Lecture 2 Review 2
More on Maxwell Wave Equations Boundary Conditions Poynting Vector Transmission Line
A. Nassiri - ANL Maxwell’s equations in differential form
∇.D = ρ Gauss' law for electrostatics ∇.B = 0 Gauss' law for magnetostatics ∂D ∇ × H = J + Ampere' s law dt ∂B ∇ × E = − Faraday' s law dt ∂ρ ∇.J = − Equation of continuity ∂t
D = εE Varying E and H fields are coupled B = µH
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2 Electromagnetic waves in lossless media - Maxwell’s equations Constitutive relations Maxwell ∂D D = εE = ε ε E ∇× H = J + r o dt B = µH = µ µ H ∂B r o ∇×E = − dt J = σE SI Units ∇ = ρ • J Amp/ metre2 .D • D Coulomb/metre2 • H Amps/metre • B Tesla ∇.B = 0 Weber/metre2 Volt-Second/metre2 Equation of continuity • E Volt/metre • ε Farad/metre • µ Henry/metre ∂ρ • σ Siemen/metre ∇.J = − ∂t
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3 Wave equations in free space
• In free space – σ=0 ⇒J=0 – Hence: ∂D ∂D ∇× H = J + = dt dt ∂B ∇×E = − dt
– Taking curl of both sides of latter equation: ∂B ∂ ∂ ∇×∇× E = −∇× = − ∇× B = −µ ∇× H ∂t ∂t o ∂t ∂ ∂D = −µo ∂t ∂t ∂ 2E ∇×∇× E = −µoε ∂t 2
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4 Wave equations in free space cont.
∂ 2E ∇×∇×E = −µoε ∂t 2 • It has been shown (last week) that for any vector A
where is the Laplacian operator 2 Thus: ∇×∇× A = ∇∇.A − ∇ A ∂ 2 ∂ 2 ∂ 2 ∇2 = + + ∂x2 ∂y 2 ∂z 2 2 2 ∂ E ∇∇.E − ∇ E = −µoε ∂t 2 There are no free charges in free space so ∇.E=ρ=0 and we get
2 2 ∂ E ∇ E = µoε ∂t 2 A three dimensional wave equation
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 5 Wave equations in free space cont.
• Both E and H obey second order partial differential wave equations:
∂ 2E ∇2E = µ ε o ∂ 2 t2 2 ∂ H ∇ H = µoε ∂t 2
What does this mean – dimensional analysis ? Volts/metre Volts/metre = µoε metre2 seconds2
-2 – µοε has units of velocity
– Why is this a wave with velocity 1/ √µοε ?
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6 Uniform plane waves - transverse relation of E and H
• Consider a uniform plane wave, propagating in the z direction. E is independent of x and y
∂E ∂E = 0 = 0 ∂x ∂y
In a source free region, ∇.D=ρ =0 (Gauss’ law) :
∂E ∂E ∂E ∇.E = x + y + z = 0 ∂x ∂y ∂z E is independent of x and y, so
∂E ∂E ∂E x = 0, y = 0 ⇒ z = 0 ⇒ E = 0 (E = const is not a wave) ∂x ∂y ∂z z z
So for a plane wave, E has no component in the direction of propagation. Similarly for H. Plane waves have only transverse E and H components.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 7 Orthogonal relationship between E and H:
• For a plane z-directed wave there are no variations along x and y:
∂ ∂A ∂ Az y H y ∂H x ∇× A = a x − + ∇× H = −a + a ∂y ∂z x ∂z y ∂z ∂Ax ∂Az a y − + ∂D ∂z ∂x = ∂A ∂A ∂t a y − x z ∂x ∂y ∂ ∂Ex Ey ∂Ez = εa + a + a ∂D x ∂t y ∂t y ∂t ∇× H = J + dt Equating terms: and likewise for : ∇×E = −µo ∂H ∂t ∂H ∂E ∂E ∂H − y = ε x y = µ x ∂z ∂t ∂z o ∂t ∂H ∂E ∂E ∂H x = ε y x = µ y ∂z ∂t ∂z o ∂t Spatial rate of change of H is proportionate to the temporal rate of change of the orthogonal component of E & v.v. at the same point in space
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8 Orthogonal and phase relationship between E and H:
• Consider a linearly polarised wave that has a transverse component in (say) the y direction only:
∂ H y ∂Ex = − − = ε Ey Eo f (z vt) ∂z ∂t ∂H x ∂E = ∂H ∂E ⇒ ε y = −εvE f ′(z − vt) ∂z x = ε y ∂t o ∂z ∂t ⇒ = −ε ′ − + = −ε − H x vEo ∫ f (z vt)dz const vEo f (z vt)
= −εvEy ε H x = − Ey µo Similarly ∂ E y ∂H x = µo ε ∂z ∂t ∂ H y = Ex ∂Ex H y µ = µo o ∂z ∂t H and E are in phase and orthogonal
