Ð ØÖÓÑ Ò Ø Ø ÈÀ ˶ ¼ ¼Ä ØÙÖ ÒÓØ × ´ÂÙÐÝ ¾¼¼ µ 8 ÓÖÝ 6 4 2 0 -2 -4 -6 -8 -8 -6 -4 -2 0 rotating dipole: t = 2 2 4 6 8 Ö ÈÓ ××ÓÒ Ô ÖØÑ ÒØ Ó È Ý× × ÍÒ Ú Ö× ØÝ Ó Ù ÐÔ Contents 1 Maxwell’s electrodynamics 1.1 Dynamical variables of electromagnetism 1.2 Maxwell’s equations and the Lorentz force 1.3 Conservation of charge 1.4 Conservation of energy 1.5 Conservation of momentum 1.6 Junction conditions 1.7 Potentials 1.8 Green’s function for Poisson’s equation 1.9 Green’s function for the wave equation 1.10 Summary 1.11 Problems 2 Electrostatics 2.1 Equations of electrostatics 2.2 Point charge 2.3 Dipole 2.4 Boundary-value problems: Green’s theorem 2.5 Laplace’s equation in spherical coordinates 2.6 Green’s function in the absence of boundaries 2.7 Dirichlet Green’s function for a spherical inner boundary 2.8 Point charge outside a grounded, spherical conductor 2.9 Ring of charge outside a grounded, spherical conductor 2.10 Multipole expansion of the electric field 2.11 Multipolar fields 2.12 Problems 3 Magnetostatics 3.1 Equations of magnetostatics 3.2 Circular current loop 3.3 Spinning charged sphere 3.4 Multipole expansion of the magnetic field 3.5 Problems 4 Electromagnetic waves in matter 4.1 Macroscopic form of Maxwell’s equations 4.1.1 Microscopic and macroscopic quantities 4.1.2 Macroscopic smoothing 4.1.3 Macroscopic averaging of the charge density 4.1.4 Macroscopic averaging of the current density 4.1.5 Summary — macroscopic Maxwell equations 4.2 Maxwell’s equations in the frequency domain 4.3 Dielectric constant i 1 1 3 4 4 6 7 10 12 14 16 17 21 21 21 22 24 28 30 34 35 38 40 42 45 47 47 47 49 51 53 55 55 55 56 57 60 61 62 63 5 Summary — oscillating dipole 5.4 Radiated power (magnetic-dipole radiation) 5.ii 4.7 Angular integrations 5.1 Plane versus spherical waves 5.3.4.4.1 Slow-motion approximation.6 Total radiated power 5.4.6 5 Electromagnetic radiation from slowly moving sources 5.5 Radiated power (electric-quadrupole radiation) 5.9 Problems 6 Electrodynamics of point charges 6.5.5 Centre-fed linear antenna 5.2 Charge-conservation identities 5.1 Angular profile of radiated power 6.5 Propagation of plane.3 Uniform motion 6.6 Problems .5 Radiation reaction 6.4 Summary 6.6 Extremely relativistic motion 6.2 Fields 6.3.3 Spherical waves — the oscillating dipole 5.3.7.2 Potentials of an oscillating dipole 5.7.4 Wave-zone fields 5.4.3 Wave-zone fields of an oscillating dipole 5.7.3.5.4.3 Linear motion 6.4.4 4.2 Plane waves 5.3 Fields of an arbitrarily moving charge 6.2 Fields of a uniformly moving charge 6.5.1 Equations of electrodynamics 5.6 Summary — electric dipole radiation 5. monochromatic waves Propagation of wave packets 4.4 Electric dipole radiation 5.3 Propagation with dispersion 4.4 Circular motion 6.7.4.4 Poynting vector of an oscillating dipole 5.3.4.2 Slow motion: Larmor formula 6.1 Potentials 6.1 Lorentz transformations 6. near and wave zones 5.3.6 Classical atom 5.4.1 Description of wave packets 4.4.7.2 Scalar potential 5.3.3 Vector potential in the wave zone 5.3.5.8 Pulsar spin-down 5.2 Propagation without dispersion 4.4.4 Radiation from an accelerated charge 6.4 Gaussian wave packet Problems Contents 65 68 68 69 69 71 73 75 75 76 78 78 79 81 83 84 84 85 85 87 88 90 90 91 94 95 96 98 99 100 101 102 103 103 105 109 109 112 114 115 116 120 121 122 122 123 124 126 127 131 132 135 4.5 Energy radiated 5.4.3 Vector potential 5.1 Wave-zone fields 5.7 Magnetic-dipole and electric-quadrupole radiation 5.7.7.3.5 Synchrotron radiation 6. It moves with a velocity v that depends on time and on position. The current density j is defined by the statement j · da ≡ current crossing an element of area da.1.1 Dynamical variables of electromagnetism The classical theory of the electromagnetic field. The dynamical system is therefore much more complicated than in mechanics.1. (1. An element of charge is a macroscopically small (but microscopically large) portion of matter that contains a net charge. (1. Although this is not a true picture of reality.1) We shall now establish the important relation j = ρv .2) We have two vectors at each position of space and at each moment of time.Chapter 1 Maxwell’s electrodynamics 1. these continuous distributions fit naturally within a classical treatment of electrodynamics.(1. Here the number of degrees of freedom is infinite. but we shall see that the formalism is robust enough to also allow for a description in terms of point particles. The quantities that describe the charge and current distributions are ρ(t. an element of charge at position x is idealized as the point x itself. x) j (t. as formulated by Maxwell. An element of charge is located at position x and has a volume dV . x) ≡ magnetic field at position x and time t.1.6) .3) ≡ velocity of an element of charge at position x and time t. (1. x) ≡ density of charge at position x and time t.1. In the mathematical description. In a classical theory these are best described in terms of a fluid picture in which the charge and current distributions are imagined to be continuous (and not made of pointlike charge carriers). but sufficiently small that the density of charge within the volume is uniform to a high degree of accuracy. The electric and magnetic fields are produced by charges and currents. 1 (1.1.1.4) ≡ current density at position x and time t. The volume of a charge element must be sufficiently large that it contains a macroscopic number of elementary charges. x) ≡ electric field at position x and time t and B (t.5) (1.7) (1.1. in which there is a finite number of degrees of freedom. involves the vector fields E (t. x) v (t. This will be our point of view here. Its charge density can be written as ρ(x) = qδ (x − r ). current is defined as the quantity of charge crossing a surface per unit time. where it is infinite.1 shows a number of charged particles approaching with speed v a small surface of area ∆a. As was mentioned previously. Figure 1. The particles within the dashed box will all hit the surface within a time ∆t.1: Point charges hitting a small surface. Comparing this with the expression given previously. The current flowing across a surface S is then j · da. ρv is the current density. the density is zero everywhere except at the position of a charge. then j (x) = q v δ (x − r ) . current = j · ∆x. there is an angle θ between the direction of the velocity vector v and the normal n ˆ to the surface. If the charge is moving with velocity v . Suppose that we have a point charge q at a position r . This situation can be described by a δ -function. For these. Let us then calculate the current associated with a distribution of charge with density ρ and velocity v . The volume of this box is (v ∆t cos θ)∆a = v · n ˆ ∆t∆a. S On the other hand. the fluid description still allows for the existence of point charges. where δ (x) ≡ δ (x)δ (y )δ (z ) is a three-dimensional δ -function.2 Maxwell’s electrodynamics θ (v∆ t )cos θ Figure 1. The current crossing the surface is then calculated as current = = = (charge crossing the surface)/∆t volume of box 1 (total charge) total volume ∆t (density)(volume of box)/∆t = ρ(v · n ˆ ∆t∆a)/∆t = ρv · (n ˆ ∆a) = ρv · ∆ a . we find that indeed. so that j is the current per unit area. The statement of the Lorentzforce law is then f = ρE + j × B = ρ E + v × B . x) = A qA δ x − rA (t) (1.2 Maxwell’s equations and the Lorentz force The four Maxwell equations determine the electromagnetic field once the charge and current distributions are specified. Maxwell’s equations can also be presented in integral form.1.1. and ∇ = (∇x .2. while the magnetic field is produced by currents and time-varying electric fields. Let f (t. ∂t (1. x) over a volume VA that encloses this charge but no other: qA = VA ρ(t. The Lorentz-force law determines the motion of the charges once the electromagnetic field is specified. x) d3 x.7) .2. let us have a collection of point charges qA at positions rA (t).4) Here.10) where d3 x ≡ dxdydz .6) The net force F acting on a volume V of the charge distribution is the integral of the force density over this volume: F (t.2. (1.2. (1. ǫ0 = 0. The Maxwell equations state that the electric field is produced by charges and time-varying magnetic fields. ∇z ) is the gradient operator of vector calculus (with the obvious notation ∇x f = ∂f /∂x for any function f ). The total charge of the distribution is obtained by integrating the density over all space: Q≡ ρ(t.1.2. x) d3 x = A qA .1. x) ≡ force density at position x and time t.2 Maxwell’s equations and the Lorentz force 3 is the current density. Then the charge and current densities of the charge distribution are given by ρ(t.1) (1.8) and j (t.9) Each qA is the integral of ρ(t. They are given by ∇·E ∇·B ∇×E ∇×B 1 ρ.3) (1. by invoking the Gauss and Stokes theorems of vector calculus. (1. = − ∂t = = µ0 j + ǫ0 µ0 ∂E . moving with velocities vA (t) = drA /dt.5) where the force density is defined to be the net force acting on an element of charge at x divided by the volume of this charge element. ǫ0 and µ0 are constants.2.1. (1. x) d3 x. (1.11) 1. (1. ∇y .2. ∂B . More generally. x) = A qA vA (t)δ x − rA (t) .2) (1. V ) = V f (t. 1. Every conceivable phenomenon involving electromagnetism can be predicted from them. We may use Gauss’ theorem to turn Eq.2) ∂t This is the differential statement of local charge conservation. and the evolution of the electric and magnetic fields.1) with respect to time to obtain ∂ρ ∂E = ǫ0 ∇ · . This is the usual expression for the Lorentz force.7 of Jackson’s text.2). ∂t µ0 The first term on the right-hand side vanishes identically (the divergence of a curl is always zero) and we arrive at Eq.2.2. Consider a volume V bounded by a two-dimensional surface S .1) can thus be written as V ∇ · j d3 x.3.1) Local charge conservation means that this statement must be true for any volume V . but the definition of Eq. ∂t ∂t where we have exchanged the order with which we take derivatives of the electric field. while the left-hand side is equal to − V (∂ρ/∂t) d3 x. (1. (1.4 Maxwell’s electrodynamics For a single point charge we have ρ = qδ (x − r ).3.] . Charge conservation means that the rate of decrease of charge within V must be equal to the total current flowing across S : − d dt ρ d3 x = V S j · da. To establish this we first differentiate Eq. Taken together. and the total force becomes F = q E (r ) + v × B (r ) . (1.3. 6. (1. For any smooth vector field u within V we have ∇ · u d3 x = u · da. j = q v δ (x − r ). (1. Those five equations summarize the complete theory of classical electrodynamics.1) into a differential statement.2.3 Conservation of charge One of the most fundamental consequences of Maxwell’s equations is that charge is locally conserved: charge can move around but it cannot be destroyed nor created. Equality of the two sides for arbitrary volumes V implies ∂ρ + ∇ · j = 0.4 Conservation of energy [The material presented in this section is also covered in Sec. and we would like to prove that this comes as a consequence of Maxwell’s equations. the Maxwell equations and the Lorentz-force law determine the behaviour of the charge and current distributions. we get 1 ∂ρ = ∇ · (∇ × B ) − ∇ · j .3. If we now use Eq. (1. (1. however small or large. Here the fields are evaluated at the charge’s position. V S The right-hand side of Eq. We have therefore established that local charge conservation is a consequence of Maxwell’s equations. 1.4) to eliminate the electric field.6) is more general.3. (1. which have not yet been involved.2) So ε is analogous to charge density. Integrating this over V gives the total work done on the charges contained in V . ≡ field energy crossing an element of surface da per unit time. In this section we formulate and derive a statement of energy conservation. This energy goes to the charge distribution.4) This last equation must be derivable from Maxwell’s equations. We start with the right-hand side of Eq. per unit time. It moves with a velocity v and the net force acting on it is f dV = ρ(E + v × B ) dV. Let ε(t. the force does a quantity of work equal to f · dx dV . The work done per unit time is then f · v dV = ρE · v dV = j · E dV . A statement of local energy conservation can be patterned after our previous statement of charge conservation. (1.4. In the next section we will consider the conservation of linear momentum.4. but Eq. x) S (t. µ0 ∂t . but we shall not pursue this here. − dt V S which is analogous to Eq.1) (1. (1.3. It is also possible to prove conservation of angular momentum.4) has not yet been derived nor the the quantities ε and S properly defined in terms of field variables. (1. is analogous to current density. and Maxwell’s electrodynamics must be compatible with it. and this statement is not correct.4.3) − dt V S V In differential form.4. This gives −j · E = − ∂E 1 E · (∇ × B ) + ǫ0 E · . the derivation should provide expressions for the quantities ε and S .4). and S (which is known as Poynting’s vector).4) and eliminate j in favour of field variables using Eq. this is ∂ε + ∇ · S = −j · E . and this takes energy away from the field. (1. (1. While the element undergoes a displacement dx. x) · da ≡ electromagnetic field energy density.4. and total energy is conserved. ∂t (1.1. (1.4 Conservation of energy 5 Other conservation statements follow from Maxwell’s equations and the Lorentzforce law.4. The reason is that the field does work on the charge distribution. Conservation of energy is one of the most fundamental principle of physics. we would write the statement d ε d3 x = S · da. Now the hard work begins.2. A correct statement of energy conservation would therefore be (field energy leaving V per unit time) = (field energy crossing S per unit time) + (work done on charges within V per unit time). But we should not expect field energy to be conserved.1). To figure out what the work term should be. The correct statement of energy conservation must therefore have the form d ε d3 x = S · da + j · E d3 x. consider an element of charge within V . If electromagnetic field energy were locally conserved. And indeed. What we have at this stage is an educated guess for a correct statement of energy conservation. 5. ε≡ 1 1 ǫ0 E 2 + B2.5 Conservation of momentum [The material presented in this section is also covered in Sec. All of this gives us −j · E = = ∂B ∂E 1 1 B· + ǫ0 E · + ∇ · (E × B ) µ0 ∂t ∂t µ0 1 1 ∂ 1 ǫ0 E 2 + B2 + ∇ · E×B . For momentum conservation we will need a vectorial quantity εa to represent the density of momentum. 1. a vectorial quantity Sa representing the flux of energy. and a tensorial quantity Tab to represent the flux of momentum (one index for the momentum component. and in this we replace ∇ × E with −∂ B /∂t.7 of Jackson’s text. Because a rate of change of momentum is a force. ∂t We have introduced an explicit component notation for vectors. 6. 2 2µ0 (1.] We should expect a statement of momentum conservation to take a form similar to our previous statement of energy conservation. µ0 (1.6) It is an important consequence of Maxwell’s theory that the electromagnetic field carries its own energy. another index for the flux direction). (1. So let εa (t. x) −Tab (t. x) dab ≡ density of a-component of field momentum. (1. ∂t 2 2µ0 µ0 We have arrived at an equation of the same form as Eq.3). In that case we had a scalar quantity ε representing the density of energy. ∂t (1.1) (1. ≡ a-component of field momentum crossing an element of surface da per unit time.5.6 Maxwell’s electrodynamics We use a vector-calculus identity to replace E ·(∇×B ) with B ·(∇×E )−∇·(E ×B ).2. and the conservation statement took the form of ∂ε + ∇a Sa = work term.5) and for the field’s energy flux (Poynting) vector.4. (1. and this shows that we do indeed have energy conservation as a consequence of Maxwell’s equations and the Lorentz-force law. using Eq. S≡ 1 E × B. (1. The calculation also gives us definitions for the field’s energy density.4) .4).3) V or the equivalent differential statement − ∂εa + ∇b Tab = fa = ρEa + (j × B )a .5.4.4. The statement of momentum conservation is (field momentum leaving V per unit time) = (field momentum crossing S per unit time) + (rate at which momentum is communicated to the charges). we have the integral statement − d dt εa d 3 x = − Tab dab + S V fa d3 x. and summation over a repeated index is understood.5.2) The minus sign in front of Tab is introduced by convention. 5. the right-hand side reads 1 (∇b Eb )Ea + (Eb ∇b )Ea − ∇a E 2 2 1 = Ea ∇b Eb + Eb ∇b Ea − δab ∇b E 2 2 1 2 = ∇b Ea Eb − δab E . and that these quantities are related by a constant factor of ǫ0 µ0 .4): f = (ǫ0 ∇ · E )E + ∂E 1 ×B ∇ × B − ǫ0 µ0 ∂t 1 1 (∇ · B )B + (∇ × B ) × B = ǫ0 (∇ · E )E + µ0 µ0 ∂ ∂B − ǫ0 (E × B ) + ǫ0 E × . 2 Doing the same for the bracketed terms involving the magnetic field. where u stands for any vector field. this gives f = −ǫ0 ∂ 1 (E × B ) + ǫ0 (∇ · E )E − E × (∇ × E ) + (∇ · B )B − B × (∇ × B ) .5. and discover the identities of the vector εa and the tensor Tab .3) to replace ∂ B /∂t with −∇ × E . The interface might be just a mathematical boundary (in . Our calculation also tells us that the field’s momentum density is given by εa ≡ ǫ0 (E × B )a .2. This means that the field’s momentum points in the same direction as the flow of energy. 1. µ0 2 This has the form of Eq. Collecting terms.4) to eliminate ρ and j from Eq. ∂t ∂t where to go from the first to the second line we have inserted a term involving ∇ · B = 0 and allowed the time derivative to operate on E × B .5) Notice that the momentum density is proportional to the Poynting vector: ε = ǫ0 µ0 S . We use this to clean up the quantities within square brackets. µ0 2 (1.6) (1. For example. (1. and that the field’s momentum flux tensor is 1 Tab = ǫ0 Ea Eb − δab E 2 2 + 1 1 Ba Bb − δab B 2 .2. 1 (∇ · E )E − E × (∇ × E ) = (∇ · E )E + (E · ∇)E − ∇(E · E ).5.4) and we conclude that momentum conservation does indeed follow from Maxwell’s equations and the Lorentz-force law.6 Junction conditions [The material presented in this section is also covered in Sec.5 of Jackson’s text. (1. I. The next step is to use Eq. (1.1. We first use Eqs.2.6 Junction conditions 7 We now would like to derive this from Maxwell’s equations. (1.1) and (1. ∂t µ0 1 To proceed we invoke the vector-calculus identity 2 ∇(u · u) = (u · ∇)u + u × (∇ × u). we arrive at fa = −ǫ0 1 ∂ (E × B )a + ǫ0 ∇b Ea Eb − δab E 2 ∂t 2 + 1 1 ∇b Ba Bb − δab B 2 . 2 In components.5.] Maxwell’s equations determine how the electric and magnetic fields must be joined at an interface. then the current density must have the form j (t.6. These distributions (also known as generalized functions) are related by the identities d θ(z ) = δ (z ). (1. these relations can be established by integrating both sides against a test function f (z ).3) where j+ is the current density above the interface.4) into Eqs. (Test functions are required to be smooth and to fall off sufficiently rapidly as z → ±∞.1)–(1. j− the current density below the interface.2.6. You can show similarly that f (z )δ (z ) = f (0)δ (z ) and f (z )δ ′ (z ) = f (0)δ ′ (z ) − f ′ (0)δ (z ) are valid distributional identities (a prime indicates differentiation with respect to z ). Differentiating the coefficients of θ(±z ) is straightforward.6. x.1) and (1. x)θ(−z ) + K (t. and σ is the surface density of charge (charge per unit surface area) supported by the interface. and this shows that dθ(z )/dz is indeed distributionally equal to δ (z ).6. x) = ρ+ (t. and K is the surface density of current (charge per unit time and unit length) supported by the interface. If the interface at z = 0 also supports a current distribution. . y )δ (z ). x) = E+ (t.3).6. x)θ(−z ). flat surface. dz d θ(−z ) = −δ (z ). We can express the fields as E (t. and the Dirac δ -function. x. = B+ (t. then the charge density must have the form ρ(t. for example. To prove this we need simply substitute Eq.2) As for any distributional identity. To begin. which is such that f (z )δ (z ) dz = f (0) for any smooth function f (z ). x) = j+ (t.2) and write ∇θ(±z ) = ±δ (z )z ˆ.) For example. or it might support a surface layer of charge and/or current. x) B (t. ∞ −∞ dθ(z ) f (z ) dz dz ∞ = − = − θ(z ) −∞ ∞ df (z ) dz dz df 0 = f (0). dz (1. we keep things simple by supposing that the interface is the x-y plane at z = 0 — a nice.2. If the interface supports a charge distribution.4). x)θ(z ) + j− (t. which is equal to 1 if z > 0 and 0 otherwise. x)θ(z ) + E− (t. Our task in this section is to formulate these junction conditions. ρ− the charge density below the interface (in the region z < 0). (1. but we must also differentiate the step functions.6. For this we use Eq. x)θ(z ) + B− (t. x)θ(z ) + ρ− (t.8 Maxwell’s electrodynamics which case nothing special should happen). x)θ(−z ) + σ (t. We now would like to determine how E and B behave across the interface. (1. (1. x)θ(−z ). E+ is the electric field above the interface. (1.1) where ρ+ is the charge density above the interface (in the region z > 0). We have introduced the Heaviside step function θ(z ). (1. We will see that these fields are solutions to Maxwell’s equations with sources given by Eqs.6.4) in an obvious notation. (1.6. y )δ (z ). 4) and the expression for ∇ × E into the third of Maxwell’s equations. ǫ0 Substituting the expression for ∇ · B into the second of Maxwell’s equations. Substituting Eq.6. where s is the distance from the interface in the direction normal to the interface. ∇ × B = µ0 j + ǫ0 µ0 ∂ E /∂t.6.6. ∇ · E = ρ/ǫ0 .9) (1. It is understood that the normal vector points from the minus side to the plus side of the interface.1) and the expression for ∇ · E into the first of Maxwell’s equations.6. (1.6.4). ∇ × E = −∂ B /∂t. this is positive above the interface and negative below the interface.5)–(1. (1. reveals that B± satisfies this equation on both sides of the interface and that the normal component of the magnetic field must be continuous: z ˆ · B+ − B− z =0 = 0.8) tell us how the fields E± and B± are to be joined at a planar interface. A straightforward calculation gives ∇·E = = = = ∇ · E+ θ(z ) + ∇ · E− θ(−z ) + z ˆ · E+ − E− δ (z ). substituting Eqs. n ˆ × B+ − B− Here. (1. then the general junction conditions are n ˆ · E+ − E− n ˆ × E+ − E− n ˆ · B+ − B− = = 1 σ. the interface’s unit normal. and the expression for ∇ × B into the fourth of Maxwell’s equations.6.5) z ˆ · E+ − E− z=0 = σ. (1. (1. ǫ0 0.11) (1.6. replaces z ˆ in the preceding equations.6. ∇×E ∇×B ∇·B ∇ × B+ θ(z ) + ∇ × B− θ(−z ) + z ˆ × B+ − B− δ (z ).6. B+ are the fields just above the interface.6 Junction conditions 9 where z ˆ is a unit vector that points in the z direction.6) Substituting Eq.7) Finally.3).6. = µ0 K . ∇ · B = 0. Then ∇s = n ˆ . . All that is required is to change the argument of the step and δ functions from z to s(x).6. reveals that E± and B± satisfy this equation on both sides of the interface and that the tangential (x and y ) components of the electric must also be continuous: z ˆ × E+ − E− z =0 = 0. reveals that E± is produced by ρ± and that the surface charge distribution creates a discontinuity in the normal component of the electric field: 1 (1.8) Equations (1.6. B− the fields just below the interface.1. reveals that B± is produced by j± and that the surface current creates a discontinuity in the tangential components of the magnetic field: z ˆ × B+ − B− z =0 = µ0 K . It is not difficult to generalize this discussion to an interface of arbitrary shape.10) (1.6. and E− . ∇ × E+ θ(z ) + ∇ × E− θ(−z ) + z ˆ × E+ − E− δ (z ).12) = 0.6. If E+ . (1. ∇ · B+ θ(z ) + ∇ · B− θ(−z ) + z ˆ · B+ − B− δ (z ). and K is the surface current density. (1. (1. σ is the surface charge density on the general interface. together with the mathematical identity ∇ × (∇Φ) = 0.5) This implies that the potentials are not unique: they can be redefined at will by a gauge transformation.1) This.] The Maxwell equation ∇ · B = 0 and the mathematical identity ∇ · (∇ × A) = 0 imply that the magnetic field can be expressed as the curl of a vector potential A(t.2) the minus sign in front of ∇Φ is conventional. Then the other inhomogeneous Maxwell equation. imply that E + ∂ A/∂t can be expressed as the divergence of a scalar potential Φ(t. is the converse true? The answer is in the negative — the potentials are not unique.7.3) (1. (1. . Bnew = B . [The material presented in this section is also covered in Secs. the last term is identically zero and we have established the invariance of the magnetic field. On the other hand. The remaining two equations will give us equations to be satisfied by the potentials. can be cast in the form ∇× E+ ∂A ∂t = 0.2 and 6. ∂t ∂t = − the last two terms cancel out and we have established the invariance of the electric field.7 Potentials Suppose that B is expressed in this form. An issue that arises is whether the scalar and vector potentials are unique: While E and B can both be obtained uniquely from Φ and A.10 Maxwell’s electrodynamics 1. 6.7. x): E=− ∂A − ∇Φ. The new electric field is given by Enew ∂ Anew − ∇Φnew ∂t ∂ ∂f = − A + ∇f − ∇ Φ − ∂t ∂t ∂ ∂f = E − ∇f + ∇ . x) is an arbitrary function of space and time. Introducing the scalar and vector potentials eliminates two of the four Maxwell equations. (1. ∂t (1.3 of Jackson’s text.7. (1. We will use the gauge invariance of the electromagnetic field to simplify the equations to be satisfied by the scalar and vector potentials. Solving these is often much simpler than solving the original equations.4) in which f (t.7. the new magnetic field is given by Bnew = ∇ × Anew = ∇ × A + ∇f = B + ∇ × (∇f ). We wish to show that this transformation leaves the fields unchanged: Enew = E . ∇ × E + ∂ B /∂t = 0. x): B = ∇ × A.7. ∂t = A + ∇f. Consider the following gauge transformation of the potentials: Φ A → → Φnew = Φ − Anew ∂f . this is numerically equal to the speed of light in vacuum.3) and (1.11) is the speed with which the waves propagate.7. they are equivalent to the two Maxwell equations that remain after imposing Eqs. it can always be imposed by redefining the potentials according to Eqs. ∂t (1.2.8). = ∇(∇ · A) − ∇2 A + ǫ0 µ0 2 + ǫ0 µ0 ∇ ∂t ∂t = ∇ × (∇ × A) − ǫ0 µ0 where we have used a vector-calculus identity to eliminate ∇ × (∇ × A).7. x) c ∂t − where c≡ √ 1 ǫ0 µ0 = − 1 ρ(t.2).7.10) = −µ0 j (t.4) with a specific choice of function f (t.7. (1.7. x) such that the new potentials Φ and A will satisfy Eq. Similarly. To see that the Lorenz gauge condition can always be imposed. x).7) govern the behaviour of the potentials. The answer is in the affirmative: If Eq.7 Potentials To derive these we start with Eq. and that they both take the form of a wave equation: 1 ∂2 + ∇2 Φ(t.8) is true then 0 = ∇ · A + ǫ0 µ0 ∂Φ ∂t .6) and (1.7.7) µ0 j = ǫ0 µ0 2 − ∇2 A + ∇ ∇ · A + ǫ0 µ0 ∂t ∂t Equations (1. (1. x) c2 ∂t2 1 ∂2 − 2 2 + ∇2 A(t. we re-express Eq.1.7.7.7. These equations would be much simplified if we could demand that the potentials satisfy the supplementary condition ∇ · A + ǫ0 µ0 ∂Φ = 0. and we ask whether it is possible to find a function f (t. (1. (1. and as we shall see below.2.7. We know that we can transform them according to Φold → Φ = Φold − ∂f /∂t and Aold → A = Aold + ∇f .7. ∂t 11 or 1 ∂Φ ∂2Φ ∂ ∇ · A + ǫ0 µ0 .7. x). We have obtained ∂2A ∂Φ . When Eq.7.8) holds we observe that the equations for Φ and A decouple from one another. for reasons that will become clear. imagine that we are given potentials Φold and Aold that do not satisfy the gauge condition. (1. (1. (1. we have added and subtracted a term ǫ0 µ0 ∂ 2 Φ/∂t2 . (1.7. (1.7.9) (1.6) where.1) and cast it in the form ∂A 1 ρ=∇·E =∇· − − ∇Φ ǫ0 ∂t =− ∂ ∇ · A − ∇2 Φ.4) in terms of the potentials: µ0 j = ∇ × B − ǫ0 µ0 ∂E ∂t ∂ ∂A − − ∇Φ ∂t ∂t ∂Φ ∂2A . x). ǫ0 (1.1) and (1.8) This is known as the Lorenz gauge condition. ρ = ǫ0 µ0 2 − ∇2 Φ − ǫ0 ∂t ∂t ∂t (1. and correspondingly increased the ease of finding solutions.9) and (1. (1. and we conclude that the Lorenz gauge condition can always be imposed.13) as well as E = −∇Φ and B = ∇ × A.7. f (x) a prescribed source function. It is easy to see that if we have such a Green’s function at our disposal. x) is a solution to the wave equation − ∂ Φold 1 ∂2 + ∇2 f = −∇ · Aold − ǫ0 µ0 .2). the wave equation becomes Poisson’s equation.1) can be expressed as ψ (x) = ψ0 (x) + G(x.2) where δ (x − x′ ) is a three-dimensional Dirac δ -function. (1. all we need to know is that solutions exist. In time-independent situations. 1.7 of Jackson’s text. (1. the source point x′ is arbitrary.7.7. then the general solution to Eq. We begin in this section with the mathematical problem of solving a generic Poisson equation of the form ∇2 ψ (x) = −4πf (x). x′ ) = −4πδ (x − x′ ). we have found that the original set of four Maxwell equations for the fields E and B can be reduced to the two wave equations (1.1) where ψ (x) is the potential. and the potentials now satisfy Poisson’s equation.8). The Lorenz gauge condition reduces to ∇ · A = 0 (also known as the Coulomb gauge condition). x′ )f (x′ ) d3 x′ . the fields and potentials no longer depend on t.7. x′ ) that satisfies a specialized form of Poisson’s equation: ∇2 G(x.8.8 Green’s function for Poisson’s equation [The material presented in this section is also covered in Sec. For timeindependent situations. (1. (Notice that we do not need to solve the wave equation for f .7. (1. supplemented by the Lorenz gauge condition of Eq. ∂t = ∇ · Aold + ∇f + ǫ0 µ0 = ∇ · Aold + ǫ0 µ0 ∂ Φold ∂t and we see that the Lorenz gauge condition is satisfied if f (t.] We have seen in the preceding section that the task of solving Maxwell’s equations can be reduced to the simpler task of solving two wave equations. Once solutions to these equations have been found. To construct the general solution to this equation we shall first find a Green’s function G(x.12 Maxwell’s electrodynamics ∂f ∂ Φold − ∂t ∂t 2 ∂ f − ǫ0 µ0 2 + ∇2 f. 2 2 c ∂t ∂t Such an equation always admits a solution. (1. and where a factor of 4π was inserted for later convenience.7.8.10) for the potentials Φ and A.7. ǫ0 ∇2 A(x) = −µ0 j (x). In this and the next section we will develop some of the mathematical tools needed to solve these equations. Introducing the potentials has therefore dramatically reduced the complexity of the equations.3) . the fields can be constructed with Eqs. (1.8.12) (1.8. 1. and the foregoing equations reduce to ∇2 Φ(x) = − 1 ρ(x).1) and (1.) To summarize. defining a radius k and angles θ and φ by the relations kx = k sin θ cos φ. In Eq. x ) = (2π )3 ′ eik·(x−x ) 3 d k. ∇2 ψ0 (x) = 0. (1. and the requirement that the Green’s function must be invariant under a translation of the coordinate system.8. All this gives G(x.8.8 Green’s function for Poisson’s equation 13 where ψ0 (x) is a solution to the inhomogeneous (Laplace) equation. . and the θ integral can easily be evaluated by using µ = cos θ as an integration variable: π 1 eikR cos θ sin θ dθ = 0 −1 eikRµ dµ = 2 sin kR .2) and (1. To construct the Green’s function we first argue that G(x. (1.6) ˜ (k) = 4π . (1. To do this it is convenient to switch to spherical coordinates in reciprocal space. k2 ′ (1. The remaining integral can easily be evaluated by contour integration (or simply by looking it up in tables of integrals). For convenience we orient the coordinate system so that the kz axis points in the direction of R ≡ x − x′ . In these coordinates the volume element is d3 k = k 2 sin θ dkdθdφ.8. Methods to solve boundary-value problems will be introduced in the next chapter. x′ ) = 2 π ∞ 0 sin kR 2 dk = kR πR ∞ 0 sin ξ dξ.3).8.8.1). x′ ) as the Fourier ˜ (k): transform of a function G G(x.8. ψ0 can simply be set equal to zero. this follows from the fact that the source term depends only on x − x′ . kR We now have G(x.6) becomes we see that Eq.8. We then express G(x. where R ≡ |R| is the length of the vector R.1).1. it is equal to π/2. x′ ) can depend only on the difference x − x′ . x′ ) depends only on x − x′ .5) where k is a vector in reciprocal space. (1. (1. so that Eq. the role of the integral is to account for the source term in Eq. ξ where we have switched to ξ = kR.8. Integration over dφ is trivial. Recalling that the three-dimensional δ function can be represented as δ (x − x′ ) = 1 (2π )3 eik·(x−x ) d3 k. and using Eqs.8. ′ (1.4) This assertion is proved by substituting Eq. (1. this expression incorporates our assumption that G(x. We then have k · R = kR cos θ. (1. (1.8.8.3) into the left-hand side of Eq. and kz = k cos θ. The role of ψ0 (x) is to enforce boundary conditions that we might wish to impose on the potential ψ (x).7) We must now evaluate this integral. x′ ) = 1 2π 2 ∞ π 2π dk 0 0 dθ 0 dφ eikR cos θ sin θ. G (1. x′ ) = 1 (2π )3 ˜ (k)eik·(x−x′ ) d3 k. ky = k sin θ sin φ.8.2) implies k2 G 4π G(x.4) to show that the result is indeed equal to −4πf (x). In the absence of such boundary conditions. we have introduced the wave.9.4 of Jackson’s text. x′ ) = −4πδ (x − x′ ). |x − x′ | Maxwell’s electrodynamics (1. x) is a solution to the homogeneous wave equation.9.8.9.8) we have recovered a familiar result: the potential of a point charge at x′ goes like 1/R. x′ )e−iω(t−t′ ) dω. c2 ∂t2 (1. ψ (t. x) + G(t. To construct the Green’s function we Fourier transform it with respect to time.8) the Green’s function for Poisson’s equation is simply the reciprocal of the distance between x and x′ . (1.3) where ψ0 (t. x) = −4πf (t.9. (1. the potential at x is built by summing over source contributions from all relevant points x′ and dividing by the distance between x and x′ . 6. x.4) does indeed solve Eq. Notice that in Eq.8) which is a generalized form of Eq.9.9.14 Our final result is therefore G(x. In terms of this the general solution to Eq. x′ ) = 1/|x − x′ |. The general solution to Poisson’s equation is then ψ (x) = ψ0 (x) + f (x′ ) 3 ′ d x.9 Green’s function for the wave equation [The material presented in this section is also covered in Sec.6) and we represent the time δ -function as δ (t − t′ ) = 1 2π e−iω(t−t ) dω. x.2) can be expressed as ψ (t. t′ . From this comparison we learn that ˜ (0.2) For this purpose we seek a Green’s function G(t. (1.9. |x − x′ | (1.9. (1. or d’Alembertian. G (1.] We now turn to the mathematical problem of solving the wave equation. x). x.8. x′ ) that satisfies G(t. (1. (1.1) can be verified by direct substitution.1) for a time-dependent potential ψ produced by a prescribed source f .8. x.8. x) = 0. ′ (1. t′ . x. ∇2 + (ω/c)2 G (1. x′ ) = 1 . (1. x′ )f (t′ . ψ0 (t.2). t′ . t′ .9. x′ ) = 1 2π ˜ (ω .5) That the potential of Eq. x′ ) dt′ d3 x′ . x. x) = ψ0 (t. G .9. (1.9) The meaning of this equation is that apart from the term ψ0 . 