Part-II Electrodynamics Michaelmas Term 2014 adc1000/ · PDF filePart-II Electrodynamics...
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Part-II Electrodynamics Michaelmas Term 2014
Anthony [email protected]
We begin with a reminder of Maxwells equations, which were discussed at length inthe IB course Electromagnetism:
E = 0
E = Bt
B = 0 (1.3)
B = 0J + 00E
t. (Ampere) (1.4)
Here E(x, t) and B(x, t) are the electric and magnetic fields, respectively, (x, t) is theelectric charge density, and J(x, t) is the electric current density (defined so that J dSis the charge per time passing through a stationary area element dS). The constants0 = 8.854187 1012 kg1 m3 s4 A2 (defined) is the electric permittivity of freespace, and 0 = 4 107 kg m s2 A2 (definition) is the permeability of free space.Via the force laws (see below), they serve to define the magnitudes of SI electrical unitsin terms of the units of mass, length and time. They are not independent since 00 hasthe units of 1/speed2 indeed, as we review shortly, 00 = 1/c
2. By convention, wechoose 0 to take the value given here, which then defines the unit of current (ampere)and hence charge (coulomb or ampere second). Since we define the metre in terms ofthe second (the latter defined with reference to the frequency of a particular hyperfinetransition in caesium) such that c = 2.99792458 m s1, the value of 0 is then fixed as1/(0c
With the inclusion of the displacement current 00E/t in Eq. (1.4), the continuityequation is satisfied, which expresses the conservation of electric charge:
t+ J = 0 . (1.5)
Maxwells equations tell us how charges and currents generate electric and magneticfields. We also need to know how charged particles move in electromagnetic fields, and
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this is governed by the Lorentz force law :
F = q(E + v B) , (1.6)
where q is the charge on the particle and v is its velocity. The continuum form of theforce equation follows from summing over a set of charges; the force density (force pervolume) is then
f = E + JB , (1.7)
in terms of the charge and current densities.
Maxwells equations are a remarkable triumph of 19th century theoretical physics. Inparticular, they unify the phenomena of not only electricity and magnetism but alsolight. To see this, consider Maxwells equations in free space ( = 0 and J = 0). Takingthe curl of Eq. (1.2), and using Eqs (1.1) and (1.4), gives1
2E = 002E
This is a wave equation for the electric field with wave speed c = 1/00. The
equivalent equation for B follows from Eq. (1.4). We see that electromagnetic wavespropagate in free space with speed that is the speed of light in vacuum.
Despite these successes, there is an apparent issue with Maxwells equations in that theyare not invariant under Galilean transformations. Recall that these are transformationsof the form
t t , x x ut , (1.9)
and so v v u, where u is the relative velocity of two reference frames. Newtonslaw dp/dt = F (where p = mv with m the mass of the particle) is invariant undersuch transformations since forces are frame-invariant in Newtonian dynamics. TheLorentz force law can be made Galilean invariant if it is assumed that the electricand magnetic fields transform under Galilean transformations as E E + uB andB B. However, there is a problem in that Maxwells equations predict waves thatpropagate at the fixed speed c, but under Galilean transformations the velocity of awave must change in the same way as particle velocities under a change of frame.One possible way out of this is to assume that Maxwells equations are not Galileaninvariant in that they only hold in one particular reference frame. The problem withthis is that, unlike the equations governing the propagation of sound waves in a gassay, it is not clear what defines the preferred frame. This interpretation began to lookincreasingly untenable with experiments (by Fizeau and Michelson and Morley in thesecond half of the 19th century) on the velocity of light in moving fluids, and this was amajor motivation for Einstein introducing the framework of special relativity in 1905.Within special relativity, it is Galilean invariance and Newtonian mechanics that are
1Recall the identity (E) = ( E)2E.
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replaced by new relativity principles, while Maxwells equations remain form-invariantunder these new (Lorentz) transformations.
In this course we shall establish the form-invariance of Maxwells equations underLorentz transformations. In doing so, we shall learn how to rewrite electromagnetismin a more covariant formalism that makes manifest its invariance under Lorentz trans-formations. With this streamlined formalism, we shall discuss the generation of elec-tromagnetic radiation by accelerating charges and develop action principles for theelectromagnetic field, which is an important step towards developing a quantum ver-sion of the theory (see Part-III courses in Quantum Field Theory).
