kostas kokkotas "field theory"

7/23/2019 Kostas Kokkotas "Field Theory"

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Kostas Kokkotas

Field Theory

Handouts from the course

July 9, 2010


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Preface

These notes are from the transparencies written for the course Field Theory.The parts related to Electrodynamics were based mainly on the book by

J.D. Jackson Classical Electrodynamics while other books like GlassicalElectrodynamics by W. Greiner

while we have also used

The Classical Theory of Fieldsby L.D. Landau & E.M. Lifshitz Eric Poissons notes Electrodynamicsby Fulvio Melia Classical Electrodynamicsby Walter Greiner

For the parts related to General Theory of Relativity we have used thebooks

General Relativity: An Introduction for Physicistsby M.P.Hobson, G. Ef-stathiou & A.N. Lasenby

A first course in General Relativityby B.F. Schutz. Gravity: an Introduction to Einsteins General Relativityby James B. Har-

tle


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Contents

1 Electrostatics-Magnetostatics (****). . . . . . . . . . . . . . . . . . . . . . . 11.1 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Green Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Laplace Equation in Spherical Coordinates . . . . . . . . . . . . . . . . . 51.4 Legendre Equation and Legendre Polynomials . . . . . . . . . . . . . . . 6

1.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Associated Legendre Functions and Spherical Harmonics . . . . . 10

1.6.1 Spherical HarmonicsYlm(, ) . . . . . . . . . . . . . . . . . . . . . . 111.6.2 Addition Theorem for Spherical Harmonics . . . . . . . . . . . 12

1.7 Multipole Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7.1 Monopole momentl = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.7.2 Dipole momentl = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.7.3 Quadrupole momentl = 2. . . . . . . . . . . . . . . . . . . . . . . . . . 161.7.4 Multipole Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.8 Energy of a Charge Distribution in an External Field . . . . . . . . 181.8.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.9 Magnetostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.10 Biot & Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.10.1 Diff. Equations for Magnetostatics & Amperes Law . . . 211.11 Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Classical Field Theory: Maxwell Equations . . . . . . . . . . . . . . . . 252.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Faradays Law of Induction (****) . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2.1 Faradays law for a moving circuit . . . . . . . . . . . . . . . . . . . 262.3 Energy in the Magnetic Field (***) . . . . . . . . . . . . . . . . . . . . . . . . 282.4 Maxwell Equations (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.4.1 Vector and Scalar Potentials . . . . . . . . . . . . . . . . . . . . . . . . 332.4.2 Gauge Transformations :Lorenz Gauge . . . . . . . . . . . . . . 342.4.3 Gauge Transformations :Coulomb Gauge . . . . . . . . . . . . 35


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VIII Contents

2.5 Green Functions for the Wave Equations (***) . . . . . . . . . . . . . . 352.5.1 Poyntings Theorem : Conservation of Energy (****) . . . 382.5.2 Poyntings Theorem : Conservation of Momentum (***) 39

2.6 Maxwell Stress Tensor (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.7 Conservation of Angular Momentum (**) . . . . . . . . . . . . . . . . . . . 41

3 Electromagnetic Waves (****). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1 Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.2 Plane Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.3 Linear and Circular Polarization of EM Waves . . . . . . . . . . . . . . 46

3.3.1 Circular Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.3.2 Elliptically Polarized EM Waves . . . . . . . . . . . . . . . . . . . . . 48

3.4 Stokes Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.5 Reflection & Refraction of EM Waves . . . . . . . . . . . . . . . . . . . . . . 50

3.5.1 E : Perpendicular to the plane of incidence . . . . . . . . . . . 533.5.2 E : Parallel to the plane of incidence . . . . . . . . . . . . . . . . . 543.5.3 Normal incidencei = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 Simple Radiating Systems (****). . . . . . . . . . . . . . . . . . . . . . . . . . 574.1 Fields and Radiation of a Localized Oscillating Source . . . . . . . 574.2 The Near Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.2.1 The Far Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.2.2 The Intermediate Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2.3 Electric Monopole Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.3 Electric Dipole Fields and Radiation . . . . . . . . . . . . . . . . . . . . . . . 604.3.1 Electric Dipole : Power Radiated . . . . . . . . . . . . . . . . . . . . 62

4.4 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.4.1 Example I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.4.2 Example II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.5 Magnetic Dipole & Electric Quadrupole Fields . . . . . . . . . . . . . . 654.5.1 Magnetic Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.5.2 Electric Quadrupole Fields . . . . . . . . . . . . . . . . . . . . . . . . . 674.5.3 Example : Electric Quadrupole Fields . . . . . . . . . . . . . . . . 694.5.4 Example : Pulsar spin-down . . . . . . . . . . . . . . . . . . . . . . . . 69

5 Radiation by Moving Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.1 Lienard - Wiechert Potentials (**) . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.1.1 Special Note about theshrinkage factor(1 n ) . . . . 765.2 Lienard - Wiechert potentials :radiation fields(****) . . . . . . . . 76

5.2.1 Some observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.3 Power radiated by an accelerated charge (****) . . . . . . . . . . . . . 78

5.3.1 Larmor Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3.2 Relativistic Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.4 Applications (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81


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Contents IX

5.5 Angular Distribution of Radiation Emitted by an AcceleratedCharge (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.6 Angular distribution of radiation from a charge in circularmotion (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5.7 Radiation from a charge in arbitrary motion (***) . . . . . . . . . . . 865.8 Distribution in Frequency and Angle of Energy Radiated ...

(****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.9 What Is Synchrotron Light? (***) . . . . . . . . . . . . . . . . . . . . . . . . . 91

5.9.1 Synchrotrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.10 Synchrotron Radiation (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.11 Thomson Scattering of Radiation (***) . . . . . . . . . . . . . . . . . . . . . 99

6 Special Theory of Relativity (****). . . . . . . . . . . . . . . . . . . . . . . . 1036.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.2 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.3 Invariance of Electric Charge; Covariance in Electrodynamics . 1076.4 Dual Field-Strength Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.5 Transformation of Electromagnetic Fields. . . . . . . . . . . . . . . . . . . 1106.6 Transformation of Electromagnetic Fields: Example . . . . . . . . . . 111

7 Dynamics of Relativistic Particles and EM Fields (**). . . . . 1157.1 Lagrangian Hamiltonian for a Relativistic Charged Particle . . . 115

7.1.1 Relativistic Lagrangian (Elementary) . . . . . . . . . . . . . . . . 1167.1.2 Relativistic Lagrangian (Covariant Treatment) . . . . . . . . 1187.1.3 Relativistic Hamiltonian (Covariant Treatment) . . . . . . . 119

7.2 Motion in a Uniform, Static Magnetic Field (****) . . . . . . . . . . 1217.3 Motion in Combined, Uniform, Static E- and B- Field (***) . . 1227.4 Lowest Order Relativistic Corrections to the Lagrangian... . . . . 1247.5 Lagrangian for the Electromagnetic Field . . . . . . . . . . . . . . . . . . . 1257.6 Proca Lagrangian; Photon Mass Effect . . . . . . . . . . . . . . . . . . . . . 1267.7 Conservation Laws : Canonical Stress Tensor . . . . . . . . . . . . . . . . 1277.8 Conservation Laws : Symmetric Stress Tensor . . . . . . . . . . . . . . . 1297.9 Conservation Laws for EM fields interacting with Charged

Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

8 A Short Introduction to Tensor Analysis . . . . . . . . . . . . . . . . . . 1338.1 Scalars and Vectors (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

8.1.1 Vector Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1348.1.2 Contravariant and Covariant Vectors. . . . . . . . . . . . . . . . . 134

8.2 Tensors: at last (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.2.1 Tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.2.2 Tensors: Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.2.3 Tensors: Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.2.4 Covariant Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1378.2.5 Parallel Transport of a vector . . . . . . . . . . . . . . . . . . . . . . . 137


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X Contents

8.2.6 Curvature Tensor (**) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.3 Geodesics (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.4 Metric Tensor (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

8.4.1 The Determinant ofg(**) . . . . . . . . . . . . . . . . . . . . . . . 1418.4.2 Christoffel Symbols (****) . . . . . . . . . . . . . . . . . . . . . . . . . 142

8.5 Geodesics in a Riemann Space (***) . . . . . . . . . . . . . . . . . . . . . . 1438.5.1 Euler-Lagrange Eqns vs Geodesic Eqns (***) . . . . . . . . . 1458.5.2 Tensors: Geodesics (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . 1458.5.3 Null Geodesics (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.5.4 Geodesic Eqns & Affine Parameter (**) . . . . . . . . . . . . . 146

8.6 Riemann Tensor (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.6.1 The Ricci and Einstein Tensors (***) . . . . . . . . . . . . . . . . 1478.6.2 Flat & Empty Spacetimes . . . . . . . . . . . . . . . . . . . . . . . . . . 148

8.7 Weyl Tensor (-) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.8 Tensors : An example for parallel transport (****) . . . . . . . . . . . 149

9 Physics on Curved Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.1 The electromagnetic (EM) force of a moving charge (***). . . . . 1519.2 The 4-current density (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.3 The EM field equations (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1529.4 Electromagnetism in the Lorenz gauge (***) . . . . . . . . . . . . . . . . 1539.5 Electric and Magnetic Fields in inertial frames (***) . . . . . . . . . 1549.6 Electromagnetism in arbitrary coordinates (***) . . . . . . . . . . . . . 1559.7 Equations of motion for charged particles (***) . . . . . . . . . . . . . 1569.8 Energy-Momentum Tensor(***) . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

