Instructor Solution Manual
MODERN ELECTRODYNAMICS
Andrew Zangwill
Georgia Institute of Technology
June 2014
A Note from the Author
This manual provides solutions to the end-of-chapter problems for the author’s Modern
Electrodynamics. The chance that all these solutions are correct is zero. Therefore, I will be
pleased to hear from readers who discover errors. I will also be pleased to hear from readers
who can provide a better solution to this or that problem than I was able to construct. I
urge readers to suggest that this or that problem should not appear in a future edition of
the book and (equally) to propose problems (and solutions) they believe should appear in a
future edition.
At a fairly advanced stage in the writing of this book, I decided that a source should be
cited for every end-of-chapter problem in the book. Unfortunately, I had by that time spent
a decade accumulating problems from various places without always carefully noting the
source. For that reason, I encourage readers to contact me if they recognize a problem of their
own invention or if they can identify the (original) source of any particular problem in the
manual. An interesting issue arises with problems I found on instructor or course websites
which were taken down after the course they serviced had concluded. My solution has been
to cite the source of these problems as a “public communication” between myself and the
course instructor. This contrasts with problems cited as a true “private communication”
between myself and an individual.
ii
Chapter 1
Mathematical Preliminaries
Chapter 1: Mathematical Preliminaries
1.1
Levi-Cività Practice I
(a) 123 = ê1 · (ê2 × ê3 ) = ê1 · ê1 = 1. The cyclic property of the triple scalar product
guarantees that 231 = 312 = 1 also. Similarly, 132 = ê1 · (ê3 × ê2 ) = −ê1 · ê1 = −1
with 321 = 213 = −1 also. Finally, 122 = ê1 · (ê2 × ê2 ) = 0 and similarly whenever
two indices are equal.
(b) Expand the determinant by minors to get
a × b = ê1 (a2 b3 − a3 b2 ) − ê2 (a1 b3 − a3 b1 ) + ê3 (a1 b2 − a2 b1 ).
Using the Levi-Cività symbol to supply the signs, this is the same as the suggested
identity because
a × b = 123 ê1 a2 b3 + 132 ê1 a3 b2
+ 213 ê2 a1 b3 + 231 ê2 a3 b1
+ 312 ê3 a1 b2 + 321 ê3 a2 b1 .
(c) To get a non-zero contribution to the sum, the index i must be different from the unequal
indices j and k, and also different from the unequal indices s and t. Therefore, the
pair (i, j) and the pair (s, t) are the same pair of different indices. There are only
two ways to do this. If i = s and j = t, the terms are identical and their square
is 1. This is the first term in the proposed identity. The other possibility introduces
a transposition of two indices in one of the epsilon factors compared to the previous
case. This generates an overall minus sign and thus the second term in the identity.
(d) The scalar of interest is S = L̂m am L̂p bp − L̂q bq L̂s as . Using the given commutation
relation,
S
= am bp L̂m L̂p − ap bm L̂m L̂p
= am bp L̂m L̂p − am bp L̂p L̂m
= am bp [L̂m , L̂p ]
= ih̄m pi L̂i am bp
= ih̄L̂i im p am bp
= ih̄L̂ · (a × b).
