The University of Queensland Department of Physics 2008
Lecture notes and tutorial problems of the undergraduate course PHYS3050/PHYS7051 ELECTROMAGNETIC THEORY III
Lecturer: Zbigniew Ficek Physics Annexe(6): Rm 436 Ph: 3365 2331 email:
[email protected] http://www.physics.uq.edu.au/people/ficek/ Consultation Hours: Wednesday 2pm – 4pm
Preface This lecture notes covers the principal elements of classical electromagnetic theory embodying Maxwell’s equations with applications mainly to situations where electric charge can be treated as a continuous fluid. The intention is to introduce students to the background of classical field theory and the applications of the electromagnetic theory to solid state physics, classical optics, radiation theory and telecommunication. The goal of this course is to provide a compact logical exposition of the fundamentals of the electromagnetic theory and the applications to various areas of physics and engineering. The treatment is quantitative throughout and an attempt has been made to imbue students with a sound understanding of the Maxwell’s equations and with the ability to apply them to modern problems in physics. The organization of the lectures is fairly standard and includes vector analysis, electrostatic, magnetostatic, mathematical techniques in the solution of the Maxwell’s equations and the Laplace’s equation, time varying fields and applications of the solution of the Maxwell’s equations. The material on vector analysis gives greater emphasis to the relationship between fields and their sources. A number of revision questions have been included at the end of each chapter. These questions have been designed not only to point out to the student the essential material of the chapter but also to test the students’ understanding of the material presented in the chapter. In addition, tutorial problems have been included at the end of the chapters that contain the most important elements of the electromagnetic theory. These problems will be discussed and solved in details during the tutorial sessions, but it is hoped that the student will attack these problems before attending the tutorial session.
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Contents 1 The Classical Theory of the Electromagnetic Field 1.1 Elementary Aspects of Electromagnetism . . . . . . . 1.2 Macroscopic Charges and Currents . . . . . . . . . . 1.2.1 Charge Density . . . . . . . . . . . . . . . . . 1.2.2 Current density . . . . . . . . . . . . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . 2 Mathematical Description of Vector Fields 2.1 Gradient of a Scalar Function . . . . . . . . . 2.2 Divergence Function . . . . . . . . . . . . . . 2.3 The Flux of a Vector Field . . . . . . . . . . . 2.4 Gauss’s Divergence Theorem . . . . . . . . . . 2.5 The Continuity Equation for Electric Current 2.6 Curl (Rotation) Function . . . . . . . . . . . . 2.7 Stokes’s Theorem . . . . . . . . . . . . . . . . 2.8 Successive Application of ∇ . . . . . . . . . . Revision questions . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . .
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13 13 16 16 18 19
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20 20 22 24 24 25 26 27 28 30 30
3 Vectorial Equations in Electromagnetism: Scalar and Vector Potentials 3.1 Maxwell’s Equations, an Example of Vectorial Equations . . . 3.2 Scalar Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 34 35 37 39 40
4 The Experimental Basis of the Development of Electromagnetic Theory 4.1 Coulomb’s Law – Force between Static Charges . . . . . . . . 4.2 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Biot-Savart Law – Force between Static Currents . . . . . . . 4.4 Current Element and Charge Element . . . . . . . . . . . . . . 4.5 The Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Amperes Circuit Law . . . . . . . . . . . . . . . . . . . . . . . 4.7 Faraday’s Law of Electromagnetic Induction . . . . . . . . . .
41 41 42 49 51 52 53 55
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Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 5 Differential Equations for the EM Field and ory 5.1 Differential Equations for the EM Field . . . ~ . . . . . . . . . . . . 5.1.1 Divergence of E ~ . . . . . . . . . . . . . . . 5.1.2 Curl of E ~ . . . . . . . . . . . . 5.1.3 Divergence of B ~ . . . . . . . . . . . . . . . 5.1.4 Curl of B 5.2 The Maxwell’s Theory . . . . . . . . . . . . Revision questions . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . .
Maxwell’s The. . . . . . . .
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61 61 62 62 63 63 64 68 68
6 Maxwell’s Equations and Prediction of Electromagnetic Waves 6.1 The Wave Equation for EM Waves in Vacuum . . 6.2 Plane Wave Solution to the Wave Equation . . . . 6.3 Harmonic Waves . . . . . . . . . . . . . . . . . . 6.4 The Transverse Nature of Plane Waves in Vacuum Revision questions . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . .
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69 70 71 73 74 78 78
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7 EM Theory and Einstein’s Special Theory of Relativity 7.1 Lorentz Transformation Equations for Space and Time . . . 7.2 Force Transformation Equations . . . . . . . . . . . . . . . . 7.2.1 The Force between Two Charges Moving with Constant Velocities . . . . . . . . . . . . . . . . . . . . . 7.3 Electric Field Lines of a Moving Charge . . . . . . . . . . . 7.4 Magnetic Field Lines of a Moving Charge . . . . . . . . . . . 7.5 Invariance of the Maxwell equations under the Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Remark on the Electromagnetic Induction . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Energy in the Electromagnetic Field: Poyntings’ Theorem
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80 . 81 . 82 . 83 . 87 . 88 . 89 . 95 . 98
99
8.1
Rate of Doing Work by the Field on the Current – Ohmic Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Electrostatic Field Energy Density . . . . . . . . . . . . . . . 8.3 Magnetostatic Field Energy Density . . . . . . . . . . . . . . . 8.4 Poynting Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Phase Relationships in EM Waves . . . . . . . . . . . . . . . . 8.6 Momentum Flux . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 EM Energy Flow: Circuit versus Field Theory . . . . . . . . . 8.7.1 Energy Flow in a Resistive Wire . . . . . . . . . . . . . 8.7.2 Energy Flow out of Battery . . . . . . . . . . . . . . . 8.7.3 Propagation of an Electromagnetic Wave along a Wire Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101 102 103 103 105 105 106 106 108 109 111 111
9 Solution of Laplace’s Equation and Boundary Value Problem112 9.1 Uniqueness of the Solution of Laplace’s Equation . . . . . . . 113 9.1.1 Dirichlet Theorem . . . . . . . . . . . . . . . . . . . . 113 9.2 Solutions of Laplace’s Equation . . . . . . . . . . . . . . . . . 114 9.2.1 Method of Separation of Variables . . . . . . . . . . . . 115 9.2.2 Solution of the Laplace Equation in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 10 General Solution of the Maxwell’s Equations 10.1 Difficulty of the Direct Solution of Maxwell’s Equations 10.2 Scalar and Vector Potentials . . . . . . . . . . . . . . . 10.2.1 Lorenz Gauge . . . . . . . . . . . . . . . . . . . 10.2.2 Coulomb Gauge . . . . . . . . . . . . . . . . . . 10.3 Solution of the Inhomogeneous Wave Equations . . . . 10.4 Rigorous Solution: Green Functions Method . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . . 11 Electromagnetic Antennas: Hertzian Dipole 11.1 Generation of electromagnetic waves . . . . . . . 11.1.1 Field of an Element of Alternating Current 11.2 Power Radiated from the Current Element . . . . 11.3 Gain of the Dipole Antenna . . . . . . . . . . . . 5
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138 . 138 . 140 . 142 . 143 . 145 . 148 . 150
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152 . 154 . 154 . 161 . 163
Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 12 Electromagnetic Theory of Polarizable Materials 12.1 Potential and Electric Field of a Single Dipole . . 12.2 Polarization Vector . . . . . . . . . . . . . . . . . ~ in a Dielectric . . . 12.3 Maxwell’s Equation for ∇ · E 12.4 Macroscopic Effects of the Polarizability . . . . . 12.5 Dense Dielectrics: The Clausius-Mossotti Relation 12.6 Dielectric in a Time Dependent Field . . . . . . . 12.7 The Complex Susceptibility and Permitivity . . . 12.8 The Loss Tangent . . . . . . . . . . . . . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . .
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166 . 167 . 168 . 171 . 173 . 176 . 178 . 182 . 185 . 187 . 187
13 Electromagnetic Theory of Magnetizable Materials 13.1 Magnetic Polarization Currents . . . . . . . . . . . . . . . ~ . . . . . . . . . . . . . . 13.2 The Magnetic Intensity Vector H 13.2.1 Static Magnetic Fields . . . . . . . . . . . . . . . . 13.3 Linear Magnetic Materials . . . . . . . . . . . . . . . . . . 13.4 Non-Linear Magnetic Materials . . . . . . . . . . . . . . . 13.4.1 Work done in magnetization of iron . . . . . . . . . 13.5 Permanent Magnetic Materials: Ferromagnets . . . . . . . ~ and H ~ fields of a ferromagnet . . . . . . . . . . . 13.5.1 B 13.5.2 Magnetic Poles . . . . . . . . . . . . . . . . . . . . 13.6 Time Dependent Magnetic Fields and Energy Loss . . . . 13.7 The complex magnetic susceptibility . . . . . . . . . . . . 13.8 Maxwell’s Equations in Dielectric and Magnetic Materials Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . .
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190 190 195 196 197 200 201 202 203 205 207 208 209 210 212
14 Poynting’s Theorem Revisited 214 ~ and H ~ . . . . . . . . . . . . . 214 14.1 Poynting Vector in Terms of E 14.2 Poynting Vector for Complex Sinusoidal Fields . . . . . . . . . 216 Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
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15 Plane Wave Propagation in Dielectric and Magnetic Media 220 15.1 Dispersive Equation and Complex Propagation Number . . . . 222 15.2 Wave Refraction and Attenuation . . . . . . . . . . . . . . . . 225 15.2.1 Propagation of an EM wave in a low loss dielectric . . 225 15.2.2 Propagation of an EM Wave in a Good Conductor . . 227 15.2.3 Skin Effect . . . . . . . . . . . . . . . . . . . . . . . . . 229 Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 16 Transitions Across Boundaries for Electromagnetic Fields 16.1 Normal Components of the Fields . . . . . . . . . . . . . . . ~ . . . . . . . . . . . . . . . . 16.1.1 Normal Component of B ~ . . . . . . . . . . . . . . . . 16.1.2 Normal Component of H ~ ~ . . . . . . . . . . . . 16.1.3 Normal Component of D and E 16.2 Tangential Components of the Fields . . . . . . . . . . . . . ~ . . . . . . . . . . . . . . 16.2.1 Tangential Component of E ~ . . . . . . . . . . . . . . 16.2.2 Tangential Component of H ~ . . . . . . . . . . . . . . 16.2.3 Tangential component of B Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
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17 Propagation of an EM Wave Across a Boundary 17.1 Representation of Plane Waves in Different Directions ~ in terms of E ~ . . . . . . . 17.1.1 Representation of B 17.2 Directions of Reflected and Transmitted Waves . . . 17.3 Angle of Reflection and Snell’s Law of Refraction . . Revision questions . . . . . . . . . . . . . . . . . . . . . .
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244 . 245 . 246 . 247 . 252 . 253
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254 . 255 . 257 . 258 . 260 . 261
18 Fresnel’s Equations ~ i normal to plane of incidence . . . 18.1 E ~ i in the plane of incidence . . . . . 18.2 E 18.3 Fresnel Equations for dielectric media Revision questions . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . .
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234 234 234 236 237 237 238 240 240 242 242
19 Applications of the Boundary Conditions and the Fresnel Equations 262 19.1 Applications in dielectrics . . . . . . . . . . . . . . . . . . . . 262
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19.1.1 Polarization by reflection . . . . . . . 19.1.2 Total internal reflection . . . . . . . . 19.2 Transmission and Reflection at a Conducting Surface . . . . . . . . . . . . . . . . . . . . . 19.2.1 Field vectors at normal incidence . . Revision questions . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . .
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267 269 271 272
20 Propagation of an EM Wave in a Rectangular Waveguide 20.1 Transverse Electric (TE) Modes . . . . . . . . . . . . . . . . 20.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . 20.3 TE Modes in a Lossless Waveguide . . . . . . . . . . . . . . 20.4 Phase and Group Velocities of Mode Propagation . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . .
273 . 275 . 278 . 280 . 283 . 286
21 Relativistic Transformation of the Electromagnetic Field 21.1 The Principle of Relativity . . . . . . . . . . . . . . . . . . . 21.2 Transformation of Electric and Magnetic Field Components 21.3 Transformation Rules in Terms of Parallel and Normal Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3.1 Rules for Parallel Components . . . . . . . . . . . . . 21.3.2 Rules for Normal Components . . . . . . . . . . . . . 21.4 Transformation of the Components of a Plane EM Wave . . . . . . . . . . . . . . . . . . . . . . 21.5 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 Transformation of Energy of a Plane EM Wave . . . . . . . Revision questions . . . . . . . . . . . . . . . . . . . . . . . . . . Tutorial problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
287 . 287 . 293 . 294 . 295 . 295 . . . . .
297 299 301 303 304
Revision Questions for the Final Examination
306
Appendix A: Proof of the Amperes law
312
Appendix B: Proof of the vector theorem, Eq. (73)
314
Appendix C: PHYS3050 Facts and Formulae
315
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Literature 1. J.D. Jackson, Classical Electrodynamics, 3rd ed. Wiley 1999. The course is aimed at this level of treatment. 2. R.K. Wangsness, Electromagnetic Fields, 2nd ed. Wiley 1986. This is an alternative textbook for the course. 3. R. Plonsey and R.E. Collin, Principles and Applications of Electromagnetic Fields, McGraw Hill 1961. This is a good general text on electromagnetic fields. 4. B.I. Bleaney and B. Bleaney, Electricity and Magnetism, 3rd ed. Oxford U.P. 1983. A very readable book ranging over circuit theory, electronics, electric properties of matter and field theory. 5. J.A. Stratton, Electromagnetic Theory, McGraw Hill, 1941. This is a very comprehensive reference on electromagnetic theory. 6. W.K.H. Panofsky and M. Phillips, Classical Electricity and Magnetism, 2nd ed. Addison-Wesley 1962. This is a good text on the relativistic representation of electromagnetism. 7. R.P. Feynman, R.B. Leighton, and M. Sands, The Feynaman Lectures on Physics, Vol. 2, Addison Wesley, 1964. A good reference to the physics of the emission of radiation.
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”To anyone who is motivated by anything beyond the most narrowly practical, it is worthwhile to understand Maxwell’s equations simply for the good of his soul” − J.R. Pierce
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1
The Classical Theory of the Electromagnetic Field
Classical theory of the electromagnetic field or Classical Electrodynamics, formulated by Maxwell more than 100 years ago, is now a well established theory with many applications in different areas of physics, chemistry and engineering. In this context ‘classical’ means ‘non-quantum’, but we would like to point out that the basic equations of electromagnetism, the Maxwell’s equations, hold equally in quantum and classical field theory. We begin the study of the electromagnetic theory with a brief review of the elementary aspects of electromagnetism that the student learnt in the first year PHYS1002 course. These aspects are in fact about the microscopic nature of the electromagnetic theory. To indicate the classical approach to the electromagnetic theory and the macroscopic nature of electromagnetic phenomena we are going to explore in this lecture notes, we first introduce the concept of macroscopic distributions of charges and currents, i.e. we define volume, surface and linear charge densities. Next, we will introduce the concept of the current density and a total (macroscopic) current.
1.1
Elementary Aspects of Electromagnetism
In many textbooks on electromagnetism, the elementary aspects of electromagnetism are introduced as postulates. In fact, they are based on experimental observations, and we will explore this fact in full details in one of the lectures. At this introductory lecture, let us review briefly the elementary aspects of electromagnetism that are: 1. Electromagnetic interactions are ONE of FOUR fundamental types:1 Type of interaction Strong interaction (nuclear) Electromagnetic Weak interaction (e.g. β decay) Gravitation 1
Relative Strength 1 10−2 10−12 10−40
At our level of discussion there is no relation between these four types of interaction, i.e. they cannot be considered as different manifestations of a single FORCE.
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2. Electromagnetic (EM) forces or interactions are due to ELECTRIC CHARGE, which is NOT in turn explicable in terms of anything else. 3. Charges are of two kinds called positive and negative. In the static limit like charges repel and unlike attract. 4. Charges are quantized in units of e ' 1.6 × 10−19 [Coulombs]. 5. In the static limit the inverse square (Coulomb) law of force holds: F~q2 =
1 q2 q 1 rˆ , 4πε0 r2
rˆ
q1 −→ q2 ,
that the charge q1 acts on the charge q2 with the force F~q2 . The force acts along the distance r and the parameter ε0 determines the property of the medium and is called the electric permittivity. The Coulomb’s law holds for microscopic charges, i.e. it holds only for charges whose the spatial dimensions are small compared with the distance separating them. The Coulomb’s law does not tell us how the first charge knows the other one is present. One usually assumes that the charge produces an electric field which then interacts with the other charges. To express this explicitly, the Coulomb force is often written as ~ q1 , F~q2 = q2 E ~ q1 is the electric field produced by the charge q1 . where E The Coulomb law can be tested to great accuracy indirectly by showing that no charge rests on the inside of a statically charged hollow conductor.2 2
S.J. Plimpton and W.E. Lawton, Phys. Rev. 50, 1066 (1936).
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If the exponent in the Coulomb’s law for point charges E∼
1 rn
is not n = 2 but n = 2(1 ± ), the potential inside the hollow conductor would be large. Since the potential inside the hollow conductor found by the experiments was less in magnitude than a small detectable potential, then ' 10−9 , the level of sensitivity of the detector. 6. Electric charge is conserved (algebraically) X
q = constant
whole universe
7. In the NON-STATIC case, i.e. the case of moving charges, the force is no longer given by Coulomb’s Law. It is given by the Lorentz equation ~ + ~v × B) ~ , F~q2 = q2 (E ~ and B ~ are the electric and magnetic fields due to q1 , and ~v is where E the velocity of the charge q2 . ~ and B ~ de8. In the electromagnetic theory, we assume that the fields E ~ B ~ and the pend on the frame of reference of the observer, that E, ~ force F must follow the required relativistic transformation law. However, we do not think that q depends on the frame of reference. i.e. q does not depend on its velocity with respect to an observer. In terms of the relativistic theory, the charge is invariant under the Lorentz transformation.3
The student has noticed that the Lorentz force involves both, the electric and ~ and how E ~ and B ~ are computed for magnetic fields. Why there must be a B 3
This is because in ordinary matter electrons move much faster than ions, their speeds depend on temperature, and electric fields are not observed to arise from changes in temperature.
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arbitrary motion of charges is the substance of electromagnetic theory. We now have the following picture of the basis of the electromagnetic theory: ELECT ROM AGN ET IC
1. CHARGE 1 =⇒
F IELD
F ORCE ON
=⇒ CHARGE 2
2. Fields are generated by charges - NOT by other fields.
1.2
Macroscopic Charges and Currents
We know from the Millikan experiment that electric charge is quantized. The electron is a point charge on the smallest scale measurable. We may then speak, on a subatomic scale, of a microscopic theory of electromagnetism. On a subatomic scale there must be very strong and rapidly varying electric and magnetic fields on spatial scales ∼ 10−8 m and temporal scales ∼ 10−10 s. When we measure the fields around a macroscopic circuit, clearly we are not looking at these fields. We are measuring fields on distance scales much larger than 10−8 m and time scales much longer than 10−10 s. In the macroscopic context, we do not distinguish individual charges. It is convenient and justifiable to regard the charge as a continuous ”fluid” distributed over the volume or surface of a charged material. Thus, we may introduce the concept of macroscopic charges and currents to indicate the macroscopic nature of electromagnetic phenomena.
1.2.1
Charge Density
Charges may be distributed throughout the volume of a material, or on the surface of the material, or may be distributed along one dimensional wires. Thus, there are different spatial distributions of charges considered in the electromagnetic theory that result in three types of charge densities: volume charge density, surface charge density and linear charge density.
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When we encounter a large number of point charges in a finite volume, it is convenient to describe the source in terms of a volume charge density, defined as Σq , ρ = lim ∆V where Σq is the algebraic sum of the charge in the volume ∆V . The limit is not to zero, but to a ∆V much larger than atomic scale size, which is still very small on the laboratory scale. If the volume charge density is represented by a continuous function ρ(~r), the total charge Q in a volume V is given by Q=
Z
ρ(~r) dV .
V
Note that the charge density is a function of the position which varies smoothly inside a charged material. An exception is a boundary between two charged materials where ρ may change discontinuously due to the presence of surface charges of a non-zero density. Thus, we may introduce the concept of surface charge density. A surface charge density is defined, analogously to the volume charge density, as Σq , ∆S where Σq is the algebraic sum of the charge on the surface ∆S. σ = lim
Then, the total charge continously distributed on a surface S is QS =
Z
σ(~r) dS .
S
In the same fasion, we may define a linear charge density on a length ∆L Σq , ∆L and then the total linear charge continously distributed on a length L is γ = lim
QL =
Z
γdL .
L
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1.2.2
Current density
For many purposes in the electromagnetic theory, it will be necessary to introduce the concept of current density. It measures the amount of current flowing through an area normal to the direction of the current. We define the current density in the following way. Let IA = ∂q/∂t is a current through the area A, as shown in Figure 1. Then the current density, normal to the area A is defined by δq I ˆ = lim n ˆ J~ = lim n A δt A ρ A v δt n ˆ = ρ~v , = lim δt A
Figure 1: Illustration of the current flow through an area A and the evaluation of the current density through the area A.
where ~v = vˆ n, and n ˆ is the unit vector normal to the area A, and the limit is taken in the same sense as for ρ. Total current through an arbitrary surface area For a surface of an arbitrary shape, the current may not be normal to the surface at all points on the surface. However, if the current density J~ is known at all very small areas, that can be approximated by flat surfaces of a macroscopic surface A, we still can obtain the total current through the area A. Consider a current density J~ through a macroscopic area A, as shown in Figure 2. To find the total current through the area, we first divide the area A into a lot of small flat areas dA, 18
Figure 2: Illustration of evaluation of the current density through an area A. The vector n ˆ is a unit vector normal to the surface element dA.
and find the current through dA as4 δI = J~ · n ˆ dA = J cos(θ)dA . Then the current through the total area A is the sum of the contributions from all the small elements of the area: IA =
Z
δI =
Z
J~ · n ˆ dA =
A
Z
~, J~ · dA
A
~ = dAˆ where dA n is a vector representing the element dA of the surface A. Note: In vector analysis it is common to represent a surface by a vector whose length corresponds to the magnitude of the surface area and whose the direction is specified by the unit vector n ˆ normal to the surface. In summary, Electromagnetic theory formulated in terms of space dependent quantities, charge densities ρ, σ or γ, and current density J~ is regarded as a MACROSCOPIC THEORY.
Revision questions Question 1. What are the elementary aspects of electromagnetism? Question 2. How do we define volume, surface and linear charge densities? Question 3. How do we define and evaluate the current density through an area? Question 4. If the current density through an area is known, how does the current through the area is evaluated?
4
In literature, very often small areas that a macroscopic area is divided to are called elementary areas or elementary surfaces.
19
2
Mathematical Description of Vector Fields
The study of electromagnetic theory requires considerable knowledge of vector analysis. In this lecture, we will introduce vector operations such as gradient, divergence and curl that we will need for our study of electromagnetic theory. As we shall see, these vector operations are very convenient to determine the properties of electromagnetic field, also considerably simplify the formulation of electromagnetic theory and allow get a better inside into electromagnetic phenomena.
2.1
Gradient of a Scalar Function
Let us first define a vector operation: Gradient of a scalar function. Assume that a function Φ represents a scalar field and that Φ is a single valued, continuous, and differentiable function of position. Let dΦ represents a change of Φ with a distance ds. The gradient of the scalar function Φ is defined as: grad Φ ≡ ∇Φ =
∂Φ n ˆ, ∂s
where n ˆ is a unit vector in the direction the rate ∂Φ/∂s has its maximum value. In other words, gradient tells us in which direction the change in Φ is maximal. ~ a change of Φ can For some other direction dX, ~ Figure 3: Illustration of the be found by projection of gradient of Φ on dX: evaluation of gradient of a scalar function Φ.
~ = dΦ = ∇Φ · dX
∂Φ ~ = ∂Φ cos θ dX . n ˆ · dX ∂s ∂s
We know from the vector analysis that it is convenient to represent a vector in a reference (coordinate) frame. Commonly used are the rectangular (cartesian) coordinates, in which we can easily find that the x component of
20
the gradient is ∂Φ ˆ ∂Φ (∇Φ)x = ∇Φ · ˆi = n ˆ·i= cos θ ∂s ∂s δΦ δΦ ∂Φ = lim = lim = . δ(s/ cos θ) δx ∂x Similarly, the y and z components of the gradient are (∇Φ)y =
∂Φ , ∂y
(∇Φ)z =
∂Φ . ∂z
Hence, in cartesian coordinates, the gradient of a scalar function Φ can be written as5 ∇Φ =
∂Φˆ ∂Φ ˆ ∂Φ ˆ i+ j+ k. ∂x ∂y ∂z
The gradient is analogous to multiplication of a vector by a scalar. The result, of course, is a vector.6 Remember: Mathematically, gradient of an arbitrary scalar function is a vector that is pointing in the direction of maximal changes of the function. Exercise in class: Consider a scalar function Φ = x. Calculate the gradient of Φ = x. Solution: Since ∂Φ =1 ∂x
and
∂Φ ∂Φ = =0, ∂y ∂z
we find that ∇Φ =
∂x ˆ ∂x ˆ ∂x ˆ ˆ i+ j+ k=i. ∂x ∂y ∂z
This exercise explicitly shows that gradient is a vector pointing in the direction of maximal changes in Φ. 5 6
Expressions for the gradient in other coordinate systems are given in the Appendix C. We do not usually take a gradient of a vector, the result would be a tensor.
21
The student should be able now to give a quick answer to simple questions as ”Is gradient a scalar or a vector?” or ”In which direction does gradient of a given scalar function point?” A practical test: Try to give the answer to the following question without any calculations: What is the direction of the gradient of a scalar function Φ = x + y?
2.2
Divergence Function
We have already defined the nabla operator and showed how it evaluates a certain property of a scalar function. We now define the divergence function, which involves the nabla operator and, as we shall see, tells us about the flow of a vector field and its sources. Definition of divergence: The divergence is the scalar function which results from operation of ∇ upon a vector F~ in a fashion analogous to the dot product of two vectors. The result is a scalar function. In Cartesian coordinates, the divergence function is written as7 ∂Fx ∂Fy ∂Fz + + . div F~ ≡ ∇ · F~ = ∂x ∂y ∂z Properties: A positive divergence means that there is a source of a vector field, and a negative divergence means the presence of a sink of the field. When ∇ · F~ = 0 everywhere, the field F~ is called solenoidal, since no starting points or sources can be assigned to the lines describing the field. In other words, it has no sources or sinks. Things to remember: Mathematically, divergence of an arbitrary vector is a scalar. Physically, a non-zero ∇ · F~ implies a source (if positive) or a sink (if negative), and if ∇ · F~ = 0, there is no source or sink – the field lines have no beginnings or ends. 7
The forms that the divergence takes in other coordinate systems are given in the Appendix C.
22
Exercise in class: Consider two vectorial fields F~1 = xˆi and F~2 = yˆi. Which of the fields is solenoidal? Solution: A vector field is solenoidal when ∇ · F~ = 0. Thus, we calculate the divergence of the fields ∂x ˆ ˆ ∂x ˆ ˆ ∂x ˆ ˆ div F~1 = (i · i) + (j · i) + (k · i) = 1 . ∂x ∂y ∂z and ∂y ˆ ˆ ∂y ˆ ˆ ∂y ˆ ˆ div F~2 = (i · i) + (j · i) + (k · i) = 0 , ∂x ∂y ∂z from which we see that the field F~2 = yˆi is solenoidal.
This exercise shows that divergence of a given vector field is different from zero only if the field amplitude changes in the direction of the field. So now it is easy to visualize the divergence of a vector field. The divergence is related to how the field changes as you move in the direction of the field. One more example for a better understanding of the meaning of divergence. Consider a vector ~r = xˆi + yˆj + z kˆ that represents the position of an arbitrary point in the Cartesian coordinates. The student can easily find that div~r = 3. What does the number 3 mean? It means that the field changes by one unit in each direction. However, this statement may not be generally true. For example, divergence of a vector F~ = 2xˆi + yˆj is also equal to 3, but in this case the field changes by two units in the x direction and one unit in the y direction.
23
2.3
The Flux of a Vector Field
A vector field propagating in space may cross some surfaces not necessary normal to the field direction. In this case, we may speak about a flux of the field through the surface. The flux is measured by the number of field lines crossing the surface.
Figure 4: A vector field F~ lines
Mathematically, the flux Ψ of a field F~ through a surface S, not necessary a closed S, as shown in Figure 4, is defined as:
crossing a surface S.
Ψ=
Z
~= F~ · dS
S
Z
F~ · n ˆ dS ,
S
where n ˆ is the unit vector normal to the surface at the point where the element of area dS is located. Thus, we calculate the flux through a surface S of arbitrary shape by dividing it up into a lot of small surfaces dS and calculate the flux through each of these small surfaces. next, we add (integrate) the fluxes to obtain the total flux through the surface S.
2.4
Gauss’s Divergence Theorem
We now introduce an important identity in vector analysis, the Gauss’s divergence theorem or Gauss’s law, which we will use frequently in our study of the theory of the electromagnetic fields. Consider a vector field F~ crossing a closed surface bounding a volume V , as shown in Figure 5. Gauss’s law, or sometimes called as the Gauss’s divergence theorem, says that Z V
∇ · F~ dV =
I
F~ · n ˆ dS .
(1)
S
The Gauss’s law states that the volume integral of the divergence of a vector field over a volume V is equal to the closed surface integral of the vector over the surface bounding the volume V .
24
Mathematically, the Gauss’s divergence theorem converts a volume integral of the divergence of a vector to a closed surface integral of the vector, and vice versa. Physically, the Gauss’s divergence theorem says that the number of the field lines flowing through the surface S is equal to the ”strength” of the field source contained inside the volume V . ˆ is the unit outward Figure 5: Illustration of the eval- Remember that n uation of the Gauss’s law for a vec- normal over S, as shown in Figure 5, tor field F~ crossing a closed sur- and S is a closed surface bounding a volface bounding a volume V . ume V . Associated with this is the pictorial representation of fields by lines of force, the direction of F~ given by the tangent to a line and the strength of F~ given by the line density per unit area. Note that from the divergence theorem Ψ 1 I ~ ~ F ·n ˆ dS = lim , ∇ · F = lim dV →0 dV dV →0 dV i.e. the divergence of a vector field is the emanating flux per unit volume.
2.5
The Continuity Equation for Electric Current
The Gauss’s divergence theorem allows us to derive the continuity equation for the electric current known as the equation of conservation of the charge. Suppose we have some charge of density ρ in a volume V enclosed by a surface S, as shown in Figure 6. Let ~v is a macroscopic velocity of the charge. Then, the rate of decrease of the total charge in the volume V is equal to the rate of transport of the charge out through the surface S. Mathematically, we can express this statement as I ∂ Z ρ dV = ρ~v · n ˆ dS . − ∂t V
S
25
Using the Gauss’ divergence theorem, we can write I
ρ~v · n ˆ dS =
S
Z
∇ · (ρ~v ) dV .
V
Thus Z V
Figure 6: Illustration of a flow of a charge through a surface S closing a volume V .
Z ∂ρ dV = − ∇ · (ρ~v ) dV . ∂t V
However, this relation holds for arbitrary V , so we can drop the integrals and obtain ∂ρ = −∇ · (ρ~v ) . ∂t
Since ρ~v = J~ , we finally obtain ∂ρ + ∇ · J~ = 0 , ∂t which is well known as the continuity equation, or the equation of conservation of charge. When stationary currents are involved, then ∂ρ/∂t = 0. In this case ∇·J~ = 0, that is for stationary currents the current density is solenoidal.
2.6
Curl (Rotation) Function
Another very important property of vector fields is curl or rotation or circulation. Curl is the operation of ∇ operator upon a vector in a fashion analogous to the cross product of two vectors. The result is a vector that in Cartesian coordinates is written as8 ∂ F~ ˆ ∂ F~ ˆ ∂ F~ +j× +k× , curl F~ ≡ ∇ × F~ = ˆi × ∂x ∂y ∂z 8
Expressions for the curl in other coordinate systems are given in the Appendix C.
26
or !
!
!
∂Fz ∂Fy ˆ ∂Fx ∂Fz ˆ ∂Fy ∂Fx ˆ − − − k ∇ × F~ = i+ j+ ∂y ∂z ∂z ∂x ∂x ∂y ˆi ˆj kˆ ∂ ∂ ∂ = ∂x . ∂y ∂z F x Fy Fz Properties: Curl is nonzero when the field increases (or decreases) in a different direction that the field pointed. If the field is pointed in the same direction as that in which is increased, the curl is zero. So the curl is related to how the field changes as you move across the field. When ∇ × F~ = 0 everywhere, the field F~ is called irrotational. Remember: Mathematically, curl of an arbitrary vector is a vector.
Exercise in class: Find which of the following vectorial fields is irrotational. rˆ F~1 = 2 , r
F~2 = 2~r ,
F~3 = ∇r3 ,
where r = |~r| 6= 0 and rˆ is the unit vector in the direction of ~r.
2.7
Stokes’s Theorem
The curl function allows us to introduce an another important identity in the theory of vectorial fields, the Stokes’s theorem that is formulated as I l
F~ · d~l =
Z
∇ × F~ · n ˆ dS ,
S
~ is the infinitesimal vector length tangent to a closed path of length l where dl bounding a surface area S.
27
The Stokes’s theorem states that the closed line integral of a vector field F~ along the contour bounding an open surface S is equal to the surface integral of the curl of the vector field over the surface.
Figure 7: Illustration of the evaluation of the Stokes’s Theorem.
We may write the Stokes’s theorem in the form ∇ × F~ · n ˆ = lim
dS→0
1 I ~ ~ F · dl . dS
This gives an intuitive meaning to any component of ∇ × F~ in terms of the line integral around a small element of surface, as shown in Figure 7. Simply, ∇ × F~ is a measure of the vorticity of the field.
2.8
Successive Application of ∇
We can introduce scalar and vector products in which the operator ∇ appears more than once. For example, since the gradient of an arbitrary scalar function Φ is a vector, we can take the divergence of the gradient ∂ 2Φ ∂ 2Φ ∂ 2Φ ∂Φˆ ∂Φ ˆ ∂Φ ˆ i+ j+ k = ∇ · ∇Φ = ∇ · + + 2 . ∂x ∂y ∂z ∂x2 ∂y 2 ∂z !
The same result is obtained if we take ∇ · ∇ as a new operator ∇2 with properties ∇2 ≡ ∇ · ∇ =
∂2 ∂2 ∂2 + + . ∂x2 ∂y 2 ∂z 2
The operator ∇2 is called the Laplacian and is a scalar. The Laplacian may also be applied to a vector, with the result ∇2 F~ =
∂ 2 F~ ∂ 2 F~ ∂ 2 F~ + + . ∂x2 ∂y 2 ∂z 2 28
It is also possible to form the curl of the gradient, which is identically zero ∇ × ∇Φ = 0 .
(2)
The divergence of the curl of a vector is also identically zero9 ∇ · ∇ × F~ = 0 .
(3)
These two properties are very useful in the vector field theory, in particular in the electromagnetic theory. The relation (2) shows that an irrotational field can always be expressed as gradient of an arbitrary scalar field. Similarly, relation (3) shows that any solenoidal field can always be expressed as a curl of an arbitrary vector field. Further discussion of the successful application of ∇ is left to the tutorial problems.
9
The proof of this and the above property is left to the student as a tutorial problem. Note that each of these properties involves two vector operations, but in writing them we usually do not put brackets that would indicate in which order the two operations should be applied. Is it clear to the student in which order the vector operations should be applied?
29
Revision questions Question 1. What is the physical meaning of the gradient of a scalar function? Question 2. Derive the explicit form of ∇ in the Cartesian coordinates. Question 3. What is the physical meaning of the divergence of a vector? Question 4. What is the physical meaning of the curl of a vector? Question 5. Write the curl of an arbitrary vector in the Cartesian coordinates. Question 6. Explain, without using any mathematics, why ∇ · ∇ × F~ = 0.
Tutorial problems Problem 2.1 Direction of a vector ~ = −ˆi + 2ˆj − 2kˆ in Cartesian coordinates, find Given a vector A ~ the expression for the unit vector Aˆ in the direction of A.
Problem 2.2 Relation between two vectors ~·B ~ = A ~·C ~ and A ~×B ~ = A ~ × C, ~ where A ~ is Show that, if A ~ ~ ~ ~ not a null vector (A 6= 0), then B = C.
Problem 2.3 Multiple applications of · and × ~ B, ~ C ~ and D. ~ Consider arbitrary vectors A, ~ × (B ~ × C) ~ equal to (C ~ × B) ~ × A? ~ Explain. (a) Is A
30
~ · (B ~ × C) ~ equal to (C ~ × B) ~ · A? ~ Explain. (b) Is A ~×B ~ =A ~×C ~ implies B ~ = C? ~ Explain. (c) Does A ~ =∇×B ~ implies A ~ = B? ~ Explain. (d) Does ∇ × A (e) Which of the following expressions do not make sense? Explain. ~ × B/| ~ B|, ~ A ~×B ~ × C, ~ A
~ · D/( ~ A ~ × B), ~ ~ × B/( ~ C ~ · D), ~ C A ~ ~ · B), ~ ∇ · (∇A), ∇2 (B ~ × B), ~ ~ · B) ~ . ∇2 (A ∇ × (A
(f ) Which of the following expressions are vectors and which are scalars? ~ ~ · ∇)B, ~ ~ · B), ~ ~ × B), ~ ∇(∇ · A), (A ∇(A ∇2 (A ~ · (B ~ × C), ~ ~ × ∇[B ~ · (C ~ × D)], ~ ~ · B), ~ ∇A A ∇2 (A ~ × B). ~ ∇ × (A Problem 2.4 Vector components and directions ~ of the point P (2, −2, 3) (a) Find the relative position vector R ~ with respect to Q(−3, 1, 4). What are the direction angles of R? ~ (The direction angles are the angles between R and the x, y and z axes repectively). ˆ ~ = ˆi + 2ˆj + 3kˆ and B ~ = 4ˆi − 5ˆj + 6k, (b) Given the two vectors A ~ in the find the angle between them. Find the component of A ~ direction of B. ~ = 2ˆi + 3ˆj − 4kˆ and B ~ = −6ˆi − 4ˆj + kˆ find (c) Given the vectors A ˆ ~ ~ ~ = ˆi − ˆj + k. the component of A × B along the direction of C
31
Problem 2.5 Flux of a vector field ~ = (a) Given the vector field A ˆ ˆ ˆ xy i + yz j + zxk, evaluate directly ~ from the definition the flux of A through the surface of a rectangular parallelepiped of sides a, b, c with the origin at one corner and edges along the positive directions of the rectangular axes, see Figure 8. ~ (b) Also evaluate ∇ · AdV directly and show that it is equal to the calFigure 8: Contour for evaluation of the flux culated flux as predicted by Gauss’s of a vector field. Theorem. R
Problem 2.6 Component in the direction of a gradient (a) A family of hyperbolas in the xy plane is given by u = xy. Find ∇ u. ˆ find the component of A ~ = 3ˆi + 2ˆj + 4k, ~ (b) Given the vector A in the direction of ∇ u at the point on the curve for which u = 3 and x = 2. Problem 2.7 Unit normal to ellipsoids The equation giving a family of ellipsoids is: u=
x2 y 2 z 2 + 2 + 2 . a2 b c
Find the unit vector normal to each point of the surface of these ellipsoids. Problem 2.8 Divergence of a radial field For fields of the form rn rˆ, (r 6= 0), find for which values of n the divergence is zero.
32
Problem 2.9 Field of a cylindrical form ˆ (ρ 6= 0), find for which For fields of the cylindrical form ρn φ, values of n the curl is zero. Problem 2.10 Line integral of a vector The vector field ~ = x2 yˆi + xy 2 ˆj + a3 e−βy cos α xkˆ , A where a, α, β are constants. ~ Evaluate directly the line integral of A around the closed path in the xy plane as shown in the Figure 9. The straight portions are parallel to the axes and the curved portion is the parabola y 2 = kx, where k = constant. Eval~ over uate the surface integral of ∇ × A the area S enclosed by C. Show that the two are related as expected by Stokes’ Theorem.
Figure 9: Contour for evaluation of the curl of a vector.
33
3
Vectorial Equations in Electromagnetism: Scalar and Vector Potentials
In this lecture, we will illustrate the application of the Gauss’s and Stockes’s theorems to vectorial equations involving ∇· and ∇× operations. These type of operations are involved in the Maxwell’s equations, the basic equations for the electromagnetic theory. We shall discuss only the vectorial nature of the equations, deferring a detailed physical interpretation of the equations to the next chapter when we consider the experimental basis for electromagnetism. Furthermore, we illustrate the Helmholtz Theorem for the unique determination of a given vector and discuss some properties of the successive application of ∇, which will allow us to introduce the concept of vector and scalar potentials to the electromagnetic field theory.
3.1
Maxwell’s Equations, an Example of Vectorial Equations
Consider the Maxwell’s equations in the differential form that are an example of differential vectorial equations involving ∇· and ∇× operations. These ~ and B ~ and are of the form differential equations are for the vector fields E ρ , ε0 ~ = 0, ∇·B ~ ~ = − ∂B , ∇×E ∂t ~ = ∇·E
~ ~ = µ0 J~ + 1 ∂ E , ∇×B c2 ∂t
(4) (5) (6) (7)
~ and B, ~ These four equations determine ∇· and ∇× of the two vectors E and we need both operations for each of the EM field vectors to completely ~ and B. ~ determine E An obvious question arises: Why do we need both ∇· and ∇× equations for ~ and B? ~ each of the vectors to completely determine E
34
The answer is in the Helmholtz Theorem, which says that an arbitrary vector F~ can always be written as Z 1 ∇ × F~ 1 Z ∇ · F~ dV + ∇× dV F~ = − ∇ 4π r 4π r V V
= F~l + F~t . In other words, if the divergence and curl of a vector field are known everywhere in a finite region, then the vector field can be found uniquely. Thus, specification of ∇ · F~ and ∇ × F~ is necessary and sufficient to determine F~ . ~ and B. ~ Hence, we need four equations of ∇· and ∇× type to determine E ~ Since F can be found from the knowledge of ∇· and ∇×, the divergence and curl are often called the sources of the field.
3.2
Scalar Potential
The electromagnetic field equations, the Maxwell’s equations, are coupled differential equations whose the solution is not in general simple and straightforward. Even the calculation of electric field produced by a macroscopic charge from the Coulombs law is usually a difficult task. Often the solutions are aided by the use of potentials. What do we mean by a potential and how helpful it is in the calculation of the fields? The idea is that the expression of the fields in terms of potentials simplifies the equations to a form that can be readily solved. In particular, the Maxwell’s equations are simplified to a form that can be solved in a reasonably simple way. A potential is a quantity from which a vector field can be derived by some process of differentiation. To show this, consider a field with zero rotation or curl, ∇ × F~ = 0. Since ∇ × ∇Φ ≡ 0, then the equation ∇ × F~ = 0 will have an integral of the form: F~ = ∇Φ .
35
Thus, the field with zero curl may be derived from the gradient of the scalar function Φ. In the field theory, the function Φ is called a scalar potential. The scalar potential function is very often used in the electromagnetic field ~ = theory. For example, the electrostatic (Coulomb) field has zero curl, ∇× E 10 0. Hence, we can always write the electrostatic field as ~ = −∇Φ . E The minus sign is inserted in the definition of Φ to agree with the definition of Φ as a potential ENERGY. The negative sign can also be understood phys~ is in the direction that a positive charge moves, ically from the fact that E hence in the direction of decreasing potential. The scalar potential is useful in the calculations of work done on charges moving in an electric field. The work done by the field in moving the unit charge q from a point A to a point B ~ as illustrated in Figin an electric field E, ure 10, is given by W q
= −
A
Figure 10: Illustration of a path =
along which a unit charge is moved from a point A to a point B in an ~ electric field E.
ZB
ZB A
~ · d~l = E
ZB
∇Φ · d~l
A B
Z ∂Φ ~ rˆ · dl = dΦ ∂r A
= ΦB − ΦA = ∆ΦAB . We see that the work done on moving the charge from the point A to the point B is equal to the potential difference between the two points.
The work is independent of the path, it depends only on the position of the start and end points. If the charge gains energy in moving from A to B, we ~ 6= 0 in general in electromagnetism, As one can see from the Maxwell’s equations, ∇×E ~ so it is not in general possible to write E = −∇Φ. 10
36
say that the point B is at higher potential than the point A, and vice versa. ~ = 0. Note an another interesting property of the field with ∇ × E From the Stokes’s law we have that I
~ · d~l = E
Z
~ ·n ∇×E ˆ dS = 0 .
It says that no work is done by the field on a test charge if it is moved along an arbitrary closed path in the field. Consequently, the work along two arbitrary paths from A to B, as illustrated in Figure 11, is Figure 11: Two arbitrary paths leading from a point A to a point B.
∆Φ(AB)1 = −∆Φ(BA)2 = ∆Φ(AB)2 .
In summary, the work done by the field is independent of the path chosen. In other words, if a charge is moved around any closed path, no net energy is requred. Thus, a field F~ with ∇ × F~ = 0 is a conservative field of force.
3.3
Vector Potential
An another useful potential in the electromagnetic theory is the vector potential. To illustrate the idea of the vector potential and how it is defined, consider a field with zero divergence, ∇ · F~ = 0. Since for an arbitrary ~ vector A: ~≡0, ∇·∇×A we have that the vector F~ can always be written as ~. F~ = ∇ × A Thus, the field with zero divergence may be derived from the curl of the ~ In the field theory, the vector A ~ is called a vector potential. vector field A. ~ = 0 always in electromagnetism so there will always We shall see that ∇ · B ~ such that B ~ = ∇×A ~ and such an A ~ is referred to as be a vector field A 37
the vector potential in electromagnetism (though there may be other electromagnetic field functions with zero divergence). ~=B ~ does not completely specify A ~ even Note also that just writing ∇ × A ~ is known everywhere. According to the Helmholtz Theorem, one needs if B ~ as well to completely determine A. ~ Equivalently, we can say to specify ∇ · A ~=B ~ still leaves us free to define ∇ · A. ~ that defining ∇ × A
Exercise in class: Applications of ∇· and ∇×. ~ B, ~ C, ~ in(a) For a given scalar function Φ and vectors A, dicate successive steps you would follow in the calculation of the following expressions: ~; ∇×∇·A ~×B ~ ·C ~ ; ∇A
~ · ∇B ~ ; A
~ ·B ~ ; A∇
~×B ~ ; ∇Φ · A
~·B ~ ×C ~ ; ∇×A
~ · ∇Φ ; A
~×∇·B ~ ; A
~·B ~ × ∇Φ ; A ~ × ∇B ~ ·C ~ . ∇·A
(b) Which of the expressions in part (a) are vectors?