9.8.1 The Energy-Momentum tensor of a perfect fluid(**) . . . 159

9.8.2 The Energy-Momentum tensor of a real fluid (*) . . . . 1599.9 Conservation of Energy and Momentum for a perfect fluid (**) 1609.10 Linear Field Equations for Gravitation (**) . . . . . . . . . . . . . . . . . 1619.11 Interaction of Gravitation and Matter (*) . . . . . . . . . . . . . . . . . . . 1639.12 Local Cartesian coordinates(-) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

9.12.1 Local geodesic coordinates and Cartesian coordinates(-) 1679.13 Geodesic deviation (-) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1689.14 Tidal forces in a curved spacetime (-) . . . . . . . . . . . . . . . . . . . . . . 169

10 Einsteins Theory of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17110.1 Newtonian Gravity (***) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17110.2 Equivalence Principle (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

10.2.1 Equivalence Principle : Dickes Experiment . . . . . . . . . . . 173

10.3 Einsteins Equations (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17410.3.1 Newtonian Limit (**) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17510.4 Solutions of Einsteins Equations (*) . . . . . . . . . . . . . . . . . . . . . . . 17510.5 Schwarzschild Solution (****) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

10.5.1 Schwarzschild Solution: Geodesics . . . . . . . . . . . . . . . . . . . 17810.5.2 Radial motion of massive particles (**). . . . . . . . . . . . . . . 179


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Contents XI

10.5.3 Circular motion of massive particles (**) . . . . . . . . . . . . . 18110.5.4 Stability of massive particle orbits (-) . . . . . . . . . . . . . . . . 18210.5.5 Trajectories of photons (-) . . . . . . . . . . . . . . . . . . . . . . . . . . 18510.5.6 Radial motion of photons(-) . . . . . . . . . . . . . . . . . . . . . . . . 18610.5.7 Circular motion of photons (-) . . . . . . . . . . . . . . . . . . . . . . 18610.5.8 Stability of photon orbits (-) . . . . . . . . . . . . . . . . . . . . . . . 186

10.6 The slow-rotation limit : Dragging of inertial frames(-) . . . . . . 18710.7 The Classical Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

10.7.1 The Classical Tests: Perihelion Advance (***) . . . . . . . . . 18810.7.2 The Classical Tests : Deflection of Light Rays (***) . . . . 19010.7.3 The Classical Tests : Gravitational Redshift (***) . . . . . 19210.7.4 The Classical Tests : Radar Delay (***) . . . . . . . . . . . . . . 193

11 Solutions of Einsteins Equations & Black Holes . . . . . . . . . . . 19711.1 Schwarzschild Solution: Black Holes (***) . . . . . . . . . . . . . . . . . . 197

11.1.1 Eddington-Filkenstein coordinates . . . . . . . . . . . . . . . . . . 19711.1.2 Kruskal - Szekeres Coordinates : Maximal Extension (*) 19911.1.3 Wormholes (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

11.2 Reissner-Nordstrm Solution (1916-18) (*) . . . . . . . . . . . . . . . . . . 20011.3 Kerr Solution (1963) (**) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

11.3.1 Infinite redshift surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20111.3.2 Penrose Process (1969) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20211.3.3 Black-Hole: Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

11.4 The Tolman-Oppenheimer-Volkov (TOV) Solution (-) . . . . . . . . 20411.4.1 TOV : A uniform density star . . . . . . . . . . . . . . . . . . . . . . . 205

A Useful constants in geometrical units . . . . . . . . . . . . . . . . . . . . . . 207


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1

Electrostatics-Magnetostatics (****)

1.1 Electrostatics

The behavior of an electrostatic field can be described by two differentialequations:

E= 4 (1.1)(Gauss law) and

E= 0 (1.2)the latter equation being equivalent to the statement that E is the gradientof a scalar function, the scalar potential:

E=

(1.3)

Eqns (1.1) and (1.3) can be combined into one differential equation for a singlescalar function(x):

2= 4 (1.4)This equation is called Poisson equation.

In the regions of space where there is no charge density, the scalar potentialsatisfies the Laplace equation:

2= 0 (1.5)

For a general distribution (x), the potential is expected to be the sumover all increments of charge d3x(x), i.e.,

(x) =

(x)

|x x|d3x (1.6)

This potential should satisfy Poissons equation. But does it? If weoperate with2 on both sides of (1.6) we get (on x not on x)


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2 1 Electrostatics-Magnetostatics (****)

2

(x) = (x)d3x2 1|x x| (1.7)But2(1/|x x|) = 0as long as x =x ! (Why ?)

The singular nature of2(1/|x x|) = 2(1/r)can be best expressed interms of theDirac -function.

Since2(1/r) = 0for r= 0and its volume integral is4 (Why?) wecan write

2

1

|x x|

= 4(|x x|) (1.8)

By definition, if the integration volume contains the point x = x

3(x x)d3x= 1otherwise is zero. This way we recover Poisson s equation

2(x) = 4(x)x=x (1.9)

Thus, we have not only shown that the potential from Coulombs law satisfiesPoissons eqn, but we have established (through the solution of Poissons eqn)the important result that :the potential from a distributed source is the superposition of the individualpotentials from the constituent parcels of charge.

We may consider situations in which is comprised ofNdiscrete chargesqi, positioned at x

i so that

(x) =N

i=1

qi3 (x xi) (1.10)

In this case the solution for the potential is a combination of terms propor-tional to1/|x x|.

1.2 Green Theorem

If in the electrostatic problem involved localized discrete or continuous distri-butions of charge withno boundary surfaces, the general solution (1.6) wouldbe the most convenient and straight forward solution to any problem.

To handle the boundary conditions it is necessary to develop some newmathematical tools, namely, the identities or theorems due to George Green(1824). These follow as simple applications of the divergence theorem

V

A d3x=

S

A n da (1.11)


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1.2 Green Theorem 3

which applies to any well-behaved vector field A defined in the volume Vbounded by the closed surfaceS.

LetA = , ( and arbitrary scalar fields). Then

() =2+ (1.12) n=

n (1.13)

where/n is the normal derivative at the surface S.When (1.12) and (1.13) substituted into the divergence theorem (1.8) pro-

duces the so-called Greens 1st identity

V 2+

d3x=

S

nda . (1.14)

If we rewrite (1.14) with and interchanged, and then subtract it from(1.14) we obtain Greens 2nd identity or Greens Theorem:

V

2 2 d3x=

S

n

n

da (1.15)

Now we can apply Poissons equation (1.8) for discrete charge, substitutingfor = 1/|x x|

V

43(x x)(x) +4(x

)

|x x|

d3x

= S

n 1

|x x| 1

|x x|

n da (1.16)

Integrating the Dirac delta function over all values ofx withinV and forxwithin the volume V yields a nonzero result

(x) =

V

(x)

|x x|d3x +

1

4

S

1

|x x|

n

n

1

|x x|

da

(1.17) The (blue) correction term goes to zero as the surfaceSgoes to infinity(becauseS falls of faster than1/|x x|) If the integration volume isfree of charges, then the first term of equation(1.17) becomes zero, and the potential is determinedonlyby the values of thepotential and the values of its derivatives at the boundary of the integration

region (the surface S). Physical experience leads us to believe that specification of the po-tential on a closed surface defines a unique potential problem. This is calledDirichlet problem orDirichlet boundary conditions.

Similarly it is plausible that specification of the electric field (normalderivative of the potential) everywhere on the surface (corresponding to a


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given surface-charge density) also defines a unique problem. The specificationof the normal derivative is known as the Newmann boundary condition.

As it turns out either one of the two conditions results in uniquesolution. It should be clear that a solution to the Poisson eqn with both and /n specified arbitrarily on a closed boundary (Cauchy boundaryconditions) does not exist since there are unique solutions for Dirichlet andNewmann condition separately.

The solution of the Poisson or Laplace eqn in a finite volume V witheither Dirichlet or Neumann boundary conditions on the boundary surface Scan be obtained by means of Greens theorem (1.15) and the so-called Greenfunctions.

In obtaining the result (1.17) we have chosen = 1/|x x| satisfying

2 1|x x|= 4(|x x|) (1.18)

The function1/|x x| is only one of a class of functions depending on thevariables x and x and called Green functions.

In general2G (x, x) = 4(|x x|) (1.19)

where

G (x, x) = 1

|x x| + F(x, x) (1.20)

andF satisfying the Laplace equation inside the volume V

2F(x, x) = 0 (1.21)

If we substitute G(x, x)in eqn (1.17) we get

(x) =

V

(x)G(x, x)d3x + 1

4

S

G(x, x)

n (x)

nG(x, x)

da

(1.22)The freedom in the definition ofGmeans that we can make the surface integraldepend only on the chosen type of BC.