1
Chapter 1
1.2
Mathematical Preliminaries
Levi-Civitá Practice II
(a) δii = 1 + 1 + 1 = 3
(b) δij ij k = iik = 0
(c) ij k j k = j k i j k = δk k δi − δk δik = 3δi − δi = 2δi
(d) ij k ij k = δj j δk k − δj k δk j = 9 − δk k = 6
1.3
Vector Identities
(a) (A × B) · (C × D)
= ij k Aj Bk im p Cm Dp = ij k im p Aj Bk Cm Dp
=
(δj m δk p − δj p δk m )Aj Bk Cm Dp
= Am Cm Bk Dk − Aj Dj Bk Ck = (A · C)(B · D) − (A · D)(B · C)
(b)
∇ · (f × g)
= ∂i ij k fj gk = ij k fj ∂i gk + ij k gk ∂i fj = fj j k i ∂i gk + gk k ij ∂i fj
= gk k ij ∂i fj − fj j ik ∂i gk = g · (∇ × f ) − f · (∇ × g)
(c) [(A × B) × (C × D)]i
= ij k {A × B}j {C × D}k = ij k j m p k st Am Bp Cs Dt
= j k i j m p k st Am Bp Cs Dt = (δk m δip − δk p δim )k st Am Bp Cs Dt
= k st Ak Bi Cs Dt − k st Ai Bk Cs Dt = Ak k st Cs Dt Bi − Bk k st Cs Dt Ai
= A · (C × D)Bi − B · (C × D)Ai
(d) (σ·a)(σ·b) = σi ai σj bj = σi σj ai bj = (δij +iij k σk )ai bj = ai bi +ik ij σk ai bj = a·b+iσ·(a×b)
1.4
Vector Derivative Identities
(a) ∇ · (f g) = ∂i (f gi ) = f ∂i gi + gi ∂i f = f ∇ · g + (g · ∇)f
(b) {∇ × (f g)}i = ij k ∂j (f gk ) = f ij k ∂j gk + ij k (∂j f )gk = f [∇ × g]i + [∇f × g]i
2
Chapter 1
Mathematical Preliminaries
(c)
[∇ × (g × r)]i
= ij k ∂j k m g rm
= k ij k m ∂j (g rm )
=
(δi δj m − δim δj )∂j (g rm )
= ∂j (gi rj ) − ∂j (gj ri )
= rj ∂j gi + gi (∇ · r) − ri (∇ · g) − δij gj
=
(r · ∇)gi + 3gi − ri (∇ · g) − gi
Therefore,
∇ × (g × r) = 2g + r
1.5
∂g
− r(∇ · g).
∂r
Delta Function Identities
(a) Let f (x) be an arbitrary function. Then, if a > 0, a change of variable to y = ax gives
∞
−∞
1
dxf (x)δ(ax) =
a
∞
dyf (y/a)δ(y) =
−∞
1
f (0).
a
However, if a < 0,
∞
−∞
1
dxf (x)δ(ax) =
a
−∞
∞
1
1
dyf (y/a)δ(y) = −
dyf (y/a)δ(y) = − f (0).
a
a
∞
−∞
These two results are summarized by δ(ax) =
1
δ(x).
|a|
(b) If g(x0 ) = 0, δ[g(x)] is singular at x = x0 . Very near this point, g(x) ≈ (x − x0 )g (x0 ).
Therefore, using the identity in part (a),
∞
∞
dxf (x)δ[g(x)] ≈
−∞
dxf (x)g[(x − x0 )g (x0 )] =
−∞
1
|g (x0 )|
δ(x − x0 ).
A similar contribution comes from each distinct zero xm . Adding these together gives
the advertised result.
3
Chapter 1
Mathematical Preliminaries
(c) We use the result of part (b). The zeroes of cos x occur at x = (2n + 1)π/2. At these
points, | − sin x| = 1; therefore,
I=
∞
exp([−(2n + 1)π/2] = exp(−π/2)
n =0
1.6
∞
exp(−nπ) =
n =0
1
exp(−π/2)
=
.
1 − exp(π)
2 sinh(π/2)
Radial Delta Functions
(a) We need to show that δ(r)/r and −δ (r) have the same effect when multiplied by an
arbitrary function and integrated over the radial part of a volume integral. If we
call the arbitrary function f (r), one of these integrals vanishes identically because
rδ(r) = 0:
∞
0
δ(r)
=
dr r f (r)
r
∞
2
dr r f (r) δ(r) = 0
0
This tells us we need to represent the arbitrary function in a smarter way. One
possibility is f (r)/r. This gives
∞
dr r2
0
f (r) δ(r)
=
r
r
∞
dr f (r) δ(r) = f (0).
0
An integration by parts shows that the proposed identity is correct:
∞
∞
∞
2 f (r) δ (r) = −
dr r
dr r f (r) δ (r) =
dr δ(r) [rf (r)]
−
r
0
0
0
∞
=
dr δ(r) [f (r) + rf (r)] = f (0).
