Electromagnetic Radiation Flashcards

Question

Derive the Poynting vector

Answer 1

Using Faraday’s law **B ·** ∂**B**/∂t = −**B ·** (∇ × **E**) Using Ampere’s law µ₀ε₀**E ·** ∂**E**/∂t = **E ·** (∇ × **B**) − µ₀**E · J** Combine **B ·** ∂B/∂t + µ₀ε₀**E ·** ∂E/∂t = −µ₀**E · 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**)

Answer 2

There is power present in an electromagnetic field. Its energy flux and direction is given by the Poyting vector

Answer 3

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**

Answer 4

*Electric* Faraday's law ∇ × **E** = −∂**B**/∂t Take curl ∇ × (∇ × **E**) = −∂/∂t (∇×**B**) Vector identity and Ampere's law ∇(∇ **· E**) − ∇²**E** = −∂/∂t (µ₀ε₀ ∂**E**/∂t) Gauss' law in vacuum/ with no charge, ∇ **· E** = 0 Wave equation ∇²**E** − µ₀ε₀ ∂²**E**/∂t² = 0 {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} *Magnetic* Ampere's law ∇ × **B** = µ₀ε₀ ∂**E**/∂t Take curl ∇ × (∇ × **B**) = µ₀ε₀ ∂/∂t (∇×**E**) Vector identity and Faraday's law ∇(∇ **· B**) − ∇²**B** = −µ₀ε₀ ∂/∂t (∂**E**/∂t) No magnetic monopoles ∇ **· B** = 0 Wave equation ∇²**B** − µ₀ε₀ ∂²**B**/∂t² = 0 {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} Two wave equations, waves travelling with speed c = 1/√µ₀ε₀'

Answer 5

Modified permittivity ε₀ → ε = εᵣε₀ Modified permeability µ₀ → µ = µᵣµ₀ **D** = ε**E** **H** = 1/µ **B** **D**-field is the electric displacement field and **H**-field is the magnetic field

Answer 6

Wave equation ∇²**E** − µε ∂²**E**/∂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 ∇²**E** − 1/c² ∂²**E**/∂t² = 0

Answer 7

Solutions are of form **E** = E₀exp(i(ωt±kz)) where **E**₀ = (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.

Answer 8

c = ω/|k| = 1/√µ₀ε₀' where k is the wavenumber (or wave-vector) where the wavelength is λ = 2π/k

Answer 9

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 {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} For an arbitrary direction k̂: **E**(**r**, t) = **E₀**exp(i(ωt−**k·r**) **B**(**r**, t) = 1/c k̂ × **E** = 1/ω **K** x **E** Using c = ω/|k|

Answer 10

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. {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} * 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

Answer 11

|B| = |E|/c **B** direction is π/2 compared to **E**

Answer 12

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)

Answer 13

U = UE + UB = ε₀E₀² cos²(ωt−kz) 〈U〉= ε₀E₀²〈cos²(ωt−kz)〉 = 1/2 ε₀E₀² = ε₀E²ᵣₘₛ where Eᵣₘₛ = E₀/√2'

Answer 14

**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.

Answer 15

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

Answer 16

Pᵣ =〈U〉 =|〈S〉|/c = 1/µ₀c |〈E × B〉|

Answer 17

There is no B-field component (**B**=0)

Answer 18

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.

Answer 19

∇²Φ − 1/c² ∂²Φ/∂t² = −ρ/ε₀ ∇²**A** − 1/c² ∂²**A**/∂t² = −μ₀**J** These are wave equations with a source term involving ρ and **J**

Answer 20

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′

Answer 21

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**.

Answer 22

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.

Answer 23

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

Answer 24

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.

Answer 25

Integrate **S** over all angles. P(t) = ∮**S ·** d**A** = ∫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

Answer 26

An oscillating current element, the simplest of all antennas.