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 9 ε ε H = − E H = E x µ y y x o µo • The ratio of the magnetic to electric fields strengths is:
2 2 Ex + E y E µ = = o = η Note: 2 2 H ε H x + H y E E 1 = = = c which has units of impedance µ B o H µoεo Volts / metre = Ω amps / metre
and the impedance of free space is:
µ 4π ×10−7 o = = 120π = 377Ω ε 1 − o ×10 9 36π
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 10 Orientation of E and H
• For any medium the intrinsic impedance is denoted by η E E η = − y = x and taking the scalar product H x H y
E.H = Ex H x + E y H y
= ηH y H x −ηH x H y = 0 so E and H are mutually orthogonal
Taking the cross product of E and H we get the direction of wave propagation
E× H = a z (Ex H y − E y H x ) A×B = a x (Ay Bz − Az By )+ 2 2 = a z (ηH y −ηH x ) a y (Az Bx − Ax Bz )+ 2 E× H = a zηH a z (Ax By − Ay Bx )
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 11 A ‘horizontally’ polarised wave
• Sinusoidal variation of E and H • E and H in phase and orthogonal ε H y = Ex µo
Hy Ex
E× H
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 12 A block of space containing an EM plane wave
• Every point in 3D space is characterised by – Ex, Ey, Ez – Which determine
• Hx, Hy, Hz • and vice versa
– 3 degrees of freedom
λ
Ex ε H y = Ex µo E× H ε H x = − E y Hy µo
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 13 Power flow of EM radiation
2 • Energy stored in the EM field in the thin box is: εE u = E 2 dU = dU E + dU H = (uE + uH )Adx µ H 2 2 2 o εE µ H uH = = + o dU Adx 2 2 2 2 ε = εE Adx H = E y µ x Power transmitted through the box is dU/dt=dU/(dx/c).... o λ dx
Ex
E× H Hy Area A
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 14 Power flow of EM radiation cont.
dU = εE 2 Adx dU εE 2 ε E 2 S = = Adx = = W/m2 Adt A(dx c) µo η • This is the instantaneous power flow – Half is contained in the electric component – Half is contained in the magnetic component • E varies sinusoidal, so the average value of S is obtained as:
2π E = E sin (z − vt) o λ E 2 sin 2 (z − vt) S = o η E 2 E 2 S = o RMS(E 2 sin 2 (z − vt))= o η o 2η
S is the Poynting vector and indicates the direction and magnitude of power flow in the EM field.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 15 Example
• The door of a microwave oven is left open – estimate the peak E and H strengths in the aperture of the door. – Which plane contains both E and H vectors ? – What parameters and equations are required?
• Power-750 W • Area of aperture - 0.3 x 0.2 m • impedance of free space - 377 Ω • Poynting vector:
E 2 S = = ηH 2 W/m2 η
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16 E 2 Power = SA = A = ηH 2 A Watts η
Power 750 E = η = 377 = 2,171V/m A 0.3.0.2
E 2170 H = = = 5.75A/m η 377
−7 B = µo H = 4π ×10 ×5.75 = 7.2μTesla
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17 Constitutive relations
-12 • permittivity of free space ε0=8.85 x 10 F/m -7 • permeability of free space µo=4πx10 H/m D = εE = ε rεoE Normally ε (dielectric constant) and µ • r r = µ = µ µ – vary with material B H r oH – are frequency dependant J = σE • For non-magnetic materials mr ~1 and for Fe is ~200,000 • er is normally a few ~2.25 for glass at optical frequencies – are normally simple scalars (i.e. for isotropic materials) so that D and E are parallel and B and H are parallel
• For ferroelectrics and ferromagnetics er and mr depend on the relative orientation of the material and the applied field:
µ µ µ Bx xx xy xz H x µ − jκ 0 At By = µ yx µ yy µ yz H y µ = jκ µ 0 microwave ij B µ µ µ H z zx zy zz z frequencies: 0 0 µo
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18 Constitutive relations cont...