0
(b) By direct calculation,
∇ · [δ(r − a)r̂] =
2
1 ∂ 2
r δ(r − a) = δ(r − a) + δ (r − a).
r2 ∂r
r
(1)
Let us look at the effect of δ (r − a) on an arbitrary test function:
∞
d
drr f (r) δ(r − a)
dr
2
0
∞
=
d δ(r − a)r2 f −
dr
dr
0
∞
δ(r − a)
d 2 r f
dr
0
∞
= −
df
2 df drδ(r − a) 2rf + r
.
= −2af (a, θ, φ) − a
dr
dr r =a
2
0
This shows that
a2
2
δ (r − a) = − δ(r − a) + 2 δ (r − a).
a
r
Combining this with (1) shows that
∇ · [δ(r − a)r̂] = (a2 /r2 )δ (r − a).
4
Chapter 1
Mathematical Preliminaries
Source: R. Donnelly, Journal of the Optical Society of America A 10, 680 (1993).
1.7
A Representation of the Delta Function
The calculation involves a change of variable,
∞
∞
sin mx
= lim
= lim
dxf (x)
m →∞ −∞
m →∞
πx
∞
1
sin y
= f (0)
.
dy
π −∞
y
dxf (x)D(x)
−∞
∞
dy y sin y m
f( )
m m π y
−∞
The assertion is proved if the integral on the far right side is equal to π. You can look up
the integral or use this trick:
∞
∞
∞
∞
∞ ∞
sin y
sin y
= 2
=
dy
dy
dy sin y
dνe−ν y = 2
dν
dy e−ν y sin y
y
y
−∞
0
0
0
0
0
∞
dν
= 2
= π.
1 + ν2
0
1.8
An Application of Stokes’ Theorem
(a) Let p = ∇ × (c × F). Then, because c is a constant vector,
pi = ij k ∂j k st cs Ft = k ij k st cs ∂j Ft = (δis δj t − δit δj s )cs ∂j Ft = ci ∂j Fj − cj ∂j Fi .
This shows that ∇ × (c × F) = c(∇ · F) − (c · ∇)F . Inserting this into Stokes’ Theorem
as suggested gives
dS n̂ · {c(∇ · F) − (c · ∇)F} =
ds · (c × F) =
S
C
or
c·
c · (F × ds)
C
dS {n̂(∇ · F) − n̂i ∇Fi } = c ·
S
F × ds.
C
This establishes the equality because c is arbitrary.
dS(n̂ × ∇) × F. Then
(b) Let K =
S
Ki
dS ij k (n̂ × ∇)j Fk =
=
S
dS ij k j st n̂s ∂t Fk
S
dS (δk s δit − δk t δis )n̂s ∂t Fk =
=
S
dS (n̂k ∂i Fk − n̂i ∂k Fk ).
S
dS {n̂i ∇Fi − n̂(∇ · F)}, which was the second equality in
This proves that K =
S
question.
5
Chapter 1
Mathematical Preliminaries
(c) This is a special case of the identity in part (a) with F = r. Therefore,
r × ds = −
C
dS {n̂i ∇ri − n̂(∇ · r)} .
S
Now, ∇ · r = 3. Also, ∇ri = êj ∂j ri = êj δij = êi so n̂i ∇ri = n̂i êi = n̂. Hence,
r × ds = −
C
1.9
dS(n̂ − 3n̂) = 2
S
dS.
S
Three Derivative Identities
(a) Consider the x-component of the gradient. We have
∂
∂
f (x − x , y − y , z − z ) = − f (x − x , y − y , z − z )
∂x
∂x
and similarly for the y and z components. This proves the assertion.
(b) Writing this out in detail,
∇ · [A(r) × r]
= ∂i ij k Aj rk = ij k [rk Aj (r)∂i r + Aj (r)∂i rk ]
= ij k Aj
rk ri
+ ij k Aj δik
r
= r̂ · A (r) × r + ij i Aj (r)
= A (r) · r × r + 0 = 0.