Answer 27

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′) = −I₀*l*ω sin(ωt′) = −I₀*l*ω sin(ω(t−r/c)) = −I₀*l*ω sin(ωt−kr)) Recall and substitute **E⊥**(**r**,t) = −1/4πε₀ qa⊥(t')/rc² = +1/4πε₀ I₀*l*ω sin(ωt−kr))/rc² or **E⊥**(**r**,t) = I₀*l*sin(θ)/4πε₀rc² ωsin(ωt−kr) Remember **B** = **E**/c **B**(**r**,t) = I₀*l*sin(θ)/4πε₀rc³ ωsin(ωt−kr)

Answer 28

P(t) = q²a²(t′)/6πε₀c³ qa(t′) = −I₀*l*ω sin(ωt−kr)). P(t) = [ (*l*ω)²/6πε₀c³ ] I₀² sin²(ωt−kr) where [] term has units of resistance

Answer 29

Rᵣₐₔ = *l*²ω²/6πε₀c³ 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

Answer 30

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 = q₀*l* p = p₀ sinωt = q₀*l* sin ωt I₀*l* = p₀ω I₀*l*ω = p₀ω²

Answer 31

q = ±q₀ sin ωt, q₀ = p₀/*l* I₀*l* = p₀ω I₀*l*ω = p₀ω² Larmor's formula P(t) = q²a²/6πε₀c³ Substitute in to get 〈P〉= I₀²*l*²ω²/12πε₀c³ ('current picture') OR 〈P〉= p₀²ω⁴/12πε₀c³ ('dipole picture')

Answer 32

* 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.

Answer 33

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 I*l* 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.

Answer 34

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 d**E**⊥(**r**, t) = I₀d*l*sinθ/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) = I₀*l*/4πε₀Rc² [2/π cos(π/2 cosθ)/sinθ] ωsin(ωt−kR) where *l* = λ/2

Answer 35

**E**(**R**,t) = I₀*l*/4πε₀Rc² [2/π cos(π/2 cosθ)/sinθ] ωsin(ωt−kR) From Poynting vector somehow get 〈S〉= 1/μ₀ (I₀²*l*²/16π²ε₀²R²c⁵) 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

Answer 36

〈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

Answer 37

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.

Answer 38

Conductors contain an 'unlimited' charge supply that is free to move. Insulators/dielectrics haave charge attached to specific atoms and molecules.

Answer 39

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

Answer 40

Total charge density ρ = ρb + ρf Gauss’s law, ε₀∇ **· E** = ρ = ρb + ρf = −∇ **· P** + ρf ∇ · (ε₀**E** + **P**) = ρf ∇ · **D** = ρf Electric displacement **D** = (ε₀**E** + **P**)

Answer 41

Polarisation proportional to **E** for linear dielectrics. **P** = ε₀χₑ**E** where χₑ is the electric susceptibility of the material. **D** = ε₀**E** + **P** = ε₀**E** + ε₀χₑ**E** = 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.

Answer 42

* Paramagnetism: Magnetisation is parallel to **B** * Diamagnetism: Magnetisation is opposite to **B** * Ferromagnetism: Magnetisation is retained even after **B** is removed

Answer 43

M ≡ Magnetic dipole moment per unit volume Bound **J**b = −∇ × **M** Surface **K**b = **M** × **n̂** Free currents not contributing to dipole moment **J**f

Answer 44

Total charge density **J** = **J**b + **J**f Ampere’s law, 1/μ₀ (∇ × **B**) = **J** = **J**b + **J**f = **J**f + ∇ × **M** ∇ × (1/μ₀**B** − **M**) = **J**f ∇ × **H** = **J**f Magnetic field **H** = (1/μ₀**B** − **M**)

Answer 45

**M** = χₘ**H** 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.

Answer 46

∇ **· D** = ρf ∇ **· B** = 0 ∇ × **E** = − ∂**B**/∂t ∇ × **H** = ∂**D**/∂t + **J**f

Answer 47

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

Answer 48

All remain the same apart from Ampere's law. ∇ × **H** = ∂**D**/∂t + **J**f Substitute **J**f = σ**E** = iωε**E** + σ**E** ∇ × **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).

Answer 49

∇ × **E** = −iωμ**H** ∇ × **H** = iωε**E** ∇ **· E** = 0 ∇ **· B** = 0 where μ = μ₀μᵣ and ε = ε₀εᵣ

Answer 50

Wave equation in source-free medium Take curl of Faradays law ∇ × **E** = −iωμ**H** ∇ × ∇ × **E** = −iωμ(∇ × **H**) ∇(∇ **· E**) − ∇²**E** = −iωμ(iωε**E**) First term is zero for source free medium. −∇²**E** = ω²εμ**E** Using εμ = 1/c² and ω²/c² = k² ∇²**E** + k²**E** = 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

Answer 51

γ = ik꜀ = iω√με꜀' (m⁻¹) in the general form, γ = α + iβ

Answer 52

∇²**E** + k꜀²**E** = 0 Replacing k꜀ = −iγ ∇²**E** − γ²**E** = 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

Answer 53

Insulator with μᵣ ≃ 1 and σ = 0 v = ω/k = 1/√εᵣε₀μᵣμ₀' = c/√εᵣ' n = c/v = √εᵣ'

Answer 54

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.

