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Contents

0 Short Review of some Mathematics, Nomenclature and Notation 3

0.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

0.1.1 Einstein’s notation for summed indices . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

0.1.2 Operations on 3-dim vectors in index and index-free notations . . . . . . . . . . . . . . . 4

0.1.3 Meet your friend, the Levi-Civita symbol . . . . . . . . . . . . . . . . . . . . . . . . . . 5

0.2 Review of Integral Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

0.3 Vectors in Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.4 Dirac delta-“Function” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

0.5 Helmholtz Theorem (Griffiths, Appendix B) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1 General Properties of Maxwell’s Equations for EM Fields and Potentials 10

1.1 First and Second-Order Maxwell Equations for the Fields and Potentials . . . . . . . . . . . . . . 10

1.1.1 Electromagnetic Fields (G7.3.3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.2 Potentials (G10.1.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Brief Review of Electromagnetic Waves (G9.1.1, 9.1.2, 9.2) . . . . . . . . . . . . . . . . . . . . . 11

1.3 Transverse and longitudinal projections of Maxwell’s Equations for the Potentials . . . . . . . . . 13

1.3.1 Choices of the divergence of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4 Initial-Value Problem with Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4.1 First-order Cauchy problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4.2 Propagation of constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.3 Gauge transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.4 Are electromagnetic fields more fundamental than potentials? . . . . . . . . . . . . . . . 16

1.5 Duality Properties of Maxwell’s Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Causal Solutions of Maxwell’s Equations: Potentials and Fields 19

2.1 Causal Solutions of a Scalar Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2 Retarded Solutions for the Maxwell Potentials and Fields (G10.1) . . . . . . . . . . . . . . . . . 20

2.3 Potentials and Fields of a Point-charge in Arbitrary Motion (G10.3) . . . . . . . . . . . . . . . . 20

2.3.1 Liénard-Wiechert potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Radiation (chapter G11) 26

3.1 Radiation from Moving Point-Charges (G11.2.1) . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.1.1 Poynting vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.1.2 Angular power distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1.3 Total power radiated by an accelerating point-charge . . . . . . . . . . . . . . . . . . . . 28

3.2 Radiating Systems of Charges in the Leading Approximation . . . . . . . . . . . . . . . . . . . . 29

3.2.1 Radiation fields of an arbitrary but localised source (G11.1.4) . . . . . . . . . . . . . . . 29

3.2.2 Power emitted by a localised charge distribution into electric dipole radiation . . . . . . . 30

3.3 Radiation from Oscillating Sources — a More General Treatment . . . . . . . . . . . . . . . . . 31

3.3.1 From the vector Potential to the Angular Power Distribution . . . . . . . . . . . . . . . . 31

3.3.2 Electric dipole radiation revisited for harmonic sources . . . . . . . . . . . . . . . . . . . 32

3.3.3 Magnetic dipole radiation fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Lecture Notes on Relativistic Electrodynamics 2019

4 Introduction to Relativity 36

4.1 A Mathematical Digression on Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.2 Galilean transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.3 Critique of Galilean relativity; postulates of Einsteinian relativity (section G12.1.2) . . . . . . . . 40

4.4 Spacetime interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.5 Lorentz transformations (G12.1.3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.5.1 Addition of velocities (section G12.1.3) . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.5.2 Four-vectors and metric tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.5.3 Relation between covariant and contravariant components . . . . . . . . . . . . . . . . . 44

4.5.4 Constructing Lorentz tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.6 Relativistic Kinematics (section G12.2.1/2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.7 Energy-momentum four-vector (section G12.2.3) . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.8 Relativistic Dynamics (section G12.2.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.9 Causal structure (section G12.1.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.10 Differential Operators in the Four-vector Formalism . . . . . . . . . . . . . . . . . . . . . . . . . 51

5 Relativistic Electrodynamics 53

5.1 Current-density 4-vector (section G12.3.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.2 4-vector Potential (section G12.3.5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.3 Faraday-Maxwell Field Tensors and Maxwell’s Equations (section G12.3.4/5 . . . . . . . . . . . 55

5.4 Motion of a Charge in an Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.5 Lorentz Transformations of Electromagnetic Fields (section G12.3.1) . . . . . . . . . . . . . . . 57

5.6 Gauge Invariance of Maxwell’s Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.7 The Liénard-Wiechert Four-vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.8 Relativistic Expression for Radiated Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.9 Energy-Momentum in Electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.9.1 Energy-momentum tensor and conservation laws . . . . . . . . . . . . . . . . . . . . . . 61

5.9.2 Explicit components of the electromagnetic energy-momentum tensor . . . . . . . . . . . 64

2


0 Short Review of some Mathematics, Nomenclature and Notation

Vectors are often naı̈vely thought of as quantities defined at each point in three-dimensional space and endowed

with a magnitude and a direction, whereas a quantity with magnitude but without direction is called a scalar.

Whilst necessary, the mere possession of these two attributes is by no means sufficient for a quantity to qualify as

a vector. And a quantity described by a single number is not necessarily a scalar!

For the moment, let us note that neither magnitude nor direction depend on the choice of coordinate system.

This is obvious in the well-known geometric representation of vectors using arrows. When we use algebraic

expressions, wesometimes need a notation that makes no explicit reference to coordinate systems. After all,

vectors enter in many fundamental equations in physics, and these should not change just because we transform

from, say, Cartesian to spherical coordinates, or because the coordinate system has been rotated or translated. We

need a more sophisticated characterisation of vectors.

0.1 Vector Spaces

Definition 0.1. A vector space V is a (possibly infinite) set of objects that is closed under additionand multiplication by a scalar:, ie. for which any two elements (vectors) u and v satisfy:

(a+ b)(u+ v) = (au + av + bu + bv) ∈ V

∀ a, b elements of some field; in what follows, we restrict these to be real. Also, (ab)u = a (bu).This operation is both commutative and associative.

Definition 0.2. If W is a subspace of V , ie. W ⊆ V , and if any w ∈ W can be written as a linearcombination:

w =∑

α

wα eα (0.1)

of a set {eα ∈ W}, then that set is said to span, or to be a set of generators of, W .If, furthermore, this set is linearly independent, in the sense that demanding that w = 0 forces allcoefficients wα in eq. (0.1) to vanish, then it is a basis of W . The number n of vectors in the largestlinearly independent set defines the dimension of W , and we often write Wn. Conversely, the numberof elements of every basis of Wn is the dimension n of Wn, and the sum in eq. (0.1) then runs from1 to n.

The set {eα} of basis vectors is said to be orthonormal if eα · eβ = 1 if α = β and 0 otherwise. Theoperation represented by the dot will be defined below.

The (real) coefficients wα in eq. (0.2) are called the components of the vector w in this basis. Thisone-to-one correspondence between Wn and Rn can be represented by a n× 1 matrix:

w 7−→

w1

w2

...

wn

In eq. (0.1), the left-hand side is explicitly basis-independent; we shall call this notation index-free, or geo-

metric. The right-hand side, in so-called index notation, makes explicit reference to a basis even though, taken

as a whole, it must still be basis-independent because of the equality.. Both notations have advantages and disad-

vantages. Fluency in both is highly recommended.

Warning! w and its components are different beasts and should never be confused. Also, always remember

that the index on eα identifies the basis vector, not a component of the vector.3


Example 0.1. (1) Rn, the set of all ordered n-tuples of real numbers is one of the most importantvector spaces.

One popular basis of R3, and the one we shall by and large use in this course, is the Cartesian

(rectangular, or standard) basis. Its three vectors are fixed and attached to a point arbitrarily chosen

as the origin (0, 0, 0).

e1 ≡ x̂ = (1, 0, 0)T

e2 ≡ ŷ = (0, 1, 0)T (0.2)e3 ≡ ẑ = (0, 0, 1)T

where the label T denotes the transpose.

(2) The set of all real-valued functions that are differentiable n times.

0.1.1 Einstein’s notation for summed indices

Now is an appropriate time to introduce the Einstein summation convention: any index which occurs twice in a

product term must be summed over. Thus, uµvµ = uνvν ≡ ∑µ uµvµ = u1v1 + u2v2 + u3v3. Such an index is

often called a dummy index and any letter can be used for it so long as that letter is not used for any other index

in the same product term!! Therefore, a dummy index appears twice and only twice in any product term.

Remaining indices, which appear only once, are called free indices. Example: in aαbνcαuβvβ , α and β are

dummy indices with summation implied, and ν is a free index.

I am sure you have noticed that I have written some component indices as subscripts instead of superscripts.

What is the relationship between uα and uα? Well, in a Euclidean space with Cartesian coordinates, ie. one inwhich the distance between infinitesimally close points can be written dl2 = dx2 + dy2 + dz2 + . . ., we haveuα = uα! In the four-dimensional spacetime introduced in a proper relativistic treatment of electromagnetism,this is no longer true, as we shall see later, and uα and uα will represent different types of components of the samevector u.

0.1.2 Operations on 3-dim vectors in index and index-free notations

From the orthonormal property of the basis vectors, it can be shown that in Cartesian coordinates (and only in

these coordinates!), the following operations on 3-dim vectors are defined:

Index-free Index Properties

Addition: w = u + v wi = ui + vi commutative, distributive

Scalar product: u · v = uv cos θ = uivi commutative, distributiveVector product: w = u× v = (uv sin θ)r̂ wi = ǫijkujvk distributivetensor product u⊗ v ui vj commutative, distributive

where wi = wi, and ǫijk is defined in the following subsection. Also, θ is the angle between u and v, and

r̂ is a unit vector perpendicular to the plane defined by u and v, and whose direction is given conventionally by

the right-hand rule. The magnitude (or norm) of a vector is simply u = (uiui)1/2; like θ, it is independent of the

coordinate system in which the vector components are expressed.

