Electromagnetism Course Notes (PHYS09060) 2017/18

Andreas Hermann School of Physics and Astronomy The University of Edinburgh

September 2017

Contents

1 Whistlestop tour of Vector Calculus 6 1.1 Gradient ...... 6 1.2 Divergence and Gauss’ Theorem ...... 7 1.3 Curl and Stokes’ Theorem ...... 7 1.4 Laplacian ...... 8 1.5 Useful Identities ...... 8 1.6 3D Taylor expansion ...... 9 1.7 Important Theorem ...... 9

2 Revision of Electrostatics 10 2.1 Charge Density ...... 10 2.2 Point charges and the δ-distribution ...... 10 2.3 Coulomb’s Law ...... 11 2.4 Electric Field ...... 12 2.5 Gauss’ law for E ...... 13 2.6 Electrostatic Potential ...... 14

3 Gauss’ Law 16 3.1 Conductors and Insulators ...... 16 3.2 Gauss’ Law in diﬀerential form ...... 16 3.3 Using Gauss’ Law ...... 17

1 3.4 Important δ-function identity and point charges (see tutorial 1.3) ...... 19

4 Poisson’s Equation and Images 20 4.1 Poisson’s Equation ...... 20 4.2 Properties of Poisson’s Equation ...... 20 4.3 A simple example: a hollow conductor ...... 22 4.4 The Method of Images ...... 22

5 Electric Dipoles and Multipoles 24 5.1 Field of an electric dipole ...... 24 5.2 Dipole interaction with external Electric Field ...... 25 5.3 Multipole expansions ...... 26

6 Electrostatic Energy and Capacitors 28 6.1 Electrostatic Energy of a general charge distribution ...... 28 6.2 Capacitors ...... 29 6.3 *Finite size disc capacitors ...... 31

7 Magnetic force, Currents and Biot Savart Law 32 7.1 Magnetic force ...... 32 7.2 Current density and current elements ...... 32 7.3 Biot Savart Law and Magnetic Fields ...... 34 7.4 Magnetic Force between Currents ...... 34 7.5 Long straight wires ...... 35

8 Divergence and Curl of B; Gauss’ and Amp`ere’slaws 36 8.1 div B and Gauss’ Law ...... 36 8.2 Magnetic Dipoles ...... 36 8.3 curl B and Amp`ere’sLaw ...... 38

9 Applications of Ampere’s` Law; Magnetic Vector Potential 40 9.1 Applications of Amp`ere’sLaw ...... 40 9.2 The Magnetic Vector Potential ...... 41 9.3 Poisson’s equation for the vector potential ...... 42 9.4 Summary of ’statics ...... 43

10 Electromotance and Faraday’s Law 44

2 10.1 Sources of steady currents ...... 44 10.2 Induced emf ...... 44 10.3 Faraday’s Law ...... 45 10.4 Connection to the Magnetic Vector Potential ...... 46 10.5 Lenz’s Law ...... 47

11 Inductance 48 11.1 Examples of Induction ...... 48 11.2 Mutual Inductance ...... 49 11.3 Self-Inductance ...... 50 11.4 Energy Stored in Inductors ...... 50

12 The Displacement Current 52 12.1 Continuity equation ...... 52 12.2 The Displacement Current ...... 52 12.3 Capacitor Paradox and Resolution ...... 53 12.4 Maxwell’s Equations ...... 54 12.5 Solution of Maxwell’s Equations in Free Space ...... 55

13 Description of Electromagnetic Waves 56 13.1 Recap of wave equations ...... 56 13.2 Plane Waves ...... 57 13.3 Electromagnetic Plane Waves ...... 57 13.4 Linear (Plane) Polarisation ...... 59 13.5 Circular Polarisation ...... 59

14 Electromagnetic Energy and the Poynting Vector 60 14.1 Poynting’s Theorem (Griﬃths 8.1.2)...... 60 14.2 Energy of Electromagnetic Waves ...... 61 14.3 Energy of discharging capacitor ...... 62 14.4 ∗Momentum of electromagnetic radiation ...... 63

15 Dielectric Materials 64 15.1 Overview ...... 64 15.2 Dielectric Materials ...... 64 15.3 Electric displacement vector and Gauss’ law in media ...... 65

3 15.4 Linear Isotropic Homogeneous Media ...... 66 15.5 Example: Dielectric in a Capacitor ...... 67

16 Magnetic Media 68 16.1 Types of magnetism ...... 68 16.2 The Magnetization Vector ...... 68 16.3 Ampere’s Law in media ...... 70

17 Summary of EM in media; boundary conditions on fields 72 17.1 Effect of Magnetic Materials on Inductance ...... 72 17.2 Electromagnetism in media: summary ...... 72 17.3 Energy densities and Poynting Vector ...... 73 17.4 Boundary conditions of fields ...... 74

18 Examples of continuity conditions; waves in media 76 18.1 Continuity conditions: examples ...... 76 18.2 Waves in media ...... 78 18.3 Waves in conductors ...... 79

19 Waves in Conductors 80 19.1 Skin Depth ...... 80 19.2 Good and poor conductors ...... 81 19.3 Phase relations of ﬁelds ...... 82 19.4 Intrinsic Impedance ...... 83

20 Waves at interfaces: normal incidence 84 20.1 Summary on plane waves and interfaces ...... 84 20.2 Interfaces between two dielectric media ...... 84 20.3 General waves at interfaces: normal incidence ...... 86 20.4 Reﬂection at Conducting Surface ...... 87

A Supplementary Material on Types of Magnetism 88 A.1 Diamagnetism ...... 88 A.2 Paramagnetism ...... 88 A.3 Ferromagnetism ...... 89 A.4 Eﬀect of Magnetic Materials on Inductance ...... 89

4 B Reﬂection at boundaries: oblique incidence 91 B.1 General Angle of Incidence ...... 91 B.2 Brewster’s Angle ...... 93 B.3 Total Internal Reﬂection ...... 94

C Retarded potentials and light cones 95 C.1 Inhomogeneous wave equations ...... 95 C.2 Solving the inhomogeneous wave equation ...... 96 C.3 Advanced and retarded potential, causality ...... 97

5 1 Whistlestop tour of Vector Calculus

You will have met vector calculus in second year mathematics courses. This year we shall see the true utility and power of vector calculus in formulating electrostatics. You need to reacquaint yourself with the div, grad, curl operators, and the concepts of line, surface and volume integrals. The following highlights some keypoints but does not replace your second year notes.

1.1 Gradient

The gradient operator (“grad”) acting on a scalar ﬁeld f(r) is a vector which in Cartesian Co- ordinates (x, y, z) reads

∂f ∂f ∂f gradf ≡ ∇f = e + e + e (1) ∂x x ∂y y ∂z z

Important things to remember:

• ∇f is a vector quantity1 • ∇f points in the direction of maximum increase of f • ∇f is perpendicular to the level surfaces of f • For a small change of spatial coordinate dr the change in f is df = ∇f · dr Z B • The line integral ∇f · dl = f(rB) − f(rA) is independent of the path from A to B A

Simple example to be memorised:2 ∇r = ˆr.

Remark Often due to the symmetry of the problem it is convenient to consider other co-ordinate systems such as spherical polar coordinates which comprise (r, φ, θ) or cylindrical polar coordinates which comprise (ρ, φ, z) (you should remind yourselves of these co-ordinate systems). In these systems the expression for the gradient (and the other operations below) look more complicated. E.g., in spherical polar coordinates ∂f 1 ∂f 1 ∂f ∇f = e + e + e ∂r r r ∂θ θ r sin θ ∂φ φ (where er = ˆr). You don’t need to remember the general formulae – you can look those up. But when the system has spherical symmetry f = f(r) (no θ or φ dependence) the gradient is simply ∂f ∇f = ∂r er and this should be remembered. This is consistent with the chain rule which states df df ∇f(r) = ∇r = ˆr (2) dr dr

1 1 1 Important example: ∇ = − ˆr = − r r r2 r3

1Vectors either underlined or boldface in these notes. 2Hat-symbols denote unit length vectors.

6 1.2 Divergence and Gauss’ Theorem

The divergence (“div”) is a scalar product of the gradient operator ∇ with a vector ﬁeld K. In Cartesians it reads ∂K ∂K ∂K ∇ · K = x + y + z (3) ∂x ∂y ∂z

The divergence represents the rate with which ﬂux lines of the vector ﬁeld K are converging towards sinks (negative divergence), or diverging from sources (positive divergence). ∂x ∂y ∂z Simple example: ∇ · r = + + = 3 ∂x ∂y ∂z

Gauss’ divergence theorem (to be memorised, and not confused with Gauss’ law in electrostatics) states that: I Z K · dS = ∇ · K dV (4) A V where A is a closed surface enclosing a volume V , dV = dx dy dz is a volume element (sometimes written d3r), and dS =n ˆdS is a vector element of area (normal to the surface).

• Gauss’ divergence theorem relates an integral over a closed surface to an integral over the volume enclosed.

• This theorem holds for any vector ﬁeld K and any closed surface A.

1.3 Curl and Stokes’ Theorem

The curl operator is a vector product of the gradient operator ∇ with a vector ﬁeld K: ∂K ∂K ∂K ∂K ∂K ∂K ∇ × K = z − y e + x − z e + y − x e (5) ∂y ∂z x ∂z ∂x y ∂x ∂y z or

ex ey ez

∇ × K = ∂ ∂ ∂ (6) ∂x ∂y ∂z

Kx Ky Kz

The curl ∇ × K(r) measures how a vector ﬁeld K(r) rotates in space near some point r. The curl is a vector and its direction is the axis of rotation by the right-hand rule and the magnitude of curl is the magnitude of the rotation

Simple example to be memorised: ∇ × r = 0

7 Stokes’s theorem (to be memorised) states that:

I Z K · dl = ∇ × K · dS (7) C A where C is a closed contour bounding a surface A.

• Stokes’ theorem relates a line integral around a closed curve to a surface integral over any open surface bounded by that curve.

• This theorem holds for any vector ﬁeld K and any closed curve C.

1.4 Laplacian

The Laplacian of a scalar ﬁeld is a scalar deﬁned as

∇2f = ∇ · (∇f) (8) and reads in Cartesian coordinates

∂2f ∂2f ∂2f ∇2f = + + (9) ∂x2 ∂y2 ∂z2

1.5 Useful Identities

These are best proved by suﬃx notation. First, there are various product identities. Generally these are as you’d expect,

1. ∇(φf) = φ∇f + (∇φ)f

2. ∇ · (φ A) = φ ∇ · A + A · ∇φ

3. ∇ × (φ A) = φ (∇ × A) + (∇φ) × A

You should be able to write these down. Others are less obvious and do not need to be memorised:

4. ∇ (A · B) = (A · ∇) B + (B · ∇)A + A × (∇ × B) + B × (∇ × A)

5. ∇ · (A × B) = B · (∇ × A) − A · (∇ × B)

6. ∇ × (A × B) = A (∇ · B) − B (∇ · A) + (B · ∇) A − (A · ∇) B

Secondly, there are some simple identities (involving two grads) that prove fundamental to Elec- tromagnetism:

• “curl grad = 0” ∇ × (∇f) = 0 (10)

where f(r) is any scalar ﬁeld.

8 • “div curl = 0” ∇ · (∇ × K) = 0 (11)

where K(r) is any vector ﬁeld.

• “curl curl = grad div - delsquared”

∇ × (∇ × K) = ∇(∇ · K) − ∇2K (12)

Note that a Laplacian acting on a vector is evaluated by component:

∂2K ∂2K ∂2K ∇2K = x + x + x x ∂x2 ∂y2 ∂z2

The ﬁrst two (10,11) are crucial to remember now. The third will become important later on.

1.6 3D Taylor expansion

As noted above the change in f due to a small change of position dr is df = ∇f · dr This is actually the ﬁrst term in the three-dimensional Taylor expansion about a point r0 which may be neatly written as

∞ X 1 0 n f(r) = (r − r ) · ∇ f(r)| 0 (13) n! r=r n=0 3 3 3 2 0 X 0 ∂f(r) 1 X X 0 0 ∂ f(r) = f(r ) + (xi − xi) + (xi − xi)(xj − xj) ··· (14) ∂x 0 2 ∂x ∂x i=1 i r=r i=1 j=1 j i r=r0

0 0 Often the ﬁrst two terms f(r) ' f(r ) + (r − r ) · ∇f(r)|r=r0 is all we require, which reduces to df = ∇f · dr.

1.7 Important Theorem

The following three statements concerning a vector ﬁeld F over some region in space are equivalent

1. ∇ × F = 0, the vector ﬁeld F is irrotational.

2. F = ∇φ, the vector ﬁeld F may be written as the gradient of a scalar ﬁeld (potential?) φ.

Z B 3. The line integral F · dl is independent of the path from A to B; IA A consequence is F · dl = 0 for any closed curve C. C

You should remind yourselves of how each implies the other. E.g., Stokes’ theorem gives 1. ⇔ 3. This theorem is the heart of electrostatics.

9 2 Revision of Electrostatics

2.1 Charge Density

At the microscopic level charge is a discrete property of elementary particles. The fundamental charge of an electron is −e, where e = 1.602 × 10−19C.

−21 The charge of a proton is +e: qp + qe < 10 e. Antimatter has the opposite charge to mat- −8 ter: qp + qp¯ < 10 e. ⇒ there is no charge in a vacuum!

Classical electromagnetism deals with macroscopic charge distributions. These are deﬁned by a charge density, ρ with units Cm−3:

ρ(r) = [Np(r) − Ne(r)]e where N are the number densities of protons and electrons.

The charge contained in an inﬁnitesimal element of volume dV centred at r is ρ(r)dV and this is sometimes referred to as a charge element. The total charge in a volume V is obtained by integration over the volume:

Z QV = ρ(r)dV (15) V where in Cartesians dV = dxdydz (sometimes we use d3r or even dτ e.g. in Griﬃths). Lower dimensional analogues are: line charges, with a charge density λ with units Cm−1; and surface charges, with a charge density σ with units Cm−2. Again the total charge can be obtained by integration: Z Z QA = σdS QL = λdl (16) A L

2.2 Point charges and the δ-distribution

In electrostatic problems it is common to introduce point charges, say Q at a particular position r 0. These are represented by a delta function:

ρ(r) = Qδ(r − r 0) (17) where δ(r − r0) = 0 for r 6= r0 but Z Z δ(r − r 0) = 1 if r 0 in V δ(r − r 0) = 0 otherwise (18) V V

Thus in (17) the charge distribution is very sharply peaked at r0 and the total charge R ρdV = Q.

• δ is a ”generalized function” or ”distribution”

• Here we are using the three dimensional delta-function, in Cartesians

δ(r − r 0) = δ(x − x0)δ(y − y0)δ(z − z0) (19)

10 • It is the limit of an appropriate series:

1 (x − x0)2 δ(x − x0) = lim √ exp − (20) ε→0 2πε2 2ε2 • Some properties: Z ∞ δ(x − x0)g(x)dx = g(x0) (21) −∞ Z ∞ d dg δ(x − x0)g(x)dx = − (x0) (22) −∞ dx dx δ(x − x0) = δ(x0 − x) (23)

2.3 Coulomb’s Law

Two point charges q1, q2 at points r1, r2 exert a force on each other given by

q1q2 F 1 = 2 ˆr12, r12 = r1 − r2 (24) 4π0r12 where ˆr12 is a unit vector, indicating that the force acts along the line connecting the two charges.

The constant 0 is known as the permittivity of free space.

−12 −1 −2 0 = 8.85 × 10 CN m (25)

A quite accurate and easily remembered number is: 1 = 9 × 109Nm2C−1 (26) 4π0

We’ll take Coulomb’s law as the empirical starting point for electrostatics. The inverse square dependence on the separation of the charges is measured to an accuracy of 2±10−16 using experiments based on the original experiment by Cavendish (Duﬃn P.31). Coulomb forces must be added as vectors using the principle of superposition. Thus the force on a point charge q at r due to a charge distribution ρ(r0) is the sum of all charge elements ρ(r0)d3r0 at positions r0 Z 0 q (r − r ) 0 3 0 F = 0 3 ρ(r )d r (27) 4π0 |r − r |

Example: Force due to a Line Charge

As an example consider the force of a line charge λ on a point charge Q. This can be obtained from the sum of the contributions: Qλdl dF = 2 ˆr 4π0r

11 Q



a r

 L/2 l dl L/2

Figure 1: diagram of integrating over line charge elements dl at angle θ to point charge

The components parallel to the line charge cancel, so we have to sum the contributions perpendicular to the line which are dF cos θ. This yields

Z L/2 Qλdl cos θ F⊥ = 2 2 −L/2 4π0 (l + a )

This integral is best solved by transforming it into an integral over dθ rather than dl i.e. we make the substitution:

l = a tan θ dl = a sec2 θdθ r = (l2 + a2)1/2 = a sec θ

Qλ Z θ0 Qλ dF⊥ = cos θdθ = 2 sin θ0 4π0a −θ0 4π0a 2 2 −1/2 where L/2 = a tan θ0 and sin θ0 = L/2 (a + L /4) (see diagram). Thus QλL F = (28) ⊥ 2 2 1/2 4π0a(a + L /4) where L is the length of the line charge, a is the distance of Q from the line.

We now look at diﬀerent limits of (28). In the limit of an inﬁnite line charge L → ∞:

Qλ F⊥ = (29) 2π0a Note that the force due to a line charge falls oﬀ with distance like 1/a

In the “far-ﬁeld” limit where a L QλL F⊥ ' 2 (30) 4π0a This is the same as the force due to a point charge q = λL at the origin

2.4 Electric Field

The force that a point charge q experiences, can be written as

F = qE (31)

12 which defines the electric field: the electric field the force per unit charge experienced by a small static test charge, q. The electric field is a vector and has units of of NC−1 or more usually Vm−1: Thus comparing with Coulomb’s force law we see that the electric field at r due to a point charge q at the origin is q E = 2 ˆr (32) 4π0r which is also known as Coulomb’s law. A positive (negative) point charge is a source (sink) for E.

Figure 2: Field lines of electric ﬁeld emanating from a positive point charge.

In a field line diagram field lines begin at positive charges (or infinity) and end at negative charges (or infinity). The density of field lines indicates the strength of the field.

2.5 Gauss’ law for E

I Q E · dS = (33) A 0 where A is any closed surface the surface, dS is a vector normal to a surface element, Q is the total charge enclosed by the surface. E(r) · dS is the Electric flux through the surface area element at point r. The left hand side is often written as I ΦE = E · dS A which is the total flux of the electric field out of the surface. Thus the total Electric flux through any closed surface is proportional to the charge enclosed – not on how it is distributed.

Simple example of Gauss’ law

First let us recover Coulomb’s law. Consider a point charge +q at the origin. Take the surface as a sphere of radius r. Now by symmetry the ﬁeld must point radially outwards. Thus E = Erer where Er has no angular dependence. Then the integral over spherical polar coordinates simpliﬁes considerably I Z 2π Z π 2 2 E · dS = dφ dθr sin θEr = 4πr Er A 0 0

13 2 and we obtain from Gauss’ law Er = q/4π0r which is Coulomb’s law. Although we have simply stated Gauss’ law as fundamental it is actually a simple consequence of Gauss’ divergence theorem and Coulomb’s Law (32). Proof: Use the identity derived in the tutorial ˆr ∇ · = 4πδ(r) r2 which implies, because the Div acts on r not the constant vector r0,

0 ! q rd− r q 0 ∇ · 0 2 = δ(r − r ) . 4π0 |r − r | 0

Then for a point charge at r0 I Z Z q q E · dS = ∇ · E dV = δ(r − r0)dV = if r0 in V , A V V 0 0 where the ﬁrst equality is the divergence theorem and the second comes from substituting the ﬁeld for a point charge and using the above identity. Similarly, by superposition of point charges, Gauss’ Law holds for any charge distribution ρ(r).

