Electromagnetism

Christopher R Prior

Fellow and Tutor in Mathematics ASTeC Intense Beams Group Trinity College, Oxford Rutherford Appleton Laboratory Contents

• Review of Maxwell’s equations and Lorentz Force Law • Motion of a charged particle under constant Electromagnetic fields • Relativistic transformations of fields • Electromagnetic waves – Waves in vacuo – Waves in conducting medium • Waves in a uniform conducting guide

– Simple example TE01 mode – Propagation constant, cut-off frequency – Group velocity, phase velocity – Illustrations

2 Reading

• J.D. Jackson: Classical Electrodynamics • H.D. Young and R.A. Freedman: University Physics (with Modern Physics) • P.C. Clemmow: Electromagnetic Theory • Feynmann Lectures on Physics • W.K.H. Panofsky and M.N. Phillips: Classical Electricity and Magnetism • G.L. Pollack and D.R. Stump: Electromagnetism

3 Basic Equations from Vector Calculus For a scalar function ϕ(x,y,z,t), ⎛ ∂ϕ ∂ϕ ∂ϕ ⎞ Gradient is normal to surfaces gradient : ∇ϕ = ⎜ , , ⎟ φ=constant ⎝ ∂x ∂y ∂z ⎠  For a vector F = (F1,F2,F3)  ∂F ∂F ∂F divergence : ∇⋅ F = 1 + 2 + 3 ∂x ∂y ∂z

 ⎛ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1⎞ curl : ∇ ∧ F = ⎜ − , − , − ⎟ ⎝ ∂y ∂z ∂z ∂x ∂x ∂y ⎠

4 Basic Vector Calculus       ∇ ⋅(F ∧ G) = G ⋅∇ ∧ F − F ⋅∇ ∧ G  ∇ ∧ ∇ϕ = 0, ∇ ⋅∇ ∧ F = 0    ∇ ∧ (∇ ∧ F) = ∇(∇ ⋅ F) − ∇2F

Stokes’ Theorem Divergence or Gauss’     Theorem ∫∫ ∇ ∧ F ⋅ dS = ∫ F ⋅ dr    S C  ∫∫∫ ∇⋅ F dV = ∫∫ F ⋅ dS   n V S dS = n dS Closed surface S, volume V, Oriented outward pointing normal boundary C 5 What is Electromagnetism?

• The study of Maxwell’s equations, devised in 1863 to represent the relationships between electric and magnetic fields in the presence of electric charges and currents, whether steady or rapidly fluctuating, in a vacuum or in matter.

• The equations represent one of the most elegant and concise way to describe the fundamentals of electricity and magnetism. They pull together in a consistent way earlier results known from the work of Gauss, Faraday, Ampère, Biot, Savart and others.

• Remarkably, Maxwell’s equations are perfectly consistent with the transformations of special relativity. Maxwell’s Equations

Relate Electric and Magnetic fields generated by charge and current distributions.

E = electric field  D = electric displacement ∇ ⋅ D = ρ H = magnetic field  B = magnetic flux density ∇ ⋅ B = 0 ρ= charge density   ∂B j = current density ∇ ∧ E = − µ (permeability of free space) = 4π 10-7 ∂t 0  -12 ε0 (permittivity of free space) = 8.854 10   ∂D ∇ ∧ H = j + c (speed of light) = 2.99792458 108 m/s ∂t

    2 In vacuum D = ε 0 E, B = µ0 H, ε 0µ0c =1

 ρ ∇ ⋅ E = Maxwell’s 1st Equation ε0 Equivalent to Gauss’ Flux Theorem:  ρ    1 Q ∇ ⋅ E = ⇔ ∫∫∫∇ ⋅ E dV =∫∫ E ⋅ dS = ∫∫∫ ρ dV = ε0 V S ε0 V ε0 The flux of electric field out of a closed region is proportional to the total electric charge Q enclosed within the surface. A point charge q generates an electric field  q  E = 3 r 4πε0r   q dS q E ⋅dS = = ∫∫ ∫∫ 2 sphere 4πε0 sphere r ε 0 Area integral gives a measure of the net charge enclosed; divergence of the electric field gives the density of the sources. 8  ∇ ⋅ B = 0 Maxwell’s 2nd Equation

Gauss’ law for magnetism: B! =0 B! dS! ∇ · ⇐⇒ · The net magnetic flux out!! of any closed surface is zero. Surround a magnetic dipole with a closed surface. The magnetic flux directed inward towards the south pole will equal the flux outward from the north pole. If there were a magnetic monopole source, this would give a non-zero integral.