• What is the relationship between ε and refractive index for non magnetic materials ? – v=c/n is the speed of light in a material of refractive index n
– For glass and many1 plastics atc optical frequencies v = = • n~1.5 µoεoε r n • εr~2.25 • Impedance is lowern = withinε r a dielectric
What happens at the boundary between materials of different n,µr,εr ?
µ µ η = o r εoε r
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 19 Why are boundary conditions important ?
• When a free-space electromagnetic wave is incident upon a medium secondary waves are – transmitted wave – reflected wave • The transmitted wave is due to the E and H fields at the boundary as seen from the incident side • The reflected wave is due to the E and H fields at the boundary as seen from the transmitted side • To calculate the transmitted and reflected fields we need to know the fields at the boundary – These are determined by the boundary conditions
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 20 Boundary Conditions cont.
µ1,ε1,σ1
µ2,ε2,σ2
• At a boundary between two media, mr,ers are different on either side.
• An abrupt change in these values changes the characteristic impedance experienced by propagating waves
• Discontinuities results in partial reflection and transmission of EM waves
• The characteristics of the reflected and transmitted waves can be determined from a solution of Maxwells equations along the boundary
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 21 Boundary conditions
The tangential component of E is continuous at a surface of discontinuity • E H µ ε ,σ – E1t,= E2t 1t, 1t 1, 1 1 • Except for a perfect conductor, the tangential component of H is continuous at a surface of discontinuity – H1t,= H2t E2t, H2t µ2,ε2,σ2
The normal component of D is continuous at the surface of a discontinuity if there is no surface charge density. If there is surface charge density D is discontinuous by an amount equal to the surface charge density. D1n, B1n µ1,ε1,σ1 – D1n,= D2n+ρs The normal component of B is continuous at the surface of discontinuity D2n, B2n µ2,ε2,σ2 – B1n,= B2n
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 22 Proof of boundary conditions - Dn
Dn1 ∆y
∆x
∆z µ1,ε1,σ1 µ2,ε2,σ2
Dn2 • The integral form of Gauss’ law for electrostatics is: D.dA = ρdV ∫∫ ∫∫∫V applied to the box gives
Dn1∆x∆y − Dn2∆x∆y + Ψedge = ρs∆x∆y hence As dz → 0,Ψedge → 0 The change in the normal component of D at a D − D = ρ n1 n2 s boundary is equal to the surface charge density
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 23 Proof of boundary conditions - Dn cont.
Dn1 − Dn2 = ρs
• For an insulator with no static electric charge ρs=0
Dn1 = Dn2
For a conductor all charge flows to the surface and for an infinite, plane surface is
uniformly distributed with area charge density rs In a good conductor, s is large, D=eE≈0 hence if medium 2 is a good conductor
Dn1 = ρs
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 24 Proof of boundary conditions - Bn
Proof follows same argument as for Dn on page 47, The integral form of Gauss’ law for magnetostatics is
– there are no isolated magnetic poles ∫∫B.dA = 0
Bn1∆x∆y − Bn2∆x∆y + Ψedge = 0
⇒ Bn1 = Bn2
The normal component of B at a boundary is always continuous at a boundary
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 25 Conditions at a perfect conductor
• In a perfect conductor σ is infinite • Practical conductors (copper, aluminium silver) have very large σ and field solutions assuming infinite σ can be accurate enough for many applications – Finite values of conductivity are important in calculating Ohmic loss • For a conducting medium – J=σE • infinite σ⇒ infinite J • More practically, σ is very large, E is very small (≈0) and J is finite
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 26 Conditions at a perfect conductor
• It will be shown that at high frequencies J is confined to a surface layer with a depth known as the skin depth • With increasing frequency and conductivity the skin depth, δx becomes thinner
Current sheet
δx δx Lower frequencies, Higher frequencies, smaller σ larger σ
It becomes more appropriate to consider the current density in terms of current per unit with:
lim Jδx = J s A/m δx → 0
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 27 Conditions at a perfect conductor cont.
∆x H H H µ ,ε ,σ ∆y y1 x1 y3 1 1 1 H y2 H y4 µ2,ε2,σ2
H x2
• Ampere’s law: ∂D H.ds = + J.dA ∫ ∫∫A ∂t J ∆x 0 0 0 sz ∆y ∆y ∆y ∆y ∂Dz H y2 + H y1 + H x1∆x − H y3 − H y4 − H x2∆x = + J z ∆x∆y 2 2 2 2 ∂t 0 As∆y → 0, ∂Dz ∂t ∆x∆y → 0, J z ∆x∆y → ∆xJ sz
H x1 − H x2 = J sz That is, the tangential component of H is discontinuous by an amount equal to the surface current density
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 28 Conditions at a perfect conductor cont.