For theDirichlet BCwe demand:GD(x, x

) = 0 for x S (1.23)Then the 1st term on the surface integral of (1.22) vanishes

(x) =

V

(x)GD(x, x)d3x 1

4

S

(x) n

GD(x, x)da (1.24)

ForNeumann BCthe simplest choice of BC on G isGNn

(x, x) = 0 for x S (1.25)


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1.3 Laplace Equation in Spherical Coordinates 5

but application o the Gauss theorem on (1.19) shows that (how?)S

GNn

da = 4= 0

which is incosistent with2G(x, x) = 43(x x).This will mean that the outflux ofG cannot be zero when there is a source

enclosed by S. Then the simplest boundary condition that we can use is

GNn

(x, x) = 4S

for x S (1.26)

Sis the total area of the boundary surface. Then the solution will be:

(x) = V (x)GN(x, x)d3x + S+ 14 S nGN(x, x)da (1.27)whereS is the average value of the potential over the whole surface

S 1S

S

(x)da (1.28)

In most casesS is extremely large (or even infinite), in which caseS 0.The physical meaning ofF(x, x): it is a solution of the Laplace eqn

insideV and so represents the potential of chargesexternalto the volume Vchosen as to satisfy the homogeneous BC of zero potential on the surface S.

It is important to understand that no matter how the source is dis-tributed, finding the Green function is completely independent or (x).

G(x, x

)depends exclusively on the geometry of the problem, is a tem-plate, potential andnot the actual potential for a given physical problem.

In other words G(x, x)is the potential due to a unit charge, positionedarbitrarily within the surface Sconsistent either with GD = 0or GN/n =4/Son the surface.

The true potential is a convolution of this template with the given(x).

1.3 Laplace Equation in Spherical Coordinates

In spherical coordinates (r,,) the Laplace equation can be written in theform

1

r22

r2(r) +

1

r2 sin

sin

+

1

r2 sin2

2

2 = 0 (1.29)

If we assume

=1

rU(r)P()Q() (1.30)


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Then by substituting in (1.29) and multiplying with r2 sin2 /UPQwe obtain

r2 sin2

1

U

d2U

dr2 +

1

r2 sin P

d

d

sin

dP

d

+

1

Q

d2Q

d2 = 0 (1.31)

We see that the therms depending on have been isolated and we can set

1

Q

d2Q

d2 = m2 (1.32)

with solutionQ= eim (1.33)

Similarly the remaining terms can be separated as:

1

r2 sin

d

d

sin

dP

d

+

l(l+ 1) m

2

sin2

P = 0 (1.34)

d2U

dr2 l(l+ 1)

r2 U= 0 (1.35)

The radial equation will have a solution

U=Arl+1 + Brl (1.36)

whilel is still undetermined.

1.4 Legendre Equation and Legendre PolynomialsThe equation (1.34) for P()can be expressed in terms ofx = cos

d

dx

(1 x2) dP

dx

+

l(l+ 1) m

2

1 x2

P = 0 (1.37)

This is thegeneralized Legendre equationand its solutions are theassociatedLegendre functions.

We will consider the solution of (1.54) for m2 = 0

d

dx

(1 x2) dP

dx

+ l(l+ 1)P= 0 (1.38)

The solution should be single valued, finite, and continuous on the interval1 x 1in order that it represents a physical potential.The solution can be found in the form of a power series

P(x) = xk

j=0

aj xj (1.39)


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1.4 Legendre Equation and Legendre Polynomials 7

wherek is a parameter to be determined.By substitution in (1.54) we get the recurrence relation (how?)

aj+2 =(k+j)(k+j+ 1) l(l+ 1)

(k+j+ 1)(k+j+ 2) aj (1.40)

while for j = 0, 1we find that

ifa0= 0 then k(k 1) = 0 (1.41)ifa1= 0 then k(k+ 1) = 0 (1.42)

These two relations are equivalent and it is sufficient to choose eithera0 or a1different from zero butnot both(why?). We also see that the series expansionis either only o dd or only on even powers ofx. By choosing either k = 0or

k= 1it is possible to prove the following properties: The series converges forx2


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Since the Legendre polynomials form a complete set of orthonormal func-tions, any functionf(x)on the interval 1 x 1can be expanded in termsof them i.e.

f(x) =

l=0

AlPl(x) (1.46)

where (how?)

Al =2l+ 1

2

11

f(x)Pl(x)dx (1.47)

Thus for problems with azimuthal symmetry i.e.m= 0the general solutionis:

(r, ) =

l=0 Alr

l + Blr(l+1)

Pl(cos ) (1.48)

where the coefficients Al [it is not the same as in eqn (1.47)] and Bl can bedetermined from the boundary conditions.

1.4.1 Example

Boundary Value Problems with Azimuthal Symmetry

Lets specify asV()the potential on the surface of a sphere of radius R,and try to find the potential inside the sphere.

If there are no charges at the origin(r = 0)the potential must be finitethere. ConsequentlyBl= 0for all l. Then on the surface of the sphere

V(r= R, ) =

l=0

AlRl

Pl(cos ) (1.49)

and the coefficients Al will be taken via eqn (1.47)

Al =2l+ 1

2Rl

0

V()Pl(cos )sin d (1.50)

If, for example V() =Von the two hemispheres then the coefficientscan be derived easily and the potential inside the sphere is (how?):

(r, ) = V

3

2

rR

P1(cos ) 7

8

rR

3P3(cos ) +

11

16

rR

5P5(cos ) . . .

(1.51)

To find the potential outside the sphere we merely replace (r/R)l by(R/r)l+1 and the resulting potential will be (how?):

(r, ) =3

2

R

r

2V

P1(cos ) 7

12

R

r

2P3(cos ) + . . .

(1.52)


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1.5 Legendre Polynomials 9

Fig. 1.1.

1.5 Legendre Polynomials

An important expansion is that of the potential atxdue to a unit point chargeatx

1

|x x| =

l=0

rl

Pl(cos ) (1.53)

wherer)is the smaller (larger) of |x| and |x| andis the angle between|x| and|x|.

Fig. 1.2.

Show that the potential is :

1

|x x| =

l=0

Alr

l + Blr(l+1)

Pl(cos ) on thez -axis


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1

|x x| = 1

r>

l=0

rl on thex-axis

1.6 Associated Legendre Functions and SphericalHarmonics

For problems without azimuthal (axial) symmetry, we need the generalizationofPl(cos ), namely, the solutions of

d

dx (1 x2) dP

dx +

l(l+ 1) m

2

1

x2

P = 0 (1.54)

for arbitrary l and m.It can be shown that in order to have finite solutions on the interval 1

x 1the parameter l must be zero or a positive integer and that the integermcan take on onlythe valuesl,(l 1), ..., 0 , ... ,(l 1),l (why?).

The solution having these properties is called an associated LegendrefunctionPml (x). For positive m it is defined as

Pml (x) = (1)m(1 x2)m/2 dm

dxmPl(x) (1.55)

IfRodrigues formulais used an expression valid for both positive and negativemis obtained:

Pm

l (x) =

(

1)m

2ll! (1 x2

)m/2 d

m

dxm Pl(x) (1.56)

There is a simple relation between Pml (x)and Pml (x):

Pml (x) = (1)m(l m)!(l+ m)!

Pml (x) (1.57)

For fixed m the functions Pml (x)form an orthonormal set in the index l onthe interval1 x 1. The orthogonality relation is 1

1

Pml (x)Pml (x)dx=

2

2l+ 1

(l+ m)!

(l m)! ll (1.58)

We have found that Qm() =eim, this function forms a complete set oforthogonal functions in the index m on the interval0 2.

The productPml Qm forms also a complete orthonormal set on the surfaceof the unit sphere in the two indices l , m.

From the normalization condition (1.58) we can conclude that the suitablynormalized functions, denoted by Ylm(, ), are :


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1.6 Associated Legendre Functions and Spherical Harmonics 11

Ylm(, ) = 2l+ 14 (l m)!(l+ m)! Pml (cos )eim (1.59)and also

Yl,m(, ) = (1)mYlm(, ) (1.60)

The normalization and orthogonality conditions are: 20

d

0

sin dYlm(, )Ylm(, ) = llmm (1.61)

An arbitrary function g(, )can be expanded in spherical harmonics

g(, ) =

l=0

l

m=l

Alm

Ylm

(, ) (1.62)

where the coefficients are

Alm=

dYlm(, )g(, ) . (1.63)

The general solution for a boundary-value problem in spherical coordinatescan be written in terms of spherical harmonics and powers ofr in a general-ization of (1.48) :

(r,,) =

l=0

lm=l

Almr

l + Blmr(l+1)

Ylm(, ) (1.64)

If the potential is specified on a spherical surface, the coefficients can bedetermined by evaluating (1.64) on the surface and using (1.63).

1.6.1 Spherical HarmonicsYlm(, )

l= 0 Y00 =

1

4

l= 1 Y11 =

3

8sin ei

Y10 =

3

4cos

l= 2 Y22 =1

415

2sin2 e2i

Y21 =

15

8sin cos ei

Y20 =

5

4

3

2cos2 1

2


22/217


l= 3 Y33 = 14

35

4sin3 e3i

Y32 =1

4

105

2 sin2 cos e2i

Y31 = 14

21

4sin (5 cos2 1) ei

Y30 =1

2

7

4sin

5cos3 3cos

(1.65)

Fig. 1.3. Schematic representation ofYlmon the unit sphere.Ylmis equal to0alongmgreat circles passing through the poles, and along l mcircles of equal latitude.The function changes sign each time it crosses one of these lines.

1.6.2 Addition Theorem for Spherical Harmonics

The spherical harmonics are related to Legendre polynomialsPl by a relationknown as the addition theorem.