Now consider an insulating sphere with a uniform charge density ρ. Again by symmetry the electric field is radial, i.e. Eθ = Eφ = 0 everywhere. Again using a spherical closed surface of radius r to calculate the electric field Er: 2 ρ 4 3 Er4πr = πr 0 3 Inside the sphere (r < a): ρ r Er = 0 3 At the surface of the sphere (r = a): ρ a 3Q a Q Er = = 3 = 2 0 3 4π0a 3 4π0a Outside the sphere (r > a): Q Er = 2 4π0r This is the same field as for a point charge Q at the centre of the sphere!

2.6 Electrostatic Potential 1 ˆr r Consider Coulomb’s law (32) and the result ∇ = − = − . For a point charge q at point r r2 r3 r0, we can thus write q E(r) = −∇ 0 4π0|r − r | and identify the Electrostatic Potential at r due to a point charge q at r0

q V (r) = 0 (34) 4π0|r − r |

14 N.B. Sometimes the symbol φ is used for electrostatic potential when there is a possible clash of notation with V for volume. Now due to superposition we can integrate Coulomb’s law to get the Electric ﬁeld for any charge distribution and similarly superposition holds for the potential which is then given, e.g. for a continuous charge distribution, by

1 Z ρ(r0)d3r0 V (r) = 0 (35) 4π0 |r − r |

Moreover, we can invoke the important theorem of section 1.7 which implies that due to the existence of the potential V , static electric ﬁelds generally obey

1. ∇ × E = 0

2. E = −∇V Z B 3. The line integral of the ﬁeld E · dl = (V (rA) − V (rB)) is independent of the path. A I 4. A consequence is E · dl = 0 for any closed curve C. C

N.B this holds only for static ﬁelds as we shall see later. N.B. The potential V is only deﬁned up to a constant which may be chosen according to conve- nience. Often we choose V = 0 at r → ∞. Property 3. above gives us the work done to move a test charge from A to B

Z B WAB = −q E · dl = q[V (rB) − V (rA)] (36) A

Note that the work done is deﬁned as the work done by moving the charge against the direction of force—hence the minus sign. The potential diﬀerence VAB = VA − VB is then the energy required to move a test charge between two points A and B, in units of Volts, V=JC−1: W V = AB (37) AB q

If we take A at inﬁnity and VA = 0 the work done to move the charge from inﬁnity to B is the potential energy of the test charge at B

U = qVB (38)

Warning Beware of confusing Electrostatic potential V and potential energy U An equipotential is a surface connecting points in space which have the same electrostatic potential. By deﬁnition r → ∞ is an equipotential with V (∞) = 0.

Property 2. above tells us that The electric ﬁeld is always perpendicular to an equipotential.

15 3 Gauss’ Law

3.1 Conductors and Insulators

A conductor: is a material in which charges can move about freely. Therefore any electric ﬁeld forces the charges to rearrange themselves until a static equilibrium is reached i.e. they feel no net force. This in turn means that

• Inside a conductor E=0 everywhere, ρ = 0, and any free charges must be on the surfaces.

• Inside a conductor V is constant and the surfaces of a conductor are an equipotential.

• The electric ﬁeld E(r) just outside a conductor must be normal to the surface. We shall show that it is proportional to the surface charge density: σ E = nˆ (39) 0

An insulator: is a material where charges cannot move around. As a consequence

• The charge density ρ(r) can have any form, the potential V (r) is in general non-uniform, and E 6= 0 inside the insulator.

• Insulators are often referred to as ‘dielectric’ materials and we shall derive macroscopic expressions of their properties later on.

3.2 Gauss’ Law in diﬀerential form

Recall that Gauss’ Law states ‘Electric ﬂux through closed surface = charge enclosed/0’

I Qenc Z ρ(r) ΦE = E · dS = = dV A 0 V 0 for any surface A enclosing the volume V . With Gauss’ divergence theorem: Z Z E · dS = ∇ · E dV A V Z Z ρ(r) ⇒ ∇ · E(r) dV = dV V V 0 Since this holds for any volume V , however small, it follows that the integrands are identical. Therefore we must have the relation

ρ(r) ∇ · E(r) = (40) 0

This is Gauss’s Law In Diﬀerential Form. It is the ﬁrst of the fundamental laws of electromagnetism i.e. Maxwell I. N.B.: for static conductors this proves the earlier claim that E = 0 ⇒ ρ = 0.

16 Side note: Expressions for the divergence in other coordinate systems: 1 ∂(ρE ) 1 ∂E ∂E Cylindrical : ∇ · E = ρ + φ + z (41) ρ ∂ρ ρ ∂φ ∂z 1 ∂(r2E ) 1 ∂(sin θE ) 1 ∂E Spherical : ∇ · E = r + θ + φ (42) r2 ∂r r sin θ ∂θ r sin θ ∂φ

These are quite cumbersome and you do not need to remember them, but they simplify in the case of cylindrical or spherical symmetry. E.g., if our system is spherically symmetric, this means that there is no distinguished direction or position – therefore the electric ﬁeld must have no angular 1 ∂(r2E ) dependence and must be radial, i.e. E = E (r)e . Then ∇ · E = r . r r r2 ∂r

3.3 Using Gauss’ Law

Griffiths (2.2.3) states “Gauss’ Law affords when symmetry permits by far the quickest and easiest way of computing electric fields”. Note well the qualifier when symmetry permits. Basically there are three kinds of symmetry that work, and for which the following gaussian surfaces for the surface integral in Gauss’ law are appropriate:

1. Spherical symmetry – concentric sphere 2. Cylindrical symmetry – coaxial cylinder 3. Plane symmetry – “pill box”

Example 1: Insulating sphere

Let us consider an insulating sphere of radius a with a uniform charge density ρ. Because of symmetry, E(r) = E(r)er. The Gaussian surface integration is over concentric spheres. Inside the sphere (r < a): I Q 4π ρ E · dS = 4πr2E(r) = = r3 A 0 3 0 ρ ⇒ E(r) = r. 30

Outside the sphere (r > a): I Q 4π ρ E · dS = 4πr2E(r) = = a3 A 0 3 0 ρ a3 ⇒ E(r) = 2 . 30 r

Test diﬀerential Gauss’ Law: 2 3 1 ∂(r Er) ρ 1 ∂r ρ r < a : ∇ · E = 2 = 2 = r ∂r 30 r ∂r 0 ! Q 1 ∂ r2 r > a : ∇ · E = 2 2 = 0 4π0 r ∂r r

17 Example 2: Line charge

Figure 3: A cylindrical surface around line charge (Griﬃths Fig 2.21).

Consider an inﬁnite line charge of density λ. By symmetry Ez = Eφ = 0, and the closed surface is chosen to be a cylinder of length L and radius ρ with the line charge as its axis.

N.B. Here, ρ is the radial coordinate of cylindrical polar coordinates. Since Eρ(ρ) is constant over the cylinder surface I λ ΦE = E · dS = Eρ2πρL = l A 0 λ ⇒ Eρ = 2π0ρ Compare the simplicity of this method to summing the Coulomb forces in the previous lecture’s notes!

Example 3: Surface charge

Figure 4: Diagram of Gaussian pillbox around surface charge sheet (Griﬃths Fig 2.22).

Consider an inﬁnite charged sheet with surface charge density σ. By symmetry, E(r) = E(z)ez, and the Gaussian surface is chosen to be a circular “pillbox” of radius, R, and height h, with its axis normal to the surface and its centre at the surface. Now we have to be careful about the nature of the charge sheet. If the surface is simply a charge sheet with nothing distinguishing the two sides, there are equal and opposite perpendicular electric ﬁelds on either side of the sheet:

2 σ 2 σ ΦE = 2EzπR = πR ⇒ Ez = (43) 0 20 (Sometimes this is referred to as a thin insulating sheet.) However, If instead the charge is on the surface of a large conducting object, we know the inside of the conductor must have E = 0, therefore the only contribution to the ﬂux comes from the electric σ ﬁeld normal to the outer surface, and thus Ez = as quoted at the beginning of the lecture. Note 0 the factor of two between the conducting surface and the thin insulating sheet!

∂E Side note: In both insulating and conductor cases ∇ · E = z = 0 for z 6= 0 but the electric ∂z ﬁeld is discontinuous across the charge sheet at z = 0 with discontinuity σ/0. We can write the

18 Electric ﬁeld for all z using Heaviside’s step function deﬁned as ( 1 for z > 0 Θ(z) = 0 for z < 0

In the current example, the ﬁelds may be written as σ σ 1 Ez = Θ(z) (conducting),Ez = Θ(z) − (insulating). 0 0 2 Then we use the identity d δ(z) = Θ(z) (44) dz and ﬁnd that ∂E σ ∇ · E = z = δ(z) (45) ∂z 0 in both cases, which is consistent with equation (40) for a source of charge at z = 0.

3.4 Important δ-function identity and point charges (see tutorial 1.3)

Consider the vector ﬁeld ˆr v = r2 2 Now this is a spherically symmetric ﬁeld with vr = 1/r so using div in spherical polars 1 ∂ 1 ∇ · v = r2 = 0 (46) r2 ∂r r2 But clearly the integral over a spherical surface radius r I 2 v · dS = 4πr vr = 4π (47)

So Gauss’ theorem which should relate the two results appears to yield a contradiction. The source of the problem is r = 0 where v diverges (is singular). The contradiction can be resolved by noting that actually (see tutorial)

ˆr ∇ · = 4πδ(r) (48) r2 then Z Z ∇ · v dV = 4π δ(r) dV = 4π V V

Identity (48) implies that ∇ · E = ρ/0 holds even for a point charge for which ρ = qδ(r) and q ˆr E = 2 4π0 r

19 4 Poisson’s Equation and Images

4.1 Poisson’s Equation

If we utilize E = −∇V in the diﬀerential form of Gauss’ Law, we get Poisson’s Equation:

ρ(r) ∇2V (r) = − (49) 0

(Recall the Laplacian operator in Cartesian coordinates: ∇2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2).

In a region of space absent of free charges, (49) reduces to Laplace’s equation:

∇2V (r) = 0 (50)

• Note that one solution is a uniform potential V = V0, but this would only apply to the case where there are no free charges anywhere.

• Generally, we have to solve Laplace’s equation subject to certain boundary conditions and this yields non-trivial solutions.

• Both Poisson’s and Laplace’s equations are among the most important partial differential equations in physics, not just EM: fluid mechanics, diffusion, heat flow etc. They can be studied using the techniques you will meet or have met in Mathematics courses, e.g. separation of variables, orthogonal polynomials etc.

4.2 Properties of Poisson’s Equation

We can deduce immediately that, if V (r) is known everywhere, one can ﬁnd ρ(r) at any point simply by diﬀerentiating V (r) twice. Example: V = a + bx2y ∂2V ∂2V ∂2V ∇2V = + + = 2by + 0 + 0 ∂x2 ∂y2 ∂z2 2 ρ(x, y, z) = −0∇ V = −20by

The inverse problem, ﬁnding V (r) from a known charge distribution ρ(r), is to actually solve Poisson’s equation. This is a harder but much more common task! It is the central problem of electrostatics. Several properties of the equation help us here.

Helpful Property 1: Linearity

2 ρ1(r) 2 ρ2(r) If ∇ V1 = − and ∇ V2 = − , then 0 0

2 ρ1(r) + ρ2(r) ∇ (V1 + V2) = − 0

20 This property is known as linearity and gives rise to the “superposition principle”, which is used to sum potentials V arising from diﬀerent point charges or integrate over the contributions from a charge distribution. Superposition also applies to E = −∇V . Warning: One still has to satisfy boundary conditions - such as V = 0 on conducting surfaces or E → 0 for r → ∞.

Helpful Property 2: Uniqueness Theorem

If a potential V (r) obeys Poisson’s equation and satisfies the known boundary conditions it is the only solution to a problem. This is known as the uniqueness theorem. Basically if one can find a solution by whatever means – for instance educated guesswork or ‘physical intuition’ – then it is the solution. Proof of Uniqueness Consider a regionFigureR, boundary 5 B. Let ρ(r) be specified within R.

Figure 5: Diagram of region and boundary for Uniqueness Theorem (c.f. Griﬃths Fig 3.4)

Boundary conditions (BC’s): these are either (i) V specified on B (Dirichlet BC’s) (ii) E = −∇V is specified on B (Neumann BC’s) Then any solution of Poisson’s equation obeying the BCs is the only solution [up to a boring additive constant in case (ii)]. NB: B could be at infinity. Proof of Theorem

Suppose there are 2 diﬀerent solutions, V1, V2. Deﬁne ψ = V1 − V2. Then ∇2ψ = 0 in R (Laplace’s equation), and either case (i): ψ = 0 on B; or case (ii): ∇ψ = 0 on B. From Laplace: ∇2ψ = 0 ⇒ ψ∇2ψ = 0

⇒ ∇ · (ψ∇ψ) − (∇ψ)2 = 0 Z h i ⇒ ∇ · (ψ∇ψ) − (∇ψ)2 dτ = 0 R Apply Gauss’ divergence theorem to ﬁrst term Z Z ψ∇ψ · dS − (∇ψ)2 dV = 0 B R First term is identical zero for either case of BC’s.

21 The remaining integrand is non-negative, so it must vanish everywhere to integrate to zero:

∇ψ = 0 ⇒ ψ = V1 − V2 = const

Hence, V1 and V2 diﬀer by only a constant (the constant is zero in case of Dirichlet BC’s).

4.3 A simple example: a hollow conductor

Figure 6: Diagram of cavity R in a conductor.

Consider a cavity R in a conductor. Claim: If ρ = 0 in the cavity, then E = 0 inside.

Proof: Inner surface of conductor is equipotential (since it is conducting), V = V0. This is our boundary condition. Now inside the cavity ∇2V = 0 since there is no charge. Thus we must solve Laplace’s equation subject to the condition that V is constant along the (closed) boundary.

But one solution is V = V0 everywhere within R and this satisﬁes the boundary condition. Uniqueness: This is the only solution. ⇒ E = −∇V = 0 for any charge-free cavity within a conductor.

4.4 The Method of Images

In general: if we guess (and then verify) the solution to Poisson’s equation then, due to uniqueness, it is the only solution. Here: The idea is to place a suitable set of “image charges” outside the region R such that they reproduce the required boundary conditions on the boundary B. Such charges do not aﬀect Pois- son’s equation within R (since the images are not in the region), but their ﬁelds will describe the correct solution.

Example: Point charge near a conducting plane

Consider a point charge, Q, at distance a from a ﬂat conducting surface at a potential V0 = 0. The task is to solve Poisson’s equation with a point charge at aez, i.e. ρ(r) = Qδ(z − a)δ(x)δ(y) and boundary condition V (z = 0) = 0 on the boundary of the physical region z ≥ 0. 2 Note we can add a homogeneous solution Vim(r) (i.e., a solution to Laplace’s equation, ∇ Vim = 0), 2 to the potential V (r): ∇ (V + Vim) = −ρ/0.

22 Q

Figure 7: Point charge near a conducting plane.

Now the potential from the point charge at aez is 1 1 V (r) = (51) 2 2 2 1/2 4π0 (x + y + (z − a) )

This potential clearly violates the boundary condition V (z = 0) = 0. Let’s consider an ‘image charge’ Qim in the unphysical region z < 0. In the physical region (z > 0), its potential Vim is a homogeneous solutions, satisﬁes Laplace’s equation.

The correct guess for the image charge is Qim = −Q at rim = −aez. This basically reﬂects the symmetry of the problem. To check that the boundary condition is actually satisﬁed we write out the potential q 1 1 V = V + V = − (52) tot im 2 2 2 1/2 2 2 2 1/2 4π0 (x + y + (z − a) ) (x + y + (z + a) ) and see that it vanishes at z = 0 as required. Vtot is hence the (unique) solution in the physical region, z ≥ 0. N.B. If the conducting sheet is at potential V0 6= 0 we simply add a constant V0 to (52).

E must be normal to the conducting surface E = Ezez. Therefore on the surface ∂V −q (z − a) (z + a) E = − = − + z 2 2 2 3/2 2 2 2 3/2 ∂z z=0 4π0 (x + y + (z − a) ) (x + y + (z + a) ) z=0 −q a σ(x, y) = ≡ . 2 2 2 3/2 2π0 (x + y + a ) 0 where you can read oﬀ the surface charge density as well. Integrating σ(x, y) over the whole surface (left as exercise) shows that the surface charge is −Q. Note that there is an attractive force between the charge distribution on the conducting surface and the point charge above it. The method of images is really an inspired guess which works when the problem has appropriate symmetry (see tutorial problems for further examples).

23 5 Electric Dipoles and Multipoles

Motivation: we want to study (approximately) the ﬁelds of non-trivial charge distributions, which may or may not be overall neutral. The simplest example is the electric dipole. An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is deﬁned as p = qa .

By convention the vector a points from the negative to the positive charge. Here we also take the origin to be at the centre and a to be aligned to the z axis (see diagram)

q r a/2

r  a r a/2

-q

Figure 8: Diagram of electric dipole aligned along z axis. Griﬃths Fig 3.37

Why are dipoles relevant? Many molecules have a permanent dipole moment p, e.g. H2O, CO. All others, and all atoms, acquire an induced dipole when placed in electric ﬁelds.

5.1 Field of an electric dipole

We ﬁrst calculate the potential from superposition, and then the ﬁeld: q 1 1 V = − (53) 4π0 r+ r− where r± are the distances from the positive/negative charge to the point r.

Now consider the “far ﬁeld limit” r a and make a Taylor expansion for small a/r (remember tutorial sheet 1):

!−1/2 1 1 a a2 = 1 ∓ cos θ + 2 r± r r 4r 1 a ' 1 ± cos θ + O((a/r)2) r 2r where O((a/r)2) indicates terms proportional to (a/r)2 or higher powers.

24 Thus we obtain in the far-field limit qa cos θ p · ˆr V = 2 = 2 (54) 4π0r 4π0r One can confirm that away from the charges this is a solution of Laplace’s equation. To get the electric field, we use a gradient identity (see tutorial sheet 1) and obtain ! 1 p · ˆr 1 3(p · r)r p E = −∇V = − ∇ 2 = 5 − 3 (55) 4π0 r 4π0 r r

The components of the electric ﬁeld E= −∇V are simplest in spherical polar coordinates: ∂V 2p cos θ 1 ∂V p sin θ Er = − = 3 Eθ = − = 3 (56) ∂r 4π0r r ∂θ 4π0r

• The above ‘far-field’ limit r a can also be interpreted as an ideal dipole, which is the limit a → 0, q → ∞ but p = qa finite. The ideal dipole is a useful approximation to the ‘physical dipole’ for which the potential is given by (53) and the field by q r − a/2 r + a/2 E = −∇V = 3 − 3 (57) 4π0 |r − a/2| |r + a/2|

• Sketches of the electric ﬁelds are slightly diﬀerent:

z z

a x x -q

Figure 9: Sketch of electric dipole ﬁeld: ideal (left) and ‘physical’ (right) Griﬃths Fig 3.37.

• Important note: for a dipole, potential is 1/r2 and ﬁeld is 1/r3, whereas for a point charge potential is 1/r and ﬁeld is 1/r2.

5.2 Dipole interaction with external Electric Field

What is the force on an electric dipole in an external ﬁeld Eext = −∇Vext? First, calculate the electrostatic potential energy of the dipole in this ﬁeld:

Udip = qVext(a/2) − qVext(−a/2) ≈ qa · ∇Vext = −p · Eext

25 where we have made a Taylor expansion to ﬁrst order in a.