Note that the vector product is neither commutative nor associative. In fact, it is antisymmetric. Also, there is

no simple expression for the tensor product in terms of magnitude and angle.

4


0.1.3 Meet your friend, the Levi-Civita symbol

The above expression for the vector product is much more compact than the one which uses a determinant. In three

dimensions the Levi-Civita symbol, ǫijk, is defined as a 27-component object which obeys the following rules:

ǫijk =− ǫjik = −ǫikj= ǫkij = ǫjki

The first line means that ǫ is zero whenever any two of its indices are the same. Now, set ǫ123 = 1 byconvention. Then the other non-vanishing components can only be 1 or −1. We say that the Levi-Civita symbolis totally antisymmetric in its indices. Because of this all-important property, it has only six non-zero components,

and only one independent component

Using the above rules, it is easy to work out the vector product. For instance, suppose you want wy = w2.

Set i = 2. Then the only values for j and k that give non-vanishing ǫ components are 1 and 3. Set j = 1;then k must be 3, again to have ǫ non zero. This gives the first term in the sum. Now, set j = 3; then k = 1,giving the second and only other term in the sum. Thus, w2 = w2 = ǫ231u

3v1 + ǫ213u1v3 = u3v1 − u1v3.

In the same way, w1 = u2v3 − u3v2, and w3 = u1v2 − u2v1, and we have regained the familiar expressionfor the components of the vector product. The third kind of product between two vectors, the tensor product, is

most easily written in component form. Indeed, u ⊗ v can be represented by a 3 × 3 matrix whose elements aregiven by uivj . An antisymmetric version of this product, called the exterior product u ∧ v, is easily constructed;it has components uivj − ujvi in any basis. In a Cartesian basis, these look like the components of the vectorproduct, except that they belong to an object with two indices (rank 2), not just one! The 3-dim exterior product

and the vector product are said to be the Hodge dual of each other, and are connected via the Levi-Civita symbol:

(u× v)i = 12ǫijkujvk − ukvj).What is also nice about this coordinate notation for the scalar and vector products is that it allows relations

involving them to be proved without sweat. Thus:

u · (v ×w) = ǫijkuivjwk

= wkǫkijuivj (cyclic permutation of indices on ǫ)

= w · (u× v).

Et voilà! For comparison, you should try the method that works out all components explicitly.

It will happen that two ǫ’s are multiplied, with one index summed over. No panic! Use the following (unproven)rule:

ǫijkǫlnk = δi

lδjn − δin δj l. (0.3)

The index (here k) that gets summed over must appear in the same position on both ǫ’s (which position does notmatter, because to go from one to the other, the permutation rules are applied twice, once to each ǫ). Note how eachfree index on the left-hand side also appears in each of the two terms on the right-hand side. This rule applies

to indices taken as algebraic symbols, not to particular values that they can take, as is obvious from the above

computation of the components of the vector product!

0.2 Review of Integral Calculus

Three important theorems govern integrals of gradients, divergences, and curls:

1. Gradient theorem:

∫

line

b

a

(∇f) · dl = f(b)− f(a);

5


2. Curl theorem (Stokes):

∫

surface

(∇× u) · da =∮

line

u · dl;

3. Divergence theorem (Green, Gauss):

∫

volume

∇ · ud3x =∮

surface

u · da.

By convention the surface element da is a vector normal to the surface and points outward when the surfaceis closed; in the curl theorem, it points in the direction of the thumb when the other fingers of your right hand curl

around the direction of circulation in the line integral.

Some of the integrals in the curl and divergence theorems have names:∮

line

u · dl is called the circulation of u

around a closed path, whereas∫

surface

u · da is called the flux of u through a surface.

Notice that in each of the theorems, the left-hand side is, loosely speaking, the integral of the derivative of an

object over a (one-, two-, or three-dimensional) region, whereas the right-hand side is the integral of the object

over the boundary of the same region. The three theorems are just particular forms of a fundamental theorem

in differential geometry. As a consequence, when a region (line interval, 2-dimensional surface) is embedded in

a higher-dimensional space, as is the case in the gradient and curl theorems, the integrals on the left are equal to

integrals over any region which has the same boundary! Furthermore, when these regions are closed, the right-hand

side vanishes.

Since∮

∇f · dl = 0, it follows that any vector field u which is the gradient of some scalar field must havezero circulation around any closed path:

∮

u · dl = 0. We say that u is conservative. The converse also holds: if∮

u · dl = 0 for any closed path in a region, Stoke’s theorem demands that ∇× u = 0 at all points in the region,and u must be the gradient of some scalar function (I omit the proof since it is rather fussy). The most useful

statement that emerges from the discussion in this paragraph is that if ∇× u = 0 over a simply-connected region(no doughnuts or surfaces with holes!), u is conservative and there exists a scalar field f such that u = ∇f .

0.3 Vectors in Curvilinear Coordinates

Other than Cartesian bases, the two most useful types of coordinate system are spherical and cylindrical.

In a spherical system, the components of a position vector are one distance and two angles, (r, θ, φ), illustratedin fig. J3.1. The transformations from spherical to Cartesian position coordinates are:

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ (0.4)

Beware: mathematicians often use φ as the polar angle and θ for the azimuthal angle in spherical coordinates!This is certainly the case in the symbolic manipulation software Maple when it knows that it is working in those

coordinates.

As for the inverse transformations, the most useful ones relate the unit vectors:

r̂ = x̂ sin θ cos φ + ŷ sin θ sin φ + ẑ cos θ

θ̂ = x̂ cos θ cos φ + ŷ cos θ sin φ − ẑ sin θ (0.5)φ̂ = −x̂ sin φ + ŷ cos φ

How are the spherical and Cartesian components of a vector u related? We can always write:

u = ux x̂+ uy ŷ + uz ẑ

= ur r̂+ uθ θ̂ + uφ φ̂

6


Insert the above expressions for the spherical unit vectors in ur = r̂ · u = ux r̂ · x + uy r̂ · y + uz r̂ · z,uθ = θ̂ ·u, and uφ = φ̂ ·u, where u is given in terms of ux, uy and uz, and work out the scalar products to obtainthe transformation law:

uruθuφ

=

sin θ cos φ sin θ sin φ cos θcos θ cos φ cos θ sin φ − sin θ− sin φ cos φ 0

uxuyuz

(0.6)

The inverse transformation can be computed either by inverting the matrix or by working out ux = x·û, uy = ŷ·u,and uz = ẑ · u, with u given this time in terms of ur, uθ and uφ. If u were dependent on the coordinates, ie.if it were a vector field, the coordinates themselves would also have to be transformed. EXERCISE: transform

u(x, y, z) = (−2,−1, 3), where (x, y, z) = (2,−1, 0), to a spherical basis.It is not hard to see that the infinitesimal displacement vector is given in spherical coordinates by:

dl = dr r̂ + (r dθ)θ̂ + (r sin θ dφ)φ̂ (0.7)

and that the volume element is just the product of the infinitesimal displacements in the three orthogonal directions:

d3x = r2 sin θ dr dθ dφ. There is no general expression for the surface element, da, as this depends on itsorientation in space.

Once we know the form of dl in spherical coordinates (or any other system for that matter) it is easy to findthe components of the gradient of a scalar field f . Just write:

df = (∂rf)dr + (∂θf)dθ + (∂φf)dφ (chain rule)

= ∇f · dl (coordinate − free form)

Working out the last line after inserting the above expression for dl gives:

(∇f)r = ∂rf, (∇f)θ =1

r∂θf, (∇f)φ =

1

r sin θ∂φf (0.8)

from which one reads off the components of the gradient operator ∇.

One might think that the divergence and the curl of a vector can now be easily found. Not quite! Unlike in

a Cartesian basis, the unit vectors of a spherical basis are not fixed at the origin; rather, they are attached to the

particular point where our vector is defined. As soon as we move away from this point, the set of basis vectors

changes its orientation unless the motion is in the radial direction. So we expect that, unlike in a Cartesian basis,

the spherical unit vectors will have non-zero spatial derivatives. These can be calculated easily from the above

expressions relating (r̂, θ̂, φ̂) to (x,y, z). In tabular form, we get:

r̂ θ̂ φ̂

∂r 0 0 0

∂θ θ̂ −r̂ 0∂φ φ̂ sin θ φ̂ cos θ −r̂ sin θ − θ̂ cos θ

Now it is a straightforward—if somewhat tedious—exercise to calculate the divergence† of a vector u in

spherical coordinates. Expand:

†For those who know such things, it is much simpler first to identify the non-zero components gij of the metric tensor for sphericalcoordinates which are the coefficients of the dr2, dθ2 and dφ2 terms in the norm of the line element given by eq. (0.7), computing

√g,

where g = r4 sin2 θ is the determinant of the diagonal gij matrix, and using the very general formula for the divergence of a vector u inany space and any basis: ∂i(

√gui)/

√g.

7


∇ · u =(

r̂∂r + θ̂1

r∂θ + φ̂

1

r sin θ∂φ

)

·(

r̂ur + θ̂ uθ + φ̂uφ

)

using the product rule for derivatives and the table above. You can then put the five terms you will get in the elegant

form:

∇ · u = 1r2∂r(r

2ur) +1

r sin θ[∂θ(sin θ uθ) + ∂φuφ] (0.9)

An even more tedious calculation yields the expression for ∇× u given inside the back cover of Jackson.A word of caution: although one uses a Cartesian basis to derive it, the identity ∇×(∇×u) ≡ ∇(∇·u)−∇2u

does hold in any coordinate system. But it is of limited interest in any basis other than Cartesian, because of the

term ∇2u. Although ∇2u = (∇2ux)x̂+(∇2uy)ŷ+(∇2uz)ẑ, it is not equal to (∇2ur)r̂+(∇2uθ)θ̂+(∇2uφ)φ̂,because the derivatives of the unit vectors are non-zero. In fact, ∇2u is best evaluated from the identity.

