PHYS 301: Electromagnetism

PHYS 301: ElectromagnetismAssoc. Prof. Mike SteelDept of Physics, Macquarie UniversityMay 21, 2010PHYS 301: Electromagnetism 1 / 123

Significance of EM (Non-examinable)Historical contextEnlightment electricity and magnetismBen Franklin1750’s Benjamin Franklin shows lightning is electrical. Two types ofcharge.1770-1800 Volta, Galvani et al: Invention of the voltaic pile (battery).Discovery of electrical action in animals.1780s Coulomb discovers inverse square law of charges.Today we’d sayF = Kq 1q 2|r 1 − r 2 | 2 ˆr 12PHYS 301: Electromagnetism Review 3 / 123

Significance of EM (Non-examinable)Historical contextEM from Maxwell and onJames Clerk MaxwellAlbert Einstein1860s Maxwell: combines all known electromagneticbehavior in five equations, (actually rather more).Embraced Faraday’s picture of fields throughoutspace.Speed of light drops out of equations.Light is electromagnetic in nature!1887 Hertz: electromagnetic waves observed.1905 Einstein: Electromagnetism violates Newton III.Birth of relativity.1912-1925 Quantum pioneers: moving charges in atoms do notdecay by radiation?Birth of quantum theory.PHYS 301: Electromagnetism Review 5 / 123

Significance of EM (Non-examinable)Why study EMSignificance of EMEM phenomena surround us everyday and run our lives.Rich source of problems that are complex, yet analytically tractable.It’s a central pillar of the culture of undegraduate physics.It makes us think about the nature of fields.It continues to be a huge active research area.To know ourselvesour nervous system is electrical.our brains generate complex magnetic fields.PHYS 301: Electromagnetism Review 6 / 123

Significance of EM (Non-examinable)Why study EMThe program of unificationEM is the archetypal model of the program of theoretical physics:Theoretical physicsReduce the world to as few laws as possible.Derive from them as many interesting consequences as possible.This program began with Newton: the apple and the moon are attractedto the earth for the same reason.It’s called “The Universal Law of Gravitation”.PHYS 301: Electromagnetism Review 7 / 123

Significance of EM (Non-examinable)Why study EMThe program of unificationIt continues today:EM and special relativity successfully combined in the 1950s: Nobelsfor Schwinger, Tomonaga and Feynman.EM and the weak interaction successfully combined in the 1970s:“electroweak” physics. Nobels for Glashow, Salam, Weinberg, Rubbia.Attempts to combine EM, weak and strong interactions into GrandUnified Theories (GUT) have proved very difficult.It has brought us giant particle accelerators, string theory, m-branesand other exotica.Combining these three with gravity would bring a Theory ofEverything (TOE). This is apparently some way off.PHYS 301: Electromagnetism Review 8 / 123

Review of vector calculusScalar fields and vector fieldsFields and the position vectorElectromagnetic theory is the origin of the field concept, due to Faradayand Maxwell.Fields assign a scalar (temperature, electric potential etc) or vector(electric field, fluid velocity etc ) to every point in space:f(x, y, z), g(r, θ, φ), F(x, y, z), G(s, φ, z).It’s convenient to make the independent variable the position vector,written r or x, regardless of the coordinate system.The position vector for a point (x, y, z) is the vector that points from theorigin to (x, y, z):r = x ˆx + y ŷ + z ẑ ≡ x î + y ĵ + z ˆk ≡ (x, y, z).We then write our fields as f(r), F(r), G(x) etc.The meaning is the same, but the notation becomes quite powerful later.PHYS 301: Electromagnetism Review 9 / 123

Review of vector calculusVector derivativesOperational definitions L Griffiths: G1.1, G1.2Previously, we’ve met the three vector derivatives in the manner they’remost often used.[ ]Defining the “del” operator ∇ = ∂∂x , ∂ ∂y , ∂ ∂z, we get the three derivativesgrad f =[ ∂f∂x , ∂f∂y , ∂f ],∂zdiv F = ∂F x∂x + ∂F y∂y + ∂F z∂z ,curl F =[ ∂Fz∂y − ∂F y∂z , ∂F x∂z − ∂F z∂x , ∂F y∂x − ∂F x∂yBut it’s not obvious why these definitions are interesting, and what to dowhen we’re not in Cartesian coordinates.].PHYS 301: Electromagnetism Review 10 / 123

Review of vector calculusVector derivativesThe Gradient: grad f(r)The gradient measures the change in space of a scalar function. Take ascalar function f(x, y, z) and at any point, walk up the steepest direction.The gradient is defined as the vector along that direction. Let’s derive theformula.The change in f along any direction du = (dx, dy, dz) isdf = ∂f ∂f ∂fdx + dy +∂x ∂y ∂z dz[ ∂f=∂x , ∂f∂y , ∂f ]· [dx, dy, dz]∂z[ =∂f∣ ∂x , ∂f∂y , ∂f ]∣ ∣∣∣|du| cos θ∂z= |grad f| |du| cos θwhere θ is the angle between the two vectors. For maximum change, umust be parallel to the first vector, giving us the gradient formula.PHYS 301: Electromagnetism Review 11 / 123

Review of vector calculusVector derivativesThe Divergence: div F(r)The divergence is supposed to measure the outflow of a vector field. Thisgives us the natural definition∮1div F(r) = lim F(r) · n da.V →0 V(Note that dim[div F] = dim[F]/L, as expected.)If V is small, we can write the changes in F with Taylor’s theorem. Takinga cube of side ∆x, ∆y, ∆z as the volume, we can evaluate the integralface by face.Everything cancels, leavingdiv F = ∂F x∂x + ∂F y∂y + ∂F z∂z .PHYS 301: Electromagnetism Review 12 / 123

Review of vector calculusVector derivativesThe Curl: curl F(r)The curl is supposed to measure the twistiness of a vector field. To makethis sensible in 3-space, we define the curl a component at a time:∮1curl F(r) · û = lim F(r) · dl,S→0 Swhere the contour C is in a plane perpendicular to the direction û. Againwe can do these integrals trivially with Taylor’s theorem. For the zcomponent,[curl F(r)] z=lim∆x,∆y→0= lim∆x,∆y→0= ∂F y∂x − ∂F x∂y[∫ x+∆xx+Cdx F x (x, y) − F x (x, y + ∆y)∫ y+∆yydy F y (x + ∆x, y) − dy F y (x, y)][−∆y ∂F x∂y ∆x + ∆x∂F y∂x ∆y]PHYS 301: Electromagnetism Review 13 / 123

Review of vector calculusInterlude: curvilinear coordinatesSpherical coordinates L Griffiths: 1.4.1In spherical coordinates, we define r = (x, y, z), withx = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ.To find the unit vectors, it’s helpful to defineˆr = r|r| , ˆφ = ẑ× ˆr| ẑ× ˆr| , ˆθ = ˆφ× ˆr.Then any vector has the form A = A r ˆr + A θ ˆθ + Aφ ˆφ, withˆr = sin θ cos φ ˆx + sin θ sin φ ŷ + cos θ ẑˆθ = cos θ cos φ ˆx + cos θ sin φ ŷ − sin θ ẑˆφ = − sin φ ˆx + cos φ ŷ.The position vector is just r = r ˆr.[Note: r ≠ r ˆr + θ ˆθ + φ ˆφ.]PHYS 301: Electromagnetism Review 14 / 123

Review of vector calculusInterlude: curvilinear coordinatesCylindrical coordinates L Griffiths: 1.4.1In cylindrical coordinates, we define as usual r = (x, y, z),withx = s cos φ y = s sin φ z = z.The unit vectors areŝ = cos φ ˆx + sin φ ŷˆφ = − sin φ ˆx + cos φ ŷẑ = ẑThe position vector isr = x ˆx + y ŷ + z ẑ= s cos φ(cos φ ŝ − sin φ ˆφ) + s sin φ(sin φ ŝ + cos φ ˆφ) + z ẑ= s ŝ + z ẑPHYS 301: Electromagnetism Review 15 / 123

Review of vector calculusInterlude: curvilinear coordinatesGradient in spherical coordinatesFor homework, we’ll see that the infinitesimal distance du becomesThen fromwe have∂f∂rdu = dx ˆx + dy ŷ + dz ẑ = dr ˆr + rdθ ˆθ + r sin θdφ ˆφ.df = grad f · du∂f ∂fdr + dθ +∂θ ∂φ dφ = grad f · [dr ˆr + rdθ ˆθ + r sin θdφ ˆφ],and we read off the gradient:grad f = ∂f∂r ˆr + 1 ∂fr ∂θ ˆθ + 1 ∂fr sin θ ∂φ ˆφ.PHYS 301: Electromagnetism Review 16 / 123

Review of vector calculusInterlude: curvilinear coordinatesDerivatives in curvilinear coordinates : div F(r)If we take different shaped volumes, we can work out the formulae in othercoordinate systems.eg. Taking a “cuboid” in cylindrical coordinates with sides ∆r, ∆z, r∆φwe can finddiv F = 1 ∂r ∂r (rF r) + ∂F z∂z + 1 ∂F φr ∂φWe’ll take these on trust, and just use the inside cover of the textbook.PHYS 301: Electromagnetism Review 17 / 123

