E⊥ = 1/4πε₀ qa⊥/c²r
Magnitude of electric field in the kink region is E⊥ ∝ 1/r.
Notice that electric field at time t depends on the motion of the charge at earlier time
τ = t − r/c
depending on observation distance r.
At large distances, plane wave emitted such
that B is perpendicular to E. Using Faraday’s law and the fact that B⊥ = E⊥/c:
B = 1/c r̂ × E
Combining E and B, Poynting vector
S = 1/μ₀ E × B
points radially outwards from the accelerated charge along r̂, with magnitude
S ∝ EB ∝ 1/r²
Radiation pattern from a displaced point charge
Based on observation angle θ
a⊥ = a sinθ
Magnitude of the electric field at r
|E(r, t)| = 1/4πε₀ q|a(t−r/c)|sinθ/c²r
Poynting flux (power flow)
|S(r, t)| = 1/16π²ε₀ q²|a²(t−r/c)|sin²θ/c³r²
S ∝ 1/r², as it should be to satisfy conservation of energy.
Note if observation angle goes to zero, aka looking along direction of motion,
sin θ → 0 and we don’t see any radiation.
Total Radiated Power from an accelerating charge
Integrate S over all angles.
P(t) = ∮S · dA = ∫2π→0 ∫π→0 S dA
No variation over φ, so integration gives 2π
P(t) = 2π ∫π→0 Sr² sinθ dθ
= 2π ∫π→0 (1/16π²ε₀c³ q²a²(t−r/c)/r² sin²θ) r²sinθ dθ
= q²a²(t−r/c)/8πε₀c³ ∫π→0 sin³θ dθ
Use trig identity for integral of sin³θ:
∫π→0 sin³θ dθ
= ∫π→0 sinθ (1−cos²θ) dθ
= [−cosθ + 1/3 cos³θ]π→0
= 2/3 + 2/3
= 4/3
Instantaneous total power emitted by an accelerated charge
P(t) = q²a²(t−r/c)/6πε₀c³
This is Larmor’s formula
What is a Hertzian dipole
An oscillating current element, the simplest of all antennas.
Fields from a Hertzian dipole
Two charges separated by distance l oscillating in z axis
Charges move up and down along z, approximate as charge per unit length μ across length l.
qa(t′) = μa(t′)l
= μ dv/dt′ l
= d(μv)/dt′ l
μv is the current associated with the oscillatory movement of the charges, and can be written in the form
I(t′) = I₀ cos(ωt)
where I₀ is the amplitude and ω is the angular frequency
Rewrite qa(t′) using the time derivative of oscillatory current
qa(t′) = −I₀lω sin(ωt′)
= −I₀lω sin(ω(t−r/c))
= −I₀lω sin(ωt−kr))
Recall and substitute
E⊥(r,t) = −1/4πε₀ qa⊥(t’)/rc²
= +1/4πε₀ I₀lω sin(ωt−kr))/rc²
or
E⊥(r,t) = I₀lsin(θ)/4πε₀rc² ωsin(ωt−kr)
Remember B = E/c
B(r,t) = I₀lsin(θ)/4πε₀rc³ ωsin(ωt−kr)
Power in a Hertzian dipole
P(t) = q²a²(t′)/6πε₀c³
qa(t′) = −I₀lω sin(ωt−kr)).
P(t) = [ (lω)²/6πε₀c³ ] I₀² sin²(ωt−kr)
where [] term has units of resistance
Ways of writing radiation resistance
Rᵣₐₔ = l²ω²/6πε₀c³
Using ω = 2πc/λ and c² = 1/(μ₀ε₀), can rewrite as:
Rᵣₐₔ = 2π/3 1/ε₀c (l/λ)²
= 2π/3 μ₀c (l/λ)²
= 2π/3 Z₀ (l/λ)²
Z₀ = μ₀c = 1/ε₀c ≃ 377ohm
is the free-space impedance
Z₀ is often given approximate form:
2πZ₀/3 ≃ 80π²
so radiation resistance may be written
Rᵣₐₔ ≃ 80π² (l/λ)²
Current and dipole moment for an oscillating dipole when treated as charge deposition and removal
Two points separated by small distance l compared to emitted wavelength λ.
Charge on each end
q = ±q₀ sinωt
Current flowing on or off the two ends
I = dq/dt = q₀ω cosωt
I₀ = q₀ω, oscillating current is equivalent to an oscillating dipole moment.
Dipole moment is p₀ = q₀l
p = p₀ sinωt = q₀l sin ωt
I₀l = p₀ω
I₀lω = p₀ω²
Time averaged power in a Hertzian dipole
q = ±q₀ sin ωt,
q₀ = p₀/l
I₀l = p₀ω
I₀lω = p₀ω²
Larmor’s formula
P(t) = q²a²/6πε₀c³
Substitute in to get
〈P〉= I₀²l²ω²/12πε₀c³ (‘current picture’)
OR
〈P〉= p₀²ω⁴/12πε₀c³ (‘dipole picture’)
Properties of the Hertzian dipole
- Non-relativistic charge motion
- Source size l ≪ λ where λ is the emitted wavelength
- Observation distance r ≫ λ (i.e. kr ≫ 1)
- Emitted power P ∝ I₀² and P ∝ ω⁴
- Frequency of emitted radiation is the same as the frequency of the charge motion. Emits monochromatic (single wavelength) radiation.
- Emits electromagnetic radiation polarised in the direction of the dipole.
Simple Half-Wave Antenna model
Two rods pointing away from each other attatched in centre to AC source. Current drops linearly to zero at the end of each rod.
Radiation pattern is identical to that of an elementary current element. Product Il is halved, so for same current I, practical antenna of length l radiates one-quarter of power that would be radiated with single current element.
Radiation resistance is
Rᵣₐₔ ≃ 20π² (l/λ)²
Formula holds strictly for very short antennas. Good approximation when l is about a quarter of the radiation wavelength.
More complex Half-Wave Antenna model
Take current as almost sinusoidal, zero at ends and max in centre.
I(z) = I₀ sin(mπ (|z|/l + 1/2))
where m is integer number of half-wavelengths contained in l, m=1 for a
half-wave antenna.
Hertzian dipole electric field expression
dE⊥(r, t) = I₀dlsinθ/4πε₀rc² ωsin(ωt−kr)
Integrate over length of the antenna
E⊥(r,t) = ∫±l/2 I(z)sinθ/4πε₀rc² ωsin(ωt−kr)dz
Include time and phase lag
Time lag is ∆t = zcosθ/c
Phase lag is kzcosθ (ω∆t = kc zcosθ/c)
Assume observer sufficiently far and two paths are parallel, r ∼ R−zcosθ.
Integral becomes
E(R,t) ≈ ωsinθ/4πε₀Rc² ∫±l/2 I(z)sin(ωt−kR+kzcosθ)dz
with solution
E(R,t) = I₀l/4πε₀Rc² [2/π cos(π/2 cosθ)/sinθ] ωsin(ωt−kR)
where l = λ/2
Power radiated from a half-wave antenna
E(R,t) = I₀l/4πε₀Rc² [2/π cos(π/2 cosθ)/sinθ] ωsin(ωt−kR)
From Poynting vector somehow get
〈S〉= 1/μ₀ (I₀²l²/16π²ε₀²R²c⁵) 4/π² cos²(π/2 cosθ)/sin²θ ω² 1/2
Substitute l = λ/2 and ω = 2πc/λ
〈S〉= I₀²/8π²ε₀²μ₀R²c³ cos²(π/2 cosθ)/sin²θ
is the average value of the flux density radiated from the antenna
Integral of cos²(π/2 cosθ)/sinθ from π to 0 with respect to θ
1.22
Average half-wave antenna power
〈P(R, t)〉= ∫ᴬ 〈S〉dA
= I₀²/8π²ε₀²μ₀c³ ∫π→0 ∫2π→0 cos²(π/2 cosθ)/sin²θ sinθdθdφ
φ integral is 2π
θ integral is 1.22
〈Pᵣₐₔ〉≃ 1.22 I₀²/4πε₀c ≃ 0.194 Z₀Iᵣₘₛ²
where Z₀ = 1/(ε₀c) and Iᵣₘₛ²= I₀²/2
Half-wave antenna radiation pattern
Simple Hertzian dipole
〈S〉∝ sin²θ
Realistic half-wave antenna
〈S〉∝ cos²(π/2 cosθ)/sin²θ
Half-wave antenna has a more directional output. (Fig 2.9, pg 42)
Can extend this basic idea and use interference to make antennas which are even more directional, such as an antenna array.
Conductors vs insulators
Conductors contain an ‘unlimited’ charge supply that is free to move.
Insulators/dielectrics haave charge attached to specific atoms and molecules.
Bound and surface charges of a dielectric
P ≡ Dipole moment per unit volume (polarisation)
Bound
ρb = −∇ · P (Gauss’s law for P)
Surface
σb = P · n̂
Free charges are any charges that are not due to the polarisation, ρf
Electric displacement from Gauss’s law
Total charge density
ρ = ρb + ρf
Gauss’s law,
ε₀∇ · E = ρ = ρb + ρf = −∇ · P + ρf
∇ · (ε₀E + P) = ρf
∇ · D = ρf
Electric displacement
D = (ε₀E + P)
Polarisaton and displacement vectors in linear dielectrics
Polarisation proportional to E for linear dielectrics.
P = ε₀χₑE
where χₑ is the electric susceptibility of the material.
D = ε₀E + P
= ε₀E + ε₀χₑE
= ε₀(1 + χₑ)E
where
ε = ε₀(1 + χₑ)
is the electric permittivity of the material.