Clearly, divergences and curls of vetors are much more complicated in spherical (and cylindrical) bases than

in Cartesian ones. Why then bother with non-Cartesian bases? Because in some important situations, those with a

symmetry, these awful expressions collapse down to very simple ones. In spherical coordinates, for instance, any

vector without angular dependence, and therefore spherically symmetric, will have a one-term divergence and zero

curl, just by inspection of the relevant expressions.

Consider the vector field r̂/r2, which plays a very important rôle in electromagnetism. Its divergence inspherical coordinates is very simple to calculate: it vanishes everywhere except at r = 0, where r̂/r2 diverges. Bycontrast, more work is needed to find the same result in a Cartesian basis, with r2 = x2 + y2 + z2. The problemat r = 0 is not an artefact of the spherical coordinates since r̂/r2 diverges in any basis. How do we know what thedivergence is at the origin then? Well, if we integrate it over a sphere of radius R centered at the origin and use thedivergence theorem to convert the volume integral to the flux of r̂/r2 through the spherical surface, we get 4π forthis flux, no matter how small we choose R!. This non-zero result can therefore only come from the origin sincewe know that the divergence vanishes everywhere else. But how should we write the divergence at the origin?

0.4 Dirac delta-“Function”

A very useful object in physics (and invented by P.A.M. Dirac, a physicist trained as an engineer) is the Dirac

delta-function, δ(x), which is zero everywhere on the x-axis, except at x = 0 where it is infinite. Clearly. thisis not a function in the ordinary sense; to a mathematician, it belongs to a class of objects called distributions. In

spite of its strange behaviour, it can be represented in terms of perfectly mundane objects. In one dimension:

δ(x) = limg→∞

1

π

sin gx

x

=1

2π

∫ ∞

−∞eikx dk

from which it is obvious that the δ-function has units of inverse x and is even in x.

The integral of the delta-function over its whole range is well defined:∫∞−∞ δ(x) dx = 1. Since f(x)δ(x− a)

is for all practical purposes f(a)δ(x − a), it follows that:

f(a) =

∫ ∞

−∞f(x)δ(x − a) dx (0.10)

for any “well-behaved” function f(x). Thus, the delta-function can be used to pick out a particular value of afunction.

Another useful property:

δ(ax) =1

|a| δ(x) (0.11)

8


These properties are readily extended to the 3-dimensional delta-function, δ(r) = δ(x)δ(y)δ(z). Indeed, wehave:

f(x0) =

∫

allspace

f(x)δ(r − r0)d3r

Now we know our divergence: ∇ · (r/r3) = 4πδ(r). This vanishes ∀ r 6= 0 and has a volume integral over allspace equal to 4π, consistent with the flux calculated above. More generally:

∇ ·(

r− r′|r− r′|3

)

= 4πδ(r − r′) (0.12)

where the differentiation is with respect to r, r′ being fixed. Now since (see eq. G1.101 and G1.102) ∇(1/|r −r′|) = −(r− r′)/|r− r′|3, we immediately find a result that will prove very useful:

∇2(

1

|r− r′|

)

= −4πδ(r − r′) (0.13)

Some care must be exercised, however, when relating 3-dim to 1-dim δ-functions in curvilinear coordinates:

δ3(r − r′) = δ(x − x′) δ(y − y′) δ(z − z′) = 1r2 sin θ

δ(r − r′) δ(θ − θ′) δ(φ − φ′) (0.14)

where the first expression is in Cartesian and the second in spherical coordinates. This is so that, in any coor-

dinate system, the integral of δ3(r − r′) over whole space is 1. There is some awkwardness about δ(r) sincer ≥ 0: normally, the limits of integration in

∫

δ(x) dx = 1 contains 0, where δ(x) is singular, but with r thelower limit lies at the singularity. An exception has to be allowed for the radial variable, and δ(r) defined so that∫∞0 2δ(r)f(r)dr = f(0). This could be achieved by including a factor of 2 in the numerator of the above expres-

sion for the delta-function in spherical coordinates so as to cancel the factor of 1/2 due to integrating only over thepositive range of r.

0.5 Helmholtz Theorem (Griffiths, Appendix B)

A function is said to have compact support if it is localised, ie., it vanishes outside a bounded volume, or if it

goes to zero faster than 1/r2 as r → ∞.The Helmholtz theorem asserts that a vector function F(r) that vanishes at infinity is uniquely determined

over space if its divergence and curl are known and if these have compact support. In that case:

F(r) = ∇u + ∇×w (0.15)

where

u(r) = − 14π

∫

∇′ · F(r′)|r− r′| d

3r′ w(r) =1

4π

∫

∇′ × F(r′)|r− r′| d

3r′

and it is understood that the integration must extend over all space.

Also, vector fields whose curl vanishes everywhere in a simply-connected space can be written as the gradient

of some scalar field, the latter being determined only up to a constant. Similarly, vector fields whose divergence

vanishes everywhere in a simply-connected space (̂r/r2 is not one of them!) can be written as the curl of somevector, the latter being determined only up to the gradient of a scalar.

The Helmholtz theorem is merely the application to 3-dim vectors of the very powerful Hodge decompositiontheorem.

9


1 General Properties of Maxwell’s Equations for EM Fields and Potentials

1.1 First and Second-Order Maxwell Equations for the Fields and Potentials

1.1.1 Electromagnetic Fields (G7.3.3)

We shall use the following microscopic version of Maxwell’s equations In SI units for the electric field E and

magnetic induction field B:

∇ · E = 4π ke ρ ∇×B −1

c2∂tE = 4π km J

(1.1)∇ ·B = 0 ∇×E + ∂tB = 0

where ke is Coulomb’s constant, and km is Ampère’s constant. Also, ke/km = c2, with c the speed of light. In SI

units, we usually write: ke = 1/4πǫ0 and km = µ0/4π.

The electric charge density ρ and current density J are not independent. Combining Gauss’ law and thedivergence of the generalised Ampère law leads to:

∂tρ = ∇ · ∂tE/(4πke) = −∇ · J (1.2)

which is the continuity equation for the sources. It expresses local conservation of charge. Integrating over an

arbitrary volume and using the divergence theorem gives the more intuitive dQ/dt = −∮

J · dS, which says thatany change in the net charge Q inside the volume must be accounted for by a flux of charge across its boundary. Thelocal and global formulations of this extremely important law are equivalent. The fact that Maxwell’s equations

seem to “know” about it, even though they are about fields, is not a fluke.

By taking the curl of Faraday’s law and the curl of Ampère’s law, and applying Gauss’ laws for electricity and

for magnetism, we arrive (EXERCISE) at the wave equations:

✷E = 4π ke(

∇ρ +1

c2∂tJ

)

(1.3)✷B = − 4π km ∇× J

where ✷ = ∇2 − (1/c2)∂2t is the d’Alembertian operator. The right-hand side of the equations are the sources ofelectromagnetic waves.

In any source-free region of space (vacuum), the fields obey the first-order vacuum equations:

∇ ·E = 0 ∇×B = 1c2∂tE

(1.4)∇ ·B = 0 ∇×E + ∂tB = 0

and the very simple second-order equations:

✷E = 0 ✷B = 0 (1.5)

These vacuum fields have field lines that do not end or start on electric charges (E) and do not enclose currents

(B) in the region of space where the vacuum equations hold. It is important always to remember that these second-

order equations hold for each component of the field only in Cartesian coordinates. If one does insist on using

curvilinear coordinates, the action of the Laplacian operator on the vector has to be defined using the identity, eg.,

∇2E = ∇(∇ · E) − ∇ × ∇ × E, and only then would each curvilinear component of this object enter in thed’Alembertian to form a scalar wave equation.

10


1.1.2 Potentials (G10.1.1)

The two homogeneous (source-free) Maxwell equations are equivalent to:

E = − ∂tA −∇V B = ∇×A (1.6)To see this, note that because ∇ · B = 0 everywhere, and assuming no “holes” in space where B is not defined,then B must be the curl of some vector A. Then Faraday’s law becomes ∇× (E + ∂tA) = 0. Again, since thisholds everywhere, E+ ∂tA must be the gradient of some scalar.

The inhomogeneous Maxwell equations then become the second-order Maxwell equations for the potentials

V and A:

∇2V + ∂t(∇ ·A) = − 4π ke ρ(1.7)

✷A − ∇(

∇ ·A + 1c2∂tV

)

= − 4π km J

1.2 Brief Review of Electromagnetic Waves (G9.1.1, 9.1.2, 9.2)

There are very well known solutions to the wave equation for Maxwell fields in vacuum, eq. (1.5): the solutions

to the source-free scalar wave equation. To obtain those plane-wave solutions, we look for vector functions that

depend only on one spatial variable, say x. Then we use a trick from monsieur d’Alembert to change variables tothe two independent variables ξ± = x± ct, so that the wave equation becomes ∂2ξ+ξ−F = 0, where F = {E, B},whose general solution drops out† immediately:

F = f−(x− ct) + f+(x+ ct)The differentiable, but otherwise arbitrary, functions f∓(x, 0) are propagated in opposite directions at velocity±cx̂, with f±(x ± ct, y, z) uniform on a plane perpendicular to the direction of propagation corresponding to aconstant value of the phase x± ct, thus justifying their name of plane wave.

In three dimensions, consider the very important case where F is a harmonic function of time: F(r, t) =F(r)e−iωt, where we use the more convenient complex exponential representation. This would occur when thesources themselves are harmonic: ρ(r, t) = ρ(r)e−iωt, and J(r, t) = J(r)e−iωt. From eq. (1.1), Maxwell’s fieldequations for the position-dependent part of harmonic fields become:

∇ ·E(r) = 4π ke ρ(r) ∇×B(r) = 4π km J(r) − iω

c2E(r)

(1.8)∇ ·B(r) = 0 ∇×E(r) − iωB(r) = 0

In vacuum Maxwell’s equations for the position-dependent part of harmonic fields are:

∇×B(r) + i ωc2

E(r) = 0

(1.9)

∇×E(r) − iωB(r) = 0These two equations are sufficient to determine the fields since they contain the divergence equations as identities.