Gauss’ law for magnetism is then a statement that There are no magnetic monopoles   ∂B ∇ ∧ E = − Maxwell’s 3rd Equation ∂t

Equivalent to Faraday’s Law of Induction:   ∂B  ∫∫ ∇ ∧ E ⋅ dS = −∫∫ ⋅ dS S S ∂t  d   dΦ ⇔ E ⋅ dl = − B ⋅ dS = − ∫ dt ∫∫ dt C S (for a fixed circuit C) N S The electromotive  force round a circuit ε = ∫ E ⋅ d l is proportional to the rate of   change of flux of magnetic field, Φ = ∫∫ B ⋅ d S through the circuit. Faraday’s Law is the basis for electric generators. It also forms the basis for inductors and transformers.    1 ∂E th ∇ ∧ B = µ0 j + 2 Maxwell’s 4 Equation c ∂t   Originates from Ampère’s (Circuital) Law : ∇ ∧ B = µ0 j       B ⋅ dl = ∇ ∧ B ⋅ dS = µ j ⋅ dS = µ I ∫ ∫∫ 0 ∫∫ 0 C S S Satisfied by the field for a steady line current (Biot-Savart Law, Ampère 1820):    µ I dl ∧ r B = 0 4π ∫ r3 µ I For a straight line current B = 0 θ 2π r Biot Need for Displacement Current

• Faraday: vary B-field, generate E-field • Maxwell: varying E-field should then produce a B-field, but not covered by Ampère’s Law.  Apply Ampère to surface 1 (flat disk): line

integral of B =μ0I Surface 1 Surface 2  Applied to surface 2, line integral is zero since no current penetrates the deformed Current I surface. Q dQ dE  In capacitor, E = , so I = = ε A A dt 0 dt ε0  Closed loop  ∂E  Displacement current density is j = ε d 0 ∂t      ∂E ∇ ∧ B = µ0 (j + jd )= µ0 j + µ0ε0 ∂t 12 Consistency with Charge Conservation

Charge conservation: From Maxwell’s equations: Total current flowing out of a Take divergence of (modified) region equals the rate of decrease Ampère’s equation  of charge within the volume.   1 ∂E ∇ ∧ B = µ j + 0 c 2 t   d ∂ j ⋅ dS = − ρdV  1 ∂  ∫∫ ∫∫∫ ⇒ ∇⋅∇ ∧ B = µ ∇⋅ j + ∇⋅ E dt 0 c 2 ∂t ( )  ∂ρ ⇔ ∇⋅ j dV = − dV  ∂ ⎛ ρ ⎞ ∫∫∫ ∫∫∫ ∂t ⇒ 0 = µ0∇⋅ j + ε0µ0 ⎜ ⎟ ∂t ⎝ ε0 ⎠  ∂ρ ⇔ ∇⋅ j + = 0  ∂ρ ∂t ⇒ 0 = ∇⋅ j + ∂t Charge conservation is implicit in Maxwell’s Equations

13 Maxwell’s Equations in Vacuum

In vacuum Equivalent integral forms     1 (useful for simple geometries) D = ε0 E , B = µ0H , ε0µ0 = 2 c Source-free equations:   1  ∫∫ E ⋅ dS = ∫∫∫ ρdV ∇⋅ B = 0 ε0     ∂B B dS 0 ∇ ∧ E + = 0 ∫∫ ⋅ = ∂t   d   dΦ Source equations: ∫ E ⋅ dl = − ∫∫ B ⋅ dS = −  ρ dt dt ∇⋅ E =     1 d   ε0 B ⋅ dl = µ j ⋅dS + E ⋅ dS  ∫ 0 ∫∫ 2 ∫∫  1 ∂E  c dt ∇ ∧ B − 2 = µ0 j c ∂t 14 Example: Calculate E from B   d  E ⋅dl = − B ⋅dS z ∫ dt ∫∫