Thus Udip is minimised when the dipole is parallel to the ﬁeld and maximised when antiparallel. One needs ∆U = 2p|Eext| to reverse the dipole’s direction. Moreover we can compute the force felt by the dipole through

F = −∇Udip = ∇(p · Eext) (58) ∂E ∂E ∂E F = p x + y + z (59) x x ∂x ∂x ∂x

Hence, a dipole in a uniform electric ﬁeld (Eext does not depend on r) does not experience a force. This is also clear since the two charges of the dipole experience equal and opposite force. However, there is still a torque that acts to align the dipole moment p along the direction of Eext:

T = (a/2) × (qEext) − (a/2) × (−qEext) = p × Eext (60) The work done by the torque in rotating the dipole from an aligned position through to an angle θ relative to the ﬁeld is: Z θ Z θ W = T dθ = pEext sin θdθ = pEext(1 − cos θ) (61) 0 0

However, in a non-uniform field there is a force on the dipole which is generally complex, but moves the dipoles along the field gradient. Example: a dipole along a growing field experiences a force along the field gradient. To summarise i) dipoles tend to align with an external field ii) dipoles migrate up field gradients

5.3 Multipole expansions

Let’s do a Taylor expansion of the electrostatic potential for an arbitrary bounded charge distribution conﬁned to some region R. Z 0 1 0 ρ(r ) V (r) = dV 0 4π0 R |r − r | 1 Z ρ(r0) = dV 0 2 0 02 1/2 4π0 R (r − 2r · r + r ) Z 0 0 0 2 !−1/2 1 0 ρ(r ) 2r · r (r ) = dV 1 − 2 + 2 4π0 R r r r

In the far ﬁeld, r r0, we can expand the integrand as

!−1/2 ! 2r · r0 (r0)2 ˆr · r0 3(ˆr · r0)2 − (r0)2 1 − + ≈ 1 + + + ··· (62) r2 r2 r 2r3 and hence Z " 0 0 2 0 2 # 1 0 0 1 ˆr · r 3(ˆr · r ) − (r ) V (r) ≈ dV ρ(r ) + 2 + 3 + ··· (63) 4π0 R r r 2r

26 We can tidy up the expansion to write it as 1 Q 1 ˆr · P 1 1 1 V (r) = + + X Q rˆ rˆ + ··· (64) 4π r 4π r2 4π r3 2 ij i j 0 0 0 i,j where Z Q = dV 0 ρ(r0) − total net charge V Z P = dV 0 r0ρ(r0) − total net dipole moment V Z 0 0 0 0 2 0 Qij = dV (3rirj − (r ) δij)ρ(r ) − components of quadrupole tensor V

• The ﬁrst term (‘monopole term’) would be the dominant term if Q 6= 0, and this would quite reasonably represent the far-ﬁeld charge distribution as that of a single point charge at the origin.

• When the total charge Q vanishes, (as in the case of a dipole) it is the next term (‘dipole term’) which is dominant.

• If the second moment term also vanishes (i.e. P = 0, see ﬁgure) then the third term ˆrT Qˆr (‘quadrupole term’) becomes the dominant term. Note is the quadrupole moment. r3

Figure 10: 4 charges forming an electric quadrupole Griﬃths Fig 3.27

Note that only the smallest non-vanishing multipole moment is independent of the choice of coordinate system. Each term in the expansion (64) is separately a solution of Laplace’s equation (away from the charges). Thus one can use the idea of the method of images and use image dipoles, quadrupoles, etc., at some point r0 outside of the physical region to solve Poison’s equation inside the physical region, and ﬁx certain boundary conditions on the boundary of the physical region.

An example is a conducting sphere in a uniform external ﬁeld E0. The boundary condition is that E is radial at the surface of the sphere which is an equipotential. This is achieved by using an image electric dipole at the centre of the sphere - see tutorial. The result is that on the surface of the sphere there is an induced charge distribution which is positive on one side and negative on the other.

27 6 Electrostatic Energy and Capacitors

6.1 Electrostatic Energy of a general charge distribution

Here we provide a proof that the electrostatic energy density (energy per unit volume)

1 u = |E|2 (65) E 2 0 is a completely general result for any electric ﬁeld.

First, assume an assembly of n − 1 point charges at positions rj. They give a potential at r (here the subscript j labels the charge):

1 n−1 q V (r) = X j 4π |r − r | 0 j=1 j

Now, bring up another charge qn (the nth one) from inﬁnity to point rn. This requires work:

q n−1 q W = q V (r ) = n X j n n n 4π |r − r | 0 j=1 n j

In particular, for charge 1 the work W1 is zero since there is no potential yet, for charge 2 the work W2 is q2q1/(4π0r12), for charge 3 the work W3 is q3 times the potential due to charges 1 and 2, etc.

Hence, the total energy UE of the charges, which is equal to the total work required to assemble all n charges, is given by summing over i = 1, ··· , n

n 1 n q q 1 n n q q U = X W = X X i j = X X i j E i 4π |r − r | 8π |r − r | i=1 0 i=1 j

Make sure you understand the factor of 1/2 which appears in the second equality when we allow both sums to go over all charges. We can write the ﬁnal equality as 1 U = X q V (66) E 2 i i i q where V = X j is the potential at r due to the other charges j. i 4π |r − r | i j6=i 0 i j

This can be generalized in the limit of a continuous charge distribution to an integral over the charge density ρ: 1 Z U = ρ(r)V (r)dr (67) E 2

It turns out that we can write this integral in another way. First recall Maxwell I, ρ ∇ · E = . 0

28 Then (67) becomes Z U = 0 V (r)(∇ · E) dr (68) E 2 Now use a product rule from section 1 to substitute the integrand in (68):

∇ · (V (r)E) = V (r)(∇ · E) + (∇V ) · E = V (r)(∇ · E) − |E|2

This gives, and we use Gauss’ divergence theorem Z U = 0 ∇ · (V (r)E) + |E|2 dr E 2 I Z = 0 (V (r)E) · dS + 0 |E|2 dr 2 S 2 Then taking the region as all space and the boundary S at infinity, where we have boundary conditions V (∞) = 0 and E = −∇V = 0 means that the first integral is zero. Therefore the final result comes from the second integral: Z 0 2 UE = |E| dV (69) 2 all space and from this we get (65) for the energy density at a point.

• Warning: The general result (69) was derived for a continuous charge distribution since we began from (67). When there are point charges we have to be careful with self-energy contributions which should be excluded from the integral (67), otherwise they lead to divergences.

• Equation (67) leads us to think that Electrostatic energy lies in the charge distribution whereas from (69) we might infer that the energy is stored in the electrostatic ﬁeld. Which picture is correct? In fact these interpretations are tautologous.

• Finally, since (69) is quadratic in the ﬁeld strength we do not have superposition of energy density.

6.2 Capacitors

A capacitor is formed when two neighbouring conducting bodies (any shape) have equal and opposite surface charges. Suppose we have two conductors, with charge Q and −Q. Since the potential is constant on each conductor, the potential difference between the two is V = V1 − V2. In general to actually find V (r) and E can be difficult (need to solve Poisson’s equation between conductors). However there is a unique well-defined value of the capacitance, defined as the ratio of the charge on each body to the potential difference between the bodies:

Q C = (70) V

Capacitance is measured in Farads = Coulombs/Volt. A capacitor is basically a device which stores electrostatic energy by charging up.

29 A

Figure 11: Diagram of Parallel Plate Capacitor Griﬃths ﬁg 2.52

Example: the Parallel Plate Capacitor

Two parallel plates of area A have a separation d. They carry surface charges of +σ and −σ, which are all on the inner surfaces of the plates because of the attractive force between the charges on the two plates. The normal to the plate is taken in ez direction (positive up). To obtain the Electric ﬁeld we use Gauss’ Law. First take a pillbox that straddles the inner surface of the upper plate say. Now inside the conducting plate E = 0. Therefore between the two plates the ﬁeld (normal to the surface of the plate) is

(+σ) σ Ein = (−ez) = − ez 0 0 Now take a pillbox that straddles the outer surface of the upper plate. Since there is no charge on the upper surface and since inside the plate E is zero, we must have that E = 0 at the outer surface. σ Eout = 0 Ein = − ez 0 Therefore, the potential diﬀerence between the plates is

Z d σd Qd V = Ezdz = = 0 0 A0 Q A C = = 0 (71) Vd d Note that C is a purely geometric property of the plates! Now, let us calculate the energy stored in the capacitor using the formulas (67) and (69) derived in the last section. First use (67), based on the charge distribution: The integral simpliﬁes to a sum of two contributions from the upper plate which has charge Q and potential V1 and the lower plate which has charge −Q and potential V2. Thus

Q QV CV 2 Q2 U = (V − V ) = = = (72) 2 1 2 2 2 2C where we have used the deﬁnition of capacitance (70). On the other hand we can, using (69), integrate over the electric ﬁeld, which is constant between the plates. Z 2 2 0 2 0 σ d Q UE = |E| dV = Ad = 2 2 0 20 A

30 and using the deﬁnition of capacitance obtain the result (71)

Q2 U = , E 2C which is of course the same as (72). Other capacitor geometries (see problem sheet) will lead to diﬀerent, usually non-uniform ﬁeld distributions E(r), but the concept of stored energy by charge separation remains the same.

6.3 *Finite size disc capacitors

So far we have assumed the capacitor plates are effectively infinite in size. In that case, the electric field between the two plates was uniform. However, when should you worry about the finite size of capacitor plates? For a finite-size capacitor it is possible that there are edge effects where the field can bulge out of the capacitor and also results in non-uniformity of the field within the capacitor. To get a feeling for when such effects become important let us compute the potential and field of a finite-size disc of surface charge density σ. We perform a two-dimensional integral over the charge distribution (in the xy-plane) to obtain the potential at a height z along the axis of the disc 1 Z σdS V = 2 2 1/2 4π0 S (ρ + z ) In plane polar co-ordinates, dS = ρdρdφ and therefore σ Z 2π Z R ρ V (z) = dφ dρ 2 2 1/2 4π0 0 0 (ρ + z ) σ h iR σ h i = 2π (ρ2 + z2)1/2 = (R2 + z2)1/2 − z 4π0 ρ=0 20 Now the z component of the field is ∂V σ z E = − = − − 1 z 2 2 1/2 ∂z 20 (R + z ) and we see that there is a z dependence – the field is clearly not uniform. Two limits are interesting: the limit R z corresponds to a point charge:

2 ! 2 σ R −1/2 σR Q E(R z) = 1 − (1 + 2 ) ez ' 2 ez = 2 ez 20 z 40z 4π0z where Q = σπR2 is the total charge. The limit R z reduces to the uniform field of an infinite plane of charge: σ E(R z) = ez (73) 20 However, when R ≈ z (i.e., A ≈ d2), you need to consider the finite size of the disc – edge effects for a capacitor are limited to the regions where the gap distance is comparable to the dimensions of the plates.

31 7 Magnetic force, Currents and Biot Savart Law

7.1 Magnetic force

The magnetic ﬁeld B is deﬁned by the force on a charge q moving with velocity v:

F = q(E + v × B) (74)

This is the Lorentz Force Law. The second term is the magnetic force. The unit of magnetic ﬁeld is the Tesla (T) which is NA−1m−1. Actually this is a pretty big unit and a Gauss = 10−4T is more commonly used. Since v × B acts perpendicular to the direction of motion, magnetic forces do no work.

7.2 Current density and current elements

The ﬁrst thing to note is that a moving charge by itself does not really constitute a current. Instead we need a moving density of charges. For the moment we will consider steady currents so that at any point we have a constant density of charged particles moving past the point (clearly we will need sources of current somewhere but let’s not worry about that for the moment). Also we can have zero net charge but a steady current, if the densities of positive and negative particles are the same but their velocities are diﬀerent. A current consists of n charges q per unit volume moving with average velocity v. These charges form a local current density:

J(r) = n(r)qv(r) (75)

The total current I passing through a surface is obtained by integration: Z I = J · dS (76) A where as usual dS points normal to the surface.

Units: The unit of current is the Ampere (A), which is a base SI unit, 1A = 1Cs−1. The unit of bulk current density J is A/m2. We can also have surface current densities usually denoted K or j (units A/m) and line current densities usually denoted I (units A). Calling all of these ‘densities’ is a bit confusing since none has units of current per unit volume but that is the way it is! What we shall see is that steady currents play the key role in magnetism as do electric charges in electrostatics, that is

Stationary charges ⇒ constant electric ﬁelds – electrostatics Steady currents ⇒ constant magnetic ﬁelds – magnetostatics

Example: Rotating disc An insulating disc of radius R, carrying a uniform surface charge density σ, is mechanically rotated about its axis with an angular velocity vector ω (in the ez direction). As a result it has a current density on its surface:

K = σv = σω × r = σrωeφ (77)

32 z

  R

Figure 12: Diagram of rotating charged disc

Note that the current density increases linearly with r. Check that you understand how the direction comes from the right hand rule.

Conductivity The conductivity σ describes the intrinsic conduction properties of a bulk material in response to an electric ﬁeld. The current density is:

J = σE (78)

Note that σ is a property of a particular material, and that it depends on temperature. A typical value of σ for a metal is 108 Ω−1m−1. We deﬁne the resistivity as ρ = σ−1. A typical value of ρ for an insulator is 1016 Ωm, and it is zero in a superconductor. Remark: Actually (78) makes a crucial assumption that J and E are parallel which is not necessarily the case for some non-isotropic materials where, for example, current can only ﬂow in certain directions. In such cases one needs a conductivity tensor σij. Current elements A current element, denoted here dI is a vector (units A · m) with

dI(r) = J(r)dV - current element in bulk dI(r) = K(r)dS - current element on surface dI(r) = I(r)dl - current element along a wire

Warning: you need to take care with current elements e.g. KdS 6= KdS since the left hand side points in the direction of the current vector on the surface, but the right hand side points normal to the surface. On the other hand Idl = Idl since a line current element always points along the wire in the direction dl. A current element is a moving charge element. For bulk current densities

dI(r) = J(r)dV = ρ(r)v(r)dV = ρ(r)dV · v(r) = dq · v(r) (79)

The force on a steady current element is therefore

dF = dI × B (80)

33 In particular, if the current element comes from a bulk current density dI = J dV we have:

dF = J × B dV (81)

7.3 Biot Savart Law and Magnetic Fields

Previously, we saw that a charge q creates an electric ﬁeld E. Here, we learn that a current element dI at r0 creates a magnetic ﬁeld dB at a position r:

µ dI(r0) × (r − r0) dB(r) = 0 (82) 4π |r − r0|3

This is known as the Biot-Savart Law. The law was established expermentally; we shall take it as our axiomatic starting point. It plays the same role for magnetostatics as Coulomb’s law does in electrostatics.

The constant µ0 is known as the permeability of free space:

−7 −1 −7 −2 µ0 = 4π × 10 Hm = 4π × 10 NA . (83)

The direction of the field is perpendicular to dI(r0) and r − r0, the vector from the current element to the point r. The direction can be remembered from the usual right hand rule for vector products. Note that superposition holds for magnetic fields, therefore the magnetic field can be calculated by integration over current elements:

µ Z dI(r0) × (r − r0) B(r) = 0 (84) 4π |r − r0|3 e.g. for integration over a volume containing a distribution of current density:

Z 0 0 µ0 J(r ) × (r − r ) 0 B(r) = 0 2 dV (85) 4π V |r − r |

7.4 Magnetic Force between Currents

Let’s plug the Biot Savart Law into expression (80) for the magnetic force on a current element:

µ0 dF 1 = 2 dI1 × (dI2 × ˆr12) (86) 4πr12 which corresponds to the force between a pair of current elements. Here r12 = r1 − r2 points from dI2 to dI1. Equation (86) is also referred to as Biot-Savart law and is the equivalent of Coulomb’s force law for the force between two point charges. If the current elements are due to bulk current densities we have µ dF = 0 J × (J × ˆr )dV dV (87) 1 4πr2 1 2 12 1 2 where the force is attractive for parallel currents, and repulsive for antiparallel currents.

34 I ez r r r = zez

O r = e

Figure 13: Diagram for calculating B for an inﬁnite straight wire, see Griﬃths Fig. 5.18

7.5 Long straight wires

We consider a long straight wire which we choose to be along the z axis so that a point r0 on the 0 0 wire is given by r = z ez. We want to compute the ﬁeld using (84). It is best to use cylindrical polars: we choose the origin along the wire so that r is ⊥ to ez i.e. r = ρeρ where ρ is the radial distance of the point from the wire. 0 0 0 0 0 0 0 Now dI = Idz ez and dr = dz ez so dr × (r − r ) = dr × r = ρdz eφ and we ﬁnd

µ Iρ Z ∞ dz0 B(r) = 0 e φ 2 02 3/2 4π −∞ |ρ + z |

To evaluate the remaining integral we use trigonometric substitution z0 = ρ tan θ so that dz0 = ρ sec2 θ dθ and ρ2 + z02 = ρ2 sec2 θ. Then we obtain

π/2 µ0I Z µ0I B(r) = eφ cos θdθ = eφ 4πρ −π/2 2πρ

We now consider the force on a current element of a second parallel wire (at distance d from the first) coming from the magnetic field B1(r) due to the first wire. Again we choose coordinates so that this current element lies at r = ρeρ in cylindrical polars µ dF = dI (r) × B (r) = I dze × 0 I e 2 1 2 z 2πd 1 φ µ I I = − 0 1 2 dze 2πd ρ There is an attractive force per unit length between two parallel infinitely long straight wires. This is the basis of the definition of the Ampère.

35 8 Divergence and Curl of B; Gauss’ and Amp`ere’slaws

8.1 div B and Gauss’ Law

We can rewrite the Biot-Savart Law for B due to a bulk current density using the expression for ∇(1/r) as

Z 0 Z µ0 0 (r − r ) 0 µ0 0 1 0 B(r) = J(r ) × 0 3 dV = − J(r ) × ∇ 0 dV (88) 4π V |r − r | 4π V |r − r |

To obtain ∇ · B we note that ∇ is with respect to the r coordinates and J(r0) depends on r0 only. We thus ﬁnd 1 1 ∇ · J(r0) × ∇ = J(r0) · ∇ × ∇ = 0 |r − r0| |r − r0| where the last equality follows since ‘curl-grad = 0’. Therefore ∇ · B = 0 (89)

This result is the second fundamental law of electromagnetism (Maxwell II).

• A magnetic ﬁeld has no divergence, which is equivalent to:

• There are no magnetic monopoles. There are no point sources of magnetic field lines, instead the magnetic fields form closed loops around conductors where current flows.

Now Gauss’ divergence theorem states that Z I ∇ · B dV = B · dS V A Thus the net magnetic ﬂux through any closed surface A must be zero,

I B · dS = 0 (90) A which is sometimes referred to as Gauss’ law for magnetic ﬁelds.

8.2 Magnetic Dipoles

Since there are no magnetic monoples, we should identify what the magnetic equivalent of an electric dipole is, i.e. a magnetic dipole. It turns out this is a current loop. Consider a circular current loop radius a carrying steady current I in the clockwise direction with axis in the ez direction. We consider the contribution to the magnetic ﬁeld at r along the axis of the loop due to the current element dI(r0) at r0 using the Biot-Savart law

µ dI(r0) × (r − r0) dB(r) = 0 (91) 4π |r − r0|3

36 z

m A a

Figure 14: Simple current loop with axis along z axis

0 0 0 0 2 2 1/2 We choose coordinates so that r = zez, r = aeρ , dI = Idleφ and |r − r | = (z + a ) . Then

0 0 0 dI(r ) × (r − r ) = Idl(zeρ + aez)

Now we see that the dB is not along ez but when we integrate around the current loop the radial components cancel. Therefore we consider µ Iadl dB = 0 z 4π(z2 + a2)3/2

Note that this does not depend on the angle φ0 around the ring. Therefore, when we integrate around the ring we simply get a factor 2πa and

µ Ia2 B = 0 (92) z 2(a2 + z2)3/2

• At the centre of the loop (z = 0): µ I B = 0 z 2a • In the far ﬁeld, at a large distance from the loop (z a):

µ Ia2 B ' 0 z 2z3

To extend this calculation to an arbitrary position r (at all angles θ relative to the axis of the loop) is tedious, but it can be shown that the field of a loop is a magnetic dipole field. I.e. in the far field limit of r a one finds µ B (r) = 0 [3(m · ˆr)ˆr − m] (93) dip 4πr3 where the magnetic dipole moment, m, is the product of the current and the vector area of the loop: 2 m = Iπa ez = IAez (94) In fact this result holds for a small current loop of any shape and with magnetic dipole moment, m, defined as Z m = IA = I dS (95) where A is the vector area of the loop.