Theaddition theoremexpresses a Legendre polynomial of orderl in theanglein terms of products of the spherical harmonics of the angles , and, :


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1.6 Associated Legendre Functions and Spherical Harmonics 13

Fig. 1.4.

Pl(cos ) = 4

l(l+ 1)

lm=l

Ylm(, )Ylm(, ) (1.66)

where is the angle between the vectors x and x, x

x = x

x cos and

cos = cos cos + sin sin cos( ).Pl(cos )is a function of the angles,with the angles

, as parametersand it maybe expanded in a series (1.63) :

Pl(cos ) =

l=0

lm=l

Alm(, )Ylm(, ) (1.67)


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The addition theorem offer the possibility to extend the expansion validfor a point charge (axially symmetric distribution) to an arbitrary chargedistribution.

By substituting (1.66) into (1.53) we obtain

1

|x x| = 4

l=0

lm=l

1

2l+ 1

rl

Ylm(, )Ylm(, ) (1.68)

This equation gives the potential in a completely factorized form in the coor-dinatesx and x. This is useful in any integration over charge densities, whenone is the variable of integration and the other the observation point.

1.7 Multipole Expansion

A localized distribution of charge is described by the charge density (x),which is nonvanishing only inside a sphere a around some origin.

The potential outside the sphere can be written as an expansion in spher-ical harmonics

(x) =

l=0

lm=l

4

2l+ 1qlm

1

rl+1Ylm(, ) (1.69)

Fig. 1.5.

This type of expansion is called multipole expansion;Thel = 0term is calledmonopole term,Thel = 1are calleddipole termsetc.


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1.7 Multipole Expansion 15

The problem to be solved is the determination of the constants qlm interms of the properties of the charge distribution (x).

The solution is very easily obtained from the integral for the potential

(x) =

(x)

|x x|d3x (1.70)

Using the expansion (1.68) for1/|x x| i.e.

(x) =

l=0

lm=l

4

2l+ 1

1

rl+1Ylm(, )

Ylm(

, )(x)rld3x (1.71)

Consequently the coefficients in (1.69) are :

qlm=

Ylm(, )(x)rld3x (1.72)

and called the multipole moments of the charge distribution (x)

1.7.1 Monopole moment l = 0

Here, the only component is

q00 =

(x)r0Y00(

, )d3x = 1

4

(x)d3x =

q4

(1.73)

Observed from a large distancer, any charge distribution acts approximately

as if thetotal chargeq(monopole moment) would be concentrated at onepoint since the dominant term in (1.68)

(x)= 4q001

rY00+ = q

r+ . . . (1.74)

NOTE:The moments with m 0 are connected (for real charge density) too themoments with m


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InSpherical representation one obtains

q11 =

(x)rY11(

, )d3x

=

3

8

(x)r (sin cos i sin sin ) d3x

in Cartesian represantation

=

3

8

(x)(x iy)d3x =

3

8(px ipy) (1.77)

also

q10 = (x)rY10(, )d3x = 3

4 (x)r cos d3x

=

3

4

z(x)d3x=

3

4pz (1.78)

1.7.3 Quadrupole moment l= 2

q22 =

(x)r2Y22(

, )d3x

= 1

4

15

2

(x) [r sin (cos i sin )]2 d3x

= 1

415

2 (x)(x iy)2d3xbecause (x iy)2 = 1

3

(3x2 r2) 6ixy (3y2 r2)

= 1

12

15

2(Q11 2iQ12 Q22) (1.79)

whereQij is the traceless (why?)quadrupole moment tensor:

Qij =

3xix

j r2ij

(x)d3x (1.80)

Analogously

q21 =

(x)r2Y21(, )d3x = 15

8

(x)z(x iy)d3x

= 13

15

8(Q13 iQ23) (1.81)

and


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1.7 Multipole Expansion 17

q20 = (x)r2Y20(, )d3x = 12 54 (x)(3z r2)d3x

= 1

2

5

4Q33 (1.82)

From eqn (1.60) we can get the moments with m


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1.8 Energy of a Charge Distribution in an External Field

The multipole expansion of the potential of a charge distribution can also beused to describe the interaction of the charge distribution with an externalfield.The electrostatic energy of the charge distribution (x)placed in an externalfield (x)is given by

W =

V

(x)(x)d3x (1.89)

The external field (if its is slowly varying over the region where (x)is non-

Fig. 1.6.

negligible) may be expanded in a Taylor series:

(x) = (0) + x (0) +12

3i=1

3j=1

xixj2

xixj(0) + . . . (1.90)

SinceE = for the external field

(x) = (0) x E(0) 12

3i=1

3j=1

xixjEjxi

(0) + . . . (1.91)

Since E= 0for the external field we can substract1

6 r2 E(0)

from the last term to obtain finally the expansion:

(x) = (0) x E(0) 16

3i=1

3j=1

3xixj r2ij

Ejxi

(0) + . . . (1.92)


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1.8 Energy of a Charge Distribution in an External Field 19

When this inserted into (1.89) the energy takes the form:

W =q(0) p E(0) 16

3i=1

3j=1

QijEjxi

(0) + . . . (1.93)

Notice, that:

the total charge interacts with the potential, the dipole moment with the electric field, the quadrupole with the electric field gradient, etc

1.8.1 Examples

EXAMPLE 1 : Show that for a uniform charged sphere all multipole mo-

ments vanish exceptq00.If the sphere has a radius R0 and constant charge density (x) = then

qlm=

R00

rlYlm(, )r2drd =

Rl+30l+ 3

Ylm(, )d

but sinceY00 = 1/

4 we have from the orthogonality relation

qlm= Rl+30l+ 3

4

YlmY00d =

Rl+30l+ 3

4l0m0

EXAMPLE 2:Perform multipole decomposition of a uniform charge dis-tribution whose surface is a weakly deformed sphere:

R= R01 +2

m=2a2mY2m(, ) , |a2m|


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qlm Rl+30l+ 3

Ylm(, )

1 + (l+ 3)

2m=2

a2mY2m(

, )

d

Apart from the monopole momentq00 of the previous example, we find

qlm= Rl+30l+ 3

(l+ 3)2

=2

a2

YlmY2d =Rl+30

a2ml2 (l >0)

Thus the nonvanishing multipole moments are : q2m= R50a2m.

1.9 Magnetostatics

There is a radical difference between magnetostatics and electrostatics :there are no free magnetic charges

The basic entity in magnetic studies is themagnetic dipole. In the presence of magnetic materials the dipole tends to align itself in

a certain direction. That is by definition the direction of the magnetic fluxdensity, denoted by B (also called magnetic induction).

The magnitude of the flux density can be defined by the magnetictorqueN exerted on the magnetic dipole:

N= B (1.94)where is the magnetic moment of the dipole.

In electrostatics the conservation of charge demands that the chargedensity at any point in space be related to the current density in that neigh-borhood by the continuity equation

/t + J= 0 (1.95) Steady-state magnetic phenomena are characterized by no change in thenet charge density anywhere in space, consequently in magnetostatics

J= 0 (1.96)

1.10 Biot & Savart Law

Ifd is an element of length (pointing in the direction of current flow) of awire which carries a current Iandx is the coordinate vector from the element


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1.10 Biot & Savart Law 21

of length to an observation point P, then the elementary flux density dB atthe pointP is given in magnitude and direction by

dB= kId x|x|3 (1.97)

NOTE: (1.97) is an inverse square law, just as is Coulombs law of electro-

Fig. 1.7.

statics. However, the vector character is very different.

If we consider that current is charge in motion and replace Id by qvwhere q is the charge and v the velocity. The flux density for such a chargein motion would be

B= kqv x

|x

|3

(1.98)

This expression is time-dependent and valid only for charges with small ve-locities compared to the speed of light.

1.10.1 Diff. Equations for Magnetostatics & Amperes Law

The basic law (1.97) for the magnetic induction can be written down in generalform for a current density J(x):

B(x) =1

c

J(x) (x x

)

|x x|3 d3x (1.99)

This expression for B(x) is the magnetic analog of electric field in terms of

the charge density:

E(x) =1

c

(x)

(x x)|x x|3 d

3x (1.100)

In order to obtain DE equivalent to (1.99) we use the relation


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(x

x)

|x x|3 = 1|x x| as 1r= rr3 (1.101)and (1.99) transforms into

B(x) =1

c

J(x)

|x x|d3x (1.102)

From (1.102) follows immediately:

B= 0 (1.103)

This is the1st equation of magnetostatics and corresponds to E= 0in electrostatics.

By analogy with electrostatics we now calculate the curlofB

B= 1c

J(x)

|x x|d3x (1.104)

which for steady-state phenomena ( J= 0) reduces to (how?) 1

B= 4c

J (1.105)

This is the2nd equation of magnetostatics and corresponds to E=4 in electrostatics.

The integral equivalent of (1.105) is called Amperes law It can be ob-

tained by applying Stokess theorem to the integral of the normal componentof (1.105) over the open surface Sbounded by a closed curve C. ThusS

B n da= 4c

S

J nda (1.106)

is transformed into C

B d= 4c

S

J nda (1.107)

Since the surface integral of the current density is the total current Ipassing

through the closed curve C, Amperes law can be written in the form:

C B d=4

c I (1.108)

1 Remember : ( A) = ( A) 2A


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1.11 Vector Potential 23

Fig. 1.8.