(c) By definition, dA =
dA
dA
dA
dx +
dy +
dz. Therefore, since ds = x̂dx + ŷdy + ẑdz,
dx
dy
dz
dA
dA
dA
d
d
d
dA = dx
+ dy
+ dz
= dx
+ dy
+ dz
A = (ds · ∇)A.
dx
dy
dz
dx
dy
dz
1.10
Derivatives of exp(ik · r)
As a preliminary, let ψ(r) = exp(ik · r) and consider the derivative
∂ ik x ik y ik y ∂ψ
=
e e e
= ikx ψ.
∂x
∂x
6
Chapter 1
Mathematical Preliminaries
The y and z derivatives are similar. We conclude from this that
∇ψ = ikψ.
Therefore, because c is a constant vector,
∇ · A = ψ(∇ · c) + c · ∇ψ = ik · A
∇ × A = ψ(∇ × c) − c × ∇ψ = −i(c × k)ψ = ik × A
∇ × (∇ × A) = ∇ × (ik × A) = −i(k · ∇)A + ik(∇ · A) = k 2 A − k(k · A) = −k × (k × A)
∇(∇ · A) = i∇(k · A) = i [k × (∇ × A) + (k · ∇)A] = i k × (ik × A) + ik2 A = −k(k · A)
∇2 A = ∇(∇ · A) − ∇ × (∇ × A) = −k(k · A) + k × (k × A) = −k 2 A.
1.11
Some Integral Identities
(a) By direction substitution,
d 3 r F · G = d 3 r ∇ϕ · G = d 3 r [∇ · (ϕG) − ϕ∇ · G] = dS · Gϕ = 0.
The last integral is zero with the stated conditions at infinity.
(b) Following the example of part (a),
d 3 r F × G = d 3 r ∇ϕ × G = d 3 r [∇ × (Gϕ) − ϕ∇ × G] = dS × Gϕ = 0.
The last integral is zero with the stated conditions at infinity.
(c) The given vector is ∂j (Pj G) = (∇ · P)G + (P · ∇)G. Integrate the given identity over
a volume V to get
d 3 r ∂j (Pj G) = d 3 r (∇ · P)G + d 3 r (P · ∇)G.
V
Therefore,
V
dS(n̂ · P)G =
S
V
d r (∇ · P)G +
d 3 r (P · ∇)G.
3
V
V
The choice G = r produces the desired identity because (P · ∇)r = P.
7
Chapter 1
1.12
Mathematical Preliminaries
Unit Vector Practice
From Chapter 1,
r̂ = x̂ sin θ cos φ + ŷ sin θ sin φ + ẑ cos θ
x̂ = r̂ sin θ cos φ + θ̂ cos θ cos φ − φ̂ sin φ
θ̂ = x̂ cos θ cos φ + ŷ cos θ sin φ − ẑ sin θ
ŷ = r̂ sin θ sin φ + θ̂ cos θ sin φ + φ̂ cos φ
φ̂ = −x̂ sin φ + ŷ cos φ
ẑ = r̂ cos θ − θ̂ sin θ.
By direct calculation,
∂r̂
= x̂ cos θ cos φ + ŷ cos θ sin φ − ẑ sin θ = θ̂
∂θ
∂r̂
= −x̂ sin θ sin φ + ŷ sin θ cos φ = − sin θφ̂.
∂φ
∂ θ̂
= −x̂ sin θ cos φ − ŷ sin θ sin φ − ẑ cos θ = −r̂
∂θ
∂ θ̂
= −x̂ cos θ sin φ + ŷ cos θ cos φ = cos θφ̂
∂φ
∂ φ̂
=0
∂θ
∂ φ̂
= −x̂ cos φ − ŷ sin φ = − sin θr̂ − cos θθ̂.