Review of vector calculusInterlude: curvilinear coordinatesIntegrals in curvilinear coordinates L Griffiths: 1.4.1We often need to perform integrals and derivatives in spherical andcylindrical coordinates.The derivative expressions are nasty and we don’t try to remember them.The integrals are simple and we do!Volume and surface integrals in spherical coordinates:∫∫ 2π ∫ π ∫ Rf(r, θ φ) dτ = dφ dθ sin(θ) dr r 2 f∮VSf(r, θ φ) da =0∫ 2π0dφ0∫ π00dθR 2 sin(θ)Volume and surface integrals in cylindrical coordinates:∫∫ 2π ∫ Z ∫ Rf(r, φ z) dτ = dφ dz dr rf∮VSf(r, φ z) da =0∫ 2π0dφ0∫ Z00dz RfPHYS 301: Electromagnetism Review 18 / 123

Review of vector calculusVector integral theoremsThe gradient theoremLet’s try to integrate a gradient along a line:∫ ba∇T · dl = limi→∞∑i∇T · (x i − x i−1 )But we saw df = f(x + dx) − f(x) = ∇f · dx,so each element in the sum is the difference T (x i ) − T (x i−1 ).Adding them all up leaves just the first and last elements:Gradient theorem∫ ba∇T · dl = T (b) − T (a)PHYS 301: Electromagnetism Review 20 / 123

Review of vector calculusVector integral theoremsDivergence theorem (Gauss’ Law)We can turn an integral over one big surface, into asum of integrals over little boxes:∮F · n da = ∑ ∮F · n daSiS i∑= lim∆V i →0Divergence theoremi∑∆V i[ 1∆V i∮= lim ∆V i div F(x i )∆V i →0i∫= div F dτ∫VV∮div F dτ =SF · n da]F · n daS iPHYS 301: Electromagnetism Review 21 / 123

Review of vector calculusVector integral theoremsStokes’ theoremTo find an integral relation for the curl, we carve upa surface instead ∮ of a volume:F · dl = ∑ ∮F · dlCiC i∑= lim∆S i →0Stokes’ theoremi∑∆S i[ 1∆S i∮]F · dlC i= lim ∆S i curl F(x i ) · n∆S i →0i∫= curl F · n da∫SS∮curl F · n da =CF · dlPHYS 301: Electromagnetism Review 22 / 123

Review of vector calculusProducts and second derivativesTips for vector derivativesVector derivative expressions can be intimidating.Some rules make it easier to get the right answers:Don’t worry about what the expression means physically.Just follow the logical steps following from ∇ ≡ [ ∂∂x , ∂∂y , ∂ ∂z ].Be very careful about vector notation—vector quantities always havea tilde (˜) underneath, scalars do not.Check on every line, is each term still the right kind of quantity:vector or scalar?PHYS 301: Electromagnetism Review 23 / 123

Review of vector calculusProducts and second derivativesDerivatives of productsProduct derivatives appear frequently with Maxwell’s equations. Some ofthe following are not obvious, but we can check them straight-forwardly bysubstitution. (We don’t memorize these!)Since fg and A · B are both scalars, we need∇(fg) = f∇g + (∇f)g∇(A · B) = A×(∇×B) + B×(∇×A) + (A · ∇)B + (B · ∇)ASince fA and A×B are both vectors, we need∇ · (fA) = f(∇ · A) + A · (∇f)∇ · (A×B) = B · (∇×A) − A · (∇×B)∇×(fA) = f(∇×A) − A×(∇f)∇×(A×B) = (B · ∇)A − (A · ∇)B + A(∇ · B) − B(∇ · A)PHYS 301: Electromagnetism Review 24 / 123

Review of vector calculusProducts and second derivativesSecond derivatives of gradientsSimilarly, we can take derivatives of derivatives. To a gradient ∇f, we canapply div and curl . The first gives us the Laplacian, written ∇ 2 :∇ 2 F ≡ ∇ · (∇f) = ∂2 f∂x 2 + ∂2 f∂y 2 + ∂2 f∂z 2 .The Laplacian of a vector is the component-by-component result:∇ 2 F = [ ∇ 2 F x , ∇ 2 F y , ∇ 2 F z].It’s easy to show that ∇×(∇f) ≡ 0.We’ll see that this has huge physical importance.PHYS 301: Electromagnetism Review 25 / 123

Review of vector calculusProducts and second derivativesSecond derivatives of div and curlA divergence ∇ · F is a scalar, so we can only take it’s gradient ∇(∇ · F).We won’t see it much.To a curl, ∇×F, we can apply the divergence or another curl.The first of these always vanishes,∇ · (∇×F) ≡ 0.The curl gives us back derivatives we already know:∇×(∇×F) = ∇(∇ · F) − ∇ 2 F.PHYS 301: Electromagnetism Review 26 / 123

Review of vector calculusProducts and second derivativesA note on notationIt’s tricky to match the fonts in a textbook exactly: Griffiths usesbold face r and r ′ to represent absolute vector locations,a funny script lower case “r” for vector differences r − r ′ that doesn’tappear to exist in any font set on the internet!We’ll use℘ = r − r ′ ,℘ = |r − r ′ |.Notations for multiple integrals abound. Don’t be confused by thefollowing which all mean the same thing:∫∫ ∫ ∫f dτ = f dv = f dr 3 = f dx 3VVVV∫∫∫∫F · da = F · n da = F · n ds = F · n dx 2SSSSPHYS 301: Electromagnetism Review 27 / 123

Formulations of ElectrostaticsCoulomb’s LawCoulomb’s LawElectrostatics is essentially defined by Coulomb’s Law.He thought in terms of forces between point charges.F = K q 1q 2|r 1 − r 2 | ̂ 2 r 1 − r 2 = K q 1q 2℘ 2 ˆ℘.For multiple charges, the forces just add up:F qi = ∑ Kq iq j|r i − r j | 2 r ̂ i − r jj≠iThis is one expression of the laws of electrostatics.In SI units,K = 14πɛ 0.PHYS 301: Electromagnetism Electrostatics 28 / 123

Formulations of ElectrostaticsCoulomb’s LawElectric field formulationPartly in an attempt to explain interactions over distances, we introducethe idea of a field, by pulling out the test charge:∑F qi = q i Kq j|r i − r j | ̂ 2 r i − r jj≠i= q i E qiwhere the electrostatic field E qiat charge q i isq jE qi = ∑ K|r i − r j | 2 r ̂ i − r j .j≠iThis is an equivalent expression of the laws of electrostatics.PHYS 301: Electromagnetism Electrostatics 29 / 123

Formulations of ElectrostaticsCoulomb’s LawContinuous charge distributionsBut most charges are not isolated: they belong to distributions.So it’s more natural to write the continuous expressionsE(r) =E(r) =E(r) =14πɛ 0∫V14πɛ 0∫S14πɛ 0∫Cρ(r ′ )|r − r ′ | ̂r 2 − r ′ dτ ′ ≡ 1 ρ(r4πɛ 0∫V) ′ ˆ℘ dτ℘2 σ(r ′ )℘ 2 ˆ℘ da ′ (surface charge density σ(x) )λ(x ′ )℘ 2 ˆ℘ dx ′ (line charge λ(x))Observe: the field is evaluated at the unprimed coordinate r. We integrateover the charge distribution with the primed coordinate r ′ .Be careful: the charge distributions ρ, σ, λ must have different units.(What are they?)PHYS 301: Electromagnetism Electrostatics 30 / 123

Formulations of ElectrostaticsCoulomb’s LawDifferential formulation of electrostaticsA consequence of Coulomb’s law is Gauss’ Law:∮E · ˆn da = 1 Q encl ,ɛ 0Sbut normally in physics, we like differential laws.Using the divergence theorem, Gauss’ law translates to∫div E(r ′ ) dτ ′ = 1 ∫ρ(x ′ ) dτ ′ ,ɛ 0Vand since this is true for any V , the integrands are equal:div E = ρ ɛ 0.Is this another statement of the laws of electrostatics?VPHYS 301: Electromagnetism Electrostatics 31 / 123

Formulations of ElectrostaticsCoulomb’s LawCoulomb’s Law and Gauss’s LawGauss’s Law is a consequence of Coulomb’s Law, not equivalent to it.To any field satisfying Gauss’s Law, we can add a nonzero curl,but such a field is not an electrostatic field.ExampleCharge q at x = 0.For some scale length l, letE =q4πɛ 0 r 2 ˆr.E ′ = E +q4πɛ 0 lˆφr .Then div E ′ = 0 but this is clearly not an electrostatic field.We need an extra condition.PHYS 301: Electromagnetism Electrostatics 32 / 123

The scalar potentialThe curl of an electrostatic fieldME tell us that the missing rule is curl E = 0, but it’s classier to prove thisstraight from Coulomb’s law, ie purely from electrostatics.Using ∇ 1|x − x ′ | = − x − x′|x − x ′ | 3 , (differentiating with respect to x, not x′ )we have E(x) = 14πɛ 0∫V= 1= ∇and so curl E = ∇×ρ(x ′ ) (x − x′ )|x − x ′ | 3 dτ ′[ρ(x ′ ) −∇ 1 ]|x − x ′ dτ ′|∫ρ(x ′ ])·4πɛ 0 V |x − x ′ | dτ ′{∇ − 1 ∫·4πɛ 04πɛ 0∫V[− 1Vρ(x ′ })|x − x ′ | dτ ′ = 0.PHYS 301: Electromagnetism Electrostatics 33 / 123