37 The ideal magnetic dipole field has the same form as the ideal electric dipole field: 2m cos θ µ m sin θ B = µ B = 0 (96) r 0 4πr3 θ 4πr3 However the physical versions of electric and magnetic dipole look a bit different:

Figure 15: Sketch of physical magnetic dipole lines c.f. Griﬃths ﬁg 5.55

It can be shown (Griﬃths 6.1) that the potential energy of the dipole in the ﬁeld is:

U = −m · Bext (97) and that an external magnetic ﬁeld creates a torque on a magnetic dipole:

T = m × Bext (98)

So one can think of a compass needle (magnetic moment along the needle) aligning with the Earth’s magnetic field. One could think of a magnetic dipole being composed of two monopoles (the ‘Gilbert Model’) and this gives the correct results for torque and energy (see Griffiths Chapter 6). However, this picture is basically wrong as the fundamental difference between a magnetic dipole and an electric dipole is that it is impossible to separate the N and S poles of a bar magnet, for example. You might wonder what current loops have got to do with ordinary bar or fridge magnets. The point is that in most atoms the electrons orbiting the nucleus act as current loops, so atoms can have magnetic dipole moments. Moreover, an electron has ‘spin’, an instrinsic magnetic moment.

8.3 curl B and Amp`ere’s Law

Consider again the Biot-Savart law in form (88) and take its curl: Z µ0 0 1 0 ∇ × B(r) = − ∇ × J(r ) × ∇ 0 dV (99) 4π V |r − r | where the curl is with respect r coordinates so can be taken inside the integral which is over r0 coordinates. Now using a product rule from lecture 1 and remembering that J(r0) does not depend

38 on the co-ordinates of r, 1 1 1 ∇ × J(r0) × ∇ = ∇2 J(r0) − (J(r0) · ∇)∇ |r − r0| |r − r0| |r − r0| 1 = −4πδ(r − r0)J(r0) − (J(r0) · ∇)∇ (100) |r − r0|

1 where we have used the now familiar result ∇2 = −4πδ(r). When we insert (100) back into r the integral (99), the second term can be shown to give zero since it can be written as a ‘boundary term’ at inﬁnity which vanishes. The ﬁrst term in (100) yields

∇ × B = µ0J (101)

This is another fundamental law of electromagnetism, i.e. Maxwell IV. The curl of a magnetic field around an axis is proportional to the component of the current density along the axis. To obtain an integral form of Ampère’slaw we use Stokes’ theorem, connecting an open surface S with its closed contour C via I Z Z B · dl = (∇ × B) · dS = µ0 J · dS C S S The integral of the magnetic field round a closed loop is related to the total current flowing across the surface enclosed by the loop:

I Z B · dl = µ0I = µ0 J · dS (102) C S

This is Ampère’s law. In a similar way to Gauss’ law in electrostatics, it is very useful for calculating magnetic fields when there is a high degree of symmetry to the problem. Example: An infinite wire of finitie radius a carries a uniform current density, J. Outside the wire, at radial distance ρ: Z Bφ2πρ = µ0 J · dS = µ0I S µ I 1 B = 0 · (103) φ 2π ρ

The field outside the wire drops off with distance as 1/ρ. This is a much easier derivation than integrating the Biot-Savart law. Inside the wire: Z 2 2 ρ Bφ2πρ = µ0 J · dS = µ0Jπρ = µ0I 2 S a µ I B = 0 ρ (104) φ 2πa2 The field inside the wire increases with radius.

39 9 Applications of Ampere’s` Law; Magnetic Vector Potential

9.1 Applications of Amp`ere’sLaw

I Z B · dl = µ0 J · dS = µ0I (105) C S Like Gauss’ law for electric fields, Ampère’slaw is the most efficient way of calculating magnetic fields when the system has appropriate symmetry. The symmetries which work, and the corresponding integration paths (”Amperian loops”), are

• Inﬁnite straight lines (see straight wire example from last lecture) → coaxial circles

• Inﬁnite planes (see next example of current sheet) → rectangular loops

• Inﬁnite solenoids (see tutorial 5.1) → rectangular loops

• Toroids (see toroidal example below) → circles

The diﬃcult part is usually working out the direction of the magnetic ﬁeld; after that Ampere’s law readily gives the answer by choosing the Amperian loop appropriately.

Field of an inﬁnite slab of current (Griﬃths Example 5.8)

An inﬁnite sheet of conductor of thickness d, carries a uniform current density J parallel to the surface of the sheet.

Figure 16: Inﬁnite current sheets and Amperian loops (c.f. Griﬃths Fig 5.33)

Let us take ez normal to the sheet and choose ex to be along the direction of the current. By B-S law B field has to be perpendicular to J (i.e. in y–z plane). Now the symmetry of the infinite plane means that any component of B in the ez direction cancels. Thus the planar symmetry implies that B is in the ey direction, i.e. parallel to the plane and perpendicular to the current. We take the contour integral around a rectangular loop parallel to the y–z plane, with length l and height h enclosing the sheet. The magnetic field outside the slab is then: µ Jd 2|B |l = µ Jld |B | = 0 y 0 y 2

40 This is a uniform magnetic field but note that the directions of the field on the two sides of the sheet are opposite to each other! µ Jd µ Jd B = − 0 e for z > d/2 B = + 0 e for z < −d/2 (106) 2 y 2 y The field inside the conducting sheet can also be calculated by choosing a loop in the y–z plane that straddles the surface of the sheet. Then using the above result for the portion outside yields the field inside the slab By = −µ0Jz |z| < d/2 where z = 0 is at the centre of the sheet. (Exercise)

Field of a toroid (Griﬃths Example 5.10)

A toroid consists of a set of coils of radius R, carrying a current I, and formed into a larger circle of radius a, so that they look like a doughnut. There are n coils per unit length around the larger circle. The toroidal symmetry is a little subtle: there is clearly symmetry with respect to rotation about z axis (no dependence on φ) but also since the current always flows in the eρ − ez plane one can deduce from the BS law that the field must always be in the eφ direction, i.e. it is circumferential (since other components cancel). Griffiths Ex 5.10 gives a proof of this for a toroid of arbitrary cross-section.

Then we take our Amperian loops to be circles perpendicular to ez. If the circle is not enclosed by the toroid, the current which cuts the circle is zero. Therefore B = 0 outside the toroid. If the Amperian loop is a circle enclosed by the toroid of radial distance from z-axis ρ, then

Bφ2πρ = µ0n2πaI Note that the rhs is constant since the same number of turns is always enclosed by such a loop. The ﬁeld inside the toroid coils is: µ nIa B = 0 (107) φ ρ Note that this is not uniform, but depends on ρ, the radial distance from the z-axis.

9.2 The Magnetic Vector Potential

Just as the theorem of 1.7 was the heart of electrostatics, the following theorem is the heart of magnetostatics: Theorem The following three statements concerning a vector ﬁeld B over some region in space are equivalent:

1. ∇ · B = 0 – the vector ﬁeld is “solenoidal”

2. B = ∇ × A – the vector ﬁeld may be written as the curl of a vector potential Z 3. The surface integral of the ﬁeld B · dS is independent of the shape of the surface S for a S I given boundary curve; in particular, B · dS = 0 for any closed surface A. A

41 We do not prove all the equivalences (see Griﬃths 1.6) but it is clear that 2. implies 1. since ‘div curl =0’: ∇ · B = ∇ · (∇ × A) = 0 (108)

Thus starting from the key property of the magnetic field, ∇ · B = 0 (no monopoles), we find from 2. that we may always write the magnetic field as the curl of a vector potential

B = ∇ × A (109)

This is our key result (c.f. E = −∇V for a static electric ﬁeld). Finally, from 3. and using Stokes’ theorem we obtain Z I ΦB ≡ B · dS = A · dl (110) S C The magnetic ﬂux through a surface is given by the integral of the magnetic vector potential around the loop enclosing that surface.

9.3 Poisson’s equation for the vector potential

Amp`ere’slaw can be written in the form:

∇ × B = ∇ × (∇ × A) = µ0J

Using a vector operator identity for “curl-curl” (see week 1) this becomes:

2 ∇ A − ∇(∇ · A) = −µ0J (111)

In the same way as we are free to choose the value of the scalar potential in electrostatics to be V (∞) = 0 (adding a constant does not change the ﬁeld E), we are free to choose the divergence of the magnetic vector potential (adding a divergence term does not change the ﬁeld B). This property is known as gauge invariance. The choice of ∇·A = 0 is known as the Coulomb gauge. It leads from (111) to Poisson’s equation for the magnetic vector potential: 2 ∇ A = −µ0J (112)

Equation 112 implies three equations, one for each component of the vector potential:

2 2 2 ∇ Ax = −µ0Jx ∇ Ay = −µ0Jy ∇ Az = −µ0Jz (113)

Assuming that J goes to zero at inﬁnity we can read oﬀ the solution using our knowledge of the solution of Poisson’s equation for such a boundary condition: µ Z J(r0) A(r) = 0 dV 0 (114) 4π |r − r0| The equivalent of the expression the electrostatic potential from a point charge can be written down for the magnetic vector potential at r due to a current element Idl0 or JdV 0 at r0: µ I(r0)dl0 µ J(r0)dV 0 dA(r) = 0 = 0 (115) 4π|r − r0| 4π|r − r0|

42 Note that the direction of dA is parallel to the current element whereas dB is perpendicular by Bios-Savart law. Example: Vector potential of a magnetic dipole (see tutorial)

µ m × ˆr A(r) = 0 (116) 4π r2

9.4 Summary of ’statics

∂ρ Electrostatics: Stationary charges = 0 and Coulomb’s law lead to ∂t ρ ∇ · E = (M-I) (117) 0 ∇ × E = 0 (M-III) (118)

(M-III) leads to E = −∇V , which in turn leads to Poisson’s equation for the scalar potential V ρ ∇2V = − 0

∂J Magnetostatics: Steady current loops = 0 and Bios-Savart law lead to ∂t ∇ · B = 0 (M-II) (119)

∇ × B = µ0J (M-IV) (120)

(M-II) leads to B = ∇×A, which in turn (in the Coulomb gauge) leads to a vector Poisson equation for A 2 ∇ A = −µ0J

In the following we shall see how (M-III) and (M-IV) need to be modiﬁed when time-varying ﬁelds are present.

43 10 Electromotance and Faraday’s Law

10.1 Sources of steady currents

So far we have considered steady currents and in particular steady current loops. But where do they originate from? Turns out they actually don’t exist – at least not without a little help. To see this let’s assume a loop of conducting wire where Ohm’s law J = σE holds. Then using Stokes’ theorem I I Z J · dr = σ E · dr = σ ∇ × E · dS = 0 C C S since ∇ × E = 0. Therefore I must be zero for the loop. The way out is to have a “battery” somewhere in the loop which causes a jump in the electric potential so that the integral from one terminal A of the battery to the other terminal B is

Z B E = E · dl (121) A where E = ∆V is the potential supplied by the battery. E is known as ”electromotance” or, con- fusingly, as the electromotive force (emf) – although it’s not a force, it’s electrostatic potential diﬀerence. I In that situation, J · dl = IL = σ∆V and we get the elementary form of Ohm’s law V = IR with R = L/σ and L the length of the loop. Examples for emf sources include: batteries, fuel cells, voltaic cells, etc. Generally we can express the emf by considering the force on a unit charge f = f + E where f is s s the force due to some external source (and is therefore non-conservative) and E is the force from the electric ﬁeld (which is conservative). Then I I E = f · dl = f · dl s C

10.2 Induced emf

a v B

Figure 17: Rectangular conducting loop moving in magnetic ﬁeld (c.f. Griﬃths Fig.7.10)

Consider a rectangular loop of conducting wire l moving with velocity v perpendicular to a uniform magnetic ﬁeld (which points into the page). The segment of the loop along a − b is perpendicular to both B and v, and charges along this segment thus experience a force in the ab direction I f = qv × B ⇒ E = f · dl = vBh, (122) mag mag

44 where h is the length from a to b. Aside: you may worry that a magnetic force f = qv × B should do no work! This is because mag if the charge q moves dl = vdt then dW = f · dl = q(v × B) · vdt = 0. So actually it is the mag mag person pulling the loop that is doing the work to produce a current, i.e. when the current flows in ab it creates a magnetic force df = dI × B to the left on an element of the wire and this must mag be balanced by a pulling force to the right. Now consider the magnetic flux through the loop Z ΦB = B · dS = BA = Bhx where A is the area and the flux integral is simply current×area here because dS k B in this example. dΦ dx B = Bh = −Bhv dt dt Therefore comparing with (122) we see

dΦ E = − B (123) dt

This result actually holds for general loops, B and v (for proof see Griffiths fig. 7.13). It is known as the flux rule or Faraday’s law of induction. E is known as an induced electromotive force (emf).

10.3 Faraday’s Law

As we have seen in (123) the induced emf can be understood in terms of the time variation of the R magnetic flux through the current loop. Note that the magnetic flux φB = S B · dS can also be changed by varying the field B, the surface R dS or the angle between B and dS.

I I I

v v changing B B B

1) 2) 3)

Figure 18: Summary of Faraday’s experiments (c.f. Griﬃths Fig. 7.20)

In a famous series of experiments Faraday found that an emf can be induced by:

1. Pulling a current loop through a magnetic ﬁeld.

2. Moving the magnet and area containing a ﬁeld across a current loop.

3. Changing the strength of the ﬁeld while keeping it in place.

Note: Scenario 2., as we clearly see from relativity, must yield the same result as scenario 1 – in both cases the loop moves relative to the magnet with the same velocity. But actually this has enormous consequences for the physics: in setup 2. the current loop is stationary and the charges

45 do not move. Therefore, for the charges to feel a force, there must be an electric field present, i.e. we deduce that the changes of the magnetic field induce an electric field. Now let us derive the differential form of Faraday’s law. Starting from (123), we use our definition of emf (121) to find dΦ I d Z Z ∂B E = − B ⇔ E · dl = − B · dS = − · dS dt dt ∂t where in the last equality we have assumed that only the magnetic field is changing while the flux surface is kept constant. Now use Stokes’ theorem I Z Z ∂B E · dl = (∇ × E) · dS = − · dS C S S ∂t which holds for an arbitrary surface S, thus leaving us with

∂B ∇ × E = − (124) ∂t

This is the full third fundamental law of electromagnetism, Maxwell-III. The curl of an electric ﬁeld around an axis is proportional to the time variation of the magnetic ﬁeld along the axis.

10.4 Connection to the Magnetic Vector Potential

The two equations ∂B ∇ × E = − B = ∇ × A ∂t can be combined to give: ∂ ∇ × E = − (∇ × A) ∂t Rearranging this: ∂A ∇ × E + = 0 (125) ∂t One possible solution to this equation is clearly: ∂A E = − (126) ∂t The electric field due to induction can be expressed as the time-derivative of the magnetic vector potential. A general solution to Maxwell-III is obtained by remembering “curl-grad = 0 ” and thus realising that any gradient field added to E will not change (125) and thus Faraday’s law. This is equivalent to adding a static electric field, which is the general solution to the homogeneouse equation ∇×E = 0 and expressed as the gradient of the scalar potential:

∂A E = − − ∇V (127) ∂t

Note that the equation E = −∇V only applies to electrostatic situations when there are no time- varying ﬁelds.

46 We see that the general solution to Maxwell-III for the electric field contains two contributions, one conservative and the other coming from the vector potential. Thus we can think of two kinds of electric field: those coming from a static charge distribution and which may be written as ∂A E = −∇V ; and those coming from a changing magnetic field and which may be written E = − . ∂t

10.5 Lenz’s Law

Determing the sign of the ﬂux in Faraday’s law or, equivalently, the direction of the induced current, often proves troublesome. But there is a simple rule known as Lenz’s Law that gives the right answer (and is based on conservation of energy):

The induced emf always acts to oppose the change that causes it or “Nature abhors a change in ﬂux!”

• In the case of the moving current loop at the start of this lecture, there is a force on the current in the wire f = IlB which acts to decelerate the wire.

• Additional work has to be done to move a current loop into or out of a magnetic ﬁeld, or through a non-uniform ﬁeld. This is to overcome the induced emf.

• For a rotating current loop (see next lecture), there is a torque m × B due to the magnetic moment of the loop. This always acts to slow down the rotation.

• A time-varying magnetic ﬁeld produces eddy currents in conducting loops. The dipole ﬁelds of these loops act to reduce the time-variation.

47 11 Inductance

11.1 Examples of Induction

As we saw last lecture an e.m.f. can be induced by changing the magnetic field strength, changing the area of a current loop in a magnetic field, or moving a current loop into or out of a magnetic field. Here we consider some common examples of rotation of a current loop about its axis in a uniform magnetic field.

AC generator

A generator has a coil of area A rotating about its diameter in a uniform magnetic ﬁeld with angular velocity ω. In this case it is only the angle between the ﬁeld and the loop that is varying.



Figure 19: AC generator

Therefore the magnetic ﬂux through the loop varies as

ΦB = AB cos ωt (128) dΦ E = − B = −ABω sin ωt (129) dt This system generates an alternating current (AC) with frequency ω. The current is π/2 out of phase with the flux, so the peak current is obtained when the flux is zero, i.e. when the loop is parallel to the magnetic field. There is zero current when the loop is perpendicular to the field.

Rotating disc of charge

An insulating disc with a uniform surface charge σ rotates around its axis. There is a uniform magnetic ﬁeld parallel to the axis of the disc. The magnetic force on an element of charge, dq, on the disc at radius, r, is:

dF = dqvBer = dqrωBer (130) where er is the radial basis vector on the disc. This magnetic force is equivalent to a radial electric ﬁeld: 0 E = F /q = rωBer (131)

48 and there is an induced emf between the centre and outer edge of the disc:

Z ωBa2 E = E0 · dr = (132) 2 This emf acts outwards to try and move the charge to the outside of the disc. If the disc were a conductor this would actually happen. So what happens to the flux rule for this type of problem? There is no change in magnetic flux here. But, it is also not clear if there is any current loop to consider a flux through. Thus, as it stands, the flux rule E = −dΦB/dt only works when there is a fixed current loop.

11.2 Mutual Inductance

Consider two current loops I1,I2 at rest. The current I1 will lead to a magnetic ﬁeld B1, which will lead to a magnetic ﬂux through loop 2:

Z Φ2 = B1 · dS2 ≡ M21I1 (133)

M21 is the mutual inductance of the two loops; it relates the ﬂux through loop 2 to the current in loop 1.

Now let us use the vector potential and Stokes’ theorem to obtain an explicit form for M12 Z I Φ2 = (∇ × A1) · dS2 = A1 · dl2 2 µ I I I dl · dl = 0 1 1 2 4π 1 2 |r1 − r2| where we have used the formula for the vector potential from section 9 equation (10). Thus I I µ0 dl1 · dl2 M12 = (134) 4π 1 2 |r1 − r2| where the integrals are taken round both current loops. This is known as the Neumann formula but it is not very useful for most practical applications. What it does reveal is that

M12 = M21 = M (135) which is a remarkable result: the ﬂux through loop 1 when there is current I in loop 2 is the same as the ﬂux through loop 2 when there is current I in loop 1, whatever the geometry of the loops! The relative geometry of the two conductors enters the size of M, which is a purely geometric quantity (a double integral around the loops)

Now let us introduce a time-dependent current I1 in loop 1. The changing ﬂux through loop 2 then gives rise to an emf dΦ dI E = − 2 = −M 1 . 2 dt dt

By Lenz’s law this emf induces a current I2 such that its magnetic ﬁeld opposes the change in current I1.

49 11.3 Self-Inductance

The above discussion similarly applies to the source loop itself, i.e. a changing current in a loop induces a “back emf” which opposes the change in current. The self-inductance of the loop, L, is deﬁned as the ratio of the induced emf to the rate of current change: dI E = −L . (136) dt It can also be written as:

ΦB = LI (137)

The unit of inductance is 1 Henry (H), which is 1 Vs/A.