1.11 Vector Potential

The basic differential laws in magnetostatics are

B= 4c

J and B= 0 (1.109)

The problems is how to solve them. If the current density is zero in the region of interest, B = 0

permits the expression of the vector magnetic induction B as the gradient ofa magnetic scalar potential B = M, then (1.109) reduces to the Laplaceequation forM.

If B= 0everywhere,B must be the curlof some vector fieldA(x),called thevector potential

B(x) = A(x) (1.110)

and from (1.102) the general form ofA is:

A(x) =1

c

J(x)

|x x|d3x +(x) (1.111)

The added gradient of an arbitrary scalar function shows that, for agiven magnetic induction B, the vector potential can be freely transformedaccording to

A A + (1.112)This transformation is called a gauge transformation. Such transformationsare possible because (1.110) specifies only thecurlofA.

If (1.110) is substituted into the first equation in (1.109), we find

( A) = 4c

J or ( A) 2A= 4c

J (1.113)

If we exploit the freedom implied by (1.112) we can make the convenientchoice of gauge (Coulomb gauge) A = 0. Then each component of thevector potential satisfies the Poisson equation


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2

A= 4

c J (1.114)

The solution for A in unbounded space is (1.111) with =constant:

A(x) =1

c

J(x)

|x x|d3x (1.115)


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2

Classical Field Theory: Maxwell Equations

2.1 Introduction

Electrostatics and Magnetostatics deal with steady-state problems inelectricity and inmagnetism.

The almost independent nature of electric and magnetic fields phenomenadisappears when we consider time-dependent problems.

Time varying magnetic fields give rise to electric fields and vice-versa. Wethen must speak ofelectromagnetic fieldsrather thanelectricor magneticfields.

The interconnection between electric and magnetic fields and their essen-tial sameness becomes clear only within the framework ofSpecial Theoryof Relativity

2.2 Faradays Law of Induction (****)

Faraday (1831), observed that a transient current is induced in a circuit if:

1. A steady current flowing in an adjacent circuit is turned on or off2. The adjacent circuit with a steady current flowing is moved relative to

the first circuit3. A permanent magnet is thrust into or out of the circuit

The changing magnetic flux induces an electric field around the circuit, theline integral of which is called the electromotive forceEand causes a currentflow according toOhms law: J = E ( is the conductivity).

The magnetic induction in the neighboring of the circuit is B and themagnetic fluxlinking the circuit is defined by

F =

S

B n da (2.1)


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26 2 Classical Field Theory: Maxwell Equations

Fig. 2.1.

Theelectromotive force around the circuit is defined by

E=

C

E d (2.2)

whereE is the electric field at the element d of the circuit C.Thus Faradays observation is summed up in the mathematical law:

E= k dFdt

(2.3)

That is, the induced electromotive force around the circuit is proportionalto the time rate of change of magnetic flux linking the circuit.

The sign is specified by Lenzs law, which states that the induced current(and accompanying magnetic flux) is in such a direction to oppose the changeof flux through the circuit. For SI units k = 1, Gaussian units k = 1/c.

2.2.1 Faradays law for a moving circuitC

E d= kddt

S

B n da (2.4)

This is eqn (2.3) in terms of integrals. We can observe that :

The induced electromotive force is proportional to the total timederivative of the flux.

The flux can be changed by changing: the magnetic induction or theshapeor the orientationor the position of the circuit.


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2.2 Faradays Law of Induction (****) 27

Fig. 2.2.

If the circuit Cis moving with a velocity v in some direction the total timederivative in eqn (2.4) must take into account the motion e.g. convectivederivative

d

dt=

t+ v and

dB

dt =

B

t + (v ) B= B

t + (B v) + v ( B)

The flux through the circuit may change because:

the flux changes with time at a point the translation of the circuit changes the location of the boundary.

Eqn (2.4) can be written as (why?):C

[E k (v B)] d= k

S

B

t n da . (2.5)

In a comoving coordinate systemC

E d= k

S

B

t n da (2.6)

whereE is the electric field in the comoving frame. Thus

E =E + k (v B) (2.7)

With the present choice of units for charge and current, Galilean covariance

requires thatk = 1/c(why?).Finally, the transformation of the electromotive force integral into a surface

integral leads to (how?)S

E +1

c

B

t

n da= 0 (2.8)


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Since the circuitCand the bounding surfaceSare abritrary, the integrantmust vanish at all points in space. Thus the differential form of Faradayslawis:

E +1c

B

t = 0 . (2.9)

Note that for time-independent electrostatic fields : E= 0.KELVIN-STOKES THEOREM :

relates the surface integral of the curl of a vector field over a surface S inEuclidean 3-space to the line integral of the vector field over its boundary

S

F dS=

S

F d (2.10)

da

Fig. 2.3.

2.3 Energy in the Magnetic Field (***)

Typically in magnetostatics, the creation of a steady-state configuration ofcurrents and associated magnetic fields involves an initial transient periodduring which the currents and fields are brought from zero to the finalvalues.


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2.3 Energy in the Magnetic Field (***) 29

For such time-varying fields there are induced electromotive forces which

cause the sources of current to do work. Since the energy in the field is by definition the total work done to establish

it, we must consider these contributions.

Suppose that we have a circuit with a constant current Iflowing in it. If the flux through the circuit changesan electromotive force Eis induced

around it. In order to keep the current constant, the sources of current must

do work.To determine the rate we note that the time rate of change of energy of

a particlewith velocity v acted on by a force F is

dE

dt =v

F

With a changing flux the added field E on each conducting electron of chargeqand drift velocityv gives rise to a change in energy per unit time ofqv Eper electron.

Summing over all the electrons in the circuit, we find that the sources dowork to maintain the current at the rate

dW

dt = IE=1

cI

dF

dt

Thus, if the flux change through a circuit carrying a current I isF, thework done by the source is :

W=1c

I F

We consider the problem of the work done in establishing a generalsteady-state distribution of currents and fields.

We can imagine that the build-up process occurs at an infinitesimal rateso that J= 0(Jis the current).

The current distribution can be broken up into a network of elementaryloops, the typical ones of which is an elemental tube of current of cross-sectional area following a close path C and spanned by a surface Swith normal n

We can express the increment of work done against the induced electro-motive force in terms of the change of the magnetic induction through theloop:

(W) =J

c

S

n Bda (2.11)

the extra is because we consider only one elementary circuit.


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Fig. 2.4. Distribution of current density broken up into elementary current loops

If we express B in terms of the vector potential A i.e. B = Athenwe have

(W) = Jc

S

( A) n da (2.12)

With application of Stokess theorem this can be written

(W) =J

c

C

A d (2.13)

but Jd Jd3x since l is parallel to J. Evidently the sum over all suchelemental loops will be the volume integral.

Hence the total increment of work done by an external source due to achange A(x)in the vector potential is

W=

1

c A J d3x (2.14)By using Amperes law

H= 4c

J

we can get an expression in terms of the magnetic fields. Then

W= 1

4

A ( H) d3x (2.15)

which transforms to (how?)

W = 1

4 [H ( A) + (H A)] d3x (2.16)If the field distribution is assumed to be localized, the 2nd integrant van-

ishes (why?) and we get

W = 1

4

H Bd3x (2.17)


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2.4 Maxwell Equations (****) 31

which is theanalog of the electrostatic equation for the energy change

W = 1

4

E Dd3x (2.18)

where D = E + 4P is the electric displacement and P the electricpolarization (dipole moment per unit volume).

If we bring the fields from zero to the final values the total magneticenergywill be (why?)

W = 1

8

H B d3x (2.19)

which is the magnetic analog of the total electrostatic energy

W = 1

8 E D d3x (2.20)The energy of a system of charges in free space i.e. electrostatic energy

is:

W=1

2

(x)(x) d3x (2.21)

The magnetic equivalent for this expression i.e. the magnetic energy is

W= 1

2c

J A d3x (2.22)

2.4 Maxwell Equations (****)

Maxwells equations are based on the following empirical facts:

1. The electric charges are sources of the vector field of the dielectric dis-placement densityD. Hence, for the flux of the dielectric displacementthrough a surface enclosing the charge we have

1

4

S

D nda= Q =

V

d3x (2.23)

This relation can be derived fromCoulombs force law.2. Faradays induction law:

E= C

E dl= 1

c

t

SB nda (2.24)

3. The fact that there areno isolated monopoles impliesS

B nda= 0 (2.25)


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4. Amperes law: C

H dl= 4c

S

J nda (2.26)

The basic laws of electricityand magnetismcan be summarized in differ-ential form:

Coulombs law D=4 (2.27)Amperes law ( J= 0) H= 4

c J +

1

c

D

t (2.28)

Faradays law E +1c

B

t = 0 (2.29)

no free magnetic poles B= 0 (2.30)where E and B are the averaged E and B of the microscopic or vacuumMaxwell equations. The two extra field quantities D and H usually called theelectric displacementand magnetic field(Bis then called themagneticinduction and M is the macroscopic magnetization)

Da = Ea+ 4

Pa

b

Qabxb

+ . . .

D= E + 4P + . . . (2.31)

Ha = Ba 4(Ma+ . . . ) H= B 4M + . . .(2.32)

The quantitiesP,M,Qabrepresent the macroscopically averagedelectricdipole, magnetic dipole and electric quadrupole moment densitiesof the material medium in the presence of of applied fields.