The scalar potentialThe curl of an electrostatic fieldSo, as well as establishing that curl E = 0 for electrostatics, we find thatthe electric field is the gradient of a function we call the scalar potential.withE = −∇V,V = 14πɛ 0∫VThis simplifies calculations enormously.ρ(x ′ )|x − x ′ | dτ ′ .Griffith takes a different route. He defines the potential as the line integralof the field, and shows that it can be calculated as we have above.Both approaches are valid as the following set of results show.PHYS 301: Electromagnetism Electrostatics 34 / 123

The scalar potentialProperties of a irrotational field L Griffiths: 1.6The following are all equivalent:curl F = 0∮F · dl = 0C∫ ba∫ bF · dl = −F · dl(Stoke’s theorem)∫ abF · dlis path independent.aF = −∇U for some scalar function U. (Helmholtz Theorem)We found that a scalar potential existed for the electric field.HT shows that a potential exists for any rotationless field (and gives us anexpression for it).PHYS 301: Electromagnetism Electrostatics 35 / 123

The scalar potentialUsing these results for the electric field,∫ xaE · dl =∫ xa−∇V · dl= V (a) − V (x) (by the gradient theorem).We can add any constant to the potential ∇C = 0,so we set V (a) = 0 as a reference, and findV (x) = −∫ xaE · dl,in accordance with Griffith’s definition of potential.Each form has its uses, depending on whether we know ρ(x) or E(x).PHYS 301: Electromagnetism Electrostatics 36 / 123

The scalar potentialReview of perfect conductors L Griffiths: 2.5.1Much of the potential theory we seeqs involves perfect conductors.We define a conductor as a material in which charges are completely freeqsto move. Metals are a great approximation (at least in electrocstatics).Recall these properties.1 E = 0 inside a conductor. If if wasn’t the charges would move to cancel it.2 ρ = 0 inside a conductor. ρ = ɛ 0 div E.3 All charge lies on the surface. No where else to go.4 Conductors are equipotentials. Line integral of the electric field vanishes.5 E at the surface is normal with E = σ/ɛ 0 ˆn, σ = −ɛ 0∂V∂n .Seeqs (1) and L Griffiths: 2.5.3.Note:∂V≡ ∇V · ˆn is the normal derivative.∂nRead L Griffiths: 2.5.2 for interesting consequences to do with charges inconducting cavities.PHYS 301: Electromagnetism Electrostatics 37 / 123

Laplace’s equationA PDE for the potential L Griffiths: 2.3.3, 3.1For lots of problems, we can’t solve either integral for V .A more powerful approach is to find a differential equation for it.Combining∇ · E = ρ ɛ 0and E = −∇V,we havePoisson’s Equation∇ 2 V = − ρ ɛ 0,and in charge-freeqs regions,Laplace’s Equation∇ 2 V = 0.PHYS 301: Electromagnetism Electrostatics 38 / 123

Laplace’s equationSolving Laplace’s equation in 1DIf our problem is uniform in y and z, (eg. parallel plate capacitor), it’s apretty easy problem!d 2 Vdx 2 = 0,meansV = mx + b.But the properties of this trivial solution are instructive:There are two unknown constants. We neeqsd boundary conditions tosolve the problem, say V (L) and V (0).Once we apply the boundaries, the solution is unique.There is a mean value property: [V (x + a) + V (x − a)] /2 = V (x).The maximum and minimum values occur at the boundaries.PHYS 301: Electromagnetism Electrostatics 39 / 123

Laplace’s equationLaplace’s equation in 3D: a simple problemExampleFind the potential inside and outside a conducting sphere, radius a, held ata potential V 0 .Solution must have spherical symmetry, so V (r) = V (r) with(1 ∂r 2 r 2 ∂V )= 0.∂r ∂rWe guess the solution V (r) = A + B/r, for constants A, B.Inside r < a: B/r blows up, so we must have V (r) = V 0 .Outside r > a: We expect the potential to vanish at infinity, soA = 0.At V (a) = V 0 , so B = V 0 a, giving V (r) = V 0 a/r.PHYS 301: Electromagnetism Electrostatics 40 / 123

Laplace’s equationSolving Laplace’s equation in multiple dimensionsWe found answers, but are they the right ones?Does nature find a different trickier solution that we didn’t manage toguess?Is it sensible to specify the potential on the surface of a sphere?In 2D and 3D, we can’t write down a simple formula that applies for everyproblem. Four big questions:How do we find solutions?How do we apply boundary conditions?On a line, a plane, points?Do we specify V (x) on boundaries, or some derivative?How many solutions will there be? 0, 1, > 1 ?How do properties of the 1D solution carry over if at all?PHYS 301: Electromagnetism Electrostatics 41 / 123

Laplace’s equationProperties of solutions to Laplace’s equationThe Mean Value Property L Griffiths: 3.1.4Solutions to Laplace’s equation satisfy a Mean Value property.Theorem (Mean Value Property)Away from boundaries:V (r) = 14πR∮sphere2 V (r ′ ) dsNote: There’s no missing lim R→0 in the above equation.This is true macroscopically for any R.CorollarySolutions to Laplace equation have extrema on the boundaries.(If not, we could find a small R that violated the MVP.)PHYS 301: Electromagnetism Electrostatics 42 / 123

Laplace’s equationProperties of solutions to Laplace’s equationProof of The Mean Value Property L Griffiths: 3.1.4We prove this for an individual charge q outside the sphere.V (r) =q4πɛ 0 ℘ with ℘ = √ R 2 + z 2 − 2Rz cos θ.ThenV ave = 14πR 2 ∫ 2π= . . .= q4πɛ 0 z0∫ π0dφdθ R 2 qsin θ √4πɛ 0 R 2 + z 2 − 2Rz cos θ .Adding up the effect for all charges gives us the required result.PHYS 301: Electromagnetism Electrostatics 43 / 123

Laplace’s equationProperties of solutions to Laplace’s equationUniqueness theorems L Griffiths: 3.1.5A uniqueness theorem tells us how to specify boundary conditions in orderto get exactly one solution to a PDE.Note: not every physical system will have exactly one solution.Nonlinear problems often have more.Theorem (First uniqueness theorem for Laplace’s equation)If at all points on the boundary S of a volume D,the potential V is specified (Dirichlet conditions)orthe normal derivative of the potential ∂V/∂n is specified (Neumannconditions)then there is a unique solution to Laplace’s equation throughout D.(For boundaries at infinity we normally assume V → 0.)This seeqsms rather abstract, but is very useful, and physically natural.We’re always going to assume that at least one solution exists.PHYS 301: Electromagnetism Electrostatics 44 / 123

Laplace’s equationProperties of solutions to Laplace’s equationProof of 1st uniqueness theoremSuppose there are two solutions to Laplace’s equation V 1 and V 2 in avolume bounded by surfaces S 1 , S 2 , . . ..Let φ = V 1 − V 2 , and consider the function φ∇φ.By the divergence theorem,∫∮∇ · (φ∇φ) dτ = [φ∇φ · ˆn] da.volS 1 +S 2 +...If φ = 0 or ∇φ · ˆn = 0 on all of S i , then ∫ V∇ · (φ∇φ) dτ = 0.But ∇ · (φ∇φ) = φ∇ 2 φ + |∇φ| 2 , and ∇ 2 φ = 0, so ∫ V |∇φ|2 dτ = 0.So ∇φ = 0, and φ = C must be a constant function.This is a nice proof because it allows mixed boundary conditions: we canhave a combination of V = V 0 or ∂V/∂n = f at different parts of theboundary.Griffith offers a simpler but less general argument.PHYS 301: Electromagnetism Electrostatics 45 / 123

Laplace’s equationApplications of uniqueness theoremsApplications of uniqueness theoremsUniqueness theorems allow us to guess solutions and be confident in theanswer.Example (Potential inside a closed conductor )Results:Conductors are equipotentials.So potential on inner surface is a constant V 0 .The function V (x) = V 0 throughout the cavity satisfies Laplace’sequation and matches the boundary conditions.V (x) = V 0 is a solution, so it must be the solution.If the potential is constant, the field inside is zero.Interior is shielded from charges outside the cavity(“Faraday cage”.)This is why AM reception is bad on the harbour bridge.PHYS 301: Electromagnetism Electrostatics 46 / 123

Laplace’s equationApplications of uniqueness theorems1st uniqueness theorem for Poisson equationThe same argument holds for solutions to Poisson’s equation for a givencharge distribution.The difference V 3 of two solutions to Poisson’s equation satisfies Laplace’sequation since∇ 2 V 3 = ∇ 2 V 2 − ∇ 2 V 1 = ρ/ɛ 0 − ρ/ɛ 0 = 0.So if we specify the interior charge distribution, and the boundaries asbefore, V 1 and V 2 can differ only by a constant.PHYS 301: Electromagnetism Electrostatics 47 / 123

Laplace’s equationApplications of uniqueness theoremsPhysical interpretation of the boundary conditionsThe standard boundary conditions have natural interpretations:Dirichlet conditions: Specify V on the boundaries (and at infinity).Connect the conductor to a “battery” with output voltage V 0 .Neumann conditions: Specify ∂V/∂n on the boundaries.“Paste charge” with surface density σ = −ɛ 0 ∂V/∂n to the boundary.Happily, these correspond to parameters we can specify in real life.PHYS 301: Electromagnetism Electrostatics 48 / 123