Inductance of a Solenoid

For a long solenoid (length l, radius a, with n loops per unit length) there is a uniform magnetic ﬁeld inside, along the axis of the solenoid:

Bz = µ0nI (138)

This result can be shown using Amp`ere’sLaw (see tutorial 5.1). The ﬂux through all nl loops is: Z 2 ΦB = B · dS = µ0nIπa nl A and the self-inductance of the solenoid is therefore purely dependent on the number of turns and the size of the solenoid (volume VS):

2 2 2 L = µ0n πa l = µ0n VS. (139)

11.4 Energy Stored in Inductors

To create a current in a loop, at every instance work is done against the induced emf, which is related to the self-inductance L: dI dU = −EIdt = LI dt. M dt Integrating this with respect to time gives:

1 U = LI2 (140) M 2

For two coils with a mutual inductance: 1 1 U = L I 2 + L I 2 + M I I M 2 1 1 2 2 2 12 1 2 Example of solenoid For a long solenoid the self inductance and magnetic ﬁeld are:

2 L = µ0n VS B = µ0nIez inside solenoid

50 The energy stored in the solenoid is:

1 2 2 1 2 UM = µ0n I VS = |B| VS 2 2µ0 This can be written in terms of a magnetic energy density associated with the magnetic ﬁeld:

|B|2 uM = 2µ0 Note that this treatment of the energy density of a magnetic ﬁeld in an inductor is very similar to the treatment of the energy density of an electric ﬁeld in a capacitor. We can write the result (140) for a single current loop in a form that uses the magnetic vector potential and the current density, by utilizing (137):

1 1 I I 1 I U = LI2 = Φ I = A · dl = A · Idl . M 2 2 B 2 2 The generalisation to volume current elements is

1 Z U = A · J dV (141) M 2

We can develop (141) further by using Amp`ere’slaw and a product rule from lecture 1

µ0A · J = A · (∇ × B) = B · (∇ × A) − ∇ · (A × B) = B · B − ∇ · (A × B)

Consequently Z Z 1 2 UM = B dV − ∇ · (A × B) dV 2µ0 1 Z I = B2dV − (A × B) · dS 2µ0 S The second integral is a boundary term which vanishes when we take the volume over all space and assume B and A vanish as 1/r2 and 1/r respectively at r → ∞. Therefore

Z 1 2 UM = |B| dV (142) 2µ0 allspace

In a similar way to the electrostatic energy UE, we can think of the magnetic energy being stored either in the (localised) current distribution (141) or throughout all space in the magnetic ﬁeld (142).

51 12 The Displacement Current

In this lecture we complete the discussion of the fundamental laws of electromagnetism, and introduce electromagnetic waves for the ﬁrst time.

12.1 Continuity equation

Consider a conserved quantity – like electric charge: experimentally it is known that electric charge is always conserved. Other conserved quantities could be mass, momentum, energy, or probability. Now consider a volume V and the rate of change of the total charge Q in that volume. In case where there is no creation or spontaneous loss of charge inside the volume (i.e., it is conserved) we have ∂Q I − = J · dS (143) ∂t A where the right hand side is a ﬂux integral, which expresses the total current out of the volume, and therefore the left hand side has a negative sign. Writing the left hand side as a volume integral over charge density ρ and the right hand side as a volume integral, Gauss’ divergence theorem gives ∂ Z Z − ρ dV = ∇ · J dV. ∂t V V Since this must hold for an arbitrary volume V (however small) we deduce the diﬀerential form:

∂ρ = −∇ · J (144) ∂t

The divergence of the current density at any point is proportional to the rate of change of the charge density at that point.

This is the continuity equation which is a statement of local conservation (here for charge). In fact it holds for any conserved quantity, and is one of the most general and useful equations in physics.

12.2 The Displacement Current

Let us return to the diﬀerential form of Amp`ere’slaw

∇ × B = µ0J (145) and take the divergence of both sides:

∇ · ∇ × B = µ0∇ · J

Now since the divergence of a curl is always zero we ﬁnd, using the continuity equation, ∂ρ −∇ · J = = 0 ∂t This result is obviously wrong for non-static charge distributions ρ(r, t).

52 To satisfy the continuity equation generally we need to modify Ampère’slaw (Maxwell-IV) by the addition of a displacement current term to go along with J, i.e. when we take the divergence of the modified Maxwell-IV we want to obtain ∂ρ ∇ · (∇ × B) = µ ∇ · J + = 0 (146) 0 ∂t Using Gauss’ law (Maxwell-I) we can replace ρ with the divergence of the electric field: ∂ ∇ · (∇ × B) = µ ∇ · J + (∇ · E) . 0 0 ∂t The order of the time derivative and the divergence of the electric field can be reversed, and the divergence operation removed from all terms to leave:

 ∂E ∇ × B = µ J + (147) 0 0 ∂t

This is the Ampère-Maxwell law (Maxwell-IV) which holds for both static and time-varying charge distributions and fields. Maxwell’s stroke of genius was to include the displacement current term – often called the Maxwell correction – albeit for different reasons than we have given here! In any case we conclude that: The effect of a time-varying electric field is to produce an additional contribution to the curl of the magnetic field. Is the displacement actually a current? Answer is not really (see next subsection) but it does, of course, have the dimensions of a current.

12.3 Capacitor Paradox and Resolution

Figure 20: Capacitor paradox (c.f. Griﬃths ﬁg 7.42)

This is an illustrative demonstration of the origin of the displacement current. Consider the circuit in Figure 20, which has a parallel plate capacitor charging up and current I(t) ﬂowing in the wire. If we want to compute B(t) by taking an Amperian loop in the form of a circle around the wire (outside of the capacitor), then the surface S that we should take to compute I Z B · dl = µ0 J · dS S

53 does not appear to be well-defined. Taking S = S1 as the surface of the disc in the plane of the loop gives R J · dS = I; but taking S = S as an extended surface which goes through the gap between S1 2 the plates and which does not cross the wire gives R J · dS = 0 since there is no current flowing S2 between the plates. But really Ampère’slaw should hold independently of the surface bounded by the fixed loop. If, on the other hand, we consider Maxwell-IV with the Maxwell correction, we replace the old Ampère’slaw in integral form by the new version Z I Z ∂E (∇ × B) · dS = B · dl = µ0 J + 0 · dS . S S ∂t Now we know (at least quasistatically) that between the plates of the capacitor, E(t) is normal to Q(t) the plates and E(t) = ez. 0A ∂E Z ∂E Therefore, 0 is the vector I/Aez . Thus, between the plates 0 · dS = I and gives the ∂t S ∂t Z 2 same contribution as does J · dS outside the plates – see tutorial sheet 6. Thus the capacitor S1 paradox is resolved. Also we see that the diplacement current is not a real current as no current flows between the capacitor plates.

One ﬁnal thing to notice about the displacement current term is that, due to the factor 0 ' 9 × 10−12C2/Nm2, it is typically much smaller than the current term. Thus, when there is a current ﬂowing, the current term dominates the displacement current term.

12.4 Maxwell’s Equations

The laws of electromagnetism are summarised in four first-order differential equations (MI-IV) known as Maxwell’s equations: ρ ∇ · E = (148) 0 ∇ · B = 0 (149) ∂B ∇ × E = − (150) ∂t ∂E ∇ × B = µ J + (151) 0 0 ∂t M-I and M-II are Gauss’ Laws for electric and magnetic fields M-III is Faraday’s law of induction M-IV is Ampère-Maxwell law including the displacement current

In the electrostatic limit, B˙ = 0, Poisson’s equation is obtained from M-I & M-III: ρ E = −∇V and ∇2V = − 0 In the magnetostatic limit, E˙ = 0, Poisson’s equations for the magnetic vector potential are obtained from M-II & M-IV: 2 B = ∇ × A and ∇ A = −µ0J The continuity equation is obtained from M-I & M-IV: ∂ρ ∇ · J = − (152) ∂t

54 12.5 Solution of Maxwell’s Equations in Free Space

In a vacuum there are no charges and currents, ρ = 0 and J = 0. We take the curl of M-III

∂(∇ × B) ∇ × (∇ × E) = − ∂t where inserting M-IV yields ∂2E ∇ × (∇ × E) = − µ . 0 0 ∂t2 Similarly, taking the curl of M-IV and using M-III leads to

∂2B ∇ × (∇ × B) = − µ . 0 0 ∂t2 Now we make use of the vector identity (to be memorised) for a vector ﬁeld K:

∇ × (∇ × K) = ∇(∇ · K) − ∇2K (153)

In the absence of charges, M-I becomes ∇ · E = 0 and from M-II ∇ · B = 0. We are thus left with two wave equations:

∂2E ∇2E = µ (154) 0 0 ∂t2 ∂2B ∇2B = µ (155) 0 0 ∂t2 We have decoupled the four (ﬁrst order) Maxwell’s equations for B and E in the vacuum, at the price of now having second order equations. But we know that the solution of these second order wave equations (to be revised next lecture) will be electromagnetic waves. The velocity of the electromagnetic waves is the speed of light: 1 c2 = = (3 × 108ms−1)2 (156) 0µ0 Maxwell’s equations predict that light, radio waves, X-rays etc. are all types of waves associated with oscillating electric and magnetic ﬁelds in a vacuum.

N.B. There are no charges present in a vacuum, and the waves propagate without the presence of matter!

55 13 Description of Electromagnetic Waves

13.1 Recap of wave equations

Let us recall (see Second Year) the wave equation in 1D (i.e. one spatial dimension x and one time dimension t) for a scalar ﬁeld u, ∂2u 1 ∂2u = (157) ∂x2 c2 ∂t2 This is a linear second-order PDE, and solved with given initial (and/or boundary) conditions, e.g. u(x, t = 0) andu ˙(x, t = 0). Now, as can readily be checked by substitution into (157), a possible solution is any function f of the form u(x, t) = f(kx − ωt) (158) where the wave velocity c is given by ω c = (159) k A convenient solution of special interest is f = A exp i(kx − ωt) = A cos(kx − ωt) + iA sin(kx − ωt) (160) These are sinusoidal waves and A is the constant amplitude (which may be complex). N.B. The real (cosine) and imaginary (sine) parts are independent solutions. Moreover it is a monochromatic wave since there is a single angular frequency ω. These are the basis of Fourier methods where we build up waves of arbitrary shape by superposition of sines and cosines. It is often convenient to work with complex solutions, but if (for physical reasons) we want to get a real solution from (160) we simply take its real part u(x, t) = Re [A exp i(kx − ωt)] = ReA cos(kx − ωt) − ImA sin(kx − ωt) (161)

Important things to remember are: k is the wavenumber; ω = 2πν is the angular frequency, and ν is the frequency; λ = 2π/k is the wavelength; the whole wave propagates to the right (increasing x)with speed c; and at any ﬁxed x, the wave oscillates with period T = 2π/ω = 2π/kc. The 1D equation (157) generalises easily to 3D

1 ∂2u ∇2u = (162) c2 ∂t2 where the second derivative w.r.t. x has been replaced by the Laplacian operator. The solution (158) generalises to u(r, t) = f(k · r − ωt) (163) where the wavevector is k = (kx, ky, kz) and the velocity is again ω c = (164) k where k = |k|. We can also write (163) as u(r, t) = g(ˆn · r − ct) (165) wheren ˆ is the unit vector in the direction of k and g is an arbitrary function.

56 13.2 Plane Waves

The generalisation of the 1D sinusoidal solution (160) is the 3D plane wave solution u(r, t) = A exp i(k · r − ωt) (166)

To see that (166) is a solution to (162) note that ∂ ∂ ∂ ∇ exp ik · r = e + e + e exp i(k x + k y + k z) x ∂x y ∂y z ∂x x y z = ik exp ik · r

The gradient is along k, and the amplitude therefore constant in a plane perpendicular to k. Hence the name ”plane wave”: it has the same (possibly complex) value in the plane where k · r = ωt + constant.

Moving on, ∇2 exp ik · r = ∇ · ∇ exp ik · r = i∇ · (k exp ik · r) = ik · ∇ exp ik · r = −k2 exp ik · r

On the other hand, one can check that ∂2 exp i(k · r − ωt) = −ω2 exp i(k · r − ωt) ∂t2 and thus with ω = ck the wave equation is fulﬁlled. Finally we can generalise to the 3D wave equation for a vector ﬁeld F

1 ∂2F ∇2F = (167) c2 ∂t2 for which a plane wave solution is

F = F 0 exp i(k · r − ωt) (168) where F 0 is a constant (complex) vector. Key things to remember about this plane wave solution are ∇ · F = ik · F (169) ∇ × F = ik × F (170) ∇2F = −k2F (171)

13.3 Electromagnetic Plane Waves

In vacuo Maxwell’s equations with ρ = 0, J = 0 read ∇ · E = 0 (172) ∇ · B = 0 (173) ∂B ∇ × E = − (174) ∂t ∂E ∇ × B = µ (175) 0 0 ∂t

57 and reduce to the decoupled wave equations

∂2E ∇2E = µ (176) 0 0 ∂t2 ∂2B ∇2B = µ (177) 0 0 ∂t2 Clearly we have plane wave solutions

E = E0 exp i(k · r − ωt) B = B0 exp i(k · r − ωt) (178) 1 for both E and B, moving at the speed of light c = ω/k = √ . However Maxwell’s equations 0µ0 imply additional constraints on our plane wave solutions. First M-I and M-II (172,173) imply

ik · E0 = 0 ik · B0 = 0 i.e. E0 and B0 and hence E and B are perpendicular to the direction of propagation k. That is, EM waves are transverse. Substituting the solution (178) in M-III we ﬁnd

ik × E0 exp i(k · r − ωt) = iωB0 exp i(k · r − ωt) or more compactly 1 B = (k × E ) . (179) 0 ω 0

Now since k and E0 are orthogonal we can take magnitudes k |B | = |E | (180) 0 ω 0

Now we should choose the directions of B0 and E0. (179) tells us that the magnetic ﬁeld is perpendicular to the electric ﬁeld, and both are perpendicular to the direction of propagation of the wave.

It is conventional to take the direction of propagation k in the ez direction, k = kez, therefore E0 and B0 lie in the x–y plane. We can now introduce various polarisation states:

• Linearly (or plane) polarized - direction of E is ﬁxed. There are two orthogonal plane polarisation states with E in x or y direction (others are superpositions).

• Circularly polarized - direction of E rotates clockwise or anticlockwise around the z axis in the x–y plane.

An unpolarised electromagnetic wave is a random mixture of polarisation. So E has a random directions as a function of z.

58 x c

E0 E z

B B0 y

Figure 21: Plane polarisation in x direction (c.f. Griﬃths ﬁg 9.10)

13.4 Linear (Plane) Polarisation

Let us first consider the case where B0 and E0 are real. Then, since they lie in the x–y plane it is conventional to take E0 = E0ex and B0 = B0ey. This is referred to as linear polarisation in the x direction, i.e. the electric field is always in the x direction and magnetic field is always in the y direction and k is in the z direction as usual. Polarisation in the y direction would have E0 = E0ey, B0 = −B0ex. More generally we can take E0 = E0nˆ

E0 · ex = E0nˆ · ex = E0 cos θ wheren ˆ is the polarisation vector and θ is the polarisation angle.

13.5 Circular Polarisation

Now consider taking E0 as a complex vector

E0 iφ E0 = √ (ex ± iey)e (181) 2

Then we ﬁnd that the real part of E is given by

E0 h i Re E = √ ex cos(k · r − ωt + φ) ∓ ey sin(k · r − ωt + φ) (182) 2

Choosing rhe minus sign in (181) gives a polarisation vector that rotates anticlockwise about ez, i.e. at time ωt = k · r + φ, ReE is in the ex direction but, as time increases, the polarisation vector rotates towards −ey. This also referred to left circular polarisation or positive helicity Likewise the choice of plus sign in (181) yields a polarisation vector that rotates clockwise about ez. This is referred to as right circular polarisation or negative helicity.

59 14 Electromagnetic Energy and the Poynting Vector

14.1 Poynting’s Theorem (Griﬃths 8.1.2)

Recall from sections 6 and 11 that the total energy stored in static electromagnetic fields is: Z 1 1 2 2 U = UM + UE = B + 0E dV (183) 2 µ0 Let us now derive this more generally, using the full Maxwell equations. Consider some distribution of charges and currents. What is the work done on a moving charge q? In a small time interval dt, the charge will move vdt and, according to the Lorentz force law, the work done on the charge will be dU = F · dl = q(E + v × B) · vdt = qE · vdt where as usual the magnetic forces do no work. Now let q = ρdV (the usual definition of charge density) and ρv = J (the usual definition of current). Then, dividing through by dt and integrating over a volume V containing the charges, we find that the power delivered to the system (i.e. the rate at which the charge carriers’ energies change) is

dU Z = E · J dV (184) dt V

Thus E · J is the power delivered per unit volume and this generalises the expression we used in section 11.4. We now use M-IV to express 1 ∂E E · J = E · (∇ × B) − 0E · µ0 ∂t Furthermore, we can use a product rule from lecture 1 to write

E · (∇ × B) = B · (∇ × E) − ∇ · (E × B) ∂B = −B · − ∇ · (E × B) ∂t where we also used M-III in the last line. Putting it all together, and noting

∂B 1 ∂B2 ∂E 1 ∂E2 B · = E · = , ∂t 2 ∂t ∂t 2 ∂t yields 1 ∂ 2 1 2 1 E · J = − 0E + B − ∇ · (E × B) 2 ∂t µ0 µ0 Finally, we can integrate over the volume V containing the currents and charges and use the divergence theorem on the second term to obtain from (184) Poynting’s Theorem:

Z I dU ∂ 1 2 1 2 1 = − 0E + B dV − (E × B) · dS (185) dt ∂t V 2 µ0 µ0 S

Let us interpret the individual terms in expression (185): The left hand side is the power delivered to the volume, the rate of energy gain of the charged particles;

60 The first term on the right hand side is the loss rate of electromagnetic energy stored in fields within the volume; The second term is the flux rate of energy transport out of the volume, across its surface S. Thus Poynting’s theorem reads: energy lost by fields = energy gained by particles+ energy flow out of volume. Hence we can identify the vector 1 S = E × B (186) µ0 as the energy flux density (energy per unit area per unit time) and it is known as the Poynting vector (it ‘Poynts’ in the direction of energy transport).

We can also write Poynting’s theorem as a continuity equation for the total energy U = UEM + UMech. The left hand side of (185) is the rate of change of mechanical energy thus d(U + U ) I EM Mech = − S · dA dt S (to avoid a nasty clash of notation with S as Poynting vector we use dA rather than dS as vector element of area). As usual, expressing energy as a volume over energy densities uEM ,uMech and using the divergence theorem on the right hand side we arrive at

∂ (u + u ) = −∇ · S (187) ∂t EM Mech which is the continuity equation for energy density. Thus the Poynting vector is an energy current and represents ﬂow of energy in the same way that the charge current J represents the ﬂow of charge.

14.2 Energy of Electromagnetic Waves

As we saw last lecture a monochromatic plane wave in vacuo propagating in the ez direction is described by the ﬁelds:

E = exE0 cos(kz − ωt) B = eyB0 cos(kz − ωt) (188)

where E B = 0 0 c The electromagnetic energy density stored in the ﬁelds associated with the wave is: 1 1 2 2 uEM = uE + uM = B + 0E 2 µ0 √ Now since |B| = |E|/c and c = 1/ µ00 we see that the electric and magnetic contributions to the total energy density are equal and (for a linearly polarised wave)

2 2 2 uEM = 0E = 0E0 cos (kz − ωt)

The Poynting vector for monochromatic waves then becomes

1 2 2 S = (E × B) = c0E0 cos (kz − ωt)ez = uEM cez µ0

61 Note that S is just the energy density times the velocity of the wave cez – as it should be for the ˆ energy flux density. For a general propagation direction k, S = uEM ck. N.B To compute the Poynting vector it is simplest to use a real form for the fields B and E rather than a complex exponential representation. The time average of the energy density is defined as the average over one period T of the wave 2 Z T 0E0 2 huEM i = cos (kz − ωt)dt T 0 2 2 0E0 T 1 2 1 B0 = = 0E0 = T 2 2 2 µ0 The energy density of an electromagnetic wave is proportional to the square of the amplitude of the electric (or magnetic) field.