ε/ε₀ = 1 + χₑ
is the relative permittivity or dielectric constant.
Three types of magnetisation
- Paramagnetism:
Magnetisation is parallel to B - Diamagnetism:
Magnetisation is opposite to B - Ferromagnetism:
Magnetisation is retained even after B is removed
Bound and surface currents of a dipole
M ≡ Magnetic dipole moment per unit volume
Bound
Jb = −∇ × M
Surface
Kb = M × n̂
Free currents not contributing to dipole moment
Jf
Magnetic (H) field from Ampere’s law
Total charge density
J = Jb + Jf
Ampere’s law,
1/μ₀ (∇ × B) = J = Jb + Jf = Jf + ∇ × M
∇ × (1/μ₀B − M) = Jf
∇ × H = Jf
Magnetic field
H = (1/μ₀B − M)
Relation between magnetic field and dipole moment
M = χₘH
where χₘ is the magnetic susceptibility.
B = μ₀(H + M) = μ₀(1 + χₘ)H
H = 1/μ₀μᵣ B
μᵣ = (1 + χₘ)
μ ≡ μ₀μᵣ = μ₀(1 + χₘ)
where μ₀ is the magnetic permeability of free space and μᵣ is the relative permeability.
Maxwell’s equations in media
∇ · D = ρf
∇ · B = 0
∇ × E = − ∂B/∂t
∇ × H = ∂D/∂t + Jf
Complex dispersion relation
Plane wave travelling in Z direction
E = (Eₓ, 0, 0)
B = (0, Bᵧ, 0)
Eₓ = E₀exp(i(ωt−kz))
Bᵧ = B₀exp(i(ωt−kz))
Use Faraday’s law
−ikE₀exp(i(ωt−kz)) = −iωB₀exp(i(ωt−kz))
B₀/E₀ = k/ω
Using Ampere’s and Ohm’s laws
−ikB₀exp(i(ωt−kz)) = −(iωεᵣε₀μᵣμ₀ + μᵣμ₀σ) E₀exp(i(ωt−kz)).
Combine
k² = ω²εᵣε₀μᵣμ₀ − iωμᵣμ₀σ
Maxwell’s equations in a conducting medium
All remain the same apart from Ampere’s law.
∇ × H = ∂D/∂t + Jf
Substitute Jf = σE = iωεE + σE
∇ × H = (iωε + σ)E
= iω(ε + σ/iω)E
= iω(ε − iσ/ω)E
= iωε꜀E
where ε꜀ is the electric permittivity of the conducting medium.
ε꜀ = ε − i σ/ω = ε′ − iε′′
σ = ωε′′
where ε′ and ε′′ are functions of frequency.
Can also write
μ = μ′ − iμ′′
Ampere’s law becomes
∇ × H = iωε꜀E
All Maxwell equations for nonconducting media will apply to conducting media if ε is replaced by ε꜀ (complex permittivity).
Maxwells equations for plane waves in a source-free medium
∇ × E = −iωμH
∇ × H = iωεE
∇ · E = 0
∇ · B = 0
where μ = μ₀μᵣ and ε = ε₀εᵣ
Helmholtz equation for electromagnetism
Wave equation in source-free medium
Take curl of Faradays law
∇ × E = −iωμH
∇ × ∇ × E = −iωμ(∇ × H)
∇(∇ · E) − ∇²E = −iωμ(iωεE)
First term is zero for source free medium.
−∇²E = ω²εμE
Using
εμ = 1/c² and ω²/c² = k²
∇²E + k²E = 0
This is the Helmholtz equation for electromagnetism.
For conductors, wave number k will be complex, k꜀
k꜀ = ω√με’
ε꜀ = ε − i σ/ω
Propagation constant
γ = ik꜀ = iω√με꜀’ (m⁻¹)
in the general form,
γ = α + iβ
Time-varying fields in media
∇²E + k꜀²E = 0
Replacing k꜀ = −iγ
∇²E − γ²E = 0
Differential equation with general solutions of form E ∝ E₀exp(−γz ).
E = E₀exp(i(ωt−kz))
= E₀exp(i(ωt+iγ))
= E₀exp(i(ωt+i(α+iβ)z))
= E₀exp(i(ωt+iαz−βz))
= E₀ exp(iωt) exp(−αz) exp(−iβz)
- exp(iωt) is time dependency of the waves
- exp(−αz) is exponential decay of wave amplitude
- exp(−iβz) is a phase factor
Phase velocity and refractive index in a dielectric
Insulator with μᵣ ≃ 1 and σ = 0
v = ω/k = 1/√εᵣε₀μᵣμ₀’ = c/√εᵣ’
n = c/v = √εᵣ’
Types of dispersion in a dielectric
n(ω) = ck(ω)/ω = √εᵣ(ω)’
In the n(ω) vs ω spectrum we see
* Normal dispersion, as ω increases, so does n
* Anomalous dispersion, as ω increases, n decreases. Absorbtion occurs.
Anomalous dispersion facts
- Many materials have anomalous dispersion in the infrared region of the spectrum
- Some materials have anomalous dispersion in the visible region of the spectrum; these are materials that we see having colour.
- Most materials have anomalous dispersion in the ultraviolet region of the spectrum
- At f > 10¹⁶Hz, n ≃ 1 for nearly all materials
The last implies:
* X-rays readily penetrate most materials with little absorption
* A constant refractive index for X-rays means one cannot make a lens for X-rays
* Gamma rays (f > 10¹⁹Hz) do not refract through any dielectric (n = 1)
How does the oscillator model for dielectrics explain different types of dispersion
Equation of motion for each electron is
ẍ + αẋ + ω₀²x = q/m [E + ẋB]
where α is the damping constant,
ω₀ = √k/m’ is the characteristic angular frequency of electron oscillations within the dipole
and k is the spring constant.
General solution
x = (q/m)E / (ω₀²−ω²)+iαω
Can show that electric susceptibility is complex
χₑ = χᵣ + iχᵢ
Because μᵣ ≡ 1 for a dielectric,
n = √εᵣ
εᵣ = 1 + χₑ
n = √1 + χₑ’
Taylor expand
n = √1 + χᵣ − iχᵢ’
≃ 1 + 1/2 χᵣ − 1/2 iχᵢ
= nᵣ − inᵢ
nᵣ = 1 + χᵣ/2
nᵢ = χᵢ/2
{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}
- nᵢ represents damping that reaches a maximum when ω = ω₀ᵢ
- around frequency range defined by bandwidth Γ, refractive indices undergo violent changes
- rays with shorter wavelength are refracted less (anomalous dispersion)
- outside Γ, dispersion behaviour is normal and shorter wavelengths are refracted more than longer wavelengths (normal dispersion),
Electric susceptibility of a gas
χₑ ≪ 1
Microscopic field in a dense material
Small sphere in the vicinity of one atom of radius R, total field is due to the field generated by the atom itself, Eₐₜₒₘ, plus background field Eₗₒ꜀ₐₗ created at that point by all other atoms
Total field is
E = Eₗₒ꜀ₐₗ + Eₐₜₒₘ
Number density of atoms is N
4/3 πR³ = 1/N
Background field causes dipole moment
p = αEₗₒ꜀ₐₗ
where α is the polarizability of the atom.
Eₐₜₒₘ = −1/4πε₀R³ p
Total field is
E = Eₗₒ꜀ₐₗ − 1/4πε₀R³ p
= (1 − α/4πε₀R³) Eₗₒ꜀ₐₗ
Clausius-Mossotti Relation
α = 3ε₀/N εᵣ−1/εᵣ+2
Lorentz-Lorenz Equation
n²−1/n²+2 = Nα/3ε₀
Current density in a conductor
J = σE
Dispersion relation in a conductor
k꜀² = εᵣμᵣ ω²/c² (1 − iσ/ωεᵣε₀)
What do imaginary parts of wavenumber represent?
Absorbtion
Condition for a poor conductor
σ/ωεᵣε₀ ≪ 1
Dispersion relation for a poor conductor
σ/ωεᵣε₀ ≪ 1 for poor conductors
Can use Taylor expansion √1+x’ ≃ 1+x/2 in the dispersion equation
k꜀ = k₀ √εᵣμᵣ’ √(1 − iσ/ωεᵣε₀)’
where k₀ = ω/c
k꜀ ≃ ω/c √εᵣμᵣ’ (1 − iσ/2ωεᵣε₀)
Characteristic damping/decay length in a poor conductor (skin depth)
δ = 1/|kᵢ| = 2/σ √εᵣε₀/μᵣμ₀’
Ratio between one wavelength and the skin depth for a poor conductor
δ/λ = ωεᵣε₀/πσ
Poor conductor, σ/ωεᵣε₀ ≪ 1, so δ/λ ≫ 1
Therefore, the electromagnetic field
can penetrate a poor conductor many wavelengths before it completely decays.
Condition for a good conductor
σ/ωεᵣε₀ ≫ 1
Dispersion relation in a good conductor
σ/ωεᵣε₀ ≫ 1
k is dominated by the imaginary term in the dispersion relation
k = ω/c √εᵣμᵣ’ (−i σ/ωεᵣε₀)¹’²
= √μᵣμ₀ε₀σω/ε₀’ (exp(−iπ/2))¹’²
= √μᵣμ₀σω’ exp(−iπ/4)
Can rewrite
exp(−iπ/4) = 1/√2’ (1−i)
k = √μᵣμ₀σω/2’ (1−i)
Damping/decay length in a good conductor
δ = 1/|kᵢ| = √2/μᵣμ₀σω’
Oscillatory wavelength in a conductor
λ = 2π/kᵣ = 2πδ
How deep can a wave penetrate into a good conductor?