Laplace’s equationApplications of uniqueness theoremsMethod of images, L Griffiths: G3.2The power of the uniqueness theorems: any solution which matches theboundary conditions is the solution.Example (Charge above a grounded conducting plane)What is the potential?Soln. must have V = 0 at z = 0 and V → 0 as |r| → ∞.Replace plane with a charge −q at r = (0, 0, −d), so[]V (x) = 1q√4πɛ 0 x 2 + y 2 + (z − d) − q√ .2 x 2 + y 2 + (z + d) 2This obviously satisfies V = 0 at z = 0.PHYS 301: Electromagnetism Electrostatics 49 / 123

Laplace’s equationApplications of uniqueness theoremsDerived resultsThe method of images solution provides many results that would be trickyotherwise:Surface charge: find σ = −ɛ 0 ∂V/∂n.Total charge by integrating σ over the plane.Attractive force on original charge: find force due to image charge.Energy of charge configuration: this is trickier, because the imagecharge is not really there. We mustn’t count energy twice.PHYS 301: Electromagnetism Electrostatics 50 / 123

Laplace’s equationSeparation of variablesSeparation of variables, L Griffiths: G3.3In Separation of Variables, we assume the solution to a PDE is expressibleas a product of functions of the coordinates:V (x, y, z) = X(x)Y (y)Z(z) V (r, θ, φ) = R(r)Θ(θ)Φ(φ)V (r, φ, z) = R(r)Φ(φ)Z(z) . . . any of 11 other coordinate systemsFor Cartesian coordinates Laplace’s equation is,X ′′ (x)Y (y)Z(z) + Y ′′ (y)X(x)Z(z) + Z ′′ (z)X(x)Y (y) = 0X ′′ (x)X(x) + Y ′′ (y)Y (y) + Z′′ (z)Z(z)= 0,which impliesfor k + l + m = 0.X ′′ (x)X(x) = kY ′′ (y)Y (y) = lZ ′′ (z)Z(z) = mPHYS 301: Electromagnetism Electrostatics 51 / 123

Laplace’s equationSeparation of variablesSeparation of variables examplesThe art lies in picking the right coordinate systems, and applying the rightboundary conditions.Example (Griffith 3.3)BCs: V (x, y = 0) = 0, V (x, y = a) = 0,V → 0 as x → ∞, V (x = 0, y) = V 0 (y).By SoV, we find(V = Ae kx + Be −kx) (C sin ky + D cos ky) ,but applying first 3 BCs leaveswith k = mπ/a.V = e −kx sin ky,PHYS 301: Electromagnetism Electrostatics 52 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functionsWe often expand vectors in terms of a set of basis vectors, eg.v = a î + b ĵ + c ˆk.Functions behave just like vectors, but in an infinite-dimensional space.We can find lots of families of basis functions to expand other functions:f(x) =∞∑a i φ i (x).i=0The basis functions φ i (x) must obey some kind of orthogonality relation,eg.∫ b{ 1, i = jφ i (x)φ j (x) dx =0, i ≠ jaThey must also be complete—there is some set of coefficients a i that willwork for any target function f(x).PHYS 301: Electromagnetism Electrostatics 53 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions: Fourier seriesThe best known example of this is a Fourier series.For any function f(x) on the domain [−a, a], we can writef(x) =f(x) =f(x) =∞∑i=0∞∑i=1∞∑i=1a m cos mπxab m sin mπxaa m cos mπxa+b m sin mπxa,if f(x) is even,if f(x) is odd,if f(x) is neither.The handout shows how the orthogonality relations let us find a m and b m .PHYS 301: Electromagnetism Electrostatics 54 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions: Laplace equationWe have family of solutions V = exp(−kx) sin ky, for k = mπ/a.The complete solution is a sum of theseV (x, y) =∞∑c m exp(−mπx/a) sin(mπx/a).m=1The coefficients are determined by the final BC, V (x = 0, y) = V 0 (y),and the fact that the functions sin(nπx/a) are orthogonal on x ∈ [0, a].Note that the most rapidly varying components in y decay fastest in x.Why?The picture shows an irritating property of Fourierexpansions: the Gibbs phenomenon. Fourier seriesconverge pointwise but not uniformly.PHYS 301: Electromagnetism Electrostatics 55 / 123

Laplace’s equationSeparation of variablesAside: The Legendre differential equationConsider the 2nd-order ODEddx[(1 − x 2 ) dPdx]+ l(l − 1)P = 0,for any l, and look for solutions on x ∈ [−1, 1].The only solutions that don’t blow up at x = ±1 are the LegendrepolynomialsP 0 (x) = 1P 1 (x)=xP 2 (x) = (3x 2 − 1)/2 P 3 (x)=(5x 3 − 3x)/2, . . .which we can write with the Rodrigues’ formulaP l (x) = 1 d l2 l dx l (x2 − 1) l .These functions are complete and orthogonal with∫ 1P l (x)P m (x) dx = 22l + 1 δ lm.−1PHYS 301: Electromagnetism Electrostatics 56 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions for spherical coordinatesSuppose we have the potential V (R, θ) specified on a sphere.We look for a solution V (r, θ) = R(r)Θ(θ) to Laplace’s eqn in s.c.:∇ 2 V = 1 (∂r 2 r 2 ∂V )+∂r ∂rto reach1 dR dr1r 2 sin θ(r 2 dR )+ 1dr Θ sin θ(∂sin θ ∂V )o +∂θ ∂θ(dsin θ dΘ )= 0.dθ dθ1 ∂ 2 Vr 2 sin θ ∂φ 2 ,Choosing our constant k = l(l + 1), we have the separated equations(dr 2 dR )(1 d− l(l + 1)R = 0,sin θ dθ )+ l(l + 1)Θ = 0.dr drsin θ dθ dθPHYS 301: Electromagnetism Electrostatics 57 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions for spherical coordinatesIt is easy to check that the radial equation has solutionsR(r) = Ar l +Br l+1 .If set x = cos θ in the polar equation, then we recover Legendre’sdifferential equation, with solutions P l (cos θ).So the complete solution isV (r, θ) =∞∑l=0(A l r l + B )lr l+1 P l (cos θ).Even before solving any particular geometry, we now know that differentspatial frequency components of the potential on the spherical surfacegenerate contributions to the potential that decay at different rates.(In general, the potential on the sphere might depend on φ. In that case,the solution comes out in terms of the spherical harmonics Y lm (θ, φ),which we will not pursue.)PHYS 301: Electromagnetism Electrostatics 58 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions for spherical coordinatesAs before, we find the values of A l , B l by using an orthogonality relation.Since the Legendre polynomials satisfy,then we have∫ π0∫ 1−1P l (x)P m (x) dx = 22l + 1 δ lm,P l (cos θ)P m (cos θ) sin θ dθ = 22l + 1 δ lm,PHYS 301: Electromagnetism Electrostatics 59 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions for spherical coordinatesExample (Potential inside a sphere with V (R, θ) = V 0 (θ) = k sin 2 θ 2 .)At r = 0, we must set B l = 0 so V (r, θ) = ∑ ∞l=0 A lr l P l (cos θ).Left-multiplying by ∫ π0 dθ sin θP m(cos θ) and evaluating at r = R we findA m = 2m + 12R m∫ π0dθ P m (cos θ)V 0 (θ) sin θ.Since V 0 (θ) = k 2 [P 0(cos θ) − P 1 (cos θ)], so A 0 = k/2, A 1 = −k/2R andV (r, θ) = k 2 r0 P 0 (cos θ) − k2R r1 P 1 (cos θ) = k [1 − r ]2 R cos θ .Note that the constant part of the applied potential gives rise to aconstant potential inside the sphere.PHYS 301: Electromagnetism Electrostatics 60 / 123

Laplace’s equationSeparation of variablesExpansion in orthogonal functions for spherical coordinatesExample (Potential outside a sphere with V (R, θ) = V 0 (θ) = k sin 2 θ 2 .)As r → ∞, we must set A l = 0 so V (r, θ) = ∑ ∞ B ll=0Pr l+1 l (cos θ).Left-multiplying by ∫ π0 dθ sin θP m(cos θ) and evaluating at r = R we findB m = 2m + 1 ∫ πR m+1 dθ P m (cos θ)V 0 (θ) sin θ.20Since V 0 (θ) = k 2 [P 0(cos θ) − P 1 (cos θ)], so B 0 = kR/2, B 1 = −kR 2 /2 andV (r, θ) = kR2r P 0(cos θ) − kR22r 2 P 1(cos θ) = kR [1 − R ]2r r cos θ .Note that the constant part of the applied potential gives rise to a termfalling off as 1/r. The formulae for A l and B l are subtly different andwould be hard to remember safely. The key is to know how to apply theorthogonality relation.PHYS 301: Electromagnetism Electrostatics 61 / 123

Laplace’s equationSeparation of variablesUsing boundary conditions correctlySometimes the boundary conditions are not quite so obvious.Example (Neutral conducting sphere in applied field E 0 ẑ.)BCs:r → ∞: Must match applied field, so V → −E 0 z + C.r = R: Sphere is conducting, so is an equipotential. May as wellchoose V (R) = 0.PHYS 301: Electromagnetism Electrostatics 62 / 123

Laplace’s equationApplications of Laplace’s equationPoling of Lithium Niobate for frequency conversion of light.PHYS 301: Electromagnetism Electrostatics 63 / 123