14.3 Energy of discharging capacitor

E Q Q

I R

Figure 22: Discharging capacitor in a circuit with a resistor

Consider a discharging circular parallel plate capacitor (plates area A) in a circuit with a resistor R. Ohm’s law gives Q V = = IR d C or dQ Q Q I = − = ⇒ Q = Q e−t/RC I = 0 e−t/RC dt RC 0 RC Now assume a ‘quasistatic’ approximation where we treat the instantaneous ﬁelds as though they were static: Q Q E = nˆ = 0 e−t/RC nˆ A0 A0 We take the normal to the plates (direction of E) asn ˆ. Now we can compute B through Amp`ere- Maxwell noting that the cylindrical symmetry implies that B is circumferential. The Amperian loop is a circle of radius r between the capacitor plates, where J = 0: I Z ∂E 2 ∂ Q0 −t/RC B · dl = µ0 J + 0 · dS = µ0πr 0 e S ∂t ∂t A0

As the left hand side is 2πrBφ, we obtain µ I(t)r B = − 0 e 2A φ

62 The Poynting vector is therefore given by

2 1 Q0 −t/RC r −t/RC I0 CR −2t/RC S = E × B = − e I0 e ez × eφ = 2 re er µ0 A0 2A 2A 0 Thus the Poynting vector (and the direction of energy ﬂow) points radially out of the capacitor.

14.4 ∗Momentum of electromagnetic radiation

Let us re-interpret the Poynting vector from a quantum perspective. EM radiation can be thought of as photons travelling with speed c, with energy

ε =hω ¯ = hν and momentum ε p =hk ¯ = kˆ. c For n photons per unit volume travelling at speed c we can interpret the average Poynting vector as average energy density nε multiplied by velocity vector ckˆ ˆ ˆ hSi = nεck = huEM ick.

Again thinking of the energy transport as effected by photons, we should have an accompanying momentum flux Pe . Pe is defined as the momentum carried across a plane normal to propagation, per unit area and unit time. For each photon p = ε/c (along kˆ) hence

P˜ = S/c

. If light (EM radiation) is absorbed on a surface (normal incidence), its momentum is absorbed, which creates a force per unit area equal to the incoming (normal) momentum ﬂux. This causes radiation pressure

prad = Pe · nˆ = S/c ⇒ prad = huEM i

If light is instead reﬂected by the surface, twice the momentum is imparted and prad doubles – but so does huEM i, and this result still holds. To understand radiation pressure classically, let’s go back to the example of a linear polarised wave propagating in ez direction: the electric ﬁeld moves charges on the surface where the radiation is absorbed; the Lorentz force qv × B (with v in the E direction) is then in the ez direction and creates the pressure.

63 15 Dielectric Materials

15.1 Overview

So far we have developed Maxwell’s equations and they offer a complete and general description of electrodynamics. However, as input we have to provide the charge and current densities ρ and J with microscopic precision. In the real world (not in vacuo) this would be a huge task, as materials are made up of atoms/molecules, and therefore contain contain about Avogadro’s number (∼ 1023) charge carriers and currents (through electronic orbits). This is the atomic level of description. Instead, we want to develop a macroscopic description of media in terms of smoothly varying quantities which are averaged in some way: these turn out to be the density ρf and current J f of free charges. The bound charges, which are bound up in the atomic structure, are dealt with by defining new fields – D, the Electric Displacement Field, and H, the Auxiliary (magnetic) Field. With these, as we shall see, we end up with a complementary macroscopic form of Maxwell’s equations which is a nice example of an effective theory derived from the microscopic theory.

15.2 Dielectric Materials

Roughly speaking, we can classify materials as conductors or dielectrics (insulators). A perfect conductor will have an ‘unlimited’ supply of free charges whereas a perfect dielectric will have no free charges and instead all charges are bound up in atoms/molecules.

Figure 23: Polarization of Dipoles in a Dielectric, due to an electric ﬁeld along z.

Let us consider the effect of an electric field on a dielectric. The field will induce a dipole moment in two ways:

1. The charge distribution of some atoms/molecules will be distorted;

2. Polar molecules (CO, H2O) will attempt to rotate to align with the external ﬁeld.

These eﬀects polarize the material and result in an induced dipole moment for each atom or molecule

hp i = αE (189) at where α is the atomic (molecular) polarizability. We take a time average h·i in (189), as pat will not be constant due to thermal fluctuations. All these atomic dipole moments give rise to a dipole moment per unit volume P or Polarization, P = nhp i, where n is the number density of the at atoms or molecules. More generally, we define the polarization field P through the net dipole

64 moment dp in a small volume dV ,

dp = P dV (190)

Let us now consider the ﬁeld due to the polarized molecules (we don’t care for now how the medium became polarized). Recall that for a single dipole at r0 the potential at r is

1 (r − r0) · p V (r) = 0 3 4π0 |r − r | This generalises by superposition to the potential due to the Polarization ﬁeld P (r0) Z 0 0 1 (r − r ) · P (r ) 0 V (r) = 0 3 dV 4π0 V |r − r | We now use a familiar identity, but this time w.r.t. the primed coordinates, 1 (r − r0) ∇0 = |r − r0| |r − r0|3 With this, we use a gradient product rule and Gauss’ divergence theorem to obtain Z 1 0 0 1 0 V (r) = P (r ) · ∇ 0 dV 4π0 V |r − r | Z Z 1 0 P 0 1 0 0 = ∇ · 0 dV − 0 ∇ · P dV 4π0 V |r − r | V |r − r | I 0 Z 1 P · dS 1 1 0 0 = 0 − 0 ∇ · P dV 4π0 S |r − r | 4π0 V |r − r | Now the ﬁrst term on the right hand side is equivalent to the potential due to a surface charge distribution on S, i.e. P · dS → σbdS or

σb = P · nˆ (191) wheren ˆ is normal to the surface. The second term on the right hand side is equivalent to the potential due to a volume charge distribution ρb given by

ρb = −∇ · P (192)

The subscript b refers to the fact the charges are bound (to the atoms). Thus the effect of polarization is to create bound charge distributions (191), (192). These charge distributions will themselves create electric fields, which will in turn alter the polarization and so on. In order to deal with this infinite regression we shall introduce an additional macroscopic field.

15.3 Electric displacement vector and Gauss’ law in media

The key idea is to divide up the charge distribution into bound and free charges

ρ = ρb + ρf

65 . Then Gauss’ law (Maxwell-I) becomes

ρ ρ ρ ∇ · P ∇ · E = f + b = f − 0 0 0 0 or ∇ · (0E + P ) = ρf (193) Now let us deﬁne the Electric displacement as

D ≡ 0E + P (194)

Gauss’ law in media, or the ﬁrst macroscopic Maxwell equation, then becomes

∇ · D = ρf (195)

N.B. The macroscopic Maxwell-I makes it tempting to think of D as the electric ﬁeld resulting from free charges with the bound charges somehow taken into account. However this can be misleading because generally i) there is no Coulomb law for D, and ii) for static ﬁelds ∇ × D = ∇ × P 6= 0 thus there is no scalar potential for D.

15.4 Linear Isotropic Homogeneous Media

So far, so good, but at the expense of introducing a new field D in addition to E. There still remains the problem of determining E and P from D. However, things become simpler when we consider an ideal type of medium which is linear, isotropic and homogeneous (LIH). Isotropic means there is no preferred direction which implies through symmetry that P is k to E. Linear means that the applied E field results in a generally small polarization of molecules through distortion and rotation, and we expect a linear response to the field:

P = χE0E (196)

χE is the susceptibility – large χE means a large response to the applied ﬁeld and the medium is easier to polarize.

Homogeneous means the medium has the same properties at all points in space so that χE has no spatial dependence. Using (196) results in

D = 0E + P = 0(1 + χE)E

or D = 0rE (197)

where r = 1 + χE is the relative permittivity (or dielectric constant) of the medium and is a dimensionless constant. In vacuum r = 1; for most insulators r = 1.05 − 1.3; some crystals have high r, e.g. mica: er = 7. For dipolar ﬂuids, e.g. deionized water: r = 80. The important point is: for LIH we have a linear constitutive relation (197) between E and D.

66 15.5 Example: Dielectric in a Capacitor

The space between the two plates of a capacitor can be ﬁlled with an insulating material rather than with a vacuum. This leads to induced polarization (bound) charges on the surfaces next to the plates. These change the capacitance in a way that depends on the geometry of the insulator and the plates. For a parallel plate capacitor, the electric ﬁeld is simply the superposition of the

Figure 24: Parallel plate capacitor with polarized dielectric.

ﬁeld from the free charges on the plates and bound charges at the surface of the dielectric 1 E = E0 + EP = (σf − σb)ˆn (198) 0 wheren ˆ is normal to the plates. The electric ﬁeld E as a function of the free charge density on the plates σf is reduced by the polarization of the dielectric between the plates. N.B. the total free charge on a plate is still Q = Aσf . Also the electric displacement turns out to be simply

D = σf nˆ .

This can be checked by the macsoscopic version of Gauss’ Law, which gives I Z D · dS = ρf dV = (Qf )enc (199) S V

Taking a Gaussian pillbox with area A straddling a plate (see Figure 24) one ﬁnds that A|D| = Aσf .

The parallel plate capacitance is given in terms of the potential diﬀerence Vd, which remains

Z 2 Vd = − E · dl 1 . When we integrate along the normal from plate 1 to plate 2, Dd Vd = Ed = 0r and we obtain Q Aσ AD A C = = f = = r 0 = C (200) Ed Ed Ed d r 0 where C0 is the capacitance without the dielectric present. For any geometry of capacitor there is an increase in the capacitance due to the presence of a dielectric between the plates. Note that it is not necessarily by just a factor r – see tutorial.

67 16 Magnetic Media

16.1 Types of magnetism

When an external magnetic ﬁeld is applied to a material, it produces a magnetization of the atoms of the material. There are several diﬀerent types of magnetization:

• Diamagnetism - the orbital angular momentum of the atomic electrons is increased slightly due to electromagnetic induction. This magnetization is opposite to the external magnetic ﬁeld.

• Paramagnetism - if the atoms of a material have intrinsic magnetic moments, they align with the applied ﬁeld, due to U = −m · B. This magnetization is parallel to the external magnetic ﬁeld.

• Ferromagnetism - in a few materials the intrinsic magnetic moments of the atoms matom spontaneously align (without an applied ﬁeld) due to mutual interactions of a quantum nature called ‘exchange interactions’ and form domains with moments matom all in the same direction. This magnetization can form permanent magnets.

16.2 The Magnetization Vector

In analogy with the polarization vector for dielectrics, the magnetization vector, M, is the key macroscopic field for magnetic media. The magnetization field M is defined by the net magnetic dipole moment dm (equivalent to small current loop) in a small volume dV , dm = MdV (201)

Thus M is the magnetic dipole moment per unit volume with units Am−1. An array of small

Figure 25: Magnetization loops, microscopic and effective. magnetic dipoles can be thought of as producing macroscopic current loops on the surface of the material. These currents circulate round the direction of M, with a surface magnetization current density j (see Figure 25). M Likewise, spatial variation of the magnetization can be expected to produce a bulk magnetization current. To quantify these effects let us calculate the field of a macroscopic magnetised object. Recall that the magnetic vector potential at r of an ideal magnetic dipole at r0 is

µ m × (r − r0) A(r) = 0 (202) 4π |r − r0|3

68 This generalises, when we replace m by MdV 0 and integrate the magnetization over some volume V , to Z 0 µ0 M × (r − r ) 0 A(r) = 0 3 dV (203) 4π V |r − r |

Now we recall that (r − r0) 1 = ∇0 (204) |r − r0|3 |r − r0| and use the curl product rule to obtain Z µ0 1 0 0 0 M 0 A(r) = 0 ∇ × M(r ) − ∇ × 0 dV 4π V |r − r | |r − r |

We can rewrite the second integral as a surface integral (see tutorial sheet 9) to obtain Z I µ0 1 0 0 0 µ0 1 0 0 A(r) = 0 ∇ × M(r )dV + 0 M(r ) × dS (205) 4π V |r − r | 4π S |r − r | Now, the ﬁrst term on the right hand side is equivalent to the potential due to a volume current in V

J M = ∇ × M (206) and the second term is equivalent to the potential due to a surface current on S (normaln ˆ)

j = M × nˆ (207) M

We use the subscript M to indicate that these are eﬀective magnetization currents resulting from the superposition of microscopic current loops. While the surface current (207) occurs for any (even constant) magnetization, the volume currents (206) relates to how the magnetization curls about a point – and thus a spatial variation in the magnetization ﬁeld. Example: Bar magnet “A cylindrical bar magnet has uniform magnetization M along its axis. To what current distribution is this equivalent?”

Now M is uniform, therefore ∇ × M = 0 and J M = 0. The surface current density j = M ×nˆ = Me ×e = Me has magnitude M and is ‘solenoidal’, mag z ρ φ i.e. resembling a solenoid with current ﬂowing circumferentially. Example: Toroidal magnet “A long cylindrical bar magnet of uniform M is bent into a loop. What is the equivalent current distribution?” The curl in cylindrical polars (ρ, φ, z) reads:

1 ∂M ∂M ∂M ∂M ∇ × M = z − φ e + ρ − z e ρ ∂φ ∂z ρ ∂z ∂ρ φ 1 ∂ ∂M + (ρM ) − ρ e ρ ∂ρ φ ∂φ z

The direction of M is circumferential, M = Meφ. In the curl formula, the only non-zero term is 1 ∂ M ∇ × M = (ρM)e = e = J ρ ∂ρ z ρ z M

69 M

Figure 26: Magnetization currents in bar magnet and toroidal magnet (JM for toroid is not quite right in ﬁgure)

Alongside the solenoidal (circumferential around the toroid) surface current jmag = M, we now also have a bulk magnetization current. N.B. The surface current j = M × nˆ has constant magnitude, so there seems to be a larger net mag current on outer than the inner surface. The bulk current J M makes up this diﬀerence.

16.3 Ampere’s Law in media

The Amp`ere-Maxwell law still holds for the full current density J ∂E ∇ × B = µ J + 0 0 ∂t

The key idea is to divide this into three contributions, J = J f + J M + J P .

J f is the current of free charges, i.e. the conduction current.

J M = ∇ × M is the magnetization current we have just met.

J P = is the polarization current – this new term comes from electric dipoles moving around.

To rationalize J P we use the deﬁnition ρP = −∇ · P and the continuity equation

ρ˙P = −∇ · J P , from which we deduce ∂P J = (208) P ∂t

We would like the Amp`ere-Maxwell law in terms of J f only: ∂E ∇ × B = µ J + J + J + 0 f M P 0 ∂t ∂P ∂E = µ J + ∇ × M + + 0 f ∂t 0 ∂t ∂D = µ J + ∇ × M + 0 f ∂t

70 where we have used the definition of D. Now shift ∇ × M onto left, divide by µ0: B ∂D ∇ × − M = J f + µ0 ∂t We thus define a new vector field B H = − M (209) µ0 which obeys ∂D ∇ × H = J + (210) f ∂t This is the Ampère-Maxwell law in media. The integral form of Ampère-Maxwell reads I Z H · dl = (J f + ∂D/∂t) · dS C S where C is a closed circuit bounding S. We run into difficulties in terminology for B, H. It is actually simplest and easiest to call them ‘magnetic field B’ (units Tesla) and ‘magnetic field H’ (units Am−1). But be warned – in some texts B is the ‘magnetic field’ and H is the ‘auxiliary field’; in others B is the ‘magnetic flux density’ and H is the ‘magnetic field strength’ – which is really confusing! Since Maxwell-II and Maxwell-III do not need to be modified as they contain no source terms J or ρ, we now basically have all the macroscopic Maxwell’s equations which hold in media. See next lecture for summary.

The magnetic susceptibility χM describes the relationship between magnetization and applied ﬁeld, by relating M to H. We will assume again a linear, isotropic, homogenous (LIH) medium. Then the relation may be written M = χM H (211)

Warning—some books, e.g. Grant & Phillips, use χBB = µ0M

The equivalent of the dielectric constant is known as the relative permeability of a material, µr:

B = µrµ0H (212)

where µr = χM + 1. The limit of no magnetization is χM = 0 and µr = 1.

In contrast to dielectrics, the magnetic susceptibility χM can be either positive or negative, and µr < 1 or µr > 1.

71 17 Summary of EM in media; boundary conditions on ﬁelds

17.1 Eﬀect of Magnetic Materials on Inductance

First we have to finish off our description of magnetism with a look at how inductance is affected by magnetisation currents (c.f. change in capacitance by polarisation of a dielectric). Solenoid filled with magnetic material Consider a long solenoid of n turns per unit length, length ` and cross sectional area A, filled with a linear magnetic material, in which M obeys M = χmH (e.g. ferrite with χm = 900). How does this change the self inductance L?

Recall the definition L = ΦB/I. This stems from Faraday’s law, Maxwell-III, and is therefore unchanged by media. Ampère’slaw in the static situation ∂D/∂t = 0 becomes I Z ∇ × H = J f ⇒ H · dl = J f · dS = nÌ in integral form, where I is the usual conduction current, and the loop is taken as shown in Figure 27. Note the symmetry: H is axial within the solenoid and vanishes outside for large `. I M

Figure 27: Solenoid with conducting core: Amperian loop to determine H.

Hence H = nI, and M is axial, with magnitude M = χmnI. Then B also must be axial, with magnitude B = µ0(H + M) = (χm + 1)µ0nI. 2 ⇒ ΦB = nA`B = (χm + 1)µ0n A`I 2 ⇒ L = ΦB/I = (χm + 1)µ0n Vs

If the ﬁlling material were ferrite, L would be 901 times larger than in vacuum (vacuum case: χm = 0). For a ferromagnetic material there is a very large increase in self inductance. On the other hand for diamagnetic/paramagnetic materials there is a small decrease/increase in the self-inductance.

For ferromagnetic materials the energy stored in an inductor increases by a large factor µr ≈ 103 − 106: See section 17.3 for energy stored in ﬁelds.

17.2 Electromagnetism in media: summary

Maxwell’s equations for macroscopic ﬁelds D, B, E, H read

∇ · D = ρf (213)

72 ∇ · B = 0 (214) ∂B ∇ × E = − (215) ∂t ∂D ∇ × H = J + (216) f ∂t Definitions of D,H are D = 0E + P B = µ0(H + M) (217) and the free charges and currents fulfil a continuity equation, ∂ρf /∂t = −∇ · J f . In LIH media, the relations between the fields simplify to

P = χE0E M = χmH (218)

D = 0rE ≡ E B = µ0µrH ≡ µH (219) where r = 1 + χE, and µr = 1 + χm. Although it may seem annoying to have to learn a second set of Maxwell’s equations, equations (213–216) are in some ways simpler than the microscopic ones since there are no constants.

17.3 Energy densities and Poynting Vector

Recall that the power delivered (through EM ﬁelds) to a system of charge carriers is E · J f per unit volume, so that the energy density u obeys du = E · J (220) dt f

Our aim is to express J f through field-related quantities. We first use macroscopic Maxwell-IV: ∂D E · J = E · (∇ × H) − E · f ∂t We then use a curl product rule on the first term (see Lecture 1) to write ∂D E · J = H · (∇ × E) − ∇ · (E × H) − E · ∂t ∂B ∂D = −H · − ∇ · (E × H) − E · ∂t ∂t ∂ 1 1 = − E · D + B · H − ∇ · (E × H) ∂t 2 2 Note that we used here E · D˙ = E˙ · D and B · H˙ = B˙ · H, which is true for linear static media. Then, integrating over a volume V of the medium and using Gauss’ divergence theorem on the second term as usual, we obtain from (220) for the total energy rate (c.f. section 14)

dU d Z 1 1 I = − E · D + B · H dV − (E × H) · dS (221) dt dt V 2 2 S

From the ﬁrst term we identify the electric and magnetic energy densities as 1 1 u = B · H u = E · D (222) M 2 E 2 and from the second term we identify the macroscopic Poynting vector as S = E × H. (223)

73 17.4 Boundary conditions of ﬁelds

There are often sharp interfaces between media. We want to know how the ﬁelds behave at such interfaces, but these boundaries acquire nonzero values of surface polarization charges σP and surface magnetisation currents j . mag In keeping with the derivation of Maxwell-I to -IV in macroscopic form, we want to avoid considering these, and think about free charges and currents only. We will exploit the integral versions of equations (213–216), with appropriate integration regions straddling the interfaces.

dS n t

Figure 28: Gaussian surface (left) and Amperian loop (right) for deriving continuity conditions for normal and tangential ﬁeld components.