1. differential operator ≡− 1 ∂2 + ∇2 . G(t. (1.9.3) yields ˜ (ω . x′ ) = −4πδ (t − t′ )δ (x − x′ ).4) (1. x.9.7) Substituting these expressions into Eq. and this gives us the condition g (ω. (1. (1. the volume integral is of order Gε where R ˜ contributes nothing in the limit ε → 0. ˜ should depend on the spatial variables R ≡ |x − x′ |. the surface integral 2 would contribute a term of order 1/ε . This integral. however. G ′ ˜ (ω . ε) + O(ε) for the surface integral.9. (1. (1. as was expressed by Eq. (There are ways of regularizing the integral so as to obtain a meaningful answer. x′ ) = G 4π (2π )3 eik·(x−x ) 3 d k.1. (1. we can use Gauss’ theorem to get at x′ . k2 − (ω/c)2 ′ which is a generalized form of Eq. two possible solutions to this equation are g± (ω. |x − x |) . (1. R) = 1. x′ ) = g (ω.9) with g representing a function that stays nonsingular when the second argument. because of the singularity at k2 = (ω/c)2 . 12. x.10) We also recall that g (0. is not defined. 0) = 1. on the other hand. This implies that G ˜ 1/ε. R) = e±i(ω/c)R . (1. (1. This would eventually lead to ˜ (ω . R =ε ˜ were to behave as 1/ε3 . x′ ) At this stage we might proceed as in Sec. and not its direction). x. (1. then dG/dR ˜ If G would be of order 1/ε4 .8 and Fourier transform G with respect to the spatial variables. In this equation. taking ˜ depends on x only through R. is equal to 4πε2 ˜ dG dR .9. G ˜ 3 and it ˆ = R/R. One method involves deforming the contour of the ω integration so as to avoid the poles. x.8) becomes Acting with this on G g ′′ + (ω/c)2 g = 0. Since G operator becomes 1 d 2 d ∇2 → 2 R .8) and integrate both sides over a sphere of small radius ε centered ˜ = ∇ · ∇G ˜ . (1.9. unless G happens to be as singular as 1/ε3 . The surface integral. approaches zero.9) into Eq.11 of Jackson’s text.7).9.8. 1. R dR dR ˜ = g/R yields g ′′ /R and Eq. G Take Eq.9).10). the Laplacian its right-hand side to be zero. We shall therefore reject this method of solution.9 Green’s function for the wave equation 15 ˜ (ω .9.9. Since ∇2 G R =ε ˜·R ˆ da + (ω/c)2 ∇G R<ε ˜ d3 x = −4π. Setting G = g/R returns −4πg (ω.8). and the left-hand side would never give rise ˜ cannot be so singular. That ˜ should behave as 1/R when R is small is justified by the following discussion. With the boundary condition of Eq. So we conclude that G ˜ must be of order hand side is dominated by the surface integral.12) . We may now safely take R = 0 and substitute Eq. G |x − x′ | (1. This is described in Sec.) ˜ will be of the form We can anticipate that for ω = 0.9. and that the leftto the required −4π . That G through R only can be motivated on the grounds that three-dimensional space is ˜ can only depend on the vector R ≡ x − x′ ) and both homogeneous (so that G isotropic (so that only the length of the vector matters.9.9.11) with a prime indicating differentiation with respect to R.9. (1. t′ . and that into Eq. it encodes the property that information about the source travels to x at the speed of light.9. is known as the retarded Green’s function. (1. 1. The time translation.9. x) c2 ∂t2 ǫ0 (1. Substituting the Green’s functions into Eq. x) = ψ0 (t.14) for the fundamental solutions to the wave equation. and we obtain ψ± (t. which is nonzero when t − t′ = −R/c. x) depends on the behaviour of the source at a later time t + R/c.9. x. Although both solutions to the wave equation are mathematically acceptable.15) to be the solution to the wave equation (1. which gives the general solution to Poisson’s equation.9. (1. x) as the only physically acceptable solution. (1. |x − x′ | δ t − t′ ∓ |x − x′ |/c . however. ψ− (t. and it properly enforces causality: the cause (source) precedes the effect (potential). the function G− (t. x′ 3 ′ d x |x − x′ | (1.13) The time integration can be immediately carried out. x.16 Maxwell’s electrodynamics Substituting this into Eq. causality clearly dictates that we should adopt the retarded solution ψ+ (t. depends on the behaviour of the source at an earlier time t − R/c — there is a delay between the time the information leaves the source and the time it reaches the observer. x) + δ t − t′ ∓ |x − x′ |/c f (t′ .14) looks very similar to Eq. ψ0 = 0.1). x) = − ρ(t.1) .9. we obtain G± (t.9.6). Recall that ψ0 (t. It means that the potential at time t depends on the conditions at the source at a shifted time t′ = t ∓ |x − x′ |/c. (1. x) + f t ∓ |x − x′ |/c. Eq. x. (1. The function G+ (t. t′ .10 Summary The scalar potential Φ and the vector potential A satisfy the wave equations − 1 1 ∂2 + ∇2 Φ(t. This solution to the wave equation is known as the retarded solution. |x − x′ | (1.9). t′ . x) + f t − |x − x′ |/c. On the other hand. x. x) is a solution to the homogeneous equation. is very important. x′ 3 ′ d x |x − x′ | (1. Except for the shifted time dependence. here the behaviour is anti-causal.9. x′ ) dt′ d3 x′ . x′ ) = or G± (t.3). ψ+ (t. which is nonzero when t − t′ = +R/c.9). x) = ψ0 (t. is known as the advanced Green’s function. x′ ). ′ These are the two fundamental solutions to Eq.9. x′ ) = 1 2π 1 e±i(ω/c)R −iω(t−t′ ) e dω = R 2πR e−iω(t−t ∓R/c) dω. So we take ψ (t.9. x).4) gives ψ± (t. and this unphysical solution to the wave equation is known as the advanced solution. x′ ).8.10. t′ . The second term on the right-hand side is the time required by light to travel the distance between the source point x′ and the field point x. x) = ψ0 (t. The “+” solution. 6) ∂t This efficient reformulation of the equations of electrodynamics is completely equivalent to the original presentation of Maxwell’s equations.11 and Problems 17 1 ∂2 + ∇2 A(t. x) = −µ0 j (t. The speed of propagation of the waves.4) j t − |x − x′ |/c. x) = 1 4πǫ0 ρ(t. an alternative to the Lorenz gauge adopted in the text. ∂t 2. We have seen that the charge density of a point charge q located at r (t) is given by ρ(x) = qδ x − r (t) . c = 1/ ǫ0 µ0 . a) Prove that the Coulomb gauge condition. is numerically equal to the speed of light in vacuum.10. |x − x′ | ∂Φ . It will be the starting point of most of our subsequent investigations. E=− 1.1. B = ∇ × A. x′ 3 ′ d x |x − x′ | (1.2) 1 ∂Φ =0 (1. where v = dr /dt is the charge’s velocity.5) 4π |x − x′ | In static situations. can always be imposed on the vector potential. ∇2 Φ = −ρ/ǫ0 .10. In this problem we explore the formulation of electrodynamics in the Coulomb gauge. the electric and magnetic fields are recovered by straightforward differential operations: A(t. x) = A0 (t. The Coulomb gauge is especially useful in the formulation of a quantum theory of electrodynamics. x′ ) 3 ′ d x.10. The retarded solutions to the wave equations are Φ(t. x) c2 ∂t2 when the Lorenz gauge condition − ∇·A+ (1. Then show that in this gauge. the time dependence of the sources and potentials can be dropped. Once the scalar and vector potentials have been obtained. so that it can be expressed as Φ(t. x) + and 1 4πǫ0 ρ t − |x − x′ |/c. ∂ρ + ∇ · j = 0. and that its current density is j (x) = q v (t)δ x − r (t) .10. x) + ∂A − ∇ Φ. Prove that ρ and j satisfy the statement of local charge conservation. ∇ ·A = 0. ∂t Show also that the vector potential satisfies the wave equation A = −µ0 j + ǫ0 µ0 ∇ .10. (1. x) = Φ0 (t.3) c2 ∂t √ is imposed. x′ 3 ′ µ0 d x. the scalar potential satisfies Poisson’s equation.11 Problems 1. (1. so that x(0) = x ˙ (0) = 0. jl (t. introduce the Fourier transform j (t. x) = 1 (2π )3 j (t. It is supposed that the force starts acting at t = 0. This shows that G(x. k) × k ˆ. j (t. Then prove that the Fourier transform of the longitudinal current is ˆ · j (t. x′ ) 3 ′ d x . assume that k points in the direction of the z axis. prove that the Fourier transform of the transverse current is ˆ × j (t. 4. where ψ is the potential and f the source. k) = k thus. x) = − 1 ∇ 4π Maxwell’s electrodynamics ∇′ · j (t. x′ ) = is a solution to 1 ik|x−x′ | e 2ik d2 + k 2 G(x. x′ ) = δ (x − x′ ). The differential equation d2 x + ω 2 x = f (t) dt2 governs the motion of a simple harmonic oscillator of unit mass and natural frequency ω driven by an arbitrary external force f (t). where jt ≡ j − jl is the “transverse current”. For this purpose. Then show that jl = jz z ˆ and jt = jx x ˆ + jy y ˆ. dx2 where k is a constant. Then prove that the wave equation for the vector potential can be rewritten as A = −µ0 jt . (d2 /dx2 + k 2 )ψ (x) = f (x).18 b) Define the “longitudinal current” by jl (t. the transverse current has components in the directions orthogonal to k only. and that it satisfies ∇ · jt = 0. x′ ) is a Green’s function for the inhomogeneous Helmholtz equation in one dimension. k) = k ˆ = k/|k| is a unit vector aligned with the reciprocal position where k vector k. . x) = 1 ∇× 4π ∇× and show that this satisfies ∇ × jl = 0. x′ ) 3 ′ d x. k) of the current density. Prove that G(x. thus. k)eik·x d3 k. jt (t. Similarly. the longitudinal current has a component in the direction of k only. with an overdot indicating differentiation with respect to t. k) k ˆ. |x − x′ | d) The labels “longitudinal” and “transverse” can be made more meaningful by formulating the results of part b) and c) in reciprocal space instead of ordinary space. such that j (t. |x − x′ | c) Prove that the transverse current can also be defined by jt (t. and that prior to t = 0 the oscillator was at rest. 3. For a concrete illustration. y ) = (ct)2 − ρ2 where ρ ≡ x2 + y 2 and θ(s) is the Heaviside step function. described by f (t′ .) Show that the solution to the wave equation in this case is given by ψ (t. a) Take the source of the wave to be a momentary point source. x. ψ (t.1) we construct solutions to the inhomogeneous wave equation ψ = −4πf in a few simple situations. x. (Notice that this also describes a point source in a space of two dimensions. x. then x(t) = (2ω )−1 t sin ωt. In this problem we consider a modified . so that the oscillator is driven at resonance. x′ )f (t′ . Check your results by verifying that if f (t) = θ(t) cos(ωt). take the source to be a sheet source described by f (t′ . dt2 Then find the solution x(t) to the differential equation (in the form of an integral) that satisfies the stated initial conditions. x) = where G(t. c) Finally. x. r G(t. 5. z ) = δ (t − r/c) . (Notice that this also describes a point source in a space of a single dimension. Show that the solution to the wave equation in this case is given by ψ (t. t′ . t′ ) for this differential equation. Notice that the wave is no longer well localized. t′ . y. b) Now take the source of the wave to be a line source described by f (t′ .11 Problems 19 Find the retarded Green’s function G(t. δ (t − t′ − |x − x′ |/c) |x − x′ | where r ≡ x2 + y 2 + z 2 . x′ ) dt′ d3 x′ . the electromagnetic interaction is mediated by a massless photon. but that the wavefront still travels outward at the speed of light. x′ ) = δ (x′ )δ (y ′ )δ (z ′ )δ (t′ ). 6. In this problem (adapted from Jackson’s problem 6. x) = 2πcθ(t − |x|/c). x′ ) = δ (x′ )δ (t′ ).1.) Show that the solution to the wave equation in this case is given by 2cθ(t − ρ/c) . In the quantum version of Maxwell’s theory. A particular solution to this equation is ψ (t. the wave is now completely uniform. x′ ) = δ (x′ )δ (y ′ )δ (t′ ). which must be a solution to d2 G + ω 2 G = δ (t − t′ ). Notice that apart from a sharp cutoff at |x| = ct. x′ ) = is the retarded Green’s function. The wave is therefore a concentrated wavefront expanding outward at the speed of light. The field equations are supplemented by the same Lorentz-force law. is this true also in the modified theory? . It is based on the following set of field equations: ∇ · E + κ2 Φ ∇·B ∇×E ∇ × B + κ2 A 1 ρ. In the original theory the potentials have no direct physical meaning. f = ρE + j × B . ∂t where κ is a new constant related to the photon mass m by κ = mc/ . ∂t Find the new expressions for ε and S . upon quantization. as in the original theory. b) Find the modified wave equations that are satisfied by the potentials Φ and A. c) Verify that in modified electrostatics. ∂t = = µ0 j + ǫ0 µ0 ∂E . and where the fields E and B are related in the usual way to the potentials Φ and A. the scalar potential outside a point charge q at x = 0 is given by the Yukawa form Φ= q e−κr . ǫ0 = 0. ∂t The Lorenz gauge condition must therefore always be imposed in the modified theory.20 Maxwell’s electrodynamics classical theory of electromagnetism that would lead. 4πǫ0 r d) Prove that the modified theory enforces energy conservation by deriving an equation of the form ∂ε + ∇ · S = −j · E . ∂B = − . to a massive photon. they should reduce to the old expressions when κ → 0. a) Prove that the modified theory enforces charge conservation provided that the potentials are linked by ∇ · A + ǫ0 µ0 ∂Φ = 0. the distance between the field point x and the charge. (2.4) The role of Φ0 (x) is to enforce boundary conditions that we might wish to impose on the scalar potential. ∇2 Φ(x) = − 1 ρ(x). the equations of electromagnetism decouple into a set of equations for the electric field alone — electrostatics — and another set for the magnetic field alone — magnetostatics.1. ∇2 Φ0 (x) = 0.1.2) where Φ0 (x) satisfies Laplace’s equation. The equations of magnetostatics will be considered in Chapter 3. (2. We have R2 = (x − bx )2 + (y − by )2 + (z − bz )2 and differentiating both sides with respect to. 4πǫ0 |x − b| (2.1. x gives 2R∂R/∂x = 2(x − bx ).1. The general solution to Poisson’s equation is Φ(x) = Φ0 (x) + 1 4πǫ0 ρ(x′ ) 3 ′ d x.3) gives Φ(x) = 1 q .3) (2.1 Equations of electrostatics For time-independent situations.1. or ∂R/∂x = (x − bx )/R. ǫ0 (2.Chapter 2 Electrostatics 2.2. E = −∇Φ. Eq.1) and the relation between potential and field.2.2 Point charge The simplest situation involves a point charge q located at a fixed point b. (2. say. |x − x′ | (2. 2.1) and in the absence of boundaries. Derivatives of R with respect 21 .2) To compute the electric field we first calculate the gradient of R ≡ |x − b|. The charge density is ρ(x) = qδ (x − b). The topic of this chapter is electrostatics. The equations of electrostatics are Poisson’s equation for the scalar potential. 3. or ∇ This gives E (x) = x−b q .2) (2.3) .3 Dipole Another elementary situation is that of two equal charges.1) This is an exact expression. |x − b| |x − b|3 (2. 2. The calculation of the electric field involves ∇R−1 = −R−2 ∇R. and we have established the useful vectorial relation x−b .3) ∇|x − b| = |x − b| Notice that ∇R is a unit vector that points in the direction of R ≡ x − b. We align the charges along the direction of the unit vector p ˆ.4) The electric field goes as q/R2 and points in the direction of the vector R = x − b. The charge density of this distribution is given by ρ(x) = qδ (x − εp ˆ) − qδ (x + εp ˆ). We can simplify it if we take the observation point x to be at a large distance away from the dipole.2. and place the origin of the coordinate system at the middle point.2. see Fig.2. − |x − εp ˆ| |x + εp ˆ| (2.22 Electrostatics z p +q y x −q Figure 2. one positive.3.1. 4πǫ0 |x − b|3 (2. and the potential is Φ(x) = q 4πǫ0 1 1 . r ≡ |x| ≫ ε. separated by a distance d ≡ 2ε.3. (2.1: Geometry of a dipole. 2. (2. to y and z can be computed in a similar way. This well-known result is known as Coulomb’s law.5) x−b 1 =− . the other negative. 4πǫ0 r3 Keeping terms of order ε only.4). 4πǫ0 r3 (2. The integral is then − δ (x′ )∇′ a 1 d3 x′ = − |x − x′ | δ (x′ ) xa (x − x′ )a 3 ′ d x =− 3. or 1 1 = 1 ± εp ˆ · x/r2 + O(ε2 /r2 ) .5). Substituting this into Eq.3. the charge distribution has a vanishing total charge.3. The potential of a dipole is then Φ(x) = 1 p·x . In general.2. For any function f of a vector x we have f (x + εp ˆ) = f (x) + εp ˆ· ∇f (x) + O(ε2 ). the potential becomes Φ= The vector p ≡ 2qεp ˆ is the dipole moment of the charge distribution.3) and setting Φ0 = 0 yields Φ(x) = − pa 4πǫ0 ′ ∇′ a δ (x ) 3 ′ d x. faster than in the case of a single point charge. |x − x′ |3 r and we have recovered Eq. Notice that in these manipulations we have inserted a result analogous to Eq. In the present situation the definition implies p = (+q )(εp ˆ) + (−q )(−εp ˆ) = 2qεp ˆ. This has to do with the fact that here. the potential falls off as 1/r2 . (2.3).1) then becomes ρ(x) = −2qεp ˆ · ∇δ (x) + O(ε2 ). this is defined as p= A qA x A . . This is the charge density of a point dipole. (2. the minus sign does not appear because we are now differentiating with respect to the primed variables. (2. It is instructive to rederive this result by directly expanding the charge density in powers of ε.3 Dipole 23 We can then approximate |x ∓ εp ˆ| = = (x ∓ εp ˆ) · (x ∓ εp ˆ) r2 ∓ 2εp ˆ · x + ε2 = r 1 ∓ εp ˆ · x/r2 + O(ε2 /r2 ) .3. |x − x′ | (2.5) when r satisfies Eq.4) where xA is the position vector of the charge qA .2.3. |x ∓ εp ˆ| r ˆ) · x 1 (2qεp . (2.3.1. (2. and we can extend this result to Dirac’s distribution: δ (x ± εp ˆ) = δ (x) ± εp ˆ · ∇δ (x) + O(ε2 ). (2. for a collection of several charges.6) where summation over the vectorial index a is understood. or ρ(x) = −p · ∇δ (x). The charge density of Eq. The standard procedure when dealing with derivatives of a δ -function is to integrate by parts.3. It should be noticed that here. as was stated previously. A typical situation in electrostatics features a charge distribution ρ(x) between two boundaries on which the potential Φ(x) is specified (see Fig.5) we can calculate the electric field of a point dipole.6). 2.10 of Jackson’s text.7) This is an exact expression for the electric field of a point dipole.3. Introducing the unit vector r ˆ = x/r = (x/r. − 4πǫ0 r3 r5 and a similar calculation can be carried out for the y and z components of ∇Φ. 3 4πǫ0 r (2. we find that the electric field is given by E (x) = 1 3(p · r ˆ)r ˆ− p . ∇x r = x/r. The outward unit normal to both boundaries is denoted n ˆ.3.4 Boundary-value problems: Green’s theorem [The material presented in this section is also covered in Secs. and the region V in between that contains a charge distribution ρ. So ∇x Φ = 1 px 3(p · x)x .8) (2.2: An inner boundary Sin on which Φ is specified.2.24 Electrostatics n Sout V S in n Figure 2. an outer boundary Sout on which Φ is specified. For a physical dipole of finite size. 1. (2. for which the charge density is given by Eq. this expression is only an approximation of the actual electric field.2) — the potential is then said to satisfy Dirichlet boundary conditions. z/r). y/r.8 and 1.3. (2.3. From the potential of Eq. 2. 4πǫ0 r3 r4 We have ∇x (p · x) = px and according to Eq.3). Taking a derivative with respect to x gives ∇x Φ = 1 ∇x (p · x) 3(p · x) − ∇x r . (2. the approximation is good for r ≫ ε. A concrete situation might involve an inner .] After the warmup exercise of the preceding two sections we are ready to face the more serious challenge of solving boundary-value problems. as stated. which is essentially an application of Gauss’ theorem. But we can make our situation fit the formulation of the theorem by digging a narrow “channel” from Sout to Sin .3). Sout .4 Boundary-value problems: Green’s theorem 25 da Sout V da S in Figure 2. V Sout Sin and noting the opposite orientations of da and n ˆ on Sin (see Fig. We would like to derive an expression for Φ(x) in the region between the boundaries. ∇ · b d3 x = b · da. x′ ). V S where b is any vector field defined within V and da is an outward surface element on S . To get there we will make use of Green’s identity. the situation that interests us. To arrive at this expression we will need to introduce a new type of Green’s function. and integrating b over both sides of the channel produces a zero result because da points in opposite directions. We therefore have ∇ · b d3 x = b · da + b · da. Gauss’ theorem states that for a volume V enclosed by a surface S . We now choose the vector b to be b = φ∇ψ − ψ ∇φ.2. the outer boundary Sout .3: The union of the inner boundary Sin . and the narrow “channel” between them forms a closed surface S that encloses the region V. 2. The theorem. says nothing about a region V bounded by two boundaries Sin and Sout . and the channel is a closed surface. we write this as ∇ · b d3 x = ≡ b·n ˆ da − b·n ˆ da V Sout Sin Sout −Sin b·n ˆ da. It is important to notice that the unit vector n ˆ points out of both Sin and Sout . this should account for both the charge density and the boundary data. The union of Sin . boundary Sin that is also a grounded conducting surface (on which Φ = 0) and an outer boundary Sout that is pushed to infinity (so that Φ = 0 there also). . its properties will be identified along the way. which we will call a Dirichlet Green’s function and denote GD (x. 2.3. as depicted in Fig. For convenience we write it in the equivalent form φ∇′2 ψ − ψ ∇′2 φ d3 x′ = φ∇′ ψ − ψ ∇′ φ · n ˆ da′ .26 Electrostatics where φ(x) and ψ (x) are two arbitrary functions. x′ )ρ(x′ ) d3 x′ V = Sout −Sin Φ(x′ )∇′ GD (x.4. We now pick φ(x′ ) ≡ Φ(x′ ) ≡ scalar potential and ψ (x′ ) ≡ GD (x. x′ )ρ(x′ ) d3 x′ V Sout −Sin Φ(x′ )∇′ GD (x.1) V Sout −Sin This is Green’s identity. where the properties of the “Dirichlet Green’s function” will be described in detail below.4. We have ∇ · b = φ∇2 ψ − ψ ∇2 φ and Gauss’ theorem gives φ∇2 ψ − ψ ∇2 φ d3 x = φ∇ψ − ψ ∇φ · n ˆ da. To sidestep this problem we demand that the Dirichlet Green’s function satisfy the condition GD (x. (2. For now we assume that it satisfies the equation ∇′2 GD (x. V Sout −Sin where all fields are now expressed in terms of the new variables x′ . x′ )∇′ Φ(x′ ) · n ˆ ′ da′ . ∇′2 Φ(x′ ) = −ρ(x′ )/ǫ0 . x′ )∇′ Φ(x′ ) · n ˆ da′ .4. x′ )ρ(x′ ) d3 x′ V Sout −Sin Φ(x′ )n ˆ · ∇′ GD (x. (2. Making these substitutions in the equation following Eq. It is not clear. and the two surface integrals take care of the boundary conditions. its normal derivative.4. x′ ) = −4πδ (x − x′ ).4) . x′ ) ≡ Dirichlet Green’s function. (2. which also appears within the surface integrals. how we might go about evaluating these integrals: the problem does not seem to be well posed. x′ ) − GD (x. Apart from a remaining simplification. or Φ(x) = 1 4πǫ0 1 − 4π GD (x. therefore. because they are ′ multiplied by GD (x. x ) which vanishes on the boundaries. we are given no information about n ˆ · ∇′ Φ(x′ ).4. x′ ) = 0 when x′ is on the boundaries. (2. x′ )da′ .1) leads to −4π Φ(x) + 1 ǫ0 GD (x.2) while the scalar potential satisfies Poisson’s equation. But there is one problem: While the value of the scalar potential is specified on the boundaries. this is the kind of expression we were seeking. The volume integral takes care of the charge distribution within V . (2.3) Then the boundary terms involving n ˆ · ∇′ Φ(x′ ) simply go away. so that the functions Φ(x′ ) are known inside the surface integrals. x′ ) − GD (x. With this property we arrive at Φ(x) = 1 4πǫ0 1 − 4π GD (x. GD (x′ . GD (x. and we pick φ(y ) ≡ GD (x. (2. (2. ∇2 GD (x. To establish Eq. • the inner boundary is spherical (Sin is described by the statement |x′ | = R.5). and its boundary values on Sin and Sout are zero. (2.2) and (2. (2.8.2).5): the Dirichlet Green’s function vanishes whenever x or x′ happens to lie on the boundaries. Notice that this is a specialized form of our original problem: GD (x.4. x′ ) is the potential produced by a point charge of strength 4πǫ0 located at x. The right-hand side. (2.5).6) which is identical to Eq. x′ ).4. ∇2 GD (x.4.4.4.4. x) = GD (x. Notice that the boundary conditions have been generalized in accordance to Eq. gives zero because both φ and ψ are zero when y is on Sin and Sout . we have found that the Dirichlet Green’s function satisfies Eq.4.4.4.4) gives us the means to calculate Φ(x) in very general circumstances. (2. G(x. This produces Eq. The problem of finding this potential has been reduced to that of finding a Dirichlet Green’s function that satisfies Eqs.4. x′ ) = −4πδ (x − x′ ). x′ ) = 0 when x or x′ is on the boundaries. 2 3 φ∇2 y ψ − ψ ∇y φ d y = V Sout −Sin φ∇y ψ − ψ ∇y φ · n ˆ day .3).2) ′ 2 we have ∇2 y φ = −4πδ (x − y ) and ∇y ψ = −4πδ (x − y ). (2. x) = GD (x. We now use Green’s identity to show that the Dirichlet Green’s function is symmetric in its arguments: GD (x′ .5) we take y to be the integration variables in Green’s identity. To summarize. In the following sections we will consider a restricted class of boundary-value problems.8. and satisfies the boundary conditions of Eq. then Eq. (2. possesses the symmetry property of Eq.4). the form of the Green’s function will depend on the shapes and locations of these boundaries. x′ ) satisfies Eq. . x′ ) = |x − x′ |−1 . We already know that GD (x. (1. (2. x ) − GD (x .2.5) This implies that it is also a solution to the standard equation satisfied by a Green’s function. x′ ).4. In the absence of boundaries. 1. (2. (2.4. for which • there is only an inner boundary Sin (Sout is pushed to infinity and does not need to be considered). (2. Once the Dirichlet Green’s function has been found. x)] after integration over d3 y .3). The left-hand side of ′ ′ Green’s identity gives −4π [GD (x.6). the potential produced by a charge distribution ρ(x) in a region V between two boundaries Sin and Sout on which Φ(x) is specified can be obtained by evaluating the integrals of Eq. where R is the surface’s radius).4.2) and vanishes when x′ lies on Sin and Sout . on the other hand. Then according to Eq. (2.4 Boundary-value problems: Green’s theorem 27 This is our final expression for the scalar potential in the region V . If we can solve this problem and obtain the Dirichlet Green’s function. y ) and ψ (y ) ≡ GD (x′ . GD (x. y ). x′ ) = −4πδ (x − x′ ). The task of finding a suitable Dirichlet Green’s function can only be completed once the boundaries Sin and Sout are fully specified. x′ ) reduces to the Green’s function constructed in Sec.4. x′ )ρ(x′ ) d3 x′ . m is limited to the values −ℓ. φ).2. With these restrictions (introduced only for simplicity).6. this is done by solving Eq. (2. θ.6) in spherical coordinates. φ) = R(r)Y (θ.4) where ℓ(ℓ + 1) is a separation constant. Particular spherical harmonics are Y00 = 1 √ . The solutions to the angular equation (2.4.5. −m . Eq.4) reduces to Φ(x) = 1 4πǫ0 GD (x. 2. −ℓ + 1.3) (2. and 3.4. 4π . While ℓ ranges from zero to infinity. ℓ. This is given by E = −∇Φ(|x| = R). φ). φ).1.5. which are labeled by the two integers ℓ and m.28 Electrostatics • the inner boundary is the surface of a grounded conductor (Φ vanishes on Sin ). · · · .8) where E is the electric field just above Sin .4.5. To proceed we will need to find a concrete expression for the Dirichlet Green’s function. ℓ − 1.4) are the spherical harmonics Yℓm (θ. (2.2) + 1 ∂2Y = −ℓ(ℓ + 1)Y.5) where Pℓm (cos θ) are the associated Legendre polynomials. (2. For m < 0 we use ∗ Yℓm = (−1)m Yℓ.5.5 of Jackson’s text. (2.7) In these situations.6) where an asterisk indicates complex conjugation.5 Laplace’s equation in spherical coordinates [The material presented in this section is also covered in Secs.4. Laplace’s equation takes the form ∇2 ψ = ∂ψ 1 ∂ r2 r2 ∂r ∂r + r2 1 ∂ψ ∂ sin θ sin θ ∂θ ∂θ + r2 1 ∂2ψ 2 ∂φ2 = 0. For m ≥ 0. 3.9). θ. For the spherical boundary considered here. and we obtain the decoupled equations dR d r2 dr dr and 1 ∂ ∂Y sin θ sin θ ∂θ ∂θ = ℓ(ℓ + 1)R (2. φ) = 2ℓ + 1 4π (ℓ − m)! m P (cos θ)eimφ . |x′ |>R (2.5. (ℓ + m)! ℓ (2. the electric field vanishes inside the conductor and there is an induced distribution of charge on the surface. σ = ǫ0 E · n ˆ. 3. Yℓm (θ. The surface charge density is given by Eq.5.5. (1.1) To solve this equation we separate the variables according to ψ (r. sin θ (2. sin2 θ ∂φ2 (2.] When formulated in spherical coordinates (r. it returns f (θ. 4π 29 Y11 Y10 Y22 Y21 Y20 = − = = 1 4 = − = 1 2 Two fundamental properties of the spherical harmonics are that they are orthonormal functions. (2. 2π .5. φ) dΩ. φ) = ℓ m Yℓm (θ. (2.5.8) for some coefficients fℓm . φ′ ) ℓ m ∗ f (θ′ .7) where dΩ = sin θ dθdφ is an element of solid angle. 8π 5 (3 cos2 θ − 1). φ′ )Yℓm (θ. φ) = ℓ=0 m=−ℓ δ (θ − θ′ )δ (φ − φ′ ) .8).5.5. φ) dΩ′ . (2.5 Laplace’s equation in spherical coordinates 3 sin θ eiφ . (2. φ′ )Yℓm (θ′ . sin θ (2. To avoid confusion we change the variables of integration to θ′ and φ′ : f (θ. one for θ and the other for φ.5.5. these can in fact be calculated as ∗ fℓm = f (θ. By virtue of Eq.10) is known as the completeness relation for the spherical harmonics. φ)Yℓm (θ. ′ 1 √ eikx 2π ∗ 1 √ eikx dk = δ (x − x′ ). φ′ ) and integrated over the primed angles. φ) = ℓ=0 m=−ℓ fℓm Yℓm (θ. ∞ ℓ ∗ Yℓm (θ′ .10) where the factor of 1/ sin θ was inserted to compensate for the factor of sin θ′ in dΩ′ (the δ -function is enforcing the condition θ′ = θ).5.2.5. This is analogous to a well-known identity. φ′ )Yℓm (θ. and that they form a complete set of functions of the angles θ and φ. φ) dΩ = δℓℓ′ δmm′ . φ) (2. The statement of orthonormality is Yℓ∗ ′ m′ (θ.7). (2. It is interesting to see what happens when Eq. 8π 3 cos θ. φ) can be represented as a sum over spherical harmonics: ∞ ℓ f (θ.5.8) means that the spherical harmonics form a complete set of basis functions on the sphere.9) Equation (2. More precisely stated. Equation (2. The quantity within the large square brackets is such that when it is multiplied by f (θ′ . φ) f (θ′ . φ′ ) dΩ′ = ∗ Yℓm (θ′ . This must therefore be a product of two δ -functions. φ). φ)Yℓm (θ.9) is substituted into Eq. The statement of completeness is that any function f (θ. 2π 15 sin θ cos θ eiφ . 4π 15 sin2 θ e2iφ . 2). φ) = ℓ=0 m=−ℓ aℓm rℓ + bℓm r−(ℓ+1) Yℓm (θ.5. ψ is to be well behaved at r = 0. (2. (2. ψ is to vanish at r = ∞. .1) can be represented as δ (x − x′ ) = = δ (r − r′ ) δ (θ − θ′ )δ (φ − φ′ ) r2 sin θ δ (r − r′ ) ∗ Yℓm (θ′ .5. φ)δ (r − r′ )δ (θ − θ′ )δ (φ − φ′ ) drdθdφ = f (r′ .5. φ).2) G(x.6. The general solution is Rℓm (r) = aℓm rℓ + bℓm r−(ℓ+1) . r2 ℓm (2. φ′ )Yℓm (θ. From these observations we deduce that the only solution to Laplace’s equation that is well behaved at the origin and vanishes at infinity is the trivial solution ψ = 0. R ∝ rℓ or R ∝ −(ℓ+1) r . for example.11) and summing over all possible values of ℓ and m: ∞ ℓ ψ (r.8 we found that 1 . φ) δ (r − r′ ) δ (θ − θ′ )δ (φ − φ′ ) 2 r sin θ drdθdφ r2 sin θ f (r.5. x′ ) for a spherical inner boundary. 2. for example. If. The general solution to Laplace’s equation is obtained by combining Eqs.5. x′ ) = |x − x′ | We will obtain an alternative expression for this. The solutions to the radial equation (2. (2. φ). the volume element is d3 x = r2 sin θ drdθdφ. We already know the answer: In Sec. θ.30 Electrostatics in which the integral over dk replaces the discrete summation over ℓ and m. θ. then bℓm ≡ 0.6.1) in the absence of boundaries.9 of Jackson’s text. then aℓm ≡ 0.6. (2. in this section we attempt something simpler and solve ∇2 G(x. We then have. on the other hand.6. x′ ) = −4πδ (x − x′ ) (2. 1. (2.5.11) where aℓm and bℓm are constants. The factor of 1/(r2 sin θ) was inserted because in spherical coordinates.6.12) The coefficients aℓm and bℓm are determined by the boundary conditions imposed on ψ (r. φ′ ). (2.6 Green’s function in the absence of boundaries [The material presented in this section is also covered in Secs. and the Jacobian factor of r2 sin θ must be compensated for. one that can be generalized to account for the presence of spherical boundaries.10). φ). θ. 3. and 3.3) are power laws. θ′ . the basis functions (2π )−1/2 eikx are then analogous to the spherical harmonics. (2.5). θ. f (x′ ) = = = f (x)δ (x − x′ ) d3 x f (r. If. (2.5.3) where we have involved Eq.] Keeping in mind that our ultimate goal is to obtain the Dirichlet Green’s function GD (x. The δ -function on the right-hand side of Eq. r′ )θ(r′ − r) + r2 g> (r. r′ ).6. θ′ (r′ − r) = −δ (r − r′ ). r′ ) = ′ ′ + r′2 g> (r′ . r′ ) can be obtained by joining the solutions at r = r′ : g (r. and φ′ . (2.3) we expand the Green’s function as G(x. x′ ) into Eq. Differentiating this equation with respect to r gives ′ ′ g ′ (r.6. (2. r′ )θ(r′ − r) + g> (r. (2. r′ ): dgℓm d r2 dr dr − ℓ(ℓ + 1)gℓm = −4πδ (r − r′ ). φ′ )Yℓm (θ.3) produces an ordinary differential equation for gℓm (r. (2.6. We let g< (r. Multiplying by r2 yields ′ ′ r2 g ′ (r. φ). r′ ) − g< (r′ . r′ )θ(r′ − r) + g> (r. and not on m: gℓm (r.4) ∇2 G(x.5. r′ )Yℓm (θ′ . r′ ) − g< (r′ . (2. θ′ .6. d dg< r2 − ℓ(ℓ + 1)g< θ(r′ − r) dr dr d dg> + r2 − ℓ(ℓ + 1)g> θ(r − r′ ) dr dr dg> ′ ′ dg< ′ ′ + r ′2 (r . and f (r)δ (r − r′ ) = f (r′ )δ (r − r′ ).1) on the Green’s function yields ∗ gℓm (r.1).6.6.5). r′ )θ(r − r′ ) + g> (r′ . To solve Eq. 1.5) This equation implies that the radial function depends only on ℓ. these are treated as constant parameters when substituting G(x.5) we first observe that when r = r′ . r′ ) be the solution to the right of r′ (for r > r′ ). Inspired by Eq. r ) δ (r − r′ ) dr dr d δ (r − r′ ).6) back into Eq.6. where a prime on g indicates differentiation. r′ ) − g< (r′ .6. We then assume that g (r. r′ ) = r2 g< (r. r′ ) = g< (r.3). x′ ) = ℓm ∗ using the spherical harmonics Yℓm (θ. (2. For convenience we have made the “radial function” depend on x′ through the variables r′ . r′ ) δ (r − r′ ). r′ )θ(r − r′ ).6) where θ(r − r) and θ(r − r ) are step functions. r′ )θ(r − r′ ) + r′2 g> (r′ . r′ ) be the solution to the left of r′ (for r < r′ ) and g> (r. φ). φ′ )Yℓm (θ.2. (2.6. these were first encountered in Sec.6. To simplify the notation we will now omit the label “ℓ” on the radial function. r ) − (r . ′ ′ (2. gℓm Yℓm r2 and setting this equal to −4πδ (x − x′ ) as expressed in Eq.6 Green’s function in the absence of boundaries 31 as we should. Applying the Laplacian operator of Eq.5. r′ ) θ(r′ − r) + r2 g> + r′2 g> (r′ . r′ ) δ (r − r′ ) ′ ′ (r. r′ ) = g< (r. the right-hand side of the equation is zero and g is a solution to Eq. r′ ) dr Substituting this and Eq. r′ ) θ(r − r′ ) r2 g< (r. We have involved the distributional identities θ′ (r − r′ ) = δ (r − r′ ). φ) for angular functions and gℓm (r. r′ ) δ ′ (r − r′ ). (2. (2. we obtain −4πδ (r − r′ ) = . r′ ) − g< (r′ . r′ )Yℓm (θ′ . r′ ) − g< (r′ . φ′ ) for radial functions. x′ ) = ℓm dgℓm 1 d r2 r2 dr dr − ℓ(ℓ + 1) ∗ (θ′ . + r′2 g> (r′ . r′ ) ≡ gℓ (r. (2. r′ ) δ (r − r′ ) ′ ′ ′ and taking another derivative brings r2 g ′ (r. 11) gives g (r.10) 2ℓ + 1 rℓ+1 We observe a nice symmetry between these two results. Because g< is a solution to Eq. as was already understood. (2. it does not matter that g> diverges at r = 0 because it is restricted to the domain r > r′ > 0.11) ℓ+1 2ℓ + 1 r> g> (r. Eq. we find that the radial function can be expressed in the compact form ℓ 4π r< gℓ (r. First. r′ ) = a(r′ )rℓ .8) gives a = 4πr′−(ℓ+1) /(2ℓ + 1).6. we arrive at 1 = |x − x′ | ∞ ℓ ℓ 4π r< ∗ Yℓm (θ′ .5).6. r′ ) is therefore continuous at r = r′ .11) gives g (r. To determine the “constants” a and b we use the matching conditions. r′ ). r> = r.11) it must be a linear combination of terms proportional to rℓ and 1/rℓ+1 .6.6.6. ℓ +1 2ℓ + 1 r> (2. r′ ) = where r< denotes the lesser of r and r′ . (2. r′ ) − g< (r′ . (2. (2. Not relying on this identity would mean having to evaluate an integral that involves 1 = |x − x′ | 1 r2 − 2rr′ [sin θ sin θ′ cos(φ − φ′ ) + cos θ cos θ′ ] + r′2 . Thus.6. r′ ) = 0 and (2. as it should. (2. r ) − (r . But we will see that when 1/|x − x′ | appears inside an integral.5).6. (2. (2.6. r> = r′ . At first sight the usefulness of this identity might seem doubtful.7) gives b = ar′2ℓ+1 . Second.6.10). which is well behaved at r = ∞. r′ ). For g> we select instead g> (r. and we set g< (r.6. according to Eq. . r′ ) = . r′ ) = g> (r.6).11) into Eq. (2. (2.6.4) and taking into account Eq. (2. φ′ )Yℓm (θ. similarly. (2. r′ ) be solutions to the homogeneous version of Eq. when r > r′ we have r< = r′ . (2. φ).12) can be very useful indeed. the representation of Eq.32 Electrostatics For the right-hand side to match the left-hand side we must demand that g< (r.6. r′ ) = b(r′ )/rℓ+1 . (2. After all.3). and (2. Combining Eqs. while r> is the greater of r and r′ . if r < r′ then r< = r.6. r ) = − ′2 .6. r′ ) = and 4π rℓ 2ℓ + 1 r′ℓ+1 (2. So g< (r.6.6.6. (2. we have turned something simple like an inverse distance into something horrible involving special functions and an infinite double sum.9). r′ ) = g< (r. (2. Eq. but its first derivative must be discontinuous to account for the δ -function on the right-hand side of Eq.2). (2. (2. But only the first term is well behaved at r = 0. and Eq.6. r′ ) and g> (r.5.9) 4π r′ℓ . It does not matter that g< diverges at r = ∞ because this function is restricted to the domain r < r′ < ∞. and that these solutions be matched according to g> (r′ . Similarly.12) ℓ=0 m=−ℓ This very useful identity is known as the addition theorem for spherical harmonics.8) dr dr r The function g (r. and Eq.7) dg< ′ ′ 4π dg> ′ ′ (r .5. Substituting Eq.6. To evaluate the radial integral we must be careful with the meaning of r> ≡ max(r. the potential depends on r only. and we have R 4π 0 1 r ′2 ρ(r′ ) dr′ = 4π r> r R 0 R ρ(r′ )r′2 dr′ = 0 Q .14) the charge enclosed by a sphere of radius r. φ′ ) dΩ′ = √ 4πδℓ. if r > R then q (r) = Q. (2.12) into Eq. (2. Let us consider an example to illustrate the power of the addition theorem.15) While the expression (2. After substituting Eq. We want to calculate the electrostatic potential both inside and outside a spherical distribution of charge with density ρ(x′ ) = ρ(r′ ). Then r > r′ . φ) 4πǫ0 R 0 r ′2 1 ρ(r′ ) dr′ = 4π r> 4πǫ0 R 0 r ′2 ρ(r′ ) dr′ . so that r < R. r> = r. In the first term we recognize r q (r) ≡ ρ(x′ ) d3 x′ = 4π |x′ |≤r 0 ρ(r′ )r′2 dr′ . r (2. φ) 2ℓ + 1 R 0 ℓ r< ρ(r′ )r′2 dr′ ℓ+1 r> ∗ Yℓm (θ′ .6.3) and writing d3 x′ = r′2 dr′ dΩ′ . At this stage we happily realize that the infinite sum over ℓ and m involves but a single term.6.6 Green’s function in the absence of boundaries 33 and such integrals typically cannot be evaluated directly: they are just too complicated. so that r > R. Then part of the integration covers the interval 0 ≤ r′ < r and the remaining part covers r ≤ r′ ≤ R. The internal potential can thus be expressed as Φin (x) = 1 q (r) + 4π 4πǫ0 r R ρ(r′ )r′ dr′ .15) for the internal potential depends on the details of the charge distribution — the particular form of the function ρ(r′ ) — a nicer . ℓm To evaluate the angular integral we multiply and divide by (4π )−1/2 ≡ Y00 (θ′ . In the first part r′ is smaller than r and r> = r. So R 4π 0 1 r ′2 ρ(r′ ) dr′ = 4π r> r r R ρ(r′ )r′2 dr′ + 4π 0 r ρ(r′ )r′ dr′ . (2. φ′ ): ∗ Yℓm (θ′ . r′ ). where we have involved Eq.2.0 .0 δm.5. but the distribution is confined to a sphere of radius R. r where Q ≡ ρ(x′ ) d3 x′ = 4π In this case we have ρ(r′ )r′2 dr′ is the total charge of the distribution.1. r> and as we might have expected from the spherical symmetry of the problem. (2. Suppose first that x is outside the charge distribution.6. We therefore have Φ(x) = √ 1 4π 4πY00 (θ. 4πǫ0 r (2. we obtain Φ(x) = 1 4πǫ0 4π Yℓm (θ. There are no boundaries in this problem. Suppose next that x is inside the charge distribution. (2. φ′ ) dΩ′ .6.6.2. φ′ )Y00 (θ′ . in the second part r′ is larger than r and r> = r′ .13) Φout (x) = the same result as in Eq.2). φ′ ) dΩ′ = √ 4π ∗ Yℓm (θ′ .7). 