• From Maxwell’s equations: – If in a conductor E=0 then dE/dT=0 – Since ∂H ∇×E = −µ dt
Hx2=0 (it has no time-varying component and also cannot be established from zero)
H x1 = J sz
The current per unit width, Js, along the surface of a perfect conductor is equal to the magnetic field just outside the surface: H and J and the surface normal, n, are mutually perpendicular:
J s = n× H
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 29 Summary of Boundary conditions
At a boundary between non-conducting media n×(E − E ) = 0 Et1 = Et2 1 2 n×(H − H ) = 0 Ht1 = Ht2 1 2 ≡ n.(D − D ) = 0 Dn1 = Dn2 1 2 n.(B − B ) = 0 Bn1 = Bn2 1 2 At a metallic boundary (large σ) n×(E1 − E2 ) = 0
n×(H1 − H2 ) = 0
n.(D1 − D2 ) = ρs
n.(B1 − B2 ) = 0 At a perfectly conducting boundary n×E1 = 0
n× H1 = J s
n.D1 = ρs
n.B1 = 0
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 30 Reflection and refraction of plane waves
• At a discontinuity the change in µ, ε and σ results in partial reflection and transmission of a wave • For example, consider normal incidence:
j(ωt−βz) Incident wave = Eie j(ωt+βz) Reflected wave = Er e
Where Er is a complex number determined by the boundary conditions
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 31 Reflection at a perfect conductor
• Tangential E is continuous across the boundary • For a perfect conductor E just inside the surface is zero – E just outside the conductor must be zero Ei + Er = 0
⇒ Ei = −Er
Amplitude of reflected wave is equal to amplitude of incident wave, but reversed in phase
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 32 Standing waves
• Resultant wave at a distance -z from the interface is the sum of the incident and reflected waves
ET (z,t) = incident wave + reflectedwave j(ωt−βz) j(ωt+βz) = Eie + Er e = − jβz − jβz jωt e jφ − e jφ Ei (e e )e sinφ = jωt 2 j = −2 jEi sin βz e
and if Ei is chosen to be real
ET (z,t) = Re{− 2 jEi sin βz (cosωt + j sinωt)}
= 2Ei sin βz sinωt
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 33 Standing waves cont...
ET (z,t) = 2Ei sin βz sinωt
• Incident and reflected wave combine to produce a standing wave whose amplitude varies as a function (sin βz) of displacement from the interface • Maximum amplitude is twice that of incident fields
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 34 Reflection from a perfect conductor Reflection from a perfect conductor
• Direction of propagation is given by E×H If the incident wave is polarised along the y axis:
Ei = a y E yi
⇒ Hi = −a x H xi then E× H = (− a y ×a x )E yi H xi
= +a z E yi H xi That is, a z-directed wave.
For the reflected wave and Ε × H = −a z EyiH xi Er = −a y E yi
So H r = − a x H xi = H i and the magnetic field is reflected without change in phase
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 36 Reflection from a perfect conductor
e jφ + e− jφ • Given that cosφ = 2
j(ωt−βz) j(ωt+βz) HT (z,t) = Hie + H r e jβz − jβz jωt = Hi (e + e )e jωt = 2Hi cos βz e
As for Ei, Hi is real (they are in phase), therefore
HT (z,t) = Re{2Hi cos βz (cosωt + j sinωt)}= 2Hi cos βz cosωt
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 37 HT (Reflectionz,t) = 2Hi cos β zfromcosωt a perfect conductor
• Resultant magnetic field strength also has a standing-wave distribution • In contrast to E, H has a maximum at the surface and zeros at (2n+1)λ/4 from the surface:
resultant wave resultant wave E [V/m] H [A/m]
z [m] z [m]
free space silver free space silver z = 0 z = 0
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 38 Reflection from a perfect conductor
ET (z,t) = 2Ei sin βz sinωt
HT (z,t) = 2Hi cos βz cosωt
E and H are /2 out of phase( ) • T T π sinωt = cos(ωt − π / 2) • No net power flow as expected – power flow in +z direction is equal to power flow in - z direction
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 39 Reflection by a perfect dielectric
• Reflection by a perfect dielectric (J=σE=0) – no loss • Wave is incident normally – E and H parallel to surface • There are incident, reflected (in medium 1)and transmitted waves (in medium 2):
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 40 Reflection from a lossless dielectric Reflection by a lossless dielectric