2 Review of electrostatics
We begin with a brief review of electrostatics. Here, we deal with static charge distri-butions, i.e., /t = 0 and J = 0. Maxwells equations for the electric and magneticfields decouple in static situations, and those for the electric field become
E = 0, E = 0 . (2.1)
Since E has vanishing curl, we can write it in terms of the gradient of the electrostaticpotential (x) as E = . Combining with E = /0 gives the Poisson equation:
2 = /0 . (2.2)
A typical problem in electrostatics is to determine the field within some region S dueto specified charges within S, and subject to appropriate boundary conditions on the(closed) boundary S. In such situations, Poissons equation has a unique solutionsubject to Dirichlet boundary conditions ( specified on S), or a unique solution upto an irrelevant constant for Neumann boundary conditions (the normal derivative of specified on S). Of course, these boundary conditions must result from other chargesoutside the region of interest.
Here, we are interested in the field of localised charged distributions with no boundarysurfaces. In this case, we can assume that the potential decays to zero at spatial infinity.Since the Poisson equation is linear, we can solve it with the help of an appropriateGreens function, G(x; x):
(x) = 10
G(x; x)(x) d3x . (2.3)
The Greens function has to satisfy 2G(x; x) = (3)(x x), where the Dirac delta
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function appears on the right, and tend to zero as |x| . Homogeneity and isotropy2demand that the Greens function only depends on |x x|. Taking x = 0, G is thena function of only r |x|. Away from the origin, we have
)= 0 , (2.4)
for which the solution that decays at infinity is G = A/r. We can fix the integrationconstant A by integrating 2G = (3)(x) over some region S that encloses the origin.Using the divergence theorem, we have
G dS = 1 . (2.5)
Taking the region S to be a sphere of radius R, the integral on the left evaluates to(A/R2) 4R2 and so A = 1/(4). The appropriate Greens function is thus
G(x; x) =1
4|x x|, (2.6)
and the solution of Poissons equation is
|x x|d3x . (2.7)
For a point charge at the origin, (x) = q1(3)(x), and the potential = q1/(40r).
The associated electric field isE =
x , (2.8)
where x = x/r is a unit vector in the direction of x. The Lorentz force law (1.6) thengives the force on a charge q2 in the field of q1 as
40|x1 x2|3(x2 x1) , (2.9)
where we have generalised to have q1 at x1 and q2 at x2. This is simply Coulombs(inverse-square) law. Note that the force is a power-law and therefore has no preferredscale. Ultimately, this reflects the fact that, in a quantum treatment, the gauge bosonthat carries the electromagnetic force (the photon) is massless. If the photon did havemass, this would introduce a scale into the problem and the potential of a point chargewould typically have the Yukawa form, er/l/r for some scale-length l inverselyproportional to the mass. As an example of such force carriers, the gauge bosons thatcarry the weak and strong forces are massive with sub-atomic associated length-scales.
2Under rigid transformations of the charge density, i.e., translations and rotations, the potentialshould transform similarly. For example, if the charges are all translated by a, the potential shouldtranslate to (x) (x a).
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2.1 Multipole expansions
We now ask the question how does the potential and field due to a localised chargedistribution vary at distances large compared to the size of the distribution? We mightexpect that if there is a non-zero total charge Q, asymptotically the field will look likethat due to a point charge Q. This is indeed correct, but what happens if the totalcharge vanishes? To answer this, we consider the multipole expansion of Eq. (2.7).
Consider a charge distribution localised within a region of extent a. We aim to calculatethe fields at distances from the charges that are large compared to a. As we shallsee, these are generally given in terms of simple low-order multipoles of the chargedistribution. Choose the origin close to the charges, so that all charges are within a ofthe origin. Observe at x where r |x| a. We now deal with the 1/|x x| factor inEq. (2.7) by expanding in the small quantity |x|/r. Using
|x x|2 = x x 2x x + x x , (2.10)
|x x|1 = 1r
(1 2x x