I ≈ exp(−x/δ)
δ = √2/μᵣμ₀σω’
if x = λ
I ≈ exp(−2π )
After one wavelength, amplitude has dropped to exp(−2π) ≃ 0.0018 of its initial value.
E and B fields in a conductor
E = E₀exp(i(ωt−kz))
B = B₀exp(i(ωt−kz))
∇ × E = − ∂B/∂t
k × E₀ = ωB₀
Rewriting,
B₀ = k/ω × E₀
- In a dielectric, k is real valued, B₀ ⊥ E₀ ⊥ k, electric and magnetic fields oscillate in phase since kᵢ = 0
- In a conductor k is complex, B₀ is not perpendicular to E₀ and k, and there is a phase difference φ.
tan φ = kᵢ/kᵣ
B/E ≃ k/ω = √μᵣμ₀σω/2ω²’ (1−i)
= √μᵣμ₀σ/2ω’ (1−i)
= 1/c √μᵣσ/2ε₀ω’ (1−i)
Factor 1/√2’ (1−i) = exp(−iπ/4), so 45◦ phase shift between E and B.
Note that, given μᵣ ≃ 1 and σ ≫ ω,
B ≫ E/c
Magnetic field component of an EM wave in a conductor is much larger than it would be in either a vacuum (B = E/c), or in a dielectric (B = nE/c for refractive index n)
What is the Drude Model
Applies to conductors.
Conduction electrons move due to the applied field E. If they were completely free to move, they would keep accelerating. Doesn’t happen because they lose forward momentum due to collisions with lattice ions. Collisions result in friction proportional to their velocity.
Define frequency of collisions as γ, the number of collisions per second.
Electrons are free to move (unbound), so we write an equation of motion without any restoring force:
mẍ + mγẋ = qE(t)
This is the Drude Model.
Properties of plasmas
- A plasma is a gas of free ions and electrons, with equal number density such that overall the plasma is neutral.
- Electrons are much lighter than the protons or ions.
- Under an externally-applied electric field, the electrons will move much more than the positive ions will. We can therefore treat a plasma in terms of the electron motion only.
- We can treat a plasma the same way as conductors. We’ll assume the positive charges are stationary in both cases.
Equation of motion for electrons in a plasma
−eE = m d²x/dt² = −mω²x
where the restoring force between electron and ion is
m d²x/dt² = −kx
and
1/4πε₀ = ω²m using ω = √k/m’
Polarization vector in a plasma
Displacement x between an electron and ion gives rise to electric dipole, p
x = e/mω² E
p = −ex
N electrons per unit volume, polarisation vector is
P = Np = − Ne²/mω² E
ignoring effects between induced dipole moments.
Displacement vector of a plasma
P = Np = − Ne²/mω² E
D = ε₀E + P = ε₀ (1 − Ne²/mε₀ω²) E
= ε₀ (1 − ω²ₚ/ω²) E
where (1 − ω²ₚ/ω²) is the equivalent permittivity of plasma.
Plasma angular frequency
ωₚ = √Ne²/mε₀’
Plasma frequency
fₚ = ωₚ/2π
= 1/2π √Ne²/mε₀’ (Hz)
Depends only on plasma density. Signals below fₚ cannot penetrate.
Frequency and time associated with plasma collisions
γ꜀ = 1/τ꜀
Drift velocity in a plasma
vdrift = aτ꜀ = −eE/m 1/γ꜀
DC conductivity in plasmas
J = σ₀E = −Ne vdrift
DC conductivity
σ₀ = Ne²/m 1/γ꜀
Relates the macroscopic quantity, conductivity, σ₀ and a microscopic quantity, collision frequency, γ꜀.
Recall plasma frequency is given by ωₚ² = Ne²/mε₀, leading to
σ₀ = ε₀ωₚ²/γ꜀
Frequency-dependent conductivity of a plasma, σ(ω)
σ(ω) = σ₀/1+(iω/γ꜀)
Derive electric permittivity of a plasma
P = ε₀χₑE = ε₀χₑE₀exp(iωt)
where we know
P = Np = N(−e)x
Substitute x from the solution of Drude’s model
P = (Ne²/m) 1/−ω²+iγ꜀ω E = ε₀χₑE
resulting in the electric susceptibility
χₑ = (Ne²/ε₀m) 1/−ω²+iγ꜀ω
and the relative permittivity
εᵣ = 1 + χₑ = 1 + ωₚ²/−ω²+iγ꜀ω
This reduces back to collisionless permittivity when γ꜀ = 0.
Probability a given electron in a plasma will experience a collision at time t
P(t) = 1/τ꜀ exp(−t/τ꜀)
Condition foor low frequency in plasmas, and their skin depth
ω ≪ γ꜀
Can ignore ω² term in permittivity.
Also need to remember that it is a good conductor, so σ₀ ≫ 1
End up with
εᵣ ≃ −iσ₀/ωε₀
δ = √2/μ₀σ₀ω’
Condition for high frequency waves in plasma
ω ≫ γ꜀
E changes sign more quickly than a collision can occur, situation describes a collisionless plasma.
Dispersion relation in a high frequency plasma
ω ≫ γ꜀
Collisionless plasma
εᵣ ≃ 1 − ωₚ²/ω²
εᵣ is real-valued, so no absorption of energy.
εᵣ = n² = c²/vₚ² = k²c²/ω² = 1 − ωₚ²/ω²
ω² = k²c² + ωₚ²
k = ± 1/c (ω²−ωₚ²)¹’²
Conventional radio bands
- LW (Long Wave): f < 300kHz, λ > 1000m
- MW (Medium Wave): f ∼ 500 − 1600kHz, λ ∼ 300m
- SW (Short Wave): f ∼ 1.6 − 30MHz, λ ∼ 90 − 10m
- VHF (Very High Frequency):f ∼ 30 − 300MHz, λ ∼ 10 − 1 m; television transmissions
- UHF (Ultra High Frequency):f ∼ 300MHz−3GHz, λ ∼ 1 − 0.1m
- L band:1-2 GHz (mobile phones, GPS)
- S band: 2-4 GHz (mobile phones, radar, microwave ovens, particle accelerators)
- C band: 4-8 GHz (satellite communication)
- X band: 8-12 GHz (high-resolution radar, deep-space communications)
- K band: 18-26 GHz
Propagation of E component transverse to the interface
Consider path C in and out of boundary (Fig 4.2, pg 77).
Integral form of Faraday’s law
∮꜀ E · dl = −∂/∂t ∮꜀ B · dS
Integrate along C anticlockwise
∮꜀ E · dl = Eₜ₂∆w − Eₜ₁∆w + Eₙ₁∆h/2 + Eₙ₂∆h/2 − Eₙ₂∆h/2 − Eₙ₁∆h/2
= −∂B/∂t ∆h∆w
Across the interface,
Eₜ₂ − Eₜ₁ = −∂B/∂t ∆h
where ∆h → 0
Transverse component of E is continuous across a boundary.
Can also express in terms of the displacement vector
Dₜ₁/ε₁ = Dₜ₂/ε₂
where ε₁ = ε₀εᵣ₁ and ε₂ = ε₀εᵣ₂
Propagation of E component normal to an interface
Closed cylindrical surface S across the boundary (Fig 4.3, pg 78) such that
aₙ₁ = −aₙ₂
are the normal unit vectors for the top and bottom faces
Gauss’ law
∮ₛ D · ds = Q
aₙ₂ · (Dₙ₁ − Dₙ₂ )∆S = ρₛ ∆S
aₙ₂ · (Dₙ₁ − Dₙ₂ ) = ρₛ
where ρₛ is the surface charge density at the interface.
The normal component of E or D is discontinuous across a boundary.
Propagation of B component transverse to an interface
Ampere’s law
∮꜀ H · dl = ∮ₛ J · dS + ∂/∂t ∮ₛD · dS
where J is the surface current density.
Hₙ₁∆h/2 + Hₙ₂∆h/2 + Hₜ₂∆w − Hₙ₂∆h/2 − Hₙ₁∆h/2 − Hₜ₁∆w
= J_z∆h∆w + ∂D/∂t ∆h∆w
Hₜ₂ − Hₜ₁ = J_z ∆h (+ ∂D/∂t∆h → 0)
where J_z is the surface current directed out of the page.
Second term goes to zero because ∆h → 0 and in a good conductor of large finite conductivity oscillating current is confined to a thin layer at the metal-vacuum boundary.
Layer thickness (depth of penetration) decreases as conductivity and frequency increases. Depth of current (skin depth) goes to zero as the conductivity goes to infinity. This is the sheet current.
δ = √2/μᵣμ₀σω’
Hₜ₂ − Hₜ₁ = J_z∆h
Surface unit vectors are
aₙ₁ = −aₙ₂
aₙ₂ × (Hₜ₂ − Hₜ₁ ) = J_z
Transverse component of B is discontinued across a boundary.
Propagation of B component normal to the interface
Gauss’s law
∮ₛ B · dS = 0
Bₙ₁ − Bₙ₂ = 0
where Bₙ₁ = B₁ · aₙ₁
Normal component of the magnetic field is continuous.
