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