Laplace’s equationApplications of Laplace’s equationOptical cloakingOne of the hottest topics in optics research in the last 2 years is Cloaking.We can now actually make simple electromagnetic cloaks usingmeta-materials, and the physics is largely Laplace’s equation.We’ve already seeqsn how to achieve shielding of an object: put it inside acavity.To avoid detection of an object we neeqsd to cloak it—the surroundingfield lines must be completely undisturbed.(From Schurig et al, Science 314, 977 (2006).)PHYS 301: Electromagnetism Electrostatics 64 / 123

Laplace’s equationApplications of Laplace’s equationCloaking outside the cloakerHere’s another way where the cloaked object is outside the cloaker!Nicorivici et al, Optics Express 15, 6314 (2007).We build a structure consisting of a cylinderof vacuum surrounded by a negative indexcylinder.This structure supports very strongresonances—very intense E fields nearby.If a dipole comes near, it would have almostinfinite negative energy.To avoid this, the system automaticallyadjusts to produce zero field at the dipole,and the far field lines are undisturbed.PHYS 301: Electromagnetism Electrostatics 65 / 123

MagnetostaticsComparison of electrostatics and magnetostaticsMagnetostatics from first principlesThe magnetic analog to Coulomb’s law is the Biot-Savart law 1 , describingthe magnetic field set up by steady currents:B(r) = µ 04π∫VI dl ′ × (r − ˆ r ′ )|r − r ′ dτ ′ for currents in a wire|B(r) = µ ∫0 J(r ′ )×(r − ˆ r ′ )4π |r − r ′ | 2 dτ ′ for general volume currentsVLike Coulomb’s law, this is a law of nature.We infer it from experiment, and it defines magnetostatics.It is equivalent to ∇ · B = 0 and ∇×B = µ 0 J.The cross product in the LoBS makes calculation much tougher than forCoulomb’s Law.1 This is one of those where people seem to say “The Law of Biot and Savart”.PHYS 301: Electromagnetism Magnetostatics 66 / 123

MagnetostaticsComparison of electrostatics and magnetostaticsComparison of electrostatics and magnetostaticsFor electrostatics we have∇ · E = ρ ɛ 0∇×E = 0For magnetostatics we have∇ · B = 0∇×B = µ 0 JThe differences are obvious:But to what extent can we exploit the techniques we’ve learnt forelectrostatics in magnetic problems?PHYS 301: Electromagnetism Magnetostatics 67 / 123

MagnetostaticsComparison of electrostatics and magnetostaticsPotentials for the magnetic fieldIntroducing the scalar potential V was fruitful for electrostatics.In general, we can’t have a scalar potential W with B = ∇W . If we try,we find:µ 0 J = curl B = curl grad W = 0,so a scalar potential doesn’t allow non-zero currents.Instead we pursue the implications of div B = 0.PHYS 301: Electromagnetism Magnetostatics 68 / 123

MagnetostaticsThe vector potentialConsequences of div B = 0 L Griffiths: 1.6.2Recall that that existence of the scalar potential was guaranteed bycurl E = 0:∮curl E = 0 ≡ E · dl = 0 (Stoke’s theorem)≡≡C∫ ba∫ bE · dl = −E · dl∫ abE · dlis path independent.a≡ E = −∇V for some scalar V (r). (Helmholtz Theorem)and that the Helmholtz theorem explicitly told us thatV (r) = 1 ∫(div E)(r ′ )4π V |r − r ′ dτ ′ = 1 ρ(r| 4πɛ 0∫V′ )|r − r ′ | dτ ′ .PHYS 301: Electromagnetism Magnetostatics 69 / 123

MagnetostaticsThe vector potentialConsequences of div B = 0 L Griffiths: 1.6.2The same approach works for div B = 0:∮div B = 0 ≡ B · ˆn da = 0 (Divergence theorem)∫S∫≡ B · ˆn da = − B · ˆn da for S 1 + ¯S 1 = closed surface SS 1 ¯S∫1≡ B · ˆn da depends only on boundary of S 1S 1≡ B = curl A for some vector A(r). (Helmholtz Theorem)The object A is called the Vector Potential.The HT gives us a formula for A(r), but we can figure it out ourselveswith a little work. . .PHYS 301: Electromagnetism Magnetostatics 70 / 123

MagnetostaticsThe vector potentialVector potential and gauge freedom L Griffiths: 5.4.1Since B = curl A, and curl grad ≡ 0, then we can add any gradient to agood potential A 0 :A 1 = A 0 + grad ψ,and A 1 is also a good potential.We can use this gauge freedom to arrange that ∇ · A = 0.This is called the Coulomb gauge.In this gauge,µ 0 J = ∇×B = ∇×∇×A = ∇(∇ · A) − ∇ 2 Aand so∇ 2 A = −µ 0 J.PHYS 301: Electromagnetism Magnetostatics 71 / 123

MagnetostaticsThe vector potentialA formula for A(r)Compare∇ 2 A = −µ 0 J and ∇ 2 V = − ρ ɛ 0.Identifying V = A i , and ρ = J i µ 0 ɛ 0 , each component of A satisfiesPoisson’s equation!We already know the solution to this equation:A(r) = µ ∫0 J(r ′ )4π |r − r ′ | dτ ′VPHYS 301: Electromagnetism Magnetostatics 72 / 123

MagnetostaticsThe vector potentialCalculations with A(r)Since finding A comes down to solving the Poisson equation, it repeats alot of what we already know and we can move on.Here’s a simple example to complement the infinite parallel plane capacitorwe looked at for electrostatics:Example (Infinite plane sheet of current)Find the vector potential A and magnetic field B for an infinite uniformcurrent sheet K(x) = K ˆx flowing in the x − y plane.The resulting magnetic field is along ŷ and independent of distance fromthe plane.PHYS 301: Electromagnetism Magnetostatics 73 / 123

MagnetostaticsThe vector potentialPhysical meaning of the vector potentialExpressed as the line integral of the electric field, the scalar potential had anatural interpretation as the work per unit charge, to move charge around:W = −q∫ rbr aE · dl = q [V (r b ) − V (r a )] .The vector potential doesn’t have a corresponding interpretation.It can’t, because magnetic forces do no work [F = qv×B].We can only intepret the energy stored in magnetic fields, once we get totime-varying fields.PHYS 301: Electromagnetism Magnetostatics 74 / 123

MagnetostaticsThe vector potential“A(r), (huah), what is it good for?”Absolutely nothing?The vector potential is tricky to calculate and gives infinite answers foreven the simplest problems (eg. infinite straight wire), it appears to haveno physical meaning in classical physics. 2However,We can build a useful multipole expansion for A(r).In electrodynamics, it does make things much easier, as we’ll see.It is central to the formulation of quantum electrodynamics (QED).2 In advanced quantum mechanics, we meet the Aharanov-Bohm effect,which is truly sensitive to A .PHYS 301: Electromagnetism Magnetostatics 75 / 123

ElectrodynamicsMaxwell’s equationsWe can reduce all electromagnetic phenomena tofour fundamental equations plus a force law.PHYS 301: Electromagnetism Electrodynamics 76 / 123

ElectrodynamicsMaxwell’s equationsMaxwell 1Gauss’ Law∮SE · da = Q encɛ 0div E = ρ ɛ 0(ME 1)Positive (negative) charges are sources (sinks) for the electric field.PHYS 301: Electromagnetism Electrodynamics 77 / 123

ElectrodynamicsMaxwell’s equationsMaxwell 3Faraday’s Law∮∫E · dl = − dC dtcurl E = − ∂B∂tSB · da(ME 3)Changing magnetic fields generate electric fields.(Why is there no other term here?)PHYS 301: Electromagnetism Electrodynamics 78 / 123

ElectrodynamicsMaxwell’s equationsMaxwell 4Ampere’s Law∮C∫dB · dl = µ 0 I enc + µ 0 ɛ 0 E · dadt S∂Ecurl B = µ 0 J + µ 0 ɛ 0∂t(ME 4)Magnetic field lines wind around currents.Changing electric fields also generate magnetic fields.(No evidence for this in Ampere’s time or even Maxwell’s. See below)PHYS 301: Electromagnetism Electrodynamics 79 / 123

ElectrodynamicsMaxwell’s equationsThere are no magnetic monopolesAFAWK, magnetic monopoles don’t exist.There is no “positive” or “negative” magnetic charge.Field lines can’t stop or start anywhere.They are continuous everywhere.There are no magnetic sources or sinks:“Gauss Law” for Bdiv B = 0 (ME 2)PHYS 301: Electromagnetism Electrodynamics 80 / 123

ElectrodynamicsMaxwell’s equationsMaxwell’s equations in vacuumCollecting the above, we end up with the axioms of our theory.“Maxwell’s equations”div E = ρ ɛ 0(ME 1)div B = 0 (ME 2)curl E = − ∂B∂t(ME 3)∂Ecurl B = µ 0 J + µ 0 ɛ 0∂t(ME 4)and the Lorentz force law connects the field equations to charges:F = q (E + v×B) .From now on, charges can move, currents can change in time, and thefields can respond appropriately in time.PHYS 301: Electromagnetism Electrodynamics 81 / 123