1 Q . the radial function can be expressed as gℓ (r. This gives the additional conditions gℓ (r = R.6). r′ ) = gℓ (r. It will be a simple matter to modify the treatment presented in Sec. r′ ) = a(r′ . the Dirichlet Green’s function can be expanded as GD (x. (2. (2. As in Eq. r′ ) = 0. the Green’s function must vanish on the inner boundary. x′ ) = ℓm ∗ gℓ (r. (2.3). q (r) becomes Q. the electric field at a radius r depends only on the charge q (r) enclosed by a sphere of that radius. the total charge.9 of Jackson’s text. 4πǫ0 r2 (2. or g< (r = R. and the field is the same as if all this charge were concentrated in a point at r = 0. (2.8). and 1 q 1 dq dΦin = − 2+ − 4πρr . we see that the last two terms cancel out.1) As in Eq. x′ ) for the specific situation described near the end of Sec. 2.6 to account for the inner boundary.16) both inside and outside the charge distribution. Since g< (r. (2. r′ = R) = 0.7) and (2.5) recall that R is now the smallest possible value of both r and r′ .6.4) (r .7. r ) − (r . R)/rℓ+1 . r′ )θ(r′ − r) + g> (r. (2.] We now return to the task of finding the Dirichlet Green’s function GD (x.6 is that now. 2.5. g> (r. r′ )Yℓm (θ′ .6.5. dr 4πǫ0 r r dr But since dq/dr = 4πρr2 according to Eq. 3. r′ ) is no longer required to be well behaved at r = 0. φ′ )Yℓm (θ. we must set it equal to g< (r.6.3).6. R)rℓ + α(r′ . recall that when r > R.7.7.4).6.7.4. r′ = R) = 0.6. r ) = − ′2 .3) dg< ′ ′ 4π dg> ′ ′ (2.7. φ). 2. (2.4. dr dr r The only difference with respect to what was done in Sec.34 Electrostatics conclusion applies to the electric field.7 Dirichlet Green’s function for a spherical inner boundary [The material presented in this section is also covered in Sec. the general solution to Eq. in which we have a spherical inner boundary at |x| = R and no outer boundary.14). r′ ) we choose instead g> (r.3) that must also satisfy the matching conditions of Eqs. r′ )θ(r − r′ ). The electric field is then E (x) = 1 q (r) r ˆ. We have recovered the well-known result that for any spherical distribution of charge. (2. r′ ) − g< (r′ . according to Eq. 2. R)/rℓ+1 . The field is obtained by differentiating Φin with respect to r.2) where g< and g> are solutions to Eq. r′ ) = g< (r. g> (r′ . . For g> (r. r′ ) = 0 and (2. r′ ) = b(r′ . (2. (2. 2.2) in the preceding section.1): GD (x.6.8) where r< denotes the lesser of r and r′ . (2. while r> is the greater of r and r′ (notice that r< r> ≡ rr′ ).8) into Eq. (2. and the terms within the brackets vanish.7.7).8 Point charge outside a grounded. and the same conclusion applies.7) Combining Eqs. We see that gℓ (r. x′ ) = ℓm 1 4π R2ℓ+1 ℓ ∗ r< − ℓ+1 Yℓm (θ′ .9) This is slightly more complicated than Eq.8). (2. but still manageable.7.2 of Jackson’s text. r′ ) = 4π R2ℓ+1 1 ℓ r< − ℓ+1 ℓ +1 2ℓ + 1 r> r< = 2ℓ+1 4π r< − R2ℓ+1 .4.4) . Φ(x) = 1 4πǫ0 GD (x. The charge density induced on the surface of the conductor is given by Eq. (2. r′ ) vanishes on the inner boundary: When r = R it must be that r < r′ .3)–(2. α = −aR2ℓ+1 . φ′ )Yℓm (θ.6) (2.1) and the Dirichlet Green’s function was obtained as GD (x.7). (2. we find that the radial function can be expressed in the compact form gℓ (r. θ0 = 0. so that r< = r = R. ℓ +1 2ℓ + 1 r r (2.7.8 Point charge outside a grounded. where E is the electric field just above the conductor and n ˆ≡r ˆ is the surface’s unit normal. φ′ )Yℓm (θ. ℓ +1 2ℓ + 1 r> r< (2.] The electrostatic potential outside a grounded. Collecting these results gives g< (r. σ = ǫ0 E · n ˆ . The charge density is ρ(x′ ) = qδ (x′ − z0 ). Because E = −∇Φ. and b = a(r′2ℓ+1 − R2ℓ+1 ). r′ ) = 1 R2ℓ+1 4π rℓ − ℓ+1 ′ ℓ +1 2ℓ + 1 r r R2ℓ+1 4π 1 r′ℓ − ′ℓ+1 .7. we shall not pursue this here. φ) < ℓ+1 ℓ+1 2ℓ + 1 r> r< (2. 2. 2ℓ + 1 (rr′ )ℓ+1 (2.8.7. (2. (2.7.5) for the “constants” produces a = 4πr′−(ℓ+1) /(2ℓ + 1). (2. The charge’s position vector is z0 = z0 z ˆ and its spherical coordinates are r0 = z0 .4.7.8. (2. when r′ = R it must be that r′ < r.6). spherical conductor of radius R is given by Eq. spherical conductor 35 which is well behaved at r = ∞.7. x′ )ρ(x′ ) d3 x′ . φ). (2.7. r′ ) = and g> (r. at a distance z0 > R from the centre of the conducting sphere. this works out to be ∂Φ .3) σ = −ǫ0 ∂r r=R We place a point charge q on the z axis.2. Notice that the Dirichlet Green’s function reduces to 1/|x − x′ | when R = 0. Solving Eqs. It would be straightforward to generalize this derivation to account also for the presence of an outer boundary.8.2). x′ ) = ℓm R2ℓ+1 1 4π ℓ ∗ r − Yℓm (θ′ .8. on the other hand. and (2.7. and φ0 is undetermined. so that r< = r′ = R. The Dirichlet Green’s function is obtained by substituting Eq. spherical conductor [The material presented in this section is also covered in Sec. |x′ |>R (2.12).7. the fact that it does not depend on φ reflects the axial symmetry of the problem. 4π After cleaning up the algebra.7) where the double factorial notation means (2n)!! = (2n)(2n − 2)(2n − 4) · · · (2) or (2n − 1)!! = (2n − 1)(2n − 3)(2n − 5) · · · (1). 4π so that the potential is actually independent of φ0 and involves a single sum over ℓ. These are ordinary polynomials of order ℓ.8. P3 (x) = 2 1 P4 (x) = (35x4 − 30x2 + 3). It is a property of the spherical harmonics that Yℓm (0. P1 (x) = x.0 . while r> stands for the greater of the two. 1 Pℓ′ (x)Pℓ (x) dx = −1 2 δℓ′ ℓ . (2. Before moving on we pause and recall the main properties of the Legendre polynomials. while if ℓ is odd it contains only odd terms. φ0 )Yℓm (θ. Pℓ (1) = 1.8. z0 ) 4πǫ0 ℓ r< 4π R2ℓ+1 q ∗ − ℓ+1 ℓ+1 Yℓm (0. but for now we calculate the charge density σ (θ) . 1 P2 (x) = (3x2 − 1). (2. the final expression for the potential is Φ(x) = q 4πǫ0 ∞ ℓ+1 ℓ r< R2ℓ+1 /z0 Pℓ (cos θ).1) gives Φ(x) = = Electrostatics q GD (x.6) and they possess the special values P2n (0) = (−1)n (2n − 1)!! . they reduce to ordinary Legendre polynomials. Yℓ. − ℓ+1 rℓ+1 r> (2.36 Substituting this into Eq.5) ℓ=0 We see that the potential depends on r and θ only. (2.0 (θ. (2n)!! P2n+1 (0) = 0. ℓ +1 4πǫ0 2ℓ + 1 r> r r < > ℓm where r< now stands for the lesser of r and z0 .5) is expressed as an infinite sum over ℓ. 2 1 (5x3 − 3x). It is another property of the spherical harmonics that for m = 0.8. if ℓ is even the function Pℓ (x) contains only even terms. For example.8. 8 The Legendre polynomials are orthogonal functions. φ0 ) = 2ℓ + 1 δm.8. 2ℓ + 1 (2. P0 (x) = 1. The potential of Eq. φ) = 2ℓ + 1 Pℓ (cos θ). We will evaluate this sum in a moment. φ). From Eq. .9 (lower curve). we should recover q ′ = σ da.5 3 Figure 2. and z0 /R = 1. Plotted is 4πR2 σ/q ′ as a function of θ for z0 /R = 1. and Eq. ℓ+1 ℓ z0 (2. 2. Plots of the charge density are presented in Fig.5 1 1.2. r> = z0 .3) we then obtain σ=− or σ (θ) = where will be shown presently to be the total induced charge on the surface of the conductor. ℓ=0 q ′ ≡ −q (R/z0 ) (2. (2. spherical conductor 37 30 25 20 15 10 5 0 0.5 theta 2 2.6 (middle curve).4 for selected values of R/z0 .4: Surface charge induced by a point charge above a spherical conductor.8 Point charge outside a grounded. Its derivative is f ′ (r) = ℓrℓ−1 + (ℓ + 1)R2ℓ+1 /rℓ+2 and evaluating this at r = R gives f ′ (R) = (2ℓ + 1)Rℓ−1 .8. on the surface of the conductor.9) (2ℓ + 1)(R/z0 )ℓ ℓ=0 0 Pℓ (cos θ) sin θ dθ. r< = r. (2.3 (upper curve). For r close to R we have r < z0 .8. Then 1 σ da = q ′ 2 ∞ π ℓ+1 q 4π (2ℓ + 1) ℓ ∞ Rℓ−1 P (cos θ). The charge density is peaked at θ = 0 and its width increases with increasing values of z0 /R: the closer the charge is to the surface the narrower is the charge distribution. which reduces to da = 2πR2 sin θ dθ after integration over φ.8.8) q′ 4πR2 (2ℓ + 1)(R/z0 )ℓ Pℓ (cos θ).8. z0 /R = 1. That q ′ is truly the total surface charge can be shown by evaluating the integral of the charge density over the surface of the conductor.5) becomes Φ(x) = ℓ 2ℓ+1 q 4πǫ0 1 ℓ ℓ+1 z0 rℓ − R2ℓ+1 Pℓ (cos θ). To evaluate this we let da = R2 dΩ = R2 sin θ dθdφ. rℓ+1 Let f (r) = r − R /r be the function of r that appears inside the brackets. If the ring is placed in the x-y plane. The δ -function in r′ indicates that the ring is infinitely thin.38 1 Electrostatics The integral over θ is equal to −1 Pℓ (x)P0 (x) dx = 2δℓ.8. because z0 equivalent to 1 q′ Φq′ (x) = (2.9.8.8.8.1) ρ(x′ ) = 2 ).5) can be decomposed as Φ(x) = Φq (x) + Φq′ (x). (2. (2.6). as required. 2πa2 where q = ρ(x′ ) d3 x′ is the ring’s total charge.8.1) into Eq. so that z0 Notice that z0 The sums over ℓ can now be evaluated. (2.9. .8.1) along with Eq.9 Ring of charge outside a grounded. It is a powerful. and we obtain σ da = q ′ .2) gives Φ(x) = q 1 4πǫ0 2πa2 × 4π Yℓm (θ. its charge density is described by q δ (r′ − a)δ (θ′ − π (2. here q ′ = −q (R/z0 ) was defined in Eq. The potential of Eq. (2. where Φq (x) = q 4πǫ0 ∞ ℓ r< P (cos θ) ℓ+1 ℓ r> (2.8. (2. This observation is the basis for the method of images. By analogy. (2.12) is absence of boundaries. φ) 2ℓ + 1 ∞ 0 ℓ r< R2ℓ+1 δ (r′ − a)r′2 dr′ − ℓ+1 ℓ+1 ℓ+1 r> r> r< ℓm ′ ∗ Yℓm (θ′ . (2.8.9) and ′ z0 ≡ R2 /z0 .8.14) Φq (x) = 4πǫ0 |x − z0 | because Φq is the potential produced by a charge q at a position z0 = z0 z ˆ in the ′ < r we have that Eq.0 by virtue of Eq. φ′ )δ (θ′ − π 2 ) dΩ . and Φq′ (x) = q′ 4πǫ0 ∞ ′ ℓ ) (z0 Pℓ (cos θ) rℓ+1 (2.10) (2. (2. /R = R/z0 < 1. but tricky.11) ℓ=0 is the potential that would be obtained in the absence of a boundary.8.8. 2. Substituting Eq.8. the potential Φq′ is the same as if all this charge were concentrated ′ at z0 . method. (2.15) ′|. This is the potential of a fictitious image charge q ′ at a position z0 ′ inside the conductor.8. (2. which consists of finding solutions for the potential in the presence of conducting surfaces by guessing the correct positions of the correct number of image charges. as was previously stated. while the δ -function in θ′ tells us that it is placed in the x-y plane.12) ℓ=0 is the modification introduced by the boundary condition Φ(r = R) = 0.13) ′ ′ < R ≤ r. For Φq we already know that Eq. While in reality the charge q is distributed on the surface of the conductor. spherical conductor In this section we consider a uniformly charged ring of radius a > R.11) is equivalent to q 1 . 4πǫ0 |x − z0 ′ ′ ′ where z0 = z0 z ˆ. 4) is the total charge on the conducting surface. Integration over θ′ then gives 2πYℓ.3) where Φq is the potential of a ring of total charge q and radius a. As in the previous section the potential of Eq. − ℓ+1 rℓ+1 r> (2. aℓ+1 where the function f (r) = rℓ − R2ℓ+1 /rℓ+1 was previously encountered in Sec.0 = 2π 2ℓ + 1 Pℓ (0)δm. Once again we will see that the sum over ℓ can be evaluated.0 . For r close to R and therefore smaller than a. φ′ ) dφ′ = 2πYℓ. and is expressed as an infinite sum over ℓ. spherical conductor 39 where dΩ′ = sin θ′ dθ′ dφ.9.0 .8. r′ ). and Eq. we have r< = r. (2. so that integrating gives zero unless m = 0. Plots of the surface charge density are presented in Fig.5 for selected values of R/a. while r> stands for the greater of the two.0 (θ) ℓ+1 − ℓ+1 ℓ+1 2ℓ + 1 r> r< r> ℓ 2ℓ + 1 Pℓ (0). (You should work through the details and make sure that these statements are true.2) can be decomposed as Φ(x) = Φq (x) + Φq′ (x). 2. r< = min(r. (2. ℓ=0 (2.0 (θ) and Pℓ (cos θ).0 (θ′ )δm.0 ( π 2 )δm.3) gives σ=− or σ (θ) = where q ′ ≡ −q (R/a) (2.9 Ring of charge outside a grounded.9. r> = a. (2. To perform the φ′ ′ ∗ integration we recall that Yℓm ∝ e−imφ .2) Once more the potential is independent of φ. while Φq′ is the potential of a fictitious image ring of total charge q ′ and radius a′ = R2 /a. the radial integration gives a2 ℓ r< R2ℓ+1 − ℓ+1 ℓ+1 ℓ+1 r> r> r< where r< now stands for the lesser of r and a.) . r′ ).8. and Eq. (2. 2. aℓ+1 q′ 4πR2 (2ℓ + 1)Pℓ (0)(R/a)ℓ Pℓ (cos θ). and r> = max(r. we arrive at Φ(x) = q 4πǫ0 ∞ Pℓ (0) ℓ=0 ℓ r< R2ℓ+1 /aℓ+1 Pℓ (cos θ).9.9. 4π Invoking once more the relation between Yℓ.9. We first calculate the charge density on the surface of the conductor. We recall that f ′ (R) = (2ℓ + 1)Rℓ−1 .2) becomes Φ(x) = q 4πǫ0 ∞ Pℓ (0) ℓ=0 f (r) Pℓ (cos θ). 4π On the other hand. So 2π 0 ∗ Yℓm (θ′ .5) q 4π (2ℓ + 1)Pℓ (0) ℓ ∞ Rℓ−1 Pℓ (cos θ).9. With these results the potential becomes Φ(x) = q 4πǫ0 ℓ r< 4π R2ℓ+1 Yℓ.2. But to proceed along this road quickly becomes cumbersome.1) (2. (2.5 1 1.10. The charge density is peaked at θ = π 2 and its width increases with increasing values of a/R: the closer the ring is to the surface the narrower is the charge distribution.10.4) By virtue of Eq. the electrostatic potential outside a bounded distribution of charge can be characterized by a (potentially infinite) number of multipole moments. Among them are Q ≡ pa Qab ≡ ≡ ρ(x) d3 x ≡ total charge ≡ monopole moment. 4.5) .5: Surface charge induced by a ring of charge outside a spherical conductor.] We now leave boundary-value problems behind.2) ρ(x) 3xa xb − r2 δab d3 x ≡ quadrupole moment tensor.5 1 0.10.5 3 Figure 2.(2.6).−m = (−1)m qℓm . (2.1 of Jackson’s text. and a/R = 1. 2. Notice that the quadrupole moment is a symmetric and tracefree tensor: Qba = Qab and Qaa = 0 (summation over the repeated index is understood).5.3 (upper curve). (2.40 Electrostatics 2. (2.5 theta 2 2. and derive the fact that in the absence of boundaries. φ) d3 x.10 Multipole expansion of the electric field [The material presented in this section is also covered in Sec.10.5 0 0. they satisfy the identity ∗ qℓ. Plotted is 4πR2 σ/q ′ as a function of θ for a/R = 1. it therefore possesses 5 independent components. Consider the complex quantities qℓm ≡ ∗ ρ(x)rℓ Yℓm (θ. ρ(x)xa d3 x ≡ dipole moment vector. a/R = 1. and we shall find a better way of packaging the components of these multipole-moment tensors. up to an infinite number of tensorial indices.5 2 1.6 (middle curve).10.3) Additional multipole moments can be constructed in a similar way.9 (lower curve). 10 Multipole expansion of the electric field 41 Because m ranges from −ℓ to ℓ. qℓm 2ℓ + 1 rℓ+1 (2. the quantities of Eq.9) This is an expansion of the potential in inverse powers of r. (2. For ℓ = 0 we have only one. 12 2π 1 15 Qxz − iQyz .10. (2.10. φ) . so that 1 q00 = √ Q.6) We see that indeed. (2.6. 4π (2. This gives q22 q21 q20 15 1 Qxx − Qyy − 2iQxy . and we will see that the q2m s give the components of the quadrupole moment tensor. For higher values of ℓ it is best to stick with the definition of Eq.10. the q2m s are directly related to the quadrupole moment tensor. . and we will see that the q1m s give the components of the dipole moment vector. r> = r. This gives q11 = − 3 px − ipy .4) give a convenient packaging of the multipole moments. and we arrive at Φ(r.10.10.12): Φ(x) = 1 4πǫ0 4π Yℓm (θ. = − 6 2π 5 1 Qzz . For ℓ = 2 we have five independent quantities. φ) 2ℓ + 1 rℓ+1 ∗ ρ(x′ )r′ℓ Yℓm (θ′ . How are the multipole moments qℓm related to the potential? The answer comes from Eq. (2. For ℓ = 1 we have three independent quantities.10.2. and r2 Y20 = 1/ 2 2 2 1/ 2 2 2 (5/16π ) r (3 cos θ − 1) = (5/16π ) (3z − r ). ℓm We recognize the integral over the primed variables as the definition of the qℓm s. θ.8) and we see that indeed.7) and we see that indeed. ∗ To evaluate Eq.3) in which we set Φ0 (x) = 0 (because we are no longer considering boundaries) and substitute Eq. = 2 4π = (2. φ) = 1 4πǫ0 ∞ ℓ ℓ=0 m=−ℓ 4π Yℓm (θ.1. ℓ+1 ℓm r> ℓm Because we are looking at the potential outside the charge distribution. q00 is directly related to the total charge. r is larger than r′ and we can set r< = r′ . φ′ ) d3 x′ . For ℓ = 1 we have ∗ ∗ rY11 = −(3/8π )1/2 (r sin θ)e−iφ = −(3/8π )1/2 (x − iy ) and rY10 = (3/4π )1/2 r cos θ = 1/ 2 (3/4π ) z .10. for a given ℓ there are 2ℓ + 1 independent real quantities within the qℓm s. φ′ ) d3 x′ . tor. This gives Φ(x) = 1 4πǫ0 4π Yℓm (θ. Thus. and the multipole moments play the role of expansion coefficients. (2.4) for ℓ = 0 we recall that Y00 = (4π )−1/2 . and we will see that q00 is proportional to the total charge. 8π q10 = 3 pz . 4π (2. to all orders. the q1m s are directly related to the dipole moment vec∗ = (15/32π )1/2 (r sin θe−iφ )2 = (15/32π )1/2 (x − iy )2 .4). For ℓ = 2 we have r2 Y22 2 ∗ 1/ 2 −iφ ∗ r Y21 = −(15/8π ) (r sin θe )(r cos θ) = −(15/8π )1/2 (x − iy )z . φ) 2ℓ + 1 ρ(x′ ) ℓ r< Y ∗ (θ′ . 9) comes from the monopole moment q00 . 2. φ0 ) are the spherical coordinates of the point x0 .10. (2. The next term in the expansion of the electrostatic potential comes from the quadrupole moment. (2.4) gives ∗ qℓm = qr0 ℓ Yℓm (θ0 .12) In a situation in which Q = 0. θ0 . however.10) 4πǫ0 r The next term comes from the dipole moment. we have 1 q1.8) and the explicit expressions for the spherical harmonics of degree ℓ = 2. 4πǫ0 2 r (2. We shall consider only axisymmetric situations and set m = 0. (2.m Y1. For example. This is a special case of a general result: Only the lowest nonvanishing multipole moments have values that are independent of the choice of origin for the coordinate system. and in this case we need three quantities.10. the total charge Q. When this vanishes also. where (r0 . 4πǫ0 r3 and Φdipole (x) = (2.3. the lowest nonvanishing multipole moment. When the total charge vanishes. all other moments change under a translation of the coordinates.11) as in Eq. (2.10. in this case the potential is characterized by a single quantity. or total charge Q: 1 Q Φmonopole (x) = . a point charge q at x = 0 has a charge density described by ρ(x) = qδ (x) and multipole moments given by √ ℓ=0 q/ 4π . (2. it is easy to show that Φquadrupole (x) = xa xb 1 1 Qab 5 . in this case the characterization of the potential involves the five independent components of the quadrupole moment tensor.10. the potential is more complicated and is approximated by Φquadrupole .42 Electrostatics The leading term in Eq.10. qℓm = 0 otherwise As expected. We shall not provide a proof of this statement. .11 Multipolar fields To conclude this chapter we construct plots of electric field lines that would be produced by a pure multipole of order ℓ.m = m=−1 3 px x + py y + pz z 4π 1 p·x . But the same point charge placed at another position x0 possesses a very different set of multipole moments.5). φ0 ). the electrostatic potential far outside the charge distribution is well approximated by Φmonopole . In this case the density is ρ(x) = qδ (x − x0 ) and Eq. the dipole moment vector. We have agreement only for q00 . the components of p. It is important to be aware of the fact that the multipole moments qℓm depend on the choice of origin for the coordinate system. (2.10. Using Eq. only the monopole moment is nonzero. Using Eq.10.7) and recalling the explicit expressions for the spherical harmonics of degree ℓ = 1. the potential is approximated by Φdipole . These curves are integral curves of the vector field E (x).4) to get ℓ (1 − µ2 )Pℓ′ dr = dµ. but varies from one curve to the next.11. (2. the field lines are determined by solving the differential equations dx = E (x). dλ (2. and we have dx ∂ x dr ∂ x dθ dr dθ ˆ .11. 2. θ) = 1 qℓ ˆ.11.11 Multipolar fields 43 The potential for such a multipole is obtained from Eq.10.11. or r(θ) = r0 sin2 θPℓ′ (cos θ) 1/ℓ . The field lines are therefore determined by integrating Eq. (2. (2. or (ℓ + 1)Pℓ (cos θ) (ℓ + 1)Pℓ (µ) dr = dθ = − dµ. Thus.11.3) In the case of an axisymmetric field the curves can depend on r and θ only. = + = r ˆ+ r θ dλ ∂r dλ ∂θ dλ dλ dλ This shows that dr/dλ = Er and rdθ/dλ = Eθ . φ) = qℓ Pℓ (cos θ). To accomplish this it is useful to recall the differential equation satisfied by the Legendre polynomials. .2) qℓ Pℓ (cos θ) .11. The electric field lines are represented by parametric curves x(λ) for which λ is an arbitrary parameter.11.9). in which we set [4π/(2ℓ + 1)]qℓ0 Yℓ0 (θ. which means that the tangent vector to the curves is defined to be everywhere equal to E . 4πǫ0 rℓ+1 (2. (ℓ + 1)Pℓ (cos θ)r ˆ + sin θPℓ′ (cos θ)θ ℓ 4πǫ0 r +2 (2.11. r sin θPℓ′ (cos θ) (1 − µ2 )Pℓ′ (µ) (2. and using Eq. dµ dµ We solve this algebraically for Pℓ and substitute into Eq.11. and we obtain ℓ ln r = ln (1 − µ2 )Pℓ′ + constant. (2.4). r (1 − µ2 )Pℓ′ (µ) ′ This can be integrated at once. This gives Φ(r.5) where r0 is constant on each curve.1) in which a prime indicates differentiation with respect to the argument. θ) = and the corresponding electric field is E (r. The field lines are plotted for selected values of ℓ in Fig.2.2) produces dr ∝ (ℓ + 1)Pℓ and rdθ ∝ sin θPℓ′ . dPℓ d (1 − µ2 ) + ℓ(ℓ + 1)Pℓ = 0.4) where µ = cos θ. Equation (2.6.5) gives the mathematical description of the electric field lines of an axisymmetric multipole of degree ℓ. (2. 44 Electrostatics 4 2 2 0 4 –2 4 2 2 4 –4 –2 –2 Ð ÐÒ × Ð ¿ –4 4 Ð 2 2 4 –4 –2 –4 1 –1 3 2 –3 4 2 –2 –2 ÅÙÐØ ÔÓÐ –2 Ð ¾ –4 4 Ð 2 1 –1 2 4 2 2 4 –4 –2 –2 –2 –2 ½ –4 Figure 2. for selected values of the multipole order ℓ. Ð Ð –4 –2 –4 –4 ½ .6: Electric field lines of an axisymmetric multipole. Suppose that at a spherical boundary surface S of radius R. z0 . 2. the total electric dipole moment of this charge distribution. θ. 8π where ρ0 and r0 are constants.2. and b = 2. In this problem we consider the electrostatic potential due to a uniformly charged disk located outside a grounded. φ) = V sin θ cos θ cos φ. . Find V as a function of q .12 Problems 1. There are no boundary surfaces in this problem. and R. The surface of the conductor is therefore at a constant potential V = 0. while Φ2 is the potential due to another charge distribution ρ2 and σ2 . z0 . 2. Calculate the electrostatic potential Φ(x) inside a charge distribution whose density is given by 3 ρ0 (r/r0 )2 sin2 θ for r < r0 . The conductor. but kept in careful isolation. a) Verify that q is indeed the disk’s total charge. its final net surface charge must also vanish.8). and provide a plot of this function. however. a = 1. make sure to include a sufficient number of terms in the infinite sum over ℓ. with q denoting the disk’s total charge. 4. so that its charge density is described by ρ(x) = 3q δ (θ − π 2) .8 by placing an additional charge −q at position −z0 = −z0 z ˆ below the conducting sphere. as it initially supports a zero net charge on its surface. For r > r0 the density is zero. the electrostatic potential is assigned the values Φ(R. ρ(x) = 2. A point charge q is brought to a distance z0 > R measured from the centre of a conducting sphere of radius R (as in Sec. conducting sphere. (Use the values q = 1. c) Calculate the surface charge density σ (θ) on the surface of the conducting sphere.12) Prove Green’s reciprocation theorem: If Φ1 is the potential due to a volume charge density ρ1 within a volume V and a surface charge density σ1 on the conducting surface S bounding the volume V . then ρ1 Φ2 d3 x + V S σ1 Φ2 da = V ρ2 Φ1 d3 x + S σ2 Φ1 da. 2 a2 + ab + b2 6. Find the potential for r > R. It is convenient to place the disk in the equatorial plane of the spherical coordinate system.2.12 Problems 45 2. 3 2π b − a3 a < r < b. 3. Express your result in terms of q . The conductor has a radius R and the potential over its surface is zero. 5.) d) Show that the total charge on the surface of the conductor is given by q′ = − 3q (a + b)R . and R. R = 1. where V is a constant. We generalize the situation considered in Sec. and that S is the only boundary present in the problem. Suppose that there are no charges outside this surface. Calculate the vector p. the conducting sphere still has a radius R. and the sphere is still grounded. b) Calculate the potential in the region R < r < a. is not grounded. The disk has an inner radius a > R and an outer radius b > a. (Jackson’s Problem 1. The first charge +q is still at +z0 = z0 z ˆ. 46 Electrostatics 7. What are the components of the quadrupole moment tensor for this charge distribution? 5λ 1 + 3 cos(2θ) δ (r − R). 10. while its lower hemisphere is maintained at a potential −V . Show that the potential outside the conducting sphere can be expressed as ∞ Φ(r. Calculate aℓ explicitly for ℓ = (1. but there is no need to include this into the mathematical model. and there are no charges outside the conductor. θ) = 4πa3 where p is a constant — it is the dipole moment of the charge distribution on the thin shell. b) Calculate the potential in the region R < r < a. a) Verify that the shell’s total charge is zero.) e) Calculate the density of charge σ (θ) on the surface of the conductor. A dielectric sphere of radius R (the inner boundary) is maintained at a potential Φ = V0 cos θ. A conducting sphere of radius R is maintained at a potential V . You may assume that there are no boundaries. θ) = V ℓ=1 aℓ (R/r)ℓ+1 Pℓ (cos θ). It is surrounded by a thin spherical shell of radius a > R with a density of charge given by ρ= where p is a constant. where the numerical coefficients aℓ are zero if ℓ is even. 5). Calculate the electrostatic potential Φ(x) outside a thin. 3. c) Calculate the potential in the region r > a. and that p = pz ˆ is the dipole moment vector of this charge distribution. θ) = where λ is a constant. It is surrounded by a thin spherical shell of radius a > R on which there is a surface charge density proportional to cos θ. 4πa3 . 8πR4 3p cos θ δ (r − a). 8. What is the total charge on the conductor? 9. and that the volume charge density is given by ρ(r. (The two hemispheres are separated by a thin layer of insulating material. and equal to (2ℓ + 1 1) 0 Pℓ (x) dx if ℓ is odd. Suppose that the upper hemisphere of a conducting sphere is maintained at a potential +V .) The conducting surface has a radius R. ρ(r. hollow sphere of radius R on which there is a surface density of charge proportional to 1 + 3 cos(2θ). d) What is the total charge of this configuration? What is the total dipole moment vector? (The total charge configuration includes the free charge on the spherical shell and the induced charge on the surface of the conductor. Find the value of V0 which makes the scalar potential Φ(x) vanish for r > a. The volume density of charge is given by 3p cos θ δ (r − a). (3.1.4) for this situation we will use spherical coordinates and express the Green’s function as in Eq. To evaluate the integral of Eq. the tools introduced in Chapter 2 to solve Poisson’s equation can be taken over directly to magnetostatics. the part of Maxwell’s equations that involve the magnetic field are decoupled from those involving the electric field. ∇2 A(x) = −µ0 j (x). (3. and they can be dealt with separately. 5.1) is very similar to the electrostatic equation (2. at z = 0.Chapter 3 Magnetostatics 3.8.3) (3. φ′ )Yℓm (θ.6. B = ∇ × A.1. 1 = |x − x′ | ∞ ℓ ℓ 4π r< Y ∗ (θ′ . µ0 4π j (x′ ) 3 ′ d x.1. The equations of magnetostatics are Poisson’s equation for the vector potential.1 Equations of magnetostatics For time-independent situations. (3.2 Circular current loop [The material presented in this section is also covered in Sec. (2.2) We recall that the current density must satisfy the statement of charge conservation.1. We therefore write the solution as A(x) = recalling the results of Sec.1) the Lorenz (or Coulomb) gauge condition ∇ · A = 0.1) in the absence of boundaries.1) ℓ=0 m=−ℓ where r< is the lesser of r and r′ . it will be sufficient to solve Eq. it will suffice to keep our discussion of magnetostatics quite brief. In particular. And because we already have gained much experience solving such equations. φ). (3.4) 3. |x − x′ | (3.5 of Jackson’s text.] As our first application we calculate the magnetic field produced by a current I flowing inside a circular loop of radius R. and the relation between potential and field.12). ℓ+1 ℓm 2ℓ + 1 r> (3. We place the loop in the x-y plane.1.1. 1.1. while r> is the greater of the two.1. ∇ · j = 0. (3.1). Because Eq.2. 47 . 0) Yℓ. r< = r′ . To obtain the vector potential we substitute Eqs. The sphericalharmonic functions can then be expressed as Yℓ.2. 0) sin φ ∞ ℓ=1 R 8π 2 Yℓ. 0) r 2ℓ + 1 ∞ ℓ+1 Yℓ.1) and (3. φ′ )eiφ dφ′ . ′ ℓm The φ′ integral evaluates to Yℓm ( π 2 . (3. R (3.2. 0) the potential simplifies to Ay − iAx = µ0 I 4π ∞ e−imφ eiφ dφ′ = 2πYℓm ( π 2 .1.4). 0)eiφ .48 Magnetostatics The density of the negative charge carriers inside the wire is given by ρ(x) = − λ δ (r − R)δ (θ − π 2 ).4) into Eq. the sum begins at ℓ = 1 because there is no spherical harmonic Y01 .1 (θ. 0).4) where I ≡ λv is the current.1 (θ. (3. and ˆ is not a constant vector: its r> = r.2) and (3. φ). (3. Its expression in where φ terms of the constant vectors x ˆ and y ˆ is obtained by noting that the position vector of a point in the plane is given by r = r cos φx ˆ + r sin φy ˆ.2) where λ is the magnitude of the linear charge density (minus the wire’s total charge per unit length).1 ( π 2 . 0) cos φ. It is convenient to consider the complex combination Ay − iAx .1 (θ. We must pay attention to the fact that φ expression in terms of x ˆ and y ˆ must be substituted inside the integral before integration is attempted.5) ℓ=1 . normalized vector φ j (x) = ρv = I ˆ φ δ (r − R)δ (θ − π 2 ).1 ( π 2 2ℓ + 1 r Yℓ. From Eqs. (3.3) ˆ is a unit vector that points in the direction of increasing φ. and this allows us to extract the real and imaginary parts of the right-hand side. which is generated by jy − ijx = I iφ′ ′ I δ (r − R)δ (θ′ − π (cos φ′ + i sin φ′ )δ (r′ − R)δ (θ′ − π 2 ) = Re 2 ).1 (θ.1 (θ.1 ( π 2 2ℓ + 1 r ℓ+1 Yℓ. φ) = Yℓ. (3. We obtain Ax = − and Ay = µ0 I 4π µ0 I 4π ∞ ℓ=1 R 8π 2 . φ) 2ℓ + 1 rℓ+1 2π 0 ′ 2π 0 ′ ∗ π Yℓm ( 2 . the total charge is given by ρ(x) d3 x = −2πRλ. and ℓ+1 ℓ=1 R 8π 2 .2. R (3.1 . dividing by r produces the ˆ. 0)δm.2.2. The charge carriers move in the negative φ direction so as to produce a positive current. Their velocity vector is ˆ = −v − sin φx v = −v φ ˆ + cos φy ˆ .2. 0) Yℓ. We notice that the φ′ dependence of the current density is contained entirely in the exponential factor ′ eiφ . After evaluating the r′ and θ′ integrals the vector potential becomes Ay − iAx = µ0 I 4π R 4π Rℓ+2 Yℓm (θ.2.2. so that r > r′ . so that a vector pointing in the φ direction is ∂ r /∂φ = −r sin φx ˆ + r cos φy ˆ.1 (θ.1 ( π 2 . 0) r 2ℓ + 1 ℓ+1 Yℓ.3) we obtain the current density. For concreteness we take x to be outside the loop. and these combine to form the vector µ0 ˆ A(x) = Iφ 4π 8π 2 R Yℓ. R we have now expressed this in terms of the primed variables. The magnetic field is therefore ˆ)r ˆ− m µ0 3(m · r . 3 4πǫ0 r So the magnetic field far outside a current loop has a dipolar behaviour. (3. A point on this strip moves in the φ direction. Then we notice that z ˆ × x/r = −(y/r)x ˆ + (x/r)y ˆ= ˆ.7) which points in the z direction and whose magnitude is the product of the current I and the area πR2 enclosed by the loop. The curl of A gives the magnetic field. (3.3 Spinning charged sphere As a second application we calculate the magnetic field produced by a hollow sphere of radius R.1). First we define the vector m = (πR2 I )z ˆ.6) can be written as − sin θ sin φx ˆ + sin θ cos φy ˆ = sin θ φ A(x) ≃ µ0 m × x . Notice that this expression for the magnetic field is very similar to what was obtained in Sec. 3.3. Since this point is at a distance R sin θ from the z axis. 0) = − 3/(8π ) sin θ and Eq. So Eq. A(x) ≃ (πR2 I ) 2 sin θφ (3. When r ≫ R the vector potential is well approximated by the leading term of the expansion. The speed is therefore ˆ = ΩR sin θφ ˆ. In this case we have Y11 (θ. and the current density is given by j (x) = (3. Here the charge density is given by ρ(x) = q δ (r − R). its z component is equal to ∇x (x/r3 )+∇y (y/r3 ) = 2/r3 − 3x5 /r5 − 3y 2 /r5 = −1/r3 + 3z 2 /r5 .8) and this shall be our final expression. uniformly charged. (3. and the velocity vector is v = v φ v (x) = Ω × x.2.1) where q is the sphere’s total charge. that is rotating with an angular velocity Ω = Ωz ˆ. 4π r3 (3. 