Similarly, the charge and current densities and J are macroscopic aver-ages of the free charge and current densities in the medium.

The macroscopic Maxwell equations are a set of8 eqns involving the com-ponents of4 fields E, B, D and H.

The 4 homogeneous eqns can be solved formally by expressing E and Bin terms of the scalar potential and the vector potential A

The inhomogeneous eqns cannot be solved until the derived fieldsD andH are known in terms ofE and B. These connections which are implicitin (2.32) are known as constitute relations, e.g. D = E and H = E/(: electric permittivity & magnetic permeability).

All but Faradays law were derived from steady-state observations andthere is no a priori reason to expect that the static equations hold un-changed for time-dependent fields.

The above equations without the red term in Amperes law are inconsis-tent. While Amperes law ( J= 0) is valid for steady - state problems, thecomplete relation is given by the continuity equation for charge and current

J + t

= 0 . (2.33)


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2.4 Maxwell Equations (****) 33

Maxwell replacedJ in Amperes law by its generalization

J J + 14

D

t (2.34)

Maxwell called the added term displacement current, without it therewould be no electromagnetic radiation (Can you repeat his steps?).

Maxwells equations, form the basis of all classical electromag-netic phenomena. When combined with theLorentz force equation

F= q

E +v

c B

(2.35)

andNewtons 2nd law of motion, provide a complete description of the clas-sical dynamics of interacting charged particles and EM fields.

2.4.1 Vector and Scalar Potentials

In electrostatics and magnetostatics we have used the scalar potential andthe vector potential A to simplify certain equations.

Since B= 0, we can define B in terms of a vector potential A :B= A (2.36)

Then Faradays law E + 1c Bt = 0 can be written

E +1

c

A

t

= 0 (2.37)

Thus the quantity with vanishing curl can be written as the gradient of ascalar potential :

E +1

c

A

t = or E= 1

c

A

t (2.38)

The definition ofB and E in terms of the potentials A and satisfies inden-tically the 2 homogeneous Maxwell equations. While A and are determinedby the 2 inhomogeneous Maxwell equations.

If we restrict our considerations to the vacuum form of the Maxwell equa-tions the inhomogeneous form of Maxwell equations can be write in terms ofthe potentials as:

2 +1c

t( A) = 4 (2.39)

2A 1c2

2A

t2

A +1

c

t

= 4

c J (2.40)

These eqns are equivalent to Maxwell eqnsbut they are still coupled. Thanksto the abritrariness in the definition of the potentials we can choose transfor-mations of the form


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A

A =A + (2.41)

= 1c

t (2.42)

Thus we can choose a set of potentials (A, )such that

A +1c

t = 0 (2.43)

This way we uncouple the pair of equations (2.39) and (2.40) and leavetwo inhomogeneous wave equations, one for and one for A

2 1c2

2

t2 = 4 (2.44)

2A 1c2

2A

t2 = 4

c J (2.45)

This set of equations is equivalent in all respects to the Maxwellequations.

2.4.2 Gauge Transformations : Lorenz Gauge

The transformations

A A =A + (2.46)

=

1

c

t

(2.47)

are calledgauge transformationand the invariance of the fields under thesetransformations is calledgauge invariance.

The relation between A and :

A +1c

t = 0 (2.48)

is called Lorenz condition 1. [Prove that there will always exist potentialssatisfying the Lorentz condition].

The Lorenz gauge is commonly used because:

It leads to the wave equations (2.44) and (2.45) which treat andAon equal footings

It is a coordinate independent concept and fits naturally into theconsiderations of special relativity.

1The condition is from the Danish mathematician and physicist Ludvig Valentin Lorenz (1829-1891) and notfrom the Dutch physicist Hendrik Lorentz (1853-1928)


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2.5 Green Functions for the Wave Equations (***) 35

2.4.3 Gauge Transformations : Coulomb Gauge

In this gauge A= 0 (2.49)

From eqn (2.39) we see that the scalar potential satisfies the Poisson eqn :

2= 4 (2.50)

with solution

(x, t) =

(x, t)

|x x|d3x (2.51)

The scalar potential is just the instantaneousCoulomb potential due to thecharge density (x, t). This is the origin of the name Coulomb gauge.

From eqn (2.40) we find that the vector potential satisfies:

2A 1c2

2A

t2 = 4

c J +

1

c

t (2.52)

Finally, if we define thetransverse (or solenoidal) current:

Jt = 1

4

J

|x x|d3x (2.53)

andlongitudinal (or irrotational)currentJl for which Jl= 0which maycancel the contribution of the term with the potential .

Then, the wave equation for A can be expressed entirely in terms of the

transverse currentJt [Can you explain it?]

2A 1c2

2A

t2 = 4

c Jt (2.54)

The Coulomb or transverse gauge is often used when no sources arepresent. Then = 0, and A satisfies the homogeneous wave equation. Thefields are given by

E= 1c

A

t , and B= A (2.55)

2.5 Green Functions for the Wave Equations (***)

The wave eqns (2.44), (2.45) and (2.54) all have the same structure,

2 1c2

2

t2 = 4f(x, t) (2.56)


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To solve (2.56) it is useful to find a Green function (as in electrostatics)

In order to remove the time dependence we introduce a Fourier transformwith respect to frequencyWe suppose that both (x, t)and f(x, t)have a Fourier integral repre-sentation

(x, t) = 1

2

(x, )eitd ,f(x, t) = 1

2

f(x, )eitd (2.57)

with the inverse transformations,

(x, ) =

(x, t)eitdt , f(x, ) =

f(x, t)eitdt (2.58)

When the representation (2.57) are inserted into (2.56) it is found that theFourier transform (x, ) satisfies the inhomogeneous Helmholtz waveequationfor each value of and k = /c2 + k2(x, ) = 4f(x, ) (2.59)Equation (2.59) is an elliptic PDE similar to Poisson eqn to which it reducesfork = 0.

The Green function G(x, x)appropriate to (2.59) satisfies the inhomoge-neous equation 2 + k2Gk(x, x) = 4(x x) (2.60)If there are no boundary surfaces, the Green function can only depend onR = x

x, and must be spherically symmetric, that is depend only on

R= |R|. This means that in spherical coordinates Gk(R)satisfies1

R

d2

dR2(RGk) + k

2Gk = 4(R) (2.61)

In other words, everywhere except R = 0 the quantity RGk(R)satisfiesthe homogeneous equation

d2

dR2(RGk) + k

2 (RGk) = 0

with solution :RGk(R) = Ae

ikR + BeikR

The general solution for the Green function is :

Gk(R) = AG(+)k (R) + BG

()k (R) (2.62)

where

G()k (R) =eikR

R (2.63)


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withA + B= 1and the correct normalization condition at R

0

limkR0

Gk(R) = 1

R (2.64)

Thefirst termof (2.62) represents a diverging spherical wavepropagatingfrom the origin, while the second term represents a converging sphericalwave.

To understand the different time behaviors associated with G(+)k andG

()k

we need to construct the corresponding time-dependent Green functions thatsatisfy

2 1c2

2

t2

G()k (x, t; x

, t) = 4(x x)(t t) (2.65)

note that the source term for (2.59) is4(x x)eit

Using the Fourier transforms (2.57) the time-dependent Green functionsbecome (how?)

G()(R, ) = 1

2

eikR

R eid (2.66)

where=t t is the relative time.The integral actually is a function and the Green function becomes:

G()(R, ) = 1

R

R

c

(2.67)

or

G()(x, t; x, t) = 1

|x

x

|t

t

|x x|c (2.68)

The infinite space Green function is thus a function only of therelativedistance R and the relative time between the source and the observationpoint.

The Green function G(+) is called theretarded Green functionand G()

is called theadvanced Green functionParticular integrals of the inhomogeneous wave equation (2.56) are

()(x, t) =

G()(x, t; x, t)f(x, t)d3xdt

to either of these maybe added solutions of the homogeneous equation in orderto specify a definite physical problem.

EXAMPLES

(x, t)=

1

|x x|

t

t |x x|

c

(x, t)d3xdt

A(x, t)=

1

|x x|

t

t |x x|

c

J(x, t)

c d3xdt


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2.5.1 Poyntings Theorem : Conservation of Energy (****)

For a continuous distribution of charge and current, the total rate of doingwork by the fields in a finite volume V is

V

J Ed3x (2.69)

This represents a conversion of EM energy into mechanical or thermal energyand it must be balanced by a corresponding rate of decrease of energy in theEM field within the volume V.

In order to derive this conservation law explicitly we use the Maxwell eqnsto express (2.69) in other terms. We use Ampere law to eliminateJ

V

J Ed3x= 1

4

V

cE ( H) E Dt d3x (2.70)If we use the vector identity,

(E H) = H ( E) E ( H)together with the Faradays law, we get

V

J Ed3x=14

V

c (E H) + E D

t + B H

t

d3x (2.71)

The terms with the time derivatives can be interpreted as the time derivativesof the electrostatic and magnetic energy densities.