Summarise the boundary conditions at the surface of a conductor
n̂ · E = ρₛ/ε₀ → normal discontinuous
n̂ × E = 0 → transverse continuous
n̂ · B = 0 → normal continuous
n̂ × B = μ₀J_z → transverse discontinuous
Wave impedance in general, and in vacuum
Z ≡ μ₀μᵣω/k
For vacuum, ω/k = c and μᵣ = 1
Z₀ ₍ᵥₐ꜀ᵤᵤₘ₎ = μ₀c = √μ₀/ε₀ = 377 Ω
Form of waves during reflection at normal incidence in a dielectric
Eᵢ = Eᵢ₀exp(i(ωt−kz))
Bᵢ = Bᵢ₀exp(i(ωt−kz))
Eᵣ = −Eᵣ₀exp(i(ωt+kz))
Bᵣ = Bᵣ₀exp(i(ωt+kz))
Eₜ = Eₜ₀exp(i(ωt−kz))
Bₜ = Bₜ₀exp(i(ωt−kz))
k changes sign for reflected field components.Must change sign of either Eᵣ or Bᵣ so reflected wave propagates in the correct direction (Poynting vector in the right
direction). Here, we chose to change the sign of Eᵣ
Reflected and transmitted field amplitudes, normal reflection in a dielectric
₀ means ‘at boundary’
* Eₜ₁ = Eₜ₂, so
Eᵢ₀ − Eᵣ₀ = Eₜ₀
* Hₜ₁ = Hₜ₂ (no current at the boundary), so
Bᵢ₀/μ₁ + Bᵣ₀/μ₁ = Bₜ₀/μ₂
* We know B = E/v = E√ε₀εᵣμ₀μᵣ’ so
B/μᵣ = E√ε₀εᵣμ₀μᵣ’ 1/μᵣ = 1/c √εᵣ/μᵣ’ E
* Substitute in
√ε₁/μ₁’ Eᵢ₀ + √ε₁/μ₁’ Eᵣ₀ = √ε₂/μ₂’ Et₀
* No magnetic properties in dielectrics,
μ₁ = μ₂ = 1
* Take √ε₁’ = n₁ and √ε₂’ = n₂, and use boundary condition Eᵢ₀ − Eᵣ₀ = Eₜ₀
n₁(Eᵢ₀ + Eᵣ₀) = n₂Eₜ₀
= n₂(Eᵢ₀ − Eᵣ₀)
* Reflected electric wave amplitude
Eᵣ₀/Eᵢ₀ = n₂−n₁/n₂+n₁
* For magnetic field, use boundary conditions, Bᵢ₀ + Bᵣ₀ = Bₜ₀, Eᵢ₀ + Eᵣ₀ = Eₜ₀, as well as the general dielectric relation B = 1/c √ε’ E * Reflected magnetic field amplitude
Bᵣ₀/Bᵢ₀ = n₂−n₁/n₂+n₁
* Transmitted fraction of wave amplitudes, use
Bᵣ₀/Bᵢ₀ + Bₜ₀/Bᵢ₀ = 1 and
Eᵣ₀/Eᵢ₀ + Eₜ₀/Eᵢ₀ = 1
get
Bₜ₀/Bᵢ₀ = Eₜ₀/Eᵢ₀ = 2n₁/n₂ + n₁
Do NOT confuse reflected and transmitted field amplitudes with intensity.
Intensity of transmitted and reflected waves, normal to a dielectric
〈Sᵢ〉= 1/μ₀〈Eᵢ × Bᵢ〉
where Eᵢ = Eᵢ₀ cos(ωt−kz)ẑ
and Bᵢ = Bᵢ₀ cos(ωt−kz)ẑ
B = 1/c √εᵣ’ E
〈Sᵢ〉= ε₀c/2 Eᵢ₀² n₁
where 1/2 is average of cos² and n₁ = √ε₁’
Can write Poynting vector using coefficients calculated for the field amplitudes,
〈Sᵣ〉= ε₀c/2 Eᵢ₀² (n₂−n₁)²/(n₂+n₁)² n₁
〈Sₜ〉= ε₀c/2 Eᵢ₀² 4n₁²/(n₂+n₁)² n₂
Reflected and transmitted portions of intensity are:
* T =〈Sₜ〉/〈Sᵢ〉
= (4n₁²/(n₂+n₁)² n₂) / n₁
= 4n₁n₂/(n₂+n₁)²
* R =〈Sᵣ〉/〈Sᵢ〉
= ((n₂−n₁)²/(n₂+n₁)² n₁) / n₁
= (n₂−n₁)²/(n₂+n₁)²
- R + T = 1
Phase changes at boundaries
- If n₂ > n₁ (e.g. light incident normally onto water or glass), there is a phase shift of π on reflection.
- If n₂ < n₁ (e.g. light passing from water to air), there is no phase change.
Refractive index of a conductor
n₂ = √εᵣ’
= √−iσ/ωε₀’
≃ √σ/2ωε₀’ (1−i)
Reflection and transmission coeffiscients for low frequency plasmas
Refractive index
n₂ ≃ √−iσ/ωε₀’
= √σ/2ωε₀’ (1−i)
= α(1−i)
with α = √σ/2ωε₀’ ≫ 1 for good conductors.
R = |n₂−n₁|² / |n₂+n₁|²
=|( (α−n₁)−iα / (α+n₁)−iα )|²
using |a−ib| = √a²+b²’
= (1− n₁/α)²+1 / (1+ n₁/α)²+1
= ( 2 − 2n₁/α + n₁²/α² ) / ( 2 + 2n₁/α + n₁²/α² )
Ignore the last terms as α ≫ 1
R = (1 − n₁/α) (1 + n₁/α)⁻¹
Taylor expand 1−x/1+x ≃ 1−2x+…
≃ 1 − 2n₁/α
In a conductor, ωε₀/σ ≪ 1, and
* R ≃ 1 − 2n₁√2ωε₀/σ’
* T = 1 − R ≃ 2n₁√2ωε₀/σ’
Reflection and transmission for high frequency waves
Dispersion relation in a good conductor
k = ± 1/c √ω²−ωₚ²’
- ω ≪ ωₚ
Evanescent wave, does not penetrate. εᵣ < 0, k imaginary, T = 0 and R = 1 - ω > ωₚ
Metal transparent. εᵣ > 0, k is real, T→1, R→0 at high frequencies.
Derive reflection and refraction laws
Incoming wave kᵢ incident from medium (1) into medium (2) at angle θᵢ to the surface normal.
Using notation √−1’ = j
Eᵢ = Eᵢ₀exp(j(ωt−kᵢ·r)
Eᵣ = Eᵣ₀exp(j(ωt−kᵣ·r)
Eₜ = Eₜ₀exp(j(ωt−kₜ·r)
and
Bᵢ = 1/v k̂ᵢ × Eᵢ
Phases must be equal at the boundary.
[kᵢ · r]|z=0
= [kᵣ · r]|z=0
= [kₜ · r]|z=0
Write each wave in form
kᵢ · r = kᵢₓx + kᵢᵧy + kᵢ₂z
kᵣ · r = kᵣₓx + kᵣᵧy + kᵣ₂z
kₜ · r = kₜₓx + kₜᵧy + kₜ₂z
At z = 0, gives
kᵢ · r = kᵢₓx + kᵢᵧy
kᵣ · r = kᵣₓx + kᵣᵧy
kₜ · r = kₜₓx + kₜᵧy
Choose axes such that kᵢᵧ = 0 (plane of incidence parallel to x̂).
kᵣᵧ = kₜᵧ = 0 too.
No y components,
kᵢₓx = kᵣₓx = kₜₓx
All components of the wave-vectors parallel to the boundary must be the same size.
kᵢₓ = k₁ sin θᵢ
kᵣₓ = k₁ sin θᵣ
kₜₓ = k₂ sin θₜ
All equal to each other.
From this, get
* Reflection | θᵢ = θᵣ
* Refraction | nᵢ sinθᵢ = nₜ sinθₜ
where we have used k₁ = n₁ω/c and k₂ = n₂ω/c.
S- and P- polarisation
Orthogonal components
* S-polarisation has E is normal to the plane of incidence
* P-polarisation has E is in the plane of incidence
Derive the Fresnel equation for S-polarised light
S-polarised component of E is perpendicular to plane of incidence, so no cosθ terms in continuity equations.
* First boundary condition using E‖ is continuous:
Eᵢ₀ − Eᵣ₀ = Eₜ₀
* Second boundary condition using H‖ is continuous and H‖ = B‖/μᵣ:
Bᵢ₀/μ₁ cosθᵢ − Bᵣ₀/μ₁ cosθᵢ = Bₜ₀/μ₂ cosθt
where μ₁ and μ₂ are relative permeabilities, set to μ₁ = 1 and μ₂ = 1.
Use H = nE/c to find
n₁/c Eᵢ₀ cosθᵢ − n₁/c Eᵣ₀ cosθᵢ = n₂/c Eₜ₀ cos θₜ
Eᵣ/Eᵢ = − (n₁ cosθᵢ − n₂ cosθₜ)/(n₁ cosθᵢ + n₂ cosθₜ)
Using cos θₜ = √1−sin²θₜ’
and Snell’s law sinθₜ = (n₁/n₂) sinθᵢ
get reflected component Fresnel equation
ρ⊥ ≡ Eᵣ₀/Eᵢ₀ |⊥
= (n₁cosθᵢ − n₂cosθₜ) / (n₁cosθᵢ + n₂cosθₜ)
= (cosθᵢ − √n²−sin²θᵢ’) / (cosθᵢ + √n²−sin²θᵢ’)
where n = n₂/n₁
The Fresnel equation for the transmitted component can be deduced by τ⊥ = 1 − ρ⊥.
Derive the Fresnel equation for P-polarised light
Take μ₁ = μ₂ = 1.