Ei = η1Hi jωµ µ Er = −η1H r η = = σ + jωεoε r ε Et = η2 Ht • Continuity of E and H at boundary requires:
Ei + Er = Et
Hi + H r = Ht Which can be combined to give 1 1 1 Hi + H r = (Ei − Er ) = Ht = Et = (Ei + Er ) η1 η2 η2
1 1 Er η2 −η1 (Ei − Er ) = (Ei + Er ) ρ = = η η ⇒ E 1 2 Ei η2 +η1 ⇒ η2 (Ei − Er ) = η1(Ei + Er )
⇒ Ei (η2 −η1 ) = Er (η2 +η1 ) The reflection coefficient
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 42 Reflection by a lossless dielectric
Ei + Er = Et
• Similarly Hi + H r = Ht
Et Er + Ei Er η2 −η1 η2 +η1 2η2 τ E = = = +1 = + = Ei Ei Ei η2 +η1 η2 +η1 η2 +η1
2η2 τ E = η2 +η1
The transmission coefficient
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 43 Reflection by a lossless dielectric
• Furthermore: H r Er = − = ρH Hi Ei
Ht η1Et η1 2η2 2η1 = = = τ H Hi η2 Ei η2 η2 +η1 η2 +η1
And because µ=µo for all low-loss dielectrics
E ε − ε n − n ρ = r = 1 2 = 1 2 = −ρ E + H Ei ε1 + ε 2 n1 n2 E 2 ε 2n τ = r = 1 = 1 E + Ei ε1 + ε 2 n1 n2 2 ε 2n τ = 2 = 2 H + ε1 + ε 2 n1 n2
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 44 Energy Transport - Poynting Vector
Electric and Magnetic Energy Density: y For an electromagnetic plane wave E
Ey (x,t) = E0 sin(kx − ωt)
Bz (x,t) = B0 sin(kx − ωt) x B where B0 = E0 c z
The electric energy density is given by
1 2 1 2 2 uE = ε E = ε E sin (kx − ωt ) and the magnetic energy is 2 0 2 0 0 1 2 1 2 uB = B = E = uE Note: I used E = cB 2µ0 2µ0c
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 45 Energy Transport - Poynting Vector cont.
Thus, for light the electric and the magnetic field energy densities are equal and the total energy density is 2 1 2 2 2 utotal = uE +uB = ε0E = B = ε0E 0 sin (kx − ωt ) µ0 Poynting Vector 1 : S = E ×B µ0 The direction of the Poynting Vector is the direction of energy flow and the magnitude E 1 E 2 1 dU = = = y S EB µ0 µ0c A dt z x Is the energy per unit time per unit area B (units of Watts/m2).
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 46 Energy Transport - Poynting Vector cont.
Proof: 2 dU total = utotalV = ε0E Acdt so 2 2 1 dU 2 E E 0 2 S = = ε0cE = = sin (kx − ωt ) A dt µ0c µ0c
Intensity of the Radiation (Watts/m2):
The intensity, I, is the average of S as follows:
1 dU E 2 E 2 I = S = = 0 sin2 (kx − ωt ) = . A dt µ0c 2µ0c
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 47 Ohm’s law J = σE
Skin depth
Current density decays exponentially from the surface into the interior of the conductor
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 48 Phasors
Fictitious way of dealing with AC circuits
jωt V i (t ) = Re{Ie } I = s R + jωL
R=6 Ω
Measurable + Phasor (not real) quantity L=0.2 mH νs(t) -
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 49 Phasors cont.
Phasors in lumped circuit analysis have no space components
Phasors in distributed circuit analysis (RF) have a space component because they act as waves
± jβx jωt V cos(ωt ± βx ) ν(x,t) = Re { V0e }= e 0
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 50 Displacement Current
∂ Observe that the vector field 1 E appears to form a continuation of the c ∂t conduction current distribution. Maxwell called it the displacement current, and the name has stuck although in no longer seem very appropriate.
We can define a displacement current density Jd , to be distinguished from the conduction current density J, by writing 4π curl B = ( J + J ) c d and define 1 ∂E J d = 4π ∂t
It turns out that physical displacement current lead to small magnetic fields that are difficult to detect. To see this effect, we need rapidly changing fields (Hertz experiment).
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 51 Displacement Current
Example: I=Id in a circuit branch having a capacitor C
2a ○○ S V
R d V (t ) Q (t ) E (t ) = = d Cd The displacement current density is given by 1 ∂E (t ) 1 ∂Q (t ) I (t ) J d = = = 4π ∂t 4πCd ∂t 4πCd
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 52 Displacement Current
The direction of the displacement current is in the direction of the current. The total current of the displacement current is A ⋅I I d = A.J d = = I 4πC ⋅d
Thus the current flowing in the wire and the displacement current flowing in the condenser are the same.