38
Revision questions Question 1. Explain why in electromagnetism we need two divergence and two ~ and magnetic B ~ fields to comcurl equations for the electric E ~ and B? ~ pletely determine the fields E Question 2. What is the Helmholtz’s theorem telling us about a vector F~ ? Question 3. Why are the scalar and vector potentials useful in electromagnetism? Question 4. Explain, which vectors can be expressed in terms of a scalar potential and which can be expresses in terms of a vector potential. Question 5. Explain, when a field F~ is a conservative field of force?
39
Tutorial problems Problem 3.1 If the potential Φ satisfies the Laplace equation, ∇2 Φ = 0, show that the vector ∇Φ is both solenoidal and irrotational. Problem 3.2 Show from the form of Maxwell’s equations that it should be ~ and Φ such that possible to define electromagnetic potentials A the fields can be calculated from ~ ~ = −∇Φ − ∂ A , E ∂t ~ ~ B = ∇×A .
Problem 3.3 In an application of the electromagnetic theory to nuclear physics, ~ B, ~ A ~ the so-called vectorial mesons are described by three vectorial fields E, and a real scalar field Φ. The fields satisfy the following equations ~ = −µ2 Φ , ∇·E ~ ~ = ∂ E − µ2 A ~, ∇×B ∂t ~ ~ = − ∂ A − ∇Φ , E ∂t where µ2 is a positive constant. Show that the above equations can be converted into two differential equations (i)
∂Φ =0, ∂t ∂2Φ ∇ 2 Φ − µ2 Φ − 2 = 0 . ∂t
~+ ∇·A
(ii)
40
4
The Experimental Basis of the Development of Electromagnetic Theory
In this lecture, we will analyse experiments that led to the development of the electromagnetic theory. We will show that the Coulomb Law for the electrostatic force between two point charges is the experimental basis for the development of the electric theory and the Biot-Savart Law for the magnetic force between two static currents is the experimental basis for the development of the magnetic theory. We also illustrate two useful laws of electromagnetism, the Gauss’s and Amperes Circuit laws, that are very helpful in calculations of the electric and magnetic fields produced by some symmetric distributions of charges and currents.
4.1
Coulomb’s Law – Force between Static Charges
In 1785, Coulomb investigated the nature of the force between charged bodies, and the results of his experiments can be formulated mathematically in what is known as Coulomb’s law F~2 =
1 q q0 rˆ . 4πε0 r2
We can write Coulomb’s law in terms of the electric field produced by the charge q that acts on the charge q 0 : ~ , F~q0 = q 0 E where ~ = E
1 q rˆ 4πε0 r2
is the electric field produced by the charge q. The electric field is an example of a vector field. In principle, we can always calculate an electric field using Coulomb’s law. However, there is an alternative way we can find the electric field. In particular, the field may be represented by means of the flux con~ from a point charge q may be readily calculated by cept. The total flux of E ~ · dS ~ over a surface enclosing q. integrating E
41
4.2
Gauss’s Law
Consider a macroscopic charge q closed by a surface S, as shown in Figure 12. We will show that the flux of the electric field produced by the charge q is proportional to the charge q, and is independent of the shape of the surface closing the charge. Consider first the flux through a spherical surface, for which we find after a simple algebra, that the flux is I I q ~ rˆ · n ˆ dS ΨS1 = E·n ˆ dS = 4πε0 r02 S1
S1 q q q I dS = 4πr02 = . = 2 2 4πε0 r0 4πε0 r0 ε0 S1
Thus, we see that the flux through the spherical surface is proportional to the charge enclosed by the surface. An obviFigure 12: Illustration of the ous question arises: Is this relation valid calculation of the Gauss’s law. only for the case of spherical surfaces or maybe it is valid for an arbitrary surface? In order to answer this question, consider the flux through an arbitrary surface I I q rˆ · n ˆ dS , ΨS2 = dΨS2 = 4πε0 r2 S2
S2
where S2 is an arbitrary surface enclosing the charge q. However, from the inverse square law and some geometry, we find dΨS2 =
q q dS cos θ q rˆ · n ˆ dS = = dΩ , 2 2 4πε0 r 4πε0 r 4πε0
where dΩ is the solid angle subtended by dS at q. We see from Figure 12 that the element of solid angle dΩ is independent of where we cut the bundle of electric lines of force. Thus I I q dΨS2 = dΨS1 and then dΨS2 = dΨS1 = . ε0 42
Hence ΨS2 =
I
dΨS2 =
q . ε0
(8)
Equation (8) is the statement of the Gauss’s Law. It says that the flux ~ through a closed surface equals 1/ε0 times the total of the elcetric field E charge contained within the surface. Furthermore, the Gauss’s Law applies not only to a single charge contained within a surface S, but also for some arbitrary number of point charges q1 , q2 , q3 , · · · , qn or a continuous distribution of charges within the surface S. We can write this statement in the following way Figure 13:
Illustration of the Gauss’s law for a number of point charges enclosed by a surface S.
I
dΨ =
I
~1 + E ~2 + E ~ 3 + · · ·) · n (E ˆ dS
= Ψ 1 + Ψ2 + Ψ3 + · · · 1 X q1 q2 q 3 + + + ··· = q, = ε0 ε0 ε0 ε0 P
where q is the algebraic sum of all charges within the surface S. For a contiuous distribution of charges inside the surface S X
q=
Z
ρdV ,
V
where ρ is the charge density in the volume V enclosed by the surface S. This general property is what one could expect, the Gauss’s law applies to an arbitrary number of charges due to the additive nature of the fields produced by each charge separately. Gauss’s Law for a source field charge outside S
43
If the source charge q is outside S, as illustrated in Figure 14, the surface integral vanishes since the total solid angle subtended at q by the surface is zero. Proof: Consider first the flux of the electric field through a surface element dS1 seen from the charge q under a solid angle dΩ: dΨ1 =
ˆ 1 dS1 q rˆ · n . 4πε0 r12
However dS1 rˆ · n ˆ 1 dS1 = − 2 = −dΩ , 2 r1 r1 where minus sign is from the fact that at the surface dS1 the angle between rˆ and n ˆ 1 is larger than 90◦ . Similarly, the flux of the electric field through a surface element dS2 seen from the position of the charge q under the same solid angle dΩ is dΨ2 =
q rˆ · n ˆ 2 dS2 q = dΩ , 2 4πε0 r2 4πε0
where now we have positive sign since at the surface dS2 the angle between rˆ and n ˆ 2 is smaller than 90◦ . Thus, we find that
Figure 14: Illustration of the calculation of the Gauss’s law for the source charge outside the surface S.
dΨ1 + dΨ2 = 0 , and integrating over all S, we obtain that the total flux through the surface S is Ψ=
I
dΨ = 0 .
S
The physical interpretation of this result is that field lines originating from an external charge and entering the surface S must also leave this surface.
44
In summary, the Gauss’s law says that the total electric flux through a surface S enclosing a charge q is: Ψ=
q , ε0
where q =
X
charges INSIDE S .
Using the definition of the flux, we often write the Gauss’s law as ~ · dS ~= q . E ε0 S
I
The power of the Gauss’s law lies in the fact that we are free to apply it to any closed surface whose shape can be chosen arbitrary such that the evaluation of the surface integral becomes a simple straightforward task. The Gauss’s law is particularly useful in the calculation of the electric field produced by certain symmetrical charge distributions. If the distribution of the charges does not correspond to any simple symmetry, the Gauss’s law is not much helpful in the calculations. We illustrate this in the following example.
Example of an application of the Gauss’ Law An infinitely long line is positively and uniformly charged with a constant linear charge density ρl . Use (a) Coulomb law, (b) Gauss’s law to find the electric field about the line. (a) If we have to calculate the field due to a static macroscopic distribution of charge using the Coulomb law, we divide the macroscopic charge into ~ The field infinitesimal (point) charges dq which produce an electric field dE. ~ of the point charge dq is given by the Coulomb field dE ~ = dE
1 dq rˆ . 4πε0 r2
Then the total electric field is found by vector addition ~ = E
Z
~ . dE
Since we are adding vectors, a caution must be employed. We use the following procedure, which is general and can be employed to any system: 45
~ produced by the infinites1. Write the expression for the electric field dE imal (point) charge dq. ~ into components 2. Choose a reference frame and resolve the vector dE dEx , dEy , and dEz . 3. Calculate each component of E separately by integration, e.g. calculate R Ex = dEx . ~ from its components 4. Find the resultant E ~ = Exˆi + Ey ˆj + Ez kˆ . E Return now to our example of the charged infinitely long line and lets try to solve the example problem following the above procedure. Take a small element dl of the line containing a point charge dq. Electric field produced by the point charge at A distance r from dl is given by the Coulomb field ~ = dE
1 dq rˆ . 4πε0 r2
We see from the figure that r=
h , sin θ
l = h cot θ .
Since the density of the charges on the line is Figure 15: Illustration of an ap- constant, the charge on the line element dl is plication of the Coulomb law to the calculation of the electric field of an infinitelt long line charge.
dq = ρl dl ,
where dl = −
h dθ . sin2 θ
Then ρl h sin2 θ ˆi + sin θˆj − 2 dθ cos θ 4πε0 h2 sin θ ρl = − cos θˆi + sin θˆj dθ , 4πε0 h !
~ = dE
where we have decomposed the unit vector rˆ into two (x, y) components rˆ = cos θˆi + sin θˆj . 46
Integrating over θ from θ = π to θ = 0, as the line is infinite and we go in the direction of increasing l (from x = ∞, θ = π to x = −∞, θ = 0), we obtain i ρl h (− sin 0 + sin π) ˆi + (cos 0 − cos π) ˆj 4πε0 h ρl ˆ = j. 2πε0 h
~ = E
Thus, the electric field produced by the charged line depends inversely on the distance from the line and points in the direction perpendicular to the line. (b) Let us now calculate the field by the direct application of the Gauss’s law. The electric field near the uniformly charged line must be radially directed because of the symmetry of the problem. The field must have cylindrical symmetry because the problem is unchanged by rotating the line about its axis. The field must also be independent of position along the line because the distance to either end is infinite. This is the ideal situation for the application of Gauss’s Law. Due to the cylindrical symmetry of the field, we can apply a cylinder surface of radius h and length L centered about the line of charge, as shown in Figure 16. Such a surface is often referred to as a Gaussian surface. Figure 16: Application of the Gauss’s
According to the Gauss’s law ~ · dS ~= q , E ε0 S
law to an infinitely long line charge.
I
where q = ρl L is the charge on the line closed by the cylinder surface. The flux through the cylinder surface splits into three fluxes I S
~ = ~ · dS E
Z A
~ A+ ~ · dS E
Z B
~ B+ ~ · dS E
47
Z C
~ C . ~ · dS E
~ A, E ~ C, E ~ B , and the magnitude of E ~ ⊥ dS ~ ⊥ dS ~ k dS ~ is constant Since E along the surface B, the flux through the cylinder reduces to that over the surface B only Z S
~ = ~ · dS E
Z B
EdSB = 2πhLE ,
and since the cylinder is symmetrically positioned about the line of charge, the magnitude of E is constant over the surface B. Then, according to the Gauss’s law 2πhLE =
ρl L , ε0
which gives E=
ρl . 2πε0 h
Note how simple are the calculations of the electric field using the Gauss’s law. However, we were able to solve this problem because we knew the direction of the field at any point around the line.
48
4.3
Biot-Savart Law – Force between Static Currents
There is a force not only between electric charges but also between electric currents. This force has a different nature than that one due to electric charges, it is due to magnetic fields produced by the currents. It has been first noticed by Oersted in 1819, and few years later, in 1827 by Ampere who showed that quantitatively the magnetic forces in macroscopic circuits can be accounted for by what has come to be known as the Biot-Savart Law. To illustrate the presence of magnetic forces between electric currents, consider an experiment involving two Figure 17: Example of an experiment long parallel wires carrying currents I 1 illustrating the existance of the magnetic and I and separated by a distance d, 2 force between two current carrying wires. as shown in the Figure 17. Experimental observations: • If I1 kI2 then the force F is attractive. • If I1 antik I2 then F is repulsive. • If one of the wires is rotated through 90◦ then F = 0. • The force is proportional to the currents I1 and I2 . All the above observations can be conbined into a single equation for the force acting on the current I2 : dF~2 =
µ0 I2 d~`2 × (I1 d~`1 × rˆ) , 4π r2
where µ0 = 4π × 10−7 [H/m] in SI units, is the permeability of the vacuum.
49
We can write the force as ~ , dF~2 = I2 d~`2 × dB where ~ = dB
µ0 I1 d~`1 × rˆ , 4π r2
which is known as the Biot-Savart law for magnetic field produced by the current element I1 d~`1 . The Biot-Savart law allows to compute magnetic field produced by an arbitrary current distribution Id~` µ0 I Z d~` × rˆ ~ . B= 4π l r2
(9)
The method requires integration of small current elements. Note, all three quantities appearing under the integral change during the integration, which complicates the evaluation of the integral. ~ by using the following procedure. We can simplify the calculations of B If we replace rˆ/r2 by −∇(1/r), the integrand becomes −d~` × ∇(1/r) . Next, using a vector identity we find that
1 1 d~` 1 ~ ~ ~ = ∇ × d` − d` × ∇ = −d` × ∇ , ∇× r r r r
since ∇ × d~` = 0. Hence, we can write µ0 I Z d~` ~ B =∇× . 4π l r Thus, the magnetic field can be expressed as ~ =∇×A ~. B 50
(10)
~ = 0 always. Thus, in this case we can first calculate A: ~ We see that ∇ · B µ0 I Z d~` ~ , A= 4π l r
(11)
~ which involves only two variables d~` and r, and then using (10), we find B. ~ is easier to calculate than the original expression (9) for B. ~ The integral for A Since the curl operation is readily performed, we may use (11) as an inter~ in a simpler way. mediate step for finding B ~ is called a vector potential, As we have already mentioned, the vector A ~ in electromagand will see later in the course many useful applications of A netic theory.
4.4
Current Element and Charge Element
The Biot-Savart law shows that the magnetic field is produced by a current element. Here, we will show that in fact the magnetic field is produced by moving charges. A charge dq moving with a velocity v is equivalent to an element of current Id`: dq = Idt = I
d` , v
where dt is the time for all the charge in d` to pass out of the volume, as illustrated in Figure 18. Figure 18: Current element of a charge Hence moving with a velocity ~v .
dq~v = Id~` . The force between the two current elements can then be written as dF~2 =
µ0 (dq1~v1 × rˆ) dq2~v2 × , 4π r2 51
or ~1 , dF~2 = dq2 ~v2 × dB with ~1 = dB
~v1 × rˆ µ0 dq1 2 . 4π r
This shows that magnetic field is produced by a moving electric charge. Moreover, it shows that the magnetic field can act with a force only on moving charges. No force of a magnetic field on stationary charges.
4.5
The Lorentz Force
When a charge is moving in electric and magnetic fields, the force acting on the charge is no longer the Coulomb force. It is the Lorentz force that is ~ and the Biot-Savart Law F~M = obtained putting Coulomb’s Law F~E = q E ~ together: q~v × B ~ + ~v × B) ~ . F~EM = F~E + F~M = q(E Thus, a motion of electric charges is modified by both the electric and mag~ if it netic forces. If the charge is stationary, the force depends only on E, moves, there is an additional force proportional to ~v .
52
4.6
Amperes Circuit Law
Amperes circuit law is a useful relation between currents and magnetic fields. This law allows us to calculate magnetic field produced by certain current distributions in a very effective way. The Amperes law says that for an arbitrary closed path around a current carrying conductor, as shown in Figure 19, the component of magnetic field tangent to the path is proportional to the net current passing through the surface bounded by the path I
Figure 19: Contour for evaluation of the Amperes circuit law.
~ · d~` = µ0 I , B
where the integration is over a closed loop and I is the total current through the loop.11
The Amperes law can be applied in highly symmetric situations to find the magnetic field more easily than by computing with the Biot-Savart law. In either case, the result is the same. In case that lack the proper symmetry, the Amperes law is not easily applied. The following example illustrates the use of Ampere’s law. Worked Example: An application of the Amperes Law An infinitely long wire caries a constant current I. Use (a) Biot-Savart law, (b) Amperes law to find the magnetic field about the wire. (a) First, we divide the current into the small current el~ due to an ements Id~`. The element of magnetic field dB ~ element I d` at a point A distance ~r = rˆ r is found from the Biot-Savart formula: ~ = µ0 I d~` × rˆ , dB 4π r2 11
The prove of the Amperes law is complicated and will not be presented at the lecture. The student, if interested in details of the prove is referred to the Appendix A.
53
~ are in the same φˆ direction normal where we note that all dB to the direction of the current. So we see from this symmetry that the field lines are circles concentric with the current. Furthermore, along any such circular path the field is constant in magnitude. Let us calculate the magnitude of the magnetic field. Since d~` × rˆ = dl sin θφˆ ,
r=
h , sin θ
l = h cot θ ,
we have dl = −
h dθ sin2 θ
and then 2 µ0 I ~ = µ0 I sin θ − h dB sin θdθ φˆ = − sin θdθφˆ . 2 4π h2 4πh sin θ
Integrating the above equation over the length of the wire, we obtain 0 µ0 I ~ = − µ0 I B sin θdθφˆ = (cos 0 − cos π) φˆ 4πh π 4πh µ0 I ˆ = φ. 2πh
Z
(b) Let us now calculate the field using the Amperes law. Since the field lines are circles concentric with the current,
54
and along any such circular path the field is constant in magnitude, this is the ideal situation for the application of Amperes law: I
~ · d~` = µ0 I ⇒ 2πhB = µ0 I ⇒ B = µ0 I . B 2πh
Note how much simpler are the calculations of the magnetic field using the Amperes law.
4.7
Faraday’s Law of Electromagnetic Induction
Faraday discovered electromagnetic induction by changing magnetic field. If we consider a closed stationary circuit located in a varying magnetic field, as shown in Figure 20, the induced electromotive force around this circuit is equal to the negative time rate of change of the magnetic flux through the circuit E =−
dΦ , dt
where Φ is the total magnetic flux through the circuit.12 From the definition of the flux Φ=
Z
~ , ~ · dS B
S
and from the fact that the emf force E is equal to the work done per unit charge, we have I E ~ + ~v × B) ~ · d~` . = (E q `
Figure 20: Illustration of a closed circuit located in a magnetic field.
In a stationary circuit ~v = 0 (and any~ ⊥ d~` since ~v k d~`). Thus way ~v × B
12
The electromotive force is induced by varying the flux of the magnetic field that is proportional to the magnetic field and the area of the loop. Thus, the electromotive force can also be induced in a constant magnetic field by varying the area S of the loop.
55
~ · d~` = 0. ~v × B Hence E=
I
~ · d~` E
`
and finally I `
~ · d~` = − ∂ E ∂t
Z
~ · dS ~ , B
S
or I `
~ · d~` = − E
Z S
~ ∂B ~ . · dS ∂t
This is the Faraday’s law written in the integral form. The student can easily show, using the Stokes’s theorem, that the Faraday’s law can be written as ~ ~ = − ∂B , ∇×E ∂t which is called the differential form of the Faraday’s law. In summary: The Faraday’s law tells us that time-varying magnetic fields give rise to electric fields. This shows that the fields are related to each other, and we then must speak of electromagnetic fields, rather than separate electric and magnetic fields.
56
Revision questions Question 1.
State the Gauss’s Law and explain under what conditions the Gauss’s law is useful in calculating the electric field produced by a macroscopic charge.
Question 2. State the Biot-Savart Law and show that magnetic field is produced by moving charges. Question 3. State the Ampere’s Law and explain under what conditions the Ampere’s law can be successfully applied for the calculation of the magnetic field produced by a macroscopic currents. Question 4. What is the Faraday’s law of electromagnetic induction and derive its differential form.
57
Tutorial problems Problem 4.1 Field of a capacitor (a) An infinite flat sheet has a uniform charge density of σ [Cm−2 ]. Use Gauss’ Theorem plus an argument from symmetry to find the electric field strength everywhere. (b) A capacitor that has plates (area A) of diameter very much greater than the spacing d between them has charge densities +σ and −σ on the plates, respectively. Use the result in (a) to find the total electric field strength due to the charges on the plates, within and outside the capacitor. (c) Calculate the force per unit area on one plate due to its attraction by charges of opposite sign on the other plate.
Problem 4.2 Properties of a radial vector field If F~ is the vector rn rˆ, (r 6= 0) show that: ∇ × F~ ∇ · F~
= 0, = (n + 2)rn−1 .
Problem 4.3 The Poisson equation Note from the above tutorial problem 4.2 that the divergence of the radial field is zero for n = −2. The student may immediately comment that this result means that the Coulomb field of this symmetry has no point source. However, we know that the Coulomb field is produced by a point charge. How then the divergence is zero if there is a point source of the field? To resolve this dilemma, show that for all r including r = 0 ∇·
rˆ = 4πδ(~r) , r2
where δ(~r) is the three-dimensional Dirac delta function.
58
Problem 4.4 Field of a non-uniformly charged sphere A sphere of radius a has a charge density increasing linearly with radius from zero at its centre to ρ0 at r = a. Use the Gauss’s law to find the electrostatic field inside as well as outside the sphere. Use the results to confirm that ~ =0, ~ = ρ and ∇ × E ∇·E ε0 everywhere. Problem 4.5 Field of charged coaxial cylinders Two infinitely long coaxial cylinders have radii a and b with b > a, as shown in Figure 21. The region between them is filled with charge of volume density given in cylindrical coordinates as ρ = A rn ,
Figure 21:
where A and n are constants. The charge density is zero everywhere else. Find the ~ everywhere, i.e. inside, electric field E between and outside the cylinders.
Problem 4.6 Field of a uniformly charged cylinder An infinitely long cylinder has a circular cross section of radius a. It is filled with a charge of constant volume density ρ. Find the electric field of this charge for all points both inside and outside the cylinder.
Problem 4.7 Charge density for a cylindrically symmetric electric field ~ = A certain electric field is given in cylindrical coordinates by E 3 ~ = 0 otherwise. Find the volume E0 (r/a) rˆ for 0 < r < a and E charge density.
59
Problem 4.8 Charge in a uniform magnetic field ~ moving (a) A charge q enters a region of uniform magnetic field B at right angles to the field. Show that the charge will undergo circular motion in the field and find an expression for the frequency of the circular motion. (b) Describe the motion of the charge if it is not moving at right angles to the magnetic field.
Problem 4.9 Magnetic field of an infinite current sheet Consider an infinite plane sheet containing an electric current which is in the same direction everywhere (the current can be considered to close at infinity if we want to justify use of closed circuit theorems). The strength of the current would be described by a current per unit length (with the ‘length’ drawn at right angles to the direction of the current). So the units of ‘surface current density’ would be amps/metre. Note that if the current is entirely confined to a plane the ordinary current density J in amps/metre2 must be considered as infinite. (a) Use an argument from symmetry plus Amp` eres circuital law to find the magnetic field due to an infinite plane with surface current density Js amps/metre. It may be helpful to think of the surface current as the limit of a large number of parallel wires. (b) Using the result of part (a), find the force per unit area between two parallel current sheets containing surface current densities I1 and I2 if the currents in the two sheets are both in the same direction.
60
5
Differential Equations for the EM Field and Maxwell’s Theory
We know now that electromagnetic forces are carried by electromagnetic fields that propagate at speed c ' 3 × 108 ms−1 . Because of the finite propagation speed we are forced to assign energy and momentum to the fields i.e. we must think of them as real physical entities as against mere mathematical conveniences (as is the case for static fields). An electromagnetic system then qualifies as static only if all the charges have been at rest longer than the time taken to traverse the system at speed c. In the 1830’s Michael Faraday carried out experiments to measure a finite electromagnetic propagation speed. He was unsuccessful due to lack of time resolution in his apparatus. Faraday would have had no reason to think that the electromagnetic speed was the same as the speed of light (then known). In the 1860’s, James Clerk Maxwell, seeking to advance Faraday’s ideas about electromagnetic fields, by a brilliant process of intuition worked out how to generalize certain differential equations deduced from static experiments. He produced a set of field equations known by his name today. Maxwell also had a theory i.e. a set of qualitative ideas underpinning his equations. The theory, unlike the equations, has not stood the test of time.
5.1
Differential Equations for the EM Field
Let us take as the source of the electromagnetic field a continuous macroscopic charge and current distribution represented in terms of the macroscopic source quantities, charge and current densities q=
Z
ρ dV ,
I=
Z
J~ · ~n dS .
Using the Gauss’s and Stokes’s Theorems, we will find differential vector ~ and B ~ from the integral forms of observational equations for the fields E results discussed in the previous chapter.
61
5.1.1
~ Divergence of E
Consider the Coulomb’s Law that can be written in the form of the Gauss’s law as Z I ρ ~ · dS ~= q = dV . E ε0 V ε0 S Applying the Gauss’s Theorem, Eq. (1), to the left-hand side of the above equation, we obtain Z
~ dV = ∇·E
Z
V
V
ρ dV . ε0
Since this relation must hold for arbitrary V , no matter how small, we can drop the integrals, and obtain ~ = ρ . ∇·E ε0 This equation is called the differential form of the Coulomb’s (Gauss’s) Law. From this equation it follows that the divergence of the electric field is zero in all regions where there is no electric charge.
5.1.2
~ Curl of E
Consider the Faraday’s flux cutting rule I `
~ · d~` = − E
Z S
~ ∂B ·n ˆ dS . ∂t
Applying Stokes’s Theorem to the left-hand side of the above equation, we get Z
~ ·n ∇×E ˆ dS = −
S
Z S
~ ∂B ·n ˆ dS . ∂t
Since this relation must hold for arbitrary S and n ˆ , it implies that ~ ~ = − ∂B . ∇×E ∂t This equation is called the differential form of the Faraday’s law.
62
5.1.3
~ Divergence of B
~ from the Biot-Savart law We calculate the divergence of B ~ = µ0 I d~` × rˆ . B 4π r2 By taking ∇· of both sides of the Biot-Savart law and applying the vector identity ~ × B) ~ =B ~ ·∇×A ~−A ~·∇×B ~ , ∇ · (A we obtain ~ = µ0 I ∇·B 4π
rˆ rˆ · ∇ × d~` − d~` · ∇ × 2 2 r r
!
.
(12)
However, ∇ × d~` = 0, since d~` is a constant vector with respect to the differentiation over x, y and z, and ∇×
rˆ ≡ 0 for any n . rn
Hence, the right-hand side of Eq. (12) is always zero, so that always ~ =0. ∇·B This equation, sometimes called Gauss’s law for magnetism, states that the ~ is zero through any closed surface. flux of the magnetic field B 5.1.4
~ Curl of B
~ from the Ampere’s circuit law We find curl of B I `
~ · d~` = µ0 I = µ0 B
Z
J~ · n ˆ dS .
S
Simply, by applying Stokes’s Theorem to the Ampere’s law, we find that the following relation Z S
~ ·n ∇×B ˆ dS = µ0
63
Z S
J~ · n ˆ dS
holds for arbitrary S. Thus ~ = µ0 J~ . ∇×B
(13)
This equation is called the differential form of the Ampere’s circuit law. However, Maxwell realized that unlike the previous three differential equations, this one, Eq. (13), could not be generally true. To see this, take its divergence and remember that ∇ · ∇ × F~ ≡ 0 for any vector function F~ , so that ~ = 0 = µ0 ∇ · J~ . ∇·∇×B Thus ∇ · J~ ≡ 0 , which means that the Ampere’s law does not include source currents. This result is in contradiction with the continuity equation of J~ varying with time. We have already seen that conservation of electric charge requires ∇ · J~ = −
∂ρ . ∂t
Thus ∇ · J~ ≡ 0 implies that ∂ρ ≡0 ∂t
i.e. ρ = const.
Hence, the Ampere law may be applied only for constant currents, i.e. accepting the Ampere law as general, we could never charge or discharge a capacitor. No wonder Maxwell was confused!
5.2
The Maxwell’s Theory
~ as follows. Since Maxwell guessed the right form of ∇ × B ∂ρ ∇ · J~ + =0 ∂t 64
and using for ρ the first Maxwell equation ~ , ρ = ε0 ∇ · E we have
~ ∂ε0 ∇ · E =0, ∇ · J~ + ∂t
~ ∂E =0. ∇ · J~ + ε0 ∂t
or
If then we, after Maxwell, write:
~ ~ = µ0 J~ + ε0 ∂ E , ∇×B ∂t instead of ~ = µ0 J~ , ∇×B we obtain
~ ~ ≡ 0 → ∇ · J~ + ε0 ∂ E = 0 , ∇·∇×B ∂t which is in accordance with conservation and motion of charge. Note from above that the term that Maxwell added ε0
~ ∂E , ∂t
has the dimensions of current density. Maxwell called it the displacement current density. Note that the displacement current density does not explicitly involve charges, they do not appear explicitly in the definition of the displacement current density. Why then we call it as a ”current”? Maxwell had a theory underpinning his equations in which the displacement current was a real physical current - due to ‘polarization of the electromagnetic ether’. This theory has not survived. Nevertheless the above term is still referred to as the ‘displacement current’.
65
With the modification of the current density, we then write the fourth Maxwell’s equation as ~ ~ = µ0 J~ + ε0 µ0 ∂ E , ∇×B ∂t which contains all the physical processes involved in the generation of magnetic fields. In summary: The differential form of the Maxwell’s equations is easier to interpret physically and is also useful in deriving the boundary conditions that the field vectors must satisfy. The Maxwell’s equations are self-consistent and no experimental evidence for requiring any further modifications has been found. The equations are used to explain and predict all physical phenomena that involve charges and/or currents.
Exercise in class: Fields within a capacitor A plane parallel capacitor is being charged with a current I. Show that the displacement current between the plates of the capacitor is equal to the conduction current I in the external charging circuit. Remember that the displacement ~ current density is, by definition, ε0 ∂∂tE so the displacement current through a surface S is: Z
ID =
ε0
~ ∂E ·n ˆ dS . ∂t
S
Assume that the external wires are perfect conductors so ~ is zero in them. Assume the space between the that E plates is a perfect insulator so no conduction current flows within the capacitor.
66
Can you see any curious consequence in this case if the displacement current is assumed to be a real physical current (flow of charges)?
67
Revision questions Question 1. Derive the differential forms for the Coulomb, Ampere and Faraday laws. Question 2. Derive the so-called Gauss’s law for magnetism. Question 3. Explain, how Maxwell resolved the difficulty with general applications of the Ampere’s law. Question 4. What is the displacement current and explain its significance in the EM theory. Question 5. Are the Maxwell’s equations independet of each other? Explain.
Tutorial problems Problem 5.1 Coulomb field Demonstrate that the Coulomb field for stationary point charge ~ = E
1 q rˆ 4πεo r2
follows from Maxwell’s equations. Problem 5.2 Magnetic field of an infinite straight wire Magnetic field at distance r from an infinite straight wire carrying constant current I in the z direction is given by ~ = µo I φˆ , B 2πr ~ in the xy plane. where φˆ is the unit vector in the direction of B Show, using only the circuit, surface and volume integrations, and the Gauss’ and Stokes’ Theorems that the magnetic field satisfies Maxwell’s equations.
68
6
Maxwell’s Equations and Prediction of Electromagnetic Waves
In the previous lecture, we have derived from the experimental laws vectorial ~ expressions for both the divergence and curl of the two basic field vectors E ~ These expressions put together completely determine the electroand B. magnetic fields and are termed the Maxwell’s equations I. II. III. IV.
ρ , ε0 ~ =0, ∇·B ~ ~ = − ∂B , ∇×E ∂t
~ = ∇·E
~ = µ0 J~ + ε0 µ0 ∇×B
~ ∂E . ∂t
As we have shown in the previous lecture, the first equation, which we shall call the Maxwell’s equation number I, is the differential form of the Gauss’s law, the second equation, which we shall call the Maxwell’s equation number II, tells us about the non-existence of magnetic charges, the third equation, which we shall call the Maxwell’s equation number III, is the differential form of the Faraday’s law, and the final equation, which we shall call the Maxwell’s equation number IV, is the differential form of the modified Ampere’s law. Maxwell’s immediate triumph with the modification of the Ampere’s law was to predict the existence of electromagnetic waves and their propagation speed. The calculated speed came (within experimental error) to be equal to the measured speed of light. This prediction obviously led to the conclusion that light was electromagnetic in nature. Thus arose a synthesis of electromagnetism and optics. In this lecture, we will show how the Maxwell’s equations predict the existence of the electromagnetic waves and will analyse their properties when the fields propagate in vacuum. Much of our discussion, however, will be about ‘how to solve the Maxwell’s equations’ for fields propagating in vacuum.
69
6.1
The Wave Equation for EM Waves in Vacuum
In a vacuum there are no sources, i.e. ρ = 0 and J~ = 0. Hence, the Maxwell’s equations reduce to the following differential equations ~ = 0, ∇·E ~ = 0, ∇·B
(14) (15)
~ ~ = − ∂B , ∇×E ∂t ~ ~ = ε0 µ 0 ∂ E . ∇×B ∂t These equations are readily to solve. The procedure is to eliminate between equations (16) and (17) to obtain differential equations for alone, using where required Eqs. (14) and (15).
(16) (17) ~ or B ~ E ~ or B ~ E
Method: Think of ∇× and ∂/∂t as linear (differential) operators. By analogy with methods of solving linear algebraic equations, applying ∇× into (16) and ∂/∂t into (17), we obtain ~ ∂∇ × B , ∂t 2~ ∂ ~ = ε0 µ 0 ∂ E . ∇×B ∂t ∂t2
= −
= −ε0 µ0
~ ∇× ∇×E
Hence ~ ∇× ∇×E
~ ∂ 2E , ∂t2
and using a vector identity
~ = ∇∇ · E ~ − ∇2 E ~ , ∇× ∇×E we obtain 2~ ~ − ∇2 E ~ = −ε0 µ0 ∂ E . ∇∇ · E ∂t2
70
~ = 0 in the vacuum, it follows at once that Because ∇ · E ~ 1 ∂ 2E , c2 ∂t2
~ = ∇2 E
(18)
where c2 =
1 . ε0 µ 0
The parameter c has the dimensions of velocity and is numerically equal to 3 × 108 ms−1 . Equation (18) is the standard form of a three-dimensional vector wave equation. Following the same procedure, we can easily show that the magnetic ~ satisfies the same equation. The proof is left to the student. field B
6.2
Plane Wave Solution to the Wave Equation
The wave equation in a vacuum is ~ = ∇2 X
~ 1 ∂ 2X c2 ∂t2
~ ≡ E, ~ B. ~ for X We will solve the wave equation in one dimension assuming that the wave propagates in the z direction. In this case, ∂/∂x = ∂/∂y ≡ 0, and then ∇2 =
∂2 ∂2 ∂2 ∂2 + + = . ∂x2 ∂y 2 ∂z 2 ∂z 2
~ and B ~ both have the same form: The differential equations for E ~ ~ ∂ 2X 1 ∂ 2X = . ∂z 2 c2 ∂t2 Such an equation has solutions of the form X = f (z ± ct) , 71
(19)
where f is an arbitrary function with the argument z ± ct. This solution represents a signal propagating with speed c as can be seen from the following discussion: Let X0 = f (z0 − ct0 ) i.e. it represents the solution X at t = t0 and z = z0 . Now examine X at time ∆t later and distance ∆z further along in z. Since a harmonic wave does not change in vacuum, we have X1 = f (z0 + ∆z − c(t0 + ∆t)) = f (z0 − ct0 ) = X0 when
∆z = c∆t ,
i.e. the signal propagates a distance ∆z = c∆t in time ∆t i.e. it propagates with speed c. Proof of solution (19): Let f represent f (z − ct). Then ∂f ∂(z − ct) ∂f ∂f = = = f0 , ∂z ∂(z − ct) ∂z ∂(z − ct) where ”0 ” means differentiation wrt z − ct. Similarly ∂ 2f ∂z 2
∂f 0 ∂f 0 ∂(z − ct) = = = f 00 , ∂z ∂(z − ct) ∂z
and ∂f ∂t
=
∂f ∂(z − ct) = −cf 0 . ∂(z − ct) ∂t
Next ∂ 2f = (−c)(−c)f 00 , ∂t2 and consequently ∂ 2f 1 ∂ 2f = . ∂z 2 c2 ∂t2 72
Thus 1 1 →c= √ , ε0 µ 0 ε0 µ 0
c2 =
which with the numerical values of the parameters ε0 ' 8.85 × 10−12 Fm−1 ,
µ0 = 4π × 10−7 Hm−1
gives c ' 3 × 108 ms−1 . The fact that the numerical value of c is equal to the velocity of light in vacuum led Maxwell to propose an electromagnetic theory of light, one of the brilliant contributions to physics in the nineteenth century.
6.3
Harmonic Waves
In vacuum, we chose a plane wave representation for the propagating EM wave. For a harmonic wave c = νλ =
ω , k
ω = 2πν,
k=
2π , λ
where ν = frequency (Hz), λ = wavelength (m), ω = radian frequency (radians s−1 ), and k = propagation constant (m−1 ). f (z − ct) = f (z −
ω t) = f1 (ωt − kz) . k
A plane wave is represented by the electric field ~ =E ~ 0 ei(ωt−kz) , E
(20)
whose the propagation is characterized by the frequency ω and the wave vector ~k. Since the phase of the wave described by Eq. (20) is by definition the argument of the exponent, we see that the surfaces of the constant phase are planes whose normal is the z-axis: z i(ωt − kz) = iω t − c
73
given at any definite time by z = const. These surfaces of constant phase travel with a constant velocity c often referred as the phase velocity of the wave.13 Therefore, the waves described by Eq. (20) are called plane waves.
6.4
The Transverse Nature of Plane Waves in Vacuum
We now investigate the relations between the amplitudes and phases of the electric and magnetic fields in a plane harmonic wave. While it is true that the magnetic field satisfies the same wave equation as the electric field, it is not independent of the latter, since one must satisfy the Maxwell equations III and IV. ~ = 0 always in electromagnetism, and Since ∇ · B ~ = ∇·B
∂Bx ∂By ∂Bz ∂Bz + + =0+0+ ∂x ∂y ∂z ∂z
for a plane wave propagating in the z direction, we have ∂Bz =0. ∂z However, for a plane wave ∂Bz = −ikBz . ∂z Hence, the rhs must be zero, which means that either k = 0 (zero frequency) or Bz = 0 (transverse wave). ~ For a propagating wave k 6= 0, so the wave is transverse in B. ~ = 0 and then by the same argument we conclude that In a vacuum ∇ · E ~ In other cases in electromagnetism the plane wave is also transverse in E. 13
Note that some textbooks on electromagnetism, for engineers in particular, use the letter j instead of i for the imaginary number. We will use i throughout this lecture notes.
74
(e.g. for a plasma in a magnetic field) a plane wave may not be purely trans~ verse in E. ~ ⊥B ~ for a plane EM wave in a vacuum. In addition: E Proof: Consider a harmonic wave propagating parallel to the z axis. In this case the field components are of the forms ~ =E ~ 0 ei(ωt−kz) , E
~ =B ~ 0 ei(ωt−kz) . B
In order to prove the statement, we will use the Maxwell’s equation III ~ ~ = − ∂B , ∇×E ∂t and expand it in Cartesian coordinates remembering that ∂/∂t ≡ iω ˆi ∂ ∂x E
x
ˆj ∂ ∂y
kˆ
∂ ∂z
Ey Ez
ˆ . = −iω(Bxˆi + By ˆj + Bz k)
(21)
For electromagnetic plane waves propagating along the z axis in a vacuum: Ez = Bz = 0,
∂ ∂ ∂ = = 0, = −ik . ∂x ∂y ∂z
Hence, Eq. (21) reduces to
ˆi ˆj kˆ 0 0 −ik = −iω(Bxˆi + By ˆj) . Ex Ey 0
Expanding the determinant, and comparing the left and right-hand sides, we find x component : y component :
k ikEy = −iωBx ⇒ Bx = − Ey , ω k −ikEx = −iωBy ⇒ By = Ex . ω 75
~ and B? ~ What does it tell us about the relation between E ~ · B, ~ and use the above result, we find If we consider a scalar product E ~ ·B ~ = (ˆiEx + ˆjEy ) · (ˆiBx + ˆjBy ) E k −k Ey + ˆj Ex ) = (ˆiEx + ˆjEy ) · (ˆi ω ω k k = − Ex Ey + Ex Ey = 0 . ω ω ~ ⊥ B. ~ This means that E In addition, note that E ω = =c. B k Thus, we may conclude that in the electromagnetic theory when E and B are related, their ratio is always a velocity characteristic of the problem in hand. This is as far as Maxwell took the subject. It was for others like Heinrich Hertz 1884 to show how to solve Maxwell’s equations with source terms ρ, J~ included (i.e. the generation of electromagnetic waves). We will consider this later. You should be aware that we have not derived Maxwell’s equations from the static limits like Coulomb’s Law and the Biot-Savart Law. The solutions to Maxwell’s equations include the static limits as special cases but many more. Maxwell’s equations have the status of postulates suggested by experimental results. In summary, we have the following important results for related electric and magnetic fields propagating in vacuum: 1. The electric and magnetic fields propagate in a form of plane waves, so-called electromagnetic (EM) waves. ~ and B, ~ i.e. both fields are 2. The plane EM wave is transverse in E perpendicular to the direction of propagation. 76
3. The electric and magnetic fields are perpendicular to each other. 4. The ratio E/B is constant and equal to the velocity of the wave, that is equal to the speed of light in vacuum.
Exercise in class: Electric and magnetic fields of a plane wave For a given electric field polarized in x direction and propagating in z direction ~ = E0 sin (ωt − kz) ˆi , E ~ from III and show that the fields represent calculate B a plane electromagnetic wave propagating in vacuum, i.e. they satisfy the Maxwell’s equations (14)−(17).
77
Revision questions Question 1. Show that the Maxwell’s equations for the EM fields propagating in vacuum can be reduced to two independent second order differential equations, the so-called the wave equations. Question 2. State the properties of an EM wave propagating in vacuum. Question 3. How do we define phase velocity of an EM wave? Question 4. The wave equations for the electric and magnetic fields propagating in vacuum are independent of each other. How then the electric and magnetic fields are related to each other? Explain.
Tutorial problems Problem 6.1 Fields of a plane wave A plane electromagnetic wave propagates in the z direction in a vacuum. Its electric field is polarized in the x direction and so has the form: ~ = E0 sin(ωt − kz)ˆi E (a) If the wave amplitude is 10 volts m−1 and its frequency is 1 MHz, find numerical values for E0 , ω, k and the wavelength. ~ directly from the above expression for E. ~ (b) Calculate ∇ × E ~ ~ Plot E and ∇ × E as functions of z for t = 0 (on the same plot). (c) Hence calculate the magnetic field of the wave using the Maxwell equation: ~ ∂B ~ . = −∇ × E ∂t
78
~ varies in phase with E. ~ (i) Show that B (ii) Calculate the ratio E/B. (d) Show that in the stated circumstances, the remaining Maxwell’s equations should be: ~ = 0, ∇·E
~ = 0, ∇·B
and
~ ~ = ε0 µ0 ∂ E . ∇×B ∂t
~ and B ~ of a right-circularly polarized (e) Write expressions for E plane wave having the same amplitude and frequency as the above one. Problem 6.2 EM wave in free space An EM wave traveling in free space (ρ = 0, J~ = 0) in the positive z direction is described by ~ = E0 cos (ωt − kz) E ~ = B0 cos (ωt − kz) B
ˆi, ˆj,
where k, E0 , and B0 are constants. Under what circumstances do ~ and B ~ fields satisfy all of Maxwell’s equations? these E
79
7
EM Theory and Einstein’s Special Theory of Relativity
The purpose of this lecture is to test the laws of electromagnetism against the principle of special relativity. We will demonstrate how one might infer the law of Biot-Savart and in general relate the magnetic field to electric field from application of special relativity to Coulomb law. Special relativity, formulated in 1905, grew out of Einstein’s meditation on electromagnetic theory and the properties of space and time. Historically, the insights of Einstein’s theory follow after electromagnetism. Logically however, special relativity contains more general statements about nature than electromagnetism. Electromagnetic field theory is just one of a possible set of field theories that are compatible with the Einstein theory of space and time. It is evident that relativistic effects are important if we would have to calculate the field of a charge moving with a speed comparable to that of light. What is not so obvious is that special relativity offers insights into aspects of electromagnetic theory even in the case of the low speed charges we consider in this course. Two such aspects are: (1) The unity of the electromagnetic field i.e. the field is a single entity ~ and B, ~ each with three with 6 components (represented by two vectors E components). (2) Understanding the nature of causal relationships in electromagnetic theory, e.g. ~ =− ∇×E
~ ∂B . ∂t
~ causes a spatially changing E? ~ Does this mean that a time changing B
80
7.1
Lorentz Transformation Equations for Space and Time
We will analyse how the fundamental laws of electromagnetism change when charges move with a constant velocity. In particular, we will show that the Maxwell’s equations are derived from the Coulomb’s law. Thus, the student will learnt that the whole electromagnetism follows naturally from electrostatics. In our calculations, we will follow the principle of special relativity. The principle of special relativity 1. The laws of physics are the same in all inertial reference frames. 2. The speed of light in vacuum is independent of the uniform motion of the observer or source. The constancy of the velocity of light, independent of the motion of the source, gives rise to the relations between space and time coordinates in different inertial reference frames known as Lorentz transformations. Consider a stationary reference frame S and a inertial frame S 0 moving with a velocity ~v parallel to the x axis, as shown in Figure 22. Let x, y, z are coordinates in the S frame and x0 , y 0 , z 0 are the coordinates in the S 0 frame. The time and space Figure 22: Two inertial reference coordinates in S 0 are related to those in S frames in relative motion. A stationary frame S and a frame S 0 moving with a by the 1D Lorentz transformations: velocity ~v parallel to the x axis.
x0 = γ(x − vt) , y0 = y , z0 = z , vx 0 t = γ t− 2 , c
81
where v2 γ = 1− 2 c
!− 1
2
is the Lorentz factor. The above transformation corresponds to a situation of ~v parallel to the x axis. Later in the course, Chap. 21, we will consider the general case of the velocity ~v of the frame S 0 in an arbitrary direction.