* B is perpendicular to plane of incidence, boundary condition is
Bᵢ₀ + Bᵣ₀ = Bₜ₀
Recall B = n E/c
n₁Eᵢ₀ + n₁Eᵣ₀ = n₂Eₜ₀
* Parallel E component is continuous at boundary,
Eᵢ₀ cos θᵢ − Eᵣ₀ cosθᵢ = Eₜ₀ cosθₜ
Eₜ₀ = n₁/n₂ (Eᵢ₀ + Eᵣ₀)
Using cosθₜ = √1−sin²θₜ’
and Snell’s Law
get reflection component Fresnel equation
ρ‖ = Eᵣ₀/Eᵢ₀ |‖
= (n₂cosθᵢ − n₁cosθₜ) / (n₁cosθₜ + n₂cosθᵢ)
= (n²cosθᵢ − √n²−sin²θᵢ’) / (n²cosθᵢ + √n²−sin²θᵢ’)
Only valid for a non-magnetic medium,
μ₁ = μ₂ = 1
All Fresnel equations
ρ⊥ = Eᵣ₀/Eᵢ₀ |⊥
= (n₁cosθᵢ − n₂cosθₜ) / (n₁cosθᵢ + n₂cosθₜ)
τ⊥ = Eₜ₀/Eᵢ₀ |⊥
= (2n₁cosθᵢ) / (n₂cosθₜ + n₁cosθᵢ)
ρ‖ = Eᵣ₀/Eᵢ₀ |‖
= (n₂cosθᵢ − n₁cosθₜ) / (n₁cosθₜ + n₂cosθᵢ)
τ‖ = Eₜ₀/Eᵢ₀ |‖
= (2n₁cosθₜ) / (n₁cosθₜ + n₂cosθᵢ)
Impedance’s relation to field strengths
Z₀ = μ₀c
In free space,
H = E/Z₀
General dielectric medium,
Z = E/H = μᵣμ₀ ω/k = μᵣμ₀/√μᵣμ₀εᵣε₀’
= √μᵣμ₀/εᵣε₀’
If μᵣ = 1,
Z = Z₀/n
What is the Brewster angle
The angle for which only S-polarised light is reflected.
ρ‖ = 0
n⁴cos²θᵢ = n² − sin²θᵢ
= n² − 1 + cos²θᵢ
Satisfied if
cos θᵢ = 1/√1+n²’
aka
tan θᵢ = n
θ_b = tan⁻¹ n
What is total internal reflection?
Wave is moving from a ”dense” to a ”rare” medium (eg glass to air), with
n₁ > n₂
While θ₁ < θ꜀, the refraction angle, θ₂, exceeds the incidence angle, θ₁. Refracted wave bends away from the surface normal.
As θ₁ increases, will reach θ₂ = π/2. This is the citical angle.
n₁sinθ₁ = n₂(sinθ₂=1)
θ₁ ≡ θ꜀ = sin⁻¹(n₂/n₁)
Above this angle, all light reflects, have totl internal reflection.
Field in rarer medium during total internal reflection
n₁sinθ₁ = n₂sinθ₂
⇒ at crit. angle, n₂/n₁ = n = sinθ꜀
Transmitted wave at angle θₜ
Eₜ = Eₜ₀exp(i(ωt−k·r))
Considering 2D wave with x and z components
Eₜ = Eₜ₀ exp(iωt −ikₓx −ik₂z)
= Eₜ₀ exp(iωt −iksinθₜx −ikcosθₜz)
sinθₜ = sinθ₁/(n₂/n₁)
⇒ cos θₜ = √1 − sin²θₜ/n₂’
where n = n₂/n₁
At limit n = sinθ꜀
cos θₜ = √1 − sin²θₜ/sin²θ꜀’
and when θ₁ > θ꜀
⇒ cos θₜ = ±i √sin²θₜ/sin²θ꜀ − 1’
Transmitted wave
Eₜ = Eₜ₀ exp(iωt) exp(−iksinθₜx) exp(−kzQ)
where Q = √sin²θ₁/n² − 1’
- First term due to time-varying nature of the wave
- Second term shows existence of a wave propagating at the interface surface
- Last term shows surface wave amplitude decaying with decay length
δ = 1/k_zQ
= 1/k_z√sin²θ₁/n²’−1
Field confined to immediate neighbourhood of the surface, but exhibits phase progression along the interface and represents a wave propagating along the x direction.
This is a surface/evanescent wave, with phase velocity vₚ.
vₚ = ω/kₓ
= ω/ksinθₜ
= ω/k n₂/n₁sinθᵢ
= c/n₁ 1/sinθᵢ
What is frustrated total internal reflection
Two optically dense media sandwich a thin layer of a rarer medium. If central layer is sufficiently thin, the exponentially decaying evanescent wave still has some amplitude on the other side. Therefore, there will be some transmission across the thin, rare layer.
Simply, what is scattering
Beam of radiation passing through matter, electric field of the wave causes charges (electrons) to move. Any accelerated charge must emit EM radiation.
This process of re-radiation is scattering. As a result of scattering, the forward-directed beam of radiation is weakened, and energy lost from the forward direction is redistributed into other angles. Magnitude of the process is very tiny, but is still a widespread and important phenomenon in nature
What is Thomson scattering
Elastic scattering by a free, charged particle.
Low energy lmit of Compton scattering.
Particle kinetic anergy and photon frequency remain unchanged.
Power reduction and evolution of energy flux for Thomson scattering
Field of a monochromatic wave
Eₓ = E₀ cos(ωt−kz)
Charge vibrating under effect of Eₓ has acceleration
aₓ = q/m E₀cos(ωt)
where we have used the definition of the Lorentz force.
Remember from Hertzian dipoles that accelerating charges radiate in a doughnut-shaped pattern.
The total power radiated is Larmor’s formula
P(t) = q²a²(t′)/6πε₀c³
where t′ = t − r/c is the delayed time.
Substitute acceleration
P(t) = q⁴/6πε₀m²c³ E₀² cos²(ω(t − r/c ))
This is scattered power by one electron.
Time-averaged power is
P(t) = c〈U〉= q⁴E₀²/12πε₀m²c³
Incident radiation flux is Poynting vector
〈S〉 = 1/2 ε₀cE₀²
Introduce scattered power formula in terms of Poynting vector
〈P〉= 8π/3 (q²/4πε₀mc²)²〈Sᵢₙ꜀ᵢₔₑₙₜ〉
Circular area is the Thomson scattering cross-section,
σ = 8π/3 (q²/4πε₀mc²)²
In scattering volume Adz, if density of scatters is N, will have total N Adz scatters, each of scattering cross-section σ.
Total power reduction is
Ad〈S〉= −N Adzσ〈S〉
Evolution of energy flux over length z.
d〈S〉/〈S〉= −Nσdz
integrate
→〈S〉=〈S₀〉exp(−Nσz)
where S₀ is energy flux at entrance of the material
What is Rayleigh scattering
Light passing through gas of neutral atoms, bound electrons are set into vibration and
scattering takes place.
Acceleration experienced by a bound electron is different to that of a free electron and so scattering is different.
Power scattered in Rayleigh scattering
(why is the sky blue)
General EoM for a bound electron
d²x/dt² + β dx/dt + ω₀²x = q/m [E + dx/dt × B]
where x is position vector of the bound electron.
For simplicity, ignore damping constant β (assume ω not in anomalous dispersion, far from resonant frequency of bound electron, ω₀)
Particle velocity is subrelativistic, so can neglect magnetic field component of Lorentz force.
EoM
d²x/dt² + ω₀²x = q/m E₀cos(ωt)
Solved by
x = (q/m)/(ω₀²−ω²) E₀cos(ωt)
a = d²x/dt² = (q/m)ω²/(ω₀²−ω²) E₀cos(ωt)
Substite into Larmor’s formula and time average
〈P〉= q²/12πε₀c³ (q²/m²ω²E₀²)/(ω₀²−ω²)²
Arrange to isolate classical radius of an electron
〈Pₛ꜀ₐₜₜₑᵣₑₔ〉= 8π/3 (q²/4πε₀mc²)² (ω²/ω₀²−ω²)² 〈Sᵢₙ꜀ᵢₔₑₙₜ〉
Scattered power for bound electrons is frequency-dependent.
When ω₀ ≫ ω, power scattered is proportional to the 4th power of frequency. Blue light is scattered much more than red light. Atmospheric gasses have ω₀ ≫ ω, so sky is blue.
Free electron scattering is frequency-independent, so solar corona is white.
What is a transmission line
Two parallel conductors separated by an insulator
What is a waveguide
A hollow tube - made of either conductor or dielectric - whose walls reflect electromagnetic waves. The tube doesn’t need to be circular.
Assumption in a transmission line
l ≫ λ
Derive the Telegrapher’s Equations
Consider single frequency of oscillation of both V and I. For short section δz of transmission line, per unit length
C = C̃ δz
L = L̃ δz
Across inductance there is a change in voltage
∆V = ∂V/∂z δz = −L̃ δz ∂I/∂t
Current leads to charging due to capacitance
− ∂I/∂z δz = ∂/∂t (CV) = ∂/∂t (C̃ δz V)
Rearrange each for Telegrapher’s Equations
∂I/∂z = −C̃ ∂V/∂t
∂V/∂z = −L̃ ∂I/∂t
Wave equations for transmission lines
Differentiate Telegrapher’s equations with respect to z and t, then vice versa.
* ∂²I/∂z² = −C̃ ∂²V/∂z∂t
∂²V/∂z∂t = −L̃ ∂²I/∂t²
⇒ ∂²I/∂z² = L̃C̃ ∂²I/∂t²
* ∂²I/∂z∂t = −C̃ ∂²V/∂t²
∂²V/∂z² = −L̃ ∂²I/∂z∂t
⇒ ∂²V/∂z² = L̃C̃ ∂²V/∂t²
These are wave equations with solutions
V = V₀exp(i(ωt−kz))
I = I₀exp(i(ωt−kz))
where wave velocity is
v = ω/k = 1/√L̃C̃’
and is the same for both the I and V solutions
Impedence of a transmission line
v = ω/k = 1/√L̃C̃’
v is independent of ω, so no dispersion of waves. Zero dispersion, different frequencies propagate at same velocity so any shape of disturbance will keep its shape as it propagates.