∂φ
1.13
Compute the Normal Vector
By definition,
n̂ =
(x/a2 )x̂ + (y/b2 )ŷ + (z/c2 )ẑ
∇Φ
=
.
|∇Φ|
(x2 /a4 ) + (y 2 /b4 ) + (z 2 /c4 )
When a = b = c, the foregoing reduces to
n̂ =
1.14
xx̂ + yŷ + zẑ
x2
+
y2
+
z2
=
r
= r̂.
r
A Variant of the Helmholtz Theorem I
Following our proof of the Helmholtz theorem,
ϕ(r) =
V
1
d r ϕ(r )δ(r − r ) = −
4π
3 V
1
1
= ∇·
d r ϕ(r )∇
|r − r |
4π
3 Using an elementary vector identity gives
8
2
V
d 3 r ϕ(r )∇
1
.
|r − r |
Chapter 1
Mathematical Preliminaries
ϕ(r) = ∇ ·
1
4π
d 3 r
∇ ϕ(r )
ϕ(r )
−
|r − r |
|r − r |
∇
V
.
On the other hand, for any scalar function ψ,
d r ∇ψ =
3
V
dS ψ.
S
Using this to transform the first term above gives the desired result,
ϕ(r) = −∇ ·
1
4π
d 3 r
V
1
∇ ϕ(r )
+∇·
|r − r |
4π
dS
S
ϕ(r )
.
|r − r |
Source: D.A. Woodside, Journal of Mathematical Physics 40, 4911 (1999).
1.15
A Variant of the Helmholtz Theorem II
From the textbook discussion of the Helmholtz theorem,
Z(r)
1
= − ∇
4π
V
+
1
∇
4π
1
∇ · Z(r )
+
∇×
d r
|r − r |
4π
3 d 3 r ∇ ·
V
d 3 r
V
Z(r )
1
∇×
−
|r − r |
4π
∇ × Z(r )
|r − r |
d 3 r ∇ ×
V
Z(r )
.
|r − r |
The first two integrals are zero because ∇ · Z = 0 and ∇ × Z = 0 in V . The divergence
theorem transforms the third term into an integral over S. Chapter 1 of the text states a
corollary of the divergence theorem that similarly transforms the fourth term into a surface
integral. The final result is
Z(r) =
1
∇
4π
dS S
1
n̂(r ) · Z(r )
−
∇×
|r − r |
4π
S
dS n̂(r ) × Z(r )
.
|r − r |
Knowledge of Z at every point of the surface S permits us to compute the required factors
n̂(r ) · Z(r ) and n̂(r ) × Z(r ).
1.16
Densities of States
This problem exploits the delta function identity
δ[g(x)] =
1
n
|g (xn )|
δ(x − xn ),
where
9
g(xn ) = 0, g (xn ) = 0.
Chapter 1
Mathematical Preliminaries
√
(a) Here, g(kx ) = E − kx2 = 0 when kx = ± E. Moreover, g (kx ) = −2kx . Therefore,
∞
g1 (E) =
−∞
√
√ 1
1
δ(kx − E) + δ(kx + E) = √ .
dkx √
2 E
E
(b) It is simplest to switch to polar coordinates in two dimensions, so
g2 (E) =
2π
d k δ(E − k ) =
2
2
∞
dk kδ(E − k ) = 2π
dφ
0
∞
2
0
0
√
k
dk √ δ(k − E) = π.
2 E
(c) It is simplest to switch to spherical coordinates in three dimensions and write
∞
d k δ(E − k ) = 4π
3
g3 (E) =
2
0
1.17
4π
dk k δ(E − k ) = √
2 E
2
∞
2
√
√
dk k 2 δ(k − E) = 2π E.
0
Dot and Cross Products
(a)
bi
= bj n̂j n̂i + ij k n̂j k m b n̂m
= bj n̂j n̂i + k ij k m n̂j b n̂m
= bj n̂j n̂i + (δi δj m − δim δj )n̂j b n̂m
= bj n̂j n̂i + n̂j bi n̂j − n̂j bj n̂i
= bi n̂j n̂j = bi .
(b) The given formula is b = b + b⊥ where b is a vector parallel to n̂ and b⊥ is a vector
perpendicular to n̂.