ElectrodynamicsMaxwell’s equationsThe displacement current L Griffiths: 7.3.2Maxwell realised that the displacement current term was necessary forconsistency.The divergence of the old Ampere’s Law gives:0 ≡ ∇ · ∇×B = µ 0 ∇ · J,but by conservation of charge we must haveFrom Gauss’s Law,∇ · J = − ∂ρ∂t ≠ 0.∂ρ∂t = ∂ ( )∂t (ɛ ∂E0∇ · E) = ∇ · ɛ 0 ,∂tso adding in this term restores consistency.Ampere never noticed because his ∂E/∂t = 0.PHYS 301: Electromagnetism Electrodynamics 82 / 123

ElectrodynamicsMaxwell’s equationsThe displacement current IIAnother demonstration of its importance comes if we try to apply theintegral form of Ampere’s law to a charging capacitor:∮∫dB · dl = µ 0 I enc + µ 0 ɛ 0 E · ˆn da.dtCWithout the extra term, we don’t get a consistent answer for the magneticfield induced by the current.SPHYS 301: Electromagnetism Electrodynamics 83 / 123

ElectrodynamicsMaxwell’s equationsConsequences of the displacement currentWith this term in placewe predict the existence of electromagnetic waves (see week 5)wave speed matches the known speed of light.Light is electromagnetic in origin.transmission by radio waves discovered soon after by HertzPHYS 301: Electromagnetism Electrodynamics 84 / 123

Potentials in electrodynamicsPotentials in electrodynamicsIn electrodynamics, we have ∇ · B = 0 so B = ∇×A is still allowed.With dynamics, ∇×E ≠ 0 and we can’t guarantee that E = −∇V .It’s easy to see, however, that the following pair of potentials work:E = −∇V − ∂A∂tB = ∇×A.This is allowed, because from Faraday’s law we have0 = ∇×E + ∂B (∂t = ∇× E + ∂A ),∂tand thus E + ∂A∂tcan be written as a gradient.PHYS 301: Electromagnetism Electrodynamics 85 / 123

Potentials in electrodynamicsEvolution equations for the potentials L Griffiths: 10.1From ME, we find that the potentials satisfy∇ 2 V + ∂ ∂t (∇ · A) = − ρ ,ɛ 0(∇ 2 ∂ 2 ) ()∂V− µ 0 ɛ 0∂t 2 A − ∇ ∇ · A + µ 0 ɛ 0 = −µ 0 J.∂tThese equations look unpleasant, butwe’ve reduced 8 equations in 6 unknowns, to 4 equations in 4unknownswe have the source terms on the rightthere are elements on the left that look familiarwe can try to simplify them using gauge transformationsPHYS 301: Electromagnetism Electrodynamics 86 / 123

Potentials in electrodynamicsGauge transformations in electrodynamics L Griffiths: 10.1.2In statics, we saw that we could add a constant k 0 to V , and a gradient∇ψ to A, without changing the fields.Since V and A are coupled in dynamics, we can’t expect to change thepotentials independently.If we try the general transformationA ′ = A + α V ′ = V + β,and demand that the E and B fields are unchanged, we find that for anyscalar function λ(r, t), we can make the following gauge transformation:A ′ = A + ∇λV ′ = V − ∂λ∂t .The art is in choosing λ(r, t) appropriately so as to simplify the particularproblem. In practice, we usually don’t find λ explicitly. It’s enough toknow that we could find it if someone asked us to.PHYS 301: Electromagnetism Electrodynamics 87 / 123

Potentials in electrodynamicsThe Coulomb gauge L Griffiths: 10.1.3There are three gauges in common use. In the Coulomb gauge, we setdiv A = 0. (As in magnetostatics, this is always possible.) The gauge’sname follows from the equation for the scalar potential∇ 2 V = − ρ ɛ 0, with solution V (r, t) = 14πɛ 0∫Vwhich are just the familiar equations of electrostatics.(Though note that ρ can now depend on time.)The vector potential satisfies an ugly equation in this gauge:(∇ 2 ∂ 2 )− µ 0 ɛ 0∂t 2 A = −µ 0 J + µ 0 ɛ 0 ∇ ∂V∂t ,ρ(r ′ , t)|r − r ′ | dτ ′so it’s mainly used for quasi-static problems like the hydrogen atom.Note that V responds instaneously to distant changes in ρ in this gauge.The vector potential ensures that the fields remain consistent withrelativity.PHYS 301: Electromagnetism Electrodynamics 88 / 123

Potentials in electrodynamicsThe Lorentz gaugeIn the Lorentz gauge, we set∇ · A + µ 0 ɛ 0∂V∂t = 0.We find the potentials separate and are manifestly consistent withrelativity or Lorentz invariant(∇ 2 − µ 0 ɛ 0∂ 2)∂t 2 V = − ρ ɛ 0(∇ 2 ∂ 2 )− µ 0 ɛ 0∂t 2 A = −µ 0 J.The required λ for the Lorentz gauge satisfies an equation of the sameform: (∇ 2 ∂ 2 )∂V 0− µ 0 ɛ 0∂t 2 λ = ∇ · A 0 − µ 0 ɛ 0∂t ,which we can in principle solve.PHYS 301: Electromagnetism Electrodynamics 89 / 123

Potentials in electrodynamicsSolution to Lorentz gauge equationsWith a little work, we can show that plausible solutions for the potentialsin the Lorentz gauge areV = 1A = µ 04πwhere the retarded time t ′ r is4πɛ 0∫V∫Vρ(r ′ , t ′ r)|r − r ′ | dτ ′J(r ′ , t ′ r)|r − r ′ | dτ ′t ′ r = t − |r − r′ |.cThus the potentials at r feel the effect of charge and currents at spacetimelocations which lie on the “light cone” for r. That is, the influence frompoints r ′ appears to travel at the speed of light to affect the potentials atother locations at other times. (Such observations helped to lead Einstein toformulate special relativity. Through quantum field theory, we now understand this interms of the electromagnetic field being “mediated” by the exchange of photons. )PHYS 301: Electromagnetism Electrodynamics 90 / 123

Potentials in electrodynamicsThe temporal gaugeIn the temporal gauge, we set V ′ = 0, by choosingλ(r, t) =∫ t0V 0 (r, t ′ ) dt ′ ,so that (∇ 2 − µ 0 ɛ 0∂ 2∂t 2 )A − ∇ (∇ · A) = −µ 0 J.This is useful for solving radiation problems like Cerenkov emission orbremsstrahlung.PHYS 301: Electromagnetism Electrodynamics 91 / 123

Energy and momentum in the electromagnetic fieldReview of energy in electrostatic fields L Griffiths: 2.4For a distribution of point charges, we can think of energy as being storedin the charge,W = 1 ∑q i V (r i ),2or in the fields,W = ɛ 02i∫V|E| 2 dτ.The real value of the latter is the concept of an energy densityU e = ɛ 02 |E|2 ,which comes alive when we think about EM waves.PHYS 301: Electromagnetism Electrodynamics 92 / 123

Energy and momentum in the electromagnetic fieldReview of energy densities in staticsIn electrostatics, we saw that the energy of a charge distribution could beviewed in two complementary fashions:W = 1 ∫ρV dτ = ɛ ∫0|E| 2 dτ.22VThe first equality views the energy as being stored in the Coulombinteraction between the charges. The second equality places the energy inthe electrostatic field itself.Although we skipped it, one can also express the energy involved inassembling a current configuration in terms of the resulting magnetostaticfield asW = 1 |B| 2 dτ.2µ 0∫VThis energy is the work done by electric forces to overcome the back-emfthat opposes the creation of steady currents. (So the B is static, but theenergy is associated with a dynamical setup period. B forces do no work.)PHYS 301: Electromagnetism Electrodynamics 93 / 123V

Energy and momentum in the electromagnetic fieldEnergy in dynamics L Griffiths: 8.1.2From F = q(E + v×B), we find the instantaneous work done a charge qmoving at velocity v isdW = F · dl = q(E + v×B) · vdt = q dt E · v.Generalising q → ρdτ and ρv → J, we get∫dW= J · E dτ,dtso J · E is the work done on all charges per unit volume per unit time.VPHYS 301: Electromagnetism Electrodynamics 94 / 123

Energy and momentum in the electromagnetic fieldEnergy balance and the Poynting vectorConsider a volume V , boundary S, containing some charge (discrete orcontinuous) and fields (both from this charge and possibly externalsources).Using Maxwell’s equations and some vector identities we can reachdWdt= − d dt∫V12(ɛ 0 |E| 2 + 1 µ 0|B| 2 )−∮SE×Bµ 0· ˆn daThe first term on the right is the rate of energy extraction from the fieldsin the volume V . So the second must represent the energy entering thevolume through its boundary surface.The quantity S = (E×B)/µ 0 is called the Poynting vector and its surfaceintegral represents the flux of electromagnetic energy.The equation is called the Poynting theorem.PHYS 301: Electromagnetism Electrodynamics 95 / 123

Energy and momentum in the electromagnetic fieldDifferential Poynting theoremAs an integral equation, the Poynting theorem describes a rigorous energybalance: electromagnetic energy entering a volume must either add to theenergy stored in the fields or be converted to other forms by interactionwith the charges.It is very frequently useful to interpret the integrands as energy densitiesand flux densities. This does not have the same rigour and at times theinterpretation implies rather strange results, as we’ll see.The differential form of the Poynting theorem is∂∂t (u mech + u em ) = −∇ · S,where we label all non-electromagnetic energy of the charges asmechanical and defineu em = 1 (ɛ 0 |E| 2 + 1 )|B| 2 .2 µ 0PHYS 301: Electromagnetism Electrodynamics 96 / 123