Substitute solutions for a single ω into the Telegrapher’s equations:
kV₀ = ωL̃I₀
kI₀ = ωC̃V₀
Divide
V₀/I₀ = L̃/C̃ I₀/V₀
V₀/I₀ = √L̃/C̃’ = Z
Z is characteristic impedance of the transmission line. If L̃ and C̃ are real, Z is real; V and I oscillate together in phase.
Coaxial cable
Two concentric cylindrical conductors separated by a dielectric, such as a central core (e.g. wire) surrounded by a metallic sheath.
Voltage applied between core and sheath produces radial electric field E.
Flowing current up/down the core/sheath produces toroidal magnetic field.
Outside sheath, Gauss’s and Ampere’s Laws tell us there is no E or B
For core of radius a and sheath of radius b, we obtain from fields that
C̃ = 2πεᵣε₀/ln(b/a)
L̃ = μᵣμ₀/2π ln(b/a)
where εᵣ and μᵣ are relative permittivity and permeability of the dielectric between the conductors.
Velocity of waves in the transmission line
v = 1/√L̃C̃’ = 1/√μᵣμ₀εᵣε₀’ = c/√μᵣεᵣ’
Impedance
Z = √L̃/C̃’ = 1/2π ln(b/a) √μᵣμ₀/εᵣε₀’
Wave velocity is reduced to the same value it would have in the dielectric if there were no conductors.
Stripline/Microstrip
Thin metal strip contained within a dielectric and surrounded by two larger conducting boundaries.
Nowadays, most common example is the “microstrip”, a single thin conducting strip laid on top of a dielectric, with a larger conducting “ground plane” on the other side. E.g. printed circuit boards (PCBs).
Assume current in the microstrip (equal and opposite to that in the ground plane) forms uniform magnetic field within the dielectric (assume strip width d ≫ g, where g is the gap between microstrip and ground plane)
Magnetic flux over length l of microstrip is
Φ = BA = μᵣμ₀I/d gl
Inductance per unit length
L̃ = 1/l Φ/I = μᵣμ₀ g/d
Capacitance along length l
C = εᵣε₀ ld/g
giving capacitance per unit length
C̃ = εᵣε₀ d/g
Wave velocity
v = 1/√L̃C̃’ = 1/√μᵣμ₀εᵣε₀’ = c/√μᵣεᵣ’
Same as for coaxial cable.
Wave speed in transmission lines depends only on the material properties of the dielectric; v is independent of geometry.
Impedence
Z = √L̃/C̃’ = g/d √μᵣμ₀/εᵣε₀’
Impedance does does depend on geometry, but may in general be split into a geometry and material factor.
Compare Z for coaxial and microstrip:
Zcoaxial = 1/2π ln(b/a) √μᵣμ₀/εᵣε₀’
Zmicrostrip = g/d √μᵣμ₀/εᵣε₀’
Material factor is
√μᵣμ₀/εᵣε₀’
= Z₀ √μᵣ/εᵣ’
where Z₀ = √μ₀/ε₀’ ≃ 377Ω is impedance of free-space.
Power flow along a transmission line
Microstrip, uniform fields.
Voltage V across dielectric gives uniform electric field
E = − V/g x̂
Current I generates magnetic field
B = −μᵣμ₀ I/d ŷ
E and B fields both transverse to direction of motion. Mode of oscillation is called TEM - transverse electric and magnetic mode.
E ⊥ B ⊥ k
E₀/B₀ = v = c/√μᵣεᵣ’
Power transfer
S = 1/μᵣμ₀ E × B
P = I₀V₀ cos²(ωt−kz)
〈P〉= 1/2 I₀V₀ = 1/2 V₀²/Z = 1/2 I₀² Z
where Z is characteristic impedance of the transmission line
Reflection and transmission in a transmission line
Transmission line with impedance Z, end is attached to a load with impedance ZL
In general, there will be a reflected wave at the boundary, with voltage and current coefficients
ρᵥ ≡ Vᵣ/Vᵢ = (ZL−Z) / (ZL+Z)
ρᵢ ≡ Iᵣ/Iᵢ = (Z−ZL) / (ZL+Z)
with same form as optics.
Special cases:
* ZL = Z gives ρᵥ = ρᵢ = 0.
Matched line, no reflection.
* ZL = 0, giving ρᵥ = −1 and ρᵢ = 1.
Shorted line, V cancels at the end of the line (V = 0).
* ZL = ∞ gives ρᵥ = 1 and ρᵢ = −1.
Open circuit, I cancels at the end of the line (I = 0)
Why use a waveguide instead of a coaxial cable
Attenuation (loss of electric field amplitude) of a high-frequency signal in a waveguide is much less than if it were sent along a coaxial cable. We use waveguides to transmit power at very high frequencies, eg hundreds of MHz.
What is a waveguide
A conducting box open at each end, where we cannot define a single V and I at any particular z along the waveguide.
What is a cavity
A closed, conducting box which stores electromagnetic energy.
Derive modes in a cavity
Closed, conducting box with rectangular surfaces. Sides have lengths a, b and d.
By convention, dimensions labelled such that b > a.
At any conducting surface, E‖ = 0 and B⊥ = 0. Therefore, if there is an EM wave within the cavity, solutions must be standing waves.
Standing waves = two travelling waves in opposite directions:
exp(i(ωt−kx)) − exp(i(ωt+kx)) = −2i sin(kx) exp(iωt)
and similarly for y and z directions.
Electric field components must have form
Eₓ = Eₓ₀ cos(kₓx) sin(kᵧy) sin(k₂z) exp(iωt)
Eᵧ = Eᵧ₀ sin(kₓx) cos(kᵧy) sin(k₂z) exp(iωt)
E₂ = E₂₀ sin(kₓx) sin(kᵧy) cos(k₂z) exp(iωt)
where Eₓ = 0 at y = 0, y = b, z = 0 and z = d
Similar relations hold for Eᵧ and E₂.
To satisfy these requirements,
kₓ = nₓπ/a
kᵧ = nᵧπ/b
k₂ = n₂π/d
where the mode numbers, nₓ, nᵧ and n₂, are integers.
In a cavity, at least two of nₓ, nᵧ and n₂ must be non-zero for there to be a finite value of electric field inside the cavity.
The overall wavenumber is
|k| = √kₓ²+kᵧ²+k₂²’
Dispersion relation is
ω²/v² = |k|² = kₓ² + kᵧ² + k₂²
where v (= c in case of vacuum) is the speed of the wave within the cavity.
Within cavity, no charge, so ∇ · E = 0
kₓEₓ₀ + kᵧEᵧ₀ + k₂E₂₀ = 0.
ω = πc √nₓ²/a² + nᵧ²/b² + n₂²/d²’
nₓ, nᵧ and n₂ must be integers, only certain values of ω are allowed.
Choosing a < b and a < d, lowest possible frequency
ωₘᵢₙ = πc √1/b² + 1/d²’
which occurs for mode nₓ = 0, nᵧ = 1, n₂ = 1, Labelled (0,1,1)
Energy flow and the relation between fields in a cavity
Lowest (0,1,1) mode
Eₓ = E₀ sin(kᵧy) sin(k₂z) exp(iωt)
Eᵧ = E₂ = 0
where E₀ is the peak field.
Magnetic field obtained using
∇ × E = −∂B/∂t
Must have time dependence, also has phase angle. By inspection,
∂B/∂t = iωB
so that
∇ × E = −iωB
E and B are 90◦ out of phase with eachother.
Therefore, at some point within the cavity
|E| ∝ cos(ωt+φ)
|B| ∝ sin(ωt+φ)
with 90◦ phase shift expressed as cos → sin
and φ term is phase shift with respect to other locations in the cavity.
Poynting vector at this point in cavity varies as
|S| ∝ |E × B|
∝ cos(ωt+φ) sin(ωt+φ)
= 1/2 sin(2(ωt+φ))
Energy flow at any given location varies at twice the frequency of the fields.
Time-averaged Poynting vector:
〈S〉=〈sin(2(ωt+φ))〉= 0
No flux.
For (0,1,1) mode, magnetic fields are
Bₓ = 0
Bᵧ = i k₂/ω E₀ sin(kᵧy) cos(k₂z) exp(iωt)
B₂ = −i kᵧ/ω E₀ cos(kᵧy) sin(k₂z) exp(iωt)
Note E · B = 0, so electric and magnetic fields are perpendicular everywhere in the cavity.
Waveguide wave vector constraints
No boundaries in z direction, so no constraint on k₂ values. Have mode numbers for x and y, labelled m for x and n for y.
kₓ = mπ/a
kᵧ = nπ/b
Dispersion relation
kₓ² + kᵧ² + k₂² = ω²/c²
k₂ can take any value, whereas kₓ and kᵧ must take discrete values to obey boundary conditions.
Cutoff frequency in a waveguide
k₂ can take any value, but kₓ and kᵧ must take
discrete values to obey boundary conditions. ω can vary continuously with a corresponding change in k₂, but not all values of ω are possible. A minimum value of ω is required
to keep a real value of k₂.
ω ≥ c √kₓ² + kᵧ² ‘ = πc √m²/a² + n²/b² ‘
This is the cutoff frequency.
ω²/c² = |k|² = kₓ² + kᵧ² + k₂²
⇒ m²π²/a² + n²π²/b² = k² − k₂²
This is the waveguide equation.