How about the magnetic field inside the capacitor? Since the is no real current in the capacitor, 1 ∂E curl B = c ∂t
Integrating over a circular area of radius r,
1 ∂E curl B ⋅da = ⋅da ∫ c ∫ ∂t S (r ) S (r )
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 53 Displacement Current
l .h.s = ∫curlB ⋅da = ∫B ⋅ds = 2πB ⋅ r S (r ) C (r ) 1 ∂ πr 2 ∂E r .h.s = E ⋅da = c ∂t ∫ c ∂t S (r ) πr 2 ∂V πr 2 1 ∂Q πr 2 I 4πI r 2 = = = = cd ∂t cd C ∂t cd C c a 2 Thus the magnetic field in the capacitor is 4πI r 2 2Ir 2πB ⋅ r = → B(r ) = c a 2 ca 2 4πI 2I 2πB ⋅ r = → B(r ) = c cr (at the edge of the capacitor) This is the same as that produced by a current flowing in an infinitely long wire.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 54 Wave in Elastic Medium
xn-2 xn-1 xn xn+1 xn+2
un-2 un-1 un un+1 un+2
The equation of motion for nth mass is 2 ∂ un m = −k(un −un −1)+ k(un +1 −un ) = k(un −1 − 2un +un +1) ∂t 2
By expanding the displacement un±1(t)=u(xn±1,t) around xn, we can convert the equation into a DE with variable x and t.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 55 Wave in Elastic Medium
∂u(x ,t ) 1 ∂2u(x ,t ) u (t ) = u(x ± ∆x t ) = u(x t )+ n (± ∆x )+ n (± ∆x )2 + n ±1 n , n , 2 ∂x n 2 ∂x n
∂2u(x ,t ) ∂2u(x ,t ) m ∂2u(x ,t ) ∂2u(x ,t ) m n = k∆x 2 n → n = k∆x n 2 2 2 2 ∂t ∂x n ∆x ∂t ∂x n
Define K ≡k ∆x as the elastic modulus of the medium and ρ = m/ ∆x is the mass density. In continuous medium limit ∆x 0, we can take out n.
∂2u(x ,t ) ∂2u(x ,t ) ρ = K ∂t 2 ∂x 2
We examine a wave equation in three dimensions. Consider a physical quantity that depends only on z and time t.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 56 Wave along z-axis
∂2Ψ(z ,t ) ∂2Ψ(z ,t ) = ν2 ∂t 2 ∂z 2 We prove that the general solution of this DE is given by Ψ(z ,t ) = f (z −vt )+ g (z +vt ) f and g are arbitrary functions.
Insert a set of new variables, ξ = z −vt and η = z +vt Then ∂ ∂ξ ∂ ∂η ∂ ∂ ∂ = + = + ∂z ∂z ∂ξ ∂z ∂η ∂ξ ∂η and ∂ ∂ξ ∂ ∂η ∂ ∂ ∂ = + = −ν + ν ∂t ∂t ∂ξ ∂t ∂η ∂ξ ∂η
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 57 Wave along z-axis
∂ ∂ 2 ∂ ∂ 2 + Ψ = − Ψ ∂ξ ∂η ∂ξ ∂η
∂2 thus Ψ = 0 ∂η∂ξ
From this equation: ∂ ∂Ψ ∂Ψ = 0 → = F (ξ) ∂η ∂ξ ∂ξ ∂Ψ = F (ξ) → Ψ = F (ξ)dξ + g (η) ≡ f (ξ)+ g (η) ∂ξ ∫
Thus Ψ(z ,t ) = f (z −vt )+ g (z +vt )
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 58 Radiation
F H E ρ β
Charges and currents Great Distance Approximate plane waves
H E H E β Great Distance Aperture fields Approximate plane waves
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 59 Radiation Antennas
Transmission line fed dipole Transmission line fed current loop
Slots in waveguide Waveguide fed horn
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 60 Radiation
In the time domain the electric scalar potential φ (r2,t) and the magnetic
vector potential A(r2,t) produced at time t at a point r2 by charge and current
distribution ρ(r1) and J(r1) are given by 1 ρ(r ,t − r c ) φ(r ,t ) = 1 12 dv 2 πε 4 0 ∫ r12 and v µ J (r ,t − r c ) A(r ,t ) = 0 1 12 dv 2 π 4 ∫ r12 v
Sinusoidal steady state 1 ρ(r )e − jβr12 φ(r ) = 1 dv 2 πε 4 0 ∫ r12 v − jβr e 12 is the phase retardation factor µ J (r )e − jβr12 A(r ) = 0 1 dv 2 π 4 ∫ r12 v
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 61 We start with B = curl A and E = −grad φ − jωA