2.2.3. the length traveled is dl = (R sin θ) dφ. and finally. or v = dl/dt = (R sin θ)Ω. qΩ ˆδ (r − R). These results combine to form the vector 3z x/r5 − z ˆ/r3 . its x component is equal to −∇z (x/r3 ) = 3xz/r5 .3. given by ℓ = 1. in a time dt the angle increases by dφ = Ω dt.2) This is the velocity of the charge distribution. 4πR2 (3.6) 4π r We will write this in a different form. we consider a narrow strip at an angle θ with respect to the z axis (see Fig. this vector is known as the magnetic moment of the current distribution.3 Spinning charged sphere 49 We see that the vector potential is expressed as an expansion in powers of R/r.2.5) becomes µ0 1 ˆ . (3. with m playing the role of dipole moment.2. 3.3 for the electric field of a dipole: Edipole = 1 3(p · r ˆ)r ˆ− p .3) . and multiplying by |m| gives 3(m · x)x/r5 − m/r3 . its y component is equal to −∇z (y/r3 ) = 3yz/r5 .2.9) B (x) ≃ 4π r3 where r ˆ = x/r is a unit vector that points in the radial direction. Apart from a common factor of (µ0 /4π )|m|. To calculate the velocity vector v of a point x on the surface of the sphere. sin θφ 4πR (3.2.3. φ′ ). φ) 3 r R3 /r2 . the double sum over ℓ and m therefore contains a single term.1: A hollow sphere rotating at an angular velocity Ω around the z axis. If we let 1 1 m ≡ q ΩR 2 z ˆ = qR2 Ω.5) while the potential outside the sphere is Aout (x) = µ0 m × x 4π r3 (r > R). we find that the angular integration is nonzero if and only if ℓ = m = 1. The radial integration reduces to ∞ r< r r<R ′ ′2 ′ . Again it is convenient to consider the complex combination Ay − iAx .1) into the expression for Ay − iAx that comes from Eq.4). jy − ijx = 4πR The angular function happens to be equal to − 8π/3Y11 (θ′ . (3.3. ˆ = − sin φx We recall that φ ˆ + cos φy ˆ. (3. After multiplication by r the vector within brackets becomes −y x ˆ + xy ˆ=z ˆ × x. Substituting this and Eq. and the vector potential is A= µ0 q Ω − sin θ sin φx ˆ + sin θ cos φy ˆ 4π 3R r R3 /r2 .50 Magnetostatics R sin θ θ Figure 3.6) .4) 3 3 we find that the vector potential inside the sphere is given by Ain (x) = µ0 m × x 4π R3 (r < R). 3 2 2 δ (r − R)r dr = R /r r>R r > 0 and we obtain Ay − iAx = µ0 q Ω 4π − 4π 4πR 3 8π Y11 (θ. The angular function within square brackets is sin θeiφ . (3.3.3. (3. (3. which is generated by ′ qΩ sin θ′ eiφ δ (r′ − R).2.1. 3. Let the distribution of current be confined to a volume V bounded by a surface S . known as the magnetic moment. (3. this is a dipolar field. (3.4.2. We take x to be outside of V . Here we will show that these expressions are in fact very general: they are the leading terms in an expansion of the vector potential and the magnetic field in powers of 1/r. and we recognize that there is no current flowing across S . From this we conclude that the magnetic field Bout outside the sphere has a dipolar behaviour. and that it takes the form of Eq.4.7) 4π R3 This states that inside the sphere.4.6 of Jackson’s text. |x − x′ | We recall that the current density satisfies ∇ · j = 0.4.4 Multipole expansion of the magnetic field [The material presented in this section is also covered in Sec. 2.] For both the current loop (Sec.2. We arrive at Bin (x) = µ0 2m .3) 1 x′ × j (x′ ) d3 x′ .6). starting from the exact expression B (x) = A(x) = µ0 4π j (x′ ) 3 ′ d x. (3. 3. 5. As we shall show below.3. (3.2) and the rotating sphere (Sec.4) ˆ× r ˆ · x′ j (x′ ) d3 x′ = − r 2 V V To simplify the vector potential we expand 1/|x − x′ | in powers of 1/r.3. except for the fact that here.2) 3 4π r where r ˆ = x/r. (3.1) A(x) = 4π r3 where the vector m. but it will be good enough to keep only a small number of terms. To calculate the magnetic field inside the sphere we express the vector potential as Ain = (µ0 /4π )(m/R3 )(−y x ˆ + xy ˆ) and take its curl. where da is the surface element on S . And in both cases we found the magnetic field to be described by µ0 3(m · r ˆ)r ˆ− m .9). these equations imply the integral statements j (x′ ) d3 x′ = 0 V (3. this implies that j · da = 0. For this we proceed directly: |x − x′ | = = (x − x′ ) · (x − x′ ) r2 − 2x · x′ − r′2 and ˆ · x′ /r + · · · = r 1 − 2r = r 1−r ˆ · x′ /r + · · · . Its x and y components are both zero. Notice also that Aout takes the same form as in Eq. characterizes the distribution of current. the magnetic moment m is a different vector.4 Multipole expansion of the magnetic field 51 Notice that these are exact results — no approximations were involved in the calculation of the vector potential.3) we found that the vector potential at a large distance from the current distribution was given by µ0 m × x . This multipole expansion is valid for points x that lie outside of the current distribution.8). (3. We examine the vector potential at a large distance r ≫ r′ . and its z component is proportional to ∇x x + ∇y y = 2. while x′ is necessarily inside. the magnetic field is constant and points in the same direction as the magnetic moment (which corresponds to the rotation axis). 3. . We could do this very systematically by involving the addition theorem for spherical harmonics (Sec. V and if we define the magnetic moment of the current distribution by m= 1 2 x′ × j (x′ ) d3 x′ .5).52 so that Magnetostatics 1 1 = 1+r ˆ · x′ /r + · · · . Eq. in which we drop the primes on the integration variables. To establish Eq.) We have e · x jz d3 x = ex xjz d3 x + ey V V yjz d3 x + ez V zjz d3 x V for the left-hand side of the identity. which integrates to (xjz + zjx ) d3 x = V S xz j · da = 0. 3.4). We consider the z component of this equation. We have A(x) = 1 µ0 1 − r ˆ× 4π r2 2 x′ × j (x′ ) d3 x′ + · · · . 3.3. Proceeding similarly with the x and y components. (At the same time we replace the vector r ˆ. Consider now ∇ · (xz j ) = xz (∇ · j ) + j · ∇(xz ) = xjz + zjx . (3. (3.4. (3. Integrating this equation over V gives jz d3 x = V V ∇ · (z j ) d3 x = S z j · da.7). (3.4. It is a straightforward exercise to show that for the current loop of Sec. (3. Similarly.4) for the rotating sphere of Sec. Eq. and we have V jz d3 x = 0.3). . We still have to establish the identities of Eqs.4.3).4. A(x) = ˆ 1 µ0 m × r +O 3 4π r2 r is a very general expression. The first term on the right-hand side vanishes by virtue of Eq.5) V then we recover Eq.4. we arrive at Eq. (3. (3. Consider ∇ · (z j ) = z (∇ · j ) + j · ∇z = jz . in which we drop the primes on the integration variables. ′ |x − x | r µ0 1 4π r 1 r2 Substituting this into the exact expression for A(x) gives A(x) = j (x′ ) d3 x′ + V V r ˆ · x′ j (x′ ) d3 x′ + · · · . but the idea is the same.4).4.2. (3.5) reproduces Eq. (3.4. (3.5) reduces to Eq. (3. (3.4.4.2.3) and (3. which is constant with respect to the primed variables. where we have invoked Gauss’ theorem.1).4.4.4) requires a bit more work. But the right-hand side is zero because no current crosses the surface S . We begin with the z component of Eq.4.3.3). and the second term is altered by Eq. and that the magnetic moment of an arbitrary distribution of current is given by Eq. (3. by an arbitrary constant vector e. which follows because ∇ · j = 0 and ∇z = z ˆ. We see that indeed. Inside the first integral we recognize the y component of the vector x × j . Calculate m.3. 3. and that the direction of m is normal to the plane. we go back to the left-hand side of Eq. (3.4) and write e · x jz d3 x = 1 ex 2 1 + ey 2 1 = − ex 2 xjz d3 x − 3 V zjx d3 x V V V yjz d x − zjy d3 x V 1 (zjx − xjz ) d3 x + ey 2 V V (yjz − zjy ) d3 x. while the second integral contains its x component. 3. Express it in terms of the sphere’s magnetic moment vector m. 4. and this yields zjz d3 x = 0.5) gives |m| = (current)(area enclosed by loop). (3.4). 2 V z (x × j )x d3 x V V and we have arrived at the right-hand side of Eq.3.7). Take the density of charge inside the sphere to be 3q/(4πR3 ) (for r < R). Eq. (3. (3. V Using these results. Show that for the current loop of Sec.4. V V On the other hand.5 Problems 53 giving rise to the integral identity xjz d3 x = − yjz d3 x = − zjx d3 x. 3.4. Prove that for a current loop of arbitrary shape that lies in a plane.5 Problems 1. (3.5) reduces to Eq.4.4. Eq.2. ∇ · (z 2 j ) = z 2 (∇ · j ) + j · ∇z 2 = 2zjz .2.5) reproduces Eq. Calculate the vector potential A(x) inside the spinning sphere of the preceding problem. And show that Eq. the sphere’s magnetic moment vector. where q is the sphere’s total charge. 2. V V Similarly. So e · x jz d3 x = − 1 ex (x × j )y d3 x − ey 2 V 1 = − e × (x × j ) d3 x . 3.3. A uniformly charged sphere of radius R is rotating around the z axis with an angular velocity Ω. zjy d3 x. (3. (3.4. .4) for the rotating sphere of Sec. 54 Magnetostatics . The resulting form of Maxwell’s equations involves only macroscopic densities of free charges and currents. Our first task will be to reformulate Maxwell’s equations in a way that conveniently incorporates the electromagnetic response of the medium. (4. ∇ × b − ǫ0 µ0 ∂t which are the same as in Chapter 1. The microscopic density of charge (including both free and bound charges) is denoted η . The microscopic electric field is denoted e.1. The macroscopic fields will be denoted E . and in which the microscopic fluctuations of all the variables involved have been smoothed out. ∇·e = 1 η. the fundamental. D .] Our goal in this chapter is to describe how electromagnetic waves propagate in dielectric materials.2) (4. The macroscopic density of free charge will be denoted ρ.1. But it is convenient to write them in a different form. These quantities are related by the microscopic Maxwell equations. in principle they require no modification even in the presence of a nontrivial medium.1 Macroscopic form of Maxwell’s equations [The material presented in this section is also covered in Sec. and the microscopic magnetic field is denoted b. the precise relationships will be given below. microscopic variables will be assigned different symbols in this chapter. and macroscopic electromagnetic fields. 6. while the microscopic density of current (including both free and bound currents) is denoted J . in which the charges that are free to move within the medium are distinguished from the charges that are tied to the medium’s molecules. 55 . 4.1. while the macroscopic density of free current will be denoted j .4) ∇·b = ∂b ∇×e+ = 0.1 Microscopic and macroscopic quantities Maxwell’s equations.1. as they were written down in Chapter 1.6 of Jackson’s text. These are obtained from the microscopic quantities by removing the contribution from the bound charges (those that are tied to the molecules and become part of the medium’s electromagnetic response) and by macroscopic smoothing.3) (4.1. To distinguish them from macroscopic quantities. are the fundamental equations of classical electrodynamics.1) (4. ∂t ∂e = µ0 J .Chapter 4 Electromagnetic waves in matter 4. ǫ0 0. The smoothing function is required to be normalized. precise relationships will also be given below. for example. x) . and we would like to replace this by a quantity that varies smoothly to reflect only macroscopic changes of density. This important result follows by straightforward manipulations. and H . x′ )w(x − x′ ) d3 x′ . For example.6) and the same is true of time derivatives. = ∂x = f (t. In this averaging operation it is convenient to replace the sharply defined function V −1 θ(R − |x − x′ |) by a better-behaved function w(x − x′ ) that smoothly goes from V −1 when x′ = x to zero when x′ is far away from x. precise relationships will be given below. and f (x) can be expressed as 1 V f (x′ )θ(R − |x − x′ |) d3 x′ . For example.1. (4. The macroscopic electric field E is obtained from e by macroscopic smoothing. then the condition is |x − x′ | < R. If the volume is a sphere of radius R. this volume is imagined to be microscopically large (so that all microscopic fluctuations will be averaged over) but macroscopically small (so that variations over relevant macroscopic scales are not accidentally discarded).56 Electromagnetic waves in matter B . where the domain of integration consists of all points x′ that lie within V (x). 4.1.5) where w is a suitable smoothing function (assumed to be normalized) and the integration is over all space. Similarly. with a now infinite domain of integration. whose detailed specification is not required. x) ≡ f (t. the macroscopic magnetic field B is obtained from b by macroscopic smoothing. x) ≡ f (t. so that a derivative with respect to x is minus a derivative with respect to x′ . The property that w does not have sharp boundaries implies that the operations of averaging and differentiation commute. One simple-minded way of smoothing out a microscopic quantity f is to average it over a volume V (x) centered at some position x. the charge density η switches from zero to infinity whenever a point charge is encountered. The averaging operation is an integration of f over the volume V (x): f (x) = V −1 f (x′ ) d3 x′ . In the third line . and D accounts for the electric polarizability of the medium. We define the macroscopic average of a microscopic quantity f (t. x) = ∇a f (t. Consider. in the sense that w(x − x′ ) d3 x′ = 1. x′ ) In the second line we use the fact that w depends on the difference x − x′ . (4. ∇a f (t.2 Macroscopic smoothing Microscopic quantities fluctuate widely over atomic distances. a specific choice would be a Gaussian function of width R centered at x − x′ = 0. and H accounts for the magnetization of the medium. ∂ f ∂x ∂ w(x − x′ ) d3 x′ ∂x ∂ = − f (t. x) to be F (t. and we would like to smooth out these fluctuations to define corresponding macroscopic quantities. Such a function.1. x′ ) ′ w(x − x′ ) d3 x′ ∂x ∂f = w(x − x′ ) d3 x′ ∂x′ ∂f . will be called a smoothing function. 4.1 Macroscopic form of Maxwell’s equations 57 we integrate by parts, noting that the boundary terms at infinity do not contribute because w vanishes there; the result is by definition the average of the microscopic quantity ∂f /∂x. Following the usage introduced in Eq. (4.1.5), we define the macroscopic electric field to be E (t, x) = e(t, x) , (4.1.7) and the macroscopic magnetic field to be B (t, x) = b(t, x) . (4.1.8) Because differentiation and averaging are commuting operations, the macroscopic form of Maxwell’s equations are ∇·E ∇·B ∂B ∇×E+ ∂t ∂E ∇ × B − ǫ0 µ0 ∂t = = 1 η , ǫ0 0, (4.1.9) (4.1.10) (4.1.11) (4.1.12) = 0, = µ0 J . The homogeneous equations (4.1.10) and (4.1.11) are the same as their microscopic version, but the inhomogeneous equations (4.1.9) and (4.1.12) involve a macroscopic averaging of the source terms. 4.1.3 Macroscopic averaging of the charge density We distinguish between the charges that can move freely within the medium — the free charges — and the charges that are tied to the molecules — the bound charges. We will consider the bound charges to be part of the medium and remove them from the right-hand side of Maxwell’s equations, which will then involve the free charges only. The influence of the bound charges will thus be taken to the left-hand side of the equations; it will be described by new macroscopic quantities, P (electric polarizability of the medium) and M (magnetization of the medium). From these we will define the new fields D ≡ ǫ0 E + P and H ≡ B /µ0 − M . In Fig. 4.1 we have a picture of the charge distribution in the medium. Let xA (t) be the position vector of the free charge qA labeled by “A”, and let vA (t) = dxA /dt be its velocity vector. Let xn be the position vector of the centre of mass of the molecule labeled by “n”; for simplicity we assume that the molecules do not move within the medium. Finally, let xB (t) be the position vector of the bound charge qB labeled by “B ” within a given molecule, relative to the centre of mass of this molecule, and let vB (t) = dxB /dt be its velocity vector. The density of free charge is ηfree (t, x) = A qA δ x − xA (t) . (4.1.13) The density of bound charge within the molecule labeled by “n” is ηn (t, x) = B qB δ x − xn − xB (t) . B qB (4.1.14) = 0. The (4.1.15) We assume that each molecule is electrically neutral, so that density of bound charge is then ηbound (t, x) = n ηn (t, x), 58 Electromagnetic waves in matter xA xn xn xB Figure 4.1: The left panel shows the free charges (small solid disks) and molecules (large open circles) in the medium. The right panel shows the distribution of bound charges within a given molecule. and the total density of charge is η = ηfree + ηbound . We first calculate the macroscopic average of the molecular charge density. We have ηn = = B ηn (t, x′ )w(x − x′ ) d3 x′ qB δ (x′ − xn − xB )w(x − x′ ) d3 x′ = B qB w(x − xn − xB ). The function w(x − xn − xB ) is peaked at x = xn + xB . Because the scale R over which w varies is very large compared with the intra-molecular displacement xB , we can approximate w by Taylor-expanding about x − xn : 1 a b ′ w(x − xn − xB ) = w(x − xn ) − xa B ∇a w (x − xn ) + xB xB ∇a ∇b w (x − x ) + · · · , 2 where to avoid confusing the notation we make the vectorial index on xB a superscript instead of a subscript. (Recalling that repeated indices are summed over, we have that xa B ∇a stands for xB · ∇.) Substituting this into the previous expression yields ηn = B qB w − B qB xa B ∇a w + 1 2 B b qB xa B xB ∇a ∇b w + · · · , where w now stands for w(x − xn ). We simplify this equation by recalling that each molecule is electrically neutral, so that the first sum vanishes. In the second term we recognize the expression for the molecular dipole moment, pn = B qB xB , or pa n (t) = B qB xa B (t), (4.1.16) and in the third term we recognize the expression for the molecular quadrupole moment, b Qab qB xa (4.1.17) n (t) = 3 B (t)xB (t). B Notice that this actually differs from the definition made in Sec. 2.10: here our quadrupole moment is not defined as a tracefree tensor. Combining these results 4.1 gives Macroscopic form of Maxwell’s equations 59 1 ab ηn (t, x) = −pa n (t)∇a w (x − xn ) + Qn (t)∇a ∇b w (x − xn ) + · · · 6 (4.1.18) for the molecular charge density. The macroscopic average of the density of bound charge is obtained by summing ηn over all molecules. This gives ηbound = − pa n ∇a w + pa nw n n 1 6 n Qab n ∇a ∇b w + · · · Qab n w + ··· Qab n δ (x − xn ) + · · · , = −∇a = −∇a 1 + ∇a ∇b 6 n pa n δ (x n 1 − xn ) + ∇a ∇b 6 n because w ≡ w(x − xn ) is the macroscopic average of δ (x − xn ). The quantities Pa (t, x) ≡ and Qab (t, x) ≡ Qab n (t)δ (x − xn ) ≡ macroscopic quadrupole density (4.1.20) n pa n (t)δ (x − xn ) ≡ macroscopic polarization (4.1.19) n have a straightforward physical interpretation. Consider, for example, Eq. (4.1.19). The quantity within the averaging brackets is n pa n δ (x − xn ), the sum over all molecules of the product of each molecule’s dipole moment with a δ -function centered at the molecule. This is clearly the microscopic density of molecular dipole moments, and Pa is its macroscopic average. So the vector P (t, x), called the macroscopic polarization, is nothing but the macroscopic average of the dipole-moment density of the medium’s molecules. Similarly, Qab (t, x) is the macroscopic average of the molecular quadrupole-moment density. In terms of these we have 1 ηbound (t, x) = −∇a Pa (t, x) + ∇a ∇b Qab (t, x) + · · · 6 for the macroscopic average of the density of bound charge. The average of the free-charge density is simply ρ(t, x) ≡ ηfree (t, x) ≡ qA δ (x − xA ) , (4.1.22) (4.1.21) A and the macroscopic average of the total charge density is therefore 1 η = ρ − ∇ · P + ∇a ∇b Qab + · · · . 6 Substituting this into Eq. (4.1.9) yields 1 ∇a ǫ0 Ea + Pa − ∇b Qab + · · · 6 = ρ. (4.1.23) The vector within the large brackets is the macroscopic field D (t, x), and we arrive at ∇ · D = ρ, (4.1.24) ∂ 1 pa w − Qab ∇b w + ∇w × mn ∂t n 6 n a + ···. (4.1.4 Macroscopic averaging of the current density Following the definitions of Eqs.1. and that apart from a factor of ǫ0 .1. 6 Notice that this macroscopic Maxwell equation involves only the density of free charges on the right-hand side. x) Jbound (t.26) (4. . = B B where w stands for w(x − xn ) in the second line. which is described by the molecular dipole-moment and quadrupolemoment densities. and we can write D = ǫ0 E + P . To put the second term in a recognizable form we define the molecular magnetic moment by 1 qB (xB × vB ) (4. x) = A qa vA (t)δ x − xA (t) . (4. dpa n /dt = (d/dt) B qB xB = a B qB vB . recalling the definition Qab n =3 Combining these results gives a = Jn for the molecular quadrupole moment.13)–(4.29) mn (t) = 2 B and we note that the vector ∇w × mn has components ∇w × mn a = = = 1 2 1 2 B b a b qB xa B vB − vB xB ∇b w b a b qB xa B vB + vB xB ∇b w − a b q B vB xB ∇b w B B 1 d ab ∇b w − Q 6 dt n a b B qB xB xB B a b q B vB xB ∇b w.25) is almost always negligible.60 with the definition Electromagnetic waves in matter 1 (4.27) (4. x) Jn (t. Jn (t. The macroscopic average of the molecular current density is a Jn = B a vB w(x − xn − xB ) a q B vB w− a b q B vB xB ∇b w + · · · . x).28) = B = n and the total microscopic current density is J = Jfree + Jbound . the effective electric field D is the electric field E augmented by the electrical response of the medium. qB vB (t)δ x − xn − xB (t) .15) we introduce the current densities Jfree (t.30) and summing over n yields a Jbound = ∂ ∂t n 1 pa n w − ∇b 6 n Qab n w + ∇× mn w n a + ···. (4. 4. (4.1.1.1. In practice the term involving Qab in Eq.25) Da = ǫ0 Ea + Pa − ∇b Qab + · · · .1. In the first term we recognize a the time derivative of the molecular dipole moment.1.1.1. (4. = j.4. ∂t The vector within brackets is the macroscopic field H (t. and j is the macroscopic free-current density.5 Summary — macroscopic Maxwell equations The macroscopic Maxwell equations are ∇·D = ρ = 0.38) (4. (4.35) 4.1. x).32) for the macroscopic average of the density of bound current.1.1.1.1. The fields D and H are defined by D = ǫ0 E + P (4. (4.12) gives ∇ × B /µ0 − M − ∂D = j.37) (4. Substituting Eq.1. x) ≡ n mn δ (x − xn ) + ···. (4.34) into Eq. x) ≡ qA vA δ (x − xA ) .25).36) Notice that this macroscopic Maxwell equation involves only the density of free currents on the right-hand side. and that apart from a factor of µ0 .1. (4. which is described by the molecular magnetic-moment density.1. x) ≡ Jfree (t. (4. and we arrive at ∇×H − with the definition H = B /µ0 − M .1.41) .1. ∂t (4.33) A and the macroscopic average of the total current density is therefore J =j+ ∂ D − ǫ0 E + ∇ × M + · · · . The average of the free-current density is simply j (t.31) as the macroscopic average of the magnetic-moment density of the medium’s molecules. ∂D =j ∂t (4.39) (4.34) having used Eq.1. we finally arrive at the expression a = Jbound 1 ∂ Pa − Qab ∂t 6 + (∇ × M )a + · · · (4.1. Here ρ is the macroscopic free-charge density.40) ∇·B ∂B ∇×E+ ∂t ∂D ∇×H − ∂t = 0. the effective magnetic field H is the magnetic field B diminished by the magnetic response of the medium.1.1.1. n mn (t)δ (x − xn ) ≡ macroscopic magnetization (4.1 Macroscopic form of Maxwell’s equations ∂ ∂t 1 pa n δ (x − xn ) − ∇b 6 a 61 Qab n δ (x − xn ) = n n + ∇× Introducing M (t. The inverse transformation is ˜ (ω.2.10) . = B (4.1. To begin we will look for monochromatic waves oscillating with an angular frequency ω .2.2. so that we can set M = 0 is our equations. We want to form plane-wave solutions to these equations.2) (4. µ0 pn (t)δ (x − xn ) (4. x) D (t. and leave unexplored issues of reflection and transmission at boundaries. solutions with different frequencies can be added to form wave-packet solutions. that the relation between D and E is itself linear). The restriction to monochromatic waves is not too severe.2.44) n the macroscopic magnetization (mn = moment).5) (4.2 Maxwell’s equations in the frequency domain [The material presented in this section is also covered in Sec. 7. We shall assume that the medium is uniform.1) (4. We suppose that the medium is not magnetized.3) (4. = P ˜ (x)e−iωt . = D ˜ (x)e−iωt . x) B (t.8) with the understanding that only the real parts of these complex fields have physical meaning. x)eiωt dt.2. We further suppose that the waves are propagating in the absence of sources. so that we can also set ρ = 0 and j = 0.9) ˜ (ω.] We now begin our discussion of electromagnetic waves propagating in a dielectric medium. x) P (t. x). we can express the electric field as 1 E (t.43) is the macroscopic polarization (pn = B qB xB is the molecular dipole moment) and M= mn (t)δ (x − xn ) (4. We therefore set E (t.6) (4.1. isotropic. and unbounded.2.2. and examine how the waves interact with the medium. 1 2 B qB x B × vB is the molecular magnetic 4. where D = ǫ0 E + P .2. Because the field equations are linear (assuming. x)e−iωt dω.1 of Jackson’s text. Maxwell’s equations therefore reduce to ∇·D ∇·B ∂B ∇×E+ ∂t ∂D ∇ × B − µ0 ∂t = = 0 0. this is a Fourier representation of in terms of the frequency-domain field E the electric field.2. (4.4) = 0. as we shall verify below.1. x) = √ 2π ∞ −∞ ˜ (ω. x) ˜ (x)e−iωt .42) (4. = E ˜ (x)e−iωt .62 and H= where P = n Electromagnetic waves in matter 1 B − M. E (4. For example. x) = √1 E 2π ∞ E (t. −∞ (4.2. = 0.7) (4. 14) ˜ ˜ + iωµ0 D ∇×B and we have the relation = 0. x) can be defined a similar and frequency-domain fields D way.1).2) for the frequency-domain displacement. we incorporate this effect through a phenomenological damping parameter γ > 0. They can only be solved once an explicit relationship (called a constitutive relation) is introduced ˜ and the electric field E ˜ . xn )e−iωt .11) (4. ˜ = ǫ0 E ˜+P ˜. What we have. x).] We suppose that the medium is sufficiently dilute that the electric field felt ˜ itself. the molecular dipole moment is zero. Notice that the applied electric field. P ˜ (ω. = 0.3. all sharing the same natural frequency ω0 . therefore. 7. the field exerted by other by any given molecule is just the applied field E molecules being negligible.1) where overdots indicate differentiation with respect to t.5 of Jackson’s text. 4.3. (4. To solve the equations of motion ˜B e−iωt and we substitute this into in the steady-state regime we write ξB (t) = ξ Eq. The equations of motion for the charge qB are therefore 2 ¨B + γ ξ ˙B + ω0 ˜ (ω.12) (4. This yields ˜B (ω ) = ξ qB /mB ˜ 2 − ω 2 − iγω E (ω. is evaluated at the molecule’s centre of mass. at least on average. When an electric field is applied. The displacement of charge qB from its equilibrium position is denoted ξB (t). This charge undergoes a motion that is governed by the applied electric field (which tends to drive ξB away from zero) and the intra-molecular forces (which tend to drive the charge back to its equilibrium position).13) (4. mB ξ ξ = qB E (4.15) These are the sourceless Maxwell equations in the frequency domain. xn ) ω0 (4. and its derivation is based on complicated molecular processes.2. in the next section we shall consider a simple classical model that nevertheless captures the essential features. however.3 Dielectric constant [The material presented in this section is also covered in Sec.2. some of the molecular charges (the electrons) move away from their equilibrium positions. B ˜ (ω. energy will gradually be removed from the oscillators (to be carried off by the radiation). which is macroscopic and varies slowly over intra-molecular distances. .2.3. (4. is a system of simple harmonic oscillators of natural frequency ω0 that are driven at a frequency ω by an applied force. and the oscillations will be damped. The constitutive relation between the polarization P describes the medium’s response to an applied electric field. x). and a dipole moment develops. Since the motion of the charges will produce electromagnetic radiation. D (4. The frequency-domain fields satisfy the equations ˜ ∇·D ˜ ∇·B ˜ − iω B ˜ ∇×E = = 0 0.3 Dielectric constant 63 ˜ (ω.2. We assume that in the absence of an applied field. We model these forces as harmonic forces.4. While quantum mechanics is required for a proper treatment.2. 3) for the molecular dipole moment in the frequency domain. (4. = = 1 Ze 2 − ω 2 − iγω m ω0 n Ze2 1 ˜ (ω. x) = Ze N ˜ (ω.3. respectively. xn )δ (x − xn ) E δ (x − xn ) .3. and that e and m are the electron’s charge and mass. x) P E 2 m ω0 − ω 2 − iγω (4. We express it as The relationship between D 2 1 ˜ = ǫ0 1 + Ze N ˜.7) Equation (4. Its macroscopic average is therefore the medium’s density N (the number of molecules per macroscopic unit volume). which we take to be uniform. x).3. We therefore arrive at p ˜n (ω ) = Ze2 1 ˜ (ω.2) then gives p ˜n = B 2 ω0 2 qB /mB ˜ (ω. xn ) E 2 m ω0 − ω 2 − iγω (4. xn ). that N is the molecular number density. where Z is the number of electrons per molecule. (4.2. x) E 2 m ω0 − ω 2 − iγω n The quantity within the averaging brackets is the microscopic molecular density. We notice that the relationship between ˜ and E ˜ is linear. 4. We recall that Z is the number of electrons per molecule. and frequency-dependent) dielectric constant 2 ωp ǫ(ω ) =1+ 2 ǫ0 ω0 − ω 2 − iγω (4. The polarization vector is then ˜ P = n 2 p ˜n δ (x − xn ) ˜ (ω. as was anticipated in Sec.6) is the product of a simplistic classical model. x) = ǫ(ω )E ˜ (ω. Because the dipole moment vanishes at equilibrium.3. where xB is the equilibrium position of charge qB . D (4. P ˜ ≡ ǫ0 E ˜+P ˜ and E ˜ also is linear. E − ω 2 − iγω 2 Since all the contributing charges qB are electrons. and m the electron’s mass. except for the fact . But a proper quantum-mechanical treatment would produce a similar result. e the electronic charge. this reduces to pn = B qB ξB . We have arrived at 2 1 ˜ (ω.3. D E 2 ǫ0 m ω0 − ω 2 − iγω or as ˜ (ω. we can write B qB /mB = 2 Ze /m. Substituting Eq.3.64 Electromagnetic waves in matter The molecule’s dipole moment is pn = B qB (xB + ξB ).5) having introduced the (complex.3.4) for the frequency-domain polarization.6) and the medium’s plasma frequency ωp = Ze2 N ǫ0 m 1/ 2 . 14) become ˜ ∇·E = 0 (4. monochromatic waves 65 0.6). To fit both curves on the same graph we removed 1 from Re(ǫ/ǫ0 ).2: Real and imaginary parts of the dielectric constant.4 Propagation of plane.5 w 2 2. monochromatic waves [The material presented in this section is also covered in Sec.5 3 Figure 4. Re(ǫ/ǫ0 ) decreases with increasing ω .1) . that the model is a bit crude. 7.] After making the substitution of Eq. and is therefore a constant slightly larger than unity.3. At low frequencies the dielectric constant behaves as ǫ/ǫ0 ≃ 1 + (ωp /ω0 )2 .2 we plot the real and imaginary parts of the dielectric constant. (4. the frequency-domain Maxwell equations (4. however. the electrons are driven at almost their natural frequency.4. and therefore approaches unity from below. In the rest of this chapter we will continue to deal with the simple expression of Eq. It is good to keep in mind. The imaginary part of the dielectric constant achieves its maximum at resonance. The real part of the dielectric constant goes through 1 at resonance. (4. that the electrons actually oscillate with a discrete spectrum of natural frequencies ωj and damping coefficients γj . At high frequencies it behaves as ǫ/ǫ0 ≃ 1 − (ωp /ω )2 .11)–(4. When ω is very close to ω0 . In Fig. as functions of ω/ω0 . then.2. and Im(ǫ/ǫ0 ) is no longer small.2.3.5). You will notice that for most frequencies. 4. and Im(ǫ/ǫ0 ) is very small.5 1 1. when ω is not close to ω0 .4. A more realistic expression for the dielectric constant is then ǫ(ω ) fj 2 = 1 + ωp 2 − ω 2 − iγ ω ǫ0 ω j j j where fj is the fraction of electrons that share the same natural frequency ωj . 4. Near resonance. however. Re(ǫ/ǫ0 ) increases with increasing ω . and resonance occurs. this behaviour is associated with anomalous dispersion. In realistic models the number of resonances is larger than one: there is one resonance for each natural frequency ωj .1 of Jackson’s text.4 Propagation of plane. this typical behaviour is associated with normal dispersion. (4. To help A right-moving wave is described by the particular solution ψ identify the physical properties of this wave we substitute Eq. while kI = 2 ω ωc γω 2 2 c 2 (ω0 − ω )2 + γ 2 ω 2 .2) (4. = 0. To see how monochromatic electromagnetic waves propagate in a dielectric ˜. so that ψ wave equation simplifies to d2 ˜(ω.1)–(4.4.66 Electromagnetic waves in matter ˜ ∇·B ˜ − iω B ˜ ∇×E = 0. ∇ × (∇ × E ˜) = substituting Eq.4. for example. consider any component of E ˜=ψ ˜(z ).4.5) (4.6) We have introduced a complex wave number k (ω ). and call it ψ medium. and we recall that c ≡ (ǫ0 µ0 )−1/2 is the speed of light in vacuum. Similar manipulations return ˜ (ω. (4.3.4) ˜ ˜ + iωǫ(ω )µ0 E ∇×B = 0.6) and get 2 ωp ω2 k2 = 2 1 + 2 .9) c ω0 − ω 2 − iγω . (4. For example. (4. x) = 0. both E isfy a frequency-domain wave equation.3) (4. Then suppose that ˜ or B ˜ . or as in Eq.4. where kR = 2 2 − ω2 ) ωp (ω0 ω 1+ 2 − ω 2 )2 + γ 2 ω 2 c 2 (ω0 2 ωp ω 1+ 2 − ω 2 − iγω c 2 ω0 ˜ = eikz .4. taking the curl of Eq.10) is the real part of the complex wave number. ˜ and B ˜ satIt is easy to show that as a consequence of Eqs.9) if the wave possesses a spectrum of frequencies. The the wave is a plane wave that propagates in the z direction.4). ˜ e−iωt We recall that the time-domain fields are reconstructed.4) and making use of the vectorial identity ∇ × (∇ × E 2 ˜ ˜ ∇(∇ · E ) − ∇ E puts this in the form ˜ (ω. as E = E if the wave is monochromatic.6) into Eq.4. (4. z ) = 0. Taking the square root gives k= or k = kR + ikI .7) the statement that the magnetic field also satisfies a wave equation. + k2 ψ dz 2 (4. ∇2 + k 2 B (4. (4.4. (4. ∇2 + k 2 E where k2 ≡ ω 2 ǫ(ω ) .4. x) = 0.2. (4.4.4.4.4.8) To be concrete we suppose that the second term within the large brackets is small. c2 ǫ0 (4. (4.4.3) gives ˜ ) − iω ∇ × B ˜ = 0.4. A positive damping constant means that some of the wave’s energy is dissipated by the oscillating electrons (which emit their own electromagnetic radiation).12) kR (ω ) n(ω ) known as the phase velocity. Another feature of Eq. so that vp < c.4.16) and (4. (4.4.4. (4.4. (4. which implies B0 (ω ) = and ˆ b=z ˆ× e ˆ. which merely confirms Eqs. The right-moving wave is therefore described by ˜(ω. n ≃ 1 − 2 (ωp /ω ) < 1 for high frequencies. and where E0 and B0 are complex amplitudes. As we shall see. so that vp > c.3) then produces B0ˆ b = (k/ω )E0 (z ˆ× e ˆ). From our previous expression for kR we obtain n=1+ 2 − ω 2 )2 + γ 2 ω 2 2 (ω0 2 2 − ω2 ) ωp (ω0 . We write ˜ (ω. On the 1 2 other hand. To form a signal one must modulate the wave and superpose solutions with different frequencies.5) and (4. 