Energy and momentum in the electromagnetic fieldFinal remarksWe tend to find similar continuity equations whenever a conservedquantity is involved:We met the charge continuity equation earlier.In fluid mechanics, conservation of mass leads to a density continuityequation,In quantum mechanics, there is a probability current that is conserved.For completeness, we note that as well as energy, we can associatemomentum with the electromagnetic fields, with the momentum densityp em = ɛ 0 (E×B) = µ 0 ɛ 0 S.Since momentum is a vector, the associated conservation equation involvesthe “Maxwell stress-energy tensor”, and is rather unpleasant.PHYS 301: Electromagnetism Electrodynamics 97 / 123

Energy and momentum in the electromagnetic fieldPoynting vector in DC electric circuitsApplying the Poynting theorem to DC circuits givessome surprising results.It’s easy to show thatThe energy leaves the battery in the normaldirection to the current flow.The energy enters a resistor in the normaldirection to the current flow.So how does the energy get from one to the other?PHYS 301: Electromagnetism Electrodynamics 98 / 123

Energy and momentum in the electromagnetic fieldPoynting vector in DC electric circuitsThe secret is that the surface of the wires iscovered in charge.To see this, think of an ideal wire attachedto a battery with no load. The wire is at thesame potential so the charge will spreadalong the wire due to repulsion.For an elegant account, see I. Galili and E. Goihbarg, Am. J. Phys. 73, 141 (2005)[http://dx.doi.org/10.1119/1.1819932] and references therein.PHYS 301: Electromagnetism Electrodynamics 99 / 123

Electromagnetic wavesReview of vacuum wave equation and solutionsReview of vacuum wave equation and solutionsWe return to Maxwell’s equations in free space: ρ = 0 and J = 0.It is easy to show that the electric and magnetic fields satisfy the 3Dscalar wave equation:(∇ 2 − µ 0 ɛ 0∂ 2∂t 2 )E = 0,(∇ 2 − µ 0 ɛ 0∂ 2∂t 2 )B = 0.(These are “scalar” wave equations since the components of the E and Bvectors are not coupled.)If we assume no variation in the x and y directions, then we must solve( ∂2∂z 2 − µ ∂ 2 )0ɛ 0∂t 2 E = 0,( ∂2∂z 2 − µ ∂ 2 )0ɛ 0∂t 2 B = 0.PHYS 301: Electromagnetism Electromagnetic waves 100 / 123

Electromagnetic wavesReview of vacuum wave equation and solutions1D wave solutionsIt is convenient to use the complex or “phasor”notation where the physical E and B are given byE = Re[Ẽ]and so find the plane wave solutionsB = Re[˜B],Ẽ = ˆxE 0 e i(kz−ωt) ˜B = B 0 e i(kz−ωt) ,where E 0 is the wave amplitude, ω = 2πν is the (angular) frequency, andthe wavenumber k = 2π/λ. From Faraday’s law we findB 0 = E 0c ŷ,so that the electric and magnetic field contributions to the wave areorthogonally polarized and travel in phase. EM waves in vacuum areapparently transverse.PHYS 301: Electromagnetism Electromagnetic waves 101 / 123

Electromagnetic wavesReview of vacuum wave equation and solutionsVelocity of EM wavesThe wave speed or phase velocity isv p = ω k = 1 √µ0 ɛ 0.We then observe that this speed coincides with the measured velocity oflight in vacuum, and thus we identify light (and radio etc) aselectromagnetic waves, and write c = 1/ √ µ 0 ɛ 0 .For any wave problem, a dispersion relation relates ω and k. For planewaves in vacuum, the dispersion relation is simplyω = c|k|.PHYS 301: Electromagnetism Electromagnetic waves 102 / 123

Electromagnetic wavesReview of vacuum wave equation and solutionsEM coupling constants in the SI system of unitsIn modern SI, we define c =299 792 458 ms −1 , and the second as9 192 631 770 periods of a certain caesium 133 atomic emission line.The metre is thus a derived unit.Since µ 0 = 4π × 10 −7 NA −2 exactly, and ɛ 0 = 1/(µ 0 c 2 ), ɛ 0 is also exactlydefined, and SI electromagnetism appears to contain no measuredconstants at all.But there must be at least one physical constant describing the couplingstrength of the electromagnetic field, (just asG ≈ 6.673 × 10 −11 m 3 kg −1 s −2 is a measured quantity describing thecoupling between mass and the gravitational force).For EM in SI, this coupling is hiding inside the derived value of the metre.PHYS 301: Electromagnetism Electromagnetic waves 103 / 123

Electromagnetic wavesReview of vacuum wave equation and solutionsGeneral form of solutionSince the waves are transverse, there is a second solution where E ispolarized along ŷ:Ẽ = ŷE 0 e i(kz−ωt) ˜B = − ˆx E 0c ei(kz−ωt) .For a general wave, we introduce the wavevector k = (k x , k y , k z ) and haveas one solutionẼ(x, t) = E 0 e i(k·x−ωt) , ˜B(x, t) = B 0 e i(k·x−ωt) .Writing k = |k| û = k û, we findk · E 0 = 0 and B 0 = 1 c û×E 0,and so k, E and B form a right-handed set.The dispersion relation is ω = c|k| = c|k|.(For the other polarisation state, we can use E 1 = û×E 0 .)PHYS 301: Electromagnetism Electromagnetic waves 104 / 123

Electromagnetic wavesEnergy and momentum in EM wavesEnergy flow in wave solutionsFinding the Poynting vector is a little tricky because we need the crossproduct of the real fields, not the complex ones:S(x, t) = 1 µ 0E×B= 1 Ẽ + Ẽ ∗× ˜B + ˜B ∗µ 0 2 2= û ɛ 0c|E 0 | 2[1 + cos[2(k · x − ωt)]] .2Thus the energy flow is parallel to the wavevector but contains anoscillatory term at twice the wave frequency. (Because the energytransport is largest at the wave anti-nodes).In practice, we usually care about the time-averaged Poynting vector〈S(x)〉 ≡ 1 T∫ T0S(x, t ′ ) dt ′ = û ɛ 0c|E 0 | 2.2PHYS 301: Electromagnetism Electromagnetic waves 105 / 123

Electromagnetic wavesEnergy and momentum in EM wavesEnergy density in wave solutionsSimilarly, since|E| 2 = Ẽ + Ẽ∗2· Ẽ + Ẽ∗2= |E 0| 2[1 + cos(2(k · x − ωt))],2the energy density of a plane wave contains oscillatory terms.Averaging over space and time, the mean energy density isu em = ɛ 02 |E|2 + 1 |B| 22µ[ 0ɛ0=4 |E 0| 2 + 1 ]4µ 0 c 2 |E 0| 2 {1 + cos[2(k · x − ωt)]}= ɛ 02 |E 0| 2 {1 + cos[2(k · x − ωt)]} ,and so energy is distributed evenly between the fields.PHYS 301: Electromagnetism Electromagnetic waves 106 / 123

Electromagnetic wavesEnergy and momentum in EM wavesOther wave quantitiesThe irradiance (or intensity) I is the magnitude of the time-averagedPoynting vector:I = 〈|S(x)|〉 = ɛ 0c|E 0 | 2,2and the time-averaged momentum isExample〈p em 〉 = 〈S(x)〉c 2 = û ɛ 0|E 0 | 2.2cFind the pressure exerted by sunlight (I ≈ 1300 Wm −2 ) on a black surfaceand a white surface.PHYS 301: Electromagnetism Electromagnetic waves 107 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsEM waves in lossless dielectricsWe can characterise a low-loss dielectric (like fused silica glass) by apermittivity function ɛ ≠ ɛ 0 , with D = ɛE.Only Ampere’s law changes, and the wave equation becomes)(∇ 2 − µ 0 ɛ ∂2∂t 2 E = 0.The plane wave solutions are unchanged, but now the wave speed isv p = 1 √µ0 ɛ = c nwith dispersion relation ω = ck/n, where n = √ ɛ/ɛ 0 is the refractiveindex.At optical frequencies, typical values of n are 1.33 (water), 1.445 (fused silica opticalfibres) and 3.4 (GaAs). At microwave frequencies, n ≈5–9 is not uncommon.Note that n < 1 is perfectly acceptable and exhibited by all materials at certainfrequencies. This is not in conflict with special relativity, since signals do not travelat the phase velocity v p, and the signal velocity is always safely below c.PHYS 301: Electromagnetism Electromagnetic waves 108 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsEM waves in conductorsMany materials (not just metals) are well described as conductors withJ = σE for a conductivity σ. Ampere’s law and the wave equationrespectively become:∇×B = µ 0 σE + µ 0 ɛ ∂E∂t ,∇ 2 E = µ 0 ɛ ∂2 E∂t 2+ µ 0σ ∂E∂t .The waves still have the same form, but the wavenumber ˜k = k + iκ isnow complex:with dispersion relationẼ = E 0 e i(˜kz−ωt) ˜B = B 0 e i(˜kz−ωt) ,˜k 2 = (k + iκ) 2 = µ 0 ɛω 2 + iµ 0 σω.PHYS 301: Electromagnetism Electromagnetic waves 109 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsProperties of waves in conductorsThe imaginary part of the wavevector κ leads to anattenuation or damping of the wave as itpropagates:Ẽ(z) = e −κz E 0 e i(kz−ωt) .This is associated with the electric field doing work on the charges in thematerial. The decay length d = 1/κ is called the skin depth.If we write ˜k = Ke iφ , and take E 0 = E 0 ˆx, we findB 0 = ˜kω E 0 ŷ= e iφ K ω E 0 ŷ,so while the fields remain orthogonal, they are no longer in phase.The B field lags behind by a phase factor φ = tan −1 (κ/k).PHYS 301: Electromagnetism Electromagnetic waves 110 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsWave dispersion in conductorsThe dispersion relation has the solution√ [√] 1µ0 ɛ( σ) √ [√] 122µ0 ɛ( σ) 22k = ω 1 + + 1 , κ = ω 1 + − 1 .2 ɛω2 ɛωThe phase velocity isv p =ωRe[˜k] = ω √ [√]2( σ) −122k = 1 + + 1 ,µ 0 ɛ ɛωso these waves are dispersive—the velocity depends on frequency.The ratio B 0 /E 0 = K/ω is also frequency dependent and the magneticwave now carries more energy than the electric wave.PHYS 301: Electromagnetism Electromagnetic waves 111 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsDispersive behaviourThese plots show the frequency dependence of various parameters for wavepropagation in a conductor with ɛ = 1.5.Seeking the asymptotic behavior givesGood conductor (σ ≫ ɛω):d → √ 2/(µ 0 σω) = λ loc /(2π)φ → π/4v = √ 2ω/(µ 0 σ)Poor conductor (σ ≪ ɛω):d → (2/σ) √ ɛ/µ 0φ → σ/(2ɛω)v → c/n (independent ofconductivity)PHYS 301: Electromagnetism Electromagnetic waves 112 / 123