(c)
(B × C)i
= ij k Bj Ck /ω 2
= ij k j m k st c am as bt /ω 2
=
(δk l δim − δk m δi )k st c am as bt /ω 2
= k st [ck ai − ci ak ]as bt /ω 2 .
Therefore,
ω 2 (B × C) = c · (a × b)a − a · (a × c)c = c · (a × b)a.
Hence,
Ω = A · (B × C) =
a · (b × b)c · (a × b)
|a · (c × c)|2
ω2
1
=
=
= .
3
3
3
ω
ω
ω
ω
10
Chapter 1
1.18
Mathematical Preliminaries
Sij and Tij
(a) It is sufficient to write
ij k Sij =
1
1
ij k Sij + ij k Sij .
2
2
Relabel the dummy indices in the second term to get
ij k Sij =
1
1
1
1
1
ij k Sij + j ik Sj i = ij k Sij − ij k Sj i = ij k (Sij − Sj i ).
2
2
2
2
2
This will be zero if Sij = Sj i .
(b) We have yi = bk Tk i = ik s bk ωs . Therefore, Tk i = ik s ωs . Notice that this representation
requires that Tik = −Tk i . Now, multiply by ipq and sum over i:
ipq Tk i
= ipq ik s ωs
= ωs (δpk δq s − δps δq k )
= ωq δpk − ωp δq k .
This is true for all values of p, q, and k. Choose p = k and sum over k:
ik q Tk i = ωq δk k − ωp δq k = 3ωq − ωq = 2ωq .
Therefore,
ωq =
1
ik q Tk i .
2
This is not an unreasonable result because Tij = −Tj i implies that T has only three
independent components, just like ω:
0
T = −ω1
−ω2
1.19
ω1
0
−ω3
ω2
ω3
0
.
Two Surface Integrals
(a) A corollary of the divergence theorem is
d 3 r ∇ψ. Put ψ = const. to get the
dSψ =
S
V
desired result.
(b) By the divergence theorem,
dS · r =
S
d r∇·r = 3
3
V
d 3 r = 3V.
V
11
Chapter 1
1.20
Mathematical Preliminaries
Electrostatic Dot and Cross Products
Begin with
ϕ = ij k aj rk ipt bp rt = (δj p δk t − δj t δpk )aj rk bp rt = (a · b)r2 − (a · r)(b · r).
Therefore,
Ei = −∂i [rs rs (a · b) − am bp rm rp ] = −2ri (a · b) + am bp (δim rp + rm δip ),
or
E = −2r(a · b) + a(b · r) + b(a · r).
Now, ∇ · r = 3 and
∇ · [a(b · r)] = ∂k ([ak bi ri ]) = ak bi δik = a · b.
Therefore,
ρ = 0 ∇ · E = 0 [−6(a · b) + 2(a · b) = −40 (a · b).
1.21
A Decomposition Identity
1
ij k (A × B)k
2
=
1
ij k k m A Bm
2
=
1
k ij k m A Bm
2
=
1
(δi δj m − δim δj )A Bm
2
=
1
(Ai Bj − Aj Bi ).
2
Therefore,
1
1
ij k (A × B)k + (Ai Bj + Aj Bi ) = Ai Bj .
2
2
12
Chapter 2
The Maxwell Equations
Chapter 2: The Maxwell Equations
2.1
Measuring B
The Lorentz force on the particle moving with velocity υ 1 is
F1 = qυ 1 × B.
Taking the cross product with υ 1 gives
υ 1 × F1 = qυ 1 × (υ1 × B) = q υ1 (υ 1 · B) − Bυ12 .
Therefore,
B=−
υ 1 × F1
(υ 1 · B)υ 1
+
.
2
qυ1
υ12
B=−
(υ 2 · B)υ 2
υ 2 × F2
+
.
qυ22
υ22
Similarly,
(1)
The dot product of υ 1 with the preceding equation is
υ1 · B = −
υ 1 · (υ 2 × F2 ) (υ 2 · B)(υ 2 · υ 1 )
+
.
qυ22
υ22
The last term above vanishes if υ 1 ⊥ υ 2 and the result can be substituted into (1) to get
an explicit formula for B:
B=−
υ 1 · (υ 2 × F2 )
υ 1 × F1
−
υ1 .
qυ12
qυ12 υ12
Source: J.R. Reitz and F.J. Milford, Foundations of Electromagnetic Theory (AddisonWesley, Reading, MA, 1960).