7.2
Force Transformation Equations
Let us investigate the force between two charges as viewed by observers in S and S 0 . We will evaluate the force from the point of view of the observer in S 0 . Consider a particle which is moving with velocity ~u = uxˆi + uy ˆj + uz kˆ in the S frame and is acted on by a force with components Fx , Fy and Fz . Then, according to the special theory of relativity, the force in the S 0 frame is vuz vuy Fz , Fx0 = Fx − 2 vux Fy − 2 x c (1 − c2 ) c (1 − vu ) c2 2
1
2
1
Fy0
(1 − uc2 ) 2 Fy = , vux Fy = x 1 − c2 γ(1 − vu ) c2
Fz0
(1 − vc2 ) 2 Fz . = vux Fz = x 1 − c2 γ(1 − vu ) c2
Suppose Fx , Fy , Fz represents the velocity independent Coulomb force. Then in the S 0 frame (source of the field now moving) the force is no longer velocity independent. Thus, the observers disagree about the magnitude of the force. They, in fact, disagree not only about the magnitude but also about the origin of the force. In electromagnetic theory we say that there is now a ~ that determines the magmagnetic force and we define a magnetic field B netic force. We now present a detailed calculation that illustrates how the form of the Maxwell’s equations is determined by nature obeying Einstein’s special theory of relativity. 82
7.2.1
The Force between Two Charges Moving with Constant Velocities
Invariance of electric charge In ordinary matter, electrons move with much greater speeds than protons yet there is no associated electric field. This implies that electric charge is independent of velocity unless electromagnetic laws modified in some complicated way (see discussion by King in Physical Review Letters 5, 562 (1960)). Consider charges q1 and q2 moving with velocities ~u and ~v in an inertial frame S. No loss of generality occurs if ~v is taken parallel to the x axis. Now consider a frame S 0 moving with velocity ~v along x axis, that q2 is stationary in S 0 . Assume that at time t = 0, the frames S and S 0 overlap. Figure 23: Two moving charges seen in two From the principle of relativity, in the S 0 frame Coulombs law holds.
different reference frames.
The force on q1 seen in S 0 is therefore F~ 0 =
1 q 1 q2 0 ~r . 4πε0 r0 3
We shall transform this expression to find the force observed in the S frame in which the source of the field (q2 ) is moving. We shall see that what we normally call the MAGNETIC FIELD arises as a natural consequence of relativistic invariance with no extra assumptions. We start with the x component of the force, that is Fx0 =
1 q1 q2 0 x , 4πε0 r0 3 83
and similarly for y and z. We need transformations of x0 to x and r0 to r. The transformation of x0 to x is found from the Lorentz transformations that at t = 0 are x0 = γx , y0 = y , z0 = z , vx t0 = −γ 2 . c To do the complete transformation of the force component, we also need transformation from r0 to r. It can be found as follows. It is seen from Figure 23 that there is an “axial symmetry”, so that the transformation could be expressed in terms of the angle θ. Thus, we can write r02 = x02 + y 02 + z 02 = γ 2 x2 + y 2 + z 2 ! " ! # v2 2 y2 + z2 2 2 2 2 2 =γ x + 1− 2 y +z = γ x + γ2 c # " v2 2 2 2 2 2 2 = γ x +y +z − 2 y +z c ! ! 2 v 2 2 v2 2 2 2 2 2 = γ r − 2 r sin θ = γ r 1 − 2 sin θ , c c where sin θ =
q
(y 2 + z 2 )/r. Hence v2 r = γr 1 − 2 sin2 θ c
!1 2
0
.
Thus, substituting the transformations into Fx0 , we obtain Fx0 =
1 4πε0
q1 q2 γx
γ 3 r3 1 −
v2 c2
sin2 θ
3
= q1 gx ,
2
where g=
1 4πε0
q2 γ 2 r3
1− 84
v2 c2
2
sin θ
3 2
.
Similarly for the remaining two components, we find Fy0 =
q1 gy , γ
q1 gz . γ
Fz0 =
In summary, the force transformations are a) x component Fx0 = Fx −
vuy
vux c2
c2 1 −
Fy
−
vuz
c2 1 −
vux c2
Fz
.
Thus q1 gx = Fx −
vuy
c2 1 −
vux c2
Fy0 =
Fy
−
vuz
c2 1 −
vux c2
Fz
.
b) y component Fy γ 1−
vux c2
,
from which we find
Fy = γ 1 −
vux vux Fy0 = γ 1 − 2 2 c c
q1 gy , γ
which we can write as vux Fy = q1 gy 1 − 2 c
vux Fz = q1 gz 1 − 2 c
.
c) z component
.
Substituting for Fy,z in x equation Fx = q1 gx + q1 gy
vuy vuz + q1 gz 2 . 2 c c
Note: Here is the germ of the magnetic field. The last two terms are typical second order relativistic terms ≈ v 2 /c2 . In a nonrelativistic calculation we 85
would have Fx = Fx0 . We now combine results a), b) and c) into a single vector equation for the force F~ . First, note that vux = ~u · ~v . Next, we can write the x component as vux vux vuy vuz = q1 gx 1 − 2 + q1 gx 2 + q1 gy 2 + q1 gz 2 c c c c vux v = q1 g 1 − 2 x + q1 g 2 (~u · ~r) c c
Fx
and with the y and z components vux = q1 g 1 − 2 y , c vux = q1 g 1 − 2 z , c
Fy Fz
these three components combine into !
~v · ~u ~v F~ = q1 g 1 − 2 ~r + q1 g 2 (~u · ~r) c c q1 g = q1 g~r + 2 [~v (~u · ~r) − ~r (~u · ~v )] , c which can be written as q1 g F~ = q1 g~r + 2 ~u × (~v × ~r) . c We can write this equation in the form of the Lorentz force
~ + ~u × B ~ F~ = q1 E
,
where ~ = g~r = E
1 4πε0
q2 γ 2 r3
1−
v2 c2
2
sin θ
3
~r ,
2
and ~ ~ = ~v × g~r = ~v × E . B c2 c2 86
(22)
Thus, a force which we see in the moving frame S 0 as the Coulomb force appears as the Lorentz force in the stationary frame S. ~ and E ~ have no independent existence. A The relation (22) shows that B pure electric field in one coordinate frame appears as a mixture of electric and magnetic fields in another coordinate frame. Note that as v → 0, γ → 1, and then ~ → E
1 q2 ~r . 4πε0 r3
Moreover, the ration of magnitudes of magnetic to electric term in the force equation is uv/c2 , i.e. magnetic forces are second order relativistic effects.
7.3
Electric Field Lines of a Moving Charge
Let β = v/c. Then, we can write the electric field as ~ = E
1 4πε0
q(1 − β 2 )
r2 1 − β 2 sin2 θ
3
rˆ .
2
For a given θ, the electric field E still varies as 1/r2 , but the field lines are crowded in the direction perpendicular to ~v , as shown in Figure 24. In the forward direction θ = 0, and then E= Figure 24: Electric field lines of the stationary charge and the charge moving with uniform velocity ~v .
1 q(1 − β 2 ) < Es , 4πε0 r2
where Es is the electric field of the stationary charge, i.e. the static electric field (at v = 0). Thus, the field amplitude is lower in the direction of motion
relative to the static field amplitude.
87
In the perpendicular direction θ = π/2, and then E=
1 1 q 2 −2 > Es . 1 − β 4πε0 r2
Thus, the field amplitude is larger in the transverse directions relative to the to the static field amplitude.14 In summary, the electric field lines radiate from the present position of the charge and are crowded in the direction perpendicular to the direction of motion of the charge.
7.4
Magnetic Field Lines of a Moving Charge
From the relation between electric and magnetic fields, we find ~ ~ = ~v × E = 1 B c2 4πε0
q(1 − β 2 ) ~v × ~r c2 r 3
1−
β2
2
sin θ
3
.
2
In order to find the direction of the magnetic field, we refer to the spherical polar coordinates, and find that ~ = Br rˆ + Bθ θˆ + Bφ φˆ = Bφ φˆ , B ˆ φˆ are unit ~ ∼ ~v × ~r, and rˆ, θ, since B vectors. Thus, the magnetic field lines form concentric rings about ~v , and there is symmetry about the plane θ = π/2. In Figure 25: Magnetic field lines of the the non-relativistic case of v 1 the charge moving with uniform velocity ~v . factor β → 0, and then the magnetic field reduces to ~ = B
1 q ~v × ~r , 4πε0 c2 r3
14
An interesting question: What is the electric field distribution of a charge moving with velocity v = c?
88
which is the Biot-Savart law. Applied to a continuous line current I: ~ = B
~ × ~r 1 I dl . 4πε0 c2 r3
The constant 1/(ε0 c2 ) is normally written µ0 - the magnetic permeability of free space.
7.5
Invariance of the Maxwell equations under the Lorentz transformation
According to the principles of relativity, the laws of physics are invariant under the Lorentz transformation. In this lecture, we will show that the basic equations for the electromagnetic theory, the Maxwell’s equations, are invariant under the Lorentz transformation, i.e. they represent laws of physics. (1) Equation for the total electric flux Consider the total electric flux through a surface S closing a moving charge q: ΨE =
I
~ ·n E ˆ dS .
S
We will use the axial symmetry and break sphere up into rings, as shown in Figure 26, lying between θ and θ + dθ. The radius of a ring seen from the center under the angle θ is r sin θ, and the width of the ring is rdθ. ~ has a radial symmetry, E ~ kn Since E ˆ everywhere, and then the flux dΨE is ~ ·n dΨE = E ˆ dS = EdS . Figure 26: Evaluation of the total electric flux through a surface S closing a moving charge.
(23)
However, the area of the ring (stripe) between θ and θ + dθ is equal to
dS = 2π(r sin θ) rdθ = 2πr2 sin θdθ . 89
Hence, substituting this result into Eq. (23) and the explicit form of E, we find that the flux through the ring is 1 4πε0
dΨE =
q(1 − β 2 ) r2
1−
q(1 − β 2 ) 2ε0
=
β2
2
sin θ
2πr2 sin θdθ
3
2
sin θdθ 1−
β2
2
sin θ
3
.
2
This gives the total flux through the surface S: ΨE =
Z
π q(1 − β 2 ) Z dΨE = 2ε0 θ=0
sin θdθ 1 − β 2 sin2 θ
3
.
2
To calculate the integral, put cos θ = x, so that sin θdθ = −dx, and then I = − = − where a =
dx
Z
(1 − β 2 + β 2 x2 )3/2 1 β3
dx
Z
(a2
+ x2 )3/2
dx 1 Z =− 3 2 3/2 1−β β + x2 β2
,
√ 1 − β 2 /β. Performing the integration, we obtain I=−
1
x
a2 β 3
(a2
+ x2 )1/2
,
and finally including the limits of the integral, we get I = −
Z−1
dx
+ β 2 x2 )3/2 1 −1 1 1 2 = − 2 3√ 2 + 2 3√ 2 = . a β a +1 a β a +1 1 − β2 1
(1 −
β2
Thus, the total flux through the surface is ΨE =
q(1 − β 2 ) 2 q = , 2 2ε0 1−β ε0 90
which is the same as for a stationary charge. Hence, the electric field produced by a moving charge satisfies the Gauss’s law. If we apply the Gauss’ theorem to ΨE , we get ~ = ρ . ∇·E ε0 Thus, we conclude that the Gauss’ law (Maxwell’s equation I) is invariant under the Lorentz transformation.
(2) Magnetic flux through a closed surface Consider the magnetic flux through a closed surface ΨM =
I
~ ·n B ˆ dS .
~ is If we choose a sphere centered on q to calculate the integral, we find that B perpendicular to n ˆ everywhere on the sphere, i.e. is tangential to the surface of the sphere. Thus, the flux through the sphere ΨM = 0. ~ = 0 always. It is easy to see that Similarly, we can show that ∇ · B 1 ~ ∇ · ~ v × E c2 1 ~ ~ =0, = 2 E · ∇ × ~v − ~v · ∇ × E c
~ = ∇·B
~ since ∇ × ~v = 0 as ~v is constant, and ~v is perpendicular to ∇ × E. Thus, we conclude that the Maxwell’s equation II is invariant under the Lorentz transformation. ~ derivative related to temporal B ~ derivative. (3) Spatial E We shall show that for a point charge under uniform velocity ~ ~ = − ∂B . ∇×E ∂t 91
~ in spherical polar coordinates To do it, we write ∇ × E "
rˆ ∂ (sin θEφ ) ∂Eθ − r sin θ ∂θ ∂φ " # θˆ 1 ∂Er ∂ (rEφ ) + − r sin θ ∂φ ∂r " # φˆ ∂ (rEθ ) ∂Er + − . r ∂r ∂θ
~ = ∇×E
#
Since the electric field has a radial symmetry, we have Eθ = Eφ = 0 , and because the electric field amplitude dependes only on r and θ, we get ~ =− ∇×E
1 ∂Er ˆ φ, r ∂θ
where Er =
1 4πε0
q(1 − β 2 ) r2
1−
β2
2
sin θ
3 2
=
K 1−
β2
2
sin θ
3
.
2
Calculate ∂Er /∂θ: − 5 3 − K 1 − β 2 sin2 θ 2 −2β 2 sin θ cos θ 2 3Kβ 2 sin 2θ = 5 . 2 1 − β 2 sin2 θ 2
∂Er = ∂θ
Hence 2 ~ = − 1 3q(1 − β ) ∇×E 4πε0 2r3
β 2 sin 2θ 1−
β2
2
sin θ
5
φˆ .
2
~ To calculate ∂ B/∂t, we use the theorem of partial derivatives. If y = f (x, t) 92
then from the maximum change of y ∂y ∂y dx + dt = 0 , ∂x ∂t
dy = we obtain
∂y ∂y ∂x =− . ∂t ∂x ∂t Thus ~ ~ ∂B ∂B = −v . ∂t ∂x Alternatively, to see this “physically”, remember that the field pattern moves with constant velocity ~v . Let a stationary observer measure the change in the ~ in a time interval dt. This change is the same as he would observed field B at a fixed time by moving a distance dx = −vdt, i.e. ~ dB
~ in dt ≡ dB
in dx = −vdt .
Hence ~ ~ ∂B ∂B =− , ∂x v∂t and then ~ ~ ∂B ∂B = −v . ∂t ∂x Now ~ = B
1 4πε0
q(1 − β 2 ) v sin θ c2 r 2
1−
β2
2
sin θ
3 2
and ~ ∂B ∂Bφ ˆ ∂Bφ ∂r ˆ = φ= φ. ∂x ∂x ∂r ∂x Since √ 2 y + z2 a sin θ = = , r r 93
φˆ
we can write Bφ as Ka
Bφ = r3
1−
2 3/2
β2a r2
=
Ka (r2 − β 2 a2 )3/2
,
where K=
1 q(1 − β 2 )v 4πε0 c2
and a =
q
y2 + z2 .
Next 1/2
∂ (x2 + y 2 + z 2 ) ∂r = ∂x ∂x
=
−1/2 1 2 x 2x = = cos θ , x + y2 + z2 2 r
−5/2 ∂Bφ 3Ka r 3 2 2r = − = Ka − r − β 2 a2 . ∂r 2 (r2 − β 2 a2 )5/2
Hence 1 3q(1 − β 2 ) vr2 sin θ cos θ ∂Bφ = − 5 ∂x 4πε0 c2 r5 1 − β 2 sin2 θ 2 = −
1 4πε0
3q(1 − β 2 ) v sin 2θ 2c2 r3
1−
β2
2
sin θ
5
,
2
and then ~ 1 ∂B ∂Bφ ˆ 3q(1 − β 2 ) v 2 sin 2θ ˆ = −v φ= 5 φ ∂t ∂x 4πε0 2 3 2 2 2c r 1 − β sin θ 2 =
1 3q(1 − β 2 )β 2 sin 2θ ˆ 5 φ . 4πε0 2 3 2 2r 1 − β sin θ 2
~ we see that Comparing with ∇ × E, ~ ~ = − ∂B . ∇×E ∂t Thus, we conclude that the Faraday’s Law (Maxwell’s equation III) is invariant under the Lorentz transformation.
94
7.5.1
Remark on the Electromagnetic Induction
It has been known since about 1831 when Faraday first waved a magnet near an electric circuit and played with transformers that when the magnetic flux through a circuit changes, an electromotive force (emf) E appears in it. Faraday gave the rule E =−
∂ΦM . ∂t
The nature of this “phenomenon” is, however often misinterpreted. Consider a point charge q moving near a closed circuit as shown in Figure 27. Because of the θ dependence of E, the electric field on side (1) is larger than that on side (2). Thus, there is a net driving force round the circuit. Calculate the resulting electromotive force in the circuit, which is equal to the work done on a charge q 0 in the Figure 27: Contour for evaluation of an circuit
electromotive force.
~ F~ · dl q0 Z ~ , ~ + ~u × B) ~ · dl = (E
E = Wq 0 =
Z
where u is the velocity of the charge q 0 in the circuit. ~ is perpendicular to both ~u and B, ~ it is also perpendicular Now, since ~u × B ~ and then to dl, Z
~ =0. ~ · dl ~u × B
Hence E=
Z
~ = ~ · dl E
Z
~ ·n ∇×E ˆ dS = 95
Z
~ . ~ · dS ∇×E
If we now take the conclusion that where is a spatially varying electric field there is also a time varying magnetic field, we obtain E =−
Z
Z ~ ∂B ~ =−∂ B ~ = − ∂ΨM . ~ · dS · dS ∂t ∂t ∂t
However, it is obvious in this example that the changing magnetic flux is not the CAUSE of the emf. The changing magnetic field and the electric field have a common CAUSE through the charge q. We can conclude that: Electric and magnetic fields do not produce each other - they are both due to electric charges. It is often thought in the textbooks, however that e.g. in a transformer the changing ΨM produces E. It happens because the flux cutting rule is an extremely powerful one for calculating the integrated electric field of electric currents. ~ = −∂ B/∂t ~ The Faraday’s rule E = −∂ΦM /∂t, which arises from ∇ × E should not be thought of as a casual relationship. What it means is that if a charge moving with a constant velocity produces a time varying magnetic field then that charge also produces a spatially varying electric field.
~ and temporal variation (4) Relation between spatial variation of B ~ of E. Since ~ ~ = ~v × E , B c2 and
~ ∇ × ~v × E
=
~ · ∇ ~v − (~v · ∇) E ~ + ~v ∇ · E ~ −E ~ (∇ · ~v ) , E
with ~v constant, we obtain ∇ · ~v = 0 ,
~ · ∇ ~v = 0 . E
96
Then n o ~ = 1 ∇ × ~v × E ~ = 1 − (~v · ∇) E ~ + ~v ∇ · E ~ ∇×B . c2 c2
Thus, ~ ~ ~ = v ∂E = − ∂E ~ = vx ∂ + vy ∂ + vz ∂ E (~v · ∇) E ∂x ∂y ∂z ∂x ∂t !
as vx = v
vy = vz = 0 and
∂ ∂ = −v . ∂t ∂x
Hence
~ ~ = 1 ~v ∇ · E ~ + ∂E . ∇×B c2 ∂t If we recognize that ~ = ∇·E
ρ ε0
and next that ~v ρ = J~ and 1/c2 ε0 = µ0 , we get the Maxwell’s equation IV. Thus, we may conclude that the Maxwell’s equation IV is invariant under the Lorentz transformation.
In summary: Maxwell’s equations for a point charge moving with uniform velocity are ~ = ρ , ∇·E ε0 ~ = 0, ∇·B ~ ~ = − ∂B , ∇×E ∂t ~ ~ = µ0 J~ + 1 ∂ E . ∇×B c2 ∂t 97
These equations arise from the necessity for the correct relativistic transformations between frames in uniform relative motion. This shows that the Maxwell’s equations are invariant under the Lorentz transformation. If the postulates of relativity are correct and Coulomb’s law gives the field of a stationary charge, these equations follow, and the force on a charge is
~ + ~v × B ~ F~ = q E
.
Revision questions Question 1. State the principle of special relativity. Question 2. Is the Coulomb’s law for moving charges equivalent to the Lorentz force? Question 3. Are the electric and magnetic fields invariant under the Lorentz transformation? Explain. Question 4. Are the Maxwell’s equations invariant under the Lorentz transformation? Question 5. Why the Faraday’s law is often misinterpreted? What is the correct interpretation of the the Faraday’s law?
98
8
Energy in the Electromagnetic Field: Poyntings’ Theorem
Energy may be transported through space by means of electromagnetic waves. ~ × B, ~ as We expect energy flow in the direction of propagation of the wave, E illustrated in Figure 28. In this lecture, we will derive an expression for the rate of the energy flow with a traveling EM wave. We will use the Maxwell equations to derive this expression, which at the same time will show that the Maxwell equations are consistent with the law of conservation of energy. We know from the circuit theory that the power (energy) flow is related to the product of voltage and current P =VI . We will show that in the field theory, where everything is expressed in terms of the electric and magnetic field amplitudes, the power flow across an element ~ Figure 28: Direction of the E~ × B~ vec- of area dS is given by tor of the field components of a plane ~ ×B ~ · dS ~ , c2 ε 0 E wave.
where S is a closed surface bounding a volume V containing a source of an EM field. ~ B, ~ If the field propagates in the direction determined by the cross product E× consider the source of the field, i.e. consider the expression ~ × B) ~ . ∇ · (E If we employ the vector identity ~ × B) ~ =B ~ ·∇×E ~ −E ~ ·∇×B ~ ∇ · (E and use the Maxwell equations III and IV: ~ ~ = − ∂B ∇×E ∂t
and
~ ~ = µ0 J~ + ε0 µ0 ∂ E , ∇×B ∂t 99
(24)
we then may write Eq. (24) as ~ ~ ~ · J~ − ε0 µ0 E ~ · ∂E , ~ × B) ~ = −B ~ · ∂ B − µ0 E ∇ · (E ∂t ∂t or in the form 1 ~ × B) ~ = −E ~ · J~ − ∂ ∇ · (E µ0 ∂t
1 1 B2 ε0 E 2 + 2 2 µ0
!
.
On the left-hand side, we put 1/µ0 = ε0 c2 and integrate the equation over some closed surface S enclosing a volume V . Then, we obtain Z
~ × B) ~ dV = − ε0 c ∇ · (E 2
Z V
V
Z ~ · J~ dV − ∂ E ∂t V
1 1 B2 ε0 E 2 + dV . (25) 2 2 µ0 !
Now we apply Gauss’s theorem to the left-hand side of the above equation, to convert the volume integral into the closed surface integral, and find I
~ × B) ~ ·n ε0 c2 (E ˆ dS
←− Energy flux
S
=−
Z
~ · J~ dV E
←− Rate of doing work by field on the current
V
∂ Z − ∂t V
1 1 B2 ε0 E 2 + dV 2 2 µ0 !
←− Field energy.
(26)
In words, this equation states that the instantaneous flow of power across a closed surface S is equal to the time rate of decrease of the energy stored in the field in the interior of S plus the power loss due ~ conducting charges, within the volume V . to the work on current J, The above interpretation of the three terms in this equation can be justified in the following way.
100
8.1
Rate of Doing Work by the Field on the Current – Ohmic Heating
R ~ · J~ dV . We will prove that this term Consider first the second term: − E V
is equal to the rate of doing work by the electric field on the current that, in other words, it is the ohmic heating. Simply, it shows that the power produced by a surce contained inside the volume can be converted into the ohmic power dissipated inside the volume. We start our calculations from the circuit theory where we express all the circuit variables in terms of the field variables. From the circuit theory, we know that in a resistive medium, as shown in Figure 29, the Ohm’s Law is: V = IR and R = R`/A, where R is the resistivity of the medium, I is the current flowing through the medium and A is the area through which the current is flowing. Thus, we can write the current as I = V /R = (V A)/(R`) .
Figure 29: Illustration of a current I flowing through a resistive medium of a length ` and cross section A.
~ = −∇ V , so that E = V /` and Since E then we can find the current density through the area A, and write it in terms of the field variable E as J = I/A =
1 E. R
Defining the conductivity σ = 1/R, the current density can be written as I/A = J = σE . Having the current density expressed in terms of the field variable E, we can calculate the rate of conversion (dissipation) of the electromagnetic field energy into heat H = IV =
E 2 `2 V2 = = σE 2 `A = σE 2 V , R R A` 101
where V = `A is the volume of the resistive medium. Hence, the energy dissipated per unit volume is H ~ · J~ . = σE 2 = E J = E V ~ · J~ is the rate of heating per unit volume in this case. Thus, E
8.2
Electrostatic Field Energy Density
We now show that the term electric field E, so that
1 2 V 2 ε0 E dV
R
represents energy contained in the
∂ Z 1 ε0 E 2 dV ∂t V 2 is the time rate of decrease of the energy stored in the electric field. −
Again, we start from the circuit theory from which we know that capacitors are devices where electric field energy can be stored. Thus, consider the work required to charge a capacitor of a capacitance C to a voltage V 1 W = CV 2 . 2 According to the field theory of electromagnetism this work done corresponds to conversion from some other form of energy into electrostatic field energy ε0 A 1 . E = W = CV 2 and C = 2 d Since the electric field in the capacitor is given by E = V /d, we obtain 1 ε0 A 2 2 1 E d = ε0 E 2 V , 2 d 2 where V = Ad is the volume of the capacitor. Hence E=
E 1 = ε0 E 2 V 2 is the energy per unit volume contained in or carried out by the electric field E.
102
8.3
Magnetostatic Field Energy Density
We now show that the term the magnetic field B, so that
R V
B 2 /(2µ0 )dV represents energy contained in
∂ Z 1 B2 dV − ∂t V 2 µ0 is the time rate of decrease of the energy stored in the magnetic field. As before, we refer to the circuit theory from which it is well known that a solenoid is a device where the magnetic field energy can be stored. Thus, the work required to energize an inductor of inductance L to a current I is E = 12 LI 2 and that the magnetic field within a long solenoid of self-inductance L = µ0 n2 A` is B = µ0 nI. According to the field theory of electromagnetism this work done corresponds to transformation of some other form of energy to magnetic field energy. Thus, for a solenoid of length ` and area of cross section A we have energy 1 1 1 1 B2 1 V , E = W = LI 2 = µ0 n2 A`I 2 = (µ0 nI)2 A` = 2 2 2 µ0 2 µ0 oand then energy per unit volume is E 1 B2 = . V 2 µ0 This is the energy per unit volume contained in or carried out by the magnetic field B.
8.4
Poynting Vector
Since the right-hand side of Eq. (26) represents the rate of increase of the electric and magnetic energies stored in the EM field and substracted by the ohmic power dissipated as heat, the left-hand side must be equal (in
103
consistency with the law of conservation of energy) to the power flowing into the volume through its surface. Thus, the expression ~ = ε 0 c2 ( E ~ × B) ~ N is a vector representing the power flow per unit area, and is referred to as the Poynting vector. It represents the energy flux in the electromagnetic field, i.e. the energy flow per unit area (measured normal to the flow) per unit time. The quantity N has dimension of power per square meter. It is easy to see: E has the dimension of volts per meter, ε0 c2 B has dimension of amper per meter. Thus, the Poynting vector has dimension volts × amper/(square meter) = Watts/(square meter). The energy flow equation can be converted into the form of a differential continuity equation or energy conservation law. From Eq. (25), we have ∂U ~ = −J~ · E ~ , +∇·N ∂t where 1 1 B2 U = ε0 E 2 + 2 2 µ0
(27)
is the energy density of the EM field. The physical meaning of the differential continuity equation is that the time rate of change of electromagnetic energy within a certain volume, plus the energy flowing out through the boundary surface of the volume is equal to the negative of the total work done by the fields on the source inside the vol~ is a conversion of electromagnetic energy into heat energy. ume. Thus. J~ · E Equation (27) shows that we consider the energy stored in electric and magnetic fields as distributed throughout the region of space where these fields are present with densities ε0 E 2 /2 and B 2 /2µ0 , respectively.
104
8.5
Phase Relationships in EM Waves
~ and B ~ contribute Here, we will show that only the in-phase components of E to net energy flow averaged over a whole cycle of the radiation. If somehow, the fields would oscillate with different phases, no energy can be transported. ~ and B ~ oscillating in phase. Thus, if E ~ = Consider orthogonal fields E ~ ˆ ˆ E0 cos(ωt)i and B = B0 cos(ωt)j then: ~ = ε 0 c2 E ~ ×B ~ = ε0 c2 E0 B0 cos2 (ωt) kˆ . N Since, cos2 (ωt) = 12 , we obtain
where Erms
1 N = ε0 c2 E0 B0 = ε0 c2 Erms Brms , 2 √ √ = E0 / 2 and Brms = B0 / 2.
Hence, the average Pointing vector is different from zero indicating that the ~ and B ~ oscillating in phase can transport energy. orthogonal fields E ~ = E0 cos(ωt)ˆi and B ~ = B0 sin(ωt)ˆj then: If, however, E ~ = ε 0 c2 E ~ ×B ~ = ε0 c2 E0 B0 cos(ωt) sin(ωt) kˆ . N Since, cos(ωt) sin(ωt) = 0, we have N = 0. Hence, the average Pointing vector is zero indicating that the orthogonal ~ and B ~ oscillating out of phase phase cannot transport energy. fields E ~ and B ~ fields oscillate In summary, electromagnetic waves in which the E in phase can transport energy.
8.6
Momentum Flux
It should be noted here, that EM waves can transport not only energy but also momentum.
105
To obtain an expression for the momentum carried by the electromagnetic field, we may employ the relativistic energy-momentum relationship E 2 = p2 c2 + m20 c4 . Since for electromagnetic radiation m0 = 0, we obtain p=
E . c
Thus the momentum flux of the electromagnetic field is: M=
~ ×B ~ ε 0 c2 E ~ ×B ~ , = ε 0 cE c
which leads to the conclusion that electromagnetic waves transport not only energy, whose the flow is given by the Poynting vector, but also momentum.
8.7
EM Energy Flow: Circuit versus Field Theory
We now consider simple examples illustrating applications of the Poynting vector to some familiar circuit problems to show how the field theory provides an alternative (surprising!) way of viewing the energy transfer through the circuits. 8.7.1
Energy Flow in a Resistive Wire
As a first example, consider a wire (resistor) of length `, carrying a current I, as shown in Figure 30. Let V is a potential difference applied along a resistive wire. We calculate the power dissipated in the wire using the circuit theory, and next will compare the formalism with that used in the field theory. We are particularly interested in the predicFigure 30: A resistive wire of length ` tion of the two theories of the direccarrying a current I. tion of energy flow through the circuit.
106
Circuit theory calculation: According to circuit theory the power dissipated in the wire is P = V I = I 2R , where I is the current flowing through the wire and R is the resistance of the wire. Since R is proportional to the length of the wire, a more power (energy) is dissipated for a longer wire. Thus, according to the circuit theory, energy flows along the wire. Field theory calculation: Let us now look at the same problem from the point of view of the field theory. According to electromagnetic field theory, the direction of the flow ~ × B. ~ ~ = ε 0 c2 E of energy is described by the Poynting vector N In order to find the direction of the Poynting vector, we calculate the directions of propagation of the electric and magnetic fields produced by the current. The electric field propagates along the wire, as shown in Figure 31, and is given by ~ = −∇ V (z) = V (z) zˆ . E `
Figure 31: Directions of the electric and magnetic fields produced by a wire and the direction of the resulting Poynting vector.
From the Amperes line integral theorem, we find that the the magnetic field propagates around the wire, and at points distant a from the wire is ~ = µ0 I φˆ . B 2πa Hence, the Poynting vector is ~ = ε 0 c2 E ~ ×B ~ = ε0 c2 V µ0 I rˆ = V I rˆ , N ` 2πa 2π`a where we have used the relations zˆ × φˆ = rˆ and ε0 µ0 c2 = 1.
107
Thus, the field theory predicts that energy flows into the wire from the air not along the wire. The energy is in the fields, the wire provides boundary conditions and guides the fields. The student may argue that the EM field carries only a small part of the energy dissipated in the wire. However, the total rate at which field energy flows into the wire by crossing the surface of the wire is given by I
~ · dS ~ = N
I
~ · rˆ dS = N
I
I
VI 2π`a = V I , 2π`a
N dSside
S
= N
dSside =
~ ⊥ dS ~ on the ends of the cylinder), which is in agreement with the re(N sult of the circuit theory. This result demonstrates quantitatively that all the power which heats the resistor enters through the sides not through the wires. If the energy is in the fields, it means that the electromagnetic energy goes out of a battery into the air, and then goes into the wire from the air. This is exactly the case we will show in the following example. 8.7.2
Energy Flow out of Battery
In the above example, we have shown that according to the field theory, the energy enters the resistor from the air. Then, a question arises: If the energy enters the resistor from the air, how does the energy get out to the air from a source of energy (battery)? To answer this question, consider a battery which provides energy to the resistor. As illustrated in Figure 32, inside ~ are in opposite dithe battery J~ and E rections. The magnetic field circulates ~ Figure 32: Directions of the current J around the battery, so we see that the ~ in a battery. ~ points out into the air, and the electric field E Poynting vector N not along the wire.
108
Thus, the battery sends energy out into the air, not along the wire.
8.7.3
Propagation of an Electromagnetic Wave along a Wire
One of the above examples showed that field energy flows into a wire so that it can be dissipated as heat. Consider now a different situation that one would like to transmit an electromagnetic wave through a resistive wire. It is well known from experiments, that an electromagnetic wave can be transmitted along the wire with very little loss of its energy. Why does it happen? This effect has a simple explanation in terms of the EM theory. If we consider the case of a perfect Figure 33: Directions of the electric and conductor we find that there will be a magnetic fields around a wire transmitting current wave along the wire with suran EM wave of wavelength λ. face charges induced producing a ra~ ×B ~ is parallel to the wire and the field descripdial electric field. Then E tion of energy transmission is that it is transmitted in the space around the ~ and B ~ are in phase and N ~ is always in wire. In the space around the wire E the same direction. Within a perfect conductor we will show later in Chap~ and B ~ are π/2 out of phase so the mean N averaged over a ter 12.6 that E cycle is zero. Thus, there is no net energy transmission within the perfectly conducting wire.
Exercise in class: Energizing of a capacitor Consider the energizing of a plane parallel capacitor with circular plates. Show that circuit and field calculations agree as to the rate of energizing of the capacitor, i.e. Pc = Pf where:
109
Pc = V I = rate of doing work (by current I and voltage V between the plates) in charging the capacitor according to circuit theory. Pf =
R
~ ×B ~ · dS ~ = rate of energy flow into the surface ε0 c2 E
S
of the capacitor according to field theory. From which direction does the energy flow in to the capacitor according to field theory? We will keep the calculation simple assuming the plates of the capacitor to be uniformly charged. Under what conditions is this assumption likely to be true?
Exercise in class: No Fluxes from Static Fields ~ and magnetostatic Consider a source of electrostatic field E ~ ~ ~ one may ask: field B. If it is arranged so that E ⊥ B, ~ × B? ~ Should we expect to see an energy flux of ε0 c2 E
110
Revision questions Question 1. Derive formula for the electrostatic field energy density. Question 2. Derive formula for the magnetostatic field energy density. Question 3. State the definition of the Poynting vector. Question 4. In a circle containing a resistor, what is the direction of the energy flow according to the circuit theory and the field theory?
Tutorial problems Problem 8.1 Poynting vector on the surface of a long wire Find the direction and the magnitude of the Poynting vector on the surface of an infinitely long, straight conducting wire of radius r and conductivity σ, carrying a current I. Problem 8.2 Energy flow in a plane wave A plane electromagnetic wave propagates in the z direction in a vacuum. Its electric field is polarized in the x direction and so has the form: ~ = E0 sin(ωt − kz)ˆi E (a) Find an expression for the vector energy flux in this wave. (b) On the same time scale plot the electric field strength and energy flux magnitude versus time at position z = 0 for 2 cycles of the wave. (c) What is the mean energy flux averaged over cycles?
111
9
Solution of Laplace’s Equation and Boundary Value Problem
In one of the previous lectures, we discussed an advantage of using the scalar and vector potentials in the calculation of the electric and magnetic fields. In this lecture, we will illustrate applications of the scalar potential to physical problems involving bounded fields. There is a class of problems in electromagnetism in which a field can be derived without involvement of the complete set of the Maxwell’s equations. This can be done with the help of the gradient of a scalar potential which satisfies Laplace’s equation ∇2 Φ = 0 . Proof: The condition for this to happen is that a vector field F~ has vanishing divergence and curl ∇ · F~ = 0
and
∇ × F~ = 0 .
(28)
Since, ∇× F~ = 0, we can always write F~ as F~ = ∇Φ, where Φ is an arbitrary scalar function. Then, ∇ · F~ = 0 means that ∇ · (∇Φ) = ∇2 Φ = 0 . Thus, the scalar potential Φ contains all the necessary information to completely specify the field of the properties (28). Example 1: Electrostatic problems involving Laplace’s equation ~ we see that the requirement ~ = ρ/ε0 and ∇× E ~ =−∂B Since in general ∇· E ∂t for Laplace’s equation to be relevant is that ρ = 0 and ∂/∂t = 0, i.e. a sourcefree region and static conditions. Of course there must be a source of charges somewhere or there would be no field anywhere. The typical situation where solution of Laplace’s equation is relevant in electrostatic is where we have source-free non-conducting regions between statically charged conductors.
112
Example 2: Magnetostatic problems involving Laplace’s equation ~ = 0 and ∇ × B ~ = µ0 J~ + 12 ∂ E, ~ we see that the requireSince in general ∇ · B c ∂t ment for Laplace’s equation to be relevant is that J~ = 0 and ∂/∂t = 0, i.e. a current free region and static conditions. Again, there must be currents somewhere or there would be no fields anywhere. The typical situation is to be calculating the magnetic field in the non-conducting region between constant currents.
9.1
Uniqueness of the Solution of Laplace’s Equation
As we shall see below, general solutions of Laplace’s equation are in terms of some constants which are usually found from boundary (initial) conditions for a given problem. A question arises: What boundary conditions are appropriate for the Laplace equation to ensure that a unique and well-behaved (physically reasonable) solution will exist inside the bounded region? Our experience leads to believe that specification of the potential on a closed surface defines a unique potential problem. This is called Dirichlet theorem or Dirichlet boundary conditions.
9.1.1
Dirichlet Theorem
Consider a volume V completely bounded by a closed surface S. Within S there is a potential Φ satisfying ∇2 Φ = 0. The Dirichlet theorem says that the value of Φ inside the volume is uniquely determined if the value of the potential is specified everywhere on the whole boundary. Proof: Suppose, to the contrary, that there exist two solutions Φ1 and Φ2 satisfying the same boundary condition, i.e. ∇2 Φ1 = 0 and ∇2 Φ2 = 0 within S, but Φ1 = Φ2 on S.
113
Let U = Φ1 − Φ2 is the difference between the solutions. Since Φ1 and Φ2 are known to be solutions of the Laplace equation, then from the linearity of the ∇2 operator ∇2 U = 0, i.e. U is also a solution of the Laplace equation. We will prove that U = 0 inside the volume. To show this, we introduce a vector F~ = U ∇U . Then using the vector property that ∇ · F~ = ∇ · U ∇U = U ∇ · (∇U ) + ∇U · ∇U , and the Gauss’ Divergence Theorem, we get Z
∇ · F~ dV =
V
=
Z
Z
U ∇2 U dV +
V
=
U ∇ · ∇U dV +
V
I
Z
∇U · ∇U dV
V
Z
(∇U )2 dV
V
~ . U ∇U · dS
S
Now, the right-hand side of the above equation is equal to zero because U = 0 over S. Also, the integral Z
U ∇2 U dV = 0 ,
V
because Φ1 and Φ2 both satisfy the Laplace equation throughout V . Hence Z
(∇U )2 dV = 0 .
V
Since the integral of a positive function is always positive, ∇U must be zero for the integral to be zero. Thus ∇U = 0 and consequently, inside the volume V , the function U is constant. Since U = 0 on S, so that inside V , we have then that Φ1 = Φ2 everywhere, as required.
9.2
Solutions of Laplace’s Equation
There are different methods of solving the Laplace equation • Method of Images 114
• Green functions method • Variational method • Method of lattices • Numerical Monte-Carlo simulations method • Method of separation of variables • Solution in spherical coordinates We will illustrate last two methods which can be applied to a large class of problems in electromagnetism. The other methods can be applied to specific problems. For these methods it is necessary that the boundaries over which the potential is specified coincide with the constant bounding surfaces.
9.2.1
Method of Separation of Variables
In cartesian coordinates the Laplace equation for the scalar potential can be written as ∇2 Φ =
∂ 2Φ ∂ 2Φ ∂ 2Φ + + 2 =0. ∂x2 ∂y 2 ∂z
(29)
Since x, y, z are independent variables, the solution of the Laplace equation is of the form Φ(x, y, z) = X(x)Y (y)Z(z) . Substituting this into the Laplace equation and dividing both sides by XY Z, we obtain 1 d2 Y 1 d2 Z 1 d2 X + + =0. X dx2 Y dy 2 Z dz 2 This equation can be separated into three independent equations. To show this, we write this equation as 1 d2 X 1 d2 Y 1 d2 Z = − − . X dx2 Y dy 2 Z dz 2 115
Both sides of the above equation depend on different (independent) variables, thus are equal to a constant, say −α2 :
−
1 d2 X X dx2
= −α2 ,
1 d2 Y 1 d2 Z − Y dy 2 Z dz 2
= −α2 .
The second equation can be written as 1 d2 Y 1 d2 Z 2 = α − . Y dy 2 Z dz 2 Again, both sides depend on different variables, thus are equal to a constant, say −β 2 : 1 d2 Y = −β 2 , Y dy 2 1 d2 Z α2 − = −β 2 . Z dz 2 Hence, after the separation of the variables, we get three independent ordinary differential equations 1 d2 X + α2 = 0 , X dx2 1 d2 Y + β2 = 0 , Y dy 2 1 d2 Z − (α2 + β 2 ) = 0 . Z dz 2 The solutions of these equations depend on whether α2 and β 2 are positive or negative constants. If we choose α2 and β 2 to be positive, the solutions of the differential equations for X and Y are in the form of trigonometric functions whereas the solution for Z is in the form of a hyperbolic function X(x) =
X
Ak eiαx + Bk e−iαx
,
k
Y (y) =
X
Cl eiβy + Dl e−iβy
,
l
Z(z) =
X
√ √ α2 +β 2 z − α2 +β 2 z Ep e + Fp e .
p
116
If α2 and β 2 had been chosen as negative constants, the hyperbolic and trigonometric solutions would be interchanged. The solutions can be equally well written in the form X(x) =
X
[Ak sin(αx) + Bk cos(αx)] ,
k
Y (y) =
X
[Cl sin(βy) + Dl cos(βy)] ,
l
Z(z) =
X
Ep sinh
q
α2 + β 2 z + Fp cosh
q
α2 + β 2 z
.
p
The above solutions of the Laplace’s equation are in a general form valid for an arbitrary problem. The constants α, β, Ak , Bk , Cl , Dl , Ep and Fp are found from specific boundary conditions. Consider two examples of an application of the general solutions: 1. We have a solution with known boundary conditions, find the problem. 2. We have a problem with specific boundary conditions, find the solution. Example 1. Consider the following two-dimensional solution of the Laplace’s equation Φ(x, z) = X(x)Z(z) = V0 sin(αx) sinh(αz)
(30)
with the lower boundary Φmin = 0 and the upper boundary Φmax = V0 . In what circumstance is the above the solution? Consider Φ(x, z) in some limits. Note that X = 0 for x = 0 or αx = π, i.e. x = π/α. The solution thus satisfies the boundary conditions along the ”vertical” lines for α = π/b. Since sinh(αz) = 0 for z = 0, the boundary condition along the lower boundary is satisfied.
117
For the solution to satisfy the upper boundary condition, the shape of the upper boundary must be such that V0 sin(αx) sinh(αz) = V0 , for all points x, z on the line, i.e. πz πx sinh sin b b
=1.
Since sin πx → 0 at the edges, it is b Figure 34: greatest (= 1) at the center x = b/2. πz Hence, sinh b must be equal to one at x = b/2. This happens when πz = arc sinh(1) ≈ 0.885 , b from which we find z=
0.885b = 0.282b . π
Figure 34 shows the shape of the box inside which the potential is of the form (30).
Example 2. We usually have reverse problems to the above that we have a set of electrodes which constitute equipotential lines or surfaces, and need to find the appropriate solution of the Laplace equation. In this example we will illustrate this situation and we will try to find potential inside a rectangular box whose three sides have potential equal to zero, and the remaining side has a potential V0 . Consider a two-dimensional problem with boundary conditions shown in Figure 35. This two-dimensional problem has a general solution Φ(x, z) =
X
[An sin(αx) + Bn cos(αx)]
n
× [En sinh(αz) + Fn cosh(αz)] .
118
The boundary condition of Φ = 0 at the side x = 0 can be satisfied by Bn = 0. The boundary condition of Φ = 0 at the side z = 0 can be satisfied by Fn = 0. In order to have Φ = 0 at x = b, we must have αb = nπ. Hence with the boundary conditions along three sides of Φ = 0, the general solution reduces to
Figure 35:
Φ(x, z) =
∞ X
Kn sin
n=1
nπz nπx sinh b b
,
(31)
where Kn = An En . To find Kn we apply the remaining boundary condition that Φ = V0 at z = a: ∞ X
nπx nπa Kn sin Φ(x, a) = V0 = sinh b b n=1
.
(32)
This is a Fourier series in x and in the usual way we use the orthogonality properties of sine functions to calculate Kn : Z 2π
(
0 for m 6= n π for m = n
(
0 for m 6= n π for m = n
sin(mφ) sin(nφ) dφ =
0
Z 2π
cos(mφ) cos(nφ) dφ =
0
Z 2π
sin(mφ) cos(nφ) dφ = 0 for all m and n .