d r < r 2π rE = − π r2B sinωt = −π r2B ω cosωt 0 θ dt 0 0 r 1 ⇒ Eθ = − B0ω r cos ωt ⎧B0 sinωt r < r0 2 Bz = ⎨ ⎩ 0 r > r0 d r > r 2π rE = − π r2B sinωt = −π r2B ω cosωt  0 θ dt 0 0 0 0  B 2 Also from ∂ ω r B ∇ ∧ E = − ⇒ E = − 0 0 cosωt ∂t θ 2r    1 ∂E ∇ ∧ B = µ j + then gives current density necessary 0 c2 dt to sustain the fields Lorentz Force Law

• Thought of as a supplement to Maxwell’s equations but actually implicit in relativistic formulation, gives force on a charged particle moving in an electromagnetic field:     f = q(E + v ∧ B) • For continuous distributions, use force density     f d = ρE + j ∧ B • Relativistic equation of motion

dP #v f# 1 dE dp# – 4-vector form: F = = γ · , f# = γ , dτ ⇒ c c dt dt ! " # $ – 3-vector component: Energy component:

d        dE 2 dγ (m0γ v ) = f = q(E + v ∧ B ) v ⋅ f = = m0c dt dt dt 16

Motion of charged particles in constant magnetic fields d        m γ v = f = q E + v ∧ B = qv ∧ B dt ( 0 ) ( )

d 2      (m0γ c ) = v ⋅ f = qv ⋅v ∧ B = 0 dt 1. From energy equation, γ is constant No acceleration with a magnetic field 2. From momentum equation,  d  d    B ⋅ (γ v ) = 0 = γ (B ⋅v ) ⇒ v // is constant dt dt !v constant and !v constant |=| !v also constant| !| ⇒ | ⊥| 17 Motion in Constant magnetic field  d    dv q   (m0γ v ) = qv ∧ B ⇒ = v ∧ B dt dt m0γ Constant magnetic field 2 gives uniform spiral about B v⊥ q ⇒ = v⊥B with constant energy. ρ m0γ m γ v ⇒ circular motion with radius ρ = 0 ⊥ qB

v⊥ qB at angular frequency ω = = (m = m0γ ) ρ m m γ v p Bρ = 0 = q q Magnetic rigidity Motion in Constant Electric Field d      d   (m0γ v ) = f = q(E + v ∧ B ) → (m0γ v ) = q E dt dt d  q  Solution of (γ v)= E dt m0 2 2 qE 2 ⎛ γ v⎞ ⎛ qE ⎞ is γ v = t ⇒ γ = 1+ ⇒ γ = 1+ ⎜ t⎟ m0 ⎝ c ⎠ ⎝ m0c ⎠

2 dx v m c 2 ⎡ ⎛ qEt ⎞ ⎤ γ 0 ⎢ ⎥ = ⇒ x = x0 + 1+ ⎜ ⎟ −1 dt γ qE ⎢ ⎝ m c⎠ ⎥ ⎣ 0 ⎦

1 qE 2 ≈ x0 + t for qE << m0c 2 m0 Energy gain is qEx

Constant E-field gives uniform acceleration in straight line 19 Relativistic Transformations of E and B

• According to observer O in frame F, particle has velocity v, fields are E and B and Lorentz force is f! = q E! + !v B! × ! " • In Frame F’, particle is at rest and force is f!! = q!E! !

• Assume measurements give same charge and force, so

q! = q and E! ! = E! + !v B! × q !v !r • Point charge at rest in F: ! ! q E = ×3 , B =0 4π#0 r

µ0q !v !r 1 • See a current in F’, giving a field B! ! = × = !v E! Rough idea− 4π r3 −c2 × 1 • Suggests B! ! = B! !v E! − c2 × Review of Waves ∂2u 1 ∂2u • 1D wave equation is 2 = 2 2 with general solution ∂x v ∂t u(x, t)=f(vt x)+g(vt + x) − • Simple plane wave:   1D : sin(ωt − k x) 3D : sin ωt − k ⋅ x ( )