Dispersion relation
k₂ = ± √ω²/c² − (m²π²/a² + n²π²/b²)
As ω gets smaller, k₂ drops to zero at cutoff frequency; corresponds to component of the wavelength in the direction of the waveguide becoming infinite.
Waveguide equation
m²π²/a² + n²π²/b² = k² − k_Z²
Phase and group velocities in a waveguide
Within interior of waveguide, must satisfy Maxwell’s equations such that ∇ · E = 0
Leads to
kₓEₓ₀ + kᵧEᵧ₀ + k₂E₂₀ = 0
In practice, use waveguides such that E₂₀ = 0.
Eᵧ₀ = − kₓ/kᵧ Eₓ₀
From boundary conditions
kₓ = mπ/a
kᵧ = nπ/b
−nπ/b Eᵧ₀ = mπ/a Eₓ₀
−a Eᵧ₀/mπ = b Eₓ₀/nπ = V₀
where V₀ is some reference voltage.
Peak fields in x and y directions are
Eₓ₀ = V₀ nπ/b
Eᵧ₀ = V₀ mπ/a
Phase and group velocities:
vₚ = ω/k₂ = c √kₓ²+kᵧ²+k₂²’/k₂ > c
v_g = dω/dk = c k₂/√kₓ²+ky²+k₂²’ < c
Phase velocity greater than speed of light, group velocity remains smaller.
Approaching cutoff frequency, v_g falls to zero.
Also note vₚ v_g = c²
Can visualise propagation by considering reflection from the conducting surfaces so that transverse standing wave is formed. Not straight so does not propagate at c down the waveguide.
Cutoff frequency for lowest mode in a waveguide
Cutoff frequency for each Transverse Electric mode TEₘₙ (E_Z = 0) is given by
ωₘₙ = cπ √m²/a² + n²/b² ‘
Lowest possible mode is the TE₀₁ mode, with convention a < b.
Cutoff frequency for lowest mode is
f₀₁ = c/2b
The TE₀₁ frequency is only dependent on the larger of the two dimensions of the waveguide.
Summary of Waveguide Concepts
- Electromagnetic waves do not pass through conductors, they are reflected by conductors.
- Electric fields that touch a conductor must be perpendicular to it.
- Magnetic field lines close to a conductor must be parallel to it.
- The electric and magnetic fields are standing waves in the x and y directions, while in z we expect a (travelling) plane wave.
- Eₓ must fall to zero at y = 0 and y = b, hence kᵧ = nπ/b, and using Eᵧ, kₓ = mπ/a.
All components of an EM wave in a waveguide
Eₓ = V₀ nπ/b cos(mπx/a) sin(nπy/b) exp(i(ωt−k₂z))
Eᵧ = −V₀ mπ/a sin(mπx/a) cos(nπy/b) exp(i(ωt−k₂z))
E₂ = 0
Bₓ = −k₂ mπ/ωa V₀ sin(mπx/a) cos(nπy/b) exp(i(ωt−k₂z))
Bᵧ = −k₂ nπ/ωb V₀ cos(mπx/a) sin(nπy/b) exp(i(ωt−k₂z))
B₂ = i √k²−k₂²’/ω V₀ cos(mπx/a) cos(nπy/b) exp(i(ωt−k₂z))
Properties of the TE₀₁ mode
- The fields Eᵧ = 0 and Bₓ = 0
- The fields Eₓ ∝ sin(πy/b) and Bᵧ ∝ sin(πy/b)
- E ⊥ B everywhere within waveguide
- E is perpendicular to the surface at every surface
- The electric field lines cross the narrow part of the waveguide
- The magnetic field lines form closed loops.
Visualise TE₀₂, TE₁₀, and TE₁₁ modes
See page 122, fig 5.16
Power flow in a waveguide
S = 1/μ₀ E × B
kₓ and kᵧ components reflect from surfaces, so net power flow in x̂ and ŷ is
〈Sₓ〉=〈Sᵧ〉= 0
Power along ẑ is
S_z = 1/μ₀ (EₓBᵧ − EᵧBₓ)
TE₀₁ mode, average power flow
〈S_z〉 = k₂/2ωμ₀ E₀² sin²(kᵧy)
= vg εrε₀E₀²/2 sin²(kᵧy)
Can see that that total power flow is high at the center of the waveguide.
From S₂ , total power transmitted
〈P〉= ∫a→0 ∫b→0〈S₂〉dx dy
= 1/2 k₂/ωμ₀ E₀² ab/2
The power that a waveguide can transmit is limited by the peak electric field E₀, and the peak field occurs at the centre of the long side of the wave guide.
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Can also obtain power flow considering energy densities of the fields.
〈U〉= 1/2 ε₀〈E²〉+ 1/2μ₀〈B²〉
Integrate over a and b,
〈Eₗ〉= ab/4 ε₀E₀²
where Eₗ is energy density per unit length of the waveguide.
Using this definition,
〈P〉= vg〈Eₗ〉
= c²k₂/ω ab/4 ε₀E₀²
= k₂/ωμ₀ E₀² ab/4
Cutoff frequency and wavelength for a waveguide
fₘₙ = c/2 √m²/a² + n²/b² ‘
where m, n are integers and mode m = n = 0 is not allowed. If dielectric rather than air/vacuum, c is replaced with 1/√μᵣμ₀/εᵣε₀’ as usual.
λₘₙ = 2 / √m²/a² + n²/b² ‘
Why a typical 2:1 aspect ratio for waveguides?
b = 2a
It allows for the largest possible range of frequencies in one guide where only the TE₀₁ mode is supported. In turn, allows the largest power to be transmitted.
What is a cyclotron
Electromagnet that provides a uniform magnetic field over some volume.
Typically, two pole pieces assist two current-carrying coils to generate a uniform B.
A charged particle moving at right angles to the magnetic field will experience the usual Lorentz force
F = q(E + v × B)
Since the magnetic force is always at right angles to the charge motion, the charge will move in a circular path at right angles to the field.
Cyclotron frequency
Equate Lorentz and centripetal forces
mv²/r = qvB
where m is the mass of the charge.
Circular path, so radius is
r = mv/qB
Acceleration of charge is
a = qvB/m = ω꜀v
Cyclotron frequency, doesnt actually depend on velocity!
ω꜀ = v/r = qB/m
Actual frequency (rate past certain point)
f꜀ = 1/2π qB/m
Power emitted in a cyclotron
a = qvB/m = ω꜀v
Substitute into Larmor’s formula
P = q²ω꜀²v²/6πε₀c³
P is constant with time; no extra factor 2 in the denominator.
Output power ∝ v², so P ∝ K (kinetic energy)
P is the total power emitted per charge in all directions (over all angles φ).
From sufficient distance, side on looks like a Hertzian dipole.
Frequency of emitted radiation is the same as the cyclotron frequency, and is polarised parallel to the plane of circular motion (when viewed side on)
How does a microwave oven work
Magnetrons.
Heated cathode emits electrons, which accelerate towards an anode due to a voltage of a few kV.
A perpendicular magnetic field of around 0.1 T, generated by a pair of ring-shaped permanent magnets, causes electrons to orbit within the magnetron cavity, emitting cyclotron radiation around 2.45 GHz as they do so.
Outer anode cavities resonate at the same frequency to build up an electric field.
A coupling loop transfers power (typically around 1 kW when operating) from this field to an exterior antenna that sits within a waveguide.
Waveguide carries 12.23 cm radiation into main oven cavity, with at least one of its transverse dimensions must be larger than 6.1 cm.
Multiple modes can be excited at once in the oven cavity. The rms electric field strength is roughly constant, although there are hotspots - which is why there is a turntable inside the oven.
Q-value of a microwave oven cavity
Cavity walls have finite resistance, so electric field penetrates ~δ into the cavity walls.
Power lost per oscillation period
∆W/W ≃ S꜀ₐᵥδ/V꜀ₐᵥ
where S꜀ₐᵥ is cavity surface area
and V꜀ₐᵥ is cavity volume
Q-value
Q ∼ V꜀ₐᵥ/S꜀ₐᵥδ
Typical microwave oven (30 × 30 × 20 cm³) with δ ∼ 1 μm, we find Q ∼ 60, 000. When food is present, Q drops to around ∼ 100.
Microwaves are absorbed efficiently in the food, with very little lost to cavity walls.
Safe to operate too, if door button pressed, power to magnetron is cut off, field in cavity decays away in time T ∼ Q/f꜀ ∼ 20 μs, even if it were empty.
Synchrotron Radiation
Synchrotron radiation is the relativistic equivalent to cyclotron radiation.
m = γm₀ where γ = Etot/E₀,
mv²/r = qvB
where v = βc and m = γm₀
mβc = qBr
where
r = βγm₀c/qB
Revolution frequency fᵣ
fᵣ = v/2πr = βc/2πr = 1/2π qB/γm₀
= f꜀/γ
This is an issue as fᵣ may fall significantly as charges accelerate.
Can be fixed by adjusting the frequency of the voltage applied at the “Dees” (synchrocyclotron, voltage synchronised to charge motion), or varying B-field with radius to match γ (isochronous cyclotron, protons come round at same time regardless of energy)
Power of synchotron radiation
Lorentz transform compresses electric field lines into a “pancake” with characteristic width ∼1/γ transverse to direction of motion.
Transform of classical Hertzian dipole radiation pattern results in the pattern for radiation emitted by a relitivistically-moving charge.
Radiation is compressed into typical width ∼1/γ in direction of charge motion. Compression is not symmetric: much more radiation emitted in forward direction than backward.
Opening angle of the radiation is θ ∼ 1/γ.