If we also remember that the sum/integrals

WE= 1

8

E Dd3x and WB = 1

8

H Bd3x (2.72)

represents the total EM energy (even for time varying fields). Then the totalenergy density is denoted by

u= 1

8(E D + B H) 1

2

0E

2 + 1

0B2

(2.73)

Then equation (2.71) can be written (how?):

V J Ed3x= V

u

t

+ c

4

(E

H) d3x (2.74)

Since the volume is abritrary, this can be cast into the form of a differ-ential continuity equation or conservation law

u

t + S= J E (2.75)


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The vectorS represents the energy flow and is called Poynting vector

S= c

4(E H) (E H) (2.76)

Poyntings theorem (Conservation of energy) : The physical meaning ofthe above relations is that the time rate of change of EM energy within acertain volume, plus the energy flowing out through the boundary surfaces ofthe volume per unit time,is equal to the negative of the total work done bythe fields on the sources within the volume.

In other words Poyntings theorem for microscopic field (E, B) isa statement of conservation of energy of the combined system ofparticles and fields.

If we denote the total energy of the particles within the volume VasEmech and assume that no particles move out of the volume we have

dEmechdt

=

V

J Ed3x (2.77)

and thetotal field energy within V as

Efield=

V

ud3x= 1

8

V

E2 + B2

d3x (2.78)

Poyntings theorem expresses the conservation of energy for thecombined system as:

dE

dt

= d

dt

(Emech+ Efield) =

S n Sda (2.79)

2.5.2 Poyntings Theorem : Conservation of Momentum (***)

The conservation of linear momentum can be similarly considered. Thetotal EM force on a charged particle is

F= q

E +v

c B

(2.80)

If we denote as Pmech the sum of all the momenta of all the particles in avolume Vwe can write the Newtons 2nd law (how?),

Ftot dPmechdt

=

V

E +

1c

J B

d3x (2.81)

where (J v) we converted the sum over particles to an integral over charge& current densities.

We can use Maxwell equations to eliminate and J from (2.81) by using


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=

1

4 E , J= c

4 B 1c Et (2.82)After some manipulations we can show (how?) that the rate of change

of mechanical momentum of eqn (2.81) can be written

dPmechdt

+ d

dt

V

1

4c(E B) d3x (2.83)

= 1

4

V

[E( E) E ( E) + B( B) B ( B)] d3x

We can identify the volume integral on the left as the total EM momen-tumPfield in the volume V

Pfield= 14c

V

(E B) d3x (2.84)

The integrant can be interpreted as the density of EM momentum

g= 1

4c(E B) (2.85)

The EM momentum density g is proportional to the energy-flux density Swith proportionality constant c2.

2.6 Maxwell Stress Tensor (***)

In order to establish that (2.84) is a conservation law for momentum, we mustconvert the volume integral on the right into surface integral of something thatwill be identified as momentum flow.

By defining the Maxwell stress tensor Tab as

Tab= 1

4

EaEb+ BaBb 1

2ab(E E + B B)

(2.86)

where the a-th component can be written as

[E( E) E ( E)]a =

b

xb

EaEb 1

2abE E

(2.87)

also

( T)a= b

xbTab (2.88)

Then (2.84) can be written as:

d

dt(Pmech+ Pfield)a=

b

V

xbTabd

3x (2.89)


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2.7 Conservation of Angular Momentum (**) 41

Then via the divergence theorem we get

d

dt(Pmech+ Pfield)a =

S

b

Tabnbda (2.90)

If (2.90) represents a statement of conservation of momentum,b

Tabnb

is the a-th component of the flow per unit area of momentum across thesurfaceSinto the volume V .

In other words it is the force per unit area transmitted across the surfaceSand acting on the combined system of particles and fields inside V .

2.7 Conservation of Angular Momentum (**)

The derivation of the EM angular momentum shares the same tactical ap-proach as that of the linear momentum

Let us define the mechanical angular momentum of the system as

Lmech= r pmech (2.91)

wherepmech is the mechanical momentum density. Then

dLmech

dt =r

E+1

cJ

B, (2.92)

and substitution of and J from Maxwells eqns leads to

d

dt

Lmech+

1

4cr (E B)

(2.93)

= 1

4r [E( E) E ( E) + B( B) B ( B)]

By using the definition of the Maxwell stress tensor (2.86), we can simplifyeqn (2.94) considerably

d

dt(L

mech+ L

field) = r

T (2.94)

whereLfield= r g (2.95)

has the interpretation of being the EM field angular momentum densityIn integral form since


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r

T=

(r

T) and

r= 0 (2.96)

we getd

dt

Lmech+

V

Lfieldd3x

=

S

(r T) nda (2.97)

The right-hand side of this equation represents the integrated torque densitydue to the fields over the boundary surface S.


53/217

3

Electromagnetic Waves (****)

3.1 Maxwell Equations

A basic feature of Maxwell equations for the EM field is the existence oftravelling wave solutions which represent the transport of energy from onepoint to another.

The simplest and most fundamental EM waves are transverse, planewaves.

In a region of space where there are no free sources ( = 0, J = 0),Maxwells equations reduce to a simple form given

E= 0, E +1c

B

t = 0

B= 0, B c

Et

= 0 (3.1)

whereD and H are given by relations

D= E and H= 1

B (3.2)

where is the electric permittivity and the magnetic permeabilitywhich assumed to be independent of the frequency.

3.2 Plane Electromagnetic Waves

Maxwells equations can be written as

2B c2

2B

t2 = 0 and 2E

c22E

t2 = 0 (3.3)

In other words each component ofB and E obeys a wave equation of theform:


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44 3 Electromagnetic Waves (****)

2

u 1

v22u

t2 = 0 where v=

c

(3.4)is a constant with dimensions of velocity characteristic of the medium.

The wave equation admits admits plane-wave solutions:

u= eikxit (3.5)

E(x, t)= Eeiknxit and B(x, t) = Beiknxit (3.6)

where the relation between the frequency and the wave vector k is

k=

v =

c or k k=

v

2(3.7)

also the vectors n, E and B are constant in time and space.

If we consider waves propagating in one direction, say x-direction then thefundamental solution is:

u(x, t) = Aeik(xvt) + Beik(x+vt) (3.8)

which represents waves traveling to the right and to the left with propagationvelocitiesv which is called phase velocity of the wave.

From the divergence relations of (3.1) by applying (3.6) we get

n E= 0 and n B= 0 (3.9)This means that E (or E) and B (or B) are both perpendicular to thedirection of propagationn. Such a wave is called transverse wave.

The curlequations provide a further restriction

B=

n E and E= 1

n B (3.10)

The combination of equations (3.9) and (3.10) suggests that the vectors n, Eand B form anorthonormal set.

Also, if n is real, then (3.10) implies that that E and B have the samephase.

It is then useful to introduce a set of real mutually orthogonal unit vectors(1, 2, n).

In terms of these unit vectors the field strengths E and B are

E= 1E0 , B= 2

E0 (3.11)

orE=2E

0 , B= 1

E0 (3.12)

E0 andE0 are constants, possibly complex.In other words the most general way to write the electric/magnetic field

vector is:


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3.2 Plane Electromagnetic Waves 45

Fig. 3.1.

E= (E01+ E02)e

iknxit (3.13)

B=

(E02 E01)eiknxit (3.14)

Thus the wave described by (3.6) and (3.11) or (3.12) is a transverse wavepropagating in the direction n.

Or thatEandB are oscillating in a plane perpendicular to the wave vectork, determining the direction of propagation of the wave.

Fig. 3.2.

The energy flux of EM waves is described by the real part of thecomplex Poynting vector

S=1

2

c

4E H =1

2

c

4[ER HR+ EI HI+ i(EI HR ER HI)]


56/217


whereE and H are the measured fields at the point where S is evaluated.1

Thetime averaged flux of energy is:

S= c

8

|E0|2n (3.15)

Thetotal time averaged density (and not just the energy density associ-ated with the electric field component) is:

u= 1

16

E E +1

B B

=

8|E0|2 (3.16)

The ratio of the magnitude of (3.15) to (3.16) is the speed of energy flow i.e.v= c/

. 2 (Prove the above relations)

Project:What will happen ifn is not real?What type of waves you will get?What will be the form ofE?

3.3 Linear and Circular Polarization of EM Waves

The plane wave (3.6) and (3.11) is a wave with its electric field vector alwaysin the direction 1. Such a wave is said to be linearly polarized with po-larization vector1. The wave described by (3.12) is linearly polarized withpolarization vector 2 and is linearly independent of the first. The two waves

:

E1 = 1E1eikxit , E2 = 2E2e

ikxit

with (3.17)

Bi =

k Ei

k , i= 1, 2

Can be combined to give the most general homogeneous plane waves prop-agating in the directionk = kn,

E(x, t) = (1E1+2E2) eikxit (3.18)

E(x, t) = 1|E1| +2|E2|ei(21) eikxit+i1 (3.19)1 Note : we use the magnetic induction H because although B is the applied

induction, the actual field that carries the energy and momentum in media is H.2 Note:To prove the above relations use cos2 x = 1/2and sinceER= (E+E)/2

we getE2R =E E/2.


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3.3 Linear and Circular Polarization of EM Waves 47

Fig. 3.3.

The amplitudes E1 =|E1|ei1 and E2 =|E2|ei2 are complex numbers inorder to allow the possibility of a phase difference between waves of differentpolarization.

LINEARLY POLARIZED

If the amplitudes E1 =|E1|ei1

and E2 =|E2|ei2

have the same phase(3.18) represents a linearly polarized wave with the polarization vectormaking an angle = tan1 ((E2)/(E1))(which remains constant as thefield evolves in space and time) with 1 and magnitudeE=

E21 + E

22 .