2π Wavelength is λ = #k | | ω Frequency is ν = 2π Phase and group velocities

∞ ∫ A(k)ei[ω(k )t−kx]dk −∞ Superposition of plane waves. While Plane wave sin( ω t kx ) has constant shape is relatively undistorted, pulse phase ω t kx = −1 π at peaks − 2 travels with the Group Velocity ω k∆x =0 dω − vg = dk ∆x ω v = = ⇐⇒ p ∆t k Wave packet structure

• Phase velocities of individual plane waves making up the wave packet are different, • The wave packet will then disperse with time

23 Electromagnetic waves

• Maxwell’s equations predict the existence of electromagnetic waves, later discovered by Hertz.   • No charges, no currents:  ∂D  ∂B ∂B! ∇ ∧ H = ∇ ∧ E = − E! = ∂t ∂t ∇× ∇× −∇ × ∂t   ! " ∇ ⋅ D = 0 ∇ ⋅ B = 0 ∂ = B! −∂t ∇× 3D wave equation: ! " 2 2 2 ! 2 ! 2 ! 2 ! ∂ D! ∂ E! 2 ! ∂ E ∂ E ∂ E ∂ E = µ = µ# E = 2 + 2 + 2 = µ# 2 − ∂t2 − ∂t2 ∇ ∂x ∂y ∂z ∂t

( E! )= ( E! ) ! E! ∇× ∇× ∇ ∇ · −∇ = ! E! −∇ Nature of Electromagnetic Waves

• A general plane wave with angular frequency ω travelling in the direction of the wave vector !k has the form         E = E0 exp[i(ωt − k ⋅ x)] B = B0 exp[i(ωt − k ⋅ x)]   • Phase ω t − k ⋅ x = 2π × number of waves and so is a Lorentz invariant. • Apply Maxwell’s equations:        ∇ ↔ −ik ∇⋅ E = 0 = ∇⋅ B ↔ k ⋅ E = 0 = k ⋅ B ∂  ↔ iω  ˙   ∂t ∇ ∧ E = −B ↔ k ∧ E = ωB Waves are transverse  to the direction of propagation, E , B and k are mutually perpendicular

Plane Electromagnetic Wave

26 Plane Electromagnetic Waves

  1 ∂E   ω  ∇ ∧ B = ↔ k ∧ B = − E c2 ∂t c2    Combined with k ∧ E = ωB  2 ω E ω kc speed of wave in vacuum is c deduce that  = = ⇒  = B k ω k

2π Reminder: The fact that ω t "k "x is an Wavelength λ =  invariant tells us that − · k ω Λ = ,"k ω c Frequency ν = is a Lorentz 4-vector,! the 4-Frequency" vector. 2π Deduce frequency transforms as c v ω! = γ(ω #v #k)=ω − − · c + v ! Waves in a Conducting Medium         E = E0 exp[i(ωt − k ⋅ x)] B = B0 exp[i(ωt − k ⋅ x)] • (Ohm’s Law) For a medium of conductivity σ, !j = σ E ! ∂E! ∂E! • Modified Maxwell: H! = !j + " = σE! + " ∇∧ ∂t ∂t i!k H! = σE! + iω$E! σ − ∧ • Put D = ω# conduction displacement current Dissipation current factor