2 and we see that n ≃ 1 + 1 2 (ωp /ω0 ) > 1 for low frequencies. because a monochromatic wave does not carry information.16) .4. The phase velocity of the wave can therefore exceed the speed of light in vacuum! This strange fact does not violate any cherished relativistic notion.4. Equation (4. n ≡ ckR /ω is the medium’s index of refraction. Eq.4) place on these solutions.17). (4.1) and (4. Substituting these expressions into Eqs. (4.2) informs us that the unit vectors are both orthogonal to z ˆ.4. Finally. in such media the damping constant is negative. z ) = ˆ B bB0 (ω )eik(ω)z .17) These equations state that the magnetic field is orthogonal to the electric field. monochromatic waves 67 is its imaginary part. It is interesting to note that in the case of lasers and masers. in such situations the wave always travels with a speed that does not exceed c.4. ψ (t.4.4.13) and ˜ (ω.4. (4.4.4. z )e−iωt = e−kI z e−i(ωt−kR z) .14) where e ˆ and ˆ b are real unit vectors. (4. and the wave must therefore decrease in amplitude as it travels through the medium.4.4.4.15) the fields are therefore transverse to the direction in which the wave propagates.7).4. which is guaranteed to be true if γ > 0. Having constructed solutions to Eqs. the electronic oscillations actually reinforce the wave and lead to an exponential amplification instead of an attenuation. e ˆ· z ˆ=ˆ b·z ˆ = 0. we should examine what constraints the full set of equations (4.4) gives us B0 (z ˆ×ˆ b) = −(k/ω )E0 e ˆ.11) From this expression we recognize that the wave is indeed traveling in the positive z direction.1)–(4. z ) = e E ˆE0 (ω )eik(ω)z (4. (4.4. k (ω ) E0 (ω ) ω (4. with a speed ω c vp = ≡ (4. and that its amplitude is linked to that of the electric field. z ) ≡ ψ (4.10) is that the wave is an exponentially decreasing function of z provided that kI is positive.4 Propagation of plane.4. 3) then simplifies to The procedure to follow to construct a wave packet is therefore to choose an ini˙ (0. differentiating Eq.and right-moving waves.2) is equivalent to that of Eq.6).9 of Jackson’s text. Including both left.5. z ) and ψ Eq. (4.5. (4. z ) = √ 2π ∞ A+ (k )eiω(k)t + A− (k )e−iω(k)t eikz dk. z ). that ǫ(ω ) is real: we work away from resonances and ignore dissipative effects.5.2) with respect to t produces iω (k ) A+ (k ) − A− (k ) eikz dk.] 4.5. z ) ± 1 ˙ ψ (0. obtain ψ (t. then calculate A± (k ) using tial configuration by specifying ψ (0.4. −∞ (4.5. we finally obtain A± (k ) = 1 1 √ 2 2π ∞ −∞ ˙ (0. and finally. z ) = √ 2π ∞ a+ (ω )eik(ω)z + a− (ω )e−ik(ω)z e−iωt dω. Alternatively. ψ ψ (0.4. (4. we write 1 ψ (t. z )e−ikz dz.5.2). (4. for simplicity.5) and we have an equal superposition of left.1) where a± (ω ) are complex amplitudes that determine the shape of the wave packet. In the following discussion we will choose the initial configuration to be timesymmetric. we can instead superpose solutions with different wave numbers k . ψ (0.68 Electromagnetic waves in matter 4. We let k (ω ) be the positive solution to Eq. −∞ (4.1).6).5.3) Equation (4. The complex amplitudes A± (k ) are determined by the initial conditions we wish to impose on ψ (t. ψ (4.and right-moving waves.1 Description of wave packets In this section we shall assume. (4. It follows from Eq.5.4. z ) and its time derivative. −∞ On the other hand.5. z ) = √1 ψ 2π and inversion gives ∞ −∞ ∞ ψ (0.5.8) with different frequencies ω . z ) = √ 2π −∞ and Fourier’s inversion theorem gives 1 A+ (k ) + A− (k ) = √ 2π ˙ (0.3).8 and 7. z ) ≡ 0.2) that ∞ 1 A+ (k ) + A− (k ) eikz dk.5 Propagation of wave packets [The material presented in this section is also covered in Secs. we write 1 ψ (t. iω (k ) (4. . (4. z ) by evaluating the integral of Eq. 7.5.5. z )e−ikz dz. (4. (4. (4. in the sense that ˙ (0. A wave packet is obtained by superposing solutions to Eq. z ) e−ikz dz. ∞ −∞ 1 iω (k ) A+ (k ) − A− (k ) = √ 2π Solving for A± (k ).2) and it is not too difficult to show that the decomposition of Eq. z )e−ikz dz.4) A(k ) ≡ A± (k ) = 1 1 √ 2 2π ∞ ψ (0.5. Defining ω (k ) to be the positive solution to Eq. −∞ (4. and more conveniently for our purposes. −∞ 1 1 √ 2 2π ∞ −∞ ψ (0. it represents that part of the wave packet that travels undisturbed in the negative z direction with a speed vp . that is. Eq.5. ′ dz ′ ψ (0.] Suppose further that ω (k ) varies slowly near k = k0 . so that it can be approximated by ′ ′ ′ ω (k ) ≃ ω0 + ω0 (k − k0 ) = (ω0 − ω0 k0 ) + ω0 k. 1 1 ψ (0. It is not difficult to make this change. translated in the z direction by −vp t. when n ≡ c|k |/ω is a constant independent of k .2) reduces to 1 ψ (t.5 Propagation of wave packets 69 4.8) .5. suppose that A(k ) is sharply peaked around k = k0 . this would produce a real wave function ψ (t. z ′ ) δ (z + vp t − z ′ ) + δ (z − vp t − z ′ ) . (4. z ′ )e−ikz dz ′ . z ) = = or 1 2 1 2 ∞ ∞ A(k ) eik(z+vp t) + eik(z−vp t) dk. z ).6) where vp = c/n is the wave’s phase velocity. ψ (t. z ) = √ 2π But from Eq.5.5) we have A(k ) = and substituting this gives ψ (t. on the other hand. is the remaining half of the wave packet.5.4.5. (4. and for time-symmetric initial data we have 1 ψ (t.5. The second term. z ). z ) = 4. Unlike what we shall find below. −∞ The factor within the square brackets is 2 cos(|k |vp t) = 2 cos(kvp t).5. (4. (4. and we can therefore replace the previous expression with 1 ψ (t. with k0 some arbitrary wave number. z ) = √ 2π ∞ A(k ) ei|k|vp t + e−i|k|vp t eikz dk. z ′ ) −∞ ∞ −∞ 1 2π ∞ −∞ eik(z+vp t−z ) + eik(z−vp t−z ) dk ′ ′ dz ′ ψ (0.7) 2 2 The first term on the right-hand side is (half) the initial wave packet. −∞ To be concrete.2 Propagation without dispersion Let us first see what the formalism of the preceding subsection gives us when there is no dispersion. which travels toward the positive z direction with the same speed vp . z + vp t) + ψ (0. ψ (0.3 Propagation with dispersion Let us now return to the general case of propagation with dispersion. z ) = √ 2π ∞ A(k ) eiω(k)t + e−iω(k)t eikz dk. [A proper treatment would have A(k ) be a symmetric function peaked about both k = k0 and k = −k0 . (4. The main features of wave propagation without dispersion are that the wave packet travels with the phase velocity vp = ω/k and that its shape stays unchanged during propagation. For time-symmetric initial data. Here ω is a nonlinear function of k . We then have ω (k ) = vp |k |. but we shall not do so here. z − vp t). ′ ≡ (dω/dk )(k0 ).5.01 1 0 0. These are the main features of wave propagation in a dispersive medium. z + vg t) + e−i(ω0 −ω0 k0 )t ψ (0.3. 4.5 w 2 2. (4. c n + ω (dn/dω ) For the dielectric medium described in Sec. but that it seems to exceed c near resonance.10) dk is known as the group velocity.02 1.5.9) we learn that in the presence of dispersion. 2 − ω )2 + γ 2 ω 2 (4.03 1. z − vg t). in units of c.3. 4. in terms of the index of refraction n(ω ).11) and the group velocity is plotted in Fig. And the phase factors indicate that the shape of the wave suffers a distortion during propagation. Then the wave packet can be expressed where ω0 ≡ ω (k0 ) and ω0 as ψ = 1 √ 2π ∞ −∞ ∞ −∞ A(k ) eik(z+ω0 t) ei(ω0 −ω0 k0 )t + eik(z−ω0 t) e−i(ω0 −ω0 k0 )t dk A(k )eik(z+ω0 t) dk ∞ −∞ ′ ′ ′ ′ ′ ′ ′ 1 = ei(ω0 −ω0 k0 )t √ 2π +e or where ′ −i(ω0 −ω0 k 0 )t 1 √ 2π A(k )eik(z−ω0 t) dk. (4. If the dispersion relation is expressed as k = n(ω )ω/c. You will notice that vg is smaller than c for most frequencies. This . z ) = ei(ω0 −ω0 k0 )t ψ (0.5 3 Figure 4.5. n2 = 1 + 2 (ω0 2 2 (ω0 − ω2 ) ωp .5 1 1. then dk/dω = c−1 [n + ω (dn/dω )] and 1 vg = . From Eq.70 Electromagnetic waves in matter 1. vg = ′ dω (k0 ) (4.5. as a function of ω/ω0 .04 1.3: Group velocity vg in a dielectric medium. the wave packet travels with the group velocity vg instead of the phase velocity vp = k0 /ω0 .9) ′ ψ (t. z ) = 0.12). (4.5 Propagation of wave packets 71 phenomenon is entirely artificial: when ω is close to ω0 it is no longer true that ω is a slowly varying function of k . Once more we take the initial data to be time-symmetric. for the second integral we simply make the substitution k0 → −k0 −2 in this result. In order to be able to carry out the calculations we shall take impose ψ the dispersion relation to be 1 ω (k ) = ν 1 + α2 k 2 . and b2 /(4a) = 1 (k − k0 )2 L2 . For the first integral we have a = 1/(2L2 ).12) this is an oscillating wave (with wave number k0 ) modulated by a Gaussian envelope of width L.13) where ν is a minimum frequency and α an arbitrary length scale. but Re(a) must be positive for the integral to converge.5. b. together with the dissipative effects it represents.14) then gives A(k ) = 2 2 1 L − 1 (k−k0 )2 L2 e 2 + e− 2 (k+k0 ) L .5. and that it is peaked at both k = k0 and k = −k0 . the imaginary part of ω .5. 1 ψ=√ 2π 1 A(k )ei[kz−ω(k)t] dk + √ 2π −∞ ∞ ∞ A(k )ei[kz+ω(k)t] dk. cannot be neglected near resonance.5.14) Here the constants a.4. and the approximations involved in defining the group velocity are no longer valid. These will have the form of Gaussian integrals. c = 0. We first calculate the Fourier amplitudes associated with Eq. This dispersion relation is not particularly realistic. The wave packet at times t = 0 is given by Eq. 4 (4. z ) = e−z /(2L ) cos(k0 z ). 4.5. Equation (4. −∞ which we may also express as 1 ψ=√ 2π 1 A(k )ei[kz−ω(k)t] dk + √ 2π −∞ ∞ ∞ −∞ A(−k )e−i[kz−ω(−k)t] dk. b = −i(k − k0 ). (4.5. 2 (4.5. and c can be complex.5.2). The width of the distribution around both peaks is equal to 1/L.15) We see that A(k ) is a symmetric function of k . but its quadratic form will allow us to evaluate the integrals that will appear in the course of our analysis.5. and we quote the standard result ∞ e−az −∞ 2 +bz +c dz = b2 π exp +c . . and we ˙ (0. (4. z )e−ikz dz −∞ ∞ e−z −∞ 2 /(2L2 ) e−i(k−k0 )z + e−i(k+k0 )z dz. The bottom line is that the very notion of group velocity is no longer meaningful near a resonance.5) gives A(k ) = = 1 1 √ 2 2π 1 1 √ 4 2π ∞ ψ (0.4 Gaussian wave packet As a concrete example of wave propagation with dispersion we consider an initial configuration given by 2 2 ψ (0. Equation (4. Furthermore. a 4a (4. The second (higher curve — the wave function is translated upward for ease of visualization) is obtained with dispersion. this wave spreads out as it travels with speed vg < c.5 initial wave 0 -0.4 propagated wave 0.12). The upper panel shows the initial wave profile. The first (lower curve) is obtained without dispersion by setting ωp = 0. . Notice the drop in amplitude as the initial wave packet splits into an equal superposition of left.4: Wave propagation with a high-frequency dispersion relation ω (k ) = 2 + c2 k 2 )1/2 .72 Electromagnetic waves in matter 1 0. calculated numerically with the help of a Fast Fourier Transform (ωp routine.5 -40 0. this wave propagates with speed c without altering its shape.2 -0. described by Eq.4 -40 -20 0 z 20 40 Figure 4. The lower panel shows two versions of the propagated wave. (4.and right-moving waves.6 -20 0 z 20 40 no dispersion dispersion 0.2 0 -0.5. (This problem is adapted from Jackson’s problem 7. which we write as −ak 2 + bk + c.18) (4. With this we have b2 (z − ik0 L2 )2 =− 2 .4.6 Problems 1. and Eq. (4. Notice that the features discussed in the preceding paragraphs are present in this example as well: the wave packet spreads out as it propagates with a group velocity vg that is smaller than c. Equation (4.5. and c = −(iνt + 1 a= 2 2 k0 L ).5. this result was expected after the discussion of the preceding subsection. (4. 2π −∞ Substituting Eq. For each of the forms f (z ) given below. the centre of the Gaussian peak travels at the group velocity vg .15) then gives L ψ = √ Re 2 2π ∞ −∞ e− 2 (k−k0 ) 1 2 L2 i[kz −ω (k)t] e dk + (k0 → −k0 ) . we see that the argument of the exponential function is a second-order polynomial in k . Re (1 + iνtf 2 )−1/2 exp − 2 2 2L (1 + iνtf 2 ) It is then a matter of straightforward algebra to bring this to its final form. calculate the wave-number spectrum |A(k )|2 of the wave packet. we have ψ = = ∞ 1 √ A(k )ei[kz−ω(k)t] dk + complex conjugate.) A wave packet in one dimension has an initial shape given by ψ (0.13). calculate the root-mean-square deviations ∆z and ∆k from the means (defined in terms of |ψ (0. the wave packet is assumed to be time-symmetric.14) yields ψ= 1 2 2 1 (z − ik0 L2 )2 e−iνt− 2 k0 L + (k0 → −k0 ) . .16) The wave therefore spreads out.17) dk is the group velocity at k = k0 for the dispersion relation of Eq. z ) = f (z )eik0 z . and test the inequality ∆z ∆k ≥ 1 2. with 1 2 2 (L2 + iνtα2 ). sketch |ψ (0.19. (4. where f = α/L and dω (k0 ) (4. 2π −∞ ∞ 2 √ Re A(k )ei[kz−ω(k)t] dk.5.6 Problems 73 having changed the variable of integration to −k in the second integral. 4. z ) = (z − vg t)2 1 ei[k0 z−ω(k0 )t] Re (1 + iνtf 2 )−1/2 exp − 2 2 2L (1 + iνtf 2 ) + (k0 → −k0 ) . In addition. A more realistic example of wave dispersion (calculated numerically) is displayed in Fig. 4a 2L (1 + iνtf 2 ) where f ≡ α/L. 4.5. we see that while the oscillations travel at the phase velocity vp = ω (k0 )/k0 . ψ (t.5. z )|2 and |A(k )|2 . (4. and this is a typical feature of propagation with dispersion. z )|2 and |A(k )|2 . Taking into account Eq.13).5. (4. where f (z ) is the modulation envelope. Because A(−k ) = A(k ) and ω (−k ) = ω (k ). but that its width increases with time: vg = να2 k0 = L(t) = L 1 + (f 2 νt)2 .5. b = k0 L2 + iz .5.16) informs us that the wave packet retains its Gaussian shape as it propagates through the medium.4. respectively). x) + 2 ωp ν ∞ 0 e−γτ /2 sin(ντ ) E (t − τ. indicates that delays longer than 1/γ do not matter very much: only the electric field’s recent history is important. ˜ 2. b) f (z ) = e−α 2 2 Electromagnetic waves in matter z /4 . This reveals that D at time t depends on the electric where ν ≡ ω0 4 field at all previous times: the relationship is nonlocal in time. We have seen that the relationship between the frequency-domain fields D ˜ is and E ˜ (ω.74 a) f (z ) = e−α|z|/2 . however. d) f (z ) = 1 for α|z | < 1 and f (z ) = 0 for α|z | > 1. . The exponential factor. ǫ0 ω0 − ω 2 − iγω Prove that as a consequence of this. 2 − 1 γ 2 . x) = ǫ0 E (t. D where 2 ωp ǫ(ω ) =1+ 2 . x). c) f (z ) = 1 − α|z | for α|z | < 1 and f (z ) = 0 for α|z | > 1. x) = ǫ(ω )E ˜ (ω. the relationship between the time-domain fields D and E is D (t. x) dτ . x′ ) 3 ′ d x |x − x′ | j (t − |x − x′ |/c. We have wave equations for the scalar and vector potentials. x) ǫ0 (5.1.3) is the wave operator and c ≡ (ǫ0 µ0 )−1/2 is the speed of light. (5. the wave equations have solutions Φ(t.5) ∂t which states that charge is locally conserved. Φ(t. We recall the main equations of electrodynamics. which incorporate the physically appropriate causal relationship between sources and fields.1.7) these are retarded solutions.2) to be equivalent to Maxwell’s equations.1. 1 ∂Φ ∇·A+ 2 = 0. x). where ≡− 1 ∂2 + ∇2 c2 ∂t2 (5. In the absence of boundaries. x) = −µ0 j (t.1. here we shall restrict our attention to charge and current distributions whose internal velocities are small compared with the speed of light. |x − x′ | (5. x) = and A(t.Chapter 5 Electromagnetic radiation from slowly moving sources 5.1. x) = − and A(t.1) and (5.1.1. (5.1.6) (5.4) c ∂t Maxwell’s equations imply that the charge and current densities satisfy the continuity equation. (5.1) (5.1. While Chapter 6 will be concerned with relativistic situations. 75 . x′ ) 3 ′ d x. the potentials must satisfy the Lorenz gauge condition. as we have formulated them in Chapter 1.2) 1 ρ(t.1 Equations of electrodynamics In this chapter and the next we will explore some of the mechanisms by which electromagnetic radiation is produced. x) = 1 4πǫ0 µ0 4π ρ(t − |x − x′ |/c. ∂ρ + ∇ · j = 0. For Eqs. We take each charge to be oscillating in the x direction with a velocity v (t. |x − x′ | in which we substitute Eq. µ0 (5. (5. The current density is j (t. We take the charge distribution to be uniform in the x-y plane. Notice that the charge density is actually time independent.1.2) Notice that the velocity vector is actually independent of position in the plane. 1.76 Electromagnetic radiation from slowly moving sources From the potentials we obtain the fields via the relations E=− ∂A − ∇Φ ∂t (5.10) and the energy flux vector (Poynting vector) is S (t. (5.8) and B = ∇ × A.7). and the electromagnetic waves will be described entirely by the vector potential. x) = δ (z ′ ) µ0 σv0 x ˆ 4π dt′ Γ(t − t′ . we recall from Sec.1.3) and we will calculate the associated vector potential without approximations. (5. x) cos ωt′ . and that x ˆ is actually an arbitrary direction within the plane. x) = 1 1 ǫ0 E 2 + B2 2 2µ0 (5.2. x) = v0 x ˆ cos ωt.1. We have A(t. x) = 1 E × B. x) = σδ (z ). x) = µ0 4π j (t′ . we could pick another direction by rotating the x-y coordinates around the z axis.3) expressed in terms of primed variables. (5. x′ ) δ (t − t′ − |x − x′ |/c) ′ 3 ′ dt d x .1) where σ is a constant surface density.2 Plane waves The simplest situation featuring the production of electromagnetic waves involves monochromatic plane waves. These are generated by a planar distribution of charges. (5. . but it is easier to start instead with the equivalent form A(t. The electromagnetic field energy density is given by ε(t. (5. δ (t − t′ − |x − x′ |/c) 3 ′ d x |x − x′ | is an integral that we must now evaluate.1. (5.2.11) 5. x) = σv0 δ (z )x ˆ cos ωt.2.1.3) into Eq. all oscillating in the plane with the same frequency ω .2. In principle we might proceed by substituting Eq.2. The resulting scalar potential will also not depend on time. so that the charge density is given by ρ(t. x) = where Γ(t − t′ .4 that the fields carry energy.9) And finally. This gives. Thus. we have that s is bounded below by |z |/c − ∆t. we obtain A(t. where k = ω/c is the wave number. y ). = 2 ds = √ 2 2 c (s + ∆t) c ℓ +z and the integral becomes ∞ Γ = 2πc |z |/c−∆t δ (s) ds. Because ℓ is nonnegative. the wave has a frequency ω . or Γ(t − t′ .4) 1 µ0 σv0 c (5. Eq. a wave number k = ω/c. Γ = 2π ℓ2 + z 2 0 The easiest way to integrate over √ the δ -function is to make its argument be the integration variable. the place where it is generated. For z < 0 we have instead sin ω (t − |z |/c) = sin ω (t + z/c) = sin k (z + ct). z ) = where E0 sin ω (t − |z |/c)x ˆ. Substituting this result into our previous expression for the vector potential.5. We also have ℓ dℓ ℓ dℓ . apart from an ambiguous constant associated with the integral’s infinite lower bound.2. To integrate this we make the change of variables x′ = ℓ cos θ. and we see that the wave travels in the positive z direction with speed c. Our final expression for the vector potential is therefore A(t. y ′ = ℓ sin θ.2. For z > 0 we have that sin ω (t − |z |/c) = sin ω (t − z/c) = − sin k (z − ct). after writing dx′ dy ′ = ℓ dℓdθ and integrating over dθ. Since this constant will never appear in expressions for the electric and magnetic fields.5) 2 has the dimensions of an electric field. 77 (x − x′ )2 + (y − y ′ )2 + z 2 or Γ= δ (t − t′ − x′2 + y ′2 + z 2 /c) x′2 + y ′2 + z 2 dx′ dy ′ after translating the origin of the (x′ . y ′ ) coordinates to the point (x. So let s = ℓ2 + z 2 /c − ∆t. because the domain of integration is the whole x′ -y ′ plane. and it travels with the speed of light. ω (5. −∞ The dt′ integral evaluates to ω −1 sin ω (t − |z |/c). and the wave now travels in the negative z direction. √ ∞ δ (t − t′ − ℓ2 + z 2 /c) √ ℓ dℓ.2. where ∆t ≡ t − t′ . (5. where θ(s) is the step function. this is allowed. x) = 2πcθ(t − t′ − |z |/c). E0 = .2 Plane waves Integration over dz ′ is immediate: Γ= δ (t − t′ − (x − x′ )2 + (y − y ′ )2 + z 2 /c) dx′ dy ′ . we simply set it to zero.4) describes a plane wave that propagates away from z = 0. which is unity when s > 0 and zero otherwise. x) = = µ0 (σv0 )(2πc)x ˆ 4π µ0 2πσv0 c x ˆ 4π ∞ −∞ t−|z |/c dt′ θ(t − |z |/c − t′ ) cos ωt′ cos ωt′ dt′ . 4): E (t.3 Spherical waves — the oscillating dipole [The material presented in this section is also covered in Secs. and (5.2 of Jackson’s text. on the other hand. we note that . From Eqs. and that it points in the x direction. • The fields are in phase.2. which is +1 for z > 0 and −1 if z < 0. Notice also that |B | is smaller than |E | by a factor 1/c. The only caveat is that in general. away from the source. To finish off our discussion of plane electromagnetic waves. z ) = −E0 cos ω (t − |z |/c)x ˆ. z ) = ǫ0 E0 cos2 ω (t − |z |/c). To see that Eq. we calculate the field’s energy density and flux vector.10). (5.1 and 9.11) we get S (t. the direction of propagation is not uniform: the constant vector z ˆ goes into a vector n ˆ that may change from point to point.1) does indeed form a valid solution to the wave equation. the plane wave travels in the positive z direction with speed c.1.7) we obtain 2 ε(t. (5. the wave travels in the negative z direction.6) We see that the electric field behaves as a plane wave that travels away from z = 0 with the speed of light. z )z ˆ. The magnetic field.9) This shows that the electromagnetic field energy travels with the wave. z ) = f (t ∓ z/c). 9. (5.1.3. (5.3. 5. we have E0 ǫ(z ) cos ω (t − |z |/c)y ˆ. Let us summarize the main points: • The plane wave travels in the (positive or negative) z direction with speed c. Denoting this function by ǫ(z ) ≡ θ(z ) − θ(−z ). (5.2. • The electromagnetic field energy travels with the wave.8) and (5. is obtained by taking the curl of Eq. (5. z ) = c The magnetic field also is a plane wave that travels away from z = 0 at the speed of light.1) where f (ξ ) is an arbitrary function of ξ = t ∓ z/c.78 Electromagnetic radiation from slowly moving sources The time-dependent part of the electric field is obtained from Eqs.2. and it points in the y direction.] 5.1 Plane versus spherical waves Any function of the form ψ (t. in the positive or negative z direction. (5. which is perpendicular to the direction of propagation.2. These properties are generic: they are shared by all electromagnetic waves. • The electric and magnetic fields are perpendicular to z ˆ and to each other. (5. z ) = cǫ(z )ε(t. is a solution to the wave equation ψ = 0. If ξ = t − z/c.2. Notice that E and B are both proportional to cos ω (t − |z |/c): the fields are in phase.3.2. (5.1.7) B (t. and with the speed of light. (5.6. which is perpendicular to both the direction of propagation and the direction of the electric field.6). • Their magnitudes are related by |B |/|E | = 1/c.8) and from Eq. This involves differentiating |z | with respect to z . (5. if instead ξ = t + z/c.4).2. We first compute the vector potential produced by the current density of Eq. This is also a solution to ψ = 0. That Eq. (5. r where f (u) is an arbitrary function of u = t − r/c. and examine the nature of the corresponding electromagnetic radiation. ∂r ∂r c c c c where an overdot indicates differentiation with respect to u. Here the wavefronts are spherical instead of planar. x′ ) δ (t − t′ − |x − x′ |/c) ′ 3 ′ dt d x . c ∂t r ∂r ∂r ¨.3.4).3. Note that a similar calculation would reveal that ψ = r−1 f (t + r/c) also is a solution to the wave equation. because ∇·j =− ∂ρ = −ω p · ∇δ (x) sin ωt = ∇ · −ω pδ (x) sin ωt .3 Spherical waves — the oscillating dipole 79 ∂ψ/∂t = f ′ (ξ ) while ∂ψ/∂z = ∓f ′ (ξ )/c. The charge density of an oscillating dipole is then ρ(t. p is still a constant vector. as required.3. A better approximation that incorporates these features is the spherical wave 1 (5. A time-varying charge density must be associated with a current density j .5. where p was a constant dipole moment vector. where a prime indicates differentiation with respect to ξ . its amplitude decays as 1/r. All this gives ψ = 0. x) = µ0 4π j (t′ . 5. (5. |x − x′ | . substitution into the wave operator yields ψ = −c−2 f ′′ + c−2 f ′′ = 0. which we align with the z axis: p = pz ˆ.2) ψ (t. as required.2) forms a valid solution to the wave equation follows by direct substitution: In the absence of an angular dependence the wave operator reduces to 1 ∂2 1 ∂ ∂ = − 2 2 + 2 r2 .3) Notice that in this equation. r) = f (t − r/c). (5. and the wave propagates outward in the radial direction.3.2 Potentials of an oscillating dipole The simplest electromagnetic spherical wave is produced by an oscillating dipole located at the origin of the coordinate system. which originate in a localized region of space and propagate outward with a decreasing amplitude.3. giving rise to an intensity that goes as 1/r2 . while we have r2 ∂ψ/∂r = Differentiating with respect to t yields ∂ 2 ψ/∂t2 = r−1 f ˙ −rf /c − f and 1 1 ¨ 1 ˙ 1 ¨ ∂ 2 ∂ψ r = − f˙ + 2 rf + f = 2 rf . 2.3 we saw that the charge density of a point dipole could be expressed as ρ(x) = −p·∇δ (x).3. Here we introduce a time dependence by making p oscillate with angular frequency ω : p → p cos ωt. In Sec. Plane waves are not a very good approximation to realistic waves.4) and this is the current density of an oscillating dipole. this wave travels inward instead of outward. x) = −p · ∇δ (x) cos ωt. (5. Here and in the following subsections we will calculate the potentials and fields produced by this dipole. ∂t The simplest solution to this equation is j (t. x) = −ω pδ (x) sin ωt. Once more it is convenient to start with the expression A(t. 80 Electromagnetic radiation from slowly moving sources After substituting Eq. (5.3.5) we obtain A(t, x) µ0 ωp 4π µ0 = − ωp 4π = − δ (t − t′ − |x − x′ |/c) 3 ′ d x |x − x′ | δ (t − t′ − r/c) dt′ sin ωt′ , r dt′ sin ωt′ δ (x′ ) sin ω (t − r/c) µ0 ωp . (5.3.5) 4π r This has the structure of a spherical wave, with its expected dependence on u = t − r/c and its amplitude decaying as 1/r. We may now compute the scalar potential. This calculation is a bit more laborious, and we start with A(t, x) = − Φ(t, x) = 1 4πǫ0 ρ(t′ , x′ ) δ (t − t′ − |x − x′ |/c) ′ 3 ′ dt d x . |x − x′ | or After substituting Eq. (5.3.3) we get Φ(t, x) = − where Γ(t − t′ , x) = ∇′ δ (x′ ) 1 p· 4πǫ0 dt′ Γ(t − t′ , x) cos ωt′ , δ (t − t′ − |x − x′ |/c) 3 ′ d x |x − x′ | is a vectorial integral that we must now evaluate. Integration by parts allows us to write Γ=− δ (x′ )∇′ δ (t − t′ − |x − x′ |/c) 3 ′ d x. |x − x′ | Because the dependence on x′ is contained entirely in |x − x′ |, and because ∇′ |x − x′ | = −(x − x′ )/|x − x′ |, evaluating the gradient gives Γ = − δ (t − t′ − |x − x′ |/c) 1 δ ′ (t − t′ − |x − x′ |/c) x − x′ 3 ′ + d x |x − x′ |2 c |x − x′ | |x − x′ | δ (t − t′ − r/c) 1 δ ′ (t − t′ − r/c) x + , = − r2 c r r δ (x′ ) Γ(t − t′ , x) = −r ˆ δ (t − t′ − r/c) 1 δ ′ (t − t′ − r/c) , + r2 c r or where r ˆ = x/r. Substituting this result into our previous expression for the scalar potential, we obtain Φ(t, x) = 1 p·r ˆ 4πǫ0 dt′ δ (t − t′ − r/c) 1 δ ′ (t − t′ − r/c) cos ωt′ , + r2 c r and integrating against the δ -functions gives Φ(t, x) = cos ω (t − r/c) ω sin ω (t − r/c) 1 . − p·r ˆ 4πǫ0 r2 c r One must be careful with the second term: a minus sign comes from the integration by part needed to convert δ ′ (t − t′ − r/c) into an actual δ -function, but this minus 5.3 Spherical waves — the oscillating dipole 81 sign is canceled out because the prime on the δ -function indicates differentiation with respect to t − t′ − r/c instead of t′ . We express our final result as Φ = Φnear + Φwave , where Φnear (t, x) = and Φwave (t, x) = − ˆ 1 p·r cos(ωu) 2 4πǫ0 r 1 ω sin(ωu) p·r ˆ , 4πǫ0 c r (5.3.6) (5.3.7) (5.3.8) and where u = t − r/c is retarded time. The scalar potential contains two terms. Of these, only Φwave has the mathematical structure of a spherical wave. The other term, Φnear , has the structure of a static dipole field, ˆ 1 p·r , Φdipole (x) = 2 4πǫ0 r to which a time dependence cos(ωu) is attached. Unlike Φwave , this term does not represent electromagnetic radiation. Instead, Φnear describes the field of a point dipole, and this field follows the dipole’s time dependence with the usual delay: the cos ωt dependence of the dipole is replaced by cos ω (t − r/c). As the name indicates, Φnear dominates when r is small, while Φwave dominates when r is large. The scalar potential becomes purely radiative when r is sufficiently large that Φnear can be neglected; the region of space for which this is true is called the wave zone. On the other hand, the region of space for which r is sufficiently small that Φwave can be neglected in front of Φnear is called the near zone. In the near zone we must have 1/r2 ≫ ω/(cr), or r≪ 1 λ c = = ω k 2π (near zone), (5.3.9) where k is the wave number and λ the wavelength. In the near zone, u = t − r/c ≃ t and the delay contained in the retarded time is hardly noticeable — the near-zone field responds virtually instantaneously to changes in the source. In the wave zone we have instead 1/r2 ≪ ω/(cr), or r≫ c 1 λ = = ω k 2π (wave zone). (5.3.10) Here the time delay is important, and the scalar potential behaves as a true spherical wave. In the region of space where r is comparable to c/ω , between the near zone and the wave zone (a region that is sometimes called the induction, or intermediate, zone), Φnear and Φwave are comparable, and the scalar potential behaves neither as an instantaneously-changing dipole field nor as a spherical wave. 5.3.3 Wave-zone fields of an oscillating dipole Let us calculate the electric and magnetic fields in the wave zone. We use the expressions sin(ωu) µ0 (5.3.11) A(t, x) = − ω p 4π r and 1 ω sin(ωu) Φ(t, x) = − p·r ˆ , (5.3.12) 4πǫ0 c r where u = t − r/c is retarded time. Notice that Eq. (5.3.11) is exact, while Eq. (5.3.12) is valid in the wave zone only. 82 Electromagnetic radiation from slowly moving sources To do this calculation we will need to compute the gradient of ψ (t, r) ≡ sin(ωu) . r We have ∇ψ = (∂ψ/∂r)∇r = (∂ψ/∂r)r ˆ and sin(ωu) ω cos(ωu) ∂ψ − =− , ∂r r2 c r because ∂u/∂r = 1/c. But since we are working in the wave zone, the first term can be neglected and we obtain the useful approximation ∇ sin(ωu) ω cos(ωu) =− r ˆ. r c r (5.3.13) We might also need the gradient of p · r ˆ. To see what this is, consider its x component: p · x x px x =− 2 + , ∇x p · r r r r p − (p · r ˆ)r ˆ . (5.3.14) r The fact that this is of order 1/r means that the gradient of p · r ˆ will be negligible in our forthcoming calculations. The electric field is related to the potentials by Eq. (5.1.8), ∇(p · r ˆ) = E=− ∂A − ∇ Φ. ∂t so that Substituting Eqs. (5.3.11) and (5.3.12), we obtain E= 1 ω sin(ωu) sin(ωu) µ0 2 cos(ωu) . ω p + ∇(p · r ˆ) + (p · r ˆ)∇ 4π r 4πǫ0 c r r Using Eq. (5.3.14) shows that the first term within the square brackets is of order 1/r2 and can be neglected in the wave zone, while using Eq. (5.3.13) gives E= cos(ωu) 1 ω2 µ0 2 cos(ωu) (p · r ˆ) ω p − r ˆ. 2 4π r 4πǫ0 c r Collecting the terms and recalling that ǫ0 µ0 = 1/c2 brings this to its final form Ewave (t, x) = µ0 2 cos(ωu) ω p − (p · r ˆ)r ˆ . 4π r (5.3.15) This has the structure of a spherical wave, and we notice that Ewave is transverse to the direction in which the wave propagates: Ewave · r ˆ = 0. (5.3.16) If, for concreteness, we let p point in the z direction, then p = pz ˆ, r ˆ = sin θ cos φx ˆ+ sin θ sin φy ˆ + cos θz ˆ, p · r ˆ = p cos θ, and p − (p · r ˆ)r ˆ = −p sin θ(cos θ cos φx ˆ+ ˆ, where θ ˆ is a unit vector that points in the direction cos θ sin φy ˆ− sin θz ˆ) = −p sin θθ of increasing θ. This yields Ewave (t, x) = − for the electric field. cos(ωu) ˆ µ0 2 ω p sin θ θ 4π r (5.3.17) (5.3. the electric and magnetic fields are related by Ewave = cBwave × r ˆ. It is easy to check that in the wave zone.3.3.13) gives us Bwave (t. since the vector r ˆ × p is perpendicular to both p and r ˆ.21) 5. So the fields satisfy Ewave · Bwave = 0 (5. which gives the direction of the electric field. the Poynting vector is S= c c B×r ˆ ×B = (B · B )r ˆ − (B · r ˆ)B .21) into Eq. (5. We have sin(ωu) µ0 ωp × ∇ . (5.3 Spherical waves — the oscillating dipole 83 The magnetic field is obtained by taking the curl of the vector potential. For p = pz ˆ we have r ˆ × p = −p sin θ(− sin φx ˆ + cos φy ˆ) = ˆ.3.3.18) that the fields are mutually orthogonal. and like the electric field it is transverse to the direction of propagation: Bwave · r ˆ = 0. because [p − (p · r ˆ)r ˆ] · (r ˆ × p) = p · (r ˆ × p) − (p · r ˆ)r ˆ · (r ˆ × p) = 0. and Eq. x) = − µ0 ω 2 cos(ωu) ˆ φ. x) r µ0 (5.3. (5. Bwave (t.20) in the wave zone.19) Furthermore. µ0 µ0 The second term vanishes because B is orthogonal to r ˆ. x) = µ0 ω 2 cos(ωu) (r ˆ × p) . For p = pz ˆ we can substitute Eq. (5. and we arrive at Swave (t. Using once more the same vectorial identity. 4π c r (5.3. we see from Eqs.24) 32π 2 c r2 . (5.3.18) The magnetic field also is a spherical wave. the same direction in which the wave propagates.23) This shows that the energy is transported in the radial direction. and (r ˆ × p) × r ˆ= (r ˆ· r ˆ)p − (r ˆ · p)r ˆ = p − (p · r ˆ)r ˆ. and after averaging over a complete wave cycle we get µ0 p2 ω 4 sin2 θ Swave = r ˆ.15) and (5.3. p sin θ 4π c r (5. (5.3. x) = c 2 ˆ. 16π 2 c r2 The energy flux oscillates in time.18) becomes −p sin θφ Bwave (t.22) This follows because the magnetic field is proportional to r ˆ × p. the energy flux vector associated with the electromagnetic radiation produced by the oscillating dipole. B (t.3. x) = 4π r and Eq.3. (5.5.23) and obtain S = = c µ0 µ0 4π 2 ω 4 p2 sin2 θ cos2 (ωu) r ˆ c2 r2 µ0 p2 ω 4 sin2 θ cos2 (ωu) r ˆ.3.3.4 Poynting vector of an oscillating dipole 1 We now calculate S = µ− 0 E × B . 9. θ. in which electromagnetic radiation is produced by an arbitrary distribution of charges and currents. φ) = − ω 2 p sin θ 4π r and µ0 ω 2 cos ω (t − r/c) ˆ B (t. (5. Using this rule we get cos2 (ωu) = 1 2 . θ. The average of any function f of retarded time u T over a complete wave cycle is defined by f (u) ≡ T −1 0 f (u) du. 2 π 32π 2 c 0 The integral evaluates to 4/3.4 Electric dipole radiation [The material presented in this section is also covered in Sec.] We now consider a much more general situation.25) becomes π µ0 p2 ω 4 P = sin3 θ dθ.24).25) S It is simplest to take S to be a sphere of constant r. and both fields are in phase. r. and they are both transverse to r ˆ = sin θ cos φ x ˆ + sin θ sin φ y ˆ + cos θ z ˆ. 5. The electric field is directed along ˆ = cos θ cos φ x θ ˆ + cos θ sin φ y ˆ − sin θ z ˆ while the magnetic field is directed along ˆ = − sin φ x φ ˆ + cos φ y ˆ. 4π c r These have the structure of a spherical wave. which travels outward at the speed of light. and that none of the energy propagates along the z axis (θ = 0 or θ = π ). φ) = − p sin θ φ. This is also the direction in which the field energy is transported.3. The fields are mutually orthogonal. r. the direction in which the wave is traveling. The field amplitudes are related by |B |/|E | = 1/c. The total power radiated (energy per unit time) is obtained by integrating S (energy per unit time per unit area) over any closed surface S . Then da = r ˆr2 sin θ dθdφ. and this gives Eq.3.3. (5.2 of Jackson’s text. 12πc (5.26) 5. and we obtain P = µ0 2 4 p ω . so that most of the energy is emitted in the directions orthogonal to the dipole.5 Summary — oscillating dipole We have seen that in the wave zone (r ≫ c/ω ). They are produced by a charge distribution whose dipole moment vector is given by p(t) = p cos(ωt)z ˆ. and Eq. where T = 2π/ω is the oscillation period. (5.3.3. with an arbitrary time dependence (not necessarily oscillating with a single frequency ω ).84 Electromagnetic radiation from slowly moving sources From the angular dependence of this expression we see that most of the energy is emitted near the equatorial plane (θ = π 2 ). the fields of an oscillating dipole are given by cos ω (t − r/c) ˆ µ0 θ E (t. Our only restrictions are that . The averaged power is then P = S · da. The angular profile of the energy-flux (Poynting) vector is described by sin2 θ. 4) Because rc ≪ λc . |x − x′ | V . (5. we introduce the following scaling quantities: rc tc vc ωc λc ≡ characteristic length scale of the charge and current distribution. and to define near and wave zones in the general case. 5. 