1. First condition, from ∇ · D = ρf : Divergence theorem: I ∇ · D = ρf ⇒ D · dS = (Qf )enclosed

Use as a Gaussian surface a small pillbox (or “patch”), vector area dS =n ˆdS and height h. Then as h → 0 (D2 − D1) · nˆ dS = σf dS where σf is the surface density of free charges only. In the absence of free surface charges, Dnormal is continuous. We can also write the continuity condition as

(D2 − D1) · nˆ = σf .

2. Second condition, from ∇ · B = 0: Z ∇ · B = 0 ⇒ B · dS = 0

Again use as a Gaussian surface a small pillbox (or “patch”), vector area dS =n ˆdS and height h. Then as h → 0 Z B · dS = (B2 − B1) · nˆ dS = 0

Therefore Bnormal is continuous. This is completely general. 3. Third condition, from ∇ × E = −∂B/∂t: We take a rectangular Amperian loop straddling the interface, with length ` and height h. Then I ∂ E · dl = (E1 − E2) · ˆt` = − ΦB. C ∂t There, ˆt = unit tangent vector, which satisfies ˆt · nˆ = 0. Unless B is infinite, the magnetic flux cutting through the loop vanishes, ΦB → 0, as h → 0:

⇒ (E1 − E2) · ˆt = 0

74 but ˆt is arbitrary within the surface plane, so Etangential is continuous and this is completely general. Note these are two conditions in 3D.

4. Fourth condition, from ∇ × H = J f + ∂D/∂t: I ∂D H · dl = j · ˆs ` + · ˆs ` h f ∂t where j = free surface current per unit area and ˆs = ˆt × nˆ = surface unit vector perpendicular to f the Amp`erianloop. Now take h → 0: the last term then vanishes, and I H · dl = (H − H ) · ˆt ` = j · ˆs ` 1 2 f

In the absence of free surface currents, Htangential is continuous. The general form is rarely needed and may be written in several equivalent ways:

(H − H ) · ˆt = j · ˆs 1 2 f (Htang − Htang) = j × nˆ 2 1 f (H − H ) × nˆ = −j 2 1 f

Summary of the continuity conditions

At interfaces between media,

1. Dn continuous – if σf = 0,

2. Bn continuous – always,

3. Et continuous – always, 4. H continuous – if j = 0. t f These are key results and you should know the derivations. Problems with nonzero σ or j are uncommon but in those cases: f f

(D2 − D1) · nˆ = σf replaces 1. above (Htang − Htang) = j × nˆ replaces 4. above. 2 1 f

75 18 Examples of continuity conditions; waves in media

18.1 Continuity conditions: examples

Example: Inclined dielectric slab

o The electric ﬁeld E outside a large dielectric slab of relative permittivity r is uniform and at angle θ to the normal to the slab. What is the electric ﬁeld Ei inside the slab?



r E 

Figure 29: Uniform ﬁeld refraction at interface. Similar to Griﬃths Fig 4.34.

o o i i Outside: D = 0E , and inside: D = r0E . Let ψ be the angle between the normal to the plane and Ei.

Take the normal to the slab in the ez direction and the tangent in the ex direction. Then we write the (x, z) components of the ﬁelds as

o o o o o o E = (E sin θ, E cos θ) D = 0(E sin θ, E cos θ) i i i i i i E = (E sin ψ, E cos ψ) D = 0r(E sin ψ, E cos ψ)

Now impose boundary continuity conditions: i o 1. Dn continuous. I.e., Dz = Dz: o o i i E cos θ 0E cos θ = 0rE cos ψ ⇒ E = r cos ψ

i o 2. Et continuous. I.e., Ex = Ex: sin θ Eo sin θ = Ei sin ψ ⇒ Ei = Eo sin ψ

Result, comparing the right hand sides: 1 cos θ sin θ = r cos ψ sin ψ −1 ⇒ ψ = tan (r tan θ)

76 Checks: For θ = π/2: Ei = Eo, ψ = π/2 (E is purely tangential, continuous). And for θ = 0: i o E = E /r, ψ = 0 (D is purely normal, continuous). Remarks: D k E everywhere; but the angle of both is altered within the slab. Ei is the superposition of uniform Eo with that of the polarization charges on the surface of the slab. Unless θ = 0, as in a parallel plate capacitor, D/0 is not “the E ﬁeld you would have had” without the slab which i o would simply be D = 0rE .

Example: Spherical cavity in dielectric

A large block of LIH dielectric of relative permittivity r > 1 contains a spherical cavity. The E ﬁeld far away from the cavity is uniform, with magnitude E0. What are E,D within the cavity?



Figure 30: Spherical cavity in dielectric - Griﬃths Example 4.7.

Use spherical polars with origin at the centre of the sphere and take the z axis k E0. There is therefore symmetry with respect to φ (angle about the z–axis).

At the surface of the spherical cavity, σp = P · nˆ wheren ˆ is outwards normal of material (inwards normal of sphere). Because of this and P k E, the ﬁeld inside is enhanced by σp(θ).

The charge around the cavity σp(θ) forms an effective dipole. Outside, the field lines are locally distorted by σp(θ). The problem is to determine V , the electrostatic potential, and hence E everywhere. Within the cavity, use the ansatz of a uniform E field in z direction, corresponding to

V (r < a) = −Einz = −Einr cos θ .

Outside the cavity, use the ansatz of the uniform field E0 plus a dipole field, which means A cos θ V (r > a) = −E r cos θ + 0 r2 where A is a constant to be fixed. Remember the uniqueness theorem for Poisson’s equation: if this ansatz fulfills the equation and the boundary conditions for the fields, it is (bar a constant) the only solution. Realise that these two expressions satisfy Laplace’s equation away from the boundary where there are no charges. We now just need to satisfy the boundary conditions on the fields. First recall that E = −∇V and in spherical polars ∂V 1 ∂V 1 ∂V ∇V = e + e + e r ∂r θ r ∂θ φ r sin θ ∂φ

77 The boundary conditions are: Et continuous, which requires Eθ continuous at r = a: A sin θ −E a sin θ + = −E a sin θ 0 a2 in

And Dn continuous, which requires Dr = rEr = −r∂V/∂r continuous at r = a: 2A cos θ E cos θ + = E cos θ r 0 a3 in Combining these yields 2A A E = E + = E − in r 0 a3 0 a3 from which A/a3 can be eliminated, giving:

3r Ein = E0 1 + 2r with Ein = Einez and Din = 0Ein (since the cavity has r = 1).

Note that Ein > E0 if r > 1, the ﬁeld inside is indeed enhanced. This example may be used to derive an approximate formula for atomic polarizabilities from the macroscopic relative permittivity r, the Clausius Mosotti equation – see tutorial sheet 10.

18.2 Waves in media

As a ﬁrst look at waves in media, let’s consider a non-conducting medium with ρf = 0, J f = 0. Let us write the permittivity = 0r and the permeability µ = µ0µr. As before, when we considered waves in vacuo in Lecture 13, we can reduce the macroscopic Maxwell equations to two decoupled wave equations,

∂2E ∇2E = µ (224) ∂t2 ∂2B ∇2B = µ (225) ∂t2

These are precisely the same as in Lecture 13 but having 0 replaced by and µ0 replaced by µ. Clearly, we then have plane wave solutions

E = E0 exp i(k · r − ωt) B = B0 exp i(k · r − ωt) (226) where k2 − µω2 = 0 must hold. Thus the wave (phase) velocity is ω 1 v = = √ k µ

1 and recalling c = √ µ00 2 1 2 1 2 v = c ≡ 2 c (227) µrr n where n is called the refractive index of the medium.

78 As in Lecture 13, Maxwell-I and -II imply

ik · E0 = 0 ik · B0 = 0 i.e. E and B are perpendicular to the direction of propagation k and the wave is transverse. This may seem like a trivial generalisation of waves in vacuo but the physics is remarkable – at least for LIH media we have managed to deal with all the atoms, atomic dipoles, polarisation etc by wrapping them up into and µ and the net result is simply to change the velocity of the wave.

18.3 Waves in conductors

In conductors there is free charge and currents flow in response to an electric field. As we shall see this has a serious effect on the propagation of an EM wave in a conductor. Let us start with Ohm’s Law J = σE and the linear relations

B = µH D = E to rewrite Maxwell-IV: ∂D ∇ × H = + J ∂t f ∂E → ∇ × B = µ + µσE ∂t Now as usual the time-derivative of Maxwell-III yields ∂ (∇ × B) = −∇ × (∇ × E) = ∇2E − ∇(∇ · E) ∂t The microscopic Maxwell-I reads ∇ · E = ρ/. Let us assume a uniform charge density so that ∇ρ = 0. Then ﬁnally we obtain ∂2E ∂E ∇2E = µ + µσ (228) ∂t2 ∂t We note an additional term on the right hand side whose origin is the free current in Maxwell-IV. How will this term aﬀect the wave? A similar calculation (Exercise) yields

∂2B ∂B ∇2B = µ + µσ (229) ∂t2 ∂t

Let us proceed blindly and bravely by making an ansatz of plane wave moving in the z direction, ˜ E = E0 exp i(kz − ωt). When we sub this into (228) we obtain

k˜2 = µω2 + iµσω

Clearly something has to become complex to solve this!

79 19 Waves in Conductors

19.1 Skin Depth

Last time we derived the wave equation in conductors (making use of Ohm’s law and assuming linear media), ∂2E ∂E ∇2E = µ + µσ (230) ∂t2 ∂t where σ is the medium’s conductivity. Substituting a plane wave ansatz ˜ ˜ E = E0 exp i(kz − ωt) (231) yields the ”dispersion relation” k˜2 = µω2 + iµσω . (232) To solve this we have to take a complex wavenumber k˜ = k + iκ (233) Equating the real and imaginary parts in (232) yields k2 − κ2 = µω2 (234) 2kκ = µσω . (235) µσω The second equation can be solved for κ = , then eliminating κ from (234) yields 2k µσω 2 k4 − = µω2k2 2 This is a quadratic in k2 with solution 1 1 1/2 k2 = µω2 + (µω2)2 + (µσω)2 2 2  !1/2  µω2 σ 2 =  1 + + 1 (236) 2 ω

(we have taken the positive square root so that the solution for k2 is positive). Then we can use (234) to obtain   2 2!1/2 2 µω σ κ =  1 + − 1 . (237) 2 ω Now the complex wavenumber (233) implies ˜ −κz i(kz−ωt) E = E0e e . (238) This describes an exponential decay of the wave along z, the propagation direction, i.e. attenuation of the wave. The characteristic distance over which the wave decays is known as the skin depth and is given by

1 δ = (239) κ

Thus the skin depth is the typical distance a wave penetrates into a conductor.

80 19.2 Good and poor conductors

σ In the result (237) the ratio ω is signiﬁcant. 1/ω has the dimensions of time as does /σ. Thus this quantity is a ratio of two timescales. We understand the timescale /σ by returning to the continuity equation for free charges,

∂ρ f = −∇ · J . (240) ∂t f Using Ohm’s law and Gauss’s law (plus linear media property) σ σ ∇ · J = σ∇ · E = ∇ · D = ρ . f f This gives ∂ρ σ f = − ρ ∂t f which has the solution −(σ/)t ρf (t) = ρf (0)e . So the free charge density decays on a timescale τ = which is the relaxation time. If this is small σ then any free excess charge is quickly rearranged away and the medium is a good conductor. A perfect conductor would have this timescale tending to zero i.e. σ → ∞. On the other hand, if τ is large, free charge hangs around for a long time and the medium is a poor conductor. σ Let us return to the quantity ω that appears in (237), which we may write using the relaxation time τ and period T = 2π/ω as σ 1 T = . ω 2π τ we see that is (roughly) the ratio of the oscillation period of the wave to the charge relaxation time in the conductor. If, for a given frequency ω, this ratio is large the medium is a good conductor, whereas if the ratio is small the medium is a poor conductor at that frequency. In the tutorial you are invited to work out the diﬀerent limits. One ﬁnds from (237) for the skin depth

2 1/2 δ ' for σ ω µωσ 4 1/2 δ ' for σ ω µσ2 Thus the skin depth is much smaller for a good conductor. Also note that for a poor conductor the skin depth does not depend on frequency. Typical metals are good conductors up to about 1 MHz. δ ' 1cm at 50 Hz (mains frequency) δ ' 10 µm at 50 MHz

81 Consequences / Applications of Skin eﬀect

• Shielding of electronic circuits by metal casework.

• Power lines and cable design: conductors > 1cm thick are wasted since the current resides only in the skin layer around the outside and there is a ‘dead zone’ in the centre

• Submarines can’t use radio

• Mobile phones don’t work inside metal boxes.

• Microwave oven doors: metal mesh stops radiation escaping, holes λ are OK.

19.3 Phase relations of ﬁelds

Maxwell-I and -II imply further constraints on waves in a conducting medium. As usual we ﬁnd ˜ ˜ ˜ ˜ ik · E0 = 0 ik · B0 = 0 ˜ ˜ Let the direction of propagation k be in the ez direction and E0 in the ex direction. Substituting in Maxwell-III leads to ˜ ˜ ˜ ik × E0 = iωB0 k˜E˜ ⇒ B˜ = 0 e . (241) 0 ω y

However, k˜ is complex so E˜0 and B˜0 will also be complex. Let us write 1/2 κ k˜ = Reiφ,R = k2 + κ2 , φ = tan−1 k iδE iδB E˜0 = E0e B˜0 = B0e

Putting these in (241) yields

Reiφ B eiδB = E eiδE (242) 0 ω 0 ⇒ δB − δE = φ (243)

Condition (243) means that the magnetic field lags behind the electric field by angle φ because, if we finally take the real part to get real fields we obtain

−κz E = E0e cos(kz − ωt + δE)ex (244) −κz B = B0e cos(kz − ωt + δE + φ)ey (245)

For the wave number, using (236,237)

!1/4 1/2 σ 2 R = k2 + κ2 = (µω2)1/2 1 + ω

 1/2 1/2 σ 2 κ 1 + ω − 1 φ = tan−1 = tan−1    1/2  k σ 2 1 + ω + 1

82 In particular, for a good conductor

φ → tan−1[1] = π/4 and k˜ ' (µωσ)1/2eiπ/4 (246)

19.4 Intrinsic Impedance

As we have seen ˜ i(kz−ωt) ˜ i(kz−ωt) E = ex E0e ,B = ey B0e where E˜0 and B˜0 are complex.

Whereas in vacuum E and H = B/µ0 are in phase, here there are not. The complex number

E˜ Z ≡ 0 (247) H˜0 is the Intrinsic Impedence of the medium. Its dimensions are Ω (Ohms) – units: E = V/m; H = A/m ⇒ E/H = V/A = Ω. One can think of it as the generalised resistance; when Z is real it reduces to the resistance. In a vacuum E0 E0µ0 = = cµ0 ≡ Zvac = 377Ω H0 B0 This is real since E,H are in phase. In a linear dielectric 1/2 E0 E0µ µr = = Zvac H0 B0 r √ As we have seen in (246), a good conductor has k˜ ≈ iµωσ and thus

E˜ E˜ µ ωµ µω 1/2 Z = 0 = 0 = ' e−iπ/4 H˜0 B˜0 k˜ σ which is complex.

83 20 Waves at interfaces: normal incidence

20.1 Summary on plane waves and interfaces

Consider a plane polarised wave propagating in the ez direction

i(kz−ωt) i(kz−ωt) E = E0e ,B = B0e

As we have seen, Maxwell-III implies ikez × E0 = iωB0. Usually we take E0 = E0ex (plane polarised in x direction) and kE B = 0 e . 0 ω y Note that E0,B0 can, in principle, be complex, as they are for waves in a conductor. Previously we indicated this by a tilde e.g. E˜0 but to lighten notation we won’t do that here and instead just refer iδE to E0 as the complex amplitude; the (real) amplitude is then the modulus |E0| i.e. E0 = |E0|e . Recall that the complex impedance, a materials property, is given by the ratio of complex amplitudes E µE Z = 0 = 0 . H0 B0 As we have seen a complex Z allows a phase shift between E and H

Now consider a plane polarised wave Einc propagating in the ez direction and crossing, at z = 0, an interface at normal incidence. Generally medium 1 has complex impedance Z = Z1 and medium 2 has complex impedance Z = Z2. We take coordinates: ex along Einc; ey along Hinc; ez along k1 (forming a right handed triad). We place the boundary at z = 0 so that the x–y plane is the interface between the two media. x Z1 Z2

Einc Etrans

vinc vtrans

Hinc Htrans z Eref

vref Href y

Figure 31: Wave at interface between two media similiar to Griﬃths ﬁg. 9.13

20.2 Interfaces between two dielectric media

It is simplest to start by considering two linear dielectric media where we have seen that

Zi = viµi

84 is real and there is no phase lag between E and H. We can take the amplitude EI to be real and write i(k1z−ωt) Einc = EI ex e E I i(k1z−ωt) Hinc = ey e µiv1 Likewise for transmitted and reﬂected waves (see diagram):

i(k2z−ωt) Etr = ET ex e E T i(k2z−ωt) Htr = ey e µ2v2

i(−k1z−ωt) Eref = ER ex e E R i(−k1z−ωt) Href = − ey e µ1v1

N.B. The reﬂected wave propagates in negative z direction hence sign switch in the wave number (so that wave speed is v = −ω/k) and sign switch in amplitude Href (so that −ez, E, H form a right-handed triad).

We now invoke continuity conditions eestablished in Lectures 17 and 18: ex and ey are both tangential to the interface and tangential components of E and H are continuous. Note that we assume that there no surface currents or charges, which is usually the case. Then the continuity conditions give

Etan = Ex is continuous ⇒ EI + ER = ET

Htan = Hy is continuous E E E ⇒ I − R = T µ1v1 µ1v1 µ2v2

For a given EI , solve for ET and ER. E.g., add the equations to find 2EI 1 1 = + ET µ1v1 µ1v1 µ2v2 Then the Amplitude transmission coefficient is E 2 t ≡ T = EI 1 + β and the Amplitude reflection coefficient is E 1 − β r ≡ R = EI 1 + β where β is defined as µ v β = 1 1 µ2v2

Now if the media are non-magnetic, i.e. the permeabilities µi = µ0 and we recall that vi = −1/2 −1 (µii) = c · ni , we can write

85 2v 2n t = 2 = 1 v1 + v2 n1 + n2 v − v n − n r = 2 1 = 1 2 v1 + v2 n1 + n2

Sanity check: if v2 = v1 (the two media the same), there is no reflected wave as expected. Otherwise, the reflected wave is in phase if v2 > v1 but out of phase if v2 < v1. Energy Flow: The Poynting vector for macroscopic electromagnetic fields is given by 1 S = E × H = E × B µ so the energy flux per unit volume averaged over one period or intensity of the wave is given by 1 1 v |hSi| = |hE × Bi| = E2 = E2 µ 2µv 0 2 0 So R, the ratio of reflected to incident intensity, and T , the ratio of transmitted to incident intensity are given by 2 2 n1 − n2 2v2 2 4n1n2 R = r = T = t = 2 n1 + n2 1v1 (n1 + n2) N.B. since R + T = 1 we recover energy conservation.