ELLIPTICALLY POLARIZEDIfE1 and E2 have the different phase the wave (3.18) is elliptically po-larizedand the electric vector rotates around k.

3.3.1 Circular Polarization

E1 = E2 = E0

1

2 =

/2and the wave becomes

E(x, t) = E0(1 i2) eikxit (3.20)

At a fixed point in space, the fields are such that the electric vector is constantin magnitude, but sweeps around in a circle at a frequency .

Fig. 3.4.


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The components of the electric field, obtained by taking the real part of(3.20)

Ex(x, t) = E0cos(kz t) , Ey(x, t) = E0cos(kz t) (3.21)For the upper sign (1 + i2) the rotation is counter-clockwise when theobserver is facing into the oncoming wave. The wave is called left circularlypolarized in optics while in modern physics such a wave is said to havepositive helicity.For the lower sign (1 i2)the wave is right circularly polarized or ithasnegative helicity.

3.3.2 Elliptically Polarized EM Waves

An alternative general expression for E can be given in terms of the complexorthogonal vectors

= 1

2(1 i2) (3.22)

with properties

= 0 , 3 = 0 , = 1 . (3.23)Then the general representation of the electric vector

E(x, t) = (E+++ E) eikxit (3.24)

whereE andE+ are complex amplitudes

IfEand E+havedifferent amplitudesbut thesame phaseeqn (3.24)represents an elliptically polarized wave with principle axes of the ellipsein the directions of1 and 2.

The ratio of the semimajor to semiminor axis is|(1 + r)/(1 r)|, whereE/E+ = r.

The ratio of the semimajor to semiminor axis is|(1 + r)/(1 r)|, whereE/E+ = r.

If the amplitudes have a phase difference between them E/E+ = rei,

then the ellipse traced out by the E vector has its axes rotated by an angle/2.

Note : Forr = 1we get back to a linearly polarized wave.

3.4 Stokes Parameters

The polarization content of an EM wave is known if it can be written in theform of either (3.18) or (3.24) with known coefficients (E1, E2)or (E, E+).

In practice, the converse problem arises i.e.given a wave of the form (3.6),how can we determine from observations on the beam the state of polarization?


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3.4 Stokes Parameters 49

Fig. 3.5. The figure shows the general case of elliptical polarization and the ellipsestraced out by both E and Bat a given point in space.

Fig. 3.6. The figure shows the linear, circular and elliptical polarization

A useful tool for this are the four Stokes parameters. These arequadratic in the field strength and can be determined through intensity mea-surements only. Their measurements determines completely the state of po-larization of the wave.

For a wave propagating in the z -direction the scalar products

1 E , 2 E , + E , E (3.25)are the amplitudes of radiation respectively, with linear polarizationin thex-direction, linear polarization in the y-direction, positive helicity andnegative helicity.


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The squares of these amplitudes give a measure of the intensity of eachtype of polarization.

Thephase information can be taken by using cross productsIn terms of the linear polarization bases (1, 2), the Stokes parameters

are:

s0 = |1 E|2 + |2 E|2 =a21+ a22s1 = |1 E|2 |2 E|2 =a21 a22s2 = 2 [(1 E)(1 E)] = 2a1a2cos(1 2) (3.26)s3 = 2 [(1 E)(1 E)] = 2a1a2sin(1 2)

where we defined the coefficients of (3.18) or (3.24) as magnitude times aphase factor:

E1 = a1ei1 , E2 = a2e

i2 , E+ = a+ei+ , E= ae

i (3.27)

Here s0 and s1 contain information regarding the amplitudes of linearpolarization, whereass2 ands3 say something about thephases. Knowingthese parameters (e.g by passing a wave through perpendicular polarizationfilters) is sufficient for us to determine the amplitudes and relative phases ofthe field components.

In terms of the linear polarization bases (+, ), the Stokes parametersare:

s0 = |+ E|2 + | E|2 =a2++ a2s1 = 2

(+ E)( E)= 2a+a cos( +) (3.28)

s2 = 2 (+ E)( E)= 2a+a sin( +)s3 = |+ E|2 | E|2 =a2+ a2

Notice an interesting rearrangement of roles of the Stokes parameters withrespect to the two bases.

The four Stokes parameters are not independent since they depend on only3 quantities a1, a2 and1 2. They satisfy the relation

s20 = s21+ s

22+ s

23 . (3.29)

3.5 Reflection & Refraction of EM WavesThe reflection and refraction of light at a plane surface between two media ofdifferent dielectric properties are familiar phenomena.

The various aspects of the phenomena divide themselves into two classes

Kinematic properties:


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3.5 Reflection & Refraction of EM Waves 51

Angle of reflection= angle of incidence Snells law: sin isin r =

nn where i, r are the angles of incidence and

refraction, whilen, n are the corresponding indices of refraction. Dynamic properties:

Intensities of reflected and refracted radiation Phase changes and polarization

The kinematic properties follow from the wave nature of the phenom-ena and the need to satisfy certain boundary conditions (BC). But not on thedetailed nature of the waves or the boundary conditions.

The dynamic properties depend entirely on the specific nature ofthe EM fields and their boundary conditions.

Fig. 3.7. Incident wave k strikes plane interface between different media, givingrise to a reflected wave k and a refracted wavek. The media below and above theplanez= 0have permeabilities and dielectric constants , and , respectively.The indices of refraction are n =

and n =

.

According to eqn (3.18) the 3 waves are:

INCIDENT

E= E0eikxit , B=

k Ek

(3.30)

REFRACTED

E =E0eikxit , B =

k Ek

(3.31)

REFLECTED

E =E0eikxit , B =

k Ek

(3.32)


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Thewave numbers have magnitudes:

|k| = |k| =k = c

, |k| =k =

c

(3.33)

AT the boundary z = 0 the BC must be satisfied at all points on theplane at all times, i.e. the spatial & time variation of all fields must be thesame at z = 0.

Thus the phase factors must be equal at z = 0

(k x)z=0 = (k x)z=0 = (k x)z=0 (3.34)

independent of the nature of the boundary conditions.Eqn (3.34) contains thekinematic aspects of reflection and refraction.

Note that all 3 wave vectors must lie in a plane. From the previous figurewe get

k sin i= k sin r =k sin r (3.35)

Sincek = k , we find that i = r ; the angle of incidence equals the angle ofreflection.

Snells law is:

sin i

sin r =

k

k =

=

n

n (3.36)

The dynamic properties are contained in the boundary conditions :

normal components ofD = E and B are continuous

tangential components ofE and H = [c/()]k E are continuousIn terms of fields (3.30)-(3.32) these boundary conditions at z = 0are:

[ (E0+ E0 ) E0] n= 0

[k E0+ k E0 k E0] n= 0(E0+ E

0 E0) n= 0 (3.37)

1

(k E0+ k E0 )

1

(k E0)

n= 0

Two separate situations, the incident plane wave is linearly polarized : The polarization vector is perpendicular to the plane of incidence (theplane defined by k and n ).

The polarization vector is parallel to the plane of incidence. The case of arbitrary elliptic polarization can be obtained by appropriatelinear combinations of the two results.


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3.5 Reflection & Refraction of EM Waves 53

3.5.1 E : Perpendicular to the plane of incidence

Since the E-fieldsare parallel to the surface the 1st BC of (3.38) yieldsnothingThe 3rd and 4th of of (3.38) give ( how?):

E0+ E0 E0 = 0

(E0 E0 )cos i

E0cos r= 0 (3.38)

The 2nd, using Snells law, duplicates the 3rd.(prove all the above statements)

Fig. 3.8. Reflection and refraction with polarization perpendicular to the plane ofincidence. All the E-fields shown directed away from the viewer.

The relative amplitudes of the refracted and reflected waves can be foundfrom (3.38)

E0E0

= 2n cos i

n cos i +

n2 n2 sin2 i=

2

1 + tan i tan r=

2sin r cos i

sin(i + r)

=

E0E0

=n cos i

n2 n2 sin2 i

n cos i + n2 n2 sin2 i

=1 tan i tan r1 + tan i tan r

=

sin(r i)sin(i + r)

=

(3.39)

Note that

n2 n2 sin2 i= n cos r but Snells law has been used to expressit in terms of the angle of incidence.

For optical frequencies it is usually permitted to put = .


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3.5.2 E : Parallel to the plane of incidence

Boundary conditions involved:normal E : 1st eqn in (3.38) tangential E : 3rd eqn in (3.38) tangential B : 4th eqn in (3.38)

The last two demand that

(E0 E0 )cos i E0cos r= 0

(E0+ E

0 )

E0 = 0 (3.40)

Fig. 3.9. Reflection and refraction with polarization parallel to the plane of inci-

dence.

The condition that normal E is continuous, plus Snells law, merely dubli-cates the 2nd of the previous equations.

The relative amplitudes of refracted and reflected fields are therefore(how?)

E0E0

= 2nn cos i

n

2 cos i + n

n2 n2 sin2 i

= 2 n

n

1 + tan i tan r=

2sin r cos i

sin(i + r) cos(i r)

=

E0E0

=

n2 cos i nn2 n2 sin2 i n

2 cos i +

kostas kokkotas "field theory"

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