We can decouple them by taking the curl of any one of them and combining it with the other to obtain Helmholtz

equations:(

∇2 + ω2

c2

)

E(r) = 0

(1.10)(

∇2 + ω2

c2

)

B(r) = 0

†See also section 9.1.1 in Griffiths

11


In one dimension, the fields have the generic form:

u(x, t) = a ei(kx−ωt) + b e−i(kx+ωt)

where k = ω/c = 2π/λ, where λ is the spatial period of the wave, aka its wavelength. These functions havethe magic x ± vt dependence on x and t that identifies them as harmonic waves travelling at the phase velocityv = ω/k = c.

In three dimensions the complex representation of the plane-wave solution of the Maxwell vacuum wave

equations for the fields is:

E(r, t) = E0 ei(k·r−ωt)

(1.11)

B(r, t) = B0 ei(k·r−ωt)

where the field strengths E0 and B0 are complex constants, and k is now called the wave vector in the direction

of propagation, with :

k · k = ω2/c2

But the wave equations are second-order differential equations, and not all their homogeneous solutions can

satisfy the first-order Maxwell equations! Inserting the solutions into the divergence and curl equations gives:

k ·E0 = 0 k ·B0 = 0 B0 = k̂×E0

c(1.12)

where k̂ is the unit vector in the direction of propagation. This means that both E and B are perpendicular to each

other and to the direction of propagation (hence the name transverse wave).

A little later we shall be deriving a much more general solution of the electromagnetic wave equation.

Electromagnetic fields store both energy and momentum. For fields in vacuum, and in SI units, this is often

given (see eq. G8.5) as:

ufield =1

2ǫ0E

2 +1

2µ0B2 (1.13)

This assumes real fields, but our previous results involve complex fields. Instead of just taking the real part of our

expressions (which would be correct), let us note that the real part of harmonic vector fields F satisfies:

F(r, t) = ℜ[

F(r) e−iωt]

=1

2

(

F(r) e−iωt + F∗(r) eiωt)

Then the scalar product of two such fields: is

F(r, t) ·G(r, t) = 14[F(r) e−iωt + F∗(r) eiωt] · [G(r) e−iωt + G∗(r) eiωt]

=1

2ℜ[F∗(r) ·G(r) + F(r) ·G(r) e−2iωt]

We will not be interested in tracking the fields at every instant in time, so we average out the harmonic time

dependence:

< F(r, t) ·G(r, t) > = 12ℜ [F∗(r) ·G(r)]

Therefore, when F is harmonic:

< F2(r, t) > =1

2|F(r)|2 (1.14)

This result an be applied to the time-averaged energy density (1.13) in the fields of a plane-wave:

< ufield > =1

4

(

ǫ0 |E(r)|2 +1

µ0|B(r)|2

)

=ǫ02E20 =

1

2µ0B20 (1.15)

12


assuming real field amplitudes.

The energy flux density is given (EXERCISE) by the Poynting vector S = (E × B)/µ0, averaged over timeto get rid of oscillations:

< S > =1

2µ0cE20 k̂ = < ufield > c k̂ (1.16)

1.3 Transverse and longitudinal projections of Maxwell’s Equations for the Potentials

When we try to solve for the vector potential A, we expect Maxwell’s theory to supply equations, either first-order

or second-order, for ∇ ×A and ∇ · A. It certainly does for ∇ × A. However, neither of Maxwell’s equations(1.7) for the potentials V and A can determine ∇ · A. A very instructive way to see this is to note that underphysically realistic conditions any 3-dim vector field may be written† as the sum of two vectors:

F = FL + FT = ∇u + ∇×w (1.17)

The first term, FL = ∇u, whose curl vanishes identically, is called the longitudinal part (or projection) of F;the second, FT = ∇ × w, whose divergence vanishes identically, is called the transverse part (or projection)of F. If F satisfies a wave equation, each of the single independent component of FL and the two independent

components of FT potentially carries energy all the way to infinity, in which case we say that it is a dynamical

degree of freedom.

Observe that ∇ ·F is really ∇ ·FL and contains no information about FT; also, ∇×F is really ∇×FT andcontains no information about FL.

Project the second equation (1.7). The transverse projection immediately gives:

AT = − 4π kmJT (1.18)

where we have used the fact a gradient is a longitudinal object.

Now, remembering that = ∇2 − (∂2t )/c2, take the divergence of the longitudinal projection of (1.7):

∇ ·[

AL − ∇(

∇ ·AL +1

c2∂tV

)

+ 4π km J

]

= (∇ ·AL) − ∇2(∇ ·AL) −∂t∇2Vc2

+ 4π km∇ · J

= − 1c2∂t

[

∂t(∇ ·AL) + ∇2V + 4π keρ]

But the terms in the square bracket on the second line are just the first of equations (1.7). Therefore, the longitudinal

projection of the second Maxwell equation for the 3-vector potential contains no information about ∇ ·A that isnot in the first equation. But that is really an equation for V with ∇ · Ȧ (more precisely, ∇ · ȦL) as a sourcetogether with ρ. Therefore, Maxwell’s theory cannot determine the divergence of the vector potential.

1.3.1 Choices of the divergence of A

Since the theory does not know ∇ ·A, we have to tell it what it is by making an arbitrary choice. Choosing ∇ ·Ato vanish (Coulomb condition), is equivalent to requirinng the vector potential to be purely transverse (hence the

name “transverse gauge” also given to this choice), and the equation for V becomes a Poisson-type equation withsolution:

V (r, t) = ke

∫

ρ(r′, t)

|r− r′| d3r′ (1.19)

This looks innocuous enough until we realise that any change in the source is instantaneously reflected in the

scalar potential. The Coulomb condition leads to acausal behaviour, which is also a reflection of the fact that the

†See section 0.5 in the mathematical preamble, and also Appendix B in Griffiths.

13


condition is not relativistically covariant, in the sense that it is not necessarily the same in all inertial frames. But

the equation for V is not a classical wave equation, and V does not really propagate as a wave, so one should notexpect proper causal behaviour from it.

When manifestly causal behaviour is desired, we can choose nstead a condition first proposed† by Lorenz :

∇ · A = −∂tV/c2, which turns eq. (1.7) into nice-looking wave equations of the type (potential) = sourcewith causal solution eq. (2.8). Then one can calculate the energy radiated to infinity following standard treatments

(eg. chapter 10 in Griffiths) and find that the scalar potential does make a mathematical contribution to the energy

radiated to infinity by accelerated charges. This, however, would not have happened if we had chosen the Coulomb

condition. Therefore, we should not attach any physical significance to that contribution: it arises simply out of

consistency with the choice of condition on ∇ ·A.Note also that the energy radiated to infinity can be calculated solely in terms of B and, therefore, of A, without

V being involved. In fact, since B = ∇× (AL + AT) = ∇×AT, only the transverse part of A contributes!

1.4 Initial-Value Problem with Constraints

The Initial Value Problem (IVP) consists in finding which data must be specified at a given time for the time

evolution of variables to be uniquely determined by their equations of “motion”, ie. equations (1.1) for the fields

and (1.7) for the potentials.

By initial data, one means the state of the system of variables at eavh point in space at initial time t0. The IVPtogether with the evolution equations constitute the Cauchy Problem of the theory. If the IVP can be solved, the

dynamical behaviour of the system can be uniquely predicted from its initial data.

Most often, the equations of “motion” take the form of a set of wave equations, each of the form F = f . Ifthey always told the whole story, the Cauchy problem would be solved by specifying the value of F and its first-order time derivatives at t = t0. Things are not so simple, however, when there are inherent, built-in constraintson the initial data. Those constraint equations must be discovered and solved. Also, we must find which initial

data we are allowed to specify freely.

1.4.1 First-order Cauchy problem

Consider the Cauchy Problem from the point of view of the fields. The wave equations (1.3) are standard wave

equations for six quantities. But the first-order field equations (1.1) must also be satisfied.

The two divergence equations contain no time derivatives and are thus constraints on E and B at t = t0.The divergence equation on E determines one of its component at initial time; the same applies to B from its

divergence equation. Indeed, the constraint equation on E can be rewritten ∇2u = ρ for a scalar field u, a Poisson-type equation which can in principle be solved for u at initial time so long as ρ falls off faster than 1/r at infinity.Once u has been found, ∇u yields EL at t0. As for B, because its divergence vanishes, it is always a manifestlytransverse object. Thus, we have no control over the longitudinal components of the fields; the two transverse

components for each are the only ones for which initial data can be freely specified at this stage.

Next, look at the two Maxwell field equations which contain time derivatives. Suppose we specify E and ∂tEat t = t0, which are needed to solve the 2

nd-order equations, eq. (1.3). Then the two transverse components of B

are determined at t0 by ∇×B = 4πkmJ+ ∂tE/c2. The two remaining components of ∂tB are determined, alsoat t0, by Faraday’s equation. Therefore, once we have specified two components of E and their time derivatives,Maxwell’s equations take over and determine all the other quantities and their time derivatives at t = t0. Alter-natively, we could have started with two components of B; specifying them and their time derivatives at t = t0constrains all the other field components and time derivatives. The two components of E (or B) that can be freely

specified at initial time are their transverse components.

