Slowly Varying Fields in Vacuum
3.5 The Wave Equation in Vacuum
3.5.2 Spherical Waves
To obtain the solution to the vector wave equations (3–59) or (3–60), we employ a trick that allows us to generate the vector solutions from the solutions of the scalar equation, (3–61). For that reason we consider initially solutions to the scalar equation
∇2ψ(r, θ, ϕ) = 1 c2
∂2ψ
∂t2 (3–64)
Using separation of variables (explored at length in Chapter 5) in spherical polar coordinates, we obtain the solution
ψ(r, θ, ϕ) =
[Aj(kr) +Bn(kr)] Ym (θ, ϕ)e−iωt (3–65) where Ymis a spherical harmonic of order, m; j(z) and n(z) are spherical Bessel functions and spherical Neumann functions of orderandA andB are arbitrary constants.
The vector solution can generally be generated from the scalar solution by the following stratagem. When ψ solves the scalar wave equation, then∇ψ and r×
∇ψ both satisfy the corresponding vector equation (see problem 3–12). Of these solutions, only the second,r×∇ψ =−∇×( rψ), has zero divergence—a requirement for B and also for E in a source-free region of space. If we let E =r×∇ψ, we obtain one kind of solution for the electromagnetic wave, called transverse electric (TE), or M type. Alternatively, if we take B =r×∇ψ, we obtain a transverse magnetic (TM), or E type wave. The function ψ from which the fields may be derived is known as theDebye Potential. A superposition of the two types of waves is the most general possible.
For the case of TE waves we findB from B =∇ × E/iω, while for TM waves we findE fromE =ic2∇ × B/ω. To summarize (after rescaling ψ),
ETE=ik ∇ ×rψ BTE= 1
c∇ × (∇ × rψ) ETM=∇ × (∇ × rψ) BTM=−ik
c ∇ × rψ
(3–66)
where the “constants”ik and 1/chave been introduced to give both the correspond- ing TE fields and TM fields the same dimensions (and the same energy flux).
We may write the components explicitly. Abbreviating the linear combination of spherical Bessel functions, Aj+Bn = f (the combination h(1) = j+in is
required to produce radially expanding waves), we obtain for TE,m waves
Er= 0 Br= 1
c
(+ 1)
r f(kr)Ym (θ, ϕ) Eθ=− km
sinθf(kr)Ym(θ, ϕ) Bθ= 1 cr
d[rf(kr)]
dr
∂Ym(θ, ϕ)
∂θ Eϕ=−ikf(kr)∂Ym(θ, ϕ)
∂θ Bϕ= im crsinθ
d[rf(kr)]
dr Ym(θ, ϕ)
(3–67)
For TM,mwaves, the components are:
Er=(+ 1)
r f(kr)Ym(θ, ϕ) Br= 0 Eθ=1
r
d[rf(kr)]
dr
∂Ym (θ, ϕ)
∂θ Bθ= km
csinθf(kr)Ym(θ, ϕ) Eϕ= im
rsinθ
d[rf(kr)]
dr Ym(θ, ϕ) Bϕ=ik
cf(kr)∂Ym(θ, ϕ)
∂θ
(3–68)
The fields described above are the 2-pole radiation fields. As we will see in Chapter 10, the TE waves are emitted by oscillating magnetic multipoles, while the TM waves are emitted by oscillating electric multipoles. Note that in contrast to the conclusion for plane waves that E and B are both perpendicular tok, for spherical waves only one of the two can be perpendicular tok.
Exercises and Problems
Figure 3.7:The voltmeter is attached to the loop ensuring that its leads do not add any area to the loop. (Exercise 3.18)
3-1 Show that the flux Φ threading a loop Γ may be written in terms of the vector potential as
Φ =
Γ
A·d
3-2The flux threading a single turn coil is gradually reduced, resulting in an in- duced current in the coil. Find the mag- netic field produced at the center of the coil by the induced current of a coil with resistanceR and radiusa.
3-3 In aBetatron electrons are acceler- ated by an increasing magnetic flux den- sity (as in example 3.3). How must the field at the orbit relate to the average field in order that the electron’s orbit re- main of constant radius?
3-4Find the flux in a toroidal coil such as that illustrated in Figure 1.13 carry- ing currentI. The mean radius of the coil isa, and the cross-sectional radius of the tube isb.
3-5A currentI charges a parallel plate capacitor made of two circular plates each of area A spaced at small distance d. Find the magnetic induction field be- tween the plates.
3-6 Show that the magnetic induction field encircling the capacitor of problem 3-5 at a large enough distance for the electric field to vanish is given by
Bϕ=µ0I 2πr
Thus at a sufficient distance, it is impos- sible to tell from the magnetic field that the circuit is broken by the capacitor.
3-7A pendulum consisting of a conduct- ing loop of radiusa and resistanceR at the end of a massless string of length swings through an inhomogeneous mag- netic induction fieldB perpendicular to the plane of the loop. Write an equation of motion for the pendulum.
3-8Aflip coil is a rectangular coil that is turned through 180◦ fairly rapidly in a static magnetic induction field.
Show that the total charge transported through the coil as it is flipped is inde- pendent of the speed of flipping. (As- sume a finite resistance.)
3-9The Lagrangian of a charged parti- cle moving in an electromagnetic field is L = 12mr˙2−qV(r) +qr˙ ·A( r). Show that thei-component ofp, pi =∂L/∂x˙i
yields the canonical momentum on the left hand side of (3–38).
3-10 Show that, for plane waves, ∇ × E =ik×E and∇ · E =ik·E.
3-11Obtain the wave equation forB (3–
60) to verify that it is indeed identical to that forE.
3-12 In order to ionize an atom, one needs to overcome the binding energy of the electron (of order 10 eV) in a dis- tance of several atomic radii (say 10−9 m) in order to provide a reasonable tun- nelling rate. Focused ruby pulse lasers can produce air sparks. Estimate the electric field strength and hence the irra- diance of the laser light required to pro- duce air sparks.
3-13 Show that if ψ solves the scalar wave equation, then ∇ψ and r ×∇ψ each solve the corresponding vector wave equation.
3-14Fill in the missing steps in the last step of the footnote (page 59) to show that
A[r(t)+vdt, t]−A [r(t), t] = (v·∇) Adt 3-15 Show that a gauge transformation
with Λ satisfying∇2Λ = 0 preserves the Coulomb gauge.
3-16 Find the condition that a gauge function Λ needs to satisfy in order to preserve the Lorenz gauge.
3-17 A circular disk of conducting ma- terial spins about its axis in a magnetic field parallel to the axis. Find the EMF generated between the axis and a fixed point on the rim of the disk.
3-18 Resolve the following paradox. A voltmeter has its terminals attached to diametrically opposite points of a circu- lar loop. From their points of attach- ment, the leads follow the curve of the loop and, after meeting, continue to the voltmeter as a twisted pair, as in Fig- ure 3.7. A time-varying flux threads the loop. When the voltmeter leads are con- sidered as part of the loop, the EMF measured appears to depend on whether the top half of the loop completes the circuit (Figure 3.7b) in which case the flux is zero, or the bottom half (Figure 3.7c) completes it, in which case the flux is just that through the original loop.