0
Thus, multiplying both sides of Eq. (32) by sin(mπx/b) and integrating over x = 0 → b, we get Z b 0
∞ X mπx nπa dx = Kn sinh V0 sin b b n=1 Z b mπx nπx × sin sin dx . b b 0
119
From the orthogonality of the sine functions, we find that all integrals on the right-hand side are equal to zero except for m = n. Thus, after performing the integration on the left-hand side, we are left with the term nπx − cos b
V0
b 0
b nπa Z b 2 nπx = Kn sinh sin dx , nπ b b 0
which we can write as V0 b nπa [1 − cos(nπ)] = Kn sinh nπ b Z b 2nπx 1 1 − cos dx . × b 0 2
The cos (2nπx/b) integrates to zero over the range 0 → b, so that the above equation becomes V0 b nπa b [1 − cos(nπ)] = Kn sinh , nπ b 2
from which we find the constant Kn : Kn =
2V0 1 − cos(nπ) . nπ sinh nπa b
The solution for Kn can be further simplified: Namely, if n is an even number, cos(nπ) = 1 and then Kn = 0. If n is an odd number, cos(nπ) = −1, and then 1 − cos(nπ) = 2. Hence, we finally arrive at Kn =
1 4V0 , nπ sinh nπa b
for odd n
and Kn = 0 for even n. Substituting Kn into Eq. (31), we find that the potential inside the box has the form
Φ(x, z) =
X odd n
nπz 4V0 sinh b nπx sin . nπ sinh nπa b b
120
As an exercise, check the correctness of this solution by showing that it satisfies the Laplace’s equation.
Exercise in class: Reduction of the solution of a three dimesional problem to a two dimensional problem Why for a two dimensional problem we simplify the general three-dimensional solution of the Laplace equations into Φ = X(x)Z(z) , or Φ = Y (y)Z(z) , but never Φ = X(x)Y (y) .
121
9.2.2
Solution of the Laplace Equation in Spherical Coordinates
In this lecture, we continue the discussion of boundary-value problems and will illustrate solution of the Laplace equation for general problems of spherical symmetry. In this case, we shall work in spherical coordinates (r, θ, φ), in which the Laplace equation takes the form ∂Φ 1 ∂ ∂Φ 1 ∂ 2Φ 1 ∂ r2 + 2 sin θ + 2 2 ∇ Φ= 2 =0. r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 !
!
2
The procedure we follow in the solution of the above equation is as follows: Multiplying both sides by r2 , the Laplace equation can be written as a sum of two separate parts ∂Φ 1 ∂ ∂Φ 1 ∂ 2Φ ∂ r2 + sin θ + =0. ∂r ∂r sin θ ∂θ ∂θ sin2 θ ∂φ2 !
!
The first part depends solely on r, whereas the second part depends solely on the angles θ, φ. Thus, the solution of this equation is of the separable form Φ = R(r)Y (θ, φ) . Hence, substituting Φ = R(r)Y (θ, φ) and dividing both sides by R(r)Y (θ, φ), we obtain dR 1 d r2 R dr dr
!
1 =− Y
"
1 ∂ ∂Y sin θ sin θ ∂θ ∂θ
!
1 ∂ 2Y + sin2 θ ∂φ2
#
.
Note, both sides of the above equation depend on different variables. It follows that the both sides must be equal to the same constant, say −α: !
d dR r2 + αR = 0 , dr dr ! 1 ∂ ∂Y 1 ∂ 2Y sin θ + − αY = 0 . sin θ ∂θ ∂θ sin2 θ ∂φ2 Thus, the Laplace equation splits into two independent differential equations. We will call them the radial part and angular part, respectively.
122
Angular part: Equation for Y . Multiplying both sides by sin2 θ, we get ∂ ∂Y sin θ sin θ ∂θ ∂θ
!
− α sin2 θY +
∂ 2Y =0. ∂φ2
This equation contains two separate parts, one dependent only on θ and the other dependent only on φ. Therefore, the solution will be of the form Y (θ, φ) = X(θ)Ψ(φ) . Hence, we get d dX 1 sin θ sin θ X dθ dθ
!
− α sin2 θ = −
1 d2 Ψ . Ψ dφ2
As before, both sides must be equal to a constant, say m2 : d dX 1 sin θ sin θ X dθ dθ 2 1dΨ = −m2 . Ψ dφ2
!
− α sin2 θ = m2 ,
First, we will solve the equation for Ψ, which we can write as d2 Ψ = −m2 Ψ . dφ2 It is the familiar differential equation for a harmonic motion. The solution of this equation is of simple exponent form: Ψ(φ) = A exp(imφ) , where A is a constant. To determine the constant m note that in rotation, φ and φ + 2π correspond to the same position in space: Ψ(φ) = Ψ(φ + 2π), which is satisfied when exp(imφ) = exp[im(φ + 2π)] . 123
This leads to exp(i2πm) = 1 , that is satisfied when m = 0, ±1, ±2, . . .. Hence, the constant m2 is not an arbitrary number, is an integer. Now, we will find X(θ) to complete the solution for Y . Using the solution for Ψ, the differential equation for X(θ) can be written as dX 1 d sin θ sin θ dθ dθ
!
m2 − α+ X=0. sin2 θ !
Introducing a new variable z = cos θ, we can rewrite this equation as
1−z
2
dX m2 − 2z − α + X=0, dz 2 dz 1 − z2
d2 X
!
or dX m2 d 1 − z2 − α+ X=0. dz dz 1 − z2 "
#
!
(33)
The above equation is known as the generalized Legendry differential equation, and its solutions are the associated Legendry polynomials. For m = 0, the equation is called the ordinary Legendry differential equation whose solution is given by the Legendry polynomials. Lets look into the solution procedure of the above equation. This will allow us to find α and X(θ). We assume that the whole range of z (cos θ), including the north and south poles (z = ±1), is in the region of interest. The desired solution should be single valued, finite, and continuous on the interval −1 ≤ z ≤ 1 in order to represent a physical potential.
124
The differential equation for X has poles at z = ±1. In order to find the solution of this equation, we first check what solution could be continuous near the poles. Lets check a possible solution near z = 1. Substituting x = 1 − z, then dx = −dz and in terms of x the equation takes a form dX m2 d x (2 − x) − α+ X=0. dx dx x(2 − x) "
#
!
We look for a solution in the trial form of power series in x X(x) = xs
∞ X
an x n .
n=0
Substituting this into the differential equation for X, we get 2s2 a0 xs−1 + (s + 1)(2sa1 − sa0 + 2a1 )xs + . . . ! m2 (a0 + a1 x + . . .)xs = 0 . − α+ x(2 − x) Near x ≈ 0, we can replace x(2 − x) by 2x, and obtain m2 a0 xs−1 + (. . .)xs . . . = 0 . 2s a0 − 2 !
2
This equation is satisfied for all x only if the coefficients at xs , xs±1 , . . . are zero. From this, we find that 1 s = ± |m| . 2 We take only s = + 12 |m| as for s = − 21 |m| the solution for X(x) at x = 0 would go to infinity. We require the solution to be finite at any point x. Thus, the solution that is continuous near x = 0 is of the form 1
X(x) = x 2 |m|
∞ X n=0
125
an x n ,
or in terms of z X(z) = (1 − z)
1 |m| 2
∞ X
a0n z n .
n=0
Using the same procedure, we can show that near the pole z = −1, the continuous solution is 1
X(z) = (1 + z) 2 |m|
∞ X
a00n z n .
n=0
Hence, we will try to find the solution in the form
X(z) = 1 − z 2
∞ 1 |m| X 2
bn z n .
(34)
n=0
What left is to determine the coefficiets bn . Substituting Eq. (34) into the differential equation for X(z), Eq. (33), and collecting all terms at the same powers of z n , we obtain X
{(n + 1)(n + 2)bn+2 − n(n − 1)bn
n
o
− 2(|m| + 1)nbn − (α + |m| + m2 )bn z n = 0 , from which we find a recurrence relation for the coefficients bn bn+2 =
(n + |m|)(n + |m| + 1) + α bn . (n + 1)(n + 2)
We have two separate solutions for even and odd n. For b0 6= 0, we put b1 = 0, and the solution is given in terms of even n. For b0 = 0, we put b1 6= 0, and the solution is given in terms of odd n. We cannot accept both the even and odd solutions at the same time, because in this case the solution X(z) would not be a single valued function that is would not be accepted as a physical potential. For example, for b0 6= 0, we have α = −|m| − m2 , but for b1 6= 0, we have α = −2 − 3|m| − m2 . If we would accept both of the solutions at the same time, the potential would have two different values.
126
We check now whether the series is converting when n → ∞ which would ensure that the potential is finite. Since bn+2 > bn , the series diverges for z = ±1. Therefore, in order to get the potential finite everywhere in the space, we have to terminate the series at some n = n0 . In other words, we assume that bn0 +1 = bn0 +2 = . . . = 0. The series terminating at n = n0 indicates that (n0 + |m|)(n0 + |m| + 1) + α = 0 . Introducing l = n0 + |m| , we see that l ≥ |m|, and α = −l(l + 1) ,
l = 0, 1, 2, . . .
Thus, the solution for X(z) is
Xlm (z) = 1 − z 2
1 |m| l−|m| X 2
bn z n ,
n
where the sum is over even n when l − |m| is an even number, and over odd n when l − |m| is an odd number. First few solutions X00 (z) = b0 = b0 P00 (z) , X10 (z) = b1 z = b1 P10 (z) , √ X11 (z) = b0 1 − z 2 = b0 P11 (z) , √ where P00 (z) = 1, P10 (z) = z, P11 (z) = 1 − z 2 , . . . are the associate Legendry polynomials of the order l. Useful examples of the associate Legendry polynomials, written in terms of θ, (z = cos θ): P00 (cos θ) = 1 , 127
P10 (cos θ) = cos θ , P11 (cos θ) = sin θ , 1 P20 (cos θ) = [3 cos(2θ) + 1] , 4 3 P21 (cos θ) = sin(2θ) , 2 3 [1 − cos(2θ)] . P22 (cos θ) = 2 An important property of the Legendry polynomials is orthogonality that Z 1 −1
Plm (cos θ)Pkn (cos θ) d(cos θ) = 0 for m 6= n and l 6= k , Z 1 −1
2 (l + m)! 2l + 1 (l − m)!
[Plm (cos θ)]2 d(cos θ) =
for m = n and l = k .
Finally, with the above notation, the solution for the angular part of the Laplace equation Y (θ, φ) is of the form Y (θ, φ) =
XX l
Alm Plm (cos θ)eimφ .
m
Radial part: Equation for R(r). With α = −l(l + 1), the differential equation for R takes the form !
dR d r2 − l(l + 1)R = 0 . dr dr Dividing by r and introducing a new function U (r) = rR(r), we obtain d2 U l(l + 1) − U =0. dr2 r2 Lets first check the asymptotic solution for r 1. In this case we can ignore the second term in the differential equation, and find that the asymptotic equation has a solution U (r 1) = Cr, where C is a constant. Following this asymptotic behavior, we will try the general solution of a form U (r) = rs . 128
Substituting this into the differential equation, we obtain [s(s − 1) − l(l + 1)] rs−2 = 0 . This equation is satisfied for any r when s = (l + 1)
or
s = −l .
Thus, the general solution for U that satisfies the asymptotic solution is of the form U (r) = C1 rl+1 + C2 r−l , and then R(r) = C1 rl + C2 r−(l+1) . Hence, the general solution of the Laplace equation in spherical polar coordinates is of the form: Φ(r, θ, φ) =
XX l
C1l rl + C2l r−(l+1) Alm Plm (cos θ)eimφ .
m
The solution can be written as Φ(r, θ, φ) =
X X n
C1l rl + C2l r−(l+1)
l
m
× [alm cos(mφ) + blm sin(mφ)] Plm (cos θ)} . This is the final solution for the potential of an arbitrary spherically symmetric problem. The potential is finite, determined and is continuous at each point (r, θ, φ).
Example 1: Potential outside a conducting sphere Consider an example of boundary-value problem with azimuthal symmetry: A conducting sphere of a radius a in an uniform electric field, as shown in Figure 36.
129
In order to find the potential, we first have to determine boundary conditions for the potential. There are two boundaries: One of the boundaries is the surface of the sphere and the other is the region at r → ∞. The conducting sphere is an equipotential volume (else there would be electric fields driving currents till it became so). We take, with no loss of generality, the zero potential on the sphere. Thus, the boundary conditions to be satisfied are: 1. The potential on the surface of the sphere Φ(a, θ, φ) = 0. 2. The potential at infinity is the uniform field potential (no effect of the ~ · ~r = sphere at infinity), so Φ = −E −Er cos θ at infinity. Since the applied potential is indepenFigure 36: A dielectric sphere in a dent of the angle φ, the induced potenuniform electric field. tial will also be independent of φ. This is a general property of the potential inside a bounded area that we shall explain in details at one of the tutorial sesions. Thus, using this property, we can set m = 0 in the general solution and get Φ(r, θ) =
Xh
i
C1l rl + C2l r−(l+1) Pl0 (cos θ) .
l
At infinity (r → ∞), the boundary condition is satisfied for all constants C1l = 0 except for l = 1 (remember P1 = cos θ). Thus Φ(r, θ) = C11 rP1 (cos θ) +
X l
130
C2l Pl (cos θ) . rl+1
We first find the explicit form of C11 . As r → ∞, the potential Φ(r, θ) → C11 r cos θ = −Er cos θ. Therefore C11 = −E. The remaining coefficients C2l we find from the other boundary condition that Φ(a, θ) = 0 on the surface of the sphere. Thus, at r = a Φ(a, θ) = 0 = −EaP1 (cos θ) +
X l
C2l Pl (cos θ) . al+1
We can determine the coefficients C2l using the orthogonality properties of the Legendry polynomials. Multiplying the above equation by Pk (cos θ) sin θ and integrating over sin θdθ = d(cos θ), we obtain Z 1
0 = −aE
−1
+
X l
P1 (cos θ)Pk (cos θ) d(cos θ)
C2l Z 1 Pl (cos θ)Pk (cos θ) d(cos θ) . al+1 −1
For k 6= 1 the first term vanishes by orthogonality of the Legendry polynomials. Moreover, all terms in the summation vanish except that for l = k. Thus, for k 6= 1 C2k Z 1 Pk (cos θ)Pk (cos θ) d(cos θ) . 0 = k+1 a −1 Since the integral is nonzero, then C2k = 0. For k = 1 and using the orthogonality property of the Legendry polynomials Z 1 −1
Plm (cos θ)Plm (cos θ) d(cos θ) =
2 (l + m)! , 2l + 1 (l − m)!
we find that for l = 1 and m = 0, the integral is equal to 2/3. Thus 2 C21 2 0 = −aE + 2 , 3 a 3 from which, we find C21 = Ea3 . Hence Φ(r, θ) = −Er cos θ + Ea3 131
cos θ . r2
The first term is just the potential of a uniform field E. The second term is the potential due to the induced surface charges or, equivalently, is the potential of the induced dipole moment p = 4πε0 Ea3 Φdip =
p cos θ . 4πε0 r2
This result shows that the sphere placed in an uniform electric field behaves as a dipole because of the effect of the charge distribution induced on its surface.
Exercise in class: Potential inside a sphere Let us suppose that the student would try to find Φ inside a sphere using the general solution for Φ. We have found before, using the Gauss’s law, that Φ = const. inside the sphere. The student immediately notice that the solutions appear to be mathematically different since that found by using the Gauss’s law is in a closed form while the general one is in the form of a series. The question to be settled is whether the two solutions are identical or whether, perhaps, only one of the solutions is correct?
132
Revision questions Question 1. What conditions are imposed on the field that is completely determined by a scalar potential? Question 2. State the Dirichlet theorem. Question 3. Explain the method of separation of variables on example of the Laplace equation in cartesian coordinates. Question 4. Explain the method of solving the generalized Legendry differential equation that satisfies the conditions imposed on the scalar potential.
133
Tutorial problems Problem 9.1 Electrostatic potential inside a rectangular box with symmetrical boundary conditions Find a series solution in rectangular harmonics for the electrostatic potential inside a 2-dimensional box in which the sides at y = 0 and y = b are at zero potential and the sides at z = a and z = −a are at potential V0 .
Problem 9.2 Potential inside a rectangular box Consider a two-dimensional region with boundaries at x = 0, b and z = 0, a, as shown in Figure 37. The boundary conditions are
Figure 37: ∂Φ = 0 ∂z Φ = 0 Φ = V0
at
z=0,
at
x = 0, b ,
at
z=a.
Find the potential at any point inside the two-dimensional region.
134
Problem 9.3 Potential inside an open rectangular box Find the potential everywhere inside a two dimensional region bounded from three sides by a finite plane at z = 0 and by semiinfinite planes located at x = 0 and x = b, as shown in Figure 38. The boundary conditions are Φ = 0
at
Φ = V0
x=0
and x = b ,
at z = 0 ,
Φ = 0
at
z→∞.
Figure 38: Problem 9.4 Potential inside a 3-dimensional box A 3-dimensional conducting box is defined by x = 0 to a, y = 0 to b and z = 0 to c. All the sides of the box are connected together and grounded except that side defined by z = c, which is insulated from the rest and held at potential V0 . Show that the potential within the hollow box can be written as the series Φ(x, y, z) =
sin
X X 16 V0 n
nmπ 2
m
odd odd
s
× sinh
135
sinh
mπ a
mπx a
q
2
sin
mπ 2 a
+
nπ b
+
2
nπy b nπ 2 c b
z .
Problem 9.5 Potential due to a thin circular ring (a) Show that for a system with axial symmetry such as a ring of charge, the electrostatic potential can be written as a series in spherical polar coordinates: Φ(r, θ) =
X
A` r` + B` r−(`+1) P` (cos θ) .
`
To determine the coefficients A` and B` consider the solution along the z-axis, where θ = 0. It is useful to know that P` (1) = 1 for any value of `. Thus, along the axis: Φ(z) =
X
A` z ` + B` z −(`+1)
.
`
(b) Calculate Φ(z) directly from the Coulomb potential. Expand this expression as a power series and hence find the A` and B` by comparison. Hence, show that for a ring of radius a and charge per unit length λ the potential everywhere can be written as the series:
Φ(r, θ) =
λ 1 + 2ε0
X
(−1) 1.3.5 . . . (` − 1)
`
even
2
` 2
` 2
` 2
! a`
r` P` (cos θ) ,
where ` = 2, 4, 6, . . . (even). The Legendre functions can be computed from the recurrence relation: (` + 1) P`+1 (x) = (2` + 1) x P` (x) − ` P`−1 (x) and knowing that P0 (x) = 1 and P1 (x) = x. However, the functions become rather tedious for hand calculations for large `, e.g. P10 (x) =
1 n −252 + 13860x2 − 120120x4 + 360360x6 1024 o −437580x8 + 184756x10 .
136
However, there is Matlab etc. (c) Calculate also the electric field components of the circular ~ = −∇Φ. ring using the relation E (There is a useful recurrence relation for the derivative of the Legendre function: (x2 − 1)
d P` (x) = `x P` (x) − ` P`−1 (x) . dx
See e.g. Abramowitz & Stegun ‘Handbook of Mathematical Functions’).
137
10
General Solution of the Maxwell’s Equations
In Chap. 6, we presented a method of solving the Maxwell’s equations for the EM fields propagating in vacuum. We were able to reduce the Maxwell’s ~ and B ~ alone. The equations equations into two differential equations for E were in the form of the wave equations whose the solutions are in the form of plane and transverse waves. In this lecture, we will present a method for general solution of the Maxwell’s equations in the presence of the sources, charges and currents that depend on position ~r and time t. The concept of vector and scalar potentials and gauges will be useful in the general solution.
10.1
Difficulty of the Direct Solution of Maxwell’s Equations
Consider the Maxwell’s equations for the fields in the presence of charges and currents I. II. III. IV.
~ = ρ/ε0 , ∇·E ~ =0, ∇·B ~ , ~ =−∂B ∇×E ∂t ~ = µ0 J~ + 1 ∂ E ~ . ∇×B c2 ∂t
(35) (36) (37) (38)
~ and B ~ depend on (~r, t), the charge In the Maxwell’s equations, the fields E and current densities also depend on (~r, t). It is not explicitly stated in the above equations, but we shall remember about this dependence in the following calculations. ~ and B ~ proLet us try to solve the Maxwell’s equations to find the fields E ~ duced by the source charges ρ and currents J. The Maxwell’s equations involve two fields that satisfy coupled differential equations. First, we will try to separate the Maxwell’s equations into a dif~ alone or B ~ alone. ferential equation for E
138
Assuming in the usual way that space and time operators commute, we act ∂ on III and ∇× on IV, and obtain from III: with c12 ∂t 2 1 ∂ ~+ 1 ∂ B ~ =0, ∇ × E c2 ∂t c2 ∂t2
and from IV: 1 ∂ ~ − ∇ × (∇ × B) ~ = −µ0 ∇ × J~ . ∇×E c2 ∂t ~ by subtraction of the two above equations, we get Eliminating E 2 ~ + 1 ∂ B ~ = µ0 ∇ × J~ . ∇ × (∇ × B) c2 ∂t2
Using the vector identity for double × product, and from II: ~ = −∇2 B ~ + ∇(∇ · B) ~ = −∇2 B ~ , ∇ × (∇ × B) we arrive to a wave equation ~− ∇2 B
1 ∂2 ~ B = −µ0 ∇ × J~ . c2 ∂t2
(39)
~ gives Similarly, elimination of B ~− ∇2 E
1 ∂2 ~ 1 ∂ E = ∇ρ + µ0 J~ . 2 2 c ∂t ε0 ∂t
(40)
Equations (39) and (40) are in the form of coupled wave equations known as inhomogeneous Helmholtz equations. We see that the current density J~ enters into both of these equations and enters in a relatively complicated way. ~ and for this reason these equations These equations are coupled through J, and are not readily soluble in general. In the absence of currents and charges, J~ = 0, ρ = 0, and then the above equations describe a free EM field, and can be solved separately, as it was illustrated in Chap. 6. The general solution of the wave equations, in the presence of space and time varying currents and charges is complicated and is more readily attained via the electromagnetic potentials.
139
10.2
Scalar and Vector Potentials
~ and B ~ directly by integration of Eqs. (39) Generally, we do not find fields E and (40). We rather first compute scalar and vector potentials from which the fields may be found. We will illustrate here the advantage of working with the potentials, i.e. with the the scalar and vector potentials. ~ defined such that the Maxwell’s equation II Introduce the vector potential A ~ always has zero divergence, ∇ · B ~ = 0, and remains unchanged. The field B hence we can always write ~ =∇×A ~, B
(41)
~ is identically zero. Substitute this relation to the Maxwell’s since ∇ · ∇ × A equation III, we obtain
~ ~ = ∇ × − ∂ A . ~ =−∂ ∇×A ∇×E ∂t ∂t
(42)
The two curl’s are equal, but it does not mean that the vectors under the curls are equal. Since always ∇ × ∇Φ = 0, where Φ is an arbitrary scalar function, the two vectors are equal with the accuracy to ∇Φ: ~ − ∇Φ . ~ =−∂A E ∂t
(43)
Scalar potential is a quantity from which a field can be derived by a process of differentiation, e.g. in electrostatics ~ = −∇Φ , E where Φ is the electrostatic potential. ~ In the static limit of ∂ A/∂t = 0, the scalar function Φ reduces to the familiar electrostatic potential. Equation (43) shows that the electric field depends on the specific choice of ~ and Φ and still get the same E. ~ One can the potentials. We can change A ~ so Eqs. (41) and (43) are object that Eq. (41) ensure a fixed value for A, 140
~ and Φ. However, we can define new potentials without satisfied for fixed A ~ and B ~ changing E ~0 = A ~ + ∇Λ , A ∂ Φ0 = Φ − Λ . ∂t
(44)
Proof: From Eq. (43), we find that ~0 = − ∂ A ~ 0 − ∇Φ0 E ∂t ! ∂ ∂ ~ A + ∇Λ − ∇ Φ − Λ = − ∂t ∂t ∂ ~ ~ . = − A − ∇Φ = E ∂t Similarly, fro Eq.(41), we find that ~0 = ∇ × A ~0 = ∇ × A ~ + ∇ × (∇Λ) = ∇ × A ~=B ~ . B as required. The transformation (44) is called a gauge transformation, and the invariance of the fields under such transformations is called gauge invariance. ~ In this case, how do we completely determine A? ~ = ∇×A ~ does From the Helmholtz theorem we know that the definition B ~ despite the fact that B ~ is completely defined. The not completely define A ~ vector potential A is arbitrary to the extent that the gradient of some scalar function can be added. Thus, infinite set of possible potentials Φ corresponds to an infinite set of possible vector potentials. Recall the Helmholtz Theorem which says that any vector field can be written as a sum two terms Z 1 Z ∇ · F~ 1 ∇ × F~ ~ F = − ∇ dV + ∇× dV 4π r 4π r V V = F~l + F~t , 141
where F~l is called the longitudinal part of the field and it has ∇ × F~l = 0, while F~t is called the transverse part as it has ∇ · F~t = 0. We see that ∇ · F~ and ∇ × F~ together determine F~ but neither do alone. ~ we complete the definition of A. ~ This is called Thus, if we define ∇ · A, ”choosing the gauge of the potential”. The above is an excellent illustration of the power of the Helmholtz theorem. This theorem enables us to recognize basic common properties of vector fields independent of their individual physical properties. 10.2.1
Lorenz Gauge
~ How do we define ∇ · A? It is done as follows. We take ∇· of Eq. (43) and obtain ~ − ∇2 Φ . ~ =−∂ ∇·A ∇·E ∂t ~ will satisfy the Maxwell’s equation I when Thus, the electric field E ∂ ~ − ∇2 Φ = ρ/ε0 . ∇·A ∂t ~ = ∇ × A, ~ we have From the Maxwell’s equation IV and B −
(45)
(46)
!
~ = µ0 J~ + 1 ∂ − ∂ A ~ − ∇Φ ∇ × (∇ × A) c2 ∂t ∂t
,
which can be written as 2 ~− 1∇∂Φ, ~ + ∇(∇ · A) ~ = µ0 J~ − 1 ∂ A −∇2 A 2 2 c ∂t c2 ∂t
or 2 ~− 1 ∂ A ~ = −µ0 J~ + ∇ ∇ · A ~+ 1 ∂Φ ∇A 2 2 c ∂t c2 ∂t 2
!
.
(47)
~ and Φ means that we can choose a set of potentials The freedom of choosing A to satisfy the condition ~+ 1 ∂Φ=0. ∇·A c2 ∂t 142
~ This equation is sometimes This is called the Lorenz gauge and defines ∇ · A. called the Lorenz equation. Under the Lorenz gauge, Eq. (47) reduces to ~− ∇2 A
1 ∂2 ~ A = −µ0 J~ , c2 ∂t2
and applying the Lorenz gauge to Eq. (46), we get 1 ∂ 2Φ ∇ Φ − 2 2 = −ρ/ε0 . c ∂t 2
We see the advantage of using the vector and scalar potentials. In terms of the potentials, the the Maxwell equations reduce to two uncoupled wave equations that can be solved separately. The equations also show that the Lorenz gauge is consistent with our experience on the sources of the EM fields. The source of the vector potential that is related to the magnetic field is a current density, and the source of the scalar potential that is related to charges is a charge density. 10.2.2
Coulomb Gauge
Another useful gauge of the potentials is the Coulomb gauge or transverse gauge ~=0. ∇·A The origin of the name ”Coulomb gauge” is in equation (45) that under the ~ = 0 reduces to the Poisson equation condition ∇ · A ∇2 Φ = −ρ/ε0 , that determines the Coulomb potential due to the charge density ρ. Where the name transverse gauge came from? Before we give the answer to this question, we first show that the solution of the Poisson equation is of the form 1 Z ρ dV . Φ(r) = 4πε0 r 143
It can be proved in the following way. From the Coulomb’s law ~ = E
r 1 Z ρˆ dV . 4πε0 r2
and using the relation rˆ 1 −∇ = 2 , r r we can write the electric field as Z ρ 1 ~ ∇ dV = −∇Φ , E=− 4πε0 r where 1 Z ρ dV . Φ= 4πε0 r
(48)
Since the electric field satisfies the Maxwell’s equation I, we find ~ = −∇2 Φ = ρ , ∇·E ε0 as required. ~ under the Coulomb gauge Now, we can find the wave equation for A 1 ∂2 ~ 1 ∂Φ A = −µ0 J~ + 2 ∇ . (49) 2 2 c ∂t c ∂t In this equation, the term involving the scalar potential is called ”longitudinal” as it has vanishing ∇×. This suggests that it may cancel the longitudinal ~ part of the current density J. ~− ∇2 A
Let us check if the longitudinal part of the current density can be expressed in terms of the scalar potential. According to the Helmholtz Theorem, the current density can be written as Z 1 Z ∇ · J~ 1 ∇ × J~ ~ J = − ∇ dV + ∇× dV 4π r 4π r V V
= J~l + J~t . 144
Using the continuity equation ∂ρ + ∇ · J~ = 0 ∂t and the solution of the Poisson equation, we find the longitudinal part of the current density 1 Z − ∂ρ 1 Z ∇ · J~ ∂t dV = − ∇ dV J~l = − ∇ 4π r 4π r V V 1 ∂ Z ρ ∂ = ∇ dV = ε0 ∇Φ . 4π ∂t ∂t V r Then 1 ∂Φ ∂ , µ0 J~l = µ0 ε0 ∇Φ = 2 ∇ ∂t c ∂t which is equal to the second term on the right hand side of Eq. (49). Hence, the inhomogeneous term in the wave equation (49) can be expressed ~ entirely in terms of the transverse current and then the wave equation for A reduces to 1 ∂2 ~ ~ ∇ A − 2 2 A = −µ0 J~t . c ∂t 2
This explains the origin of the name ”transverse gauge”.15
10.3
Solution of the Inhomogeneous Wave Equations
We have shown that the Maxwell equations can be reduced to two indepen~ and Φ. In fact, we have four scalar dent wave equations for the potentials A equations for (Ax , Ay , Az , Φ). Each of these equations has the same form. Therefore, it is enough to solve one of the four equations. 15
The transverse gauge is often used in atomic physics to calculate the EM fields produced by orbiting electrons. In this case, the orbiting electrons produce a current that is solenoidal.
145
We will illustrate the solution on the equation for Φ: 1 ∂ 2Φ ∇ Φ − 2 2 = −ρ/ε0 . c ∂t 2
(50)
A general solution of the above equation may be found by considering two limiting cases: (a)
Electrostatic limit: ∂/∂t ≡ 0
In this limit the wave equation for Φ reduces to the Poisson equation whose the solution is 1 Z ρ(r0 ) dV 0 . Φ(r) = 4πε0 r0
(b) Source free limit:
ρ=0
In this case, the wave equation (50) reduces to the homogeneous equation ∇2 Φ −
1 ∂ 2Φ =0. c2 ∂t2
This equation has a spherically symmetric solution of the form Φ(r, t) =
f (t − r/c) , r
where f (t − r/c) is an arbitrary function of the retarded time t − r/c. The retardation r/c is equal to the time needed for the electromagnetic wave to pass the distance from the source to a given point in space. Proof: If there are no charged (boundary) surfaces in the space, the potential can depend only on r, and must in fact be spherically symmetric. Thus, in spherical coordinates only the radial part of the Laplacian will contribute to the wave equation "
∂Φ 1 ∂ ∇ Φ= 2 r2 r ∂r ∂r 2
146
#
.
Since ∂f ∂f ∂tr 1 = = − f0 , ∂r ∂tr ∂r c where tr = t − r/c and f 0 = ∂f /∂tr , we have f0 f 1 ∂ rf 0 1 ∂ r2 + 2 =− 2 +f ∇Φ = − 2 r ∂r cr r r ∂r c " # 1 f0 r 1 00 1 0 1 = − 2 + − f + − f = 2 f 00 , r c c c c rc "
!#
!
2
where f 00 = ∂ 2 f /∂t2r . Moreover, ∂/∂t = ∂/∂tr , and then 1 1 ∂ 2Φ = 2 f 00 . 2 2 c ∂t cr Consequently, we obtain ∇2 Φ =
1 ∂ 2Φ 1 1 = 2 f 00 − 2 f 00 = 0 , 2 2 c ∂t cr cr
as required. Summarizing the above analysis, we can construct a general solution of the wave equation by noting that it must represent a spherical wave outside the source and reduce to the appropriate static limit. This solution is Φ(r, t) =
1 Z ρ(t − r/c) dV , 4πε0 V r
where r is the distance coordinate from the source (from the charge ρdV ) at the time when the potential wave left it. This exhibits the causal behavior associated with the wave disturbance. The argument of ρ shows that an effect observed at the point r at time t is caused by the action of the source a distant r away at an earlier or retarded time tr = t − r/c. The time r/c is the time of propagation of the disturbance from the source to the point r. Thus, the Maxwell’s equations satisfy the causality principle.
147
10.4
Rigorous Solution: Green Functions Method
The wave equations all have the basic structure 1 ∂ 2 Φ(r, t) = −4πf (r, t) , c2 ∂t2 where f (r, t) is a known (source distribution) function. ∇2 Φ(r, t) −
To solve this equation, we will introduce the Green function of the equation and solve it as an inhomogeneous Helmholtz equation. Suppose that Φ(r, t) and f (r, t) have the Fourier integrals Φ(r, ω) = f (r, ω) =
Z ∞ −∞ Z ∞
Φ(r, t)eiωt dt , f (r, t)eiωt dt .
−∞
When we insert it into the wave equation, we find that the Fourier transform Φ(r, ω) satisfies the inhomogeneous Helmholtz wave equation
∇2 + k 2 Φ(r, ω) = −4πf (r, ω) ,
where k = ω/c is the wave number. The advantage of working in Fourier components is to remove the derivative over time, and consequently to reduce the wave equation to a differential equation involving the space variables only. Green function For a unit point source the potential satisfies the Poisson equation 1 = −4πδ(r) . r The function 1/r = G(r) is called a Green function of the above differential equation. In analogy, we can define the Green function of the wave equation ∇2
1 ∂2 ∇ − 2 2 G(r, t − t0 ) = −4πδ(r)δ(t − t0 ) . c ∂t !
2
148
The Fourier transform gives
∇2 + k 2 Gk = −4πδ(r)eiωt0 .
where Gk is the Fourier transform of the Green function G(r, t − t0 ), which we are trying to find. If there are no boundary surfaces, the Green function depends only on r, and then the Laplacian operator in spherical coordinates depends only on r giving 1 d2 −iωt0 rG e + k 2 Gk e−iωt0 = −4πδ(r) . k r dr2
Everywhere except r = 0, the function rGk e−iωt0 satisfies the homogeneous equation d2 −iωt0 rG e + k 2 (rGk e−iωt0 ) = 0 , k dr2
whose the solution is rGk e−iωt0 = Aeikr + Be−ikr . In this general solution for the Green function we could equally well choose the exponential form Gk =
e±ikr eiωt0 . r
Using the inverse Fourier transform, we find G(r, τ ) =
1 Z ∞ e±ikr −iωτ e dω , 2π −∞ r
where τ = t − t0 . The integral 1 Z ∞ −iω(τ ∓r/c) e dω 2π −∞ 149
is the delta function δ(τ ∓ r/c). Thus 1 G(r, τ ) = δ(τ ∓ r/c) . r The Green function is a casual response function, and has the same property as the scalar potential of a point source.
In summary of the general solution of the Maxwell’s equations The general (retarded) solutions of the Maxwell’s equations are given in terms of the vector and scalar potentials ~ =∇×A ~ , B
~ ~ = −∇Φ − ∂ A , E ∂t
(51)
with 1 Z ρ(t − r/c) dV , 4πε0 r ~ − r/c) 1 Z J(t ~ A(r, t) = dV , 4πε0 c2 r Φ(r, t) =
(52) (53)
and the potentials satisfy the Lorenz gauge. Thus, we do not find the fields by a direct integration of the Maxwell’s equations. If the charge and current distributions are known, we first calculate the scalar and vector potentials from Eqs. (52) and (53), and then find the electric and magnetic fields from Eqs. (51).
Revision questions Question 1. Do we find the fields by a direct integration of the Maxwell’s equations? Explain.
150
Question 2. Explain the usefulness of the scalar and vector potentials in the solution of the Maxwell’s equations. Question 3. Why do we use gauges in the solution of the Maxwell’s equations? Question 4. Explain how do we solve inhomogeneous wave equations. Question 5. What is the Green function of a given differential equation?
151
11
Electromagnetic Antennas: Hertzian Dipole
The electromagnetic fields of charges in uniform motion are effectively bound to the charges. The fields of accelerated (oscillating) charges on the other hand can propagate as electromagnetic (EM) waves at the speed c and can have a life of their own (until absorbed by some other charges). In this lecture we will show how electromagnetic waves are generated by oscillating charges. This will also illustrate an application of the general solution of the Maxwell’s equations in calculations of the electric and magnetic fields produced by a source system containing time varying charges and currents. First, we will show that this problem can be solved with the help of only ~ Next, we will apply this concept to the problem of the vector potential A. generation of electromagnetic waves. Consider the retarded solutions of the Maxwell’s equations ~ =∇×A ~ , B
~ ~ = −∇Φ − ∂ A , E ∂t
(54)
with 1 Z ρ(t − r/c) Φ ≡ Φ(r, t) = dV , 4πε0 r ~ ≡ A(r, ~ t) = A
~ − r/c) 1 Z J(t dV . 4πε0 c2 r
The above solution holds for the Lorenz gauge in which ~=− ∇·A
1 ∂Φ . c2 ∂t
Assume that the position and time variations of the charges and currents can be separated, so that the charges and currents vary sinusoidally in time ρ(r, t) = ρ(r)eiωt
iωt ~ t) = J(r)e ~ and J(r, .
In this case, the Lorenz gauge takes the form ~ = − iω Φ , ∇·A c2 152
which gives c2 ~. Φ=− ∇·A iω
(55)
~ Thus, the scalar potential can be expressed in terms of the vector potential A. In other words, the scalar potential can be eliminated from the field equations ~ As a result, we can express both E ~ and B ~ leaving only the dependence on A. ~ in terms of the vector potential A alone. Substituting Eq. (55) into Eq. (54), we obtain 2 ~ ~ − ∂A , ~ =∇×A ~. ~ = c ∇(∇ · A) B (56) E iω ∂t
Hence, both the electric and magnetic fields can be found from the vector ~ potential A. This result may seem rather strange at first, since normally we should expect to need both the scalar and vector potentials in order to completely determine the EM fields. The explanation and in fact an another way of saying the same thing is that time varying charge must satisfy the continuity equation ∂ρ = −iωρ , ∇ · J~ = − ∂t so that ∇ · J~ ρ=− . iω We see that the time varying charge density can be expressed in terms of the current density. Then the scalar potential becomes 1 Z ρ(t − r/c) 1 Z ∇ · J~ Φ(r, t) = dV = − dV 4πε0 r 4πε0 iω r Z ~ 1 J c2 ~. = − ∇· dV = − ∇ · A 4πε0 iω r iω Thus, specification of J~ alone is sufficient to completely determine all sources ~ in terms of J~ conof moving (oscillating) charges, and hence a solution for A tains all the necessary information to completely specify the time-varying EM fields.
153
11.1
Generation of electromagnetic waves
We have learnt that the introduction of the concept of displacement current by Maxwell led to the prediction of electromagnetic waves in vacuum. Now, we inquire into the sources of electromagnetic waves, i.e. how to generate EM waves of different wavelengths and different properties, how to control these properties and how to propagate these waves in desired directions. 11.1.1
Field of an Element of Alternating Current
Consider a linear element ∆l carrying an alternating current I = I0 exp(iωt), as shown in Figure 39. The current element may be viewed as two charges Q and −Q oscillating back and forth and can be served as an antenna, i.e. a source of electromagnetic waves. Assume that ∆l is much smaller than the wavelength λ = 2πc/ω corresponding to the frequency of the oscillation. In this case, we can ignore the phase variation of the current along ∆l. Of course, this is an approximation that may not hold for many realistic antennas. However, an understanding of the properties of such an antenna is of great interest since, in principle, all radiating structures can be considered as a sum of small radiating eleFigure 39: A linear alternating cur- ments. Moreover, many practical anrent element ∆l oriented in the z direction. The generated fields are calcu- tennas working at low frequencies are very short compared with the wavelated in the ~r direction. length. For the fields around a small current element, ∆l λ, there are three spatial regions (zones) of interest: • The near field (static) zone:
∆l r λ
• The intermediate field (induction) zone: • The far field (radiation) zone:
∆l r ∼ λ
∆l λ r 154
We will see that the fields have different properties in the different zones. In the near zone the fields have the character of static fields, with a strong dependence on the properties of the source. In the far field zone, the fields are transverse to the radius vector and fall of as r−1 , typical of radiation fields, and are independent of the source. ~ and B ~ fields around the current element. We start by Let us calculate the E considering the retarded current element that cab be written as ~ , ~ − r/c)dV = I(t ~ − r/c)dl = I0 ei(ωt−kr) dl J(t where k = ω/c. Since the current is the same at any point of the antenna (∆l λ), it ~ and then the vector potential appears as a constant for the integartion in A, becomes ~= A
I0 ei(ωt−kr) ~ ∆l , 4πε0 c2 r
~ ~ = R dl. where ∆l In the near zone, where r λ (or kr 1), the exponent exp(−ikr) can be replaced by unity. In the far field zone kr 1, the exponential oscillates rapidly, and in this region it is sufficient to approximate exp(−ikr) ≈ 1−ikr. In the intermediate zone, all powers of kr must be retained. We now calculate directions and magnitudes of the electric and magnetic fields produced by the current element. The fields are most easily evaluated in spherical polar coordinates, if we choose the direction of the current ele~ along the z-axis, i.e. ∆l ~ = ∆l~k. We then have that in the spherical ment ∆l ~ has components coordinate system, the vector potential A Ar = Az cos θ =
I0 ∆l cos θ ei(ωt−kr) , 4πε0 c2 r
Aθ = −Az sin θ = −
I0 ∆l sin θ ei(ωt−kr) , 4πε0 c2 r
Aφ = 0 . 155
(57)
~ and B, ~ we have to calculate ∇ · A ~ According to Eq. (56), to find the fields E ~ that in spherical polar coordinates are given by and ∇ × A, ~= ∇·A
1 ∂ (Aθ sin θ) 1 ∂Aφ 1 ∂ (r2 Ar ) + + , 2 r ∂r r sin θ ∂θ r sin θ ∂φ
and rˆ ∂ (Aφ sin θ) ∂Aθ θˆ 1 ∂Ar ∂ (rAφ ) − + − r sin θ ∂θ ∂φ r sin θ ∂φ ∂r "
~ = ∇×A
#
φˆ ∂ (rAθ ) ∂Ar + − r ∂r ∂θ "
"
#
#
.
Since Aφ = 0 and there is no φ dependence of Ar and Aθ , i.e. ∂Ar,θ /∂φ = 0, the above equations reduce to ~ = ∇·A
1 ∂ (r2 Ar ) 1 ∂ (Aθ sin θ) + , 2 r ∂r r sin θ ∂θ "
~ = ∇×A
∂ (rAθ ) ∂Ar φˆ − . ∂r ∂θ r
(58)
#
(59)
~ Magnetic field B Calculate first direction and magnitude of the magnetic field produced by the ~ = ∇ × A, ~ we easily find from Eq. (59) that the magnetic antenna. Since, B field of the current element is ~ =∇×A ~ = Bφ φˆ , B where "
1 ∂ (rAθ ) ∂Ar Bφ = − r ∂r ∂θ
#
,
(60)
and Br = Bθ = 0. Conclusion: The magnetic field produced by the antenna is perpendicular to the radius vector at all distances.
156
Calculate the magnitude Bφ . Substituting Eq. (57) into Eq. (60), we obtain
Bφ
i(ωt−kr) −I0 ∆l ∂ sin θe ei(ωt−kr) ∂ cos θ = + 4πε0 c2 r ∂r r ∂θ
ei(ωt−kr) I0 ∆l i(ωt−kr) ik sin θe + sin θ = 4πε0 c2 r r "
#
.
Hence "
#
I0 ∆l ik 1 Bφ = + sin θei(ωt−kr) . 4πε0 c2 r r2 A number of interesting general conclusions follow from this equation. In the first place, we note that the magnetic field is composed of two terms: the near zone term ∼ 1/r2 and the far zone term ∼ 1/r. Secondly, we note that in the limit of ω → 0, the magnetic field is composed of only the near zone term that is the familiar Biot-Savart formula. Finally, the most important is that the far zone term is only present for an oscillating field (ω 6= 0) and therefore it represents a radiation field arising from accelerated (oscillating) charge. ~ Electric field E We now calculate direction and magnitude of the electric field produced by the antenna, which as we have shown before can be found from the vector potential 2 ~ ~ − ∂A , ~ = c ∇(∇ · A) E iω ∂t
where ~ ∂A ~ = iω Ar rˆ + Aθ θˆ . = iω A ∂t
~ Substituting Eq. (57) into Eq. (58), we get We first calculate ∇ · A. ~ = ∇·A
1 ∂ (r2 Ar ) 1 ∂ (sin θAθ ) + 2 r ∂r r sin θ ∂θ 157
! i(ωt−kr) i(ωt−kr) 1 ∂ r cos θe 1 ∂ I0 ∆l 2 e − sin θ = 4πε0 c2 r2 ∂r r sin θ ∂θ r (
I0 ∆l = 4πε0 c2 I0 ∆l = − 4πε0 c2
i cos θ h i(ωt−kr) 2 sin θ cos θei(ωt−kr) i(ωt−kr) e − ikre − r2 r2 sin θ # " 1 ik cos θei(ωt−kr) . + r2 r
)
~ Since in spherical coordinates Next, we take gradient of ∇ · A. ∇ = rˆ
∂ 1 ∂ φˆ ∂ + θˆ + , ∂r r ∂θ r sin θ ∂φ
we obtain for the components of the gradient: h
h
h
~ ∇ ∇·A
r
~ ∇ ∇·A
"
i
~ ∇ ∇·A
I0 ∆l cos θ ik k 2 ik i(ωt−kr) 2 = − − + − 2 e − 4πε0 c2 r3 r2 r r # " 2ik k 2 i(ωt−kr) I0 ∆l cos θ 2 + − e , = 4πε0 c2 r3 r2 r "
i
θ
i φ
#
#
I0 ∆l 1 ik = + 2 sin θ ei(ωt−kr) , 2 3 4πε0 c r r = 0.