2π ωc c λc = . This implies that the distribution of charge and current is localized 3 within a region whose volume is of the order of rc . ωc The characteristic length scale is defined such that if ρ(t. rc ≡ characteristic velocity of the source.1 Slow-motion approximation.4.4. we will see that the results to be derived in this section are very close to those obtained in the preceding section.4) to simplify our expressions. ≡ characteristic time scale over which the distribution changes. (5. (5. The characteristic time scale is defined such that ∂ρ/∂t is of order ρ/tc throughout the source. x′ ) and j (t.3) (5. 5.5. the wave zone can also be defined by the condition r ≫ λc /(2π ). Φ(t. We can now define the near and wave zones: near zone: wave zone: |x − x′ | ≪ λc c = . ≡ tc 2π ≡ ≡ characteristic frequency of the source. The slow-motion approximation means that vc = rc /tc is much smaller than the speed of light: vc ≪ c. x′ ) 3 ′ d x. 85 These conditions will allow us to formulate useful approximations for the behaviour of the electric and magnetic fields. ρ oscillates with a frequency Ω. We will compute the potentials and fields in these zones. |x − x′ | ≫ 2π ωc (5. using Eqs.4. then tc ∼ 1/Ω and ωc ∼ Ω. or rc ≪ λc .2) The source is therefore confined to a region that is much smaller than a typical wavelength of the radiation.4. respectively.4.2 Scalar potential We begin by calculating the scalar potential. If. x′ ) are the charge and current densities.1) This condition gives us rc = vc tc ≪ ctc = λc . near and wave zones To make the slow-motion approximation precise.1)–(5. for example. Although the present context is very general.4.4. then |x′ | is at most of order rc throughout the source. tc 2πc ≡ ≡ ctc ≡ characteristic wavelength of the radiation. x) = 1 4πǫ0 ρ(t − |x − x′ |/c. • the charges are moving slowly.4.4 Electric dipole radiation • the source is confined to a bounded region V of space. Now relative 2 rc /r. The third term would be smaller still. because it involves the time derivative of the total charge ρ(t. (5.86 Electromagnetic radiation from slowly moving sources where V is the region of space occupied by the source. (5. u = t − r/c (5.4. x′ ) d3 x′ . which is conserved. c dt V V V The second term vanishes. c where an overdot indicates differentiation with respect to t. x′ ) d3 x′ + · · · . the ǫ term is of order rc /r ≪ 1 and can be neglected. We have 1 ρ(t − |x − x′ |/c) = ρ(u) + ρ ˙ (u) r ˆ · x′ + ǫ + · · · . This gives ρ(t − |x − x′ |/c) where is retarded time and ǫ is of order Let us now Taylor-expand the charge density about the retarded time u instead of the current time t.6) . x′ ) 1+O ′ |x − x | λ2 c d3 x′ . and what we have is a potential that adjusts instantaneously to the changes within the distribution. and we neglect it. = ρ(t − r/c + r ˆ · x′ /c + ǫ/c) = ρ(u + r ˆ · x′ /c + ǫ/c). Here |x − x′ | is large and we can no longer Taylor-expand the density as we did previously. x) = 1 4πǫ0 |x − x′ |2 ρ(t.3). x′ ) 3 ′ d x − |x − x′ | 1 ρ ˙ (t. Instead we must introduce another approximation technique. r is much larger than r′ .5) V We see that the near-zone potential takes its usual static expression. We see that relative to r ˆ · x′ . c where an overdot now indicates differentiation with respect to u. x) = = 1 4πǫ0 1 4πǫ0 ρ(t. To witness radiative effects we must go to the wave zone. We then have Φ(t.4.4. so that |x − x′ | = = (x − x′ ) · (x − x′ ) r2 − 2x · x′ + r′2 1/ 2 2 2 = r 1 − 2r ˆ · x′ /r + O(rc /r ) 2 2 = r 1−r ˆ · x′ /r + O(rc /r ) 2 = r−r ˆ · x′ + O(rc /r). We have obtained Φnear (t. The time delay between the source and the potential has disappeared. Relative to the first term. The electric field it produces is then a “time-changing electrostatic field”. Because ρ(t − ε) = ρ(t) − ρ ˙ (t)ε + 1 2ρ 1 ρ(t − |x − x′ |/c) = ρ(t) − ρ ˙ (t)|x − x′ | + · · · . x′ ) d3 x′ + · · · c 1 d ρ(t. the second term is of order |x − x′ |/(ctc ) = |x − x′ |/λc . and by virtue of Eq. except for the small correction of order |x − x′ |2 /λ2 c and the fact that charge density depends on time. In the near zone we can treat |x − x′ |/c as a small quantity and Taylor-expand the charge density about the ¨(t)ε2 + · · · we have current time t. this is small in the near zone. This near-zone field does not behave as radiation. We use the fact that in the wave zone. x′ ) 3 ′ d x − |x − x′ | ρ(t. Our final expression for the potential is therefore Φwave (t. x′ ) d3 x′ + · c ∂u V In the first integral we recognize Q.4 Electric dipole radiation 87 to ρ(u). The second term. We see that the radiative part of the scalar potential is produced by a time-changing dipole moment of the charge distribution.8) (5. ρ(u. |x − x′ | |x − x′ | λc V In the near zone we approximate the current density as j (t − |x − x′ |/c) = j (t) 1 + O .4.4.8) when the dipole moment is given by p(t) = p0 cos(ωt). = 1+r ′ |x − x | r r and it is sufficient to keep the leading term only. The first term on the right-hand side of Eq.10) the dipole moment. x′ ) d3 x′ (5.7) V r ˆ · x′ ρ ˙ (u) + · · · d3 x′ c r ˆ ∂ ρ(u. You should check that Φwave (t. (5. x′ )x′ d3 x′ (5. In the second integral we recognize p(u). and this is small by virtue of Eq. where p0 is a constant vector. the time-derivative term is of order rc /(ctc ) = vc /c. is radiative: it depends on retarded time u = t − r/c and decays as 1/r. x) ≃ 1 r ˆ· p ˙ (u) 4πǫ0 cr reduces to Eq. x′ )x′ d3 x′ + · · · . (5. on the other hand.3. it is nonzero whenever dp/dt is nonzero. this is actually independent of u by virtue of charge conservation.9) is the total charge and p(u) = V ρ(u. x) = where Q= V 1 r ˆ· p ˙ (u) 1 Q + + ···. We therefore have ρ(t − |x − x′ |/c) = ρ(u) + in the wave zone.4. the dipole moment vector of the charge distribution. monopole potential associated with the total charge Q.3 Vector potential The exact expression for the vector potential is A(t. (5.8) is the static. The wave-zone potential is then Φ(t.4.5.4. the total charge of the distribution. this does depend on retarded time u. Inside the integral for Φ we also have 1 1 1 2 2 ˆ · x′ /r + O(rc /r ) = 1 + O(rc /r) . 5. x′ ) 3 ′ d x. we shall simply omit it in later calculations. x) = µ0 4π j (t − |x − x′ |/c. x) = = 1 1 4πǫ0 r 1 1 4πǫ0 r ρ(u) + V r ˆ · x′ 2 2 ρ ˙ (u) + O vc /c c (5. This term does not depend on time and is not associated with the propagation of radiation. 4πǫ0 r 4πǫ0 cr ρ(u.4.4.1). The other components of the equation are established with similar manipulations. x′ ). and there are no radiative effects in the near zone. where S is the two-dimensional surface bounding V . x′ ) 1+O ′ |x − x | λc d3 x′ .12) we start with the identity ∇ · (z j ) = z (∇ · j ) + j · (∇z ). x) = and ˆ· p ˙ (u) 1 r 1 + O(vc /c) 4πǫ0 cr (5.4.88 and we obtain Electromagnetic radiation from slowly moving sources Anear (t. where p0 is a constant vector. dt This is the same thing as the z component of Eq.4. After integration over the source we have V jz d3 x − d dt ρz d3 x = V V ∇ · (z j ) d3 x = S z j · da. x) instead of (u. ∇ · j = −∂ρ/∂t. Because no current is crossing this surface. x) = . (3.4.3).12) V The vector potential is therefore Awave (t. x) = µ0 4π V |x − x′ | j (t.4. x) = µ0 1 4π r j (u. the right-hand side vanishes and we have jz d3 x = V d dt ρz d3 x = V dpz .13) in the wave zone. we have instead j (u. and as we shall prove presently.4. except for the correction of order |x − x′ |/λc and the fact that the current density depends on time.13) reduces to reduces to Eq. This has the structure of a spherical wave. But here the current density depends on time.12). (5.4.15) 4π r where u = t − r/c and r ˆ = x/r. 5. The potential responds virtually instantaneously to changes in the distribution.4. They therefore give rise Awave (t. (5.5) when the dipole moment is given by p(t) = p0 cos(ωt).4. In the wave zone we have instead j (t − |x − x′ |/c) = j u + r ˆ · x′ /c = j (u) 1 + O(vc /c) and Awave (t. and we see that the radiative part of the vector potential is produced by a time-changing dipole moment. except for the fact that ρ and j are expressed in terms of the variables (t. the volume integral of j vanishes — recall Eq.4. x′ ) 1 + O(vc /c) d3 x′ . x′ ) d3 x′ = p ˙ (u). are generated by time variations of the dipole moment vector p of the charge and current distribution. (5. (5. in which we substitute the statement of charge conservation.14) µ0 p ˙ (u) 1 + O(vc /c) .4. (5. (5. (5.11) We see that here also.4 Wave-zone fields The potentials Φwave (t. the vector potential takes its static form. V In static situations. To establish the z component of Eq.3. x) = µ0 p ˙ (u) + ··· 4π r (5. You may verify that Eq. r ˆ· p ¨r ˆ+ O c r so that ∇ r ˆ· p ˙ =− (5.16) To obtain this result we made use of the fact that p ˙ depends on x through u = t − r/c. the direction in which the wave propagates. and neglect terms that decay faster. To see what this is we examine its x component: ∇×p ˙ x = ∇y p ˙z − ∇z p ˙y 1 y 1 z ¨z + p ¨y = − p c r c r 1 ˆ× p ¨ x.5. and that it is orthogonal to both r ˆ and the electric field.19) Notice that the wave-zone magnetic field behaves as a spherical wave. to the radiation emitted by an arbitrary source. 4π r 4πǫ0 c2 r or Ewave (t.4. 4π cr (5.3.17) Notice that the wave-zone electric field behaves as a spherical wave.4 Electric dipole radiation 89 to electric-dipole radiation.4. Notice finally that the fields are in phase — they both depend on p ¨(u) — and that their magnitudes are related by |B |/|E | = 1/c. c The magnetic field is then Bwave (t.4. when computing the gradient of the scalar potential we can neglect ∇r−1 = −r ˆ/r2 . The electric field is then E = − ∂A − ∇Φ ∂t µ0 p ¨ 1 (r ˆ· p ¨)r ˆ = − + . the leading-order contribution. so that acting with ∇ amounts to taking a derivative with respect to u and multiplying by −c−1 ∇r = −c−1 r ˆ. . = − r ˆ· p ¨ +O c r r = ∇x 1 1 .18) so that 1 ∇×p ˙ (u) = − r ˆ× p ¨(u). x) = − ˆ× p ¨(u) × r ˆ ¨(u) − r ˆ· p ¨(u) r ˆ µ0 r µ0 p =− . in our slow-motion approximation. which we calculate as ∇x (r ˆ· p ˙) y z x p ˙x + p ˙y + p ˙z r r r 1 1 1 x y z 1 = p ¨x − ∇x r + p ¨y − ∇x r + p ¨z − ∇x r + O r c r c r c r 1 x 1 . 5. and that it is transverse to r ˆ. To get the magnetic field we need to compute ∇ × p ˙ (u). = − r c (5. 4π r 4π r (5.4. For example. We now compute the electric and magnetic fields in this approximation. we only need ∇(r ˆ· p ˙ ).5 are therefore shared by all electric-dipole radiation fields. To get the electric field we keep only those terms that decay as 1/r. x) = − ˆ× p ¨(u) µ0 r . The properties listed in Sec. 19).22) 5.4.6 Summary — electric dipole radiation To leading order in a slow-motion approximation. x) = − and Ewave = cBwave × r ˆ. (5. (4π )2 c To evaluate the integral we use the trick of momentarily aligning the z axis with the instantaneous direction of p ¨(u) — we must do this for each particular value of u. The field energy flux vector is Swave = c 2 ˆ. (5.20) Bwave r Swave = µ0 the same result as in Eq. Then r ˆ× p ¨(u) = |p ¨(u)| sin θ and P = µ0 p ¨(u) (4π )2 c 2 sin2 θ dΩ = µ0 p ¨(u) (4π )2 c 2 0 2π π dφ 0 sin3 θ dθ.3.4. The power’s angular distribution is described by dP µ0 2 = p ¨(u) sin2 Θ(u).5 Energy radiated 1 ˆ in the wave zone The Poynting vector is S = µ− 0 E × B . 6πc . (5. and they are mutually orthogonal. Substituting Eqs.4.17) and (5.4. The last term vanishes by virtue of Eq.4. the wave-zone fields of an arbitrary distribution of charges and currents are given by Bwave (t.21) This is the total power radiated by a slowly-moving distribution of charge and current.4.4.90 Electromagnetic radiation from slowly moving sources 5. The fields are generated by time variations of the source’s dipole moment vector. Bwave r µ0 and the total power radiated is given by P = S · da = µ0 2 p ¨(u) . and since E = cB × r — as follows from Eqs.20) yields µ0 2 P = r ˆ× p ¨(u) dΩ.19) — we have S = (c/µ0 )(B × r ˆ) × B = (c/µ0 )[|B |2 r ˆ − (B · r ˆ)B ]. (5. they are each transverse to r ˆ.4. and we obtain c 2 ˆ. (5. The fields behave as spherical waves. The fact that the Poynting vector is directed along r ˆ shows that the electromagnetic field energy travels along with the wave. 6πc (5.4.4.19) and (5. The energy crossing a sphere of radius r per unit time is given by P = S · da. dΩ (4π )2 c where Θ(u) is the angle between the vectors r ˆ and p ¨(u).23). The angular integration gives (2π )(4/3) and we arrive at P = µ0 2 p ¨(u) . where da = r ˆr2 dΩ and dΩ = sin θ dθdφ. (5. x)x d3 x. p(t) = V µ0 r ˆ× p ¨(t − r/c) 1 + O(vc /c) 4π cr ρ(t. 1: Electric field lines of an oscillating dipole. x) = I sin k (ℓ − |z |) δ (x)δ (y )z ˆ cos ωt. are represented in Fig. p(t) = p0 cos(ωt) z ˆ. from z = −ℓ to z = ℓ.2) The current is an even function of z (it is the same in both arms of the antenna) and its goes to zero at both ends (at z = ±ℓ).3. 5. 5. The wire runs along the z axis.] As an example of a radiating system we consider a thin wire of total length 2ℓ which is fed an oscillating current through a small gap at its midpoint. p(t) = p0 [cos(ωt) x ˆ+ sin(ωt) y ˆ]. and it is usually represented by j (t.5. 5. To be consistent we must be sure that vc ≪ c for this (5. and the gap is located at z = 0 (see Fig. using the electricdipole approximation.4 of Jackson’s text.5 Centre-fed linear antenna [The material presented in this section is also covered in Sec. (5. where k = ω/c.1.5.) For such antennas. and I is the current’s peak value. 5. at selected moments of time.5.5 15 Centre-fed linear antenna 15 91 10 10 5 5 0 0 -5 -5 -10 -10 -15 0 2 4 6 8 oscillating dipole: t = 0 10 12 14 -15 0 2 4 6 8 oscillating dipole: t = 1 10 12 14 15 15 10 10 5 5 0 0 -5 -5 -10 -10 -15 0 2 4 6 8 oscillating dipole: t = 2 10 12 14 -15 0 2 4 6 8 oscillating dipole: t = 3 10 12 14 Figure 5.1) . The current at the gap (at z = 0) is I sin kℓ. 2. the current typically oscillates both in time and in space. for ω = 1.11. The power’s angular profile is proportional to sin2 Θ(u). p(t) = p0 cos(ωt) z ˆ. The curves were obtained by following the method detailed in Sec. The electric field lines of an oscillating dipole. We want to calculate the total power radiated by this antenna. Witness the transition between near-zone behaviour and wave-zone behaviour which occurs near r = λ = 2πc/ω ≡ 1. The field lines of a rotating dipole. are represented in Fig.2. 9. where Θ(u) is the angle between the vectors r ˆ and p ¨(u). for ω = 1. p(t) = p0 [cos(ωt) x ˆ + sin(ωt) y ˆ]. The oscillating current is provided by a coaxial feed.2: Electric field lines of a rotating dipole. . z=l antenna coaxial feed z = −l Figure 5.92 Electromagnetic radiation from slowly moving sources 8 8 6 6 4 4 2 2 0 0 -2 -2 -4 -4 -6 -6 -8 -5 -4 -3 -2 -1 0 1 rotating dipole: t = 0 2 3 4 5 -8 -6 -4 -2 0 rotating dipole: t = 1 2 4 6 8 6 4 2 0 -2 -4 -6 -8 -8 -6 -4 -2 0 rotating dipole: t = 2 2 4 6 8 Figure 5. at selected moments of time.3: Centre-fed linear antenna. Witness the transition between near-zone behaviour and wave-zone behaviour which occurs near r = λ = 2πc/ω ≡ 1. 4. that ℓ ≡ rc ≪ λc ≡ 2π/k — refer back to Eq.7) P = 12π For a fixed frequency ω . x) d3 x. For this it is efficient to turn to Eq. (5. x) = I0 1 − |z |/ℓ δ (x)δ (y )z ˆ cos ωt. we arrive at our final result µ0 c 2 I0 kℓ . We can therefore approximate sin[k (ℓ − |z |)] by k (ℓ − |z |). (5. According to Eq.5. (5. (4π )2 c dP dΩ µ0 c I0 kℓ 32π 2 p ¨(u) = −(I0 c)(kℓ)z ˆ sin(ωu) (5.3) in which we substitute Eq.5. which means that k |z | is small throughout the antenna. none of the energy propagates along the axis. or after using Eq.2).5. we have that Θ(u) = θ. (5.5. this reduces to = 2 sin2 θ.5) After averaging over a complete wave cycle. and evaluating the integral gives p ˙ (t) = I0 ℓz ˆ cos ωt.5. In other words. (5. (5. so long as the condition kℓ ≪ 1 is satisfied. and Eq. (5.6) we learn that most of the energy is radiated in the directions perpendicular to the antenna.2). In this approximation the current no longer oscillates in space: it simply goes from its peak value I0 at the gap to zero at the two ends of the wire. the power increases like the square of the feed current I0 . (5. where θ is the angle between the vectors r ˆ and z ˆ. Using sin2 θ dΩ = 8π/3. Taking an additional derivative and making the substitution t → u = t − r/c yields p ¨(u) = −I0 ωℓz ˆ sin(ωu). where I0 = Ikℓ (5.4.1).5. the angular distribution of the power is given by dP dΩ = = µ0 2 p ¨(u) sin2 θ 2 (4π ) c µ0 (I0 c)2 (kℓ)2 sin2 (ωu) sin2 θ. the second time derivative of the dipole moment vector. Jackson gives an exact treatment of the antenna of Eq. In Sec.4 of his book. . therefore.12). Since the vector p ¨(u) is always aligned with the z axis.5. we must demand that kℓ ≪ 1. or equivalently.5. For a fixed current. (5.4) is the value of the current at the gap. (5. (5.3).4. the power increases like the square of the frequency.5. To compute the power radiated by our simplified antenna we first need to calculate p ¨(t).5.22). 9.5 Centre-fed linear antenna 93 distribution of current. p ˙ (t) = j (t.1) becomes j (t.6) To obtain the total power radiated we must integrate over the angles. From Eq.5. We have ℓ p ˙ (t) = I0 z ˆ cos ωt −ℓ 1 − |z |/ℓ dz. the atom emits electromagnetic radiation.5).3) Combining the last three equations. 4πǫ0 a2 for the second derivative of the dipole moment vector. dt The orbital energy of a classical atom is Eorb = − 1 e2 . This gives p ¨(t) = 1 e e2 a ˆ(t) 4πǫ0 m a2 (5. the system’s dipole moment is p(t) = −ea(t) where a(t) is the electron’s position vector.6.4.4) (5. µ0 2 p ¨(u) .6. as it sets the atomic .1) then gives P = e6 µ0 . the total energy radiated by the atom per unit time is given by Eq. me2 (In this step we make a leap and momentarily leave the classical realm. 4πǫ0 2a (5. it eventually goes to zero.6. since the Bohr radius is very much a consequence of quantum mechanics. Because this is nonzero. =− dt 3 (4πǫ0 )2 m2 c3 a2 (5. (5. (5.6.6 Classical atom The classical picture of a hydrogen atom contains a proton at the centre and an electron on a circular orbit around it.6. the orbital radius is equal to Bohr’s radius a0 = 4πǫ0 2 . it is easy to show that the orbital radius must decrease.5) To obtain the lifetime we integrate Eq. As is well known. we have p ¨(t) = −ea ¨ = −(e/m)F . The atom therefore loses energy to the radiation. P = 6πc Substituting Eq.1) 1 e2 a ˆ. at a rate given by da 1 e2 4 .94 Electromagnetic radiation from slowly moving sources 5.6. In this section we calculate the lifetime of the classical atom.21). Supposing for simplicity that the proton is at rest. we therefore have dEorb = −P. But we may still think of a0 as giving an appropriate initial condition. the classical model leads to an unstable atom. The Coulomb force exerted by the proton is F =− where a ˆ = a(t)/a. According to the electricdipole approximation. starting with the initial condition that at t = 0. and as a result. Since the electron is accelerated by the Lorentz force F exerted by the proton.6. the orbital radius a gradually decreases. 6πc (4πǫ0 )2 m2 ca4 (5.2) Energy conservation dictates that the energy carried away by the radiation must come at the expense of the atom’s orbital energy. (5. or when higher accuracy is required. the magnetic moment vector m. and we obtain τ ≡ 0 (dt/da) da as a0 (4πǫ0 )5 c3 6 . we know that the electron is accelerated. The results of this section. The effect is there.6. where vc is a typical internal velocity of the source. however. In some cases. but it is well hidden.6.) Integration of Eq. that it emits electromagnetic waves. but the point remains that the size of this ellipse cannot change with time. Ignoring the fact that the classical laws of electrodynamics do not apply to a real hydrogen atom. and that the orbital radius cannot possibly stay constant. 8πα5 c e2 1 ≃ ≡ fine structure constant (4πǫ0 ) c 137 (5. we need to compute the next term in the expansion. We will see that at the next-to-leading-order. The paradox is therefore this: The electron’s equations of motion do not account for the decrease in orbital radius. We recall the definitions of these objects: p(t) = V ρ(t.6. mc 1 λC ≃ 1.7 Magnetic-dipole and electric-quadrupole radiation τ 0 95 dt = length scale. 5. x′ )x′ d3 x′ . we will show that the electron’s radiation reaction is actually contained in Maxwell’s theory. Equation (5.1) .8) This result shows that under classical laws.7) The lifetime can then be expressed as τ= (5. we seem to find an inconsistency in our description of the classical atom. the hydrogen atom would be extremely short lived.) On the other hand.5. This is our task in this section. Is classical electrodynamics internally inconsistent? Later on.7. τ= 4me10 To put this in friendlier terms we introduce α≡ and λC ≡ 2π ≃ 2. On the one hand.8). that the radiation carries energy away from the system. and it is a subtle matter to reveal it. 9. leave us with a curious paradox. the dipole moment p either vanishes or does not depend on time. by the way.6. the fact that the electron feels the Coulomb field of the proton guarantees that the electron’s motion will be circular and that its orbital radius will not change with time. and the electric dipole moment tensor Qab .] Electric-dipole radiation corresponds to the leading-order approximation of the electromagnetic field in an expansion in powers of vc /c.6) (5. (5.43 × 10−12 m ≡ electron’s Compton wavelength. in Chapter 6. gives an order-of-magnitude estimate of the time required by an atom to make a transition from an excited state to the ground state.6. (5. the wave-zone fields depend on the dipole moment vector p. they unambiguously predict that the radius should be constant. (The most general orbit allowed by the Coulomb force is actually an ellipse.5).3 of Jackson’s text. (5. especially Eq.5) is elementary.6. and the leading term is actually zero. In such cases.7 Magnetic-dipole and electric-quadrupole radiation [The material presented in this section is also covered in Sec.6 × 10−11 s. and in which summation over the repeated index b is understood. 2π ωc (5. x′ ) d3 x′ .6) in the wave zone. 5.4) where rb ≡ xb /r are the components of the unit vector r ˆ = x/r. (5. nor any other result derived in this subsection.3) ′ ′2 3 ′ ρ(t. where u = t − r/c and w is a vector that will be determined.7.7.1 Wave-zone fields To make our calculations more efficient we first derive a few useful results that follow from the fact that in this section. 5. Later we will find that w is not just a function of retarded time u. In the foregoing manipulations we will introduce the vector Q(t. this is will be incorporated at a later stage. In this subsection we rely on the wave-zone condition of Eq. ∂u ∂r ˆ In the first term we substitute ∇x u = −c−1 ∇x r = −c−1 rx and we recover our previous expression. Given the form of the vector potential. We will not need an expression for the scalar potential. its curl is ∇×A= w(t − r/c) µ0 ∂ w(t − r/c) µ0 ∇× = (∇r) × . where V is the volume that contains the charge and current distribution.7. but that it depends also on r ˆ: w = w(u. So throughout this discussion we shall assume that r≫ λc c = . x) = µ0 w(u) 4π r (5. then we also have that rc ≪ λc /(2π ) ≪ r. as we shall see. For example.7. r ˆ).4.7.7) 4π cr for the wave-zone magnetic field. We can anticipate that the wave-zone potentials will have the form of a spherical wave. (5. this volume is bounded by the surface S . we place ourselves exclusively in the wave zone. and the derivative of w(u)/r with respect to r is −c−1 w ˙ (u)/r. but we do not rely on the slow-motion assumption. (5. This does not affect our result for the magnetic field. To see this.7. let us calculate more carefully the x derivative of the vector w: ∇x w = ∂w ∂w ∇x u + · ∇x r ˆ. which is determined by the vector potential. x′ ) 3x′ a xb − r δab d x . x) = − (5.7.1. But if vc /c ≪ 1.96 Electromagnetic radiation from slowly moving sources 1 2 V m(t) Qab (t) = = V x′ × j (t. 4π r 4π ∂r r The gradient of r is the vector r ˆ. r ˆ) = Qab (t)rb . We then recognize that ∇x r ˆ is of order 1/r and that the . up to terms of order 1/r2 that can be neglected in the wave zone. as was shown in Sec. in the wave zone E can be obtained directly from B .5).5) Notice that this condition is independent of any assumption that concerns vc : there is a wave zone whether or not the source is moving slowly.2) (5. x) defined by Qa (t.7. we shall verify that the vector potential is given by A(t. We therefore have ˆ× w ˙ (u) µ0 r B (t. (5. x) = ˆ× w ˙ (u) µ0 r c |B |2 r ˆ= 2 µ0 16π c r2 2 r ˆ. Equation (5. and here it becomes important to specify the dependence of w on the radial vector r ˆ. × ˆ× r r ∂r c r for the wave-zone electric field. To get the electric field we go back to one of Maxwell’s equations. and that they are insensitive to an eventual component of w along r ˆ. dΩ 16π 2 c (5. 16π 2 c where dΩ = sin θ dθdφ is an element of solid angle.7.8) it is easy to show that the wave-zone Poynting vector is given by S (t.10) gives the radiation’s angular distribution. once more.9) Evaluating this on a sphere of constant r gives dP = S · da = or µ0 dP 2 = r ˆ× w ˙ (u) . . that irrespective of the exact expression for the vector w(u.10) µ0 2 r ˆ× w ˙ (u) dΩ. x) × r ˆ 4π r (5.5. that their amplitudes differ by a factor of c. we have that j = 0 in the wave zone. We will use this observation later. and discard from w any longitudinal component. and µ0 c r ˆ× w ˙ (u) ∂E . This yields ∇× and we obtain ˆ× r ˆ× w ¨ (u) µ0 r ∂E = . ∇ × B = µ0 j + 1 ∂E . We conclude that the dependence of w on r ˆ is invisible to ∇ so long as all computations are carried out in the wave zone. (5.7.7. r ˆ).7. c2 ∂t Because the current density vanishes outside the volume V . Notice that the fields and the radiation’s angular profile depend only on r ˆ× w ˙ .7 Magnetic-dipole and electric-quadrupole radiation 97 second term can be neglected in the wave zone. x) = − ˆ× w ˙ (u) × r ˆ µ0 r = cB (t. ∂t 4π r Integrating this with respect to t gives E (t.8) 1 r ˆ ∂w ˙ r ˆ× w ¨ (u) r ˆ× w ˙ (u) = (∇r) × =− r . and we can also pull the 1/r factor in front of the derivative operator. we can treat r ˆ as a constant vector when evaluating the curl. To be more concrete we shall have to calculate the vector w. that they are in phase.7) and (5. the fields are both transverse (orthogonal to r ˆ) and mutually orthogonal. From these results we infer.7. = c2 ∇ × B = − ∇× ∂t 4π r By the same argument as the one just given. Using Eqs.7. and that the field energy travels in the radial direction. 7.7. (5. r ˆ) the vector related to the electric quadrupole moment which we introduced in Eq. 5.2 Charge-conservation identities Before we get to this we need to establish the useful identities j (t. x′ )(r ˆ · x′ ) d3 x′ = m(t) × r ˆ+ Q 6 6 dt ρ(t. Consider the z component of Eq. which covers static situations (and which was derived in Sec.4). V We obtain zjy d3 x + V V d dt ρyz d3 x V in a similar way. = −xz ∂t Integrating both sides over V and using Gauss’ theorem (together with the statement that j · da = 0 on the bounding surface S ) gives V ∇ · (xz j ) xjz d3 x = − yjz d3 x = − zjx d3 x + V d dt ρxz d3 x. V Let us now substitute these results into our previous expression for Iz .12). and Q(t.4.4. Equation (5. Consider now the relation = xz ∇ · j + xj · ∇z + z j · ∇x ∂ρ + xjz + zjx . ∂t 1 d 2 dt ρz 2 d3 x.12) is a generalization of Eq.7.4.12). refer back to Eq. (5. V in which we have changed the integration variables from x′ to x and replaced the constant vector r ˆ by the arbitrary constant vector e. 2 dt V V V = ρz 2 d3 x V . and we derive it using similar methods. x)x d3 x. x′ )r′2 d3 x′ . (3.7. We write the left-hand side as Iz = e · jz (t.12) V where p(t) is the electric dipole moment.3). x′ ) d3 x′ = p ˙ (t) V (5.98 Electromagnetic radiation from slowly moving sources 5. Consider now ∇ · (z 2 j ) = z 2 ∇ · j + 2z j · ∇z = −z 2 integrating this over V leads to zjz d3 x = V ∂ρ + 2zjz . Iz is given by Iz = ex V xjz d3 x + ey V yjz d3 x + ez V zjz d3 x.11) was already derived near the end of Sec. 3. Equation (5. (5.11) and V 1 d 1 ˙ (t.4). m(t) the magnetic dipole moment. We have Iz 1 d ex ρxz d3 x xjz d3 x − zjx d3 x + 2 dt V V V 1 d 1 d ρyz d3 x + ez yjz d3 x − + ey zjy d3 x + 2 dt V 2 dt V V 1 1 = − ex (x × j )y d3 x + ey (x × j )x d3 x 2 2 V V 1 d 3 ex ρxz d x + ey + ρyz d3 x + ez ρz 2 d3 x . r ˆ) + r ˆ j (t.7. More explicitly.3.7. (5.7. e) = Qzx ex + Qzy ey + Qzz ez = ex V ρ(3xz ) d3 x + ey V ρ(3yz ) d3 x + ez V ρ(3z 2 − r2 ) d3 x = 3[ ] − ez ρr2 d3 x. 5. x′ )(r ˆ · x′ ) d3 x′ + · · · . but we shall now keep it. is otherwise general: it is not yet limited by a slow-motion assumption. however. x′ d3 x′ (5. x) = µ0 4π j (t − |x − x′ |/c.12).7 Magnetic-dipole and electric-quadrupole radiation 99 In this last expression.7. We will. which are appropriate in the wave zone.13) 4π r V for vector potential. x) = µ0 1 4π r j (u. neglect all remaining terms. the first two terms are equal to −ex my + ey mx = (m × e)z . Substituting the expansion for j into Eq. (5. The bracketed terms are related to Qz : Keeping in mind that we use e as a substitute for r ˆ. x′ ) d3 x′ + V 1 ∂ c ∂u V j (u.7. This term was neglected in our discussion of electric-dipole radiation.5. V where [ ] stands for the bracketed terms in our previous expression for Iz . Notice that this expression. though restricted to the wave zone. Using these new results we finally obtain Iz = (m × e)z + 1 d Qz + ez 6 dt ρr2 d3 x .13) gives A(t.7. the second term is smaller by a factor of order c−1 rc /tc ≡ vc /c.7.3 Vector potential in the wave zone We are now ready to calculate the vector w. V which is the same statement (after the substitutions x → x′ and e → r ˆ) as Eq. . A(t. which contribute at order (vc /c)2 . we have Qz (t. (5. We now incorporate the slow-motion approximation by Taylor-expanding the current density: ∂j 1 ˆ · x′ ) (u) + · · · . |x − x′ | V In this we substitute the approximations 1 1 ≃ |x − x′ | r and t − |x − x′ |/c ≃ u + r ˆ · x′ /c. j (u + r ˆ · x′ /c) = j (u) + (r c ∂u Relative to the leading term. This gives µ0 1 A(t. We begin with the exact expression for the vector potential. x) = j u+r ˆ · x′ /c. x′ ) 3 ′ d x. where u = t − r/c is retarded time. and we have w=p ˙ (u) + 1¨ 1 d2 1 m ˙ (u) × r ˆ+ Q (u.7.7.7.17) Such a source will emit magnetic-dipole radiation. r ˆ). But we have seen in Sec. The second and third terms give rise to magnetic-dipole radiation and electric-quadrupole radiation. We have obtained w(u.16) where Q(3) ≡ ∂ 3 Q/∂u3 and we have used the vectorial identity r ˆ × (m ¨ ×r ˆ) = m ¨ − (r ˆ· m ¨ )r ˆ. And according to Eq. (5. r ˆ) .7.3). (5. Substituting this into Eq.15) an expression that was already anticipated in Eq. To calculate the radiated power we shall need r ˆ× w ˙ =r ˆ× p ¨(u) + 1 1 m ¨ (u) − r ˆ· m ¨ (u) r ˆ + r ˆ × Q(3) (u. respectively.100 Electromagnetic radiation from slowly moving sources According to Eq. x′ )r′2 d3 x′ . (5. r ˆ) = 0. The first term on the right-hand side of Eq. this is the leading term in the expansion of the fields in powers of vc /c.7. r ˆ) = p ˙ (u) + 1¨ 1 m ˙ (u) × r ˆ+ Q (u.10) gives µ0 dP m ¨ am ¨ b δab − ra rb . r ˆ) + r ˆ m(u) × r ˆ+ Q 6 6 du ρ(u. x′ )r′2 d3 x′ .11). 5. We can therefore discard this last term in our expression for w. Eq. c2 which we rewrite in component form as = r ˆ× w ˙ 2 = 1 m ¨ am ¨ b δab − ra rb . c 6 (5.7. (5.4 Radiated power (magnetic-dipole radiation) Let us calculate the power radiated by a distribution of charges and currents for which p ¨(u) = Q(3) (u. (5.12). (5. This means that any component of w along r ˆ will not contribute to the wave-zone fields. and where summation over both repeated indices is understood.1 that the electric and magnetic fields.7. = dΩ 16π 2 c3 .7.14) and in terms of this the wave-zone vector potential is given by A(t.7. the first integral is equal to p ˙ (u).17). these contribute at order vc /c beyond the leading term. x) = µ0 w(u.7. 4π r (5. r ˆ) . as well as the radiated power. depend only on r ˆ× w ˙ . 5. V The quantity within the square brackets defines the vector w.14) gives rise to electric-dipole radiation. c 6c (5.16) becomes r ˆ× w ˙ = so that r ˆ× w ˙ 2 1 m ¨ − r ˆ· m ¨ )r ˆ c 1 |m ¨ |2 − (r ˆ· m ¨ )2 . (5. the second integral is equal to 1 d 1 ˙ (u.7. c2 where ra = xa /r are the components of the vector r ˆ.7. (5. V The last term is proportional to r ˆ. Under the conditions of Eq. r ˆ) + r ˆ c 6 6c du2 ρ(u.7. Under the conditions of Eq. 36c2 a b But the vector Q(3) depends on r ˆ.7.21). On the other hand. 5. Writing also Qb = Qbd rd .4.5.7 Magnetic-dipole and electric-quadrupole radiation 101 and integrating over the angles yields P = where µ0 m ¨ am ¨ b δab − ra rb . The ratio of powers is then Pmagnetic-dipole ∼ Pelectric-dipole m ¨ cp ¨ 2 ∼ vc c 2 .10) gives 1 µ0 dP (3) Q(3) Q = δab rc rd − ra rb rc rd .7. (5. 36c2 The inner product can be expressed as = (r ˆ· r ˆ) Q(3) · Q(3) − r ˆ · Q(3) r ˆ · Q(3) = Q(3) 2 − r ˆ · Q(3) . (5. 4πc3 1 1 ra rb dΩ = δab .4). 5. because by the definition of Eq. (5.7. (5. Eq.7. 36c2 ac bd Substituting this into Eq. 6πc3 µ0 2 p ¨(u) . Our final result for the radiated power is therefore ra rb ≡ Pmagnetic-dipole = µ0 2 m ¨ (u) . m ∼ (jrc )(rc ) ∼ jrc ∼ ρvc rc . (5. we arrive at r ˆ× w ˙ 2 = 1 (3) Q(3) Q δab rc rd − ra rb rc rd .7.19).5 Radiated power (electric-quadrupole radiation) Let us now calculate the power radiated by a distribution of charges and currents for which p ¨(u) = m ¨ (u) = 0.7. Qa = Qac rc . So for slowing-moving sources. 6c 1 r ˆ × Q(3) · r ˆ × Q(3) . and ρrc and that p 4 2 m ¨ ∼ ρvc rc /tc . 6πc (5. 2 and we have r ˆ× w ˙ 2 = 1 (3) Q(3) Q δab − ra rb .19) Such a source will emit electric-quadrupole radiation. (5.7. the power emitted in magnetic-dipole radiation is typically smaller than the power emitted in electric-dipole radiation by a factor of order (vc /c)2 ≪ 1.7.16) becomes r ˆ× w ˙ = so that r ˆ× w ˙ 2 1 r ˆ × Q(3) .7. Pelectric-dipole = 3 In orders of magnitude we have that the electric dipole moment is p ∼ (ρrc )(rc )∼ 4 4 3 4 2 4 ¨ ∼ ρrc /tc . 