7 12 Copper : σ = 5.8×10 , ε = ε 0 ⇒ D =10 -8 −4 Teflon : σ = 3×10 , ε = 2.1ε 0 ⇒ D = 2.57×10 Attenuation in a Good Conductor         − ik ∧ H = σ E + iωε E ⇔ k ∧ H = iσ E − ωε E   ∂B    Combine with ∇ ∧ E = − ⇒ k ∧ E = ωµH ∂t       ⇒ k ∧ (k ∧ E ) = ωµ k ∧ H = ωµ (iσ − ωε)E      ⇒ (k ⋅ E ) k − k 2 E = ωµ (iσ − ωε)E   k 2 i since k E 0 ⇒ = ωµ (− σ + ωε) ⋅ = ωµσ For a good conductor D >> 1, σ >> ωε, k 2 ≈ −iωµσ ⇒ k ≈ (1− i) 2 ⎡ ⎛ x ⎞⎤ ⎛ x ⎞ 1 Wave form is exp⎢i⎜ωt − ⎟⎥exp⎜− ⎟, k = (1− i) ⎣ ⎝ δ ⎠⎦ ⎝ δ ⎠ δ 2 where δ = is the skin - depth ωµσ Maxwell’s Equations in a Uniform Perfectly Conducting Guide Hollow metallic cylinder with perfectly conducting boundary surfaces x z Maxwell’s equations with time dependence exp(iωt) are:      ∂B  2 ∇ ∧ E = − = −iωµH ∇ E = ∇(∇ ⋅ E)− ∇ ∧ (∇ ∧ E) y ∂t   ⇒ = iωµ ∇ ∧ H  ∂D   ∇ ∧ H = = iωε E = −ω2εµ E ∂t  ⎧  ⎫ ⎛ ⎞ ⎜ 2 2 ⎟⎪ E ⎪ ⎨ ⎬ 0 ⎜∇ + ω µε ⎟  = ⎝ ⎠⎪ ⎪   ⎩⎪H ⎭⎪ Assume E(x, y, z,t) = E(x, y)e(iωt−γ z)    (iωt−γ z) ⎧ E ⎫ H (x, y, z,t) = H (x, y)e Then ∇ 2 + (ω 2εµ + γ 2) ⎨  ⎬ = 0 [ t ] H γ is the propagation constant ⎩ ⎭ Can solve for the fields completely in terms of Ez and Hz A simple model: “Parallel Plate Waveguide”

Transport between two infinite conducting plates (TE01 mode):  E = (0,1,0)E(x) e(iωt−γ z) where E(x) satisfies d 2E ∇2E = = −K 2E, K 2 = ω2εµ + γ 2 t dx2 y ⎧sin ⎫ i.e. E = A⎨ ⎬Kx ⎩cos⎭ x To satisfy boundary conditions, E=0 on x=0 and x=a, need z nπ E = Asin Kx, K = Kn = , n integer x=0 a x=a Propagation constant is 2 nπ ⎛ ω ⎞ K K 2 2 1 where n γ = n − ω εµ = − ⎜ ⎟ ωc = a ⎝ ωc ⎠ εµ

31 Cut-off frequency, ωc

2 nπ ⎛ ω ⎞ nπx iω t −γ z nπ γ = 1− ⎜ ⎟ , E = A sin e , ωc = a ⎝ ωc ⎠ a a εµ

. ω<ωc gives real solution for γ, so attenuation only. No wave propagates: cut-off modes.

. ω> ωc gives purely imaginary solution for γ, and a wave propagates without attenuation. 1 1 2 2 2 2 2 ⎛ ωc ⎞ γ = ik, k = εµ (ω − ωc ) = ω εµ ⎜1 − 2 ⎟ ⎝ ω ⎠ . For a given frequency ω only a finite number of modes can propagate. nπ aω ω > ω = ⇒ n < εµ c a εµ π For given frequency, convenient to choose a s.t. only n=1 mode occurs.

32 Waveguide animations

• above cut-off

• lower ω

• at cut-off

• below cut-off

• variable ω

33 Phase and group velocities in the simple wave guide

1 Wave number: 2 2 2 k = εµ(ω −ωc ) < ω εµ 2π 2π Wavelength: λ = > , the free − space wavelength k ω εµ ω 1 Phase velocity: vp = > , k εµ larger than free - space velocity Group velocity: 2 2 2 dω k 1 k = εµ ω −ωc ⇒ vg = = < ( ) dk ωεµ εµ smaller than free - space velocity 34

Calculation of Wave Properties

• If a=3 cm, cut-off frequency of lowest order mode is 8 ωc 1 3×10 nπ fc = = ≅ ≅ 5GHz ωc = 2π 2a εµ 2 × 0.03 a εµ • At 7 GHz, only the n=1 mode propagates and

1 1 2 2 2 2 2 2 9 8 −1 k = εµ(ω −ωc ) ≅ 2π(7 − 5 ) ×10 /3×10 ≈103m 2π λ = ≈ 6cm k ω v = ≈ 4.3×108ms−1 > c p k k v = = 2.1×108 ms−1 < c g ωεµ 35

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