Consequence of length contraction:
Contract by 1/γ, transverse acceleration is
a = d²x/dt²
Apparent acceleration must have extra γ² term.
Larmor formula for cyclotron radiation
P = q²a²/6πε₀c³
must be modified with extra γ² for each factor of a, becoming
P = q²a²γ⁴/6πε₀c³
Radiated power is increased by a factor γ⁴, which can be enormous if γ is significant
Uses of different synchotron sources
Using electrons:
* radiated power is much higher than it would be for protons
* radiation is much more forward-directed, easier to utilise in experiments
Using protons:
* emit far less EM radiation, good for colliding particles together in storage rings for particle physics experiments
* e+e− colliders, doubling collision energy means sixteen times the energy lost to radiation, too costly to replace
Frequency of synchotron radiation
Observer only sees synchrotron radiation for a short time period per orbit. Observed pulse length is shortened 1/γ from opening angle, and shortened another 1/γ because of Lorentz contraction. Gives typical emitted frequency of synchrotron radiation w.r.t. cyclotron as
fₛ ∼ f꜀γ²
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Another explanation
Take the emitted cyclotron frequency in charge rest frame, apply relativistic Doppler shift to observer frame.
f′/f = √1+β / 1−β’
= √(1+β)(1+β) / (1−β)(1+β)’
≃ 2 / √1−β²’ = 2γ
where we have approximated β ≃ 1
Accounts for one of the γ factors.
The other arises from synchrotron radiation only seen by an observer of angle 1/γ around the instant when the charge is moving directly towards the observer. Apparent frequency is a factor γ larger.
Synchotron pulse duration
Synchrotron radiation is emitted over a wide range of frequencies due to pulse length shortening.
Consider charge of rest mass m₀ moving relativistically in a circle due to uniform magnetic field B. Pulse period (how long it takes to orbit once) is
tᵣ = 1/fᵣ = 2πγm₀/eB = γ/f꜀
where f꜀ = eB/2πmₑ
Pulse duration is
dt = 1/fₛ = 1/γ²f꜀
Pulse duration is much shorter than the period:
dt = tᵣ/γ³
Frequency components are spaced at 1/tᵣ, spaced apart in frequency by f꜀/γ. As fₛ ≫ f꜀/γ, the frequency spectrum of synchrotron radiation is essentially continuous up to fₛ. See photons with energy from zero to E ≃ hfₛ.
fₛ = cβγ³/2πr
Critical photon frequency/energy in a synchotron
Typical photon frequency:
fₛ = cβγ³/2πr
Critical frequency defined such that:
* half radiation power in photons above critical frequency
* half radiation power in photons below critical frequency
Critical frequency is also known as the half power point.
Synchrotron radiation is compose of very, very many low frequency photons and rather fewer high-energy photons.
Critical frequency:
f꜀ᵣᵢₜ = 3/2 cγ³/2πr
where β = 1 has been used.
Critical energy:
E꜀ᵣᵢₜ = 3/2 ħcγ³/r
Many more low-energy photons than high, average photon energy is lower than critical energy.
Average photon energy:
〈Eᵧ〉≃ E꜀ᵣᵢₜ/3 = Eₛ/2
= 1/2 ħcγ³/r = 1/2 hγ² qB/2πm₀
Average photon frequency:
〈fᵧ〉≃ f꜀ᵣᵢₜ/3 = fₛ/2
Radiation energy per synchotron orbit
In a uniform field B, a charge emits EM radiation with average power
P = q²a²γ⁴/6πε₀c³
Charge orbits in circle of radius
r = γm₀c/(eB)
so has acceleration
a = v²/r
where v = βc
Power can be written
P = q²β⁴c⁴γ⁴/6πε₀c³r²
= q²cβ⁴γ⁴/6πε₀r²
Radiation energy emitted during one orbit (with period tᵣ):
U₀ = P tᵣ = q²cβ⁴γ⁴/6πε₀r² 2πr/βc
= q²β³γ⁴/3ε₀r
Number of photons emitted per synchotron orbit
Nᵧ = U₀/〈Eᵧ〉
= 2U₀/Eₛ
= 2 q²β³γ⁴/ε₀r 2πr/hcβγ³
= 2/3 q²β²γ/ε₀ħc
In nearly all practical cases, dealing with electrons that have reasonable kinetic energy (~10 MeV or higher). Then q = e, m₀ = mₑ and β = 1 to a good approximation.
Relation between synchrotron rdiation and the field energy density
Power emitted per electron:
P = e²cβ⁴γ⁴/6πε₀r²
Thomson cross-section for scattering of electrons has similar form:
σT = e⁴/6πε₀²m²c⁴
Synchrotron radiation power in terms of Thomson cross-section:
P = σT m²ε₀c⁵β⁴γ⁴/e²r²
Power in terms of B-field, using
r = βγmc/eB
which leaves us with
P = σT B²ε₀c³β²γ²
In terms of energy density,
UB = 1/2μ₀ B²
synchrotron radiation power is
P = 2σT UB ε₀μ₀c³β²γ²
ε₀μ₀ = 1/c², so
P = 2σT UB cβ²γ²
Shows that the electron takes energy from the magnetic field at some rate σT, where the magnetic field has an effective Poynting flux SB = UB c
Coherent vs incoherent synchrotron radiation
Two electrons radiating separately (far apart), total power emission is 2 times single electron power emission.
If electrons in same location, total field they generate is equivalent to single charge q = 2e.
Total radiated power would be 4 times single electron power emission.
A group of N charges within λ of each other will radiate coherently, with a power P ∝ N²
If far apart, they radiate incoherently with a power P ∝ N
What is Bremsstrahlung Radiation?
Radiation from a charged particle when passing close to an atomic nucleus. It feels a strong electric field (and loses kinetic energy and slows down).
Strong electric field at right angles to particle motion causes same effect as a strong magnetic field: electromagnetic radiation is emitted.
Broad spectrum of X-ray emissions, large number of low-energy photons and small number of high-energy photons. Max. energy of the emitted photons is almost the initial kinetic energy of the electrons.
Cutoff frequency for Bremsstrahlung radiation
ν꜀ = Eₖ/h
Duane-Hunt law
Power released per electron in Bremsstrahlung radiation
Most of the time, large impact parameter b, such that |v| ≃ |v′|, electron only
slightly deflected and doesn’t change much in energy.
Distance from nucleus as a function of time:
r(t) = √b² + v²t² ‘
where r = b at t = 0.
For a nuclear charge Q, acceleration experienced by the electron is
a = 1/mₑ eQ/4πε₀r²
Emitted power as a function of time
P(t) = e⁴Q²/96π³ε₀³mₑ²c³ 1/(v²t²+b²)²
Total energy released as photons:
W = ∫±∞ P(t) dt
= e⁴Q²/192π²ε₀³mₑ²c³ 1/vb³
Bremsstrahlung emitted power
Beam of electrons, each electron will have a different closest distance b from nucleus.
Larger b values are more probable, with probability varying as ∼2πb db
Nₑ electrons in beam, number density per unit volume of nuclei in target Nᵢ, energy loss per unit length of targe is:
dE/dl = Nᵢ ∫(bmax→bmin) NₑW2πb db
= NᵢNₑe⁴Q²/96πε₀³vmₑ²c³ [1/b] (bmax→bmin)
Upper limit bmax = ∞ is fine - electrons can in principle travel very far from the nucleus. 1/∞ goes to zero.
bmin → 0 would lead to divergence and predict infinite emitted power.
Original assumption |v|≃|v′| must break down at small b, total energy emitted must be less than initial kinetic energy of electron.
P_brem = NᵢNₑe⁴Q²/96πε₀³mₑ²c³ [1/bmin]
Could choose limit to cut off calculation when ∆v ∼ v.
∆v = Qe/mₑ ∫±∞ b/(v²t²+b²)³’² dt = 2Qe/mₑbv
Limit is
bmin = 4Qe/πmₑv
Could also choose limit when quantum effects become significant,
bmin ∼ ħ/mₑv
In practice, people typically set bmin = ħ/(mv) (the quantum limit).
Quantum limit yields
P_brem ≃ NᵢNₑe⁴Q²v/48ε₀³mₑc³h Wm⁻³
Radiated power strongly depends on the charge of the nucleus.
What is Cherenkov radiation
Radiation emitted when a uniformly-moving charge is moving with velocity v꜀ which is greater than the speed of light, vₚ, within the medium.
Derive Cherenkov geometry
Charge is moving faster than electric field itself can propagate. Apply Huygen’s principle to wavefronts at each point in time. The wavefront propagates at an angle to the direction of charge motion - the individual point sources overlap at angle θ.
Distance travelled by the charge in time ∆t,
v꜀∆t = βc∆t
Distance travelled by individual wavefronts
vₚ∆t = c/n ∆t
Overall wavefront, obtained by overlapping infinitesimal wavefronts from each emitting point, perpendicular to the direction it’s moving. See pg 160, fig 6.19
cosθ = (c/n ∆t) / (v∆t)
= c/βcn
= 1/nβ
where βc is charge’s velocity, and n is the refractive index of the medium.
Assuming charge is moving with large kinetic energy, and velocity (nearly always true), β ≃ 1 and Cherenkov angle takes form
cosθ ≃ 1/n
Water, for example, has refractive index ~1.33 for visible wavelengths, so Cherenkov angle is θ ≃ 41◦
Overall, get conical Cherenkov wavefront with normal at angle θ to direction of charge motion. Will only see radiation if the material itself is transparent to it. Slower moving particles (β < 1) give rise to a narrower radiation cone.
Minimum velocity for Cherenkov radiation is when v꜀ = vₚ, aka β = 1/n
Any charge can generate Cherenkov radiation.