Hence, the radial part of the electric field, Er , is
Er
i ~ c2 h ~ − ∂A = ∇(∇ · A) r iω ∂t
r
I0 ∆l cos θ 2 2ik k 2 i(ωt−kr) I0 ∆l cos θ ei(ωt−kr) = + − e − iω 4πε0 iω r3 r2 r 4πε0 c2 r " # I0 ∆l cos θ 2 2 ik iω i(ωt−kr) = + 2+ − e . 3 4πε0 c ikr r r cr "
#
Since k = ω/c, the 1/r terms cancel and then Er simplifies to I0 ∆l cos θ 2 2 + 2 ei(ωt−kr) . 3 4πε0 c ikr r
Er =
158
(61)
Similarly, we find the polar component of E as
Eθ
i ~ c2 h ~ − ∂A = ∇(∇ · A) θ iω ∂t
θ
I0 ∆l sin θ ei(ωt−kr) I0 ∆l sin θ 1 ik i(ωt−kr) = + e + iω 4πε0 iω r3 r2 4πε0 c2 r " # 1 I0 ∆l sin θ 1 ik i(ωt−kr) e . = + 2+ 3 4πε0 c ikr r r "
#
(62)
The azimuthal component of the electric field Eφ = 0. Note that the radial part of the electric field, Eq. (61), contributes only to the near and intermediate zones, whereas the angular polar part, Eq. (62), contributes to all of the zones. Theorem: The near-zone 1/r3 part is the Coulomb type contribution. It is similar in nature to a static field surrounding a small linear-current element and an electric dipole. Proof: The Coulomb or static field is for ω → 0. In this limit the 1/r3 contribution is I0 ∆l sin θ 1 i(ωt−kr) e 4πε0 c ikr3 I0 ∆l sin θ 1 1 2 1 − ikr + = (−ikr) + · · · 4πε0 c ikr3 2 I0 ∆l sin θ 1 = − , 4πε0 c r2
Eθ =
where we have taken only the real (physical) part of the field. Since I0 = ∆q/∆t and ∆l/∆t = c, we get Eθ = −
∆q sin θ 1 , 4πε0 r2 159
as required. Electric and magnetic fields in near and far field zones Consider first the near field zone (r λ). In this limit the magnetic and electric fields are ~ near = I0 ∆l 1 sin θei(ωt−kr) φˆ , B 4πε0 c2 r2 ˆ . ~ near = −i I0 ∆l 2 ei(ωt−kr) cos θˆ r + sin θ θ E 4πε0 c kr3 Since the magnetic field is real and the electric field is imaginary, the Pointing vector involving the near-zone field components is a pure imaginary quantity. It does not represent any flow of energy. This imaginary quantity represents energy that oscillates back and forth between the source and the region of space surrounding the source. Consider now the far zone or radiation components of the magnetic and electric fields. I0 ∆l sin θ ik i(ωt−kr) ~ rad = EθR θˆ , e , E EθR = 4πε0 c r I0 ∆l sin θ ik i(ωt−kr) ~ rad = BφR φˆ , B BφR = e . (63) 4πε0 c2 r Note the following properties of the radiation components: 1. The electric and magnetic fields oscillate in phase. 2. The electric and magnetic fields are orthogonal to each other. 3. The ratio
EθR R Bφ
= c, the value for plane waves in free space.
4. The electric and magnetic fields are transverse to the radius vector at all distances. ~ = c2 ε 0 E ~ rad × B ~ rad is a real quantity and is in 5. The Poynting vector N the direction of the radius vector, indicating that the energy of the field propagates away from the current element. These properties show that in the far zone the EM field is in a form of plane waves propagating with the speed of light c.
160
11.2
Power Radiated from the Current Element
Consider now the radiation power emitted by the antenna. By the radiation power we mean that part of energy which is carried by the radiation components of the field. We will show that the radiation power is equal to the radiation losses, i.e. energy carried by the plane electromagnetic wave that propagates on its own independent of the source. The power flux at any point is given by the Poynting vector ~ = c2 ε 0 E ~ ×B ~ . N Then, the total power radiated across a sphere of radius r is W =
I
~ , ~ · dS N
S
where dS = r2 sin θ dθdφ . ~ ×B ~ which vary as 1/r2 will have net radiOnly those partial products in E ated power. The other partial products are small as they fall off more rapidly than 1/r2 . Thus, the only part of the fields entering into the expression for the radiated power is the far field zone part (radiation component) consisting of the terms varying as 1/r. ~ and B ~ oscillate in phase as sine or cosine The radiation components of E functions. Thus, an average of their product over time is E R¯B R =
where EθR
0
and BφR
0
1 R R E B , 2 θ 0 φ 0
are amplitudes of EθR and BφR .
On substituting from Eq. (63), we find that the time averaged Poynting vector is 2 2 2 ¯ = 1 c2 ε0 I0 ∆l k sin2 θ N 2 16π 2 ε20 c3 r2
I02 ∆l2 4π 2 I02 ∆l 2 sin θ = = 32π 2 ε0 c λ2 r2 8ε0 c λ "
161
#2
sin2 θ . r2
Hence, integrating over all directions, we get the total power emitted by the antenna W =
Z π Z 2π 0
0
=
Z 2π 0
¯ sin θ dθdφ r2 N
I02 ∆l 8ε0 c λ "
#2
dφ
Z π
sin3 θdθ .
0
Performing the integration over φ that is equal to 2π, and over θ that is equal to 4/3, we get I 2 ∆l W = 2π 0 8ε0 c λ "
#2
4 πI02 ∆l = 3 3ε0 c λ "
#2
.
We can write the total power radiated in terms of the power absorbed in an equivalent resistance, called the radiation resistance, as "
1 2π ∆l W = 2 3ε0 c λ
#2
1 I02 = RI02 . 2
where "
2π ∆l R= 3ε0 c λ
#2
is the radiation resistance. Since 1/(ε0 c) =
q
µ0 /ε0 = 377, or 120π, we obtain for R: "
R = 80π
2
∆l λ
#2
.
For example, if ∆l/λ ≈ 0.1, then R = 0.8π 2 ≈ 8 ohms. This example shows that for a current element which is 10% of the wavelength long, the resistance is very small. Thus, if ∆l/λ 1, the radiation losses are negligible, that the radiated power is very small. In terms of the emitted radiation, the emitted EM waves are very weak in power.
162
With a short linear current element, an appreciable power would be radiated only if the current amplitude I0 were very large. A large current, on the other hand, would lead to large amounts of power dissipation in the conductor, and hence a very low efficiency. We can conclude that current carrying systems that have linear dimensions small compared with the wavelength radiate negligible power. An efficient antenna should have dimensions comparable to or greater than the wavelength.
11.3
Gain of the Dipole Antenna
A further property of the dipole antenna that is worthy of consideration is the directional property of power radiated in different directions. The gain or directivity function of a transmitting antenna is the ratio of the Poynting flux to the flux due to an isotropic radiator emitting the same total power W : N (θ, φ) , Niso
gT (θ, φ) = where Niso =
W 4πr2
is the energy flux uniform in all directions. Since for the infinitesimal dipole: I 2 ∆l N (θ, φ) = 0 8ε0 c λ "
#2
sin2 θ , r2
and πI02 ∆l W = 3ε0 c λ "
163
#2
,
we obtain gT (θ, φ) = gT (θ) =
3 2 sin θ . 2
The directivity function gT (θ, φ) defines a three-dimensional surface called the polar radiation pattern of the antenna. The function varies as sin2 θ, and hence the radiation is most intense in the θ = π/2 direction (perpendicular to the axis of the dipole) and zero in the directions θ = 0, π (on the axis of the dipole). The maximum gain then is 1.5 for directions defined by θ = π/2, in Figure 40: The polar radiation pattern of the equatorial plane of the dipole. the dipole antenna. The gain function is independent of φ. The spatial distribution of the radiated power can be shown on a polar diagram, such as in Figure 40, which gives the relative values of the radiated power at different positions on the surface of a sphere centered on the dipole. We can conclude, that the directivity function gT (θ, φ) is a measure of how effective the antenna is in concentrating the radiated power in a given direction.
164
Revision questions Question 1. Under what condition the fields can be expressed solely by the vector potential? Question 2. What do we mean by ”near zone” and ”far field zone”? Question 3. What are the properties of the EM field in the far field zone? Question 4. Show that the power radiated by an antenna is equal to the radiation losses. Question 5. Why emitted waves of a short antenna are very weak? Question 6. What is the spatial distribution of the power emitted by a short antenna?
165
12
Electromagnetic Theory of Polarizable Materials
Condense matter physics and electromagnetism theories used to be regarded as almost separate subjects. However, in next few lectures, we will illustrate how matter and electromagnetism are integrated and how the electromagnetic theory may be applied to condense matter physics to explain the properties of conducting materials and insulators, or dielectrics as they are often called. Materials such as conductors or insulators are composed of atoms that contain free and bounded electric charges. The charges can be redistributed by an application of external fields. We will focus on ”new” fields induced by redistributed electric charges and will show how these problems can be solved with the help of only the scalar potential Φ. We proceed our analysis by introducing the microscopic (atomic) model of materials and for this purpose simplified models of the atom will be used. In this simplified model, the atom consists of a positively charged nucleus surrounded by a spherically symmetric cloud of electrons. In the absence of an external electric field, the electrons and nucleus are in a stable equilibrium. What happens if the atom is placed in an external electric field? If an uncharged dielectric (insulator) is placed in an electric field, the field will redistribute the charges within the dielectric. An effective non-zero charge is induced by rearrangement of bound charges within the atoms or molecules of the dielectric. In dielectrics these charges are a set of molecular dipoles that, in turn, will set up a secondary (induced) field so that the net field will be modified from its original (external) value. We will show how the modification of the field led to the concept of an another field vector; the dielectric ~ The advantage of using the dielectric displacement displacement vector D. vector and its physical properties will be explored. In materials, electric dipoles may exist permanently or may be induced by the external field. Dielectrics with permanent dipole moments are usually electrically neutral due to random orientation, in the absence of electric fields, of the dipole moments. An example is H2 O. The application of an external electric field causes all molecules composing the dielectric to align themselves with the external field. This produces a macroscopic dipole moment. We will 166
study electromagnetic theory of polarizable materials in terms of microscopic objects, electric dipoles, which we will treat as building blocks of dielectric materials. As we shall see, despite this simplicity, the model satisfactorily predicts the macroscopic behavior of dielectric materials.
12.1
Potential and Electric Field of a Single Dipole
Mathematically, for the calculations of the field of distributed charges inside a dielectric material, it is convenient to deal with a separate object, the dipole, that is composed of two bounded charges of oposite signs separated by a small distance, not as just a pair of individual plus and minus charges. Suppose that two opposite sign point charges ±q are separated by a distance d. We will find the potential Φ ~ at a point A disand the electric field E tance r and angle θ under the assumption that the distance r is much larger than the separation between the charges, i.e. r d. The potential Φ produced by two separated charges of opposite sign is called the dipole potenFigure 41: A schematic model for tial. the calculation of the potential produced by an electric dipole.
We define electric dipole moment as the product of the charge times the separation. It is a vector that points from the negative charge to positive charge p~ = q d~ . Since r2 − r1 ≈ d cos θ at distances r d and d cos θ = p~ · rˆ, we get for the potential at the point A: q q − Φ = 4πε0 r1 4πε0 r2 q r2 − r1 q d cos θ 1 p~ · rˆ = = ≈ . (64) 4πε0 r1 r2 4πε0 r1 r2 4πε0 r2 ~ = −∇Φ. Hence, we can find electric field of the dipole using the relation E Since the potential of the dipole depends on r and θ, it is convenient to work 167
in the spherical coordinates in which the electric field is given by ~ = Er rˆ + Eθ θˆ + Eφ φˆ , E where the components are 1 2p cos θ ∂Φ = , ∂r 4πε0 r3 1 p sin θ 1 ∂Φ = = − , r ∂θ 4πε0 r3 1 ∂Φ =0. = − r sin θ ∂φ
Er = − Eθ Eφ Figure 42: The electric field lines of a dipole moment.
Hence ~ = E
p ˆ . 2 cos θ r ˆ + sin θ θ 4πε0 r3
One can note that the field has azimuthal symmetry. The field is proportional to (1/r)3 so that it falls rapidly with distance. Moreover, in the direction perpendicular to the dipole moment, θ = 90◦ , the field points in the opposite direction to the dipole moment at all distances. Figure 42 shows a sketch of the electric field lines of an electric dipole moment.
12.2
Polarization Vector
If there are N dipoles per unit volume of a dielectric, the total dipole moment is: P~ =
N X
p~i ,
i=1
and is called the polarization. The electric potential set up in the space by a dielectric material with an arbitrary volume distribution of electric dipoles can be calculated by using the potential produced by a single dipole, Eq. (64), and the above definition
168
of P~ . The electric potential at an arbitrary point A distance r from a volume element dV containing such dipoles is dΦ =
1 P~ · rˆ dV , 4πε0 r2
where rˆ is the unit vector from dV towards A, and we have assumed that r is much larger than the extent of the volume element dV . Let rˆ be the unit vector from A towards dV . (We want to integrate over V with the position of A fixed). In this case, we change rˆ → −ˆ r, and obtain rˆ 1 ~ P· − 2 dΦ = 4πε0 r
!
dV ,
and then the potential measured in the direction towards the volume V is Figure 43: Schematic diagram for calculations of the scalar potential produced at point A by a macroscopic material of the polarization P~ .
! rˆ 1 Z ~ P · − 2 dV . Φ= 4πε0 r V
It is difficult to interprete this form of the potential as it is not in the form of a potential known from the electrostatics.
However, the result can be transformed into a form that can be interpreted with the help of the general form of the static potentials. First, noting that −
1 rˆ =∇ , 2 r r
we have that rˆ P~ · − 2 r
!
1 = P~ · ∇ . r 169
~ = Φ∇ · A ~+A ~ · ∇Φ, we can write Next, applying a vector identity ∇ · (ΦA) the above expression as
1 ∇ · P~ P~ . P~ · ∇ =∇· − r r r
Hence, we can write the potential as a sum of two terms both involving the 1/r function
1 Z ∇ · P~ 1 Z P~ ∇ · dV + − dV . Φ= 4πε0 r 4πε0 r V
V
Note that the second term varies with r explicitly as the 1/r function, but in the first term, the 1/r function is under the nabla operator. However, we can use the Gauss’s divergence theorem to transform the first term into the surface integration. In this case, the first term becomes a simple function of 1/r, so that the potential takes the form Φ=
1 4πε0
I S
P~ · n ˆ 1 Z ∇ · P~ − dS + dV , r 4πε0 r
(65)
V
where n ˆ is the unit vector normal to the surface of the material.
Figure 44: Surface (left picture) and volume charges (right picture). On comparing with the general form of the static potential, Eq. (48), the expression (65) can be interpreted as follows. The first term on the right-hand 170
side, a surface integral, is a potential equivalent to that of a surface charge density σs = P~ · n ˆ . The second term is a potential equivalent to that of a volume charge density σV = −∇ · P~ . Where did this interpretation came from? It can be explained by using a simple picture shown in Figure 44. Surface charges exist because there are no neighboring charges at the end surfaces of the material to cancel them out. Volume charges exist because the number of dipoles per unit volume changes, that there is an incomplete cancellation of charge density from adjacent dipoles. Materials which have a non-zero volume charge density are called inhomogeneous materials. Thus, a sufficient condition for a material to be homogeneous is that the polarization of the material have a zero divergence.
12.3
~ in a Dielectric Maxwell’s Equation for ∇ · E
We have just seen that a non-zero potential outside a dielectric material is produced by surface and volume charges. A question then arises: What is the electric field inside and outside the dielectric when extra polarization charges are present? In general, the electric field in the dielectric can be found from the Maxwell’s equation I: ~ = ∇·E
ρ , ε0
where ρ is the total volume charge density inside the dielectric.16 In the dielectric it is convenient to (mentally) separate the polarization charges from whatever other charges might be there also. The other charges are usually referred to as the free charges or conducting charges to distinguish them from the bound charges in the dielectric. 16
Why the surface charge density is not taken into account in the Maxwell’s equation I?
171
We may write the Maxwell’s equation I as ~ = ρf + ρp , ε0 ∇ · E where ρf is the ”free” charge density and ρp is the polarization charge density throughout the volume. If we express ρp in terms of P~ , i.e. if we write ρp = −∇ · P~ , we obtain ~ = ρf − ∇ · P~ . ε0 ∇ · E Hence ~ + P~ ) = ρf . ∇ · (ε0 E
(66)
Now it is common practice to drop the subscript f , but one must remember that ρ now stands for the charge density not counting the polarization charges. Next, we can define a new vector ~ = ε0 E ~ + P~ , D
(67)
called the dielectric displacement field,17 and write Eq. (66) as ~ =ρ. ∇·D ~ inside a dielectric is We can read this equation that the source of the field D the free charge density ρ. 17
The reason for the name ”dielectric displacement” can be easily understood if we refer to the Maxwell’s theory. Take time derivative of both sides of Eq. (67): ~ ~ ∂E ∂ P~ ∂D = ε0 + . ∂t ∂t ∂t We know from the Maxwell’s theory that the first term on the rhs of the above equation represents displacement current density, and the second term is the polarization current ~ density. Therefore, ∂ D/∂t can be called a generalization of the displacement current ~ can be regarded as the dielectric displacement. density, and then D
172
12.4
Macroscopic Effects of the Polarizability
We turn now to a consideration of the macroscopic effects of the polarizability of dielectric materials. We will consider only ideal dielectrics. Ideal dielectrics can be divided into following categories: 1. Homogeneous – properties independent of the position. 2. Isotropic – properties independent of direction. ~ 3. Linear – polarization proportional to E. 4. Stationary – properties independent of time. Case of simple isotropic and linear dielectrics Ordinary dielectrics (glass, teflon, plastics etc.) are linear in polarization for ~ fields not strong enough to cause dielectric breakdown i.e. P~ ∝ E. For these materials, we can write ~ = χε0 E ~ , P~ = N αE where α is the polarizability of a single atom (molecule), and χ is the dielectric susceptibility. Then, we can write the dielectric displacement as ~ = ε0 E ~ + P~ = ε0 (1 + χ)E ~ = ε0 εr E ~ = εE ~ , D where εr is the relative permittivity or dielectric constant, and ε is the permittivity. Hence ~ = ∇ · (εE) ~ =ρ ∇·D or ~ = ∇·E
ρ , ε
if ε is independent of position i.e. if ε is a permittivity of a homogeneous dielectric.
173
The same result would be obtained by replacing ε0 in Coulomb’s law by ε. 1 q1 q2 rˆ . F~ = 4πε r2 The ratio ε/ε0 then represents the relative shielding of q1 from q2 by the polarization charges induced in the medium. How do we determine the permittivity of a given material? The theory of the molecular structure of a material will yield an estimate of χ and ε. The dipole moment p~ of a molecule will be proportional to ~ so we can define a molecular susceptibility χm such the local electric field E that p = χm ε0 E. Then P~ =
X
~ , p~i = N χm ε0 E
i
where N is the number of molecules per unit volume. Thus ε = ε0 εr = ε0 (1 + N χm ) and ε is a quantity directly measurable from the measurement of capacitance C=
ε0 A εA = εr = εr C 0 , d d
where C0 is the capacitance without the dielectric.
We can summarize: Filling a capacitor with dielectric multiplies its capacitance by εr .
Exercise in class: Capacitor filled with a homogeneous dielectric A plane parallel capacitor has charges +σ and −σ per unit area on its plates. The capacitor is filled with a homogeneous and linear dielectric of dielectric constant εr = 1 + χ. Show that:
174
(a) The electric field within the dielectric is: E=
σ . εr ε0
(b) The polarization charge per unit area on the surface of the dielectric adjacent to the surface of the negatively charged plate is: σs =
χσ . 1+χ
(c) The capacitance of the capacitor is C = εr C0 where C0 is the capacitance of the same capacitor without the dielectric (i.e. a vacuum or air between the plates).
175
12.5
Dense Dielectrics: The Clausius-Mossotti Relation
In the standard calculations of the polarization of macroscopic materials, it is ~ is the same at any point of the material. Howoften assumed that the field E ~ may vary with the position ever, for an extended dense materials the field E and the calculation of the dielectric constant of the material may not agree with an experimental measurement. This is what really happens. Therefore the approach of a constant field through the whole area of the material must be modified. It was Lorentz, who proposed an approach that resolved this problem. The Lorentz theory of polarizability of dense dielectric materials distinguishes be~ and the local electric field E ~ loc as seen by a tween the mean electric field E typical dipole. The typical dipole is considered to be at the center of a small ~ loc is thus the mean sphere that has been excavated from the dielectric. E ~ ~ ~ field E minus Eplug where Eplug is the field of the spherical volume excavated. Let us see what would be the field at this area if the dipole is not there. Let Let Let
~ be the mean field throughout the dielectric. E ~ Eplug be the field due to the spherical plug alone. ~ loc be the field in the spherical hole. E ~ =E ~ loc + E ~ plug E
and
~ ~ plug = − P . E 3ε0
~ minus the conThus, the field acting to polarize the molecule is equal to E tribution to the total average field from the molecule itself. Hence ~ ~ loc = E ~+ P . E 3ε0 The argument now is that each molecule is at the centre of a small hole ~ loc . If α is the molecular and the field acting on the molecule is thus E polarizability, its induced dipole is thus: ~ loc p~ = αE
176
If there are N molecules per unit volume then:
~ loc P~ = N p~ = N αE
~ ~+ P . = N α E 3ε0
~ = (εr − 1)ε0 E. ~ Substituting for P~ in the above By definition: P~ = χε0 E equation, we obtain "
~ = Nα E ~ + (εr − 1)ε0 E ~ (εr − 1)ε0 E 3ε0
#
.
Hence
(εr − 1)ε0 = N α 1 +
Nα εr − 1 = (εr + 2) , 3 3
and finally εr − 1 Nα = . εr + 2 3ε0 This is known as the Clausius-Mossotti relation for a dense dielectric. Assuming that α is known, we can compute εr - dielectric constant of the material. The Clausius-Mossotti relation works pretty well for most of dense materials. However, it has a weak point. To illustrate this, we solve the ClausiusMossotti equation for εr . Since εr − 1 = (εr + 2)
Nα , 3ε0
we obtain for εr :
εr = 1 +
2N α Nα / 1− 3ε0 3ε0
.
Note that εr =
2N α 3ε0 Nα − 3ε0
1+ 1
→ ∞ as
Nα →1. 3ε0
We see that one can adjust the number of the dipoles to get an infinite dielectric constant of a finite material. This is called the Clausius-Mossotti 177
catastrophe. This effect can be explained as follows: Removal of the plug leaves polarization charges, whose field tends to line up the dipole parallel to the field. The α → 1, i.e. the system is self-polarizing (Clausius-Mossotti catastrophe) if N 3ε0 system induces P~ without an external field.
12.6
Dielectric in a Time Dependent Field
Let us consider what will happen if a dielectric is introduced into an alternating electric field. The student immediately conclude that the alternating electric field will create oscillating dipole moments inside the material, or equivalently, an oscillating polarization of the material. This is true, but then an another question arises: How does the polarization follow the oscillating external field? To answer this question, let us consider an experiment, illustrated in Figure 45, involving a plane plate capacitor filled with a dielectric and connected to an oscillating AC current. The alternating current will create an oscillating electric field inside the capacitor which will cause the electric dipoles of the dielectric to oscillate in time. On the other Figure 45: A plane plate capacitor filled hand, the oscillating electric field with a dielectric and connected to an alterwill induce oscillating polarization nating current. charges. What are the oscillating polarization charges equivalent to? The induced polarization charges do not produce any currents inside the dielectric. There is no DC current in response to a DC electric field, but if P~ is changing with
178
~ is changing with time) there will be an AC current density: time (because E ∂ P~ ∂σs n ˆ= . J~ = ∂t ∂t Thus, ∂ P~ /∂t plays the role of polarization current density. With the alternating electric field present, the polarization P~ may lag in ~ This means there is internal friction and phase behind the driving field E. heat dissipation. If it happen, the capacitor will exhibit resistive as well as capacitive properties. Since resistivity leads to a dissipation of the energy, we proceed by considering the work done in charging the capacitor. It is defined as dQ dW = V I = Ed , dt dt where V is the voltage. Thus, we have to find how quickly the charge on the plates is changing in time. Since we are interested in the time variation of the electric field and the polarization, we express the charge in terms of the field quantities Q = CV = CEd =
εA Ed = AεE = AD , d
from which, we obtain Q = A(ε0 E + P ) . Then work in terms of the field variables is dW d d = Ed A(ε0 E + P ) = E dA (ε0 E + P ) . dt dt dt Since dA = V is the volume of the capacitor, we can write Z dW = dt V
"
d dP E (ε0 E) + E dt dt 179
#
dV ,
or Z Z d 1 dP dW 2 = ε0 E dV + E dV , dt dt 2 dt
V
V
where we took into account a possibility that the electric field and polarization can vary across the capacitor’s plates. ~ The first term in the above equation is the rate of doing work building up E field. ~ The second term is the rate of doing work on the dipoles by E. Thus, the supplied energy to the capacitor is used to build up the electric field inside the capacitor and to polarize the dielectric. Consider separately both terms. First term: If E = E0 cos(ωt), the first term takes the form Z
(−E0 cos(ωt) ε0 ωE0 sin(ωt)) dV .
V
The element of work done per unit volume and unit time is dW = −ε0 ωE02 cos(ωt) sin(ωt) . dV Averaging over a cycle, we get 2π/ω Z dW 2 = −ε0 ωE0 cos(ωt) sin(ωt) dt = 0 . dV t=0
No energy has been used to build up the electric field inside the capacitor. Work is done building up the field in one part of the cycle but the stored energy is given back in another part. Second term:
180
If P = χε0 E = χε0 E0 cos(ωt) the same zero net energy conversion averaged over a cycle will happen with this term. If there is internal friction there will be a phase difference between P and E, that P does not follow the changes in E. If we write P = χε0 E0 cos(ωt + φ) , where φ represents a phase difference between P and E, we get for the polarization P = χε0 E0 cos φ cos(ωt) − χε0 E0 sin φ sin(ωt) . Taking the time derivative, we get dP = −ωχε0 E0 cos φ sin(ωt) − ωχε0 E0 sin φ cos(ωt) . dt Hence, the work done per unit volume per cycle will be dW dV
= −
−
2π/ω Z 0 2π/ω Z
ωχε0 E02 cos φ cos(ωt) sin(ωt) dt ωχε0 E02 sin φ cos2 (ωt) dt .
0
Since the first integral is zero, we get 2π/ω Z dW 2 = −ωχε0 E0 sin φ cos2 (ωt) dt . dV
(68)
0
The integral on the rhs of the above equation is positive and dW/dV must be positive corresponding to energy dissipation (or the dielectric would keep getting energy from its interior and building up the field with it). Thus, sin φ must be negative, so −π < φ < 0. This means that the polarization lags in phase the electric field. P = χε0 E0 cos(ωt − φ) . In summary: Equation (68) shows that an energy is lost in each cycle of the oscillations. In practice, it is dissipated as a heat in the material. The energy loss is caused by the work required to change the polarization of the material.
181
12.7
The Complex Susceptibility and Permitivity
We have learnt that the polarization of a linear dielectric is proportional to the field and the proportionality is determined by the dielectric constant or susceptibility. In the previous lecture, we have modeled the fact that in real ~ by dielectrics, the polarization may lag in phase behind the driving field E ~ We may now ask: If P~ introducing the phase difference φ between P~ and E. ~ ~ is aflags in phase behind E, how does then the relation between P~ and E fected? To answer this question, it is convenient to use complex exponentials to represent amplitude and phase of an oscillating quantity. We write ~ =E ~ 0 eiωt E
and P~ = P~0 ei(ωt−φ) ,
(φ is positive) .
We can write the complex polarization in different forms P~ = P~0 e−iφ eiωt = (P~0 cos φ − iP~0 sin φ)eiωt ,
P~0 P~0 P~ = cos φ − i sin φ E0 eiωt , E0 E0 ~ 0 eiωt = ε0 χc E0 eiωt = ε0 χc E ~ . P~ = ε0 (χ0 − iχ00 )E ~ is the same as before for a constant external The relation between P~ and E field. However, now χc = χ0 − iχ00 is a complex quantity called a complex susceptibility. Thus, the phase difference between the polarization and the external field leads to a complex susceptibility of a dielectric material. In other words, the internal friction of the material results in a complex susceptibility. With the complex polarization, the dielectric displacement takes the form ~ = ε0 E ~ + P~ = ε0 E ~ + ε0 χ c E ~ = ε0 (1 + χc )E ~ , D or ~ = ε0 (1 + χ0 − iχ00 )E ~ = ε0 εr E ~ = εc E ~ , D 182
where εc is a complex permittivity, and εr is a complex relative permittivity or dielectric constant εc = ε0 (1 + χ0 ) − iε0 χ00 . How do we in practice estimate losses in a given material? In other words, how do we calculate χ00 for a given material? The imaginary part of εr , χ00 , can be determined from the following experiment involving a capacitor filled with a dielectric. As we have just shown, the imaginary part of the dielectric susceptibility represents losses, i.e. corresponds to net energy dissipation. In the circuit theory language the imaginary component of the dielectric susceptibility adds a resistive component to the capacitor. The material filling the capacitor could Figure 46: also have some ordinary ohmic conductivity (due to the presence of ‘free’ charges as well as ‘bound’ charges in the material). Let us calculate the magnitude of the total resistance of the dielectric. Simply, we will consider the Ohm’s law for the capacitor and will find the relation between an external current supplied to the capacitor and voltage. The relation will give us an information about the resistance of the capacitor. Let I be the oscillating current in the external circuit, and Ip =
Z
~ = A dP J~p · dA dt
be the polarization current in the dielectric. Let Ic =
Z
~ = AσE = Aσ V J~c · dA d
be the conduction current in the dielectric due to its finite conductivity σ.
183
If σs is the charge density on the plates, then the effective charge on the plates is Q = σ s A = qI − qp − qc , where qI is the charge supplied by I, qp is the charge removed by the polarization current Ip , and qc is the charge removed by the conduction current Ic . We can write the effective charge as Q=
Z
I dt −
Z
Ip dt −
Z
Ic dt .
Since the electric field inside the capacitor is E=
V σs = , ε0 d
we can find the surface charge density σs = ε0
V . d
and write the effective charge as Z Z Z V Q = σs A = ε0 A = Idt − Ip dt − Ic dt . d By taking a derivative in time of the both sides of the above equation, and substituting for P the relation, P = χc ε0 E = χc ε0 Vd , we get
ε0
A dV χc ε0 dV σV = I − Ip − Ic = I − A −A . d dt d dt d
(69)
Putting dV = iωV , dt and solving Eq. (69) for I, we get
χc = χ0 − i χ00 ,
ε0 A σ (1 + χ0 − iχ00 )iω + V . I= d ε0
Separating real and imaginary parts and putting C0 = ε0 A/d (the capacitance there would be if the dielectric were lossless), we find σ I = C0 + ωχ00 + iωC0 (1 + χ0 ) V . ε0 184
Since the capacitor transmits some charges through the internal dielectric, in the circuit theory this system is equivalent to a parallel circuit, as shown in Figure 47. For the parallel circuit
I=
1 + iωC R
V .
Comparing with the above result for current flow in the lossy capacitor we see that the effective capacitance 0 is C0 (1 + χ ) and the effective resistance is Figure 47:
R=
1
C0
σ ε0
+ ωχ00
,
where we remember that C0 is the capacitance in the absence of losses.
12.8
The Loss Tangent
The properties of a dielectric material are usually specified by giving its dielectric constant K, and its loss tangent tan δ. We can write the general Ohm’s law as σ (1 + χ ) − i χ + ε0 ω
I = iωC0
0
00
V .
The quantity in [ ] brackets can be defined as a complex relative permittivity εr : σ εr = 1 + χ − i χ + ε0 ω 0
00
.
This is a generalization on the previous definition of complex relative permittivity to include the effects of ohmic conductivity. Then we can define a 185
generalized permittivity and dielectric constant ε = ε0 εr . Now if we write εr = K 0 e−iδ = K 0 (cos δ − i sin δ) = K 0 cos δ(1 − i tan δ) = K(1 − i tan δ) . Then the standard form for permittivity is ε = ε0 K(1 − i tan δ) , where tan δ is the ‘loss tangent’.18 Since K 0 cos δ = (1 + χ0 ) and K 0 sin δ = (χ00 + h
tan δ =
χ00 + [1 +
σ ε0 ω χ0 ]
σ ) ε0 ω
it follows that:
i
.
In this equation, tan δ includes the effects of finite conductivity and the effects of polarization damping force.
18
In practice, it is read out on some AC bridges as an alternative to reading out the resistive property of a lossy capacitor.
186
Revision questions Question 1. How do we calculate the electric field of a microscopic dipole moment? Question 2. What are the surface and volume charge densities? Question 3. Explain the advantage of introducing the concept of dielectric displacement field. Question 4. How do we calculate the polarization vector of a dense dielectric? Question 5. In real dielectrics, does the polarization vector follow the changes of an external electric field? Question 6. How are losses in a dielectric represented in the permitivity of the dielectric? Question 7. What it meant by the loss tangent of a medium?
Tutorial problems Problem 12.1 Capacitor with 2 different dielectrics Consider a capacitor in which there are two different dielectrics of thickness d1 and d2 having dielectric constants ε1 and ε2 . The separation of the plates is then d1 + d2 . (a) Calculate the polarization charge distributions everywhere. Show that at the junction of the two dielectrics there is a surface charge density: σs =
ε1 − ε2 σ. ε1 ε2
(b) Show that the capacity of the capacitor, if the area of the
187
plates is A, is: C=
ε1 ε2 ε0 A . ε2 d1 + ε1 d2
Problem 12.2 Capacitor filled with an inhomogeneous dielectric A parallel plate capacitor has charges +σ and −σ per unit area on its plates which are separated by a distance d. The volume between the plates is filled with an inhomogeneous dielectric. The dielectric susceptibility is zero at the positively charged plate and increases linearly with distance, reaching its maximum value of unity at the negatively charged plate. If x is the coordinate measured from the positive plate toward the negative plate and α is the gradient of dielectric susceptibility, i.e. χ = αx, show that: (a) The electric field through the dielectric varies like: E=
σ ε0 (1 + αx)
0≤x≤d.
(b) There is a volume charge distribution throughout the dielectric given by: ασ ρ=− , (1 + αx)2 together with a surface charge distribution on the dielectric adjacent to the negative plate given by: σ σ1 = . 2 (c) The capacitance of the capacitor is: C=
ε0 αA . ln 2
Hint : You can see there will be a volume charge distribution within the dielectric because: ~ 6= 0 , because χ = αx . ρ = −∇ · P~ = −∇ · (χε0 E)
188
Also since this is the only charge in the region the polarization charge density ρ also satisfies the general Maxwell equation: ~ = ρ . ∇·E ε0 From these two equations you can develope a differential equation describing the variation of the electric field across the capacitor. Notice that it is ε0 we use in the equations (not ε) because we are treating the polarization charges explicitly. After solving for the electric field E(x) one can calculate the charge distribution from: ~ . ρ = −∇ · P~ = −∇ · (χε0 E)
Problem 12.3 Equivalent circuit of a lossy capacitor So-called electrolytic capacitors are made by electrolytic deposit of a thin layer of dielectric on aluminium. Consider a 1000 µF capacitor that might be used in smoothing the ripple in a DC voltage supply generated by rectifying the mains (50 Hz) supply. The dielectric has a complex dielectric constant: εr = 1 + χ0 − iχ00 = 5 − i 10−4 and a conductivity of σ = 2 × 10−13 [mho m−1 ]. What is the equivalent parallel resistance of the capacitor at frequency 50 Hz? and at 1 kHz? Problem 12.4 Field at the centre of a uniformly polarized sphere Consider a dielectric sphere with uniform polarization P~ throughout. The equivalent polarization charges will be a volume charge density −∇ · P~ which will be zero because of the uniform character of P~ together with a surface charge density P~ · n ˆ . Show then that the polarization charges will give rise to an electric field at the centre of the sphere given by: ~ ~ = −1 P . E 3 ε0
189
13
Electromagnetic Theory of Magnetizable Materials
In first lecture on electromagnetic theory of polarizable materials, we have discussed how the polarization of dielectrics by an externally applied electric field is equivalent to creation of volume and surface distributions of charges. Analogously, a magnetic field can act on molecular scale current loops existing in the building blocks of materials, atoms, to produce macroscopic effects. It was Amp` ere who first suggested that the magnetism of matter was due to the cooperative effects of currents circulating in atoms and not, as previously thought, due to a separate magnetic charge called poles. Thus, the magnetic properties of materials can be considered from an atomic viewpoint of electron’s currents, in which case a fundamental understanding of sources of the magnetic field can be developed.
13.1
Magnetic Polarization Currents
Let us illustrate the microscopic (atomic) theory of magetism from which we will then formulate the macroscopic theory. According to the atomic theory, the electrons rotating in orbital paths are equivalent to circulating currents on an atomic scale. We can define the magnetic moment of an electron current loop as µ ~ = IAˆ n, which is equal to the product of the area Figure 48: A schematic diagram of of the plane loop closed by the electron’s a current loop created by orbiting elec- orbit and the magnitude of the circulattron. ing current. The vector direction n ˆ of the moment is perpendicular to the plane of the loop and along the direction set by the right-hand rule, as shown in the Figure 48.
190
A macroscopic material body contains a lot of current loops, so we can define a macroscopic dipole moment per unit volume of the material, called magnetization ~ = M
X
µ ~i .
i
The theory of magnetism is based on the above definition of magnetization, and is formulated by a simple theorem: Theorem:
If a volume V enclosed by surface S has a magnetic dipole mo~ , which may be a function of position, ment per unit volume M the macroscopic magnetic fields so produced are equivalent to those of: ~. • A volume current density J~V = ∇ × M ~ ×n • A surface current density J~S = M ˆ.
Proof of the Theorem: The theorem is proved by showing that the vector potential due to the dipole distribution in a volume V closed by a surface S can be written in the form ~= A
~ ×n ~ ˆ 1 I M 1 Z ∇×M dV + dS , 4πε0 c2 R 4πε0 c2 R V
S
where R is a distance from the centre of the volume dV closed by a surface S. Consider a current loop of radius a, as shown in Figure 49. We will find the vector potential produced by the current loop at the point X. We start from the solution of the Maxwell’s equations, which gives the (static) vector potential of a current loop ~= A
~ 1 Z J~ I I dl dV = . (70) 4πε0 c2 r 4πε0 c2 r
Due to the radial symmetry of the loop, it Figure 49: A geometry for the cal- is convenient to work in the polar spherical culations of the vector potential produced at point X by a current loop of radius a.
191
~ coordinates, in which the current element dl can be written as ~ = adφ φˆ = −a sin φ dφ ˆi + a cos φ dφ ˆj , dl and the distance from dl to X is given by h
r = (x − a cos φ)2 + (y − a sin φ)2 + z 2
i1/2
,
which for a R can be written as 1/2
r =
x2 + y 2 + z 2 − 2ax cos φ + a2 − 2ay sin φ
≈
R2 − 2ax cos φ − 2ay sin φ
1/2
ax cos φ ay sin φ ≈ R 1− − R2 R2
!
.
Hence r
−1
=R
−1
ax cos φ ay sin φ 1+ + R2 R2
!
,
where we have used the Taylor expansion of 1/(1 − x) = 1 + x + . . .. Thus, the vector potential (70) takes the form Z 2π I ax cos φ ay sin φ ~ A = + 1+ 2 4πε0 c R 0 R2 R2 × −a sin φ dφ ˆi + a cos φ dφ ˆj .
!
Since Z 2π
sin φ dφ =
Z 2π
0
cos φ dφ =
0
Z 2π
sin φ cos φ dφ = 0 ,
0
~ simplifies to the formula for A ~= A
Z 2π I 2 2 ˆi + a2 x cos2 φ ˆj dφ , −a y sin φ 4πε0 c2 R3 0
which can be written as ~= A
Z 2π Z 2π Ia2 2 ˆ ˆ −y i sin φ dφ + xj cos2 φ dφ . 2 3 4πε0 c R 0 0
192
Next, since Z 2π
2
sin φ dφ =
0
Z 2π
cos2 φ dφ = π ,
0
we obtain ~= A
i Ia2 π h ˆ ˆj . −y i + x 4πε0 c2 R3
(71)
Using the relation ˆ = kˆ × kˆ × R
x y xˆ y ˆ z ˆ i + j + k = ˆj − ˆi , R R R R R
we can write Eq. (71) as ˆ µ Ia2 π ˆ ˆ ˆ× R k × R = k 4πε0 c2 R2 4πε0 c2 R2 1 1 = − µ ~ ×∇ , 2 4πε0 c R
~ = A
where we have used the result ˆ R 1 = −∇ , 2 R R and the fact that µ ~ is in the direction normal to the loop, i.e. in the z direction. Summarizing what we have obtained so far: We have obtained a formula for the vector potential produced by a single magnetic dipole moment µ ~. If we consider a set of dipole moments and change, for a convenience, the direcˆ into −R, ˆ as shown in Figure 50, tion of R ~ we can write that the vector potential dA produced by a set of dipole moments contained in a volume element dV is ~= dA 193 Figure 50:
1 ~ × ∇ 1 dV , M 4πε0 c2 R
~ is the magnetic dipole moment where M per unit volume. Then, the vector potential produced by the dipole moments contained in the total volume V is given by ~= A
Z 1 ~ × ∇ 1 dV . M 2 4πε0 c R
Using a vector identity ~ = ∇Φ × A ~ + Φ∇ × A ~, ∇ × (ΦA) we can write the integral function as ~ ~ ~ × ∇ 1 = −∇ 1 × M ~ = ∇×M −∇× M , M R R R R and then the vector potential takes the form ~= A
Z Z ~ ~ 1 1 ∇×M M dV − dV . ∇ × 4πε0 c2 R 4πε0 c2 R
(72)
The first term is in the form easy to interpret. Refering to the definition of ~ can be interpreted as the the vector potential, Eq. (53), we see that ∇ × M volume current density. Thus, the first term represents the vector potential produced by the volume currents existing inside the material. We still have to do something with the second term as it is in an unfamiliar form. In order to transform the second term into a familiar form, we apply a theorem that:19 Z ~ Z ~ M ×n ˆ M dV = dS . (73) − ∇× R R V S Thus, the application of the theorem to the second term in Eq. (72) leads to the vector potential of the form ~= A 19
Z Z ~ ~ 1 ∇×M M ×n ˆ 1 dV + dS , 2 2 4πε0 c V R 4πε0 c S R
Proof of the theorem is given in Appendix B.
194
or ~= A
Z ~ Z ~ 1 1 JV JS dV + dS . 4πε0 c2 V R 4πε0 c2 S R
We see by refering to the definition of the vector potential that the effective ~ per unit (Ampere) currents associated with a macroscopic dipole moment M volume are: ~ throughout the volume. (i) A current density J~V = ∇ × M ~ ×n (ii) A surface currents J~S = M ˆ. Figure 51 illustrates the source of the macroscopic surface current.
Figure 51: Source of the macroscopic surface current.
13.2
~ The Magnetic Intensity Vector H
When we were dealing with dielectric materials in the presence of electric fields, it was convenient to introduce a new vector, the displacement vector ~ in order to eliminate the necessity of taking the electric dipole polarizaD, tion P~ of the material into account explicitly. A similar procedure is used for the magnetic materials, where a new magnetic ~ is introduced to eliminate the magnetization M ~ . We will intensity vector H 195
illustrate this idea for both static and time-varying fields.
13.2.1
Static Magnetic Fields
For a static current distribution, the Maxwell’s equation IV reduces to ~ = µ0 J~ . ∇×B In a medium where there are magnetic polarization currents as well as conduction currents we can write J~ = J~c + J~m , where J~c is the conducting current, and J~m is the magnetization current. Thus ~ , J~ = J~c + ∇ × M and then ~ = µ0 (J~c + ∇ × M ~). ∇×B This equation can be written as ~ − µ0 M ~ ) = µ0 J~c , ∇ × (B or in the form
~ B ~ = J~c . ∇× −M µ0 ~ 0−M ~ has as its source only the conduction This shows that the vector B/µ current J~c . Therefore, to eliminate the necessity of dealing directly with the magnetization current J~m , we can define a new vector ~ ~ = B −M ~ , H µ0 which is called the magnetic (field) intensity vector.
196
~ the Maxwell’s equation IV for static fields in magnetic mateIn terms of H, rials takes the form ~ = J~c . ∇×H ~ (well not In dealing with magnetic materials we often know J~c but not M directly anyway.) Think e.g. of an inductor filled with some magnetizable ~ becomes useful. It is a way of avoiding a detailed material like iron. Then H ~ is the calculation of the polarization currents. The magnetic intensity H ~ magnetic analogue of the dielectric displacement D in the electric case. We may drop the subscript c and write ~ = J~ , ∇×H but we should remember that J~ is now not the total electric current density everywhere. ~ enables us to write the Maxwell’s In summary: The use of the vector H equation IV in terms of only the conducting current density in any magnetic ~. material. There is no need to deal with the magnetization M
13.3
Linear Magnetic Materials
~ is linFor most materials (excluding ferromagnetics) the magnetization M ~ and then because of the early proportional to the applied external field B, linear relation ~ ~ , ~ = B −M H µ0 ~ Thus, for linear magnetic mathe magnetization is also proportional to H. terials we write ~ = χm H ~ , M ~ M ~ , and H ~ will be in the same direction, and then at any point the vectors B, ~ ~ and we get the following relation between B and H ~ = µ0 H ~ + µ0 M ~ = µ0 (1 + χm )H ~ = µ0 µr H ~ = µH ~ . B 197
The parameter µr = (1 + χm ) is called the relative permeability, χm is the magnetic susceptibility, and µ = µ0 (1 + χm ) is called the magnetic permeability. Since ~ = µ0 (H ~ +M ~), B ~ such that and we have defined M ~ ~ ~ = µ0 χ m H ~ = µ0 χ m B = χ m B , µ0 M µ 1 + χm we get ~ = H
~ B . µ0 (1 + χm )
~ Thus, if we know the material we use, we can find H.
Example: A solenoid filled with magnetizable material In this example, we illustrate some interesting properties and relations between magnetic field vectors produced by a solenoid filled ~ is modwith a magnetizable material. We will show how the field B ified due to the presence of the material and what is the source of ~ vector. the H Consider the expression for the magnetic field produced by a long solenoid B = µ0 N I . In the presence of the material, we should include the Ampere surface currents as well as the conduction currents in the wire. Since B = µ0 I 0 , where I 0 is the total current per unit length, we have χm B NI + µ
B = µ0 (N I + M ) = µ0 (N I + χm H) = µ0
198
,
which can be written as
B 1−
µ0 χm µ
= µ0 N I ,
or B=
µ0 N I µ0 N I µ0 χm = χm . 1− µ 1 − 1+χ m
Note that if χm is positive then B is greater than it would have been in the absence of the magnetizable material. Evidently in this case the macroscopic Ampere current is in the same sense as the conduction current in the solenoid. Since 1−
χm 1 1 = = , 1 + χm 1 + χm µr
we have for the magnetic field B = µr µ0 N I = µN I .