†L. Lorenz, On the Identity of the Vibrations of Light with Electrical Currents, Philosophical Mqnzine and Journal of Science, 34,

July-December, 1867, pp. 287–301 (translated from Annalen der Physik und Chemie, June 1867)

14


One of the advantages of this Cauchy analysis is that it does not rely on any one particular solution, but is

valid for any electromagnetic field and potential. We note that since ∇ · A, which determines the longitudinalcomponent of A, is arbitrary, only the two transverse components—controlled by ∇×A—correspond to physicaldynamical degrees of freedom, consistent with the Cauchy-data analysis of the first-order Maxwell equations.

1.4.2 Propagation of constraints

The next question we must address in the Cauchy Problem is whether constraints are automatically satisfied ∀ t >t0, That is, if it can be shown that the field equations and source conservation laws lead to constraints obeyingthe homogeneous wave equation (constraint) = 0 at t = t0. We certainly expect the theory to carry its ownconstraints forward in time. Indeed, take the divergence of the wave equation for E in eq. (1.3):

(∇ · E) = 4π ke[

∇2ρ + 1c2∂t∇ · J

]

Eliminating ∇ · J with the continuity equation gives immediately:

(∇ ·E − 4π keρ) = 0

as expected. Since the source term on the right of the wave equation for B is a curl, one immediately writes:

(∇ ·B) = 0

Constraints that do not continue to hold automatically in the future must be enforced by hand at all times. This

could be the case for conditions such as the ones we have imposed. Fortunately, one can show that the Coulomb

and Lorenz conditions propagate forward in time once imposed at initial time. Take for instance the wave equation

for A in eq. (1.7). Impose ∇ ·A = 0 at initial time in both the equations for V and A. Then take the divergenceof the resulting wave equation for A and the time derivative of the resulting equation for V , and use the continuityequation to obtain:

(∇ ·A) = 0showing that if we choose ∇ · A = 0 at t = t0, it remains the same for all time. Similarly, it can be shown(EXERCISE) that imposing the Lorenz condition everywhere in space at t = t0 also leads to a wave equation for∇ ·A+ ∂tV/c2.

1.4.3 Gauge transformations

As will be discussed later in the more powerful formalism provided by Relativity, Maxwell’s equations (1.7)

for the potentials are actually invariant under the so-called gauge transformation): A → A + ∇f and V →V − ∂tf , where f is any differentiable function of position and time. It should be clear from eq. (1.6) that thefields are unaffected by a gauge trransformation; we say that they are gauge-invariant. But it is equally important

to notice that since the transformation shifts A by a gradient, which is a longitudinal object, it does not affect the

transverse components of A; these are seen to be independent of any choice, and thus entirely physical (contrary

to assertions sometimes made that the electromagnetic potential is not as physical as the fields because it is not

gauge-invariant—now we know that this only applies to AL and the 3-scalar potential V ). Maxwell’s theory isprobably the smplest gauge-invariant field theory, but practically all fundamental theories of Nature (non-linearised

General Relativity a possible exception) currently known share this property.

We also observe that, even after imposing the Lorenz condition, there is still some arbitrariness left: one can

gauge-transform both A and V to new functions that still obey the Lorenz condition. Indeed, let f be some scalarfunction that satisfies the homogeneous wave equation f = ∇2f − 1c2 ∂2t f = 0. Then add ∇2f to ∇ · A and1c2∂2t f to −∂tV/c2 to obtain:

∇ · (A + ∇f) = − 1c2∂t(V − ∂tf) (1.20)

15


This shows that the transformed potentials still satisfy the Lorenz condition! (EXERCISE: Is there still arbitrariness

left after imposing the Coulomb condition instead?)

The Lorenz condition ∇ · A = −∂tV/c2, which is imposed in almost all treatments of electromagneticradiation, can make you believe that V and the three components of A propagate to infinity, whereas I hope tohave convinced you that only the transverse components of A do.

The Lorenz condition relates the longitudinal component of A to V . Knowing one determines the other. Now,I will show you that this longitudinal component can be made to disappear without affecting Maxwell’s equations

for the fields and the two transverse components of the vector potential.

Now, for the first time, we shall have to look at actual solutions of the wave equations for A and V . To makethings as simple as possible, take plane-wave solutions A = A0e

i(kx−ωt), where the x-axis has been aligned alongthe direction of propagation, and V = V0e

i(kx−ωt). Then:

∇ ·A = ∂xAx = ikA0xei(kx−ωt), ∂tV = −iωV0ei(kx−ωt)

Inserting into the Lorenz condition with ω/k = c yields, as expected, a relation between the longitudinal compo-nent Ax and V : A0x = V0/c.

Now fold in f = f0ei(kx−ωt) into eq. (1.20) to get:

ik (A0x + ik f0)ei(kx−ωt) = i

ω

c2(V0 + iω f0) e

i(kx−ωt)

Since f0 is arbitrary, we can choose it to cancel A0x, which at the same time gets rid of V0, leaving us with thetransverse components of A!

Although the analysis under the Lorenz condition is quite a bit more involved than with the Coulomb condition,

the conclusions are the same: only the two transverse components of A propagate, in the sense that they carry

energy to infinity.

1.4.4 Are electromagnetic fields more fundamental than potentials?

In classical electromagnetism, the equation of motion for charges (aka the Lorentz force law) contains fields but

no potentials. In fact, there is nothing in classical electromagnetism that cannot be derived also from the fields.

Thus, for a long time after they were introduced, potentials were considered useful to simplify some calculations,

but non-physical. One very often quoted argument held that potentials could not be physical because, unlike the

fields, they were not gauge-invariant. But now we see from the IVP analysis of Maxwell’s theory that it is possible

to set consistent initial data for the transverse part of the vector potential, which could not be true if they were

gauge-variant

In a seminal paper published in 1959, Aharonov and Bohm2 proposed experiments aimed at observing the

potential directly. One experiment involved two beams of electrons travelling on opposite sides of a very long,

very thin solenoid. When a current flows in the solenoid, the magnetic field outside is strictly zero, but there still

exists a vector potential A so as to match the potential inside where there is a magnetic field B. The circulation∮

A · d l over the complete paths of the beams would not vanish, but would simply be the flux of the interior fieldthrough the cross-sectional area of the solenoid. Classically, this exterior vector potential cannot be observed.

The authors, however, noticed that electrons do not really obey the Lorentz force law (except in the classical

limit), but instead Schrödinger’s equation, in which the potentials are present. The wave-functions of the electrons

would interfere and switching the magnetic field on and off ought to perturb the interference pattern, even though

no force acted on the electrons. The effect that they (and others before) discussed was soon observed experimen-

tally. Interestingly enough, in the final section of the paper the authors are still a bit confused about the gauge

objection, since they think that it could be raised against them:

2http://link.aps.org/doi/10.1103/PhysRev.115.485 (free access from UofT computers).

16

http://link.aps.org/doi/10.1103/PhysRev.115.485


“ The main objection that could be raised against the above suggestion is grounded in the gauge

invariance of the theory. In other words, if the potentials are subject to the transformation Aµ →A′µ = Aµ + ∂ψ/∂xµ, where ψ is a continuous scalar function, then all the known physical quantitiesare left unchanged. As a result, the same physical behavior is obtained from any two potentials,

Aµ(x) and A′µ(x), related by the above transformation. This means that insofar as the potentials are

richer in properties than the fields, there is no way to reveal this additional richness. It was therefore

concluded that the potentials cannot have any meaning, except insofar as they are used mathematically,

to calculate the fields.”

They are reduced to say that the objection is wrong because counter-examples can be produced. In fact,∮

A·d lis gauge-invariant because

∮

(∇f) · dl = 0 (see section G1.3.3). Since the transverse A is itself gauge-invariant,only the longitudinal component AL varies, and it is really

∮

AL · dl that is unaffected by gauge transformations.Moreover, the high degree of symmetry makes knowledge of the curl sufficient to determine A everywhere in

cylindrical coordinates (ρ, φ, z): Aout = (BR2/2ρ)φ̂, and Ain = (Bρ/2)φ̂, where B is the magnitude of the

field inside and R the radius of the solenoid. The divergence of A vanishes everywhere, which ensures that A istransverse. This does not contradict the fact that ∇ ×A = 0 outside the solenoid. For A to be pure longitudinalthere, the curl would have to vanish everywhere in a simply connected space, ie., a space where every closed loop

in the space can be contracted to a point, which is certainly not the case here for loops around the solenoid. What

is important is that electrons in the quantum-mechanical regime are described by a wave-function, with a phase

that can be shown to depend on A. Phase differences, of course, are responsible for interference, and when the

beams go through a region of zero magnetic field but non-zero potential on topologically distinct paths (paths

which cannot deformed into one another while leaving their ends fixed), A disturbs the interference pattern of

the beams. (Strictly speaking, the word “path” is not so appropriate, though: in quantum mechanics paths are not

well-defined, and this is why there can be an interference effect.)

So we must conclude that electromagnetic potentials (or at least their transverse part) are physical, and that

they contain more information than fields. Only in the quantum-mechanical regime, however, can this information

be retrieved.

1.5 Duality Properties of Maxwell’s Theory

If we define E = E/c+ iB in SI units, Maxwell’s equations can be written in a very compact way:

∇ · E = 0 ∇× E − ic∂tE = 0 (1.21)

The total energy density in the fields can be written as E · E∗/2µ0.In this form, it is easy to see that Maxwell’s equations and the total field energy density are invariant under the

duality transformation:

E −→ eiφE (1.22)with φ any real constant. Choosing φ = π/2 gives E −→ −Bc and B −→ E/c. This is the reason for thesimilarity between the electric and magnetic fields† with equivalent source types.

When sources are put in, however, Maxwell’s equations are no longer invariant. Invariance can only be restored

by including a magnetic charge density ρm and a magnetic current density Jm so that now:

∇ · E = 4π km ρ ∇× E −i

c∂tE =

4π i kmc

J (1.23)

where ρ = ρec + i ρm and J = Jec + iJm. when both electric and magnetic sources are included, the dualitytransformation with φ = π/2 exchanges not only the fields, but also the electric and magnetic charges.