2.2
The Coulomb and Biot-Savart Laws
1
1
(a) Use ∇(1/r) = −r/r to write E(r) = −
. Then,
d 3 r ρ(r )∇
4π0
|r − r |
1
1
= 0.
∇ × E(r) = −
d 3 r ρ(r )∇ × ∇
4π0
|r − r |
3
Similarly, because ∇2 (1/r) = −4πδ(r),
∇ · E(r) = −
1
4π0
d 3 r ρ(r )∇2
1
1
=
|r − r |
0
13
d 3 r ρ(r δ(r − r ) = ρ(r)/0 .
Chapter 2
The Maxwell Equations
(b) Here we write
µ0
µ0
µ0
1
j(r )
3 3 j(r )
=
=
∇×
d
.
B(r) = −
r
∇×
r
d 3 r j(r )×∇
d
4π
|r − r |
4π
|r − r |
4π
|r − r |
This gives ∇ · B(r) = 0 because ∇ · ∇ × f = 0 for any f . To compute the curl of B,
let
r − r
R
= 3,
g(r − r ) =
|r − r |3
R
so
µ0
d 3 r ∇ × [j(r ) × g(r − r )]
∇ × B(r) =
4π
µ0
µ0
3 (1)
= −
d r [j(r ) · ∇] g +
d 3 r j(r )∇ · g.
4π
4π
Focus on the first integral. We know that [j(r ) · ∇] g(r − r ) = − [j(r ) · ∇ ] g(r − r ).
Therefore,
[j(r ) · ∇ ] gx (R) = ∇ · [gx (R)j(r )] − gx (R)∇ · j(r ).
(2)
The charge and current density are time-independent so the continuity equation reads
∇·j=−
∂ρ
= 0.
∂t
Accordingly, the second term on the right-hand side of (2) vanishes. Therefore, using
the divergence theorem, the x-component of the first integral in (1) is
µ0
−
4π
µ0
d r ∇ · [gx (R)j(r )] = −
4π
3 dS · j(r )gx (R) = 0.
The integral is zero because j vanishes on the surface at infinity. The y- and zcomponents are zero similarly. Therefore, (1) becomes
∇×B=
But
∇·g =∇·
µ0
4π
d 3 r j(r ) [∇ · g] .
1
1
r − r
= −∇2
= 4πδ(r − r ).
= −∇ · ∇
|r − r |3
|r − r |
|r − r |
Therefore,
∇ × B(r) = µ0
d 3 r j(r )δ(r − r ) = µ0 j(r).
14
Chapter 2
2.3
The Maxwell Equations
The Force between Current Loops
(a)
C1
r1 − r2
ds1 ·
=−
|r1 − r2 |3
C2
1
=
ds1 · ∇1
|r1 − r2 |
C1
∂
1
=
ds1
∂s1 |r1 − r2 |
∂
C1
1
= 0.
|r1 − r2 |
r1 − r2
r1 − r2
r1 − r2
(b) We use the identity ds1 × ds2 ×
.
= ds2 ds1 ·
−(ds1 ·ds2 )
3
3
|r1 − r2 |
|r1 − r2 |
|r1 − r2 |3
Substituting this equation into the given expression for F1 generates two terms. One
of them is zero by part (a). What remains is
µ0
r1 − r2
F1 =
I1 ds1 × I2 ds2 ×
.
4π
|r1 − r2 |3
C1
C2
This is the desired formula because the magnetic field at point r1 produced by a
current loop which carries a current I2 is
B2 (r1 ) =
µ0
4π
I2 ds2 ×
C2
2.4
r1 − r2
.
|r1 − r2 |3
Necessity of Displacement Current
The divergence of the suggested equation is
∇ · ∇ × B = µ0 ∇ · j + ∇ · jD .