(74)
Thus, the introduction of a magnetizable material into the solenoid is equivalent to replace µ0 by µ. We may go further and consider the following question: What is the ~ vector? source of the H It is easy to answer this question. Since B = µH , we then have from Eq. (74) that H = NI . ~ depends only on the parameters of the solenoid. Thus, H We see that the effect of filling the solenoid with a material of relative permeability µr is to multiply B by a factor µr (assuming the current in the wire remains the same).
Example: ~ depends only on the parameters of To illustrate further that H the solenoid, consider an another example which shows that in a solenoid, the magnetic intensity H is independent of the presence
199
or absence of the magnetic material. As in the preceding example, we consider a solenoid filled with a magnetic material. By the definition H=
B −M . µ0
When we remove the magnetic material, M = 0, and then H=
B µ0 N I = = NI . µ0 µ0
With the material present H=
B µN I −M = − χm H = µr N I − χm H . µ0 µ0
Hence H = (1 + χm )N I − χm H , which can be written as H(1 + χm ) = (1 + χm )N I . Thus H = NI , as before.
This gives rise to the notion of H as an inducing field and B as a resultant field. This concept is much used in the study of magnetic properties of ma~ is the field introduced to hide the properties of terials. The reason is that H ~ the material. Thus, H is the same independent of the presence or absence of the material.
13.4
Non-Linear Magnetic Materials
There are materials, i.e. iron, whose the magnetic properties are not de~ = µH. ~ They are called ferromagnetic matescribed by the linear formula B rials and have a specific property that placed in an external magnetic field 200
their magnetization undergoes a saturation. This is because all the internal current loops are lined up, which breaks the linear property. The reason for this unusual behavior is that ferromagnetic materials do not have a unique value of magnetic susceptibility because of strong magnetic nonlinearities. ~ and H, ~ and For this reason, it is difficult to provide a relation between B it is usually presented graphically in terms of the so-called hysteresis. An example of the hysteresis is shown in Figure 52.
Figure 52: Hysteresis loop of a ferromagnetic material.
13.4.1
Work done in magnetization of iron
Think of the case of the solenoid of length `, cross-section area A, and filled with magnetic material. The applied voltage V is: V = −E =
dΨ d = `N BA , dt dt
where N ` is the number of turns. The rate of doing work to power the solenoid is P = V I = A`N I
dB dB =V H , dt dt
where V = Al is the volume of the solenoid.
201
In the time dt that it takes to change B by dB, then the work done is dW = P dt = V H dB. Thus, the work done per unit volume is dW = H dB . Hence, the work done per unit volume in one cycle of the hysteresis loop of a non-linear material is given by the area of the loop.
Figure 53: An example of hysteresis loops of a ’hard’ (iron) and a ’soft’ (ferrite) ferromagnetic materials. Hard ferromagnetics retain some magnetization in the absence of external fields. This property of hard ferromagnetics makes them useful for permanent magnets.
It is sometimes useful to write Z
H dB = µ0
Z
H d(H + M ) = µ0
Z
H dH + µ0
Z
H dM
1 µ0 H 2 + µ0 H dM , 2 where the first term on the rhs is the work to establish magnetic field, and the second terms is the work by the field H to establish magnetization dM . =
Z
Which part really uses the energy: 12 µ0 H 2 or µ0 H dM ? To check it, consider properties of magnetic materials in a time varying magnetic field. R
13.5
Permanent Magnetic Materials: Ferromagnets
We have so far considered magnetic materials that are called diamagnetics and paramagnetics. In these materials the magnetization exists only in the 202
~ is a function of the external presence of an external magnetic field, i.e. M ~ ∼ B). ~ However, there is a class of materials, called ferromagnetics field, (M or permanent magnets in which macroscopic magnetization exists even in the absence of the external field. Consider first an exercise, which will illustarte several points that are of interest in designing of ferromagnetic materials.
Exercise in class. Magnetic field of a solenoid Consider a cylindrical current sheet on the surface of a cylinder of radius a, called a solenoid. The lines of current flow are circles round the surface of the cylinder. As in the case of the plane sheet we measure the strength of the current by a current per unit length with the length measured along the surface of the cylinder parallel to the axis and at right angles to the direction of current flow. (a) Assuming the magnetic field within the cylinder is uniform and parallel to the axis and that the magnetic field is zero Figure 54: Amp`ere’s loop to calcu- outside the cylinder show that within the late magnetic field inside and outside a ~ = µ0 I¯ where I¯ is cylinder the field is B solenoid. the current per unit length. (b) Can you think of the arguments from Amp` ere’s theorem for the field being uniform within the cylinder and zero outside?
13.5.1
~ and H ~ fields of a ferromagnet B
We can now easily extend the arguments from the above exercise to analyse ~ the magnetic field a long homogeneous ferromagnetic material. In this case B
203
~: inside the material is due solely to the M ~ ×n J~s = M ˆ, or in terms of the magnitudes Js = M .
~ and H ~ inside and outside the material. Let us find B Using Amp` ere’s law
H
~ · d~` = µ0 I, where I = Js ` = M `, we obtain B
B` = µ0 M `
i.e.
B = µ0 M ,
where we have used the fact that B is zero outside the magnet. Thus H=
B µ0 M −M = −M =0 , µ0 µ0
in the region where M 6= 0, i.e. inside the magnet. Outside the magnet H=
B B −0= , µ0 µ0
~ is just a scaled replica of B. ~ i.e. H
204
13.5.2
Magnetic Poles
~ = 0 always and the lines of B ~ form closed loops, we have Since ∇ · B
~ ~ = −∇ · M ~ . ~ = ∇·B −M ∇·H µ0 ~ has a source (field lines start and stop) where M ~ varies i.e. at the Thus, H ends of the magnet (magnetic poles). For a ferromagnet, we can write ~ = −∇ · M ~ = ρm . ∇·H Thus, we can think of ρm as a volume density of ‘magnetic charge’ giving ~ field. It must be stressed that this equivalence is purely mathrise to the H ematical, and does not prove a physical existence of magnetic charges. Moreover, for the ferromagnet ~ = J~c = 0 . ∇×H ~ here to those of the E ~ field Note the similar mathematical properties of H in electrostatics ~ = ∇·E
ρ ε0
~ =0. and ∇ × E
Historically, magnetostatics of permanently magnetized materials (ferromag~ field and ’magnetic charges’ or ’poles’. One nets) developed via use of the H obtains a law analogous to Coulomb’s Law for the force between magnetic poles.
205
Exercise in class: Plane magnetized material An infinite plane surface divides the universe into a vacuum on one side and a magnetic material on the other. Within the magnetic material there exists a uniform ~ which is parallel magnetic moment per unit volume M to the surface. (a) Show that while the direction of the magnetic in~ is different on the two sides of the duction vector B surface, its magnitude is given everywhere by: B=
M . 2ε0 c2
(b) Find the magnitude and direction of the magnetic ~ everywhere due to an infinite plane parallel slab field B of material of thickness d which is permanently uniformly magnetized with dipole moment per unit vol~ lying parallel to the bounding surfaces. ume M ~ everywhere. (c) Find the magnetic intensity H
206
13.6
Time Dependent Magnetic Fields and Energy Loss
~ may not stay in In a time-dependent magnetic field the magnetization M ~ This corresponds to internal friction and phase with the driving field H. heat dissipation. Let H = H0 cos(ωt) . Then the magnetisation in a lossy material varies as M = M0 cos(ωt + φ) , where M0 and φ represent the amplitude and phase of the magnetization response to the magnetizing field. Thus M = M0 cos φ cos(ωt) − M0 sin φ sin(ωt) . Now the work done in magnetization per cycle of the AC current producing the magnetizing field is W = µ0
Z
2π/ω Z
H dM = µ0
t=0
H
dM dt . dt
Since dM = −ωM0 cos φ sin(ωt) − ωM0 sin φ cos(ωt) , dt we have W = W1 + W2 . Consider the term W1 : W1 = −µ0 H0 M0 ω cos φ
2π/ω Z
cos(ωt) sin(ωt) dt .
t=0
This term is zero - it represents reversible energy conversion to and from H. Consider now the term W2 : W2 = −µ0 H0 M0 ω sin φ
2π/ω Z 0
207
cos2 (ωt) dt .
This term represents work done against internal friction during magnetizing and demagnetizing the material. Since cos2 (ωt) dt is positive, sin φ must be negative so that work is done on the material i.e. φ is negative. R
13.7
The complex magnetic susceptibility
Let H = H0 eiωt , which for a physical field can be written as
H = H0 cos(ωt) = Re H0 eiωt
.
Then M = M0 ei(ωt−φ) = M0 e−iφ eiωt or M = (M0 cos φ − iM0 sin φ)eiωt
=
M0 M0 cos φ − i sin φ H0 eiωt . H0 H0
This result can be written in terms of real and imaginary susceptibility M = (χ0 − iχ00 )H0 eiωt = (χ0 − iχ00 )H . Using this complex number notation, we find B = µ0 (H + M ) = µ0 H + µ0 (χ0 − iχ00 )H = [µ0 (1 + χ0 ) − iµ0 χ00 ]H and then 0
00
B = (µ − iµ )H = µH , 0
00
where µ = µ0 (1 + χ0 ), µ = µ0 χ00 , and µ is the complex permeability.
208
13.8
Maxwell’s Equations in Dielectric and Magnetic Materials
Let us complete the lectures on dielectric and magnetic properties of materials by incorporating our findings into the Maxwell’s equations, to see how the basic equations of electromagnetism are modified in the presence of the material. The Maxwell equation IV contains a current density term ~ ~ = µ0 J~ + µ0 ε0 ∂ E . ∇×B ∂t We think this is always true provided J~ is the total electric current density. Applying this in a region where there may be electric and magnetic polarization effects we can write the current density as ∂ P~ ~ . +∇×M J~ = J~c + J~E + J~M = J~ + ∂t Thus, the Maxwell’s equation IV takes the form ~ ~ ~ = µ0 J~c + µ0 ∂ P + µ0 ∇ × M ~ + µ 0 ε0 ∂ E , ∇×B ∂t ∂t which can be written as
~ B ~ = J~c + ∂ (ε0 E ~ + P~ ) . ∇× −M µ0 ∂t ~ and H, ~ this equation With the introduction of the new vectorial fields D takes the form ~ ~ = J~ + ∂ D , ∇×H ∂t where we must remember that J~ represents the conduction current only.
209
In summary: The Maxwell’s equations for the electromagnetic field propagating in a material are of the following form ~ = ρ, ∇·D ~ = ∇·H ~ =0, ∇·B ~ ~ = −µ ∂ H , ∇×E ∂t ~ ~ = J~ + ∂ D , ∇×H ∂t but in general ~ = −∇ · M ~ . ∇·H These equations are supplemented by appropriate constitutive relations, which ~ and the magnetic induction B ~ with the displaceconnect the electric field E ~ and the magnetic field H ~ ment field D ~ = εE ~ D
and
~ = µH ~ . B
These relations carry information about the material.
Revision questions Question 1. State the definition of magnetization. Question 2. How do we define a volume current density and a surface current density? ~ instead of B ~ for a magnetic Question 3. Explain why it is useful to use the H field in a magnetic material. Question 4. What is the form of the Maxwell’s equation IV for static fields in magnetic materials?
210
Question 5. What it is a hysteresis loop, and what does it correspond to? Question 6. What is meant by a homogeneous and isotropic material? Question 7. How do we distinguish between soft and hard ferromagnetics? Question 8. What are the modifications to the Maxwell’s equations of an electromagnetic field propagating in a material?
211
Tutorial problems Problem 13.1 Field of a uniformly magnetized sphere ~ per unit A magnetized sphere has a uniform dipole moment M volume. (a) Find and sketch a diagram of the magnetic polarization currents in and on the sphere. (b) Apply the Biot-Savart law to this current distribution to find ~ at the centre of the sphere. Show the magnetic flux density B that: ~ ~ = 2M . B 3ε0 c2 Problem 13.2 Fields in an inhomogeneous magnetic material A long solenoid of radius a and having N turns per metre carries current I. It is filled with an inhomogeneous material with a gradient in magnetic susceptibility. The susceptibility is zero at the centre and increases linearly with radial distance r from the centre of the solenoid, attaining a value of 2 at r = a. ~ M ~,H ~ throughout the solenoid. (a) Find the vectors B, ~ first by integrating the Maxwell equation (Hint: Study H ~ = J~ , ∇×H ~ is unand use the symmetries of the situation. Show that H changed by the presence of the magnetizable material). (b) Show that the self-inductance of the solenoid is the same as would be obtained by replacing the above-mentioned magnatic material with a homogeneous material having a magnetic susceptibility of 34 . (c) What are the magnetic polarization currents on the surface and within the solenoid? Calculate them in terms of I.
212
Problem 13.3 Fields in a dense magnetic material Using a method similar to that used in the case of electric fields within dense dielectrics, show that if a spherical cavity is exca~ vated in a uniformly magnetized material, the magnetic field B ~ and the magnetic intensity H in the hole and in the material are related by ~ ~ hole = B ~ material − 2M , B 3ε0 c2 ~ ~ hole = H ~ material + M . H 3
213
14
Poynting’s Theorem Revisited
We have seen in Chapter 8 how energy of the electromagnetic field may be transported through vacuum (empty space) by means of electromagnetic waves. We have shown that the direction of propagation of energy is determined by the Poynting vector. In this lecture, we will reconsider the Poynting theorem taking into account propagation of the electromagnetic field in magnetizable materials. A question we will try to answer: How the energy is propagated inside a magnetizable material?
14.1
~ and H ~ Poynting Vector in Terms of E
~ We have shown that it is useful to represent the magnetic field in terms of H ~ vector when the field propagates inside a magnetizable vector rather then B material. ~ vector for the field in vacuum Using the concept of the H ~ ~ , ~ = B = ε 0 c2 B H µ0 the Poynting vector of the EM field in vacuum takes the form ~ = ε 0 c2 E ~ ×B ~ =E ~ ×H ~ . N ~ ×H ~ also turns out to be the correct expression for the The cross product E Poynting vector when magnetizable materials are involved and is the expression most commonly quoted for it. In order to show this, we consider, as before in Section 8, a flow of the energy through a closed surface S. Using the Gauss’s theorem and a vector identity ~ × B) ~ =B ~ · (∇ × A) ~ −A ~ · (∇ × B) ~ , ∇ · (A we can write I
~ × H) ~ · dS ~= (E
SZ
=
Z
~ × H) ~ dV ∇ · (E
V
~ · (∇ × E) ~ dV − H
V
Z V
214
~ · (∇ × H) ~ dV . E
Now, substitute from the Maxwell equations ~ = − ∂B ~ , ∇×E ∂t ~ = J~ + ∂ D ~ , ∇×H ∂t and obtain I
~ × H) ~ · dS ~=− (E
S
Z V
Z Z ~ ~ ~ · ∂ D dV − E ~ · J~ dV . ~ · ∂ B dV − E H ∂t ∂t V V
The individual terms in this equation have the following interpretation: 1. The lhs is the rate of flow of field energy out of the volume V enclosed by the surface S. 2. First term on the rhs is the rate of work in establishing the magnetic field in V . 3. Second term is the rate of doing work in establishing the electric field in V . 4. Third term is the rate of doing work on the currents in V . We go further with the Poynting’s Theorem. ~ = µH, ~ we then get If we substitute for B Z
Z ~ ∂B ∂ 1 ~ dV = µH 2 H· ∂t ∂t 2
Z
Z ~ ∂ 1 2 ~ · ∂ D dV = E εE ∂t ∂t 2
∂ Z 1 B2 dV = dV , ∂t 2 µ
and
dV =
∂ Z 1 2 εE dV . ∂t 2
Hence ~ × H) ~ · dS ~=−∂ (E ∂t S
I
Z
1 2 1 B2 εE + 2 2 µ 215
!
dV −
Z
~ · J~ dV . E
Thus, the rate of flow of field energy out of volume V is equal to the rate of changing energy of the EM field plus the arte of doing work on the currents in V .
14.2
Poynting Vector for Complex Sinusoidal Fields
It is well known that the electromagnetic field (e.g. light) is a real physical quantity (observable). However, in the electromagnetic theory it is advantageous to represent the real electromagnetic field by complex sinusoidal quantities because of its mathematical simplicity. In addition, what we usually ~ ∗ · Ei, ~ which is a real quantity. measure is the average intensity of the field, hE For time varying fields, we usually write ~ =E ~ 0 eiωt E
and
~ =H ~ 0 eiωt , H
~ 0 and H ~ 0 are complex quantities including both amplitude and phase where E information. We understand that the electric and magnetic fields are given ~ and H. ~ by the REAL PARTS of E The power of the complex exponential scheme lies in the fact that for operations such as summation, subtraction, integration etc., we take real parts AFTER the operation. For example:
~ 1 + ReE ~ 2 = Re E ~1 + E ~2 ReE Re
,
~2 ~1 dE d ~ dE ~2 . + Re = Re E1 + E dt dt dt
However, some care has to be taken in evaluating the Poynting vector. ~ =E ~ × H, ~ but if we write complex expoThe Poynting vector is given by N ~ × H, ~ we note that nential expressions for E
~ = ReE ~ c × ReH ~ c 6= Re E ~c × H ~c N
,
where we put a subscript c to indicate that we are writing the fields using complex exponentials. Proof:
216
We can write the complex electric field as
~c = E ~ 0 eiωt = E ~ r + iE ~ i (cos ωt + i sin ωt) E =
~ r cos ωt − E ~ i sin ωt + i E ~ r sin ωt + E ~ i cos ωt . E
Similarly, for the magnetic field we can write
~c = H ~ 0 eiωt = H ~ r + iH ~ i (cos ωt + i sin ωt) H =
~ r cos ωt − H ~ i sin ωt + i H ~ r sin ωt + H ~ i cos ωt , H
~ r, E ~ i, H ~ r and H ~ i are real vectors. where E However, from the above, we see that ~c = E ~ r cos ωt − E ~ i sin ωt , ReE ~c = H ~ r cos ωt − H ~ i sin ωt . ReH ~c × H ~ c ) we get extra terms in addition to those Clearly, if we calculate Re(E ~ c × ReH ~ c , and then in the expression for ReE ~ c × ReH ~ c 6= Re(E ~c × H ~ c) , ReE as required. ¯~ ~ , but rather N In experiments, we do not measure N , the mean Poynting vector averaged over the cycles of oscillations. There is a useful expression for the MEAN POYNTING VECTOR in ~ c and H ~ c: terms of the complex exponential E 1 ~ 1 ~∗ ~ ¯~ ~∗ ~ ¯ ~ N = Re E c × Hc = Re Ec × Hc = ReEc × ReHc , 2 2
~ indicates an average over the whole cycle of the sinuwhere the bar over N soidal field to eliminate the rapid temporal oscillations at frequency ω. Proof:
217
Calculate the Poynting vector involving measurable (real) parts of the complex fields
~ = ReE ~ c × ReH ~c = E ~ r cos ωt − E ~ i sin ωt × H ~ r cos ωt − H ~ i sin ωt N
~r × H ~ r cos2 ωt + E ~i × H ~ i sin2 ωt − (E ~r × H ~i + E ~i × H ~ r ) cos ωt sin ωt . = E Since 1ZT 1 1ZT cos2 ωt dt = sin2 ωt dt = , T 0 T 0 2 1ZT and cos ωt sin ωt dt = 0 , T 0 we obtain for the average Poynting vector 1 ~ ¯~ ~ ~ ~ N = (E r × Hr + Ei × Hi ) . 2 On the other hand, take ~ c∗ = H
~ r − iH ~ i e−iωt , H
~c = E
~ r + iE ~ i eiωt , E
and then ~c × H ~ ∗ ) = (E ~r × H ~r + E ~i × H ~ i ) + i(E ~i × H ~r − E ~r × H ~ i) . (E c Hence 1 ~ ~ c∗ = 1 (E ~r × H ~r + E ~i × H ~ i) , Re Ec × H 2 2 ¯~ which is equal to N , as required.
~ of complex exponential fields In summary: The average Poynting vector N satisfies the relation ¯~ ¯ ~c × ~ c = 1 Re E ~c × H ~ c∗ , N = ReE ReH 2 and this is the expression for calculation of the Poynting vector of complex fields.
218
Revision questions ~ and H ~ Question 1. What is the form of the Poynting vector in terms of the E vectors? Question 2. How do we write the electric and magnetic field vectors for oscillating fields? Question 3. What is the useful form of the Poynting vector for complex sinusoidal fields?
219
15
Plane Wave Propagation in Dielectric and Magnetic Media
In this lecture, we shall examine in some details how existing radiation field is modified by the material it passes through. We will consider the propagation in a linear and homogeneous material and as we shall see, the conductivity is the most significant parameter for modifications, rather than the dielectric and magnetic constants. We have learnt that the properties of a lossy linear and homogeneous dielectric material can be described using a complex permittivity and similarly, the properties of a lossy magnetic material are described by a complex permeability. Thus, for a lossy linear and homogeneous material the Maxwell’s equations describing the EM field inside the material are of the form ~ = ρ/ε , ∇·E ~ = 0, ∇·H ~ ~ = −µ ∂ H , ∇×E ∂t ~ ~ = J~ + ε ∂ E , ∇×H ∂t
(75)
where ε, µ are complex quantities that characterize the material, ρ is a free ~ charge in the material, and J~ is conduction current only, i.e. J~ = σ E. Note from Eq. (75) that an EM field propagating in a linear and homoge~ and H ~ fields, and its properties neous material is solely described by the E depend only on the material constants ε and µ. Thus, the only factors that can modify the wave from that propagating in vacuum are the constants ε and µ. We remember that in the vacuum the EM field propagates in the form of plane transverse waves. Questions we will try to answer are: What are the properties of the EM waves propagating inside a material? Does an EM wave retain its transverse nature?
220
To answer these questions, consider a plane wave propagating in one dimension, the z direction ~ =E ~ 0 ei(ωt−kz) . E
(76)
~ to check whether the wave retaines its transverse propFirst, we shall find H erties when propagating through the lossy material. Thus, for the electric field (76), the Maxwell’s equation IV takes the form ~ = σE ~ + iωεE ~ = (σ + iωε)E ~ . ∇×H Since for a plane wave propagating in the z direction the derivatives ∂/∂x ~ and H ~ are zero, we obtain and ∂/∂y of E
ˆi ˆj kˆ ∂ 0 0 ∂z Hx Hy Hz
~ . = (σ + iωε)E
From the LHS, we see that ~ z = 0 → (σ + iωε)Ez = 0 . (∇ × H) ˆ ~ ⊥ k. Hence Ez = 0 unless (σ + iωε) = 0. Thus, E ~ = 0. Also Since Ez = 0, we have that ∇ · E ~ = −µ ∇×E
~ ∂H ∂t
~ ⊥ H. ~ can be used to show that E In addition, we can find from the above equation that Hz = 0, which means ~ ⊥ ~k. that H We can summarize our thusfar findings that the wave propagating in a lossy material retains its transverse nature. However, the losses are expected to modify somehow the wave. What kind of modifications are made by the losess?
221
~ = 0, the quantity ε occurs only in the equation for ∇ × H, ~ and Since ∇ · E it is common to proceed as follows: ~ ~ = J~ + ε ∂ E = σ E ~ + iωεE ~ , ∇×H ∂t ~ = iω σ + ε E ~ = iω ε − i σ E ~ . ∇×H iω ω Now ε−i
σ σ σ = ε0 − iε00 − i = ε0 − i ε00 + ω ω ω
= ε¯ ,
which gives ~ = iω ε¯E ~ . ∇×H Physically what has been done is to lump together the conduction and the lossy dielectric constant term. To an external observer they are inseparable. Only using some theory of the internal structure of the dielectric, they can be separated. We can summarize that the electric and magnetic fields of the propagating wave in a lossy conducting material satisfy the following equations ~ = 0 , ∇·E ~ = 0 , ∇·B ~ ~ , ~ = −µ ∂ H = −iωµH ∇×E ∂t ~ = iωεE ~ , ∇×H
(77) (78) (79) (80)
where we have left the bar off ε.
15.1
Dispersive Equation and Complex Propagation Number
We now proceed to solve the propagation equations (77)–(80). We will see that the fact that µ and ε are complex numbers, the solution leads to a dispersive equation.
222
Thus, we look for plane wave solutions, and will try to find how the propagation number k behaves. The procedure is as follows: Taking ∇× of (79) and using (80), we obtain ~ = ω 2 µεE ~ , ∇ × (∇ × E) which, after applying the double × vector identity, simplifies to ~ = ω 2 µεE ~ . −∇2 E ~ of the transverse EM wave is independent of x and y, we have Since E ~ ∂ 2E ~ . − = ω 2 µεE 2 ∂z ~ given in Eq. (76), is On the other hand, the second derivative of E, ~ ∂ 2E ~ . = − k2E 2 ∂z Hence, we obtain a dispersion equation k 2 = ω 2 µε . This dispersion equation is not as simple as it looks. We cannot just say that phase velocity is vp =
ω 1 =√ , k µε
as we usually do for EM waves propagating in vacuum, because ε and µ are complex quantities and then k is a complex number. What does an imaginary value of k mean? To give the answer to this question, we write the complex number k as k = α − iβ , where α and β are real. Then E = E0 ei(ωt−kz) = E0 ei(ωt−αz+iβz) = E0 e−βz ei(ωt−αz) . 223
Clearly from this, the phase velocity is ω , α
vp =
i.e. the real part of k plays the role of the propagation number, and β = − Im(k) is the attenuation (losses) coefficient. In practice, the losses come from (1) Conduction currents, (2) Lossy dielectric, and (3) Lossy magnetic material, which are lumped together in ε and µ. We can find the explicit forms of α and β. To do this, we write the complex number k as σ ε −i ε + ω
1 2
k = ω (µε) = ω
0
00
0
00
1
(µ − iµ )
2
,
which can be written as
k=ω
ε0 µ0 − µ00 ε00 +
σ ω
− i µ0 ε00 +
σ + ε0 µ00 ω
1 2
.
Let us introduce a short notation for the real and imaginary parts of k:
p = ε0 µ0 − µ00 ε00 +
σ ω
,
σ + ε0 µ00 . ω
q = µ0 ε00 + Then
1
k = ω (p − iq) 2 , which can be written as k=ω
2
p +q
2
1 2
−iarctan(q/p)
e
or
k = ω p2 + q 2
1
224
4
e−iθ ,
12
,
where θ = 12 arctan (q/p). Hence
α = Re(k) = ω p2 + q 2
1 4
β = −Im(k) = ω p2 + q 2
cos θ ,
1 4
sin θ ,
which in general are quite complicated and difficult to interpret. The interpretation becomes more transparent when we make some simplifications that are discussed in the following two examples.
15.2
Wave Refraction and Attenuation
We now consider two examples of the propagation of an EM wave in different materials: Low loss dielectrics and conducting materials. 15.2.1
Propagation of an EM wave in a low loss dielectric
In the first example, we illustrate the propagation of an EM wave in a low loss dielectric with no magnetic effects. For dielectrics, σ is negligible and also we disregard magnetic properties. Thus, for a low loss dielectric ε00 +
σ ε0 , ω
and µ = µ0 =
1 . ε 0 c2
Hence, the propagation number simplifies to ω σ k = √ ε0 − i ε00 + ε0 c ω
1 2
ω 0 2 σ = √ (ε ) + ε00 + ε0 c ω
225
2 ! 21
12
e−iΘ
,
where ε00 + ε0
Θ = arctan
σ ω
!
ε00 + ≈ ε0
σ ω
,
as the losses are small. Then 00 s √ ω ε0 −iΘ/2 ω ε0 −i ε +σ/ω 0 2ε = e . k= √ e c ε0 c ε0 Hence, the real part of k takes the form ω α= c
s
ε0 ε00 + σ/ω cos ε0 2ε0
!
√ ω ε0 ω√ ≈ εr . √ = c ε0 c
We can then find phase velocity of the wave and refractive index of the dielectric material ε0 ω =c 0
ε0 , vp = α ε s √ c ε0 n = = = εr > 1 . vp ε0 r
Thus, we may conclude that inside the dielectric, the EM wave propagates with a phase velocity vp < c and can be refracted from the original direction on the entry to the dielectric. Similarly, we can find β ω β = −Im(k) = c
s
ε0 ε00 + σ/ω sin ε0 2ε0
!
.
Since for small θ, sin θ ≈ θ, we get ω β≈ c
s
ε0 ε0
ε00 + σ/ω 2ε0
!
ω ε00 + σ/ω √ = . c 2 ε0 ε0
Thus, losses (absorption of the wave) are small.
226
We can summarize, that the theory predicts that the refractive index for a lossless dielectric is given by √ n = εr . Table below compares theoretical values of n with that obtained experimentally. An excellent agreement is observed, except for polar molecules. For √ example, water has a refractive index of 1.33, whereas εr = 9. The reason for this discrepancy lies, not in the inadequacy of the laws of electromagnetism, but rather in the tacit assumption that the dielectric constant is a constant independent of the frequency. The refractive index of the water exhibits a strong dependence on ω as it is composed of polar molecules that the polarization depends strongly on ω.
√
εr Air 1.00029 Argon 1.00028 CO2 gas 1.00047 Benzene 1.49 Ethanol 5.3 NaCl 2.47 Water 9.0
15.2.2
n (experimental) 1.00029 1.00028 1.00045 1.48 1.36 1.54 1.33
Propagation of an EM Wave in a Good Conductor
As a second example consider a propagation of the EM field in a good conductor with no dielectric and magnetic losses: ε00 = µ00 = 0. Examples of good conductors: Cu, Ag. Consider the Maxwell equation IV, which for an EM wave propagating in a conducting material can be written as ~ ~ = J~ + ε ∂ E = σ E ~ + iωεE ~ . ∇×H ∂t 227
(81)
Definition of a Good Conductor Good conductor is when the conduction term on the right-hand side of the Maxwell equation (81) dominates over the displacement current, i.e. when σ ωε. On physical grounds, for a good conductor the conductivity current is much larger than the polarization current. For example, for copper at ω = 1 MHz = 2π × 106 rad 5.8 × 107 σ = = 1.0 × 1012 . ωε0 2π × 106 × 8.85 × 10−12 Consider general expression for k:
k = ω
0 0
εµ −µ
00
σ ε + ω 00
−i µ
0
σ ε + + ε0 µ00 ω
00
1 2
,
that for a conductor with no dielectric losess reduces to
0 0
k = ω (ε µ − 0) − i µ
0σ
ω
1
+0
2
,
and finally
0 0
k = ω ε µ − iµ
0σ
1
2
ω
.
Now we might as well drop the dashes on ε, µ understanding that they are real quantities, and obtain σ k = ω εµ − iµ ω
1 2
√
σ = ω εµ 1 − i εω
1
2
,
that can be written as "
√ k = ω εµ 1 +
σ εω
2 # 12
228
e−i arctan
σ εω
1 2
.
Remembering that for a good conductor the [ ] brackets, and get √
σ εω
1
1. Thus, we can drop ”1” in
σ −i π σ −i π 2 √ k = ω εµ e 2 = ω εµ e 4 . εω εω √ Since cos(π/4) = sin(π/4) = 1/ 2, we obtain
1 1 √ k = ωµσ √ − i √ 2 2
r
!
r
=
ωµσ (1 − i) . 2
Then r
α = Re(k) =
ωµσ = β = −Im(k) . 2
Knowing α, we can find the phase velocity of the wave s
vp =
2ω , µσ
which shows that the velocity of propagation in a good conductor depends on the frequency, even if µ and σ are assumed independent of frequency.
15.2.3
Skin Effect
Since the propagation number is a complex number and for a good conductor β = α, there is a very heavy attenuation of a propagating wave. For a complex propagation number, we can write the electric field as E = E0 e−βz ei(ωt−αz) . We see, that the amplitude of the electric (and also magnetic) field decreases exponentially as z increases. Since we can write α=β=
2π , λ
229
the distance for attenuation ”e” fold (i.e. amplitude falls to a factor original value in a distance β1 ) is z=
1 e
of its
1 λ = , β 2π
i.e. δ=
1 λ λ = ≈ !! β 2π 6
where δ is called the skin depth in the conductor. It is the distance the wave must propagate in order to decay by an amount e−1 . This effect is sometimes called ”skin effect” as with an increasing σ the current flows in a narrower and narrower layer, until in the limit of σ → ∞ a true current exists only on the surface of the conductor.
230
Revision questions Question 1. Show that the propagation number of an electromagnetic field in dielectric and magnetic materials is a complex number. Question 2. What are the physical consequences of the propagation number of an electromagnetic field being a complex number? Question 3. What is the definition of a good conductor? Question 4. Why the theoretical and experimental values of the refractive index of water are significantly different? Question 5. What is meant by the skin effect of a propagating wave?
Tutorial problems Problem 15.1 Relations between the electric and magnetic field vectors in dielectric and conductor (a) Show that for a plane wave propagating in a dielectric along the x-direction, the electric and magnetic field vectors satisfy the relations ∂Ey ∂Bz ∂Ez ∂By = − , = , ∂x ∂t ∂x ∂t ∂Hz ∂Dy ∂Hy ∂Dz = − , = . ∂x ∂t ∂x ∂t (b) Show that for the propagation in a conductor ∂Ey ∂x ∂Hz ∂x
∂Bz , ∂t ∂Dy = − − σEy , ∂t = −
where σ is the conductivity of the conductor.
231
(c) Show that Ey and Hz each satisfies a second order differential equation of the form ∂ 2 Ey ∂ 2 Ey ∂Ey = εµ + µσ . 2 2 ∂x ∂t ∂t
Problem 15.2 Energy flow in a conductor Using the relations of Problem 15.1, part (b), show that in a conductor the rate of decrease of electromagnetic field energy in a volume element dV is equal to the net rate at which energy flows out of this volume element plus the rate of Joule heating in the element. Problem 15.3 Practical example on propagation and attenuation If a field propagates as a plane wave in the form: E = E0 ei(ωt−kz) , where ω is the (real) radian frequency and k is a complex propagation constant that can include the effects of losses due to conduction, dielectric loss and magnetic loss, in lectures we derived a dispersion equation
k=ω
ε0 µ0 − µ00 ε00 +
σ ω
− i µ0 ε00 +
σ + ε0 µ00 ω
1 2
.
Consider propagation at a frequency of 1 MHz in a medium with the following properties: 1. Conductivity σ = 10−9 mho m−1 . 2. Dielectric susceptibility χe = 3 − i10−4 . 3. Magnetic susceptibility χm = 2 − i10−3 . (a) Calculate the phase velocity and wavelength at 1 MHz. (b) Compare these values with those for free space propagation at 1 MHz.
232
(c) Calculate the attenuation coefficient at 1 MHz. How far will the wave propagate before its amplitude falls to one tenth of its initial value? (d) Is this material a good conductor at 1 MHz? Over what frequency range would it be classified as a good conductor ?
233
16
Transitions Across Boundaries for Electromagnetic Fields
In the lectures on electric and magnetic properties of materials we have confined our attention principally to the case of a single material medium completely occupying the spatial region where electric and magnetic fields existed. We now investigate the relations between the fields which hold at a boundary between two different materials. They are of great help in the calculations of propagation problems in optics, where light can propagate between different material media. Across boundaries between different materials there are sharp changes in electrical properties of ε, µ, σ. On a macroscopic scale the fields may have to be regarded as varying discontinuously across such boundaries. The source of such discontinuities will be the surface polarization charges P~ · n ˆ and cur~ rents M × n ˆ discussed previously. The question we will address in this lecture is: How the electromagnetic fields transfer through a boundary between two materials? In the analysis of the transfer properties of the electromagnetic field across a boundary between two different materials, it is convenient to decompose the fields into two components, normal and tangential to the boundary between the materials and study how each of the components transfers through the boundary.
16.1
Normal Components of the Fields
We will use the Maxwell’s divergence equations I and II to investigate the transition of normal field components at a boundary between two materials. 16.1.1
~ Normal Component of B
Consider two materials with different constants ε and µ, as shown in the ~ propagates from material (1) Figure 55. Suppose that a magnetic field B ~ will be different in different materials due to the to (2). We expect that B presence of different surface currents.
234
~ = 0, to find To check this, we first apply the Maxwell’s equation II, ∇ · B ~ transfers through the boundary between how the normal component of B the two materials. Using the Gauss’ divergence theorem, the Maxwell’s equation II can be written in the integral form as I
~ ·n B ˆ dS = 0 ,
(82)
S
where S is an arbitrary surface closing some area on the boundary plane. Figure 55: A Gauss’s surface for the derivation of the boundary condition on the normal ~ component of B.
In order to evaluate the integral, we consider a thin cylinder of area δS and thickness δ` including the boundary, as shown in Figure 55. In this case, we can distinguish three surfaces that we represent by unit vectors n ˆ1, n ˆ 2 and n ˆ 3 , and therefore the surface integral in Eq. (82) splits into three terms (B~2 · n ˆ 1 ) δS + (B~1 · n ˆ 2 ) δS +
Z
~ ·n B ˆ s dS = 0 ,
sides
where the first term is the flux of the magnetic field through the top surface, the second term is the flux through the bottom surface, and the third term is the flux through the side surface of the cylinder. ~ is finite everywhere and we are interested in the transformation of Since B the field at the boundary, we make the height δ` tend to zero, δ` → 0. Then, the integral Z
~ ·n B ˆ dS → 0 as δ` → 0 .
sides
~ over the surface has Thus, we find that the total flux of the magnetic field B contributions from the top and bottom ends only B~2 · n ˆ 1 + B~1 · n ˆ2 = 0 . 235
(83)
Now, if we introduce the notation n ˆ1 = n ˆ
then
n ˆ 2 = −ˆ n,
and we find that Eq. (83) becomes n ˆ · (B~2 − B~1 ) = 0 , or equivalently, we may write that B2⊥ = B1⊥ , ~ in material (1) normal to the where B1⊥ is the component of the field B boundary and B2⊥ is the component in material (2) normal to the boundary. ~ is continuous at all points across Thus, the normal component of B a boundary separating two different materials. On physical grounds, we can understand this result by noting that the mag~1 × n ~2 × n netic fields of polarization currents M ˆ 1 and M ˆ 2 are parallel to ~ as shown the boundary and so do not affect the normal component of B, in Figure 56. 16.1.2
~ Normal Component of H
~ = µH, ~ we have Since B B1⊥ = µ1 H1⊥ = µ2 H2⊥ = B2⊥ . Thus, we obtain that µ2 H1⊥ = H2⊥ , µ1 which shows that the nor~ is not mal component of H continuous across a boundary between two different materials since µ1 and µ2 are not equal. Figure 56: Direction of magnetizations and reOn physical grounds, this result sulting polarization currents at the boundary be~ tween two different materials. for the normal component of H is the consequence of different magnetizations of the materials, when µ1 6= µ2 . 236
16.1.3
~ and E ~ Normal Component of D
~ = 0, an identical argument to that Since in nonconducting dielectrics ∇ · D ~ will show that the normal component of D ~ is applied the above for B continuous across a boundary. ~ = εE. ~ Since the normal component of D ~ In the case of dielectrics we write D is continuous across a boundary, we have ε1 E1⊥ = ε2 E2⊥ , ~ is not continuous across the which shows that the normal component of E boundary. On the physical grounds, we can understand it as follows. The electric field of the dielectric surface charge P~ · n ˆ is normal to the boundary, as shown in Figure 57. Since the polarizations of the dielectrics are different, so the fields of the dielectrics are different. This difference causes ~ the discontinuity in the incident E Figure 57: Polarization charges at the field. boundary between two different materials.
The discontinuity of the normal component of the electric field leads to a change of the direction of propagation of the field, as it is illustrated in Figure 58. The change of the direction depends on the ratio ε2 /ε1 . When ε2 /ε1 > 1, θ2 > θ1 .
16.2
Tangential Components of the Fields
Different boundary conditions apply to the components of the fields parallel (tangential) to a boundary between two different materials. We will use the Maxwell’s curl equations III and IV to investigate the transition of tangential field components at a boundary between different materials.
237
Figure 58: Boundry between two dielectrics showing change of the direction of propagation of an electric field. 16.2.1
~ Tangential Component of E
~ we will apply the Faraday induction law, For the tangential component of E, the Maxwell’s equation III, to a closed path such as shown in the Figure 59. We choose a closed loop in which the sides perpendicular to the boundary are made infinitely short compared to the parallel sides L. Consider the Maxwell’s equation III ~ ~ = − ∂B . ∇×E ∂t
Figure 59: A closed loop for the derivation of boundary conditions for the tangential com~ ponent of E.
I
~ · d~l = − E
l
Z a
Integrating both sides of this equation over the surface a and applying the Stokes’s theorem, we obtain
~ ∂B ·n ˆ 0 da . ∂t
Hence
~ 2 · tˆ2 + E ~ 1 · tˆ1 L + E
Z
~ · d~l = − E
ends
238
Z
~ ∂B ·n ˆ 0 da , ∂t
(84)
where n ˆ o is the unit vector normal to the surface a, and tˆ1 and tˆ2 are unit vectors along the paths L on the side (1) and (2), respectively.20 As dl → 0 the right-hand side and the second term on left-hand side of ~ and B ~ are finite everywhere. In this limit, the Eq. (84) go to zero, since E area enclosed by the path approaches zero. Since, tˆ2 = −tˆ1 = tˆ, we then have
~2 − E ~ 1 · tˆ = 0 . E
~ · ~t is the component of E ~ tangential to the surface. Since this is true But E ~1 ˆ for any t, we obtain E1k = E2k , where E1k and E2k are the components of E ~ 2 parallel to the boundary. Thus, we may conclude that: and E ~ is continuous across a boundary The tangential component of E between two different dielectric materials. The continuity of a tangential component can be written in an equivalent form as
~2 − E ~1 = 0 . n ˆ× E Explanation From Figure 60, we see that the ~ can be writcross product n ˆ×E Figure 60: Illustration of the direction of the ten as ~
vector n ˆ × E.
~ n ˆ×E
= E sin(π/2 − θ) = E cos θ .
~ is equal to the projection of the vector E ~ on the In other words, n ˆ×E ~ boundry. Thus, n ˆ × E is tangential to the boundary, i.e. it is the tangential ~ component of E. 20
Do not mix the unit vector n ˆ o normal to the surface a with the unit vector n ˆ normal to the boundary. Actually, n ˆo ⊥ n ˆ.
239
16.2.2
~ Tangential Component of H
Consider the Maxwell’s equation IV: ~ ~ = J~c + ε ∂ E , ∇×H ∂t where J~c is the conduction current, i.e. not counting polarization currents. The application of the Stokes’s theorem then gives I
~ · d~l = H
Z a
l
J~c · n ˆ da + ε
Z a
~ ∂E ·n ˆ da . ∂t
~ Both terms on the right-hand side go to zero as δl → 0 because J~c and ∂ E/∂t are finite. 21 Hence
~2 − H ~ 1 · tˆ = 0 , H
or
~2 − H ~1 = 0 . n ˆ× H ~ is continuous Thus, we conclude that the tangential component of H at all points across a boundary between two different materials. 16.2.3
~ Tangential component of B
~ is not in general We may easily deduce that the tangential component of B continuous across a boundary because of the presence of the magnetic polar~ ×n ization surface currents M ˆ , which do not have a finite current density as they flow in an infinitely thin surface layer. ~ the term Thus, if we examine the corresponding Maxwell equation for ∇× B, in the integral involving J~ may stay finite as δj → 0. 21
Infinite current density can occur in a material which has infinite electrical conductivity, but here we ignore this special case.
240
This conclusion can be justified as follows. Since ~ = ε 0 c2 B ~ −M ~ , H we have
~2 − H ~ 1 = ε 0 c2 B ~2 − B ~1 − M ~2 −M ~1 H
.
Take scalar product of both sides with tˆ, and using the fact that the tan~ is continuous gential component of H across a boundary
~2 − H ~ 1 · tˆ = 0 , H
we obtain
~2 −M ~ 1 · tˆ . ~2 − B ~ 1 · tˆ = 1 M B ε 0 c2
~ 1 6= M ~ 2 , the differIt is seen that if M Figure 61: Directions of magnetizations ence between the two magnetizations and resulting magnetic fields produced at ~ generates a discontinuity in the B a boundary between two materials. field, as shown in Figure 61.
Summarizing: Field components that are continuous across a boundary between two materials: ~ • The normal component of D. ~ • The tangential component of E. ~ • The normal component of B. ~ • The tangential component of H.
241
Revision questions Question 1. Which of the Maxwell’s equations are used to analyse the transmission of the normal components of the EM field through a boundary between two materials? Question 2. Which of the Maxwell’s equations are used to analyse the transmission of the tangential components of the EM field through a boundary between two materials? Question 3. Explain, using physical grounds, why the normal component of the ~ is continuous across a boundary between two mamagnetic field B terials. Question 4. Explain, using physical grounds, why the tangential component of ~ is continuous across a boundary between two the electric field E materials.
Tutorial problems Problem 16.1 Change in direction of a field line A (static) electric field line makes an angle θ1 to the normal to the boundary between two dielectrics. The dielectric constant on the ‘incident’ side is ε1 and on the other side is ε2 . There are no local charges other than the polarization charges in and on the dielectrics. Show that the angle the electric field makes to the normal to the boundary in the second dielectric is ε2 θ2 = arctan tan θ1 ε1
.
Problem 16.2 Angles and polarization charges An infinite slab having a dielectric constant of 4 is placed in a ~ 0. vacuum which is permeated by an electric field E
242
~ 0 should make to the boundary normal (a) Find the angle that E ~ so that E inside the slab makes an angle π/4 with the boundary normal. (b) Find the surface density of polarization charge on the slab in terms of E0 .
Problem 16.3 Wave reflection from a dielectric at normal incidence Consider waves normally incident on a boundary between two dielectrics having dielectric constants 4 (on the incident side) and 9. ~ (a) Draw a diagram showing the directions and magnitudes of E ~ and B in the incident, reflected and transmitted waves. Apply ~ and B ~ directly. Rememthe general boundary conditions for E ber that B = E/vp . (b) Show that the magnitude of the amplitude reflection coefficient is 0.2.