†Compare for instance the results of problems G3.36 and G5.34.

17


If instead one chooses tan φ = −ρm/ρec, the magnetic charge and current densities are transformed to zero,and this, without changing the physical content of the theory! So, if all particles have the same universal ratio of

magnetic to electric charge, we may not be able to tell whether magnetic charge exists.

18


2 Causal Solutions of Maxwell’s Equations: Potentials and Fields

2.1 Causal Solutions of a Scalar Wave Equation

We begin by deriving a general result for the solution of a wave equation that involves the ✷ operator.

As a first step, consider functions† G(r, r′) that, by definition, satisfy:

(∇2r + k2)G(|r − r′|) = −4π δ(r− r′) (2.1)

where k is a constant. Now we adopt the notation introduced in section G1.1.4 for the distance vector from thesource point to the observation (field) point: = r− r′.

A solution of this equation for G( ) is:

G( ) =1 (

A eik + B e−ik)

(2.2)

Let us check that this is indeed a solution. We have, using equ. G1.101 and G1.102:

(∇2r + k2)(

e±ik)

=1∇2e±ik + e±ik ∇2

(

1)

+ 2∇

(

1)

·∇e±ik + k2 e±ik

=

[

−✁✁✁k2 ± ❙❙❙

2ik2− 4π δ( ) ∓ ❙❙❙

2ik2+

✁✁✁k2]

e±ik

= − 4π δ( )

Next, with the Fourier integral representation linking the time-domain to frequency-domain:

g(r, t) =1

2π

∫ ∞

−∞g(r, ω) e−iωt dω g(r, ω) =

∫ ∞

−∞g(r, t) eiωt dt (2.3)

we can transform a typical wave equation:

✷Ψ(r, t) = − 4π f(r, t)

to its Helmholtz form for the fequency-dependent function Ψ(r, ω):

(∇2 + k2)Ψ(r, ω) = − 4π f(r, ω) (2.4)

where k2 = k · k can be taken as a short form for ω2/c2.Now we are ready to find functions that satisfy:

✷r,tG( , t− t′) = − 4π δ( ) δ(t − t′) (2.5)

A very useful representation of the Dirac delta-fuction is:

δ(t− t′) = 12π

∫ ∞

−∞eiω(t

′−t) dω (2.6)

The Fourier transform of the source term: δ( ) δ(t − t′), is f(r, ω) = δ( ) eiωt′ . In the frequency domain, then,the wave equation becomes:

(∇2r + k2)G(r, r′, ω, t′) = − 4π δ( ) eiωt′

†They are an example of Green functions.

19


Assume separable solutions of the form G(r, r′)eiωt′; inserting into this equation, we get from (2.1) the solutions

G±(r, r′, ω, t′) = ei(±k +ωt′)/ . Then, transforming back to the time domain and using the representation (2.6)

for the δ-function yields the time-dependent Green functions:

G(±)( , t− t′) = 12π

∫ ∞

−∞

eiω[± /c−(t−t′)]

dω =δ(

t′ − [t∓ /c])

(2.7)

What does knowing the solution G for eq. (2.5) do for us? Well, we recognise that:

✷r,t

∫ ∞

−∞G(±)( , t− t′) f(r′, t′) d3r′ dt′ =

∫ ∞

−∞f(r′, t′)✷r,tG

(±)( , t− t′) d3r′ dt′ = = − 4π f(r, t)

has the generic form ✷Ψ(r, t) = −4πf(r, t), which shows that the general solution of a wave equation withsources can be written, up to a possible plane-wave solution of the homogeneous wave equation for Ψ, as theretarded solution:

Ψ(r, t) =

∫ ∫ ∞

−∞G(+)( , t− t′) f(r′, t′) d3r′ dt′

=

∫

f(r′, t′r) d3r′ (2.8)

where the suffix r stands for the fact that t′ must be evaluated at the retarded time t′ = t− /c. This automaticallyensures the proper causal behaviour of the solutions, in the sense that the solution at time t is only influenced bybehaviour at the source point r′ at time t− /c.

Important fine print: for the integral to exist, the source f should fall off at least as fast as 1/ at infinity. Tomake sure this is the case, we require sources to be localised in space, ie. of finite extent.

There is also an advanced solution for which the integrand is evaluated at t′ = t+ /c, but it is manifestly notcausal since it implies propagation backward in time.

2.2 Retarded Solutions for the Maxwell Potentials and Fields (G10.1)

If we impose the Lorenz condition, the electromagnetic potentials obey the wave equations:

✷V (r, t) = − 4π ke ρ, ✷A(r, t) = − 4π km J (2.9)

whose retarded solutions are, from our findings of section 2.1:

V (r, t) = ke

∫

d3r′ρ(r′, t′r)

(2.10)

A(r, t) = km

∫

d3r′J(r′, t′r)

In his section 10.2.1, Griffiths guesses those solutions from the static solutions with a plausibility argument, and

goes on to show that they indeed satisfy the wave equations. but we have derived them.

For the potentials, and only for the potentials, the solutions have the same dependence as the static potentials,

except for evaluating the integrands at retarded time before integrating. This is not true of the Maxwell fields!

2.3 Potentials and Fields of a Point-charge in Arbitrary Motion (G10.3)

2.3.1 Liénard-Wiechert potentials

We start with the 3-scalar potential of eq. (2.10):

V (r, t) = ke

∫

ρ(r′, t′r) d3r′

20


Since the volume of the charge distribution for a point-charge is vanishingly small, we can bring out the distance

term in the integral, leaving the integral of the charge density over the volume of the distribution. We might expect

the volume integral of ρ to be the charge of the particle, but this is not true if the particle is moving, because thevolume to be integrated over is not the actual volume of the charge distribution.

Indeed, ρ(r′, t′r) must be evaluated at fixed t′r if the integral is to be the total charge. The potential measured at

time t comes from what different points of the source were doing at different times. For instance, the informationthat arrives at the point of observation from the furthest point of the distribution at a given time t must have leftearlier than the one received at the same time from the closest point. If the source is static, the integral at time t isthe same as at retarded time and gives the charge. Now, let the source move with a velocity that has a component

v · ˆ along the direction of observation. Call L the actual dimension of the source along ˆ at retarded time. Wewant to find how far the furthest point of the distribution appears to be at the time information from the front and

back ends is received at an obervation point very close to the front end. Call that distance L′. During the timeL′/c it takes the signal received at time t to travel the distance L′ from the furthest point in the source, that samefurthest point travels a distance L′ − L at velocity (v · ˆ). Equating L′/c and (L′ − L)/(v · ˆ), we find thatL′ = L/(1− β · ˆ), where β = v/c.

Therefore, an approaching source appears thicker by 1/(1 − β) in the direction of its motion, and a recedingsource appears thinner by 1/(1 + β). Since the transverse directions remain unaffected, the volume of the sourceappears distorted by the same factor. Since the factor is independent of the size of the volume, it will affect any

volume, even one that is vanishingly small. The correction should not be interpreted as a relativistic effect, but as

a geometrical one.

Then there comes:

V (r, t) = keq 1

1− β · ˆ

Obtaining the spatial components of the four-potential is now straightforward. The source J(r′, t′r) is justρ(r′, t′r)v, with v uniform over the volume of the distribution for a point-particle. We find immediately from eq.(2.10):

A(r, t) = 4π km

∫

J(r′, t′r) d3r′ = 4π km cβ

∫

ρ(r′, t′r) d3r′ = 4π km c

q β

1− β · ˆ

These are the so-called Liénard-Wiechert potentials:

V (r, t) = keq 1

1− β · ˆ A(r, t) =β

cV (r, t) (2.11)

Warning: Everything on the right-hand side must be evaluated at retarded time! This includes and ˆ.

Example 2.1. (G10.3): the potentials of a point-charge moving at uniform velocity v.

Start from the definition of retarded time: tr = t− /c, toegether with: = |r− r′|, where r′ is theposition of the charge at tr. For uniform motion, take r

′ = vtr. A re-arrangement then gives:

|r− v tr| = c(t − tr)

To solve for ctr, it is always best to square this expression. We obtain (EXERCISE) a quadraticequation whose solution is:

ctr = γ2[

(ct− r · β) −√

(ct− r · β)2 + (r2 − c2t2)/γ2]

where γ ≡ 1/√

1− β2, and we have chosen the minus sign in front of the root so as to recovertr = t− r/c when v = 0.

21


Our next step involves the computation of the factor (1 − β · ˆ) in the denominator of the Liénard-Wiechert potentials given by eq. (2.11).We already know that = c(t−tr), and = r−vtr. Insertingthese in the factor, there comes:

(1− β · ˆ) = − β · = ct − r · β − ctr/γ2

There only remains to use our previous result for ctr, leading to:

(1− β · ˆ) =√

(ct− r · β)2 + (r2 − c2t2)/γ2

With this, the scalar potential (2.11) of the point-charge is:

V (r, t) = keq

√

(ct− r · β)2 + (r2 − c2t2)/γ2(2.12)

Notice how V is now written in terms only of the coordinates of the observation point and the (uni-form) velocity of the charge. A(r, t) follows immediately from eq. (2.11).

We can gain more insight about this complicated expression by expanding the terms inside the square

root. A little re-arrangement allows us to write:

(ct− r · β)2 + (r2 − c2t2)(1 − β2) = (r − βct)2 + (r · β)2 − β2r2

Now βct is the position vector of the charge at time t, so that the vector R ≡ r − βct points fromthe position of the charge at time t toward the observation point, in contrast to = r − βctr whichpoints to the same observation point at time t, but from the retarded position of the charge.