Charge conservation:
∂ρ Sinusoidal steady state div J + = 0 div J + jωρ = 0 ∂t Because ρ and J are related by the charge conservation equation, φ and A are also related. In the time domain, ∂φ Sinusoidal steady state div A + µ ε = 0 div A + jωµ ε φ = 0 0 0 ∂t 0 0 With ω ≠ 0 div A φ = − jωµ0ε0
Substituting for φ: 1 H = curl A µ0 1 1 E = grad div A − jωA c = ω = cβ jωµ0ε0 µ0ε0 jω = − grad div A − jωA β2
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 62 Near and far fields
We consider the transmission characteristics of a particular antenna in the form of a straight wire, carrying an oscillatory current whose length is much less than the electromagnetic wavelength at the operating frequency. Such antenna is called a short electric dipole. P z P r r θ y z Avoiding spherical polar coordinates θ
L I y φ Coordinates transformation IL
x strength of the radiated field x jωP = IL
The components of the dipole vector in these coordinates are
px − p sin θ = = P 0 0 pz p cos θ
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 63 Dipole radiation
The retarded vector potential is then
µ Je − jβz A = 0 dv 4π ∫ z v ω Where we used β = . We also replace Jdv by IL = j ω P and obtain c ∫ v µ e − jβz A = 0 ( jωP ) 4π z i j k 0 jωµ0 ∂ ∂ ∂ jωµ0 − jβz curl A ≈ = jβPxe 4πz ∂x ∂y ∂z 4πz − jβz − jβz 0 Pxe 0 Pze
Thus the radiation component of the magnetic field has a y component only given by − jβz Pxe H y = − jβjω 4πz
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 64 Dipole radiation
Electric field:
∂A jωµ P (− jβ)e − jβz We start with divA ≈ z = 0 z ∂z 4πz 0 µ jωP (− jβ) then = 0 z grad divA 0 4πz − jβz (− jβ)e
The first term we require for the electric field is simply
2 − jβz 0 − jω − ω µ e grad divA = 0 0 β2 4πz Pz
The second term we require for the electric field is − P 2 − jβz x − ω µ e − jωA = 0 0 4πz − Pz
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 65 Dipole radiation
Electric field:
The electric field is the sum of these two terms. It may be seen that the z components cancel, and we are left with only x component of field given by 2 − jβz ω µ0M xe E x = 4πz
Note that this expression also fits our expectation of an approximately uniform plane wave. The ratio of electric to magnetic field amplitudes is
2 E x µ0ω ω 1 µ0 = = µ0 = µ0c = µ0 = = η H y βω β µ0ε0 ε0
as expected for a uniform plane wave.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 66 Dipole radiation
We will now translate the field components into the spherical polar coordinates.
β in radial direction
P ⊗ H in y direction r θ E in x direction IL o
since P x = − P sin θ we have
2 − jβr − jβr ω µ0P sin θe − ωβP sin θe E θ = E x = and H φ = H y = 4πr 4πr
1 * The Poynting vector ( E × H ) is in r direction and has the value 2
3 2 2 µ0ω βP sin θ S r = S z = 2(4πr )2
This vector (real) gives the real power per unit area flowing across an element of area ⊥ to r at a great distance.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 67 Radiation pattern
Dipole axis
θ Dipole length is proportional to power density per unit area at some fixed distance.
Note: No radiation takes place along the dipole axis, and the radiation pattern has axial symmetry, with maximum radiation being in the equatorial plane.