Answer 55

* 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)

Answer 56

Equation of motion for each electron is ẍ + αẋ + ω₀²x = q/m [**E** + ẋ**B**] 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),

Answer 57

χₑ ≪ 1

Answer 58

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** = α**Eₗₒ꜀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**

Answer 59

α = 3ε₀/N εᵣ−1/εᵣ+2

Answer 60

n²−1/n²+2 = Nα/3ε₀

Answer 61

**J** = σ**E**

Answer 62

k꜀² = εᵣμᵣ ω²/c² (1 − iσ/ωεᵣε₀)

Answer 63

Absorbtion

Answer 64

σ/ωεᵣε₀ ≪ 1

Answer 65

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)

Answer 66

δ = 1/|kᵢ| = 2/σ √εᵣε₀/μᵣμ₀'

Answer 67

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.

Answer 68

σ/ωεᵣε₀ ≫ 1

Answer 69

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)

Answer 70

δ = 1/|kᵢ| = √2/μᵣμ₀σω'

Answer 71

λ = 2π/kᵣ = 2πδ

Answer 72

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.

Answer 73

**E** = **E₀**exp(i(ωt−kz)) **B** = **B₀**exp(i(ωt−kz)) ∇ × **E** = − ∂**B**/∂t **k** × **E₀** = ω**B₀** 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)

Answer 74

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.

Answer 75

* 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.

Answer 76

−e**E** = 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'

Answer 77

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** = N**p** = − Ne²/mω² **E** ignoring effects between induced dipole moments.

Answer 78

**P** = N**p** = − Ne²/mω² **E** **D** = ε₀**E** + **P** = ε₀ (1 − Ne²/mε₀ω²) **E** = ε₀ (1 − ω²ₚ/ω²) **E** where (1 − ω²ₚ/ω²) is the equivalent permittivity of plasma.

Answer 79

ωₚ = √Ne²/mε₀'

Answer 80

fₚ = ωₚ/2π = 1/2π √Ne²/mε₀' (Hz) Depends only on plasma density. Signals below fₚ cannot penetrate.

Answer 81

γ꜀ = 1/τ꜀

Answer 82

**v**drift = **a**τ꜀ = −e**E**/m 1/γ꜀

Answer 83

**J** = σ₀**E** = −Ne **v**drift 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꜀

Answer 84

σ(ω) = σ₀/1+(iω/γ꜀)

Answer 85

**P** = ε₀χₑ**E** = ε₀χₑE₀exp(iωt) where we know **P** = N**p** = N(−e)x Substitute x from the solution of Drude’s model **P** = (Ne²/m) 1/−ω²+iγ꜀ω **E** = ε₀χₑ**E** 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.

Answer 86

P(t) = 1/τ꜀ exp(−t/τ꜀)

Answer 87

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'

Answer 88

ω ≫ γ꜀ **E** changes sign more quickly than a collision can occur, situation describes a collisionless plasma.

Answer 89

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

Answer 90

* 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

Answer 91

Consider path C in and out of boundary (Fig 4.2, pg 77). Integral form of Faraday’s law ∮꜀ **E ·** d**l** = −∂/∂t ∮꜀ **B ·** d**S** Integrate along C anticlockwise ∮꜀ **E ·** d**l** = **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

Answer 92

Closed cylindrical surface S across the boundary (Fig 4.3, pg 78) such that **a**ₙ₁ = −**a**ₙ₂ are the normal unit vectors for the top and bottom faces Gauss’ law ∮ₛ **D ·** d**s** = Q **a**ₙ₂ **·** (**Dₙ₁** − **Dₙ₂** )∆S = ρₛ ∆S **a**ₙ₂ **·** (**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.

Answer 93

Ampere’s law ∮꜀ **H ·** d**l** = ∮ₛ **J ·** d**S** + ∂/∂t ∮ₛ**D ·** d**S** 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.