20.3 General waves at interfaces: normal incidence

Basically we now repeat the above calculation but for complex impedance so that there may be phase lag between E and H

i(k1z−ωt) Einc = EI ex e E I i(k1z−ωt) Hinc = ey e Z1

i(k2z−ωt) Etrans = ET ex e E T i(k2z−ωt) Htrans = ey e Z2

i(−k2z−ωt) Eref = ER ex e E R i(−k2z−ωt) Href = − ey e Z1 We again assume that there no surface currents or charges and the continuity conditions reduce to Etan = Ex continuous and Htan = Hy continuous:

EI + ER = ET E E E I − R = T Z1 Z1 Z2

Solving for ET and ER, for a given EI yields as before E 2Z t ≡ T = 2 EI Z2 + Z1 E Z − Z r ≡ R = 2 1 EI Z2 + Z1

86 N.B. These are now complex quantities. Energy Flow: With complex impedances we need to bit more careful with the Poynting vector. Generally we use the time-averaged Poynting vector which is given by 1 1 hSi = kˆ < |E |2 2 Z 0 and the intensity is given by its magnitude 1 1 |hSi| = < |E |2 2 Z 0

20.4 Reﬂection at Conducting Surface

”Why are metals shiny?” Let the x–y plane be a boundary between vacuum (medium 1) and a conductor (medium 2): Z1 = Zvac = 377Ω s −iµω 1 − i Z = = 2 σ σδ where δ = p2/µσω is the skin depth. 10 Z2 is complex and ω-dependent. But its typical magnitude is tiny – e.g. for Cu at 10 Hz:

−4 |Z2| = 0.036Ω = 10 Zvac and at 1014 Hz (visible light frequency)

|Z2| = 3.6Ω = 0.01Zvac

The amplitude reﬂection then becomes (note phase reversal)

Z − Z r = 2 1 ' −1 Z2 + Z1 to within (complex) terms of order 1 percent. Good conductors exhibit near perfect reﬂection (with phase reversal) – this explains why metals are shiny. The physical origin of the shininess is the skin eﬀect: the transmitted wave decays like e−z/δ and almost all the energy that is put in comes back out.

87 A Supplementary Material on Types of Magnetism

A.1 Diamagnetism

This is the only type of magnetism for materials where all atoms (or molecules) have paired electron spins, and there is no net atomic magnetic moment. We consider a simple classical atomic model where each electron is orbiting round the nucleus with velocity v at radius r. The current and magnetic moment associated with an individual electron labelled i are: ev er × v e I = m = i i = L (248) 2πr 2 2me The magnetic moment m is proportional to the orbital angular momentum L, and in the opposite direction because of the negative electron charge. This relationship is also obtained in quantum mechanics, where L and Lz take discrete values in units ofh/ ¯ 2. When an external magnetic ﬁeld is applied it changes the electrons’ angular velocities: eB ω0 = ω ± ∆ω ∆ω = (249) 2me where ∆ω is known as the Larmor precession frequency. This can either be thought of as a result of the Lorentz Force or, equivalently, as an example of electromagnetic induction.

For an electron pair, one electron has a +∆ω and the other a −∆ω. In both cases this increases the magnetic moment in the direction opposite to the magnetic ﬁeld.

The magnetization vector can be written:

2 2 Ne Zr0H M = NαM H = − (250) 6me

2 2 where N is the atom number density, αM is the atomic magnetic susceptibility, and r0 = hr i is the mean-squared electron radius. The atomic magnetic susceptibility is very small and scales with the nuclear number: 2 2 e Zr0 −29 αM = − ≈ −5 × 10 Z (251) 6me Note that the minus signs are critical. In a diamagnetic material, the magnetization acts to decrease the magnetic ﬂux density B inside a material. This is consistent with Lenz’ law.

A.2 Paramagnetism

Materials where (at least some of) the atoms or molecules have individual electron spins that are not paired, have an intrinsic magnetic moment, m. These can be understood in the same way as polar molecules that have an intrinsic electric dipole moment. In the absence of an external magnetic field the directions of the m are randomised by thermal motion. When an external field is applied, it aligns the magnetic moments parallel to the field to give a positive magnetization:

N|m|2 M = H (252) 3kT

88 Note the 1/kT dependence of the magnetic susceptibility, which is related to the thermal ﬂuctua- tions.

There is still a diamagnetic eﬀect in a paramagnetic material, so the total magnetization is given by: ! |m|2 e2Zr2 M = N − 0 H (253) 3kT 6me For a typical paramagnetic material at room temperature with Z ≈ 50, the paramagnetic term is twice the diamagnetic term. This ratio is larger for lighter atoms and for lower temperatures. It is usually safe to assume that it is greater than one. The magnetic susceptibility of a paramagnetic material is small and positive. The magnetization acts to increase the magnetic ﬂux density B inside the material.

A.3 Ferromagnetism

There are a few materials that exhibit very large magnetization effects, with χM and µr 1. 3 6 Typical values of µr are in the range 10 − 10 . For ferromagnets there is a critical temperature TC , the Curie temperature, below which the valence band electrons spontaneously align their intrinsic magnetic moments parallel to each other. Above the critical temperature the alignment is broken down by thermal effects. Iron, Nickel and Cobalt are ferromagnetic at room temperature, with TC = 1043K, 627K and 1388K. The spontaneous alignment of the magnetic moments extends over ≈ 10−11m3 (≈ 1017 atoms) in regions known as domains. They are separated by domain walls where the magnetic moments of the electrons rotate from one collective alignment to another over a short distance. In the absence of an external magnetic field all possible directions of alignment have equal energy associated with them. As the temperature is decreased through TC , each domain of the material spontaneously chooses an axis for its own alignment. An unmagnetized ferromagnet has random domain orientations, and M=0. The application of an external magnetic field moves the domain walls to align more magnetic moments with the field. This creates a large magnetization parallel to the applied field. There is a saturation level for the magnetization for a particular material. Once the domain walls have been moved by the application of the external field they can become pinned in their new position. This creates a permanent magnet where some magnetization continues to exist after the external field has been removed. There are ways to unpin the domain walls, e.g. by applying a reversed magnetic field, or by a mechanical shock.

A.4 Eﬀect of Magnetic Materials on Inductance

The eﬀect of placing a magnetic material inside a solenoid can be understood by replacing the magnetization vector by a surface magnetization current, IM: B = nIC + IM = nIC + M (254) µ0 The self-inductance of the coil is increased because of the additional ﬂux through the solenoid produced by the magnetization: Z Z Φ = B · dS = µ0(nIC + M) · dS (255) A A

89 dΦ 2 2 L = = µrµ0n πa l (256) dIC For diamagnetic/paramagnetic materials there is a small decrease/increase in the self-inductance. For a ferromagnetic material there is a very large increase.

For ferromagnetic materials the energy stored in an inductance increases by a large factor µr ≈ 103 − 106: 1 U = LI2 = µ U (257) 2 r 0 The energy density of the magnetic ﬁeld inside a magnetic material can be written as:

1 B2 1 uM = = B · H (258) 2 µrµ0 2

The energy transfers in a hysteresis cycle can be obtained by integrating this energy density around the hysteresis curve.

Note that making a small gap in the ferromagnetic core will not change the inductance or the energy stored significantly. The field strength in the gap is small H = B/µ0, and the gap makes a negligible contribution to the flux density compared to the ferromagnetic regions.

90 B Reﬂection at boundaries: oblique incidence

In Lecture 20 we analysed the case of waves impinging on an interface at normal incidence. Here we consider a general angle of incidence.

B.1 General Angle of Incidence

Figure 32: Wave at interface between two media see Griﬃths ﬁg. 9.14

As before we take an interface between two media to be the x–y plane at z = 0: medium 1 is z < 0; medium 2 is z > 0.

We can take the incident wave vector kI to be in the x–z plane which is then the plane of incidence; y out of page

i(k .r−ωt) Einc = EI e I i(k .r−ωt) Eref = ER e R i(k .r−ωt) Etrans = ET e T We also have the corresponding magnetic ﬁeld vectors e.g.

1 ˆ Hinc = k × Einc µ1v1

Now we have to ﬁt the boundary conditions at the interface. First of all we note that all the boundary conditions will be of the form

i(k ·r−ωt) i(k ·r−ωt) i(k ·r−ωt) ( )e I + ( )e R = ( )e T

So for the boundary conditions to hold for all points on the interface x–y plane we must have the exponential factors (i.e. the phases) equal

⇒ kI · r = kR · r = kT · r = φ = constant (259) and straightaway we see that kI , kR, kT , all lie in the same plane—the plane of incidence. i.e. none of them has a component in the y direction Then (259) becomes kI sin θI x = kR sin θRx = kT sin θT x But this must hold for all x and also we know from k = ω/v that

v2 n1 kI = kR = kT = kT v1 n2

91 which together imply

θI = θR angle of incidence equals angle of reﬂection

n1 sin θI = n2 sin θT Snell’s Law We now have the job of satisfying the boundary conditions (see sections 17,18) which become

1(EI + ER)z = 2(ET )z (260)

(BI + BR)z = (BT )z (261)

(EI + ER)x,y = (ET )x,y (262)

(HI + HR)x,y = (HT )x,y . (263) The ﬁnal two equations are both pairs of equation for the two transverse x,y components.

Polarisation Eﬀects

The reﬂection and transmission coeﬃcients r, t depend on the polarisation state of the incident beam. There are two basic polarisation states

A. EI in plane of incidence (EI has no y component and HI along ey)

B. HI in plane of incidence (HI has no y component and EI along ey) Other polarisation states can be decomposed into A+B by superposition. We will only work out CASE A: E in plane of incidence

Figure 33: Wave at interface between two media see Griﬃths ﬁg. 9.15

Clearly (261) is automatically satisﬁed as B has no z component. It turns out (as you can check) that (263) does not give any additional information to (260) and (262)

Thus noting sign of Ez: − for I,T but + for R

i(φ−ωt) Einc = EI (ex cos θI − ez sin θI ) e i(φ−ωt) Eref = ER (ex cos θI + ez sin θI ) e i(φ−ωt) Etrans = ET (ex cos θT − ez sin θT ) e

Then condition (260) ⇒ 1(−EI + ER) sin θI = −2ET sin θT and condition (262)⇒ (EI + ER) cos θI = ET cos θT

2 equations in 2 unknowns, solve for ET ,ER: We deﬁne as before µ v µ 1/2 β = 1 1 = 1 2 µ2v2 1µ2

92 and also cos θ α = T cos θI then we ﬁnd E α − β E 2 r ≡ R = t ≡ T = (264) EI α + β EI α + β which can also be written as Z cos θ − Z cos θ 2Z cos θ r = 2 T 1 I t = 2 I (265) Z2 cos θT + Z1 cos θI Z2 cos θT + Z1 cos θI where Zi is the usual impedance. For nonmagnetic dielectrics µ1 = µ2 = µ0, Z1 = Zvac/n1, Z2 = Zvac/n2 ⇒ n cos θ − n cos θ r = 1 T 2 I n1 cos θT + n2 cos θI use Snell’s law to eliminate n’s: sin 2θ − sin 2θ r = T I (266) sin 2θT + sin 2θI Fresnel Formula for case A (E in plane of incidence)

B.2 Brewster’s Angle

An interesting consequence of Fresnel equations (264) is that r = 0 when α = β. This occurs at special angle of incidence know as Brewster’s angle θI = θB 2 !1/2 n1 2 1 − sin θB = β cos θB (267) n2 2 n1 2 2 2 ⇒ 1 − sin θB = β (1 − sin θB) (268) n2 and ﬁnally this gives 1 − β2 sin2 θ = (269) B 2 n1 − β2 n2

In the typical case µ1 = µ2 we have β = n2/n1 and one can show that

n2 tan θB = (270) n1 ◦ (θB ' 50 for water/air) N.B. Brewster’s angle only exists for case A: in case B there is no such eﬀect.

Brewster angle microscopy: Shine ‘case-A light’ on clean surface at θI = θB: no reflected ray Now adsorb thin layer of another material: reflected ray caused solely by film ⇒ sensitive probe of film structure Polarisation by reflection: Unpolarised light source = random superposition of waves with E in plane of incidence (case A) and transverse to it (case B). Near to Brewster’s angle reflected ray is almost all polarised One can eliminate reflected ray (glare) with polaroid filter which cuts out one plane of polarised light; basis of polaroid sunspecs etc.

93 B.3 Total Internal Reﬂection

◦ Choose n1 > n2 (e.g. wave leaving dielectric into vacuum) then θT > θI ; θT = 90 at θI = θC . Snell: sin θC = n2/n1. For θI > θC : we have Total Internal Reﬂection

Figure 34: Total internal reflection see Griffiths fig 9.28

Evanescent Waves

To see what is happening for θI > θC , we persevere with the maths and note that if

n2 sin θT = sin θI > 1 n1 2 !1/2 2 1/2 n2 then cos θT = (1 − sin θT ) = i sin θI − 1 n1 clearly we can’t interpret θT as an angle any more but the maths is valid One can show that i(k ·r−ωt) i(kx−ωt) −z/α Eevanescent = ET e T = ET e e where ωn ω q k = 1 sin θ α−1 = n2 sin2 θ − n2 c I c 1 I 2 with α ' the wavelength (∼ 0.5µm). So we have attenuation in the z direction The transmission coeﬃcient t 6= 0 but no energy is carried into medium 2. Instead there is a travelling wave directed along the interface, which decays in the z direction (into medium 2): The decay is not adsorption or the skin eﬀect but can be thought of as Light tunnelling: light can tunnel across a thin layer of medium 2 via the evanescent wave.

94 C Retarded potentials and light cones

C.1 Inhomogeneous wave equations

Recall the microscopic Maxwell equations, including source terms (i.e., charge and current distributions): ρ ∇ · E = (271) 0 ∇ · B = 0 (272) ∂B ∇ × E = − (273) ∂t ∂E ∇ × B = µ J + (274) 0 0 ∂t

Previously, we had considered the equations in vacuo (no free currents or charges) and were able to write wave equations for the electric and magnetic fields. We discussed their solutions at length. Now, with sources present, can we find similar equations? Recall from electrostatics that the most general way to solve the fields of a given charge distribution ρ(r) came through the Poisson equation 2 for the scalar potential, ∇ V = −ρ/0. We will thus attempt to derive equivalent equations for the potentials in the case of time-dependent sources. Maxwell-II and -III allow us to introduce the scalar and vector potentials: ∂A B = ∇ × A,E = −∇V − (275) ∂t Plugging these definitions into Maxwell-I and -IV and using the curl product rule we obtain ∂ ρ ∇2V + (∇ · A) = − (276) ∂t 0 ! ∂V ∂2A ∇ (∇ · A) − ∇2A = µ J − ∇ − (277) 0 0 ∂t 0 ∂t2

Note (276) and (277) are four coupled diﬀerential equations. To decouple them, we make use of the gauge degree of freedom for the potentials V and A. With an arbitrary scalar ﬁeld φ we can change the potentials according to

A → A0 ≡ A + ∇φ ∂φ V → V 0 ≡ V − ∂t These changes will leave the fields E and B unchanged. We now choose φ such that ∇ · A + 0µ0∂V/∂t = 0. This is known as the Lorentz gauge (we previously used the Coulomb gauge ∇ · A = 0) and corresponds to a gauge field φ that fulfills the homogeneous wave equation, ∇2φ − 2 2 0µ0∂ φ/∂t = 0. With the Lorentz gauge used in both (276) and (277), we obtain decoupled equations for the potentials:

2 2 ∂ V ρ ∇ V − 0µ0 2 = − (278) ∂t 0 ∂2A ∇2A − µ = −µ J (279) 0 0 ∂t2 0

95 These represent inhomogeneous wave equations for the potentials V and A (the sources appear on the right hand side), which are driven by (in general) time-dependent charge and current distributions ρ(r, t) and J(r, t).

C.2 Solving the inhomogeneous wave equation

Solving equation (278) by itself or equation (279) for each component is the same task, so we focus from here on on solving (278). We begin by writing a formal Fourier series of the time dependence of the potential and source: 1 Z ∞ V (r, t) = √ dωVω(r) exp −iωt (280) 2π −∞ 1 Z ∞ ρ(r, t) = √ dωρω(r) exp −iωt (281) 2π −∞

Note the Fourier components are given by 1 Z ∞ ρω(r) = √ dtρ(r, t) exp iωt 2π −∞ Using this ansatz (or decomposition) in the wave equation for V we obtain

Z ∞ Z ∞ 2 2 1 ∇ + 0µ0ω Vω(r) exp −iωt = − dωρω(r) exp −iωt −∞ 0 −∞ Comparing the integrands (or noting that plane waves with diﬀerent frequency are orthogonal) we obtain the Helmholtz equation for each monochromatic component:

 2 2 2 ρω(r) ∇ + ω /c Vω(r) = − (282) 0

1/2 Note we have introduced c = (0µ0) , the speed of light in vacuum. We have also reduced our problem slightly (rid ourselves of the time dependence) but still need to solve this equation. Here, the method of Green’s functions comes in handy.

Excourse: Green’s functions

This is a general method to obtain solutions for linear differential equations. Let D be a linear differential operator, and we are interested in solving the differential equation

Df(x) = s(x) where s(x) is a source term, x some generalised coordinate, and f(x) the solution we want. Then, the Green’s function G of D is deﬁned by

DG(x, x0) = −δ(x − x0). (283)

In other words, G(x, x0) is the solution of D for a Dirac-delta source term. It is also called the ”kernel” of the diﬀerential equation. If G is known, the general solution f(x) can be written out as Z f(x) = − dx0G(x, x0)s(x0), (284)

96 the convolution of the Green’s function and the particular source term. We have actually encountered a Green’s function before, in electrostatics. There, we aimed to solve 2 2 Poisson’s equation, ∇ V = −ρ/0. With D ≡ ∇ , we solved ∇2G(r, r0) = −δ(r − r0) because it is (bar for a pre-factor) the potential of a point charge: 1 1 G(r, r0) = 4π |r − r0| Hence, we were able to write down the general solution of Poisson’s equation: Z 0 Z 0 0 0 ρ(r ) 1 0 ρ(r ) V (r) = − dr G(r, r ) − = dr 0 (285) 0 4π0 |r − r |

Another example for a simple diﬀerential equation would be the one-dimensional diﬀusion (or heat) equation, ! ∂ ∂2 − α u(x, t) = ρ(x, t) ∂t ∂x2 For the heat equation, the Green’s function is 1 (x − x0)2 G(xt, x0t0) = exp − . (286) 4πα(t − t0) 4α(t − t0) Here, it can be interpreted as the solution of a very localised heat spot at time t0 and location x0 (e.g., a short and focused laser pulse), which then dissipates over time and space in the form of an ever-widening Gaussian.

C.3 Advanced and retarded potential, causality

The Green’s function of the operator (∇2 + ω2/c2) is 1 1 G (r, r0) = exp ±iω/c|r − r0| (287) ω 4π |r − r0|

Note there are two solutions for Gω, and we interpret them further below. Gω is also known as the Helmholtz kernel, and allows us to write out the potential for a monochromatic source: Z 0 1 0 0 exp ±iω/c|r − r | Vω(r) = dr ρω(r ) 0 . (288) 4π0 |r − r | An integration over all frequencies gives us the full solution for the potential: 1 1 Z ∞ V (r, t) = √ dωVω(r) exp −iωt 2π 4π0 −∞ 1 Z 1 1 Z ∞ 1 0 √ 0 = dr 0 ρω(r) exp −iω(t ∓ |r − r | 4π0 |r − r | 2π −∞ c Z 1 0 1 1 0 = dr 0 ρ(r, t ∓ |r − r |) 4π0 |r − r | c

This, at last, gives us the potential at an arbitrary point (r, t) in space and time. However, the potential depends on the charge distributions either in the past or the future. To be precise, at either the

97 0 0 • retarded time t = t − 1/c|r − r | ≡ tret, or the

0 0 • advanced time t = t + 1/c|r − r | ≡ tadv.

Both are valid solutions, as Maxwell’s equations are invariant under time reversal. However, experience tells us that time moves forward, and cause and eﬀect move from the past to the future. Hence, the retarded potential as given by

Z 1 0 1 1 0 V (r, t) = dr 0 ρ(r, t − |r − r |) (289) 4π0 |r − r | c is the correct solution. It can be interpreted as follows: sources (charges) on the retarded light cone of a given point (r0, t0) (see Figure 35 left) are being summed up and contribute to the potential at (r0, t0). Sources on the advanced light cone of (r0, t0) can absorb (or interact with, or react to) the ﬁelds emitted at (r0, t0). future t t t t future tadvtadv

else- else- (r(0r,t00,t)0) where (r(0r,t0,t0)0) where

trettret r r past r r ρρ(r(,t)r,t) past

Figure 35: Left: time and space evolution of charge distribution ρ(r, t), and its inﬂuence on point (r0, t0). Right: the resulting causal structure of electromagnetism.

In a broader interpretation, information can only travel at the speed of light, and Maxwell’s theory has a relativistic causality structure: for a given point in space and time (the present), the past is everything within the retarded light cone, the future is within the advanced light cone, and the rest of the universe, the elsewhere, is not causally connected to the present at all.