Because of the non-uniform nature of the pattern we have the concept of antenna gain, which for a lossless antenna is the power flow per unit area for the antenna in the most efficient direction over the power flow per unit area we would obtain if the energy were uniformly radiated in all directions. The total radiated power is
π 2π π 2π 3 2 2 µ0ω βP 3 W = ℜe{S r }(r sin θdθdφ) = sin θdθ dφ ∫ ∫ 32π2 ∫ ∫ θ=0 φ=0 θ=0 φ=0
µ ω3βP 2 = 0 12π2
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 68 The average radiated power per unit area is 2 W µ ω3βP = 0 4πr 2 48π2r 2 Hence the antenna gain, g defined by
radiated power/unit area in the most efficient direction g = average radiated power/unit area over a large sphere
becomes
2 ω3βP 48π2r 2 3 g = = This result is the gain of a small dipole. 32π2r 2 ω3βP 2 2
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 69 Radiation resistance
Recall µ ω3βP 2 µ ωβI 2L2 W = 0 = 0 12π 12π
The radiation resistance Rr is defined as the equivalent resistance which would absorb the same power W from the same current I, i.e. R I 2 W = r 2
Combining these results we obtain
2 µ0ωβL R r = 6π
Using ω = cβ, β = 2π λ , c =1 µ0ε0 and η = µ 0 ε 0 , we find
2 η 2 2π L 2 R r = (βL ) = η R ≈ 20(βL ) Ω (η ≈120πΩ) 6π 3 λ r
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 70 Consider an arbitrary system of radiating currents y2 z2 We start with the vector potential ⊗ Far field point − jβr12 z1 r12 µ0 J (r1)e α x2 A(r ) = dv P2 2 π 4 ∫ r12 θ r v P 2 1 ψ We will regard r12 fixed. For P2 a r1 distance point, we replace r12 with r2 o µ A(r ) = 0 J (r )e − jβr12dv So 2 π 1 4 r2 ∫ v
− jβr12 Approximations for r12 in e require more care, sine phase differences in radiation effects are crucial. We use the following approximation
r2 = r1 + r12 r2 ≈ r1 cos ψ + r12 r12 ≈ r2 − r1 cos ψ
− jβr2 µ0e + jβr cos ψ + β ψ A(r ) = J (r )e 1 dv factor e j r1 cos expresses the 2 π 1 4 r2 ∫ phase advance of the radiation from v the element at P1 relative to the phase at the origin.
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 71 − jβr2 µ0e We have A(r2 ) = ℜ 4πr2 where β ψ j r1 cos is called the radiation vector. It depends on the ℜ = J (r1)e dv ℜ ∫ internal geometrical distribution of the currents v and on the direction of P2 from the origin O, but not on the distance.
− jβr2 µ0e The factor depends only on the distance from the origin O to the field point 4πr2 P2 but not on the internal distribution of the currents in the antenna.
The radiation vector ℜ can be regarded as an effective dipole equal to the sum of the individual dipole elements Jdv , each weighted by phase factor e jβ r 1 cos ψ , which depends on
the phase advance βr1 cos ψ of the element in relation to the origin, and direction OP2.
− jβr e e − jβr H θ = jβ ℜφ and H φ = − jβ ℜθ 4πr 4πr
E θ = ηH φ and E φ = −η H θ
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 72 Small circular loop
z P2 Calculate the radiated fields and power at large distance. r y Using the symmetry the results will be independent of the azimuth coordinate φ. θ The spherical polar coordinates of a point P1 at a general position on the loop are (a, π/2, φ′). a o φ′ ψ I being the angle between OP1 and OP2 with a unit vector in the direction of OP1 (cos φ′,sin φ′,0) and a unit vector in the direction of OP2 (sinθ,0,cosθ):
x We have cos ψ = sin θcos φ′
The radiation vector is then given by filamentary current ℜ(θ ) = J (r )⋅u e jβa sin θcos φ′dv ℜ(θ ) = Idr ⋅u e jβa sin θcos φ′ ,0 ∫ 1 ˆφ ,0 ∫ 1 ˆφ 2π 2π jβa sin θcos φ′ ℜ(θ,0) = ∫Iae cos φ′dφ′ ℜ(θ,0) ≈ Ia∫ (1+ jβa sin θcos φ′)cos φ′dφ′ 0 2 0 ℜφ(θ,0) = jβπIa sin θ
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 73 Electric and magnetic fields
− jβr 2 2 jβe − (βa) I sin θ − jβr (βa) ηI sin θ − jβr H θ = ℜ = e and E φ = −ηH θ = e 4πr φ 4r 4r Poynting vector 4 2 2 1 * (βa) ηI sin θ S r = − E φH θ = 2 32r 2
Total power radiated 2π π = θ φ θ 3 1 W S r r sin d rd Substituting for Sr and using sin θ = (3sin θ − sin3θ) ∫φ=0 ∫θ=0 4 πηI 2 (βa)4 W = 12
Radiation resistance
1 2 πη 4 2 4 W = ℜr I ℜr = (βa) ℜ = 20π (βa) Ω (η =120πΩ) 2 6 r
Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 74