243
17
Propagation of an EM Wave Across a Boundary
In the preceding lecture, we have derived relations for transmission of electric and magnetic fields through a boundary between two non-conducting materials. We have treated each field separately. We now proceed to consider a propagation of an EM wave across a boundary between two materials. We shall find that for the propagation of an EM wave the situation is different that we cannot treat the fields separately. The boundary conditions must be satisfied simultaneously for both the electric and magnetic fields. This leads to a modification that at a boundary between two different materials, the general boundary conditions cannot be satisfied by a transmitted wave alone. There has to be a reflected wave also. What happens is that the incident radiation is absorbed by charges in the boundary and reradiated in all directions. The waves interfere destructively except in two directions along those of the reflected and transmitted waves. The directions, amplitudes and phases of the reflected and transmitted waves can be derived from the general boundary conditions already obtained. This is an explanation often presented in texbooks on classical optics. In this lecture, we shall attack these problems from the standpoint of the electromagnetic theory of light and will show how the propagation phenomenon can be understood with the help of the Maxwell’s equations. More precisely, in this and the following lecture, we consider a number of well-known optical phenomena and will show that they can be explained in terms of the electromagnetic theory of light.
244
17.1
Representation of Plane Waves in Different Directions
Let us first set up the formalism we shall use in the study of propagation of an EM wave through a boundary between two materials. The question we will try to answer is: What is the convenient method to represent plane waves propagating in different directions? Suppose that a plane wave propagates in the z direction. Then ~ =E ~ 0 exp[i (ωt − kz)] . E Let n ˆ p is the unit normal to the phasefront (wavefront) of the propagating wave, and ~r is a position vector that is independent of n ˆ p , as shown in Figure 62. The vector ~r determines position of a wave-front as seen by an obFigure 62: Illustration of the wave front monserver O. itored by an observed O distant ~r from the point A.
We see from Figure 62 that the distance z the wave propagated in time t is z=n ˆ p · ~r . Hence, we can write ~ =E ~ 0 exp[i (ωt − n E ˆ p · ~r k)] . Thus, if at the point A appear few plane waves propagating in different directions, the observer can distinguish them by different n ˆ ’s. In other words, the vector ~r sets a reference frame for the observation of different waves at the point A. Remember, that an EM wave is composed of the electric and magnetic fields, so we have to consider the propagation of both fields. However, there is ~ and E ~ of a plane wave that we can find the a useful relation between B magnetic field from the knowledge of the electric field. 245
17.1.1
~ in terms of E ~ Representation of B
The magnetic field of a plane wave propagating in the direction determined by the unit vector n ˆ p can be found from the relation ~ . ~ = kn ˆp × E B ω
(85)
Proof: We shall show that the relation (85) arises from the Maxwell equation III: ~ ~ = − ∂B . ∇×E ∂t To prove it, we expand both sides of this equation, and obtain
ˆi ˆj 0 0 Ex Ey
kˆ
∂ ∂z
0
=−
~ ∂B ∂Bx ˆ ∂By = −ˆi −j , ∂t ∂t ∂t
where we have used the fact that the wave is transverse and propagates in the z direction. Comparing the coefficients standing at the same unit vectors, we find that the x component gives −
∂Ey ∂Bx =− ∂z ∂t
from which, we obtain ikEy = −iωBx
or
ω Ey =− . Bx k
The y component gives: ∂Ex ∂By =− ∂z ∂t from which, we obtain −ikEx = −iωBy
or 246
Ex ω = . By k
Thus
E = Ex2 + Ey2
1 2
=
1 ω 2 ω Bx + By2 2 = B . k k
~ B, ~ n ~ ×B ~ gives the direction of n Since E, ˆ p , are mutually orthogonal, E ˆp ~ gives the direc(Poynting result) of the propagation direction. Then n ˆp × E ~ Hence, written vectorially tion of B. ~ ~ =B = k n ~ . H ˆp × E µ ωµ √ If ε and µ are real, k = ω εµ, and then we have ~ = H
s
ε ~ . n ˆp × E µ
(86)
(87)
~ and H ~ to In the next few lectures, we will use the continuity conditions for E analyse different optical phenomena. With the relations (86) or (87), we will be able to limit the analysis to the electric field alone, as knowing properties ~ we can infer from Eq. (86) properties of H. ~ of E,
17.2
Directions of Reflected and Transmitted Waves
Armed with the formalism to study propagation of an EM wave between two materials, we now consider some familiar elementary results of classical optics on reflection and transmission to show that they can be derived from the Maxwell’s equations. The most useful results in this connection concern the continuity of tangential components of ~ and H ~ fields. the E However, most of the results can be derived from the fact that ~ the tangential component of E is continuous across a boundary. Figure 63: Propagation vectors for the incident, reflected and refracted beams at a 247 boundary between two materials.
Consider properties of an EM wave (radiation beam) at a boundary between two materials, as illustrated in Figure 63. Let the origin of position coordinates ~r be located on the boundary S. Then the electric fields for incident, reflected and transmitted waves are ~i = E ~ 0 exp [i (ωt − n E ˆ i · ~r k1 )] , ~r = E ~ 1 exp [i (ωt − n E ˆ r · ~r k1 )] , ~t = E ~ 2 exp [i (ωt − n E ˆ t · ~r k2 )] . These equations determine directions of the three electric fields relative to the direction of observation ~r. Note that we have assumed the presence of the reflected wave without any particular reasons. Why do we need the presence of the reflected wave in the propagation of an incident wave through the boundary? The answer to this question is provided by the requirement that the tangen~ and H ~ must be continuous through the boundary. tial components of E ~ is polarized in the plane of incidence, i.e plane containing n Suppose that E ˆ and n ˆ i . Then, the following two equations result from the continuity conditions Ei cos θi − Er cos θr = Et cos θt ,
(88)
Hi + Hr = Ht .
(89)
and
Since H=
k E, ωµ
the relation (89) takes the form Ei + Er =
k2 Et , k1
248
(90)
as for a dielectric µ1 = µ2 = µ0 . Now we see that without Er , Eq. (88) gives Ei =
cos θt Et , cos θi
while Eq. (90) gives Ei =
k2 Et . k1
Thus, without Er we would get two different values for Ei or Et , which we cannot accept as both continuity conditions Eqs. (88) and (89) must be satisfied at the same moment. Hence, we conclude that the continuity conditions ~ and H ~ will be satisfied only if Er 6= 0. for E The same argument applies to the need of the transmitted wave. Without Et , we get two equations Er =
cos θi Ei cos θr
and Er = −Ei ,
which evidently cannot be in general satisfied simultaneously. Now, we will try to answer the following question: What are the relative directions of the three waves represented here by the unit vectors n ˆi, n ˆ r and n ˆt? To answer this question, we use the boundary condition for the tangential ~ component of E:
~i + E ~r = n ~t , n ˆ× E ˆ×E i.e. n
o
~ 0 exp [i (ωt − n ~ 1 exp [i (ωt − n n ˆ× E ˆ i · ~r k1 )] + E ˆ r · ~r k1 )] n
o
~ 2 exp [i (ωt − n =n ˆ× E ˆ t · ~r k2 )] 249
.
This relation must hold over the whole surface S for all ~r (subject to n ˆ ·~r = 0). Thus, the exponential phase factors must all be the same. Otherwise, if it was true for one ~r it would not be true for other ~r’s, but we have a freedom of choosing ~r. Hence k1 n ˆ i · ~r = k1 n ˆ r · ~r = k2 n ˆ t · ~r , from which we have n ˆ i · ~r = n ˆ r · ~r ,
(91)
and n ˆ i · ~r =
k2 n ˆ t · ~r . k1
These relations will help us to prove that: Incident, reflected and transmitted waves are coplanar. In other words, n ˆi, n ˆr, n ˆ t are coplanar, the property observed in experiments. To show this, we first define planes determined by pairs of the unit vectors (ˆ ni , n ˆ ), (ˆ nr , n ˆ ) and (ˆ nt , n ˆ ), and next will show that all the pairs of the unit vectors form the same plane. It is well known from the vector analysis that a cross product between two vectors determines a plane in which these two vectors are. Thus, we will determine the cross products n ˆi × n ˆ, n ˆr × n ˆ and n ˆt × n ˆ , and then will show that the cross products are equal. In order to find the cross products, we use the following relation valid for arbitrary ~r and n ˆ , such that n ˆ · ~r = 0. ~r = −ˆ n × (ˆ n × ~r) . Proof: Since n ˆ × (ˆ n × ~r) = (ˆ n · ~r)ˆ n − (ˆ n·n ˆ )~r , 250
(92)
and n ˆ · ~r = 0 as the vector ~r lies on the plane S, we obtain n ˆ × (ˆ n × ~r) = −~r , as required. Using Eq. (92), we can then write Eq. (91) as n ˆ i · [ˆ n × (ˆ n × ~r)] = n ˆ r · [ˆ n × (ˆ n × ~r)] . Interchanging (·) and (×) products, we find (ˆ ni × n ˆ ) · (ˆ n × ~r) = (ˆ nr × n ˆ ) · (ˆ n × ~r) . This must be Figure 64: Illustration that the inci- plane S. Thus:
true
for
all ~r
in
dent and reflected beams are coplanar.
n ˆi × n ˆ=n ˆr × n ˆ. This implies that n ˆ r is in the ”plane of incidence”, i.e. the plane containing n ˆ and n ˆ i , as shown in Figure 64. Similarly, the relation k1 n ˆ i · ~r = k2 n ˆ t · ~r implies that k1 n ˆi · n ˆ × (ˆ n × ~r) = k2 n ˆt · n ˆ × (ˆ n × ~r) . As before, by interchanging (·) and (×) products, we find that k1 (ˆ ni × n ˆ ) · (ˆ n × ~r) = k2 (ˆ nt × n ˆ ) · (ˆ n × ~r) . This relation is valid for all ~r in S. Thus k1 n ˆi × n ˆ = k2 n ˆt × n ˆ. This means that n ˆ t is in the plane of incidence. Thus, we may conclude that n ˆi, n ˆ , and n ˆ t are coplanar. The coplanar property of the waves is observed in any experiment. Note that what we have just shown is an another example of a remarkable triumph of the Maxwell’s electromagnetic theory.
251
17.3
Angle of Reflection and Snell’s Law of Refraction
We now illustrate that other familiar laws of elementary optics can also be derived from the Maxwell’s equations. We have already shown that n ˆi × n ˆ=n ˆr × n ˆ, from which and from the definition of the cross product, we have that sin θi = sin θr . Hence θi = θr . Thus, the angle of incidence equals the angle of reflection, another familiar law of elementary optics. Moreover, we have shown that k1 n ˆi × n ˆ = k2 n ˆt × n ˆ, from which we have k1 sin θi = k2 sin θt , and then k2 n2 sin θi = = n12 = . sin θt k1 n1 This is the well-known law of refraction in optics, called the Snell’s law. In the case where k1 and k2 are purely real (e.g. in dielectrics), the refractive index has a simple physical interpretation n12 =
k2 ω/k1 v1 = = , k1 ω/k2 v2
i.e. the refractive index in equal to the ratio of phase velocities.
252
Revision questions Question 1. Why there are three waves, incident, reflected and refracted in a propagation of an EM wave between two dielectric materials? Question 2. What is a plane of incidence? Question 3. Incident, reflected and transmitted waves are coplanar: True or false? Question 4. State the Snell’s law, and briefly explain how it is derived from the Maxwell’s equations. Question 5. How does the refractive index depend on phase velocities of an EM wave propagating between two dielectric materials?
253
18
Fresnel’s Equations
In the propagation of an EM wave between two different materials, it is important to know how much of the energy of the incident beam is reflected and transmitted. In this lecture, we will find relations between the amplitudes of the incident, reflected and refracted beams. More precisely, we will express the amplitudes of the reflected and transmitted beams in terms of the amplitude of the incident beam. We shall see that the relations depend on the angle of propagation of the incident beam and the material constants. ~ does not give sufficient information The boundary condition on tangential E ~ r and E ~ t in terms of E ~ i . For a given E ~ i there are still two to calculate E ~ viz E ~ r and E ~ t . We unknowns in the equation for continuity of tangential E ~ i, E ~ r and E ~ t . This can be obtained from need a second relation between E the continuity of the magnetic field. ~ is not continuous across a boundary We know that tangential component of B ~ because of the presence of M × n ˆ surface currents in magnetized materials. To allow for such possibilities we can use the more general condition that ~ is continuous across a boundary, i.e. tangential component of H
~i + H ~r = n ~t . n ˆ× H ˆ×H Since ~ ~ , ~ =B = k n ˆp × E H µ ωµ where n ˆ p is a unit ray vector in the direction of propagation, we have an equation !
k1 k2 ~ i + k1 n ~r = n ~t n ˆ× n ˆi × E ˆr × E ˆ× n ˆt × E ωµ1 ωµ1 ωµ2
!
,
(93)
which together with
~i + E ~r = n ~t , n ˆ× E ˆ×E ~ r and E ~ t in terms of E ~ i. contains sufficient information to determine E
254
(94)
Since these two equations involve vectors of an arbitrary polarization, the solution of these two equations is greatly facilitated by decomposing each of the vectors into two components: electric field components parallel and normal to the plane of incidence. Then, we can solve these two cases separately that are necessary and sufficient to determine the relations between the amplitudes valid for an arbitrary polarization since a superposition of these two cases gives the solution for an arbitrary polarized incident wave. Equations (93) and (94) also provide a simple explanation of why we need reflected and transmitted fields at the boundary to obtain the correct results for the field amplitudes.
18.1
~ i normal to plane of incidence E
~ i of an EM wave acting on a boundary is Suppose that the electric field E normal to the plane of incidence, as shown in Figure 65. In this case, the incident electric ~ i is purely tangential to the field E boundary. Since the materials are ~ r and E ~t isotropic, the induced fields E will also be tangential to the the boundary. Thus the condition
~i + E ~r = n ~t n ˆ× E ˆ×E gives ~0 + E ~1 = E ~2 . E
Figure 65: Incident beam with the elec-
(95)
tric field normal to the plane of incidence.
Note then that ~0 = n ~1 = n ~2 = 0 . n ˆ·E ˆ·E ˆ·E If we want to express E1 and E2 in terms of E0 , we need an another equation ~ 0, E ~ 1 and E ~ 2 , which comes from the continuity condition of H ~ through for E the relation ! ! k1 k2 k1 ~ ~ ~ n ˆ i × Ei + n ˆ r × Er = n n ˆ t × Et . n ˆ× ˆ× ωµ1 ωµ1 ωµ2 255
~ α exp i(ωt − n Since the phase factors in E ˆ α · ~r k) are the same, we obtain i k1 h ~0 + n ~ 1 = k2 n ~2 . n ˆ× n ˆi × E ˆ× n ˆr × E ˆ× n ˆt × E ωµ1 ωµ2
~ × (B ~ × C) ~ = (A ~ · C) ~ B ~ − (A ~ · B) ~ C, ~ and the fact Using a vector identity A ~ ~ ~ that n ˆ · E0 = n ˆ · E1 = n ˆ · E2 = 0, we obtain k1 ~0 + n ~ 1 = k2 n ~2 . n ˆ·n ˆiE ˆ·n ˆrE ˆ·n ˆtE ωµ1 ωµ2
However n ˆ·n ˆ i = cos(π − θi ) = − cos θi , n ˆ·n ˆ r = cos θr = cos θi , n ˆ·n ˆ t = cos(π − θt ) = − cos θt , and then, we obtain k1 ~ ~ 2 cos θt . ~ 1 cos θi = k2 E E0 cos θi − E ωµ1 ωµ2
(96)
~ are all in the same direction, we might as well drop the vector Since E’s signs. Thus, Eqs. (95) and (96) take the form E0 + E1 = E2 , E0 cos θi − E1 cos θi =
(97) k2 µ1 E2 cos θt . k1 µ2
Eliminating E1 using Eq. (97), that E1 = E2 − E0 , we get E0 cos θi − (E2 − E0 ) cos θi =
k2 µ1 E2 cos θt , k1 µ2
which can be written as !
k2 µ1 2E0 cos θi = cos θi + cos θt E2 , k1 µ2 or 2k1 µ2 E0 cos θi = (k1 µ2 cos θi + k2 µ1 cos θt ) E2 . 256
(98)
Thus E2 =
2k1 µ2 cos θi E0 . k1 µ2 cos θi + k2 µ1 cos θt
(99)
Using the Snell’s law k1 sin θi = k2 sin θt , we can eliminate θt . This will allow us to predict the amplitude of the reflected wave from only knowing materials properties (k1 , k2 , µ1 , µ2 ) and the angle of incidence. Thus, from the Snell’s law, we have k2 cos θt =
q
k22 − k12 sin2 θi
and then substituting to Eq. (99), we get E2 =
2k1 µ2 cos θi q
k1 µ2 cos θi + µ1 k22 − k12 sin2 θi
E0 .
(100)
Similarly, eliminating E2 = E0 + E1 from Eq. (98) above, we obtain q
E1 =
k1 µ2 cos θi − µ1 k22 − k12 sin2 θi q
k1 µ2 cos θi + µ1 k22 − k12 sin2 θi
E0 .
(101)
Equations (100) and (101) are called Fresnel equations for the electric field amplitudes. ~ fields are not parallel to each other, but their relative The corresponding H magnitudes can be deducted from equations of the form ~ ~ = k n ˆ×E H µω
18.2
i.e.
H=
kE . µω
~ i in the plane of incidence E
~ i is in the plane of incidence. In this case H ~ i is tangential Suppose now that E ~ r and H ~ t are tangential too. Thus to the boundary plane and then H
~i + H ~r = n ~t n ˆ× H ˆ×H 257
becomes ~0 + H ~1 = H ~2 . H ~ is given by The continuity of tangential E
~0 + E ~1 = n ~2 . n ˆ× E ˆ×E We have two simultaneous equations. In contrast to the previous case, it is ~ now simpler to work in terms of H. ~ in terms of H ~ Thus, we express E ~ = − µω n ~ . ~ = −ω n ˆ×B ˆ×H E k k Hence µ1 µ2 ~ 0 + µ1 n ~1 = n ~2 . n ˆ× n ˆi × H ˆr × H ˆ× n ˆt × H k1 k1 k2 ~ i normal to plane of Continuing with a procedure similar to the case of E incidence, we obtain q
H1 =
H2 =
k22 µ2 cos θi − µ1 k1 k22 − k12 sin2 θi q
k22 µ1 cos θi + µ2 k1 k22 − k12 sin2 θi 2k22 µ2 cos θi q
k22 µ1 cos θi + µ2 k1 k22 − k12 sin2 θi
H0 ,
(102)
H0 .
(103)
Equations (102) and (103) are called Fresnel equations for the magnetic field amplitudes. ~ are not parallel, but their relative amplitudes can be deThis time the E’s ducted from the relation E = (µω/k)H.
18.3
Fresnel Equations for dielectric media
Consider now two specific examples of propagation of an EM wave between two dielectric materials. We shall show that in this case, the Fresnel equations can be simplified to forms containing only geometrical factors.
258
In a dielectric: conductivities σ1 = σ2 = 0, µ1 = µ2 = µ0 , k = 2π/λ = real, and vp = Since k ∼ 1/λ ∼ 1/vp ∼
√ √
ω 1 =√ . k εµ0
ε, the Snell’s law (k1 sin θi = k2 sin θt ) becomes
ε1 sin θi =
√
ε2 sin θt ,
and then sin θi = sin θt
s
ε2 v1 = = n12 . ε1 v2
Consider two examples: Example 1: In the first example we assume that E is normal to the plane of incidence. From Eq. (100), we have that in general 2k1 µ2 cos θi E2 = . E0 k1 µ2 cos θi + k2 µ1 cos θt Since for a dielectric µ1 = µ2 = µ0
and
sin θi k2 = , k1 sin θt
we obtain E2 2 cos θi 2 cos θi = = k2 sin θi E0 cos θi + k1 cos θt cos θi + sin cos θt θt 2 cos θi sin θt = . cos θi sin θt + sin θi cos θt Hence E2 2 cos θi sin θt = . E0 sin (θt + θi )
259
Similarly, we can readily show that sin(θt − θi ) E1 = . E0 sin (θt + θi )
(104)
Example 2: Consider now the case of E in the plane of incidence. Following the same procedure as above, we can show that E2 2 cos θi cos θt = , E0 sin (θt + θi ) cos (θi − θt ) and similarly tan(θi − θt ) E1 . = E0 tan (θt + θi ) It is seen from the above examples that the reflection and refraction amplitudes are different for linearly polarized waves with the electric field vector oscillation in the direction normal to the plane of incidence and for waves with the electric field vector oscillating in the plane of incidence. However, the above examples show an interesting feature that for normal incidence, (θi = 0), the reflected and refracted amplitudes are independent of the polarization.
Revision questions Question 1. Explain in general, what are the Fresnel’s equations? Question 2. Under which conditions the ratios of the reflected to the incident and transmitted to the incident beams amplitudes depend solely on the angles involved? Question 3. Does the amplitude of the reflected beam depend on polarization of the incident beam?
260
Tutorial problems Problem 18.1 Reflected and transmitted fields at a boundary between two nonconducting materials Consider a propagation of an electromagnetic wave between two different non-conducting materials. Prove the following: ~ i in the plane of incidence (a) For E E1 sin 2θt − sin 2θi = E0 sin 2θt + sin 2θi
and
E2 4 sin θt cos θi = . E0 sin 2θt + sin 2θi
~ i normal to the plane of incidence (b) For E E1 sin(θt − θi ) = E0 sin(θt + θi )
261
and
E2 2 sin θt cos θi = . E0 sin(θt + θi )
19
Applications of the Boundary Conditions and the Fresnel Equations
In this lecture, we examine some of the consequences of the Snell’s law. There are two cases possible: n2 > n1 and n1 > n2 . In the first case, an optical wave travels from an optically ”rarer” to optically ”denser” medium. In the second case, we have the inverse situation. We will consider these two cases separately for dielectrics and for conductors.
19.1
Applications in dielectrics
In propagation between two dielectrics there are interesting polarization dependent effects that the two components of the EM wave (parallel and perpendicular to the plane of incidence) are not transmitted and reflected equally. 19.1.1
Polarization by reflection
~ in the plane of incidence, as shown in Figure 66. Consider the case of E From the Fresnel equations, we have that the ratio of reflected to incident amplitude is tan(θi − θt ) E1 . = E0 tan(θi + θt ) Note from the ratio, if θi + θt = π2 then tan(θi + θt ) = ∞ and consequently E1 = 0 . However, according to Eq. (104), at the same time E1 will not be Figure 66: Geometry of reflection and ~ field polarized zero for the electric field comporefraction for the incident E nent normal to the plane of inciin the plane of incidence. dence.
262
~ i has arbitrary polarization then E ~ r will be plane polarized with E ~r Thus, if E normal to the plane of incidence. If θi + θt =
π 2
then θt =
π 2
− θi . Thus
sin θi = n21 = sin θt
s
ε2 sin θi = = tan θi . ε1 sin( π2 − θi )
Hence, the angle of incidence for total linear polarization of the reflected wave is s
θi = arctan
ε2 . ε1
It is known in the literature as the Brewster’s angle. ~ is in the plane of incidence, There is an equivalent alternative proof that if E then E1 = 0 (no reflected field polarized in the plane of incidence). Alternatively, it can be proved using the continuity conditions for the tan~ and H, ~ from which we have gential components of E E0 cos θ − E1 cos θ = E2 cos θt ,
(105)
H0 + H1 = H2 .
(106)
and
Since H= and θt =
π 2
k E, ωµ
(107)
− θ, we get E0 − E1 = E2 tan θ , E0 + E1 = n12 E2 ,
where n12 = k2 /k1 .
263
(108)
However, from the Snell’s law we have that sin θ sin θ sin θ = = = tan θ = n12 . sin θt cos θ sin π2 − θ Thus, Eqs. (108) will be satisfied simultaneously only if E1 = 0, i.e. there is no reflected field in the plane of incidence. ~ i polarized normal We now prove that the above conclusion is not true for E to the plane of incidence. In this case H0 cos θ − H1 cos θ = H2 cos θt ,
(109)
E0 + E1 = E2 .
(110)
and
From Eq. (109) we find, after applying Eq. (107), that E0 − E1 =
cos θt E2 = n12 E2 . cos θ
Thus, E1 must be present, otherwise E0 or E2 would have two different values.
19.1.2
Total internal reflection
We now illustrate an another interesting effect that has many practical applications: Total internal reflection at a boundary between two dielectrics. It is well known from experiments that in the case of propagation from an optically more dense to optically less dense medium, e.g. from water in to air, the incident beam can be completely reflected at the boundary with no transmission. Does it mean that there is no refracted beam? If this is the case, one can easily find from the conditions of continuity of the tangental ~ and H ~ that the refracted beam has to be present to satisfy components of E those two conditions. It looks that the EM theory is in a trouble, but no panic, lets see how we can handle this problem in terms of the EM theory.
264
Consider the Snell’s law sin θi , sin θt = n21
s
where n21 =
ε2 . ε1
One can see that in the case of n21 < 1, that happens when ε2 < ε1 , and when the wave is going from an optically more dense to optically less dense medium, real angles θt are obtained only for sin θi ≤ n21 . Since θt increases with θi , an interesting situation arises when θt = π/2, at which angle the refracted wave glaze along the boundary. The angle of incidence at which θt = π/2 or equivalently at which sin θi = n21 is called a critical angle for propagation. For greater θi , sin θt > 1, and then the angle of refraction θt becomes imaginary. In this case, there is no real refracted wave, only a reflected wave. What does it mean ”no real” refracted wave? The answer to this question is provided by calculating the cosine of θt , which in the case of the total internal reflection is an imaginary number cos θt = =
q
1 − sin2 θt =
v u u sin2 θi ±it
n221
v u u t1 −
sin2 θi n221
− 1 = iβ ,
We see that although sin θt is still real, cos θt becomes imaginary when sin θt > 1. What are the consequences of this property? Consider the propagation of the transmitted wave in the less optically dense medium, as illustrated in Figure 67, for which Et = E2 e−i(ωt−ˆnt ·~rk) , 265
with the propagation distance n ˆ t · ~r = z cos θt + x sin θt = iβz + αx . Then 0
0
0
0
Et = E2 e−i(ωt−iβ z−α x) = E2 e−β z e−i(ωt−α x) . Here β 0 = kβ and α0 = kα. One can see that there is attenuation of the field amplitude in the z direction but no phase propagation. Phase propagation occurs in the x direction along the boundary. Thus, for sin θt > 1, an evanescent wave exists along the boundary (in the x direction) which is attenuated exponentially in the second medium in the normal z direction. We may say that the field enters into the second material, but the wave does not. This illustrates a general method of applying the Fresnel equations. For only a limited range of circumstances will all the Figure 67: A schematic diagram for angles θi , θr , θt be real. We can howcalculation of the propagation distance ever always apply a generalized Snell’s of the transmitted wave under the total Law k sin θ = k sin θ to find sin θ and 2 t 1 i t reflection. cos θt and proceed as above. Note that the planes of constant phase are normal to the boundary (i.e. they have their normals tangential to the boundary). The phase of the transmitted wave below the boundary must match the incident wave above. The wavelength in the second medium is λ0 =
2π 2π = = 0 α kα
2π 2π sin θi λ0 n21
=
λ0 n21 , sin θi
where λ0 is the wavelength of freely propagating waves in this medium. The planes of constant amplitude in the transmitted medium are parallel to the boundary.
266
19.2
Transmission and Reflection at a Conducting Surface
Propagation of EM fields in conductors (metals) is more complicated phenomenon than in dielectrics. Consider a propagation of an EM wave in a medium composed of a dielectric and a conductor, and assume that the incident wave originates in the dielectric. From previous work, we know that in a dielectric the propagation number is k12 = ε1 µ1 ω 2 and in a conductor iσ . = ε2 µ2 ω − iωσµ2 = ε2 µ2 ω 1 − ε2 ω A characteristic property of propagation in the conductor is that in the limit of of a good conductor, σ/(ε2 ω) 1, the propagation number k2 → ∞. As we shall see this is a major factor for the propagation in the conductor to be completely different than in the dielectric. k22
2
2
Consider the Snell’s law which can be written as k1 sin θi . sin θt = k2 Assume that the metal is a good conductor. Since in a good conductor k2 → ∞, we see from the above equation that in the propagation from a dielectric to a good conductor, sin θt → 0 independent of θi . Thus, the direction of the transmitted wave is normal to the surFigure 68: Propagation of an EM face independent of the angle of inciwave from a dielectric to a conductor. dence θi . ~ and H ~ in the conductor lie tangential As a consequence, the field vectors E to the boundary and so the normal components of these vectors on the conductor side of the boundary are zero.
267
It follows that: ~ (or H) ~ is continuous across the • Since the normal component of B ~ or B ~ is zero on the dielectric boundary, the normal component of H side also. ~ of the reflected wave must be equal Thus the normal component of B and opposite to that of the incident wave. • In a good conductor k2 2 2 k1
=
K 1 −
iσ →∞ εω
as
σ 1, εω
and then we have from the Fresnel equations E2 → 0
E1 → −E0 .
and
This means that the electric field in the conductor (which is tangential to the boundary) → 0. Since Ek is continuous across the boundary, we have that Ek is zero also in the dielectric at the boundary. ~ of the reflected wave must be equal and Thus, the tangential component of E opposite to that of the incident wave.
In summary, we have two useful special boundary conditions at the surface between a dielectric and a perfect conductor: ~ = 0. 1. The tangential component of E ~ or H ~ = 0. 2. The normal component of B
268
19.2.1
Field vectors at normal incidence
We now consider the special case of normal incidence at a boundary, i.e. when the wave propagation vector coincides with the normal to the boundary. Figure 69 shows how the field vectors must look in the plane of the surface between dielectric and conductor. Since we already know that for a good conductor for which σ →∞, εω the tangential component of the transmitted electric field Et = 0, and the tangential component of the reflected magFigure 69: The field vectors in the netic field Hr = Hi , we find that the replane of the surface between dielectric lations between the field components take and conductor. the form Et = Ei − Er Ht = Hi + Hr
→ 0, → 2Hi .
In this case, the power reflection coefficient becomes αp =
E2 Er Hr = r2 Ei Hi Ei
→ 1 as
σ →∞. εω
Thus, we have the total reflection of the energy carried by the EM wave. We see an advantage of using dielectric-conductor boundary to propagate an EM wave in a material. 1. The energy of the wave is completely reflected, and in fact the total reflection is independent of the angle of incidence. 2. The polarization remains constant in the propagation. Before concluding the lecture, let us clarify a problem regarding the reflection at the normal incidence. Namely, one could think that under the normal incidence, there is only the incident and transmitted wave with no reflected 269
wave. Here, we prove the necessity of assuming the existence of the reflected wave. ~ is normal to the plane of incidence. Then, from the continuity Assume that E of the tangential components at the boundary, we have Ei − Er = Et , Hi + Hr = Ht .
(111) (112)
However s
H=
ε E= µ0
s
ε ε0
s
s
ε0 ε0 E=n E. µ0 µ0
Thus, Eq. (112) can be written as n1 Ei + n1 Er = n2 Et , and then we get two equations for the amplitudes of the electric field Ei − Er = Et , n2 Et . Ei + Er = n1 If Er is missing, we could not simultaneously satisfy both equations. Thus, there always is a reflected wave in the normal incidence.
270
Revision questions Question 1. What is a Brewster’s angle and why one could call it a polarizing angle? Question 2. Under what conditions will the reflected and transmitted amplitudes for perpendicular polarization be the same as those for parallel polarization? Question 3. Under which conditions an incident wave will be completely reflected at a boundary between two dielectrics? Question 4. What is meant by an evanescent wave? Question 5. What are the boundary conditions at a surface between a dielectric and a perfect conductor? Question 6. In the propagation of an EM wave between a dielectric and a perfect conductor, the direction of the transmitted wave is normal to the boundary surface independent of the angle of incidence: True or false? Question 7. In the propagation between a dielectric and a perfect conductor what happens to the electric field at the boundary? Question 8. Is there a reflected beam at the normal incidence to a boundary between two materials?
271
Tutorial problems Problem 19.1 Brewster’s angle for different polarizations Consider a propagation of an electromagnetic wave between two non-conducting materials. ~ i in the plane of incidence, find (a) For the polarization of the E the relation between the critical angle of incidence and the Brewster’s angle. Note, at the critical angle of incidence θt = π/2, and at the Brewster’s angle of incidence θi + θt = π/2. (b) Show that under the condition of no reflection at a boundary between two non-conducting materials, the sum of the Brewster’s angle and the angle of refraction is π/2 for ~ i normal to the plane of incidence, ε1 = ε2 = ε0 and µ1 6= µ2 . (i) E ~ i in the plane of incidence, µ1 = µ2 = µ0 and ε1 6= ε2 . (ii) E
Problem 19.2 Total internal reflection independent of polarization Show, using the Fresnel’s equations that under the total internal reflection, energy of an incident wave is completely reflected independent of polarization of the wave.
272
20
Propagation of an EM Wave in a Rectangular Waveguide
In many practical problems we want to send a signal (EM wave) to different places in an efficient way. Sending it in an uncontrolled way through the free space in air is not a good choice. Due to scattering of the wave on different objects there will be significant losses and uncontrolled changes of the direction of propagation. Therefore, we need some device which would help us to send the signal in more controlled way and where it would be protected from degradation. Conducting rectangular waveguides provide this: the waves can be propagated and delivered to the specific points (receivers) without significany losses. In this lecture, we discuss the propagation of bounded EM waves by considering the propagation of radiation through a waveguide where the radiation is fully confined in the transverse plane. We will consider the case where the bounding walls are made of a conductor, are planar and cross section is rectangular. Figure 70 shows an example of the rectangular waveguide. An EM wave propagating along the z direction may undergo a multiple reflection from the walls. Thus, we are likely to get standing waves in the x direction due to reflections between the walls x = 0 and x = a. We also get standing waves in the y direction due to reflections between the walls y = 0 and y = b. Figure 70: A schematic diagram of a rectangular waveguide.
The principles behind the propagation of an EM wave along the rectangular waveguide are analysed by solving the Maxwell’s equations for the fields inside the waveguide subject to the good conductor boundary conditions being satisfied at x = 0, x = a, y = 0 and y = b. We write the z
273
dependence of any field component in the form e−γz
Thus
∂ ≡ −γ , ∂z
where γ describes the propagation conditions, e.g. γ purely imaginary describes a wave propagating without loss. ~ H, ~ and we use We describe the electromagnetic field by the vector pair E, the Maxwell’s equations in the form ~ = ρ =0, ∇·E ε ~ ~ =0, ∇ · B = 0 or ∇ · H ~ ~ ~ = − ∂ B = −µ ∂ H , ∇×E ∂t ∂t ~ ~ ~ + ε ∂E , ~ = J~ + ε ∂ E = σ E ∇×H ∂t ∂t where σ is the conductivity of the interior of the waveguide, not the bounding metallic surfaces. We assume the interior if filled with a dielectric, but allow a possible non-zero conductivity of the dielectric. As we shall see, certain characteristic modes of propagation are found. In general a characteristic mode is one which propagates with constant polarization. In a rectangular waveguide, one may have TE (transverse electric) waves, or TM (transverse magnetic) wave, but not TEM wave. If both E and B fields are transverse, the wave would be going straight down the guide. However, such a wave would not satisfy various boundary conditions and could not be transmitted through the waveguide. ~ is transverse to the direction in which the For a TE wave, the electric field E ~ is wave is propagating that is, down the waveguide. The magnetic field B ~ and therefore there is a substantial component of B ~ that perpendicular to E points in the direction of the waveguide.
274
20.1
Transverse Electric (TE) Modes
As we have already mentioned is possible to propagate a wave with the elec~ transverse to the tric field polarized in the xy plane, a TE wave with E waveguide axis, by lifting the restriction that the magnetic field is transverse to the direction of propagation. To show this, we will look for the solution of the Maxwell’s equations with Ez = 0 that satisfies the good conductor boundary conditions. The outline of the procedure is as follows: First, using the Maxwell’s equations, we find relations between the components of the electric and magnetic fields. Next, we convert these relations into a single differential (wave) equation for Hz , the longitudinal component ~ Then we find the solution of the wave equation that satisfies the good of H. conductor boundary conditions. We proceed as follows: Since the waveguide is of a rectangular shape, we consider the Maxwell’s equations in Cartesian coordinates. ~ + µ ∂ H~ = 0, we have • From III, ∇ × E ∂t ˆi ∂ ∂x E
x
ˆj ∂ ∂y
Ey
kˆ −γ 0
~ + iωµH
=0.
Hence x component : y component : z component :
γEy + iωµHx = 0 , − γEx + iωµHy = 0 , ∂Ey ∂Ex − + iωµHz = 0 . ∂x ∂y
~ = 0, we have • From II, ∇ · H ∂Hx ∂Hy ∂Hz + + =0, ∂x ∂y ∂z 275
(113) (114) (115)
and from the fact that ∂/∂z = −γ, we get ∂Hx ∂Hy + − γHz = 0 . ∂x ∂y
(116)
~ − (σ E ~ + ε ∂ E~ ) = 0, we have • From IV, ∇ × H ∂t ˆi ∂ ∂x H
ˆj ∂ ∂y
x
Hy
kˆ −γ Hz
− (σ
~ =0, + jωε)E
which in terms of the components gives ∂Hz + γHy − (σ + iωε)Ex = 0 , ∂y ∂Hz − (σ + iωε)Ey = 0 , ∂x ∂Hy ∂Hx − =0 (Ez = 0) . ∂x ∂y
−γHx −
(117) (118) (119)
~ = ρ/ε = 0, we have • From I, ∇ · E ∂Ex ∂Ey + =0 ∂x ∂y
(Ez = 0) ,
(120)
where we have used the TE condition Ez = 0. We shall now express Ex , Ey , Hx and Hy in terms of Hz to get a wave equation for Hz . From Eqs. (113) and (114), we get Ex Ey iωµ =− = . Hy Hx γ
(121)
Using this relation, we can then express Ex and Ey in terms of Hy and Hx and then show that Eq. (115) becomes identical to Eq. (116).
276
Thus, we use Eq. (121) in Eqs. (117) and (118). In Eq. (117) substitute for Ex : ∂Hz iωµ + γHy − (σ + iωε) Hy = 0 . ∂y γ Hence ∂Hz −1 iωµ γ − (σ + iωε) γ ∂y
Hy = =
γ2
∂Hz γ ∂Hz −γ =− 2 , − iωµ(σ + iωε) ∂y k ∂y
(122)
where k 2 = γ 2 − iωµ(σ + iωε). In Eq. (118) substitute for Ey and find Hx =
−γ ∂Hz γ ∂Hz = − , γ 2 − iωµ(σ + iωε) ∂x k 2 ∂x
(123)
Using Eqs. (121), (122), and (123), we find that Eqs. (119) and (120) are automatically satisfied. Substituting Eqs. (122) and (123) into Eq. (116), we obtain two equations −γ ∂ 2 Hz , γ 2 − iωµ(σ + iωε) ∂x2 −
∂ 2 Hz γ − γHz = 0 , γ 2 − iωµ(σ + iωε) ∂y 2
which can be written as ∂ 2 Hz ∂ 2 Hz + + k 2 Hz = 0 , ∂x2 ∂y 2
(124)
where k 2 = γ 2 − iωµ(σ + iωε). We have obtained the wave equation for Hz which we will solve assuming the good conductor boundary conditions.
277
Now, we proceed to solve Eq. (124) for Hz with the boundary conditions for the field components at the surface of a good conductor. Having Hz , then we can solve Eqs. (122) and (123) for Hx and Hy . With these solutions, we then will be able to solve Eq. (121) for Ex and Ey . After that, we will know all the field components. In other words, we will know how the fields propagate through the rectangular waveguide.
20.2
Boundary Conditions
Let us proceed with the steps mentioned above. First, we solve the wave equation (124) with the known boundary conditions at the surface of a good conductor: ~ normal to boundary (in xy plane) = 0. 1. H ~ tangential to boundary (in xy plane) = 0. 2. E According to Figure 71: We must have at the boundaries along the x-axis, x = 0 and x = a, the field components Hx = 0 and Ey = 0. Looking at Eq. (118), we see that this means that the derivative ∂Hz /∂x must be equal to zero at x = 0 and x = a. Similarly, at the boundaries along the y-axis, at y = 0 and y = b, the field components Hy and Ex must vanish, Hy = 0 and Ex = 0. Looking at Eq. (117), we see that this means that the derivative ∂Hz /∂y must be equal to zero at the boundaries y = 0 and y = b. Summarizing, we see that the solucomponents at the surface of the waveguide. tion of Eq. (124), which satisfies these boundary conditions is of the form
Figure 71: Electric and magnetic field
Hz = H0 cos(kx x) cos(ky y) eiωt−γz , 278
with kx x and ky y satisfying the standing wave conditions: kx a = mπ
and
ky b = nπ .
Hence, possible values for Hz inside the waveguide are
Hz = H0 cos
mπx a
cos
nπy b
eiωt−γz ,
(125)
where m, n = 0, 1, 2 . . .. An (m, n) combination of the integer numbers represents a possible TE mode of propagation. The modes are designated in the form TEmn . What left is to determine the coefficient γ. With the solution (125), the wave equation (124) gives a condition for k: "
mπ − a
2
nπ − b
#
2
+k
2
Hz = 0 .
For a non-trivial solution, Hz 6= 0, and then mπ k = a
2
2
nπ + b
2
.
Since that on the other hand k 2 = γ 2 − iµω(σ + iεω) , we find that mπ γ = a 2
2
nπ + b
2
+ iµω(σ + iεω) .
(126)
We see that in general the propagation constant γ is a complex number. We can write γ = β + iα. Then the solution (125) takes the form
Hz = H0 cos
mπx a
cos
nπy b
e−βz ei(ωt−αz) .
Thus, α = 2π/λg , where λg is the wavelength in the waveguide at frequency ω. The phase velocity in the waveguide is vp = ω/α. The parameter β is the attenuation coefficient describing losses in the waveguide. Energy loss may be due to: 279
• Ohmic resistivity (σ of the medium finite). • Dielectric losses (described by ε having an imaginary component) • Magnetic losses (described by µ having an imaginary component)
20.3
TE Modes in a Lossless Waveguide
Suppose that the interior of the waveguide is filled with a perfect dielectric, i.e. we assume a lossless propagation for which we have σ = 0 and ε and µ both purely real. In this case, we find from Eq. (126) that mπ γ = −εµω + a 2
2
2
nπ + b
2
,
or introducing the phase velocity, we obtain ω2 mπ γ =− 2 + v0 a
2
2
nπ + b
2
√ where v0 = 1/ εµ is the phase velocity of propagation of waves in an infinite (unbounded) medium of the type filling the waveguide. It is convenient to express γ 2 in terms of the wavelength of the input wave. Since 2πf 2π ω = = , v0 f λ0 λ0 where λ0 is wavelength of the input wave, or equivalently, the infinite size medium wavelength, we can write the expression for γ 2 as 2π γ =− λ0 2
2
mπ + a
2
nπ + b
2
.
One can see that the parameter γ 2 can be negative or positive even after all the assumptions of a lossless propagation. The nature of the propagation depends on the size of the cross section of the waveguide, and changes according as: 280
1. Assume that γ 2 negative. In this case, γ is purely imaginary, and we have a propagating wave mπx nπy cos Hz = H0 cos a b
ei(ωt−kg z) ,
where we have put γ = ikg , with kg the guide propagation constant. The wave then propagates with a guide wavelength λg given by 2π = kg = λg
s
2π λ0
2
−
mπ a
2
−
nπ b
2
.
There is a maximum wavelength λ0 (= λc say) such that kg is real 1 m = 2 λc 2a
2
n + 2b
2
.
Thus, there is a minimum frequency fmn such that the TEmn mode will propagate down the waveguide s
fmn = vm
m 2a
2
n + 2b
2
.
This can be derived from the cut-off condition fmn λc = vm . Thus, the waveguide acts as a high-pass filter for any (m, n) mode. 2. Consider γ 2 positive. Then γ = β say is purely real and
Hz = H0 cos
mπx a
cos
nπy b
e−βz eiωt .
There is no phase propagation but amplitude attenuation. Note there is no energy loss mechanism available. This is an evanescent mode at frequencies less that fmn analogous to the case of total internal reflection. Field components in the TE modes. Since mπx Hz ∼ cos a
nπy cos b
281
eiωt−γz ,
we find from Eq. (122) the Hy component of the field Hy ∼
∂Hz ∂y
!
∼ cos
mπx a
sin
nπy b
eiωt−γz .
eiωt−γz .
Then from Eq. (123), we find the Hx component Hx ∼
∂Hz ∂x
!
∼ sin
mπx a
cos
nπy b
Next, from Eq. (121), we find the components of the electric field mπx Ex ∼ Hy ∼ cos a
nπy sin b
eiωt−γz ,
nπy b
eiωt−γz .
and Ey ∼ Hx ∼ sin
mπx a
cos
We see from the above equations that the mode m = n = 0 is not allowed to propagate through the waveguide, since Ex , Ey , Hx and Hy all contain sine terms so all the field components vanish for m = n = 0. All other TEmn modes are allowed.22 TM (transverse magnetic) modes Put Hz = 0 and go through the whole procedure again. Ez 6= 0 now. Equations analogous to Eqs. (122), (123), and (124) appear for components of ~ vector this time. Consequently the previous discussion about modes the E and their cut-off frequencies for TE modes is also true for TM modes. The only difference is that more modes are not allowed, i.e. in contrast to the TE propagation, TM modes with either m = 0 or n = 0 are not allowed.23
22
The mode m = n = 0 is never possible in a transmission line consisting of a single closed conductor like a rectangular waveguide. It is possible in 2-conductor lines e.g. the coaxial line or the twin wire transmission line. 23 This is an interesting difference between properties the TE and TM modes, and the student is encourage to analyze, as a tutorial problem, where the difference is coming from.
282
Special properties of the TE10 mode We now proceed to discuss interesting properties of some of the propagating modes. For example, If the transverse dimensions of the rectangular waveguide are different (a 6= b) there is a finite range of frequencies over which the TE10 mode is the only allowed mode. This means that a waveguide can be designed to allow propagation in one mode only. In other words, it can work as a frequency filter transmitting nondegenerate modes. To show this, consider the frequency of the TEmn mode s
fmn = vm
m 2a
2
n + 2b
2
,
from which we find that frequencies of two neighbouring modes TE10 and Figure 72: Electric and magnetic fields TE01 are of the TE10 mode. vm vm < f01 = , f10 = 2a 2b where we adopt the convention that a > b. Thus, in the frequency range from f10 → f01 , the TE10 mode is the only mode allowed.