Inserting r = R+βct into (r·β)2 − β2r2, a miracle occurs: any time dependence is absorbed into R,and we get: (r ·β)2 − β2r2 = (R ·β)2 − β2R2. When added to R2, there comes: R2(1−β2 sin2 θ),where θ is the angle between the direction of the velocity and the direction of R. Then the scalarpotential becomes:

V (r, t) = keq

R

1√

1 − β2 sin2 θ(2.13)

What this manipulation has done has been to transfer the explicit time dependence of the potential at

the observation point to one involving the position of the charge at time t, with the time dependenceimplicit. The [1 − β2 sin2 θ]−1/2 factor deforms the isotropic potential keq/R, enhancing it most inthe direction perpendicular to the direction of motion of the charge, or θ = π/2. This occurs when thedistance between the moving charge and the point of observation is minimum, with Vmax = keγq/R.

2.3.2 Fields

The fields are E = −∇V − ∂tA and B = ∇ ×A. Since the potentials are to be evaluated at retarded time, wewill need to know the position and time derivatives of t′ = tr = t− /c.

• Derivative of retarded time with respect to observer time

∂ttr = 1−1

c∂t = 1−

1

2 c∂t

2

= 1 +1

c

r− r′ · ∂trr′(tr) ∂ttr= 1 + ˆ · β ∂ttr

where β = ∂trr′(tr)/c is the dinensionless velocity of the particle at retarded time. Solving for ∂ttr yields:

∂ttr =1

1 − β · ˆ (2.14)

22


• Gradient of retarded time and ofSince ∇tr = −∇ /c, we can write: ∇tr = −∇ 2/2c . The gradient of the length squared of a vector issurprisingly complicated; from product rule (4) in Griffiths’ front cover, it is:

∇2 = 2

[

×∇× + ( ·∇)]

To find the first term, we note that:

(∇× )i = (✘✘✘✘✿ 0∇× r)i − (∇× r′)i = − ǫijk ∂trr′k ∂jtr = ǫikj vk ∂jtr

or ∇× = v×∇tr. Using the triple vector-product rule, we then get: ×∇× = v ( ·∇tr)−(v· )∇tr.The second term in ∇ 2 is: i∂i

j = i∂i(rj − r′j) = j − ivj∂itr, or: ( · ∇) = − ( · ∇tr)v.

Putting everything together, there comes:

∇tr = −1

c

[❳❳❳❳❳❳v ( ·∇tr) − (v · )∇tr + −❳❳❳❳❳❳( ·∇tr)v

]

= − ˆc

+ (β · ˆ)∇tr

Solving for ∇tr gives:

∇tr = −1

c

ˆ

1 − β · ˆ (2.15)

Therefore:

∇ = − c∇tr =ˆ

1 − β · ˆ (2.16)

We are now ready to tackle the calculation of the fields. First, we need the gradient of V :

∇V (r, t) = ∇

(

keq

− β ·

)

= − keq

( − β · )2∇( − β · )

Calculating ∇(β · ) is again best done in index notation. With j = rj − r′j and ∂if(tr) = ∂itr∂trf(tr):

∂i(jβj) = βj ∂i

j + j ∂iβj

= βj ∂irj − βj ∂itr ∂trr′j + j ∂itr ∂trβj

= βj δij + ( j∂trβj − βj vj) ∂itr

= βi + (j∂trβj − c βj βj) ∂itr

So:

∇(β · ) = β +( · a

c− β2 c

)

∇tr

where a = c∂trβ is the acceleration of the charge at retarded time. Putting in our results from eq. (2.15) and(2.16), we find after re-arranging:

∇V (r, t) = keq2

1

(1− β · ˆ)3[

(1− β · ˆ)β −(

1

γ2+

· ac2

)

ˆ

]

where γ2 = (1− β2)−1. Using similar methods, one also shows (EXERCISE) that:

∂tA(r, t) = keq2

1

(1− β · ˆ)3[

(1− β · ˆ)( a

c2− β

)

+

( · ac2

+1

γ2

)

β

]

23


The electric field† then is:

E = − keq2

1

(1− β · ˆ)3[

❤❤❤❤❤❤❤(1− β · ˆ)β −(

1

γ2+

· ac2

)

ˆ + (1− β · ˆ)( a

c2− ❙❙β

)

+

( · ac2

+1

γ2

)

β

]

= − keq2

1

(1− β · ˆ)3[

(1− β · ˆ) ac2

+ (β − ˆ)( · a

c2+

1

γ2

)]

= keq2

1

(1− β · ˆ)3[

ˆ − βγ2

+ (ˆ − β) · ac2

− · (ˆ − β)c2

a

]

= keq

c212

1

(1− β · ˆ)3(

c2(ˆ − β)γ2

+ ×[

( ˆ − β)× a]

)

(2.17)

where in the last line we have used the vector identity A× (B×C) = B (A ·C)−C (A ·B).Note the acceleration-dependent term which has a 1/ distance dependence, falling off more slowly than any

static electric field from localised sources. The first term, which depends only on velocity, goes like 1/ 2 and,compared to the 1/ term, becomes negligible at large distances. Accordingly, we shall write:

Enear = keq2

ˆ − βγ2 (1− β · ˆ)3 Efar = km

q ˆ×[

(ˆ − β)× a]

(1− β · ˆ)3 (2.18)

Later on, we will have to define more carefully what we mean by “far”. For the magnetic induction field, there

comes from eq. (2.11):

B = ∇×A = 1c∇× (V β) = 1

c

[

(∇× β)V + ∇V × β]

But (∇× β)i = ǫijk∂jβk = ǫijk∂jtr∂trβk = −(a×∇tr)i. With our previous results for V and ∇V , we obtain:

B = − keq

c

12

1

(1− β · ˆ)3ˆ×

[

β

γ2+

· ac2

β + (1− β · ˆ) ac2

]

=1

cˆ×

[

keq2

1

(1− β · ˆ)3(

ˆ− βγ2

+· ac2

( ˆ− β) − · ( ˆ− β) ac2

)]

=1

cˆ×E (2.19)

In the first two terms inside the round brackets in the second line, the ˆ make no contribution since they are part

of a vector product with ˆ.

The magnetic induction field of the charge is always perpendicular to the plane defined by its electric field and

the “line of sight” to the retarded point. But only the electric far-field is pependicular to ˆ. The electric near-field

has a component along the line of sight (as in the static case), but also a component along the direction of the

velocity of the charge at retarded time.

What eq. (2.19) says is that once we have calculated the electric field of a point-charge, we have also calculated

its magnetic induction field.

Example 2.2. (G10.4) Going back to a charge in uniform motion, whose potentials we have already

derived in example 2.1, it is not hard to obtain expressions in terms of the position and time coordinates

of the observation point — that is because most of the work has already been done. Only near-fields

exist, ie. those with a 1/ 2 fall-off and no dependence on acceleration. Focusing on the electric field,we have:

E = Enear = keq2

ˆ − βγ2 (1− β · ˆ)3 = ke q

− βγ2 ( − β · )3

†Griffiths defines u = c( ˆ − β), with · u = c · ( ˆ − β) = c ((1− β′ · ˆ), but we shall not use this extra quantity.24


In example 2.1 we found that: (1 − β · ˆ) =√

(ct− r · β)2 + (r2 − c2t2)/γ2, and, from thedefinitions of , we immediately get: − β = r− βctr − c(t− tr)β = r− βct. We arrive at:

E = keq

γ2r− βct

[(ct− r · β)2 + (r2 − c2t2)/γ2]3/2 (2.20)

As with the potentials, it is useful to have an equivalent expression in terms of the distance to the

charge at time t. We recognise the numerator as the vector R that points from the present positionof the charge to the observation point. Directly from example 2.1, the expression inside the square

brackets in the denominator is: R(1 − β2 sin2 θ), where θ is the angle between R and the velocity ofthe charge. Therefore:

E = keq

γ21

(1− β2 sin2 θ)3/2R

R3(2.21)

Like the potential, the isotropic part of the field gets deformed by the factor γ−2(1 − β2 sin2 θ)−3/2,which decreases it by a factor 1/γ2 in directions along the velocity, but boosts it by a factor γ inthe transverse direction. Thus, the field lines get squashed toward the transverse plane. At very high

speeds (γ ≫ 1), the field resembles more and more the transverse field of an electromagnetic wave,which then becomes a good approximation for the field of a point-charge. This limit can be very useful

whenever we deal with particles travelling at speeds very close to the speed of light.

The magnetic field is easily found, since ˆ = (r−βctr)/ = R/ +β. Then, because of eq. (2.21):

B =1

cˆ×E = 1

cβ ×E (2.22)

and, as well as being perpendicular to the electric field, which is always true, the magnetic field is also

perpendicular to the velocity of the charge if it is in uniform motion. Again, this is indistinguishable

from an electromagnetic wave if the direction of motion of the particle is replaced by that of the wave.

25


3 Radiation (chapter G11)

3.1 Radiation from Moving Point-Charges (G11.2.1)

In section 2 of these notes we derived very general expressions (eq. (2.17) and (2.19) for the fields of a point-charge

in arbitrary motion. It is now time to show that such a point-charge can radiate energy to infinity in the form of

electromagnetic waves.