Electromagnetic wavesEM waves in dielectrics and conductorsRemarks on dispersionHere, dispersion has arisen even though σ and ɛ were taken constant. Inreal materials, ɛ itself varies with frequency, so all materials are dispersive,even if their conductivity is negligible.A familiar example is splitting white light into component colours with aprism, which occurs because ɛ increases as the wavelength decreasestowards the UV.In fact, a dispersive ɛ is a direct consequence of the absorption/emission oflight by atoms and the requirement of causality, a very deep result knownas Kramers-Kronig.Thus absorption (loss) and dispersion are intimately connected. Like “Loveand Marriage”, “You can’t have one without the other.” 33 1955 Sinatra song by Cahn & Van Heusen.PHYS 301: Electromagnetism Electromagnetic waves 113 / 123

Emission of electromagnetic wavesDipole radiationOrigin of EM wavesEM radiation is produced by virtually any time-dependent chargeconfiguration. We will consider electric dipole radiation—the generation offar field EM waves by a harmonically-oscillating dipole.Suppose two spheres are connected by a fine conducting wire orientedalong z, and that a charge q oscillates between them with q = q(t).Associated with the charge oscillation is a currentI = dqdt ẑ.PHYS 301: Electromagnetism Radiation 114 / 123

Emission of electromagnetic wavesDipole radiationPotentials of oscillating dipoleWe work out the potentials in the far field using the Lorentz gauge.For the vector potential, we haveA(r) = µ ∫0 J(r ′ , t ′ r)4π |r − r ′ | dτ ′= µ 04πV∫ l/2−l/2I(z ′ , t − |r − z ′ ẑ|/c)|r − z ′ ẑ|We expand the factor |r − z ′ ẑ| for large r asdz ′ .|r − z ′ ẑ| = [ r · r − 2z ′ r · ẑ + z ′2] 1 2≈ r − z ′ cos θ,and so assuming r ≫ l (Approx. 1) we write the denominator as1|r − z ′ ẑ| ≈ 1 r .PHYS 301: Electromagnetism Radiation 115 / 123

Emission of electromagnetic wavesDipole radiationVector potential of oscillating dipoleThe current term I(z ′ , t − |r − z ′ k|/c) ≈ I(z ′ , t − r/c + z ′ cos θ/c).In order that we can take I(τ) ≈ I(τ + z ′ cos θ/c), we assumez ′ cos θ/c ≪ T , where T is the oscillation period. This is equivalent toλ ≫ l , (Approx. 2)2ie. the wavelength of emitted radiation is much longer than the dipole size.The vector potential is thusA(r) = µ ∫ l/204π ẑ I(z ′ , t − r/c)dz ′−l/2 r= µ 0l4πI(t − r/c)rẑPHYS 301: Electromagnetism Radiation 116 / 123

Emission of electromagnetic wavesDipole radiationScalar potential of oscillating dipoleWe could work out V (r, t) in the same way, but it’s easier to use theLorentz condition:∂V∂t = −c2 ∇ · A= − c2 µ 0 l4π∂ I(t − r/c)∂z rand since ∫ t0 I(t′ ) dt ′ = q(t), we findV (r, t) == l4πɛ 0zr 2 [ 1r I(t − r/c) − 1 c I′ (t − r/c)],l [z 14πɛ 0 r 2 r q(t − r/c) − 1 ]c I(t − r/c) .PHYS 301: Electromagnetism Radiation 117 / 123

Emission of electromagnetic wavesDipole radiationPotentials and fields for harmonic oscillationWe choose a harmonic oscillation q = q 0 cos ωt, and the final potentials areA = − µ 0ωp 04πr= − µ 0ωp 0V =sin [ω(t − r/c)] ẑ(4πr sin [ω(t − r/c)]cos θ ˆr − sin θ ˆθ)p [0z 14πɛ 0 r 2 r cos [ω(t − r/c)] − ω ]c sin [ω(t − r/c)] .where the dipole moment p 0 = q 0 l.From B = curl A, we eventually getB = − ˆφ µ 0p 0 ω sin θ4πrwith terms going as r −1 and r −2 .[ ωc cos [ω(t − r/c)] + 1 r sin [ω(t − r/c)] ],PHYS 301: Electromagnetism Radiation 118 / 123

Emission of electromagnetic wavesDipole radiationThe electric fieldWe find the electric field from E = −∇V − ∂A/∂t.The algebra is straight-forward but messy and yieldsE r = − 2ωp [0 cos θ sin [ω(t − r/c)]4πɛ 0 r 2 −cE θ = ωp [(0 sin θ1cos [ω(t − r/c)]4πɛ 0 ωr 3 − ω )rc 2 −]cos [ω(t − r/c)]ωr 3E φ = 0],sin [ω(t − r/c)]r 2 cwith terms going as r −1 , r −2 and r −3 . We consider the radiation zonewhere r ≫ c/ω = λ/(2π) (Approx 3), so thatB = − ˆφ µ 0p 0 ω 2 sin θ4πrccos [ω(t − r/c)],E = − ˆθ p 0ω 2 sin θ4πɛ 0 rc 2cos [ω(t − r/c)].PHYS 301: Electromagnetism Radiation 119 / 123

Emission of electromagnetic wavesDipole radiationFar-field behaviourThus in the far-field, the fields propagate outwards and in phase, andB = ˆr×E ,cwhich is exactly the relation required of a plane wave.In fact, the 1/r factor makes these spherical waves and ensures that theenergy flux through any sphere of radius r is conserved. At large r, thewaves tend to plane waves.PHYS 301: Electromagnetism Radiation 120 / 123

Emission of electromagnetic wavesDipole radiationEnergy in the far-fieldThe Poynting vector in the radiation zone isS = p2 0(4π) 2 ɛ 0ω 4r 2 c 3 sin2 θ cos 2 [ω(t − r/c)].Note that due to the sin 2 θ factor, the emission is strongest in the planeperpendicular to the dipole and vanishes along the line of the dipole.Integrating the time-averaged S over the whole sphere, we find the totalpower flux to be 〈∮ 〉P = S · da = µ 0p 2 0 ω412πc .SThe power radiated thus increases very strongly with frequency.PHYS 301: Electromagnetism Radiation 121 / 123

Emission of electromagnetic wavesDipole radiationConsequences of the far-field behaviourThis description of dipole radiation explains two obvious factors about thedaytime sky—it is blue, and it is partially polarized. The polarization iscomplete in a band at 90 ◦ to the sun and the sense of polarization isperpendicular to the plane containing the sun, the observer and the sourcepoint in the sky.Incoming light from the sun is scattered by air molecules. Scatteringconsists of the oscillation of molecules driven by radiation from the sun,and the subsequent emission of radiation by these oscillating dipoles. Theω 4 factor in the emission rate ensures that light at the blue end of thevisible spectrum is scattered much more efficiently than at the red end.Since EM waves are transverse, the scattering molecules oscillate in theplane perpendicular to the incoming wavevector. Since they can’t emitalong the line to the observer, at 90 ◦ to the sun, only one polarization canreach the observer.PHYS 301: Electromagnetism Radiation 122 / 123

Emission of electromagnetic wavesDipole radiationNear-field behaviourThe higher order terms in r −2 and r −3 play a role in the near-field.Since the integral over the Poynting vector for these terms over a wholesphere must vanish, there is no outward energy flux associated with theseterms, but they still oscillate with the dipole.Thus the field configuration there is more complicated and lookssomething like a “pulsing” static dipole field.PHYS 301: Electromagnetism Radiation 123 / 123