20.4
Phase and Group Velocities of Mode Propagation
Consider the velocity with which the wave propagates inside the waveguide. We look into the phase velocity, the velocity the wave front propagates vp = f λg =
ω = r 2 kg 2π λ0
ω −
mπ a
2
−
Thus v0
vp = r 1−
mλ0 2a
2
283
−
nλ0 2b
2
,
nπ b
2
.
where v0 is the phase velocity in the unbounded medium. We see that the phase velocity of the wave inside the waveguide is greater than in an unbounded medium, and so may be greater than the speed of light in vacuum. Alternately, we could write
mπ a
2
+
nπ b
2
=
2π λc
2
= kc2 ,
where λc is the infinite medium wavelength at the cut-off frequency for the m, n mode. Then vp = q
ω k02 − kc2
=
ω r
k0 1 −
kc k0
2
=r
v0 1−
ωc ω
2
,
(127)
provided ω > ωc i.e. in the pass-band for that mode. Note that for ω → ωc , the phase velocity vp → ∞. Moreover, if a vacuum or air fills the waveguide then v0 = c and vp > c. The group velocity Since vp > c typically, we see that the phase velocity vp is not the velocity of propagation of energy or information down the waveguide. It is propagated with the group velocity which is smaller than the phase velocity if the medium is dispersive. When dispersion is not present, phase and group velocities are equal. According to Eq. (127), the waveguide is a dispersive medium, since the phase velocity depends on frequency, so the energy or information is propagated with the group velocity vg = dω/dkg that differs from the phase velocity vp = ω/kg . Just a brief explanation how do we define the group velocity. The group velocity is the velocity of propagation of some modulation of multi-frequency wave that carries information. A single frequency harmonic wave carries no information. It is just there. A finite bandwidth is required to carry information.
284
We can illustrate the concept of group velocity by considering a sum of two cosine waves of slightly different frequencies, ω and ω + dω: cos(ωt − kz) + cos[(ω + dω)t − (k + dk)z] ! " # dω dk 2ω + dω 2k + dk = 2 cos t − z cos t− z 2 2 2 2 ! dω dk = 2 cos t − z cos(ωt − kz) , 2 2 where we have assumed that dω 2ω and dk 2k. The velocity of the amplitude modulation is vg =
dω 2 dk 2
=
dω . dk
In the rectangular waveguide 2π kg = = λg
q q
k02
−
kc2
ω 2 − ωc2
=
vm
,
where k0 = ω/vm and kc = ωc /vm . Remember, vm is the infinite medium phase velocity and ωc is the cut-off (angular) frequency of the (m, n) mode. Then differentiating 1 dkg 1 1 2 ω q = (ω − ωc2 )− 2 2ω = . dω vm 2 vm ω 2 − ωc2
Finally q
s
2 vm ω 2 − ωc2 ωc dω < vm . vg = = = vm 1 − dkg ω ω
Thus, the group velocity of a wave propagating inside the waveguide is smaller than the phase velocity. Moreover s
vg vp = vm 1 −
ωc ω
2
vm r
1− 285
ωc ω
2
2 . = vm
In a vacuum-filled waveguide vg vp = c2 . Thus, relativity is still all right.
Revision questions Question 1. Explain what is meant by TE and TM waves. Question 2. Explain briefly the procedure of finding the components of the electric and magnetic fields transmitted through a waveguide. Question 3. Why a TEM wave cannot be transmitted through a rectangular waveguide? Question 4. Explain the notation, what does it mean TEmn ? Question 5. Explain, how a waveguide can work as a frequency filter. Question 6. Define the phase and group velocities and the relation between them. Question 7. In a vacuum waveguide the phase velocity of an EM wave is greater than speed of light: True or false?
286
21
Relativistic Transformation of the Electromagnetic Field
The final part of the course is devoted to relativistic effects in the EM theory. Using the argument that the Maxwell’s equations, as physical observables, are invariant under the Lorentz transformation, we shall show how the fields transform according to the relativistic rules. Then, we will illustrate on few examples, how the fields change when one goes between different inertial frames and how this affects propagation of an EM wave. Finally, we show how frequency and energy transform according to the transformation rules and point out the evidence of the dependence of the energy on frequency.
21.1
The Principle of Relativity
We begin from the recollection of the principles of relativity. 1. The laws of physics are the same in all inertial reference frames. 2. The speed of light in vacuum is independent of the uniform motion of the observer or source. The constancy of the velocity of light, independent of the motion of the source, gives rise to the relations between space and time coordinates in different inertial reference frames known as Lorentz transformations. Consider a stationary reference frame S and a inertial frame S 0 moving with a velocity ~u parallel to the x axis, i.e. ~u = uˆi. The time and space coordinates in S 0 are related to those in S by the Lorentz transformations x0 y0 z0 ct0 −1/2
where γ = (1 − β 2 )
= = = =
γ(x − βct) , y, z, γ(ct − βx) ,
is the Lorentz factor, and β = u/c.
287
The above transformation corresponds to a situation of ~u parallel to the x axis. If the axis in S and S 0 remain parallel, but the velocity ~u of the frame S 0 is in an arbitrary direction, the generalization of the above transformations is ~r 0 = ~r + (γ − 1)
~ β~ (~r · β) ~ , − γ βct β2
ct0 = γ ct − β~ · ~r
,
where β~ = ~u/c. Proof: Decompose the vector ~r into two components: parallel and normal to ~u ~r = ~rk + ~r⊥ . Then, using the one dimensional Lorentz transformations, we have
~ ~rk0 = γ ~rk − βct , ~r⊥0 = ~r⊥ . However, we can write the parallel and normal components as ~ β~ (~r · β) β2 = ~r − ~rk .
~rk = ~r⊥ Hence
~ ~r 0 = ~rk0 + ~r⊥0 = γ ~rk − βct + ~r − ~rk = ~r + (γ − 1)
~ β~ (~r · β) ~ . − γ βct β2
The transformation of time can be proved in the similar way. Since
ct0 = γ ct − βrk
,
and βrk = β~ · ~rk =
~ (~r · β) β~ · β~ = ~r · β~ , β2 288
we obtain
ct0 = γ ct − β~ · ~r
,
as required. We will need the inverse Lorentz transformations, which are ~r = ~r 0 + (γ − 1)
~ β~ (~r 0 · β) ~ 0, + γ βct β2
ct = γ ct0 + β~ · ~r 0
.
Note that ~r is a function of ~r0 and t0 . The principle of relativity indicates, and we have showed it explicitly in Chap. 7, that the Maxwell equations are invariant under the Lorentz transformation. The same rule applies to the continuity equation. Thus, the Maxwell’s equations together with the continuity equation have the same form in two different inertial frames. If in the frame S: ~ = ρ, ~ =0, ∇·D ∇·B ~ ~ ~ = J~ + ∂ D , ~ = − ∂B , ∇×H ∇×E ∂t ∂t ∂ρ ∇ · J~ = − , ∂t then in the frame S 0 : ~ 0 = ρ0 , ~0 = 0 , ∇0 · D ∇0 · B ~0 ~0 0 ~ 0 = − ∂B , ~ 0 = J~ 0 + ∂ D , ∇0 × E ∇ × H ∂t0 ∂t0 0 ∂ρ ∇0 · J~ 0 = − 0 , ∂t where the prime variables are functions of the transformed variables, t0 and ~r 0 .
289
Of course, the EM fields together with the current and charge densities in the S 0 frame must be different from that in the S frame in order to match the same Maxwell equations. From this, an interesting question arises: What are the relations between the EM fields, charge and current densities in the S 0 and S frames? In order to answer this question, we have to find the transformation of a time derivative of an arbitrary scalar function Ψ, and the transformation of divergence ∇ · F~ . Consider the transformation of a time derivative ∂Ψ/∂(ct) ∂Ψ ∂ct0 ∂Ψ ∂x0 ∂Ψ ∂y 0 ∂Ψ = + + ∂(ct) ∂(ct0 ) ∂(ct) ∂x0 ∂(ct) ∂y 0 ∂(ct) ∂Ψ ∂z 0 ∂Ψ ∂ct0 ∂~r 0 0 + = + ∇ Ψ · . ∂z 0 ∂(ct) ∂(ct0 ) ∂(ct) ∂(ct) However ∂ct0 =γ , ∂(ct)
∂~r 0 = −γ β~ , ∂(ct)
which gives !
∂Ψ ∂ =γ − β~ · ∇0 Ψ . ∂(ct) ∂(ct0 ) Consider now the divergence ∇ · F~ =
∂Fx ∂Fy ∂Fz + + . ∂x ∂y ∂z
Since ∂Fx ∂Fx ∂ct0 ∂Fx ∂x0 = + ∂x ∂(ct0 ) ∂x ∂x0 ∂x ∂Fx ∂y 0 ∂Fx ∂z 0 + + , ∂y 0 ∂x ∂z 0 ∂x and ∂ct0 = −γβx , ∂x 290
∂x0 β2 = 1 + (γ − 1) x2 , ∂x β 0 0 ∂z ∂y = =0, ∂x ∂x we obtain ∂Fx ∂Fx ∂Fx = −γβx + αx 0 , 0 ∂x ∂(ct ) ∂x where αx = 1 + (γ − 1)βx2 /β 2 . Similarly, for Fy and Fz , and finally we get ∇ · F~ = α ¯ ∗ (∇0 · F~ ) − γ β~ ·
∂ F~ , ∂(ct0 )
where α ¯ is a 3 × 3 diagonal matrix
α ¯=
2
1 + (γ − 1) ββx2 0 0
0
0 β2
1 + (γ − 1) βy2 0 2 0 1 + (γ − 1) ββz2
,
as required. Using the above transformations, we can derive transformations for the current density J~ and the charge density ρ. In order to do it, we consider the continuity equation, that can be written as ∂cρ . ∇ · J~ = − ∂(ct) Hence ~ ∂ ~ − γ β~ · ∂ J = −γ α ¯ ∗ (∇ · J) − β~ · ∇0 cρ , ∂(ct0 ) ∂(ct0 ) !
0
or ~ − γ β~ · ∇0 (cρ) = −γ ∇0 · (¯ α ∗ J) 291
∂ ~ · J~ . cρ − β ∂(ct0 )
Since ~ , β~ · ∇0 (cρ) = ∇0 · (cρβ) we obtain i ∂ h ~ · J~ . ∇0 · α ¯ ∗ J~ − γcρβ~ = − γ cρ − β ∂(ct0 )
Thus, the continuity equation will be invariant under the Lorentz transformation if
cρ0 = γ cρ − β~ · J~ , J~ 0 = α ¯ ∗ J~ − γcρβ~ .
(128)
In order to understand the physical meaning of these equations, consider the following example. Example Assume that in the S frame there is a stationary volume charge of density ρ 6= 0. Since ρ is stationary, there are no currents in the S frame (J~ = 0). What are the charge and current densities as seen in the S 0 frame? In the S frame J~ = 0 ,
ρ 6= 0 .
According to Eq. (128), in the S 0 frame J~ 0 = −γcρβ~ ,
ρ0 = γρ .
Thus, there is a current in the S 0 frame. As seen from S 0 a given part of the charge is length contracted in the direction of motion so the charge density is correspondingly increased by the factor γ > 1. The length contracted charge density appears from S 0 to move in the opposite direction. We can understand this result: The stationary charge in the S frame moves with velocity −~u in the S 0 frame. This effect is predicted by the Galileo transformation.
292
Less obvious and more interesting is the following situation. Let’s in the S frame one observes J~ 6= 0 ,
ρ=0.
Then, according to Eq. (128), someone will see a non-zero charge density ρ0 6= 0 in the S 0 frame. This is a pure relativistic effect, which cannot be predicted by the Galileo transformation. Note, the charge ρ0 is proportional to u/c2 .
21.2
Transformation of Electric and Magnetic Field Components
To find the transformation rules for electric and magnetic field components we will use the transformations of the time and space derivatives derived above. Consider two of the Maxwell equations that in the S frame are ~ = ρ, ∇·D ~ ~ = J~ + ∂ D . ∇×H ∂t These equations should be equivalent of two equations ~ 0 = ρ0 , ∇0 · D ~0 ~ 0 = J~ 0 + ∂ D , ∇0 × H ∂t0 in the S 0 frame. Using the transformations of the time and space derivatives, we have "
#
!
∂ ∂ ~ −γ ~ = J~ , ×H α ¯ ∗ ∇ − γ β~ − β~ · ∇0 cD 0 0 ∂(ct ) ∂(ct ) " # ∂ 0 ~ ~ = cρ . α ¯ ∗ ∇ − γβ · cD ∂(ct0 ) 0
293
~ and H ~ Substituting the transformations of J~ and ρ, we find that the D vectors transform as
,
~ 0 = γ −β~ × cD ~ +α ~ H ¯ −1 ∗ H
~0 = γ α ~ + β~ × H ~ cD ¯ −1 ∗ cD
,
(129)
where α ¯ −1 is the inverse of the matrix α ¯ α ¯ −1 =
2
1 + ( γ1 − 1) ββx2
0
0 β2
1 + ( γ1 − 1) βy2 0 2 1 0 1 + ( γ − 1) ββz2
0 0
.
From the Maxwell equations ~ = 0, ∇·B ~ ~ = − ∂B , ∇×E ∂t ~ and B ~ vectors transform as we find that the E
~0 = γ α ~ + β~ × cB ~ E ¯ −1 ∗ E
~ 0 = γ −β~ × E ~ +α ~ cB ¯ −1 ∗ cB
21.3
.
(130)
Transformation Rules in Terms of Parallel and Normal Components
Suppose that the frame S 0 is moving with speed u in the direction parallel to the z axis. In this case, βx = βy = 0, βz = β 6= 0, and then the transformations take the form ~ 0 = γcDxˆi + γcDy ˆj + cDz kˆ + γβ kˆ × H ~ , cD ~ 0 = −γβckˆ × D ~ + γHxˆi + γHy ˆj + Hz kˆ , H ~ 0 = γExˆi + γEy ˆj + Ez kˆ + γcβ kˆ × B ~ , E ~ 0 = −γβ kˆ × E ~ + γcBxˆi + γcBy ˆj + cBz kˆ . cB 294
(131)
It is useful to rephrase the transformation rules in terms of components parallel and normal to ~u. The parallel components are the z components and the normal components lie in the xy plane. For example
~ =E ~⊥ + E ~ k = Exˆi + Ey ˆj + Ez kˆ , E ~ H ~ and B. ~ and the same for D,
21.3.1
Rules for Parallel Components
From Eq. (131), we readily find that ~ k0 = E ~ z0 kˆ = E ~ z kˆ = E ~k , E and similarly ~ k0 = B ~k , B
~ k0 = H ~k , H
~ k0 = D ~k . D
Thus, the components parallel to the direction of ~u are invariant under the transformations (129) and (130).
21.3.2
Rules for Normal Components
Consider now the components perpendicular to ~u: ~ 0 = cD 0ˆi + cD 0 ˆj cD ⊥ x y = γcDxˆi + γcDy ˆj + γβHxˆj − γβHyˆi = γ (cDx − βHy ) ˆi + γ (cDy + βHx ) ˆj , ~ 0 = H 0ˆi + H 0ˆj H ⊥ x y = γ (Hx + βcDy ) ˆi + γ (Hy − βcDx ) ˆj , ~ ⊥0 = Ex0ˆi + Ey0 ˆj E = γ (Ex − βcBy ) ˆi + γ (Ey + βcBx ) ˆj , 295
~ 0 = cB 0ˆi + cB 0 ˆj cB ⊥ x y = γ (cBx + βEy ) ˆi + γ (cBy − βEx ) ˆj , which can be written as
~ 0 = γ cD ~ ⊥ + β~ × H ~⊥ cD ⊥ ~0 = γ H ~ ⊥ − β~ × cD ~⊥ H ⊥ ~0 = γ E ~ ⊥ + β~ × cB ~⊥ E ⊥ ~ 0 = γ cB ~ ⊥ − β~ × E ~⊥ cB ⊥
.
(132)
We shall illustrate the transformation rules on the following examples: Example 1 - purely electric field in S ~ 6= 0 but B ~ = 0. Suppose that in S, one observes E 0 Then from the transformation rules, in S : ~ k0 = E ~k , E ~ k0 = 0 , B
~ ⊥0 = γ E ~⊥ , E ~⊥ . ~ ⊥0 = − γ ~u × E B c2
Thus ~ ~ ~0 =B ~ ⊥0 = − ~u × E⊥ = − ~u × E , B c2 c2 ~ k = 0. since ~u × E Thus what appears to be purely an electric field to one observer is seen as both an electric and a magnetic field to a second observer moving with respect to the first. Example 2 - purely magnetic field in S
296
~ = 0, but B ~ = Now suppose that in S, one observes E 6 0. 0
Then using the transformation rules, in S : ~⊥ , ~k , ~ ⊥0 = γ B ~ k0 = B B B ~ k0 = 0 , E
~⊥ . ~ ⊥0 = γ~u × B E
Thus ~0 =E ~ 0 = ~u × B ~ ⊥ = ~u × B ~ . E ⊥ We see that what appears to be a purely magnetic field for one observer will appear to be both an electric and a magnetic field to a relatively moving observer. This result could be used to calculate the emf in an electric dynamo from the point of view of an observer watching the conductor move in a magnetic field or from the point of view of an observer moving with the conductor.
21.4
Transformation of the Components of a Plane EM Wave
In this final lecture of the course, we illustrate the transformation rules of the components of a plane EM wave. We illustrate this on two examples. Example 1 Suppose that a plane wave propagates in vacuum along the z axis. Then the electric and magnetic fields of the wave are ~ = ˆiEei(ωt−kz) = ˆiE0 = ˆiEx , E ~ = ˆjBei(ωt−kz) = ˆjB0 = ˆjBy . B Hence from the transformation rules (132), in the S 0 frame moving in the same direction: ~ 0 = γ E0ˆi − βcB0ˆi = γ (E0 − uB0 ) ˆi , E
~ 0 = γ cB0ˆj − βE0ˆj = γ (cB0 − βE0 ) ˆj . cB 297
Since in vacuum cB0 = E0 , we obtain ~0 =γ 1− u E c
E0ˆi = r
1− 1−
s
u c
2 u
E0ˆi =
c−u ˆ E0 i , c+u
c
and ~ = γ 1 − u cB0ˆj = cB c 0
s
c−u cB0ˆj . c+u
Thus, we observe that the amplitudes of the fields are reduced, but the ratio ~ 0 |/|B ~ 0 | = |E0 |/|B0 | is constant and independent of u, as it should be, since |E speed of light is the same in all inertial frames. This is consistent with the principle of relativity that speed of light is independent of the motion of the observer. Example 2 Suppose that an observer S 0 moves in the direction of the electric field of an EM wave, i.e. ~u = uˆi, as shown in Figure 73. In this case ~ 0 = E0ˆi + γβcB0 kˆ = ˆi + γ u kˆ E0 , E c
and ~ 0 = γcB0ˆj . cB
Figure 73: Text
Thus, the magnetic field remains unchanged, but the electric field turns towards the direction of propagation of the wave.
298
An interesting situation occurs when the velocity u → c. In this case, we have γ
1 u u =r →∞, 2 c c u 1− c
and then ~ 0 = E kˆ , E ~ 0 becomes perpendicular to ~u. i.e. the direction of E However, the Poynting vector of the wave is still in the direction of ~u: ~0×H ~ = γ u E0 γB0 kˆ × ˆj = −γ 2 u E0 B0ˆi . E c c
21.5
Doppler Effect
Consider a plane wave propagating in vacuum ~ ~ 0 ei(ωt−~k·~r) . E(t) =E In the moving frame S 0 this wave will have a different frequency ω 0 and the wave vector ~k 0 , but the phase of the wave will remain unchanged as it is invariant under the transformation, i.e. φ = ωt − ~k · ~r = ω 0 t0 − ~k 0 · ~r 0 . Using the inverse Lorentz transformation ~ 0, ~r = α ¯ ∗ ~r 0 + γ βct ct = γ ct0 + β~ · ~r 0 , we will find the relations between ω 0 , ω and k 0 , k that ensure the invariance of the phase of the wave.
299
Thus, with the above transformations, we find ω 0 t0 − ~k 0 · ~r 0 = ωt − ~k · ~r ωγ ~ 0 = ct0 + β~ · ~r 0 − α ¯ ∗ ~k · ~r 0 − γ~k · βct c ωγ ~ 0 0~ ~ β−α ¯ ∗ k · ~r 0 = ωγt − γt k · ~u + c ωγ ~ 0 ~ ~ = γ ω − k · ~u t − β−α ¯ ∗ k · ~r 0 . c Hence
ω 0 = γ ω − ~k · ~u , ωγ ~k 0 = α ¯ ∗ ~k − 2 ~u . c Consider two special cases. In the first case, assume that the wave propagates ˆ in vacuum, ~k = (ω/c)~k. Then
ω 0 = γω 1 −
u cos θ c
,
where θ is the angle between the direction of propagation of the wave and the direction of the motion ~u. For the propagation direction θ = 0 s
ω0 = ω
c−u . c+u
The frequency in S 0 is smaller than that in S, and ω 0 → 0 as u → c. On the other hand, the frequency ω 0 → ∞ when u → −c. If the wave propagates in a material body, where vf < c, we have Figure 74: Text
~k = (ω/vf )kˆ ,
and then the frequency in S 0 is u ω = γω 1 − cos θ vf 0
300
!
.
For the case of θ = 0 ω0 = ω r
u vf
1− 1−
2 u
.
c
When u = vf , the frequency ω 0 = 0, but for c > u > vf , we obtain that the frequency ω 0 < 0.
21.6
Transformation of Energy of a Plane EM Wave
Consider an EM wave propagating in the ~k direction and an observer moving in the z direction, as shown in the Figure 75. ~ = E ˆi. Then, in the frame S 0 : Let E ~ 0 = γE ~ + γ β~ × cB ~⊥ E ~ + γ cos φ β~ × cB ~ . = γE ~ = B ˆj + Bk kˆ (the wave propagates in the plane yz), and cB = E, Since B we obtain ~ 0 = γE ~ − γ cos φ βcB ˆi E ~ . = γ (1 − β cos φ) E Consider now energy of an electric field of a plane EM wave confined in a volume ∆V . In the S frame 1 We = ε0 E02 ∆V . 4 In the S 0 frame 1 We0 = ε0 (E00 )2 ∆V 0 . 4 However ∆V 0 = ∆x0 ∆y 0 ∆z 0 .
Figure 75: An EM wave is propagating in the ~k direction and an observer is moving in the z direction.
301
Since, we have assumed that the observer (S 0 frame) moves in the direction of the z axis, and the wave propagates in the direction ~k · ~u = u cos φ, we obtain ∆V 0 = ∆x∆y∆z 0 , where ∆z 0 =
∆z
.
γ 1 − uc cos φ
Hence We0
2
1 u = ε0 E02 γ 2 1 − cos φ 4 c
u = We γ 1 − cos φ c
∆V
γ 1 − uc cos φ
.
It is interesting to compare the transformation of energy with the transformation of frequency. Since
We0 = We γ 1 −
u cos φ c
,
and u ω = ωγ 1 − cos φ c 0
,
we see that the energy and frequency transform in the similar way! This indicates that We ∼ ω. Note, that a similar proportionality was predicted in quantum physics, We = h ¯ ω. In the electromagnetic theory, the proportionality forms backgrounds of the so called quantum electrodynamics.
302
Revision questions Question 1. Are the Maxwell’s equations invariant under the Lorentz transformation? Explain. Question 2. What are the rules for transformation of the electric and magnetic field components? Question 3. What are the rules for transformation of the components of an EM wave? Question 4. What is Doppler effect? Question 5. How does the energy of an EM wave transform in the relativistic case?
303
Tutorial problems Problem 21.1 A relativistic linear dynamo A linear dynamo is constructed using a horizontal U-shaped or rectangular section of wire with one side moving at speed v = dx dt ~ as shown in in a vertical magnetic field B Figure 76.
Figure 76:
The rectangle has width `. The dynamo is relativistic in that v may be comparable to c. The general definition of emf round a circuit is: E =W =
I ~ F
q
· d~` =
I
~ + ~v × B) ~ · d~` . (E
(a) Calculate the emf induced along the moving arm: (i) In the frame of reference in which the U-shaped section is at rest. (ii) In the frame of reference in which the slide-wire is at rest. Do you arrive at the same value of emf in each frame? If not, does that bother you? (b) In practice, emf ’s are usually calculated using Faraday’s Flux Cutting Rule because it is the simplest way mathematically. The flux cutting rule E = − dΦ dt may be thought of as an integral of the ~ = − ∂ B~ . Now Maxwell’s equations are Maxwell equation ∇ × E ∂t supposed be ‘relativistically correct’, i.e. they take the same form in all inertial frames of reference. Show that the flux cutting rule holds in both reference frames mentioned above. Problem 21.2 Plane wave propagating in a material Suppose that a plane wave propagates in a material with ε 6= ε0 and µ 6= µ0 . Show that in the frame S 0 , moving with velocity u
304
in the direction of propagation of the wave (z-axis), the ratio ~ 0 |/|H ~ 0 | depends on the velocity u. |E
Problem 21.3 Electric and magnetic fields in a moving frame Suppose that in a stationary frame S one has measured an uniform ~ = E0ˆi and an uniform magnetic field B ~ = B0 ˆj, electric field E such that cB0 > E0 . Is it possible to find a frame S 0 moving along the z-axis in which ~ 0 or B ~0 ? one could not measure either E If yes, find the amplitude of the measured field and the velocity u of the frame S 0 .
305
Revision Questions for the Final Examination Question 4.1 Prove that the total electric flux through a closed surface S is proportional to the total charge inside the surface. Question 4.2 Prove the Amperes circuit law. Question 4.3 Derive the integral form of the Faraday’s law and then transform it into the differential form ~ =− ∇×E
~ ∂B . ∂t
Question 5.1 Starting from the Maxwell’s equations derive the continuity equation, i.e. show that conservation of charge is built into the Maxwell’s equations. Question 6.1
~ satisfies the same Using the Maxwell’s equations show that B ~ wave equation as E.
Question 6.2
Show, using the proof of solution of the wave equation, that f (z + ct) represents a signal propagating in the negative z direction with speed c.
Question 7.1 Show that magnetic and electric fields of a charge moving with a constant velocity ~v are related by ~ ~ = ~v × E . B c2 Question 7.2 Show that the magnetic field lines produced by a charge moving with a constant velocity ~v form concentric rings about ~v . Question 7.3 Show that the electric field produced by a moving charge satisfies the Gauss’s law. Question 7.4
Explain the statement: Electric and magnetic fields do not produce each other - they are both due to electric charges.
306
Question 8.1 Using the Maxwell’s equations derive the continuity equation for the Poynting vector. Question 8.2 Show, using the field theory calculation, that the power dissipated along a resistive wire is P = V I, the same predicted by the circuit theory.
Question 9.1 (a) Obtain a series solution of the Laplace equation in the cartesian coordinates for the electrostatic potential inside a closed three dimensional area. (b) What would be the solution of the Laplace equation for a closed two dimensional problem. Question 9.2 Consider a conducting sphere of radius R. The surface of the sphere is kept at a potential Φ(R, θ, φ) = V0 sin θ sin φ . Using the above as a boundary condition, find the potential at any point inside the sphere.
Question 10.1 Show that in general the Maxwell’s equations cannot be simpli~ and B ~ fields. fied to two separate differential equations for the E Then show that by introducing the concept of vector and scalar ~ and Φ potentials one can arrive to differential equations for A that can be separated from each other under the Lorenz gauge. Question 10.2 Explain, why the Coulomb gauge is often called ”Transverse gauge”. Question 10.3 Prove that the homogeneous wave equation ∇2 Φ −
1 ∂2Φ =0 c2 ∂t2
has the solution of the form Φ(r, t) =
307
f (t − r/c) , r
where f (t − r/c) is an arbitrary function of the retarded time t − r/c.
Question 11.1 Show that in spherical polar coordinates, the magnetic field of a ~ = ∆lI ~ 0 exp(iωt) has only an azimuthal short current element I ∆l component of the form ~ = I0 ∆l ik + 1 sin θ ei(ωt−kr) φˆ . B 4πε0 c2 r r2
Question 11.2 Show that in the far field zone of a radiating short current element, the electric and magnetic fields oscillate in phase and are orthogonal to each other. Question 11.3 (a) Given the expressions for the EM field of a Hertzian dipole, show that the total radiated power from the dipole is πI02 ∆l 3ε0 c λ
W =
2
.
(b) Show that the emitted power is equivalent to the power lost on a resistor of resistance R = 80π 2
∆l λ
2
.
Question 11.4 Show that the time averaged Poynting vector of the EM field emitted by a short current element is maximal in the equatorial plane of the element.
Question 12.1 Show that the electrostatic potential due to a distribution of electric dipoles of moment per unit volume P~ throughout a volume V enclosed by a surface S is that of a volume charge density −∇ · P~ together with a surface charge density P~ · n ˆ. Question 12.2 Illustrate the Lorentz theory of polarizability of dense dielectric materials and derive the Caussius-Mossotti relationship for a dense dielectric. Question 12.3 Show that the polarization of a dielectric driven by a time varying electric field lags in phase the driving field.
308
Then show that the phase difference between the polarization and a time-varying electric field results in a complex permittivity of the dielectric.
Question 13.1 Explain the concept and advantage of introducing the magnetic ~ intensity vector H. Question 13.2 Show that inside a ferromagnet H = 0 and explain the physical meaning of this result. Question 13.3 Derive the Maxwell’s equations for the EM fields in electric and magnetic materials.
Question 14.1 (a) Derive the special form of Poynting’s Theorem applicable in certain material media I S
~ × H) ~ · dS ~=− (E
Z
~ ~ · ∂ B dV − H ∂t
Z
~ ~ · ∂ D dV − E ∂t
Z
~ · J~ dV . E
(b) Interpret the above equation in terms of energy storage and energy flow etc. (c) State qualitative meaning of each term in the equation. Question 14.2 Prove that the useful formula for the mean Poynting vector for sinusoidal fields is 1 ~ ¯~ ~∗ , N = Re E c × Hc 2
~c = E ~ 0 exp(iωt) and H ~ c∗ = H ~ ∗ exp(−iωt). where E 0
Question 15.1 Show that the amplitude of a plane wave propagating in a nonconducting material is damped with the rate β which arises from the imaginary parts of the complex permittivity and permeability. Question 15.2 Derive the expressions for the attenuation coefficient and the phase velocity of an EM wave propagating in a conducting medium. What are the values of the quantities for a propagation inside a
309
good conductor? Question 15.3 Show that in a good conductor an EM wave propagates on the surface of the conductor.
Question 16.1 Prove the following boundary conditions at a bounding surface between two dielectrics: ~ is continuous across the bound(a) The normal component of B ary. ~ is continuous across the bound(b) The tangential component of E ary. ~ is continuous across the bound(c) The tangential component of H ary.
Question 17.1 Show, using the Maxwell’s equations that the electric and magnetic vectors of an EM wave are related by ~ ~ =B = k n ~ , H ˆp × E µ ωµ where n ˆ p is the unit vector in the direction of propagation of the wave. Question 17.2 Prove that in the reflection and refraction at a bounding surface, the direction of incident, reflected and refracted waves are coplanar. Question 17.3
Derive the familiar laws of elementary optics: (a) Angle of reflection equals to the angle of incidence. (b) Snell’s law of refraction.
Question 17.4
~ and H ~ that both Show, using the continuity conditions for E reflection and refraction takes place in the incidence of light on a boundary between two dielectrics.
310
Question 19.1 Show that under the Brewster’s angle of incidence there is no reflected electric field in the plane of the incidence. Question 20.2 Show that in a vacuum-filled rectangular waveguide vp > c, and vg vp = c2 .
Question 21.1 Find the condition under which the continuity equation for ρ and J~ is invariant under the Lorentz transformation. Question 21.2 Show that electric charge Q is invariant under the Lorentz transformation.
311
Appendix A: Proof of the Amperes Law Consider a long wire of radius a carrying current I. Let P is a point on the integration path, as shown in Figure 77.
Figure 77: The source circuit and the integration path to prove the Ampere law. The magnetic field at P is r) µ0 I I d~s × (−ˆ ~ . B= 2 4π s r Moving P by d~` is equivalent to moving the current circuit by −d~`. The solid angle subtended by −d~`, d~s at P is: (−d~` × d~s) · rˆ −d~` · d~s × rˆ d~s × −ˆ r = = d~` · . 2 2 2 r r r (The element of area normal to rˆ is −d~` × ~s · rˆ) Thus due to the path element d~`, the change in solid angle subtended at P by the circuit is: dΩ = d~` ·
I s
d~s × (−ˆ r) . 2 r
312
Hence 4π I ~ 4π ~ ~ dΩ = d~` · dB = B · d` , µ0 I s µ0 I ~ at some where integration is around the circuit s giving the magnetic field B point P as shown in the diagram. Now integrating round the closed path: I `
I ~ · d~` = µ0 I dΩ . B 4π `
If P moves round a closed path (returning to its original position but not circulating through the current loop: I
dΩ = 0 .
`
But if P circulates through the loop: I
dΩ = 4π ,
`
and then I `
~ · d~` = µ0 I 4π = µ0 I . B 4π
We conclude that the line integral of the magnetic field round a closed loop path is equal to µ0 I, where I is the current passing through the loop.
313
Appendix B Proof of the vector theorem, Eq. (73): −
Z ~ ~ M ×n ˆ M dV = dS . ∇× R R S
Z V
This is an application of the more general theorem −
Z
∇ × F~ dV =
Z
V
F~ × n ˆ dS .
S
~ be a constant vector. Then Let C ~ = (∇ × F~ ) · C ~ − (∇ × C) ~ · F~ = C ~ · (∇ × F~ ) . ∇ · (F~ × C) We will prove the general theorem by using the divergence theorem Z
~ dV = ∇ · (F~ × C)
Z
~ ·n (F~ × C) ˆ dS .
S
V
~ and using (133), we get For an arbitrary constant vector C, ~· C
Z
∇ × F~ dV
=
V
=
Z ZV
~ · ∇ × F~ dV C ~ dV = ∇ · (F~ × C)
Z
~ ·n F~ × C ˆ dS .
S
V
Hence ~· C
Z
∇ × F~ dV = −
Z
~ × F~ · n C ˆ dS .
S
V
However ~ × F~ · n ~ · F~ × n C ˆ=C ˆ and then we obtain ~· C
Z
~· ∇ × F~ dV = −C
V
Z
F~ × n ˆ dS .
S
~ we finally have Since this is true for arbitrary C, Z V
∇ × F~ dV = −
Z S
as required. 314
F~ × n ˆ dS ,
(133)
Appendix C: PHYS3050 Facts and Formulae I
Gauss0 Divergence Theorem :
F~ · n ˆ dS =
S
I
Stokes0 s Theorem :
Z
∇ · F~ dV
V
F~ · d~` =
Z
∇ × F~ · n ˆ dS
S
`
Numerical values in SI units: ε0 = 8.85 × 10−12 , µ0 = 4π × 10−7 , c = 3 × 108 [ms−1 ] . −19 For the electron: e = 1.6 × 10 [C] , m = 9.11 × 10−31 [kg] ~ + ~v × B) ~ F~ = q(E
The Lorentz force law :
F~ =
Coulomb0 s Law :
Biot − Savart Law : Gauss0 Law :
1 q1 q2 rˆ 4πε0 r2
µ0 I d~` × rˆ 4π r2 I Q ~ ·n E ˆ dS = ε0
0
~ = dB
Amp`ere s Circuital Law :
I
~ · d~` = µ0 I B
Maxwell’s Equations in vacuum: ~ = ρ , ~ =0, ∇·E ∇·B ε0 ~ ~ ~ = − ∂B , ~ = µ0 J~ + 1 ∂ E . ∇×B ∇×E ∂t c2 ∂t Maxwell’s Equations in material bodies: ~ =ρ, ∇·D
~ =0, ∇·H
~ ~ = − ∂B , ∇×E ∂t
~ ~ = J~ + ∂ D . ∇×H ∂t 315
Poynting vector: ~ = ε 0 c2 E ~ ×B ~ N Poynting’s Theorem: I
~ × B) ~ ·n ε0 c (E ˆ dS = − 2
S
Z V
Z ~ · J~ dV − ∂ E ∂t V
1 1 B2 ε0 E 2 + 2 2 µ0
!
dV
~ = ε0 E ~ + P~ but in polarizable materials where it is convenient to define D ~ ~ = B −M ~: and H µ0 I
~ ×H ~ ·n E ˆ dS = −
Z V
S
Z Z ~ ~ ~ · ∂ B dV − E ~ · ∂ D dV − E ~ · J~ dV H ∂t ∂t V
V
A theorem on the calculation of the mean Poynting vector from complex fields: h i 1 ¯~ ~c × H ~ c∗ N = Re E 2
The rate of doing work in magnetization dB dW =H dt dt and B = µ0 (H + M ). Hemholtz Theorem: An arbitrary vector F~ can be written as: Z 1 Z ∇ · F~ 1 ∇ × F~ ~ F = − ∇ dV + ∇× dV 4π r 4π r V V
= F~l + F~t Fields and potentials: ~ ~ = −∇Φ − ∂ A , E ∂t 316
~ =∇×A ~ B
Differential equation for the vector potential: 2~ ~ − 1 ∂ A = −µ0 J~ + ∇ ∇ · A ~ + 1 ∂Φ ∇A 2 2 c ∂t c2 ∂t
!
2
In the Lorentz gauge: ~ = − 1 ∂Φ ∇·A c2 ∂t the differential equations for the electromagnetic potentials are: ρ 1 ∂ 2Φ =− 2 2 c ∂t ε0 2~ ~ − 1 ∂ A = −µ0 J~ ∇2 A c2 ∂t2 and these have solutions of the form: ∇2 Φ −
1 Z ρ Φ= dV 4πε0 r ~= A
1 Z J~ dV 4πε0 c2 r
The field of a Hertzian dipole: 2 I0 ∆` cos θ 2 + 2 ei(ωt−kr) Er = 3 4πε0 c ikr r
"
#
I0 ∆` sin θ 1 1 ik i(ωt−kr) Eθ = + + e 4πε0 c ikr3 r2 r "
#
I0 ∆` sin θ ik 1 Bφ = + 2 ei(ωt−kr) 2 4πε0 c r r The mean energy flux from the Hertzian dipole: 2 ¯ = I0 ∆` N 8ε0 c λ
"
317
#2
sin2 θ r2
A series solution in 2 dimensions to Laplace’s equation in Cartesian coordinates: Φ(x, z) =
X
[Ak sin(αx) + Bk cos(αx)][Ck sinh(αz) + Dk cosh(αz)]
k
A series solution in 3 dimensions to Laplace’s equation in spherical polar coordinates: (
Φ(r, θ, φ) =
X
`
C1` r + C2` r
−(`+1)
" X
#)
[alm cos(mφ) +
blm sin(mφ)] P`m
m
`
Useful properties of trigonometrical functions: sin(α ± β) = sin α cos β ± sin β cos α cos(α ± β) = cos α cos β ∓ sin α sin β 1 (1 − cos 2α) sin2 α = 2 1 (1 + cos 2α) cos2 α = 2 Z π
sin3 θ dθ =
0
Z 2π
4 3 (
0 for m 6= n π for m = n
(
0 for m 6= n π for m = n
sin(mφ) sin(nφ) dφ =
0
Z 2π
cos(mφ) cos(nφ) dφ =
0
Z 2π
sin(mφ) cos(nφ) dφ = 0 for all m and n
0
318
(cos θ)
Properties of Legendry polynomials: Z1
Plm (cos θ) Pkn (cos θ) d(cos θ) = 0
unless m = n and l = k
−1
Z1
[Plm (cos θ)]2 d(cos θ) =
−1
P10 = cos θ ,
P0 = 1 ,
P21 =
2 (l + m)! 2l + 1 (l − m)! 1 P20 = (3 cos(2θ) + 1) , 4
P11 = sin θ ,
3 3 sin(2θ) , P22 = (1 − cos(2θ)) , 2 2 Pl (1) = 1 , for all l .
A theorem on the electrostatic potential due to a distribution of electric dipoles of moment per unit volume P~ : Φ=
1 I P~ · n ˆ 1 Z ∇ · P~ dS + − dV 4πε0 r 4πε0 r S
V
A theorem on the vector potential due to a distribution of magnetic dipoles ~: of moment per unit volume M ~ ~ ×n 1 Z ∇×M 1 I M ˆ dV + dS 2 2 4πε0 c r 4πε0 c r
~= A
V
S
A dispersion equation:
k=ω
0
0
εµ −µ
00
σ ε + ω
00
σ 0 00 −i µ ε + +εµ ω
319
0
00
1 2
The skin depth in a good conductor: s
δ=
2 ωµσ
General boundary conditions: ~ is continuous across a boundary. • The normal component of B ~ is continuous across a boundary. • The normal component of D ~ is continuous across a boundary. • The tangential component of E ~ is continuous across a boundary. • The tangential component of H Special boundary conditions at the surface between a dielectric and a perfect conductor: ~ = 0. • The tangential component of E ~ or H ~ = 0. • The normal component of B The Fresnel equations: ~ normal to the plane of incidence. Case 1: E Reflection: q
E1 =
k1 µ2 cos θi − µ1 k22 − k12 sin2 θi q
k1 µ2 cos θi + µ1 k22 − k12 sin2 θi
E0
Transmission: E2 =
2k1 µ2 cos θi q
k1 µ2 cos θi + µ1 k22 − k12 sin2 θi
E0
~ in the plane of incidence. Case 2: E Reflection : q 2 µ1 k2 cos θi − µ2 k1 k22 − k12 sin2 θi q H1 = H0 µ1 k22 cos θi + µ2 k1 k22 − k12 sin2 θi 320
Transmission: H2 =
2µ1 k22 cos θi µ1 k22
q
cos θi + µ2 k1 k22 − k12 sin2 θi
H0
In dielectric media the Fresnel equations become: ~ normal to the plane of incidence: Case 1: E E1 sin(θi − θt ) = , E0 sin(θi + θt )
E2 2 cos θi sin θt = E0 sin(θi + θt )
~ in the plane of incidence: Case 2: E tan(θi − θt ) E1 , = E0 tan(θi + θt )
E2 2 cos θi cos θt = E0 sin(θi + θt ) cos(θi − θt )
Rectangular waveguides: In the propagation direction z, the fields vary as e−γz . ~ satisfies: For TE modes, the longitudinal component of H ∂ 2 Hz ∂ 2 Hz + + k 2 Hz = 0 , 2 2 ∂x ∂y where k 2 = γ 2 − iωµ(σ + iωε) Then satisfying the boundary conditions (assuming the walls are perfect conductors) requires: γ2 =
mπ a
2
+
nπ b
2
For the lossless waveguide:
321
+ iµω(σ + iεω)
Cut-off frequency for the mn mode: s
fmn = vm
m 2a
2
n + 2b
2
Phase velocity: vm
vp = r
1−
fc 2 f
Group velocity: vg = vm
v u u t
fc 1− f
!2
VECTOR FORMULAS
∇(Φ + Ψ) = ∇Φ + ∇Ψ ~ + B) ~ = ∇·A ~+∇·B ~ ∇ · (A ~ + B) ~ = ∇×A ~+∇×B ~ ∇ × (A ∇(ΦΨ) = Φ∇Ψ + Ψ∇Φ ~ = A ~ · ∇Φ + Φ∇ · A ~ ∇ · (ΦA) ~ × B) ~ = B ~ · (∇ × A) ~ −A ~ · (∇ × B) ~ ∇ · (A ~ = ∇Φ × A ~ + Φ∇ × A ~ ∇ × (ΦA) ~ × B) ~ = A∇ ~ ·B ~ − B∇ ~ ·A ~ + (B ~ · ∇)A ~ − (A ~ · ∇)B ~ ∇ × (A ∇ · ∇Φ = ∇2 Φ ~ = 0 ∇ · (∇ × A) ∇ × ∇Φ = 0 ~ ~ = ∇(∇ · A) ~ − ∇2 A ∇ × (∇ × A) ~ · (B ~ × C) ~ = B ~ · (C ~ × A) ~ =C ~ · (A ~ × B) ~ A ~ × (B ~ × C) ~ = B( ~ A ~ · C) ~ − C( ~ A ~ · B) ~ A
322
FORMS OF VECTOR OPERATIONS IN CYLINDRICAL COORDINATES
∇Φ = ρˆ
∂Φ ∂Φ φˆ ∂Φ + + zˆ ∂ρ ρ ∂φ ∂z
~ = 1 ∂(ρAρ ) + 1 ∂Aφ + ∂Az ∇·A ρ ∂ρ ρ ∂φ ∂z ~ = ρˆ 1 ∂Az − ∂Aφ ∇×A ρ ∂φ ∂z
!
∂Aρ ∂Az + φˆ − ∂z ∂ρ
!
1 + zˆ ρ
∂(ρAφ ) ∂Aρ − ∂ρ ∂φ
1 ∂ ∂Φ 1 ∂ 2Φ ∂ 2Φ ρ + 2 2 + 2 ∇Φ = ρ ∂ρ ∂ρ ρ ∂φ ∂z !
2
VECTOR AND DIFFERENTIAL OPERATIONS IN SPHERICAL COORDINATES
~ = (Ax sin θ cos φ + Ay sin θ sin φ + Az cos θ) rˆ A + (Ax cos θ cos φ + Ay cos θ sin φ − Az sin θ) θˆ + (−Ax sin φ + Ay cos φ) φˆ = Ar rˆ + Aθ θˆ + Aφ φˆ
∇Φ =
∂Φ 1 ∂Φ ˆ 1 ∂Φ ˆ rˆ + θ+ φ ∂r r ∂θ r sin θ ∂φ
323
!
~ = ∇·A
1 ∂ (r2 Ar ) 1 ∂ (sin θAθ ) 1 ∂Aφ + + r2 ∂r r sin θ ∂θ r sin θ ∂φ θˆ 1 ∂Ar ∂ (rAφ ) rˆ ∂ (sin θAφ ) ∂Aθ − + − r sin θ ∂θ ∂φ r sin θ ∂φ ∂r "
~ = ∇×A
#
φˆ ∂ (rAθ ) ∂Ar − + r ∂r ∂θ "
"
#
#
1 ∂ ∂ 2Φ ∂Φ 1 ∂ ∂Φ 1 ∇Φ = 2 r2 + 2 sin θ + 2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 !
!
2
324