2 dΩ 16π c 36c2 ac bd .18) We might compare this result with Eq. 4π 3 These angular integrations are worked out in Sec. (5. for slowing-moving distributions.7. Qab ∼ (ρrc )(rc ) ∼ ρrc . we obtain Γ ≡ Q(3) ac Qbd δab rc rd − ra rb rc rd 1 1 (3) (3) (3) Qac Qbd δab δcd − Q(3) Q δab δcd + δac δbd + δad δbc = 3 15 ac bd 1 (3) (3) 1 1 (3) (3) 1 (3) = Qac Qac − Q(3) Qaa Qbb − Q(3) Q(3) ac Qac − 3 15 15 15 ac ca 2 1 (3) Q(3) − = ac Qac 3 15 1 (3) (3) = Q Q . Taking these results into account.16). the total power is the sum of individual contributions. because the angular profiles associated with each term are orthogonal.21) There is no interference between the different terms in Eq.6 Total radiated power In situations in which all three types of radiation contribute. Our final result for the radiated power is therefore Pelectric-quadrupole = µ0 (3) (3) Q (u)Qab (u). 5 ac ac (3) where we have used the property that Qab is a symmetric. 720πc3 ab (5.7. the power emitted in electric-quadrupole radiation is smaller than the power emitted in electric-dipole radiation by a factor of order (vc /c)2 ≪ 1. We see also that electric-quadrupole radiation is of the same order of magnitude as magnetic-dipole radiation. 5. In orders of mag(3) 2 3 5 5 3 nitude. 4πc 36c2 ac bd are evaluated in Sec.7.7.7. tracefree tensor (so that Qca = Qac and Qaa = 0).7.20) where summation over the two repeated indices is understood. (5. P = µ0 p ¨(u) 6πc 2 + 1 m ¨ (u) c2 2 + 1 (3) (3) 4 4 Q (u)Qab (u) + O vc /c 120c2 ab . .102 Electromagnetic radiation from slowly moving sources and integrating over the angles yields P = The angular integrals rc rd ≡ and ra rb rc rd ≡ 1 4π ra rb rc rd dΩ = 1 δab δcd + δac δbd + δad δbc 15 1 4π rc rd dΩ = 1 δcd 3 µ0 1 (3) Q(3) Q δab rc rd − ra rb rc rd . and Qab ∼ ρrc /tc ∼ p ¨rc /tc ∼ p ¨vc . 5. We therefore have Pelectric-quadrupole ∼ Pelectric-dipole vc c 2 . 7. in agreement with Eq. and · · · ≡ (4π )−1 (· · ·) dΩ is the average of the quantity (· · ·) over a sphere of constant r. and we must therefore have ra rb = λδab .7.7. and the pulsar is believed to be the source of energy for this supernova remnant.24).8 Pulsar spin-down As an application of magnetic-dipole radiation we consider the oblique-rotator model of a pulsar. To derive the result for ra rb rc rd we note that the right-hand side must be a four-index tensor that is symmetric in all pairs of indices. a rotating neutron star that emits pulses of electromagnetic radiation at regular intervals.23) (5. which gives λ′ δab δcd δab δcd + δac δbd + δad δbc = λ′ (3 × 3 + 3 + 3) = 15λ′ . 15 = (5.7 Angular integrations In this subsection we establish the formulae ra ra rb ra rb rc ra rb rc rd 0. If we associate the pulse frequency with the neutron star’s spin frequency. by simply substituting r ˆ = (sin θ cos φ. they would have to be equal to some vector a that does not depend on the angles θ and φ (because they have been integrated over) nor on r (because r appears nowhere on the left-hand side of the equation). 5. These relations can be derived by brute-force computation. To determine λ we simply take the trace of the left-hand side.7.7.8 Pulsar spin-down 103 5. prettier way that relies on arguments of symmetry. and it is easy to see that it must be equal to λ′ (δab δcd + δac δbd + δad δbc ). and this agrees with the result of Eq. Equating the results produces λ′ = 1/15. To determine λ′ we evaluate the double trace of the left-hand side. because there is no preferred direction in a homogeneous three-dimensional space.7. We must form this tensor with two copies of δab . sin θ sin φ. The zero result for ra rb rc follows from the fact that we do not have a vector that could be combined with δab to form a symmetric. (5. The right-hand side must therefore be zero. δab ra rb = δab ra rb = 1 = 1. Take the zero result for ra . which gives 1. cos θ) and integrating over dΩ = sin θ dθdφ.22) (5. .5. But there is an easier. for some constant λ. The only such tensor available is δab .23). The Crab pulsar is observed to spin down: the frequency of the pulses decreases with time. If these integrals were not zero.25) where ra = xa /r are the components of the vector r ˆ. and the double trace of the right-hand side. = 3 = 0. Equating the results produces λ = 1/3. three-index tensor. 1 = δab δcd + δac δbd + δad δbc . λδab δab = 3λ. But there is no such vector available. To arrive at the result for ra rb we observe that the right-hand side must be a constant tensor that is symmetric in the pair of indices a and b.24) (5.7. and the trace of the right-hand side. The most famous pulsar is located in the Crab nebula. 1 δab . (5. 104 Electromagnetic radiation from slowly moving sources z m R α Figure 5.4: The neutron star rotates with angular velocity Ω around the z axis. The magnetic moment m makes an angle of α with respect to this axis, and it rotates along with the star. then we can interpret this phenomenon in terms of the pulsar losing rotational energy, which is given by 1 (5.8.1) Erot = I Ω2 , 2 where I is the star’s moment of inertia and Ω its angular velocity. If P denotes the pulses period (which is not to be confused with the power radiated by electromagnetic waves), then Ω = 2π/P . If the neutron star is modeled as a solid sphere 2 of uniform mass density, then I = 2 5 M R , where M is the pulsar’s mass and R its radius. The loss of rotational energy translates into a decrease of Ω, or an increase of P : ˙ ˙ rot = I ΩΩ ˙ = −(2π )2 I P . (5.8.2) E P3 ˙ ≃ 4 × 10−13 s/s and P ≃ 0.03 s. For the Crab we have the observational values P If we take the neutron star to have a mass of 1.4 solar masses and a radius of 12 km, then the rate of loss of rotational energy is ˙ rot ≃ −7 × 1031 J/s. E This is comparable to what is required to power the Crab nebula. The energetics of the Crab can therefore be explained by the pulsar losing its rotational energy. What is responsible for this loss? We can imagine that the neutron star is carrying a magnetic field, and that the spinning motion of this field produces magneticdipole radiation. The energy carried off by the radiation will then come at the expense of the star’s rotational energy. For this model to work we need the field’s orientation to differ from the star’s rotational axis — otherwise we would have a stationary situation analogous to the one analyzed in Sec. 3.3. This is the obliquerotator model depicted in Fig. 5.4. We imagine that the neutron star maintains a magnetic dipole moment m that is oriented at an angle α with respect to the rotation axis. This vector rotates along with the star, also with an angular velocity Ω. If the rotation axis is aligned with the z direction, we can write m(t) = m0 sin α cos Ωt x ˆ + sin α sin Ωt y ˆ + cos α z ˆ , (5.8.3) 5.9 Problems 105 where m0 is the magnitude of the pulsar’s magnetic moment. This can be related to the magnitude of the magnetic field on the surface of the neutron star. Because the star is located well within the near zone, we can use Eq. (3.4.2) to describe the surface field: ˆ)r ˆ− m µ0 3(m · r , B= 3 4π R where R is the star’s radius. The field is maximum at the magnetic pole, where r ˆ is aligned with m. We therefore have Bmax = µ0 2m0 , 4π R3 (5.8.4) a relationship between m0 , the star’s radius, and the maximum surface magnetic field. The time-changing magnetic moment of Eq. (5.8.3) produces magnetic-dipole radiation. This radiation takes energy away from the star, at a rate given by Eq. (5.7.18), 2 ˙ rad = µ0 m ¨ (u) , E 3 6πc where u = t − r/c is retarded time. Substituting Eq. (5.8.3) gives ˙ rad = µ0 m0 Ω2 sin α 2 , E 6πc3 or ˙ rad = E 2π Bmax Ω2 R3 sin α 3µ0 c3 2 (5.8.5) (5.8.6) after involving Eq. (5.8.4). ˙ rad and |E ˙ rot | ≃ 7 × 1031 J/s, we need a To produce an equality between E magnetic field such that Bmax sin α ≃ 5 × 108 T. This is very large, but not unreasonable. A main-sequence star typically supports a magnetic field of 0.1 T. When the star collapses to form a neutron star, the magnetic field is frozen in and the magnetic flux 4πR2 B is conserved. Because R decreases by a factor of 105 during the collapse, B increases by a factor of 1010 , and fields of the order of 108 T can easily be achieved for neutron stars. 5.9 Problems 1. A point electric dipole is rotating around the z axis at an angular velocity ω , so that p(t) = p0 cos(ωt) x ˆ + sin(ωt) y ˆ, where p0 is a constant. a) Calculate the wave-zone electric and magnetic fields of this rotating dipole. ˆ and φ ˆ. Express the fields in terms of the unit vectors θ b) Calculate dP/dΩ , the angular profile of the radiated power, averaged over a complete wave cycle. c) Calculate the total radiated power. How does it compare with the power radiated by an oscillating dipole? 2. Two equal charges q are moving with constant angular velocity Ω on a circle of radius b. The charges are situated at diametrically opposite points on the circle. Calculate, to leading order in a slow-motion approximation, the total power radiated by this distribution of charges. 106 Electromagnetic radiation from slowly moving sources 3. The current density inside a centre-fed, linear antenna is given by j (t, x) = I cos(ωt) sin(kℓ − k |z |)δ (x)δ (y ) z ˆ, where I is the current’s peak amplitude, ω the oscillation frequency, ℓ the antenna’s half-length, and k = ω/c. In this problem we do not impose the slow-motion approximation, so that kℓ is not assumed to be small. a) Show that in the wave zone (kr ≫ 1), the vector potential can be expressed as µ0 w(u, θ) , A(t, x) = 4π r and find an expression for the vector w(u, θ). Here, u = t−r/c is retarded time and θ is the angle between x and the z axis. b) Calculate dP/dΩ , the angular profile of the radiated power, averaged over a complete wave cycle. For the two special cases kℓ = π/2 (halfwave antenna) and kℓ = π (full-wave antenna), provide parametric plots of this angular profile (like those displayed in Jackson’s Figure 9.7). Compare these profiles with the sin2 θ dependence obtained for electric-dipole radiation. c) From the general expression for dP/dΩ derived in part b), obtain the short-antenna limit kℓ ≪ 1. Make sure that your result matches the one derived in the text. 4. An oscillating current I (t) = I0 e−γt sin(ωt) flows in a circular loop of radius a placed in the equatorial plane of the coordinate system (r, θ, φ). The current is started at t = 0, and it decays exponentially due to dissipation within the loop; we assume that γ ≪ ω , so that the time scale for dissipation is much longer than the oscillation period. The current density is given by I (t) ˆ. δ (r − a)δ (θ − π/2) φ a Calculate P , the total power radiated by the loop, averaged over an oscillation period. Calculate also the total amount of energy radiated by the loop. j (t, x) = 5. A nearly spherical surface described by r = R(t, θ) ≡ a(1 + ǫ cos θ cos ωt), where a, ω , and ǫ are constants, supports a nearly uniform distribution of charge. It is assumed that ǫ ≪ 1, and the charge density is given by q δ r − R(t, θ) , ρ(t, x) = 4πa2 where q is the total charge. Calculate P , the electromagnetic power emitted by this distribution of charge, averaged over a complete wave cycle. You may work to leading order in the slow-motion approximation. You may also work to leading order in ǫ. 6. A ring of radius b is placed in the x-y plane, and the origin of the coordinate system is placed at the centre of the ring. Initially, the ring is at rest and it supports a linear charge density proportional to sin φ, where φ is the angle from the x-axis. The ring is then made to rotate with a uniform angular velocity Ω, so that the linear charge density becomes proportional to sin(φ − Ωt). The ring’s volume charge density is described by ρ(t, r, θ, φ) = λ δ (r − b)δ (θ − π 2 ) sin(φ − Ωt). b (Express this in terms of the angle θ from the z axis. (c) Find an exact expression for A(t. At what angle θ is most of the radiation emitted? (b) Find an exact expression for Φ(t. a charge −q is at z = +a cos ωt. 7. A current I is suddenly established in an infinite wire that extends along the z axis. 0). 4π x2 + y 2 The electric field eventually disappears.5. a and ω are constants. the vector potential A(t. Because ∇ · j = 0.9 Problems 107 (a) Calculate. without approximations. to leading order in a slow-motion approximation. For this situation calculate dP/dΩ . The current density is given by j (t. From your answer you should be able to verify that at late times. Calculate. where θ(t) is the step function. and the magnetic field settles down to the value it would have for an unchanging current. Three charges are located along the z axis: a charge +2q stays at the origin z = 0. and the final charge −q is at z = −a cos ωt. x) associated with this distribution of current. the angular profile of the radiated power. 4π x2 + y 2 By ≃ µ0 2Ix . It is assumed that aω ≪ 2πc and that the charges simply go through one another when they meet at z = 0. the total radiated power. 0). . it is consistent to set ρ(t. x) = 0. 4π t Bx ≃ − µ0 2Iy . the vector potential at the centre of the ring. the total power radiated by the ring.) Calculate also P . averaged over a complete wave cycle. x) = Iθ(t)δ (x)δ (y )z ˆ. the electric and magnetic fields that are produced by the distribution of current have the components Ez ≃ − µ0 2I . the scalar potential at the centre of the ring. What is the frequency of the electromagnetic waves emitted by this system of charges? 8. 108 Electromagnetic radiation from slowly moving sources . 3) We now would like to derive how the electric and magnetic fields transform under such a change of reference frame.9.1. z ′ ) of an event seen ′ in S are related to the coordinates (t. 11. so that the frame’s velocity vector is v = v x ˆ. ∂t′ ∂t ∂t′ ∂x ∂t′ ∂y ∂t′ ∂z ∂t′ or ∂ψ ∂ψ ∂ψ =γ +v ∂t′ ∂t ∂x 109 . z = z′. c2 (6. where γ= y ′ = y. x′ . y ′ .10 of Jackson’s text. y. To keep the discussion concrete we will restrict it to the electromagnetic field produced by a single point particle in arbitrary (relativistic) motion. The reversed transformation is obtained by reversing the sign of v : x = γ (x′ + vt′ ). (6. c2 (6.3. We note first that Lorentz transformations alter the differential operators. and 11.2) These are the Lorentz transformations between the two reference frames. We consider a reference frame S ′ that moves with constant speed v relative to another frame S .1. To set the stage we begin with a review of the Lorentz transformations of special relativity. t = γ t′ + v ′ x .1) 1 1 − v 2 /c2 . we take the motion to take place in the x direction. z ) of the same event seen in S by the well-known relations x′ = γ (x − vt). t′ = γ t − v x .1 Lorentz transformations [The material presented in this section is also covered in Secs.] In this final chapter we will look at the electrodynamics of charge and current distributions for which the characteristic speed vc is close to the speed of light. y = y′ .Chapter 6 Electrodynamics of point charges 6. z ′ = z. applying the chain rule on an arbitrary function ψ gives ∂ψ ∂ψ ∂t ∂ψ ∂x ∂ψ ∂y ∂ψ ∂z = + + + . 11. x. For example. The coordinates (t′ .1. and that the other components of the current density will be left alone. and we further expect that jx will transform as x. ∂z ′ ∂z ψ 1 ∂2ψ ∂2ψ ∂2ψ ∂2ψ + + + c2 ∂t′2 ∂x′2 ∂y ′2 ∂z ′2 ∂ ∂ ∂ ∂ 1 ψ +v +v = − 2 γ2 c ∂t ∂x ∂t ∂x ∂2ψ ∂2ψ ∂ v ∂ v ∂ ∂ ψ+ + + 2 + 2 + γ2 ∂x c ∂t ∂x c ∂t ∂y 2 ∂z 2 2 2 2 2 1 ∂ ψ ∂ ψ ∂ ψ ∂ ψ = − 2 2 + + + c ∂t ∂x2 ∂y 2 ∂z 2 = ψ.110 Electrodynamics of point charges after using Eq. But we can write it in the equivalent form ρ′ = qγ (1 − v 2 /c2 )δ (x − vt)δ (y )δ (z ) v = γ qδ (x − vt)δ (y )δ (z ) − 2 qvδ (x − vt)δ (y )δ (z ) c v = γ ρ − 2 jx . = − ′ In other words. and jy = jz = 0. however. As seen from S we have ρ = qδ (x − vt)δ (y )δ (z ).1. while jx = jy = jz = 0. It is sufficient to consider a simple special case: a point charge q moving in the x direction with a speed v (the same speed that enters in the Lorentz transformation). Rewriting ρ′ in terms of the unprimed coordinates gives ρ′ = qδ [γ (x + vt)]δ (y )δ (z ) q δ (x − vt)δ (y )δ (z ). jx = qvδ (x − vt)δ (y )δ (z ).3). = γ This expression for ρ′ involves a factor of 1/γ instead of the expected γ . =γ + ∂x′ ∂x c2 ∂t ∂ ∂ = . jx γ (jx − vρ). This shows that ′ = ρ transforms as t. (6. the particle is not moving. (6. the charge and current densities transform as ′ jx = γ (jx − vρ). ρ′ = γ ρ − v jx . As seen from S ′ . and ′ ′ ′ ρ′ = qδ (x′ )δ (y ′ )δ (z ′ ). We can expect that the Lorentz transformation will involve a linear mixture of ρ and jx . we obtain ∂ ∂ ∂ . we first work out how ρ and j should transform.5) To figure out how the electric and magnetic fields transform under a change of reference frame. This will tell us how the potentials Φ and A transform. Doing a similar calculation for all other partial derivatives. c and this is the expected expression of a Lorentz transformation. This is confirmed by the fact that the right-hand side is zero for the situation considered here. ′ jy = jy . and jx has not yet made an appearance. ′ jz = jz . that is.4) It follows from this that the wave operator is a Lorentz invariant: ′ v ∂ ∂ ∂ .6) . We conclude that under a change of reference frame. = .1.1. ∂y ′ ∂y ∂ ∂ = . c2 (6.1. =γ +v ∂t′ ∂t ∂x (6. and from this we will be able to get to the fields. All the equations satisfied by the potentials are therefore preserved under a Lorentz transformation. We can now figure out how the fields transform. To see how the potentials transform we recall the wave equations (Φ/c2 ) = −ρ/(ǫ0 c2 ) = −µ0 ρ and A = −µ0 j from Sec. ′ Ex ∂ ∂ +v ∂t ∂x ∂Ax ∂Φ = − − ∂t ∂x = Ex . (6. it is possible to show that they are in fact completely general. and we substitute Eqs. For example. You should check that the Lorenz gauge condition.4) and (6. c = .7) are the Lorentz transformations for the scalar and vector potentials. E′ = − ∂ A′ − ∇′ Φ′ . A′ z = Ax . A′ x = γ Ax − v Φ . Thus. These are designed to leave the form of the wave equations unchanged under a change of reference frame. c2 ∂t also takes the same form in S ′ . Φ′ = γ (Φ − vAx ) (6. Because is an invariant. This implies that Maxwell’s equations themselves are invariant under a change of reference frame. We simply write down how the fields in S ′ are related to the potentials. ∇·A+ 1 ∂Φ = 0.1.1.1.1 Lorentz transformations 111 While the derivation of these transformation rules was restricted to a very simple situation. ∂y ′ ∂z ′ ∂y ∂z Yet another is ′ By ∂A′ ∂A′ z x − ′ ∂z ∂x′ v v ∂ ∂ ∂ Ax − 2 Φ − γ Az + = γ ∂z c ∂x c2 ∂t ∂Ax v ∂Az ∂Az ∂Φ = γ +γ 2 − − − ∂z ∂x c ∂t ∂z v = γ By + 2 Ez . and that A must transform as j . Another example is ′ Bx = ∂A′ ∂A′ ∂Az ∂Ay y z − = − = Bx . = −γ 2 Ax − v ∂ v ∂ Φ − γ2 + c2 ∂x c2 ∂t Φ − vAx Or ′ Ey = −γ ∂ ∂ ∂ Ay − γ Φ − vAx +v ∂t ∂x ∂y ∂Ay ∂Φ ∂Ax ∂Ay − γv = −γ + − ∂t ∂y ∂x ∂y = γ Ey − vBz . these equations tell us that Φ/c2 must transform as ρ. 5. c2 A′ y = Ay .7).6.1. ∂t′ B ′ = ∇ ′ × A′ . and Ez = γvBy = vBy = (v × B ′ )z . O moves in the negative x direction with speed v . ′ ′ ′ ′ ′ ′ Ex = 0. c2 ′ Bz = γ Bz − v Ey . |r ′ − rO | = b2 + v 2 t′2 . the electric field lines are distributed isotropically around r ′ .] We can use the Lorentz transformations of Eqs. 0). . ′ Ez = γ Ez + vBy (6.1. 0). We want E and B as measured at O. As seen in the reference frame S ′ .1. b. 0. In other words.2.1) where b is the charge’s displacement above the x axis.9) Two observers in relative motion will therefore disagree as to the value of the electric and magnetic fields. and its value at O is given by the standard expression ′ q r ′ − rO ′ |3 . only the electromagnetic field as a whole has an observer-independent existence. ′ 2 2 ′ ′ ′ ′ ′ Ey = γEy .2. In another reference frame S ′ we have Ex = Ex . ′ By = γ By + v Ez . because relative to S ′ . B=0 → B′ = − 1 v × E′. while the magnetic field transforms as ′ Bx = Bx . 0). c2 (6. b. b. (6. We conclude that E and B are observer-dependent fields.8) and (6.10) A pure electric field in S is therefore perceived as a mixture of electric and magnetic ′ ′ ′ fields in S ′ . By = γ (v/c )Ez = (v/c )Ez = −(v × E )y /c2 .9) to calculate the electric and magnetic fields of a point charge q that moves with constant speed v in the x direction. E = 0 → E′ = v × B′.10 of Jackson’s text. Ey = −γvBz = −vBz = (v × B )y . (6. 4πǫ0 |r ′ − rO √ ′ ′ We have r ′ − rO = (vt′ . and ′ ′ Bz = −γ (v/c2 )Ey = −(v/c2 )Ey = −(v × E ′ )z /c2 . By = γBy .112 Electrodynamics of point charges The complete summary of these results is that the electric field transforms as ′ Ex = Ex . Bz = γBz . We shall first do the calculation in S ′ .2 Fields of a uniformly moving charge [The material presented in this section is also covered in Sec.2) As measured in S ′ . In other words.8) under a change of reference frame.1. In S ′ there is only an electric field E ′ . Ez = γEz . the charge’s position. 2 4πǫ0 (b + v 2 t′2 )3/2 ′ Ez = 0. Then Bx = Bx . ′ but the observer at O has a position rO = (−vt′ . 6.1. (6. Bx = 0. The charge’s position vector is r (t) = (vt. ′ Ey = γ Ey − vBz . and the primed components of the electric field are E′ = − ′ Ex =− q vt′ . the origin of the reference frame S . 4πǫ0 (b2 + v 2 t′2 )3/2 ′ Ey =− b q . (6.11) A pure magnetic field in S is therefore perceived as a mixture of electric and magnetic fields in S ′ .1. 11. c2 (6. which moves along with the particle. 0). Suppose next that E = 0 in S . Suppose first that ′ B = 0 in some reference frame S . the charge’s position vector is r ′ = (0. There are interesting special cases to the foregoing results.1. 4) c2 for the fields measured at O. The unprimed fields are obtained by inverting Eqs. (6. 0).2 Fields of a uniformly moving charge 113 t –3 –2 –1 1 2 3 –0. (6. We also have b2 + γ 2 v 2 t 2 = γ 2 (b2 /γ 2 + v 2 t2 ) v2 = γ 2 b2 + v 2 t 2 − 2 b2 c b2 v2 = γ 2 (b2 + v 2 t2 ) 1 − 2 2 c b + v 2 t2 v2 = γ 2 r2 1 − 2 sin2 ψ . r (t) = (vt. We now transform the field to the frame of reference S .2.8) and (6. 2 ′ 2 ′ ′ Bx = Bx = 0.9). Plots of Ex and Ey as functions of time are shown in Fig. in the S frame. 6. c .1.5 –1 –1. Because B ′ = 0 we have Ex = Ex . The coordinates of O are x = y = z = 0. The fields can also be expressed in terms of the charge’s position vector. Ey = γEy . and Bz = (γv/c )Ey .1. q γb . This is defined by v · r (t) .2.1. 4πǫ0 (b2 + γ 2 v 2 t2 )3/2 Bx = 0.5 –2 Figure 6. b. Ey = − By = 0.1: Electric field of a uniformly moving charge. We have v · r = v 2 t. E z = Ez = 0.3) v Ey (6. We therefore obtain Ex = − and q γvt .6. By = −(γv/c )Ez = 0. so cos ψ = vt/ b2 + v 2 t2 . as measured by an observer at the origin of a fixed reference frame.5) cos ψ = vr(t) √ √ where r ≡ |r | = √ b2 + v 2 t2 . For this purpose we let ψ be the angle between the vectors r and v . 4πǫ0 (b2 + γ 2 v 2 t2 )3/2 Bz = Ez = 0 (6. while the y component is always negative and achieves its maximum (in magnitude) at that time. which implies sin ψ = b/ b2 + v 2 t2 .2. which is done by ′ ′ ′ reversing the sign of v . The x component of the field goes through zero at t = 0. and the time coordinates are therefore related by t′ = γt. 6) that E is always directed along −r (t).1 of Jackson’s text. (6. and a(t) = dv /dt its acceleration vector. however. and the field is amplified by a factor γ > 1. and E should therefore point in the direction of r (t′ ). The electric field. But this is not what we have found: the field does point in the direction of r (t). . (6. and weakest in the longitudinal direction. During the time in which the signal propagates. We let v (t) = dr /dt be its velocity vector. But it is remarkable that the field registered at O does not involve a time delay associated with the propagation of a signal from charge to observer.3 Fields of an arbitrarily moving charge [The material presented in this section is also covered in Secs.3.3). Our picture. We will return to this issue later on.6) for the electric field at O. that is. when r is directed along v . a vector that points from the charge to O. is strongest in the directions transverse to the motion. The charge density is ρ(t. the particle moves to its current position.2. And it is unremarkable that the field lines should be distorted by a Lorentz contraction in the direction of the motion. When ψ = 0. it should know only of the earlier position. (6. a signal originates at the charge when it was at the earlier position. and investigate it more fully. is curiously wrong. It is unremarkable that the field lines should emanate from the position of the charge — we have the same picture in the charge’s rest frame. but instead toward an earlier position r (t′ ). we have that [1 − (v/c)2 sin2 ψ ]−3/2 = 1. In this picture.] We now allow the point charge q to move on an arbitrary trajectory r (t).5 and 14.114 Electrodynamics of point charges and this can be substituted into Eqs.3. (6. this is revealed by the presence of sin2 ψ in the expression for the electric field. But when instead ψ = π/2. The results derived in this section are at once remarkable and unremarkable. so that r is orthogonal to v . 6. We see from Eq. and this means that the electric field lines originate at the position of the moving charge. x) = q v (t)δ x − r (t) . The electric field therefore points away from the charge.2) To obtain the fields produced by this particle we shall first calculate the potentials. x) = qδ x − r (t) . the current position of the charge. and the electric field is reduced by a relativistic factor of γ −2 < 1. but the field registered at O should not yet be aware of this. which in principle can be solved for t′ .2. since the distance traveled is |r (t′ )|.2. The field lines.7) for the magnetic field.2. then [1 − (v/c)2 sin2 ψ ]−3/2 = γ 3 > 1. are not distributed isotropically around the charge.1) and the current density is j (t. as seen in S . and B= (6. 6. Allowing for such a delay would suggest that the field at time t should point not toward the charge at its current position r (t). and it propagates to O in a time ∆t ≡ t − t′ determined by the condition that the signal travels at the speed of light. The end result is E=− 1 q 4πǫ0 γ 2 1 − (v/c)2 sin2 ψ 1 v×E c2 3/ 2 r r3 (6. according to which information should be delayed. we must have t − t′ = |r (t′ )|/c. 3.4) for the scalar potential. (6.7) and this implicit equation can in principle be solved for t′ . (6.2 and 5. Φ= q 1 4πǫ0 κR .3. We have ds/dt′ = 1 + c−1 dR/dt′ . 5. (6.3) The same manipulations produce Φ(t. x) = q 4πǫ0 δ (t − t′ − |x − r (t′ )|/c) ′ dt |x − r (t′ )| ′ ˆ (t′ ) = R(t ) .5) In terms of these the argument of the δ -function is t − t′ − R(t′ )/c. With this we obtain ds/dt′ = κ. At this stage it is convenient to introduce the quantities R(t′ ) = x − r (t′ ). the delay (t − t′ ) is set by the time required by light to travel the distance R(t′ ).3 Fields of an arbitrarily moving charge 115 6. 1. x) = µ0 4π j (t′ .6) qµ0 v δ (s) dt′ ds 4π R ds qµ0 v δ (s) ds 4π κR qµ0 v .7) is clear: The signal received at position x and time t depends on the state of motion of the particle at an earlier time t′ . Substituting Eq. (6.3. R ′ R(t ) (6. κ ≡ 1 − v (t′ ) · R The integral for the vector potential is then A = = = Similarly.9.3.3.3. We shall call t′ the retarded time corresponding . To evaluate the integrals of Eqs.2) we employ a technique that was first put to use in Secs.3.3. |x − r (t′ )| (6.3) and (6.2) yields A(t.3.1 Potentials To calculate the potentials associated with the densities of Eqs.3. R(t′ ) = |x − r (t′ )|. The meaning of Eq.3.3. (6. For example. 4π κR s=0 The condition s = 0 gives R(t′ )/c = t − t′ .4) we let s ≡ t′ + R(t′ )/c − t be the new variable of integration. where to verify that dR/dt′ = −R · v /R = −R ˆ (t′ )/c. s=0 (6. and it is easy ˆ · v . |x − x′ | and integrating over d3 x′ returns A(t. x) = qµ0 4π v (t′ ) δ (t − t′ − |x − r (t′ )|/c) ′ dt . we express the vector potential as A(t. x′ ) δ (t − t′ − |x − x′ |/c) 3 ′ ′ d x dt .1) and (6.6.3. or |x − r (t′ )| = c(t − t′ ). x) = qµ0 4π v (t′ )δ x′ − r (t′ ) δ (t − t′ − |x − x′ |/c) 3 ′ ′ d x dt . (6. |x − x′ | in terms of the retarded Green’s function obtained in Sec. 7). Its dependence on x comes from the explicit dependence of R and κ on position. x) = 1 q 4πǫ0 κR qµ0 v 4π κR (6. x) = q 4πǫ0 1 δ (s) dt′ R(t′ ) v (t′ ) δ (s) dt′ . (6. x′ ) d3 x′ = q .) The factor of 1/κ accounts for the fact that ρ(t − |x − x′ |/c.3.3) and (6. x′ ) d3 x′ = = q = q κ ρ(t′ . (6. x) = and A(t. B = ∇ × A.9) indicating that the quantities within the square brackets must be evaluated at the time t′ determined by the retarded condition of Eq.10) q 1 4πǫ0 c2 (6. it is all the more curious that the effect was not seen in the preceding section. and the result is not q .3. R(t′ ) (6. in the previous expression we sample the charge density at different times while performing the integration.116 Electrodynamics of point charges to the spacetime point (t. (Stay tuned. ∂t The dependence of A on t comes from its explicit dependence on t′ and the retarded condition |x − r (t′ )| = c(t − t′ ). x) = and A(t. We shall write the potentials as Φ(t. ∂A E=− − ∇ Φ.3.3.3. and Eq. These expressions are known as the Li´ enard-Wiechert potentials. But we can simplify things by going back to earlier forms of Eqs.3. and this result clarifies the origin of κ = 1 − v · R 6. While we do have ρ(t.3.2 Fields Having obtained the potentials. (6.3. ret ˆ /c. Given that the retardation effect is present in the Li´ enard-Wiechert potentials.11) .8) ret . but there is also an implicit dependence contributed by the retarded condition. ρ(t − |x − x′ |/c. and they implement the notion that the potentials here and now should depend on the retarded position and velocity of the particle. Φ(t. x). ret (6. This mixture of explicit and implicit dependence complicates the task of taking derivatives.7) will be referred to as the retarded condition.3. x′ )δ (t − t′ − |x − x′ |/c) d3 x′ dt′ δ (t − t′ − |x − r (t′ )|/c) dt′ . we can calculate the fields by straightforward differentiation. x′ ) d3 x′ is not equal to the charge q .4). Apart from the presence of κ in the denominator. the potentials take their familiar form from time-independent situations. Instead. κR δ (s) ds This can be integrated at once. The derivative with respect to s can now be expressed in terms of d/dt′ . = 4πǫ0 c2 κ dt′ κR = In the first step we used the fact that the only t dependence is contained in s. ∇ κ R ds cκR where we have integrated the second term by parts. and integrate next (thereby setting t′ by the retarded condition). and we finally obtain ∇Φ = q 4πǫ0 1 1 1 d ∇ − κ R cκ dt′ ∇R κR δ (s) ds.6): R(t′ ) = x − r (t′ ) and κ = 1 − v (t′ ) · R We begin with the scalar potential. We now set to work on d dt′ ∇R − v /c κR = ˆ 1 dR a − κR dt′ c − ˆ − v /c d R κR . This gives ∇Φ = = = q 4πǫ0 q 4πǫ0 q 4πǫ0 ∇ ∇R ′ 1 δ (s) + δ (s) dt′ R cR ∇R ′ 1 1 δ (s) + ∇ δ (s) ds κ R cκR d ∇R 1 1 − δ (s) ds.3. It is easy to establish that ∇R = R −1 2 ′ ˆ ˆ so that ∇R = −R/R .3 Fields of an arbitrarily moving charge 117 where s = t′ + R(t′ )/c − t. Eqs. so that the dependence of the potentials on t and x is fully explicit.6. and we obtain E= q 1 1 1 d − ∇ + 4πǫ0 κ R cκ dt′ ∇R − v /c κR . And in the fourth step we replaced the s-derivative by a t′ -derivative. Collecting these results gives us an expression for the electric field: E = = q 4πǫ0 q 4πǫ0 1 − ∇ κ 1 − ∇ κ 1 1 d + R cκ dt′ 1 d 1 + R cκ dt′ 1 d ∇R v − 2 κR c κ dt′ κR ∇R − v /c δ (s) ds. This expression badly needs to be simplified. In the third step we integrated by parts. we can replace ∇s by ∇R/c.5) and (6.3. Before we proceed we recall the definitions of ˆ (t′ )/c = ds/dt′ . It will then be straightforward (if tedious) to differentiate first. We have ∂A ∂t v ′ ∂s ′ q 1 δ (s) dt 2 4πǫ0 c R ∂t v ′ q 1 δ (s) ds = − 4πǫ0 c2 κR d v q 1 δ (s) ds = 4πǫ0 c2 ds κR q 1 v 1 d δ (s) ds. and we already have obtained dR/dt = −v · R in the preceding subsection. Leaving this result alone for the time being. ret ˆ. κ2 R2 dt′ . R R and since s depends on x through R. We have ∇Φ = q 4πǫ0 ∇ 1 1 δ (s) + δ ′ (s)∇s dt′ . Their advantage is that t′ inside the integrand is no longer restricted. we move on and calculate ∂ A/∂t. In the second step we substituted ∂s/∂t = −1. (6. + 2 2 (R κ Rc = − The first and fourth terms cancel out. is R This gives 1 d c dt′ ∇R − v /c κR =− ˆ × (R ˆ − v /c) × a ˆ ˆ − v /c R R R + + . on the other hand. The acceleration field. We have ∇(1/R) × v = −(R ′ ˆ and d(R × v /κR)/dt can be expressed as ˆ − v /c) × v d (R d = ′ ′ dt κR dt ˆ − v /c R κR ×v+ ˆ − v /c R × a.3. To obtain the magnetic field we take the curl of Eq. κR . (6. ret (6. which gives ∇×A = = = = so that B= q 1 4πǫ0 c2 q 1 4πǫ0 c2 q 1 4πǫ0 c2 q 1 4πǫ0 c2 ∇ ∇s × v ′ 1 × v δ (s) + δ (s) dt′ R R ∇R × v ′ 1 1 × v δ (s) + ∇ δ (s) ds κ R cκR d ∇R × v 1 1 ×v− δ (s) ds ∇ κ R ds cκR 1 d ∇R × v 1 1 ×v− δ (s) ds.118 Electrodynamics of point charges ˆ /dt′ = d(R/R)/dt′ = where a = dv /dt′ is the particle’s acceleration.11). and also d(κR)/dt′ = d(R − v · ˆ − (a · R − v · v )/c = −v · R ˆ − a · R/c + v 2 /c. Plugging these R/c)/dt′ = −v · R results back into our previous expression gives 1 d c dt′ ∇R − v /c κR 1 ˆ /c)R ˆ − a v /c − (v · R κR2 κRc2 ˆ ˆ − v /c R ˆ /c + v 2 /c2 + R − v /c a · R − −v · R 2 2 κ R κ2 R 2 c2 ˆ 1 ˆ + R − v /c 1 − κ − v 2 /c2 = − 2 v /c − (1 − κ)R κR κ2 R 2 1 ˆ − v /c)(a · R ˆ ) − κa + 2 2 (R κ Rc ˆ − v /c ˆ − v /c ˆ ˆ − v /c R R R R = 1 − v 2 /c2 − − 2+ 2 2 2 κR R κ R κR2 1 ˆ − v /c)(a · R ˆ ) − κa . ret 1 q 1 1 1 d ∇ ×v− 4πǫ0 c2 κ R cκ dt′ ˆ × v )/R2 . represents electromagnetic radiation that propagates independently of the charge. and the expression within the square brackets ˆ × [(R ˆ − v /c) × a] = (R ˆ · a)(R ˆ − v /c) − R ˆ · (R ˆ − v /c)a = (R ˆ · a)(R ˆ − v /c) − κa. The electric field is naturally decomposed into a “velocity field” that decays as 1/R2 and an “acceleration field” that decays as 1/R. We now work on simplifying this expression. The velocity field represents that part of the electric field that stays bound to the particle as it moves.12) our final expression for the electric field of a moving point charge.3. ∇ κ R cκ dt′ κR ∇R × v κR . x) = ˆ × (R ˆ − v /c) × a ˆ − v /c R R q + 2 3 2 4πǫ0 γ κ R κ3 c2 R . it is the particle’s generalized Coulomb field. We have dR 2 ˆ ˆ ˆ −v /R + (R/R )(v · R) = −[v − (v · R)R]/R. R2 γ 2 κ2 R 2 κ2 c2 R Substituting all this into our previous result for the electric field returns E (t. ) It shows that the field points in the direction opposite to the vector −R + Rv /c evaluated at the retarded time t′ . ret (6. we arrive at B= or B (t. This is where the charge would be after a time ∆t if it were moving with a constant velocity v (t′ ). 2 2 2 R γ κ R κ Rc cκR − Collecting these results.2: where does the electric field point? Does it point toward r (t′ ). We have ˆ − v /c) × v 1 d (R ′ c dt κR = ˆ − v /c (R ˆ · a)(R ˆ − v /c) − κa ˆ R R ×v + 2 2 2 + 2 R γ κ R κ2 Rc2 ˆ − v /c R + ×a cκR ˆ×v ˆ · a)R ˆ×v a×v ˆ×v R (R R + 2 2 2+ − = − 2 2 R γ κ R κ Rc2 κRc2 ˆ×a a×v R + + cκR κRc2 ˆ×v ˆ · a)R ˆ×v R ˆ×v ˆ×a R (R R = − + 2 2 2+ + . then R = −r and the bound field points in the direction opposite to r (t′ ) + (∆t)v (t′ ).) So the field brings information to the . 