Answer 94

Gauss’s law ∮ₛ **B ·** d**S** = 0 **Bₙ₁** − **Bₙ₂** = 0 where **Bₙ₁** = **B₁** · **aₙ₁** Normal component of the magnetic field is continuous.

Answer 95

n̂ **· E** = ρₛ/ε₀ → normal discontinuous n̂ × **E** = 0 → transverse continuous n̂ **· B** = 0 → normal continuous n̂ × **B** = μ₀**J_z** → transverse discontinuous

Answer 96

Z ≡ μ₀μᵣω/k For vacuum, ω/k = c and μᵣ = 1 Z₀ ₍ᵥₐ꜀ᵤᵤₘ₎ = μ₀c = √μ₀/ε₀ = 377 Ω

Answer 97

**E**ᵢ = **E**ᵢ₀exp(i(ωt−kz)) **B**ᵢ = **B**ᵢ₀exp(i(ωt−kz)) **E**ᵣ = −**E**ᵣ₀exp(i(ωt+kz)) **B**ᵣ = **B**ᵣ₀exp(i(ωt+kz)) **E**ₜ = **E**ₜ₀exp(i(ωt−kz)) **B**ₜ = **B**ₜ₀exp(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

Answer 98

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.**

Answer 99

〈**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

Answer 100

* 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.

Answer 101

n₂ = √εᵣ' = √−iσ/ωε₀' ≃ √σ/2ωε₀' (1−i)

Answer 102

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'

Answer 103

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.

Answer 104

Incoming wave **kᵢ** incident from medium (1) into medium (2) at angle θᵢ to the surface normal. Using notation √−1' = j **Eᵢ** = **Eᵢ₀**exp(j(ωt−**kᵢ·r**) **Eᵣ** = **Eᵣ₀**exp(j(ωt−**kᵣ·r**) **Eₜ** = **Eₜ₀**exp(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.

Answer 105

Orthogonal components * S-polarisation has **E** is normal to the plane of incidence * P-polarisation has **E** is in the plane of incidence

Answer 106

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⊥.

Answer 107

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

Answer 108

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)

Answer 109

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

Answer 110

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

Answer 111

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.

Answer 112

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

Answer 113

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.

Answer 114

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

Answer 115

Elastic scattering by a free, charged particle. Low energy lmit of Compton scattering. Particle kinetic anergy and photon frequency remain unchanged.

Answer 116

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

Answer 117

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.

Answer 118

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.

Answer 119

Two parallel conductors separated by an insulator

Answer 120

A hollow tube - made of either conductor or dielectric - whose walls reflect electromagnetic waves. The tube doesn’t need to be circular.

Answer 121

*l* ≫ λ

Answer 122

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

Answer 123

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

Answer 124

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.

Answer 125

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.

Answer 126

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 g*l* Inductance per unit length L̃ = 1/*l* Φ/I = μᵣμ₀ g/d Capacitance along length *l* C = εᵣε₀ *l*d/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.

Answer 127

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

Answer 128

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)

Answer 129

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.

Answer 130

A conducting box open at each end, where we cannot define a single V and I at any particular z along the waveguide.

Answer 131

A closed, conducting box which stores electromagnetic energy.

Answer 132

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)

Answer 133

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 = iω**B** so that ∇ × **E** = −iω**B** **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.

Answer 134

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.

Answer 135

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.

Answer 136

m²π²/a² + n²π²/b² = k² − k_Z²

Answer 137

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.

Answer 138

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.

Answer 139

* 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.

Answer 140

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))

Answer 141

* 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.

Answer 142

See page 122, fig 5.16

Answer 143

**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. {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} 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

Answer 144

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 '

Answer 145

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.

Answer 146

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.

Answer 147

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

Answer 148

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)

Answer 149

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.

Answer 150

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.

Answer 151

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)

Answer 152

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

Answer 153

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

Answer 154

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 {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{} 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.

Answer 155

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

Answer 156

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

Answer 157

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

Answer 158

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.

Answer 159

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

Answer 160

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

Answer 161

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.

Answer 162

ν꜀ = Eₖ/h Duane-Hunt law

Answer 163

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

Answer 164

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/d*l* = 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.

Answer 165

Radiation emitted when a uniformly-moving charge is moving with velocity v꜀ which is greater than the speed of light, vₚ, within the medium.

Answer 166

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.

Electromagnetic Radiation Flashcards

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