3.1.1 Poynting vector

The Poynting vector, which tells us about the amount of electromagnetic energy passing through a unit area per

second, is (in SI units: S = E×B/µ0. Now, the magnetic induction field of a point-charge is related to its electricfield by: B = ˆ × E/c. Then S = E × ( ˆ × E)/(µ0c). Actually, what we are looking for is the energy flux inthe direction of observation, ˆ, that is:

S · ˆ = 1µ0c

[

E× ( ˆ×E)]

· ˆ = 1µ0c

[

E2 ˆ− (ˆ ·E)E]

· ˆ = 1µ0c

[

E2 − ˆ · E)2]

=1

µ0cE2⊥

where E⊥ is the component of the electric field perpendicular to that direction. From eq. (2.18), the near-fieldEnear depends on ˆ − β, so only the component of β perpendicular to ˆ contributes to S · ˆ. But the near-fieldfalls off like 1/ 2, much faster than the far-field, which falls as 1/ with distance. The far-field dominates thePoynting vector at large distance (sometimes called the far-zone), and we shall neglect the near-field. Then, in that

far-off region , eq. (2.18) immediately leads to:

S · ˆ = µ0 q2

16π2 c

[ ˆ× ((ˆ− β)× a)]22 (1− β · ˆ)6 (3.1)

Now ˆ×[

(ˆ− β)× a]

= (a · ˆ)( ˆ− β)− a(1− β · ˆ). Then:

[ˆ× (( ˆ− β)× a)]2 = (a · ˆ)2( ˆ− β) · ( ˆ− β) − 2 [a · (ˆ− β)] (a · ˆ)(1− β · ˆ) + a2 (1− β · ˆ)2

= (a · ˆ)2[

1 − ✘✘✘✘2ˆ · β + β2 − 2 (1 − ✟✟✟ˆ · β)]

+ 2β · a (a · ˆ)(1− β · ˆ) + a2 (1− β · ˆ)2

= 2(β · a) (a · ˆ)(1 − β · ˆ) + a2 (1− β · ˆ)2 − (a ·ˆ)2

γ2

We arrive at a very general expression for the radial component of the Poynting vector of a point-charge:

S · ˆ = µ0 q2

16π2 c

a2 (1− β · ˆ)2 + 2(β · a) (a · ˆ)(1 − β · ˆ) − (a · ˆ)2/γ22 (1− β · ˆ)6 (3.2)

where, again, the right-hand side is to be evaluated at retarded time.

There are two interesting cases:

1. β × a = 0 (velocity and acceleration collinear — linear motion)In this case, Efar (and threfore the radiation) is polarised perpendicular to ˆ in the plane defined by ˆ and

the acceleration vector.

Rather than eq. (3.2), it is simpler to insert β × a = 0 directly into eq. (3.1):

S · ˆ = µ0 q2

16π2 c

[ˆ× ( ˆ× a)]22 (1− β · ˆ)6 =

µ0 q2

16π2 c

[ˆ(ˆ· a)− a)]22 (1− β · ˆ)6

=µ0 q

2

16π2 c

a2 − (a · ˆ)22 (1− β · ˆ)6

26


If we denote by θ the angle from the direction of β to ˆ, there comes:

S · ˆ = µ0 q2

16π2 c

a2

2

sin2 θ

(1− β cos θ)6 (3.3)

We note that there is no radiation in the common direction of acceleration and velocity.

2. β · a = 0 (velocity and acceleration perpendicular to each other — happens if the charge is in circularmotion)

In eq. (3.2), the middle term in the numerator now vanishes, leaving:

S · ˆ = µ0 q2

16π2 c

a2 (1− β · ˆ)2 − (a · ˆ)2/γ22 (1− β · ˆ)6

If, at some instant, we align the z axis of the coordinate system with β and the x axis with a, we can denoteby θ the angle of ˆ with respect to β and by φ the azimuthal angle of its projection of ˆ on the x-y plane.Then, at that instant we can write:

S · ˆ = µ0 q2

16π2 c

a2

2 (1− β cos θ)4[

1− 1γ2

sin2 θ cos2 φ

(1− β cos θ)2]

(3.4)

In both cases, if we wish to follow the angular distribution through time, we must calculate β · ˆand a · ˆwhosetime dependence will contribute.

3.1.2 Angular power distribution

The energy per unit time (power) per unit solid angle detected at the point of observation is:

dE

dΩdt=

dP

dΩ= |S · r̂| 2

An often more relevant question is: how much power is emitted by the charge in the same time interval? This is

not the same, because of the geometrical effect discussed in section 2.3.2. Indeed, using eq. (2.14), there comes:

dE

dΩdt′=

dPrdΩ

=dE

dΩdt

dt

dt′=

dP

dΩ(1 − β · r̂)

where dPr/dΩ is the power radiated per unit solid angle. The angular power distributions for our two cases ofacceleration parallel or perpendicular to velocity follow immediately:

1. β × a = 0

dPrdΩ

=µ0 q

2

16π2 c

a2 sin2 θ

(1− β cos θ)5 (3.5)

When β ≈ 1 (γ ≫ 1), we have, assuming θ small:

1 − β cos θ = 1 −√

1− 1γ2

cos θ ≈ 1 −(

1 − 12γ2

) (

1 − θ2

2

)

≈ 12γ2

+θ2

2=

1

2γ2(1 + γ2 θ2)

and, with sin θ ≈ θ, we find:dPrdΩ

≈ 2µ0 q2

π ca2 γ8

(γ θ)2

(1 + γ2θ2)5

which peaks at θ = 1/(2γ). This justifies our assuming θ small. So the radiation is emitted within a narrowbeam of angular width 1/γ (see Fig. 11.13 in Griffiths) in the forward direction relative to the motion of thecharge.

27


2. β · a = 0

dPrdΩ

=µ0 q

2

16π2 c

a2

(1− β cos θ)3[

1 − 1γ2

sin2 θ cos2 φ

(1− β cos θ)2]

−→β→1

µ0 q2

2π2 cγ6

a2

(1 + γ2 θ2)3

[

1 − 4 γ2 θ2 cos2 φ

(1 + γ2 θ2)2

]

(3.6)

using the same method as in the parallel case. The radiation peaks in the forward direction (θ = 0) and isconfined to a cone of width O(1/γ). A charge in circular motion therefore emits a narrow beam tangent tothe orbit at any time.

3.1.3 Total power radiated by an accelerating point-charge

Obtaining the total radiated power from the general angular distribution that follows from eq. (3.2):

dP

dΩ=

µ0 q2

16π2 c

a2 (1− β · ˆ)2 + 2(β · a) (a · ˆ)(1− β · ˆ) − (a · ˆ)2/γ2(1− β · ˆ)5 (3.7)

is no easy task! In the cases we have considered, however, it is manageable. Start with the collinear case, eq. (3.5).

The φ integration yields 2π, and with the change of variables x = cos θ, Maple readily evaluates the θ integral:

> Int((1-xˆ2)/(1-beta*x)ˆ5,x=-1..1)=2*Pi*int((1-xˆ2)/(1-beta*x)ˆ5,x=-1..1)

> assuming beta>0, beta simplify(int(1/(1-beta*x)ˆ3 - (1/2)*(1-betaˆ2)*(1-xˆ2)/(1-beta*x)ˆ5,x=-1..1))

> *2*Pi assuming beta>0, beta


3.2 Radiating Systems of Charges in the Leading Approximation

3.2.1 Radiation fields of an arbitrary but localised source (G11.1.4)

Earlier we wrote down the retarded solutions for the 3- vector and 3-scalar potentials, eq. (2.10):

A(r, t) = km

∫

d3r′J(r′, t′)

δ(t′ − t+ /c) V (r, t) = ke∫

d3r′ρ(r′, t′)

δ(t′ − t+ /c)

It should be kept in mind that the solution for A assumes the Lorenz condition: ∂ctV + c∇ ·A = 0.Now we shall assume that the source is located around the origin and is much smaller than the distance r

between the origin and the point of observation, that is, r′ ≪ r (first apporoximation in section G11.1.4). Thismeans that we can write:

=√

r2 + r′2 − 2r r̂ · r′ = r√

1 − 2 r̂ · r′

r+r′2

r2≈ r − r̂ · r′

and, neglecting terms that fall off faster than 1/r:

1=

1

r

1

[1− 2r̂ · r′/r + r′2/r2]1/2 ≈1

r

(

1 +r̂ · r′r

)

≈ 1r

Then:

A(r, t) ≈ km1

r

∫

J(r′, t′) δ(t′ − t+ r/c) d3r′ (3.11)

where in this approximation the retarded time is t′r ≈ t− r/c+ r̂ · r′/c ≈ t− r/c.With r′i treated as a scalar, and using the divergence theorem for a localised J, we can write:

0 =

∫

∇′ · (J r′i)d3r′ =

∫

[J ·∇′r′i + r′i∇′ · J]d3r′ =∫

J i d3r′ −∫

r′i ∂t′rρd3r′

where we have used the continuity equation. Now, to leading order, ∂t′r ≈ ∂t, and it can be pulled out of theintegral as a total derivative. Thus, we arrive at:

A(r, t) ≈ km1

rdt

∫

r′ ρ(r′, t− r/c) d3r′ = km1

rdtp

∣

∣

∣

t−r/c(3.12)

where p is the electric dipole moment of the source.

The magnetic field is obtained from ∇ × A, noting that spatial derivatives of the 1/r factor will be of order1/r2 and can be dropped if we are only interested in radiative fields. Then:

B = km1

r

∫

∇× J(r′, t− r/c) d3r′

But, as is easily verified in Cartesian coordinates:

∇× J(r′, t− r/c) = 1c∇(−r)× ∂tJ = −

r̂

c× ∂tJ

In what follows, we shall also make the similar correspondences: ∇· → −(r̂/c) · ∂t and ∇ → −(r̂/c)∂t.Taking everything with an unprimed coordinate dependence out of the integral, we have:

B = − r̂× ∂ctA

For the electric field, we use the Lorenz condition ∂ctV = − c∇ ·A → r̂ · ∂tA so as to avoid the scalar potential:

E = − ∂tA − ∇V = − ∂tA + r̂ ∂ctV = − ∂tA + (r̂ · ∂tA) r̂ = r̂× (r̂× ∂tA) = − r̂×Bc29


To leading order, then, the radiation fields are (see also eq. G11.56–57), denoting by a dot the time derivative:

Brad = −1

cr̂× ∂tA = −

kmc

1

rr̂× p̈

(3.13)

Erad = − r̂×Bc =

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