6. (6. ret ˆ×v ˆ · a)R ˆ×v R ˆ×a R (R q 1 − − − 4πǫ0 c2 γ 2 κ3 R 2 κ3 Rc2 cκ2 R . (The situation is described in Fig. and an acceleration field that represents electromagnetic radiation.2? The answer is that in general. x) = − ˆ×v ˆ · a)R ˆ × v /c + κR ˆ×a (R q 1 R + 4πǫ0 c2 γ 2 κ3 R2 cκ3 R . This result is interesting: the field points not in the direction opposite to the charge’s retarded position r (t′ ). where ∆t is determined by the retarded condition ∆t ≡ t − t′ = r(t′ )/c.3. the time interval corresponds to the time required for a light signal to travel from the charge’s retarded position to the observation point.6. but in the direction opposite to the charge’s anticipated position ra ≡ r (t′ ) + (∆t)v (t′ ). as we expect it to behave largely independently of the charge. Consider our expression (6.12) for the bound electric field. 6. as we actually found in Sec. Our general results for the electric and magnetic fields allow us to resolve the issue that was first brought up near the end of Sec.2. the electric field follows neither option. If the observation point is at the origin of the coordinates (x = 0). as it was argued it might do? Or does it point toward r (t). (We exclude the radiative field from this discussion.13) A little more algebra reveals that this can also be expressed as B= 1 ˆ R c ret × E. the retarded position of the particle. its current position.3. nor in the direction opposite to the charge’s current position r (t).3 Fields of an arbitrarily moving charge 119 which allows us to use results that were generated during the calculation of the electric field.14) The magnetic field also is decomposed into a velocity field that stays bound to the particle. 6.3. and anticipated position of an accelerated charge. Because the extrapolation is based only on partial information (the values of r and v at the retarded time. . r (t′ ) = (vt′ . the charge’s anticipated position ra coincides with its current position r (t).3. but the field orients itself as though it were anticipating the motion of the particle during the time the information propagates from the charge to the observer.16) b2 + (vt′ )2 = c(t − t′ ).3.15) E=− 4πǫ0 (γκr′ )3 We recall that κ = 1 + v · r ˆ′ /c. (1 − v 2 /c2 )t′2 − 2tt′ + t2 − b2 /c2 = 0. (6.3. after squaring.12) reduces to Eq. and nothing more). b. (6.2: the information needed time to reach the observer. and the field misses the actual position of the particle.12) reduces to q r q r ′ + r′ v /c ˆ′ + v /c E=− =− . This explains the result found in Sec. Eq. 6. 6. The retarded condition is r′ = and it gives. This is a quadratic equation for t′ .2: Retarded position.3. 0) ≡ r ′ .3. We once more take the observation point to be at x = 0. (6. it is not fully accurate. b. 0). observer that does indeed originate from the retarded position. and v (t) = v (t′ ) = v x ˆ. current position. (6. but it was sufficient to correctly anticipate the actual motion of the charge. which we do not need to solve. and the field “miraculously” orients itself in this direction.120 Electrodynamics of point charges retarded position v (t’) r(t’) anticipated position ra current position Figure 6. 2 3 ′ 2 4πǫ0 γ κ r 4πǫ0 γ 2 κ3 r′3 or q γ (r ′ + r′ v /c) .2. Because a = 0. Notice that uniform motion is a special case: when the velocity is constant.3 Uniform motion Let us verify that the general expression of Eq.3) in the case of uniform motion in the x direction. We set r (t) = (vt. (6. 3. ret (6. (6.2.20) (6.4 Summary The scalar and vector potentials of a point charge in arbitrary motion described by the position vector r (t) are given by the Li´ enard-Wiechert expressions Φ(t. x) = ˆ × (R ˆ − v /c) × a ˆ − v /c R q R + 2 3 2 4πǫ0 γ κ R κ3 c2 R 1 ˆ R c .18) ret (6.3.15) involves the vector r ′ + r′ v /c.3.3.19) ˆ = R/R.3. R = |R|. The electric field is E (t.23) Both fields are naturally decomposed into velocity (bound) fields that decay as 1/R2 .3. 6. 4πǫ0 (b2 + γ 2 v 2 t2 )3/2 (6. and acceleration (radiative) fields that decay as 1/R. x) = and A(t. R ˆ (t′ )/c. (6.6. ret q 1 4πǫ0 κR (6. This is (γκr′ )2 = γ 2 (r′ + v · r ′ /c)2 = γ 2 (r′ + v 2 t′ /c)2 = γ 2 c2 (t − t′ + v 2 t′ /c2 )2 = γ 2 c2 (t − t′ /γ 2 )2 = c2 (γ 2 t2 − 2tt′ + t′2 /γ 2 ).3. κ = 1 − v (t′ ) · R and where t′ is determined by the retarded condition |x − r (t′ )| = c(t − t′ ). .3. and this vector therefore coincides with r . x) = qµ0 v 4π κR . where R(t′ ) = x − r (t′ ).16) to eliminate t′ and obtain (γκr′ )2 = c2 b2 /c2 + γ 2 t2 (1 − 1/γ 2 ) = b2 + γ 2 v 2 t 2 .3 Fields of an arbitrarily moving charge 121 The numerator of Eq. In the denominator we have (γκr′ )2 raised to the power 3/2.3. (6. Substituting these results into Eq.3). Its x component is vt′ + r′ v/c = vt′ + c(t − t′ )v/c = vt.3.21) (6.22) and the magnetic field is B= ret × E.17) the same statement as Eq.15) gives E=− γ r (t) q . (6. (6. We now use Eq.3. We have that S (t′ ) · da′ dt dt′ is the energy crossing an element of S ′ per unit t′ -time.] At large distances from the particle — in the wave zone — we can neglect the velocity fields and approximate the electric field by E (t. S (t′ ) · da′ .1 Angular profile of radiated power Because our expressions for the fields refer directly to the charge’s retarded position. The sphere S ′ has a constant radius R(t′ ) and its element of ˆ (t′ ) dΩ′ . The energy crossing an element of S ′ per unit time t is. 6. (6. and We recall that R = x − r (t′ ). The sphere has a constant radius R(t′ ) ≡ |x − r (t′ )|.2) ˆ = R/R. κ = 1 − v (t′ ) · R ˆ (t′ )/c. it is a relatively simple task to calculate the power that crosses a sphere S ′ centered at this position (see Fig. We also determine the angular distribution of the radiated power. R = |R|. R ′ ′ ′ that t is determined by the retarded condition t − t = R(t )/c. by definition. 6.4. 14. it would be much more complicated to adopt any other reference point. x) = and the magnetic field by B= 1 ˆ R c ret ˆ × (R ˆ − v /c) × a R q 4πǫ0 κ3 c2 R .1) × E.4. ret (6. and 14. it is convenient to convert this “per-unitt-time” flux to a “per-unit-t′ -time” measure. The displayed quantity can is dt/dt′ = 1 + c−1 dR(t′ )/dt′ = 1 − v (t′ ) · R . But because all quantities are expressed in terms of retarded time t′ instead of t.122 Electrodynamics of point charges S’ retarded position R(t’) current position Figure 6.3: Sphere S ′ centered at the retarded position r (t′ ) of the accelerated charge. and the conversation factor ˆ (t′ )/c ≡ κ. In this section we use these expressions to calculate how much power is radiated by the accelerated charge.4 of Jackson’s text.4.3). 14. 6. where S = E ×B /µ0 is the Poynting vector.2.3. where dΩ′ is an element of solid angle on the surface area is da′ = R2 (t′ )R sphere.4 Radiation from an accelerated charge [The material presented in this section is also covered in Secs. Integrating this over dt′ (which involves displacing S ′ in response to the particle’s motion during integration) returns the total energy radiated by the charge.3) energy crossing an element of S ′ per unit t′ -time and unit solid angle. E×B = E × (R µ0 µ0 c µ0 c ˆ .4. To ease the notation we shall now agree to drop the primes on dP ′ /dΩ′ . we shall do so in the following subsections.3) over dΩ′ gives the total power crossing S ′ .4. To gain insight into this it will help to look at specific simple situations.4. (6. in particular Eq. and R.4 Radiation from an accelerated charge 123 ˆ (t′ )R2 (t′ )κ dΩ′ . (5. the angular profile of radiation produced by a slowly-moving source is given by µ0 dP = |p ¨|2 sin2 Θ.5) This expression could have been derived on the basis of the electric-dipole approximation of Sec. we no longer have a need for the redundant notation [ ]ret . 6.4. dΩ′ µ0 c µ0 c (4πǫ0 )2 κ5 c4 or ˆ × [(R ˆ − v /c) × a] 2 µ0 q 2 R dP ′ = . |a|2 − (a · R (4π )2 c ˆ . According to the results of that section.4) The general expression for the angular profile is complicated. (6. Integrating Eq.6.22). and also set κ ≃ 1. We have arrived at dP ′ dΩ′ ˆ = κR2 S · R = (6. and we finally obtain µ0 dP ≃ q |a| dΩ (4π )2 c 2 sin2 θ.4.2 Slow motion: Larmor formula Our first special case will be the limit v ≪ c. This gives because E is orthogonal to R ˆ × [(R ˆ − v /c) × a] 2 R dP ′ q2 1 1 2 2 = κR | E | = . dΩ (4π )2 c . then |a|2 − (a · R ˆ )2 = |a|2 (1 − If θ denotes the angle between the vectors a and R 2 2 2 cos θ) = |a| sin θ. a.4.4). In this slow-motion approximation ˆ in Eq. We then have we can neglect v /c in front of R dP dΩ ≃ ≃ ≃ µ0 q 2 ˆ ˆ × a) 2 R × (R (4π )2 c µ0 q 2 ˆ )R ˆ2 a − (a · R (4π )2 c µ0 q 2 ˆ )2 . and it depends on the individual directions of v . Because it is understood that all quantities refer to the retarded time t′ . (6. 5.4. which is a definition for dP ′ = therefore be expressed as S (t′ ) · R ′ ′ ′ (dP /dΩ ) dΩ . dΩ′ (4π )2 c κ5 (6. The Poynting vector is given by S= 1 1 ˆ × E ) = 1 |E |2 R ˆ.4. (1 − v cos θ/c)5 (6.4) becomes ˆ × (R ˆ × a) 2 µ0 q 2 R dP = . 6πc (6.6) This result is known as Larmor’s formula. And because the ˆ plays the same role as r sphere S ′ is centered on the particle. Eq.4. (1 − v cos θ/c)5 (6.9) is well approximated by cos θmax ≃ 1 − 1 − 512 γ + · · ·. We now let θ be the angle between R direction of v and a. We have that v × a = 0.4.4.4. the angular profile is strongly modified — it tips forward in the direction of v (θ = 0).4. The angular function is f= sin2 θ .4. For v/c approaching unity. 6. however.10) θmax ≃ 2γ . Integration of dP/dΩ over all angles yields P ≃ µ0 2 q |a| . (6. so 43 −4 −2 that γ ≫ 1.7) For v/c ≪ 1 this reproduces the sin2 θ behaviour encountered in the preceding subsection. and θmax ≃ π/2: most of the radiation is emitted in the directions transverse to v .124 Electrodynamics of point charges where p is the dipole moment of the charge distribution. dΩ (4π )2 c κ5 ˆ ×(R ˆ ×a)|2 = |a|2 sin2 θ. This condition might hold at all times. and we recall that κ = 1−v ·R ˆ /c = We already know that |R 1 − v cos θ/c.4. so that p ¨ = q a. and this gives us the For such small angles we have cos θmax = 1 − 2 approximate relation 1 . (6. For such values of v/c. and Θ the angle between the vectors r ˆ ≡ x/|x| and p ¨. so Eq. (6. but we assume that v and a are pointing in the same direction. The electric-dipole approximation therefore produces the same result as in Eq.4).4. (6. This is the usual nonrelativistic situation. At this moment the particle is ˆ and the common moving on a straight line. γ ≫ 1. and the power radiated at small angles is very large (see Fig.5). the radiation is beamed in the forward direction. Here p = q r (t).8) Differentiating with respect to µ ≡ cos θ and putting the result to zero indicates that the maximum of the distribution is at the value of µ that solves (3v/c)µ2 + 2µ − 5v/c = 0. however.9) cos θmax = 3(v/c) 5 3 (v/c) − 75 For v/c ≪ 1. 6. (6.4. which is determined by 1 + 15(v/c)2 − 1 . this expression is well approximated by cos θmax ≃ 2 8 (v/c) + · · ·.3 Linear motion In this subsection we no longer restrict the size of v/c. 1 2 θmax + · · ·. but it is sufficient to assume that it holds only momentarily. So most of the radiation is emitted at angles close to θmax . Substituting these results gives µ0 dP = q |a| dΩ (4π )2 c 2 sin2 θ . 8γ and θmax is very small: most of the radiation is emitted in the forward direction. the vector R ˆ. so that Θ ≡ θ. For v/c close to unity. 5 –1 4000 2000 0 –2000 –4000 5000 10000 15000 20000 25000 Figure 6. . the radiation propagates mostly in the directions transverse to the charge’s velocity (the horizontal direction in the plot). Notice the different scales in the two diagrams.4 Radiation from an accelerated charge 125 1 0.4: The first plot shows the angular profile of radiation produced by a nonrelativistic particle in linear motion.5 –0.6. The second plot shows the angular profile of radiation produced by a relativistic particle in linear motion.2 0.2 0.4 –0.4 –0. the radiation is beamed in the forward direction with an angular width of order 1/(2γ ). (6. κ = 1 − v cos θ/c. 6πc (6.4 Circular motion We now take a and v to be orthogonal vectors. 6. and v points in the z direction. The final answer is therefore P = µ0 2 q |a| γ 6 .4. ˆ = v cos θ. and According to this. and ˆ = sin θ cos φ x R ˆ + sin θ sin φ y ˆ + cos θ z ˆ.4) gives ˆ·a (R ˆ)2 µ0 q 2 a2 dP 1− . v = v z ˆ. For concreteness we let a be directed along the x axis.11) becomes P = µ0 q |F | 6πc m 2 .126 Electrodynamics of point charges This is the statement that when the charge moves with a highly relativistic speed.4.4. This force is F = dp/dt. the radiation is strongly beamed in the direction of the velocity vector. a · R ˆ = a sin θ cos φ.4. For linear motion in the x direction we have px = mv/(1 − v 2 /c2 )1/2 and (mv )(v/c2 )a ma dpx ma + = . and Eq. (R The squared norm of this vector is ˆ × [(R ˆ − v /c) × a] R 2 ˆ · a)2 (R ˆ − v /c) · (R ˆ − v /c) (R ˆ · a)(R ˆ − v /c) · a + κ2 a2 − 2κ(R ˆ · a)2 (1 − 2R ˆ · v /c + v 2 /c2 − 2κ) + κ2 a2 = (R ˆ · a)2 (1 − v 2 /c2 ) + κ2 a2 = −(R ˆ·a (R ˆ)2 = κ2 a2 1 − . The total power radiated is obtained by integrating Eq. We let the angles (θ. At this moment the particle’s motion is circular. (1 − v cos θ/c)5 6 and the integral evaluates to 4 3 γ .4. (6. = 2 2 1 / 2 dt (1 − v /c ) (1 − v 2 /c2 )3/2 (1 − v 2 /c2 )3/2 This result can be expressed as F = mγ 3 a.12) This is the total power radiated by a charge in instantaneous linear motion. (6. but it is sufficient to assume that it holds only momentarily. 2 κ γ2 = Substituting this result into Eq. (6.11) This generalizes the Larmor formula of Eq. (6. φ) give the direction of ˆ relative to this set of axes. v · R ˆ × (R ˆ − v /c) × a R = = ˆ · a)(R ˆ − v /c) − R ˆ · (R ˆ − v /c)a (R ˆ · a)(R ˆ − v /c) − κa.4. This condition might hold at all times.4.7) with respect to dΩ = sin θ dθdφ.6). We therefore have a = ax the vector R ˆ. It is instructive to express this in terms of the force acting on the particle. We have P = µ0 2 q |a| 2π 2 (4π ) c π 0 sin3 θ dθ . = 2 3 2 dΩ (4π ) c κ κ γ2 . where p = mγ v is the particle’s relativistic momentum. (6.5: Angular profile of radiation produced by a relativistic charge in circular motion.4.5 Synchrotron radiation We continue our discussion of circular motion and provide additional details.4. We now assume that the charge moves at all times on a circle of radius ρ at a constant angular velocity Ω.15) This is the total power radiated by a charge in instantaneous circular motion.14) P = 6πc This is a different generalization of the Larmor formula of Eq. (6.4. The result is µ0 2 q |a| γ 4 .12) and notice the extra factor of γ 2 . 6. but not its magnitude.5.4. In this case the force changes only the direction of the vector v . (6. but it still features a strong beaming in the forward direction (direction of v ) when v is close to the speed of light.6. (6. The total power radiated is obtained by integrating dP/dΩ over the angles θ and φ. and Eq.4.16) . (6. (6. We may also express this in terms of the force F acting on the particle.6). You should compare this with Eq.14) becomes P = µ0 q |F | 6πc m 2 or γ2.13) = dΩ (4π )2 c (1 − v cos θ/c)3 γ 2 (1 − v cos θ/c)2 This angular profile is rather complex. Its position vector is r (t′ ) = ρ cos(Ωt′ )x ˆ + sin(Ωt′ )y ˆ. 6.17) (6.4. (6. So F = dp/dt = d(mγ v )/dt = mγ a. This is illustrated in Fig. dP q 2 |a|2 sin2 θ cos2 φ µ0 1 − .4 Radiation from an accelerated charge 127 400 0 –400 2000 4000 6000 8000 Figure 6. its velocity vector is v (t′ ) = v − sin(Ωt′ )x ˆ + cos(Ωt′ )y ˆ.4. and quantitatively different from the one encountered in the preceding subsection.4.4. The radiation is beamed in the (horizontal) direction of the velocity vector. (6. 4πǫ0 c2 r [1 − (v/c) cos Ωt′ ]3 (6.7. (6. dt′ c (6.6 for selected values of γ = [1 − (v/c)2 ]−1/2 . It is useful to recall that t′ is time measured by a clock that moves with the charged particle.4. with E=− q vΩ v/c − cos Ωt′ . To perform the remaining calculations we allow ourselves to neglect all terms ˆ=y linear in ρ/r. and its acceleration vector is Electrodynamics of point charges a(t′ ) = −v Ω cos(Ωt′ )x ˆ + sin(Ωt′ )y ˆ. and t′ .18) We wish to calculate the electric field that would be measured by a detector situated far away. A much cruder approximation will suffice when calculating the field’s amplitude.19) This gives the relationship between t − r/c. it is easy to see that the length of the vector R ≡ x − r (t′ ) is given by R(t′ ) = r − ρ sin Ωt′ .20) We see that κ will deviate strongly from 1 when v is close to the speed of light and when | cos Ωt′ | is close to unity. the wave deviates strongly from a simple sinusoid. and a short one associated with the pulse duration. (6. To understand these features it is helpful to consider the extreme relativistic regime.4.21) The electric field can be expressed in terms of detector time t by inverting Eq. as γ increases the dominant harmonic is shifted to higher frequencies. the retarded time measured at the detector. ˆ × (R ˆ − v /c) × a = −v Ω(v/c − cos Ωt′ )x and R ˆ. We take R ˆ. The phase modulations which originate from the discrepancy between t and t′ produce a wave that is characterized by two distinct timescales: a long one associated with the rotational frequency Ω. and we assume that r ≫ ρ. 6. Its behaviour is displayed in Fig. We shall need some level of precision in the calculation of t(t′ ). These plots show that in the relativistic regime.128 where v = Ωρ.4. in which γ ≫ 1. (6.4.1). The relation between t′ and true time t is given by Eq. in the wave zone.4.3. because it is this function which determines the phasing of the electric field. at position x = ry ˆ.20). The frequency spectrum of the radiation is presented in Fig.21) . Substituting all this into Eq. Working to first order in ρ/r. the radiation is carried by a large number of harmonics of the fundamental frequency Ω. 6.21) then gives t = t′ + r/c − (ρ/c) sin Ωt′ . while t is time measured by a clock attached to the detector. The plots reveal that for γ moderately larger than unity. We place this detector on the y axis.19). c (6. The wave-zone electric field is given by Eq. (6. In this regime the basic mechanism behind the narrowing of the pulses is very easy to identify: When v/c → 1 the denominator of Eq. which we first evaluate as a function of retarded time t′ . ˆ − v /c = (v/c) sin(Ωt′ )x R ˆ − 1 − (v/c) cos(Ωt′ ) y ˆ. From this it follows that κ= dt v = 1 − cos Ωt′ .4. We use the condition r ≫ ρ to calculate E approximately.4. and this leads to an interesting modulation of the electromagnetic wave.1) gives E = E x ˆ. these measurements of time differ substantially when v is comparable to c. Equation (6. (6.4. As we shall see. the time measured at the particle.4. or Ω(t − r/c) = Ωt′ − v sin Ωt′ . 5 1 1 0.5 -0.0.5 0 0.5 2 2 1.5 synchrotron radiation: gamma = 1. The relativistic modulation of the wave pulse for large values of v/c is clearly apparent.5 0 0.2. and a short one associated with the pulse duration.87 2 2.5 synchrotron radiation: gamma = 1.55 2 2.5 3.5 -0.5 0 0 -0.5 field kappa 3 2. the plots reveal that the radiation is characterized by two distinct time scales: a long one associated with the rotational frequency Ω.5 1 1.5 -0.0.8.5 0 0. whose deviation from 1 measures the magnitude of the relativistic time-dilation effect.5 1 1 0.5 1.6: Electric field produced by a charged particle in circular motion.5 Figure 6.5 1 1.70 2 2.5 1. The plots also show κ = dt/dt′ .5 synchrotron radiation: gamma = 1. v/c = 0.5 0. . v/c = 0.5 0 0 -0.4.5 -0.5 1 1.5 field kappa 3 3. 2 2.5 3 field kappa 2.5 3.5 field kappa 2 0.6.5 -0.83 2 2. v/c = 0.5 -0.5 1 1.5 field kappa 3 2. In particular. for selected values of the relativistic γ factor.5 0 0.5 -0.5 -0.5 synchrotron radiation: gamma = 1. as a function of detector time t.5 1.5 2 2 1.5 0 0. v/c = 0.5 0.78 2 2.5 1 1.6.5 0 0. v/c = 0.5 0 -1 -0.5 synchrotron radiation: gamma = 1.4 Radiation from an accelerated charge 129 1 field kappa 2.5 1 1.5 0 1 0.5 -0.5 2. v/c = 0.5 synchrotron radiation: gamma = 2. 8 1.7: Fourier transform of the electric field produced by a charged particle in circular motion.4 0.87.0.4 0. wc = 1.2 0 0 5 10 15 20 25 30 synchrotron radiation: gamma = 1.5 0 0 2 4 6 8 10 12 synchrotron radiation: gamma = 1.2 0 0 5 10 15 20 25 30 35 40 45 50 synchrotron radiation: gamma = 1.8 1 0.4.8 1 0. the dominant harmonic is shifted to higher frequencies.2 0. wc = 8.6 0.4 1. v/c = 0.8 0.5 3 6 2.5 4 3.2. v/c = 0.7 1.8 0 0 20 40 60 80 100 120 synchrotron radiation: gamma = 2.7 14 2. v/c = 0.6 0.83.55.78. In addition.2 0.130 Electrodynamics of point charges 10 9 8 7 4. wc = 2.5 1 0.3 0. for selected values of the relativistic γ factor.6 2 1.7 0.8.5 5 2 4 1. v/c = 0.1 0. The plots reveal that the number of relevant harmonics of the fundamental frequency Ω increases with γ .0.6 0.5 1.70.1 0 0 10 20 30 40 50 60 70 80 synchrotron radiation: gamma = 1.5 0. . wc = 1. v/c = 0. wc = 4.0 Figure 6. 4 5 1 0. wc = 5.4 0.6.5 0.2 1. v/c = 0.9 0.5 3 2 1 0 0 1 2 3 synchrotron radiation: gamma = 1. (6.4. and it travels along the y axis toward the detector. the parallel force plays no role in the energetics of the radiation. (6. Most of the radiation is therefore emitted during a very short portion of the particle’s orbit. therefore. (6.7.4 Radiation from an accelerated charge 131 approaches zero whenever cos Ωt′ = 1. and a component F⊥ that is perpendicular to it (F⊥ · v = 0).24) It is reduced by a factor γ −3 ≪ 1 relative to the orbital timescale associated with the angular velocity Ω. just as the particle crosses the x axis in the positive sense. the characteristic frequency ωc of the radiation is shifted relative to the fundamental frequency Ω. the radiation is “on” for a duration of order (γ Ω)−1 .4.20) we obtain κ≃ 1 1 + (γ Ωt′ )2 .22) 1 1 − (γ Ωt′ )2 . (6. To see how long the pulse lasts in terms of detector time t. 2γ 2 Again. From Eq. P = µ0 q |F⊥ | 6πc m 2 γ2. During the remaining portion of the orbit little radiation is emitted. 6. It is given by ωc ∼ γ 3 Ω. 2γ 2 which shows that κ is small (of order γ −2 ≪ 1) for a duration of order (γ Ω)−1 in terms of particle time t′ .6 Extremely relativistic motion To conclude this section we note that at any point on the arbitrary trajectory of a point charge. this shows that in terms of particle time t′ .23) This relation reveals that the pulse duration. we approximate 1 −2 γ Ωt′ .19) as Ω(t − r/c) ≃ 2 γ Ωt′ ≃ 2γ 3 Ω(t − r/c). We also obtain v/c − cos Ωt′ ≃ − and Eq. In this limit. and the detector must wait for the completion of a full orbit before receiving the next pulse.4. the applied force can be decomposed into a component that is parallel to the velocity vector v .6) is given by ∆tpulse ∼ 1 γ3Ω . (6. As a consequence.4. γ ≫ 1 and the total power emitted by the particle is well approximated by Eq. 4γ 4πǫ0 c2 r [1 + (γ Ωt′ )2 ]3 (6. .4. 6.4. which gives Eq.4. as measured by a clock attached to the detector (and as displayed in Fig. (6. When the motion is extremely relativistic. and the displacement of the spectrum toward higher frequencies when γ increases is seen very clearly in Fig.6.15). 6. This radiation is strongly beamed in the forward direction. −2 + O(γ −4 ) and we To express these ideas mathematically we write v/c = 1 − 1 2γ 1 ′ ′ ′ ′ 2 approximate sin Ωt ≃ Ωt and cos Ωt ≃ 1 − 2 (Ωt ) . which happens whenever Ωt′ is a multiple of 2π . The contribution from the parallel force is smaller by a factor of 1/γ 2 ≪ 1. and it can be neglected.21) reduces to E≃ q v Ω 4 1 − (γ Ωt′ )2 .4. x) = µ0 4π j (t − |x − x′ |/c.3 of Jackson’s text. elliptical). x′ ) 3 ′ d x.132 Electrodynamics of point charges 6. A+ (t. x) = µ0 4π j (t + |x − x′ |/c. How can it then produce a (presumably finite) radiation-reaction force that drives the inspiral? The answer is that the divergent part of the electron’s self-field exerts no force — it just travels along with the electron. Let us examine the vector potential first. We will see that while part of the field is not affected by this switch. The orbit gradually shrinks. and the radiation-reaction force would be acting in the opposite direction (pushing the orbit out instead of in).6 we saw that a classical electron orbiting a classical proton emits electromagnetic radiation. and the classical atom is unstable to the emission of electromagnetic radiation.1. 5. and the electron would move out instead of in. An unphysical solution to Maxwell’s equations is the advanced solution. because the electron’s own field diverges at the electron’s position. |x − x′ | . a time reversal) of the correct picture. This describes waves that are properly outgoing in the wave zone. We can figure out which part of the electron’s field is producing the radiationreaction force by examining what happens when we switch from outgoing-wave boundary conditions to incoming-wave boundary conditions. It is the electron’s self-field which provides the force that drives the electron’s inspiraling motion. dt 6πc This loss of energy translates into a decrease in the electron’s orbital radius. In this case we would have incoming waves bringing energy to the system instead of outgoing waves taking energy away. and 16. in fact. To figure out what this part is. The electron’s inspiraling motion cannot take place in this picture. The electron must also be subjected to its own electric field. µ0 dE 2 = −P = − q |a| . A− (t. 16. the classical hydrogen atom loses energy at a rate determined by Larmor’s formula. And we will see that while the unaffected part is infinite at the electron’s position. have a mirror image (more precisely. We are forced to conclude that the proton’s electric field cannot be the only field acting on the electron. let us ask what would happen if instead of imposing outgoing-wave boundary conditions on our spherical-wave solutions to Maxwell’s equations. This picture. We would. its motion must stay circular (or more generally.] In Sec. in contradiction with our previous description. These waves remove energy from the system. however. This conclusion is difficult to understand. The physical solution to Maxwell’s equations is the retarded solution. and the radiation reaction drives the electron inward. is incomplete: So long as the electron feels only the proton’s Coulomb field. It is natural to assume that the electron is subjected only to the electric field produced by the proton. 16.5 Radiation reaction [The material presented in this section is also covered in Secs. |x − x′ | where j is the electron’s current density. It is this part of the field that is responsible for the radiation reaction. The part of the self-field that is responsible for the radiation reaction is finite. The atom’s energy would then increase.2. we were to impose incoming-wave boundary conditions. the remaining part changes sign. the remaining part is finite. x′ ) 3 ′ d x. As a result. We have µ0 j (t. therefore. For the scalar potential we have the two solutions Φ± (t. and we conclude that the radiation-reaction part of the scalar potential is given by (recall that c−2 = ǫ0 µ0 ) Φrr (t. ± 3 3 6c ∂t The first and third terms are insensitive to boundary conditions.2) . change sign under a switch of boundary conditions. d x ± ′ 4π |x − x | 4πc dt and we will see that the fields this potential generates produce the radiation-reaction force that acts on the electron. on the other hand. x′ ) d3 x′ + · · · . the second term is of order (rc /tc )/c = vc /c. x) = − µ0 d 4πc dt j (t. however. (6.6. x′ )|x − x′ |2 d3 x′ + · · · . x′ )|x − x′ | d3 x′ + 2 2 2c ∂t 1 ∂3 ρ(t. |x − x′ | and we Taylor-expand the charge density: ρ(t ± |x − x′ |/c) = ρ(t) ± |x − x′ | ∂ρ |x − x′ |2 ∂ 2 ρ (t) + (t) c ∂t 2c2 ∂t2 |x − x′ |3 ∂ 3 ρ (t) + · · · . We want to evaluate A± near the electron. and determine what changes when we choose one sign over the other.5.5 Radiation reaction 133 and this describes waves that are incoming in the wave zone. x′ ) 3 ′ µ0 d A± = j (t.5. changes sign under a change of boundary conditions. c ∂t We see that the first term is the same irrespective of the choice of boundary conditions (outgoing waves versus incoming waves). x′ ) 3 ′ 1 d ρ(t. x) = − µ0 1 ∂ 3 4πc 6 ∂t3 ρ(t. it cannot. x′ )|x − x′ |2 d3 x′ + · · · . however. in the near zone. x′ ) d3 x′ d x ± 4πǫ0 |x − x′ | c dt 1 ∂2 ρ(t. These waves bring energy to the system. vanishes by virtue of charge conservation. We can therefore take |x − x′ | to be small. x′ ) 3 ′ d x. The second term. and they are associated with radiation reaction. contribute to a radiation-reaction force.1) Relative to the leading term. (6. x) = 1 4πǫ0 ρ(t ± |x − x′ |/c. ± 6c3 ∂t3 This gives Φ± = 1 ρ(t. and we shall take this to be small — we are working under a slow-motion approximation. they have nothing to do with radiation reaction. and the radiation reaction drives the electron outward. this must be the vector potential’s contribution to the reaction-reaction force. x′ ) d3 x′ + · · · . The second and fourth terms. and the force it contributes will also change sign. Because the retarded solution provides the correct sign. we must examine both solutions simultaneously. The second term. we shall write Arr (t. To isolate the part of the (retarded) vector potential that is responsible for the radiation reaction. and Taylor-expand the current density: j (t ± |x − x′ |/c) = j (t) ± |x − x′ | ∂ j (t) + · · · . and differentiating a third time yields −2(x − r ) · a ˙ + 2v · a + 4v · a = −2(x − r ) · a ˙ + 6v · a. These are the radiation-reaction potentials acting on a point charge q moving on a trajectory r (t).5. These expressions will apply. 4πc dt We have |x − r |2 = (x − r ) · (x − r ). x′ ) = qδ (x′ − r (t)) and j (t. This gives and we average W ˙ W ≡ = = 1 T T 0 ˙ dt W 1 T d 1 (a · v ) dt − T 0 dt T T 1 (a · v ) − |a|2 .5. (6. Differentiating once more gives −2(x − r ) · a +2v · v . The radiation-reaction part of the electric field is given by Arr (t.5.5. x) = − and µ0 q ∂ 3 x − r (t) 4πc 6 ∂t3 2 µ0 d q v (t). (6.5) µ0 qa ˙. x) = 4πc 3 and µ0 q a. dt ˙ over a time interval T . in particular.3) Φrr (t. and it originates from the charge’s own electric field! It is easy to verify that the radiation-reaction force does negative work on the particle and takes energy away from it. (6. T 0 T 0 µ0 2 q 6πc µ0 2 q 6πc |a|2 dt . We finally arrive at 1 µ0 q (x − r ) · a ˙ −v·a (6. and our considerations have been general: Eqs. x) = − Err = − or Because Arr ∂ Arr µ0 1 − ∇Φrr = q a ˙− a ˙ . and differentiating once with respect to t gives −2(x − r ) · v . to a point charge q if we set ρ(t.134 Electrodynamics of point charges Up to now the charge and current densities have been left unspecified. or µ0 2 ˙. x′ ) = q v (t)δ (x′ − r (t)).6) The radiation-reaction force is therefore Frr = q Err . x) = − 4πc Here we have defined a = dv /dt and a ˙ = da/dt.1) and (6. we also have Brr = 0.5.5. where v = dr /dt. 6πc We write d a ˙ · v = (a · v ) − |a|2 .7) q a Frr = 6πc This force acts on the point charge.2) give the radiation-reaction potentials for an arbitrary distribution of change and current.5. 6πc does not actually depend on x. As the charge moves with velocity v the force does work at a rate ˙ = Frr · v = µ0 q 2 a W ˙ · v. (6. Substituting this into our expressions for the potentials returns Φrr (t. ∂t 4πc 3 Err = (6.4) Arr (t. 3. The particle comes in from infinity.6 Problems 1. the total power emitted by the particle.3. the radiationreaction force of Eq. or at least the finite part that is sensitive to the choice of boundary conditions. the charge has a Lorentz γ -factor that can be large compared with unity. (It is noteworthy that this statement holds only on the average. You are asked to derive an equation for dω/dt and to express your result in terms of ω .6).7) points in the direction of −v and drives the orbit to a smaller radius. Its position is given as a function of t′ by z (t′ ) = z0 cos(ωt′ ). b) Produce a parameteric plot of the angular function for β = 0. but we do not assume that it is much smaller than c. The electron’s own field. The charge emits electromagnetic radiation. 6. Calculate P .) We have a complete resolution of the paradox stated at the beginning of this section. A charge q moves on a trajectory described by z (t′ ) = b2 + (ct′ )2 . the work done by this radiation-reaction force matches the amount of energy taken away by the radiation. and it eventually returns to infinity.4.) 2.12. where m is the charge’s rest mass.5. . (Indeed. and other relevant constants. or else that it begins and ends with a vanishing acceleration.95. 6.) Furthermore. where z0 and ω are constants. a) Show that the time-averaged angular profile of the radiation is described by 2 2 dP µ0 2 2 1 + (β/2) cos θ sin2 θ.5. It is assumed that β ≡ z0 ω/c is fairly close to unity. we may discard the boundary terms and obtain ˙ = − µ0 q 2 |a|2 = − P . as given by the Larmor formula of Eq.3. (The time scale over which ω changes is much longer than the rotational period. where T ≡ 2π/ω is the period. it is easy to check that for a circular orbit. The time average is defined by T (· · ·) ≡ T −1 0 (· · ·) dt′ . (You may need to know that the charge’s relativistic energy is E = γmc2 . The work done by the radiation-reaction force therefore matches the energy taken away by electromagnetic radiation.) A charge q moves in simple harmonic motion along the z axis. (This problem is adapted from Jackson’s problem 14. (6. A point charge q moves on a circle of fixed radius a with an angular frequency ω (t) that changes slowly with time. (6. so that the motion is relativistic.8) where P is the power radiated by the point charge. as it well should if energy is to be conserved in the process. It therefore loses energy and this causes the angular frequency ω to change with time. Show in particular that this does not depend on time t′ . W 6πc (6.) We take aω to be smaller than c.6 Problems 135 Supposing now that the motion is periodic with period T (as it would be in the case of atomic motion).6. Comment on its appearance in view of the beaming effect discussed in Sec. it turns around at z = b when t′ = 0. = ( qz ω ) 0 dΩ 32π 2 c (1 − β 2 cos2 θ)7/2 where θ is the angle from the z axis. acts on the electron and prevents it from keeping its circular motion. a. and we have energy conservation. where b is a constant. and initial speed v is incident on a fixed charge Q with an impact parameter b that is large enough to ensure that the particle’s deflection in the course of the collision is very small. But this approximation must be introduced only after you have calculated the particle’s acceleration.136 Electrodynamics of point charges 4. mass m. (This problem is adapted from Jackson’s problem 14. . Because the deflection angle is so small. Calculate the total energy radiated by the particle.) A nonrelativistic particle of charge q . it is sufficient to assume that the particle’s trajectory is a straight line and that its velocity vector is constant.7.