The left side is identically zero so, using the continuity equation and ∇ · E = ρ/0 ,
∂ρ
∂
∂E
∇ · jD = −µ0 ∇ · j = µ0
= µ0 0 ∇ · E = ∇ · µ0 0
.
∂t
∂t
∂t
Since µ0 0 = c−2 , this equation is satisfied by the standard form of the displacement current,
jD =
2.5
1 ∂E
.
c2 ∂t
Prelude to Electromagnetic Angular Momentum
The time-changing magnetic field induces an electric field in accordance with the integral
form of Faraday’s law:
d
ds · E = −
dS · B.
dt
C
S
By symmetry, the electric field is azimuthal. Specifically, if we choose C to be a circle of
radius r coaxial with the z-axis,
15
Chapter 2
The Maxwell Equations
1 dΦ
φ̂.
2πr dt
The force qE on the particle produces a torque around the z-axis so the mechanical angular
momentum of the particle is
dL
q dΦ
=r×F=−
ẑ.
dt
2π dt
Therefore, as suggested,
d
qΦ
L+
= 0.
dt
2π
E=−
2.6
Time-Dependent Charges at Rest
(a) The charge density is
ρ(r, t) =
qk (t)δ(r − rk ).
k
We find the current density using the continuity equation, ∇ · j +
∂ρ
= 0. Specifically,
∂t
∂ρ =
q̇k (t)δ(r − rk ) = −∇ · j.
∂t
k
Since ∇ · ∇|r − r |−1 = −4πδ(r − r ), a current density which does the job is
j(r, t) = −
r − rk
1 q̇k
.
4π
|r − rk |3
k
(b) We begin with Gauss’ law:
∇·E =
1 1
r − rk
1 1 2
=
qk (t)∇·
=
−
q
(t)∇
qk (t)δ(r−rk ) = ρ(r)/0 .
k
4π0
|r − rk |3
4π0
|r − rk |
0
k
k
k
The curl of the electric field is
∇×E=−
1 1
= 0.
qk (t)∇ × ∇
4π0
|r − rk |
k
∂B
. This will be satisfied if B(r, t) = B(r) is a time∂t
independent vector field. Using this and c2 = 1/µ0 0 , the Ampère-Maxwell law looks
like
Faraday’s law is ∇ × E = −
∇ × B = µ0 j +
µ0 r − rk
1 ∂E
1 r − rk
=−
q̇k
+
q̇k (t)
= 0.
2
3
2
c ∂t
4π
|r − rk |
4π0 c
|r − r |3
k
k
16
Chapter 2
The Maxwell Equations
We satisfy the equation above and also ∇ · B = 0 if B is a constant vector everywhere
in space. Given the initial conditions, we conclude that
B(r, t) ≡ 0.
(c) The current density j(r, t) in (b) shows that the changes in qk (t) at each point rk occur
because a radial stream of charge flows in and out of each point to and from infinity
as needed.
2.7
Rotation of Free Fields in Vacuum
(a) By assumption,
∇·E=0
∂B
∇×E=−
∂t
Therefore,
∇·B=0
∇×B=
1 ∂E
.
c2 ∂t
∇ · E = (∇ · E) cos θ + c(∇ · B) sin θ = 0
c∇ · B = −(∇ · E) sin θ + c(∇ · B) cos θ = 0
∇ × E
= (∇ × E) cos θ + c(∇ × B) sin θ = −
= −
c∇ × B
∂
∂t
∂B
1
B cos θ − E sin θ = −
c
∂t
= −(∇ × E) sin θ + c(∇ × B) cos θ =
= c
1 ∂E
∂B
cos θ + c 2
sin θ
∂t
c ∂t
1 ∂E
∂B
sin θ + c 2
cos θ
∂t
c ∂t
1 ∂
1 ∂E
(cB sin θ + E cos θ) = c 2
.
2
c ∂t
c ∂t
(b) If E ⊥ B, the stated transformation simply rotates the two vectors by an angle θ while
retaining their perpendicularity. Therefore, E and B also describe a plane wave in
vacuum.
2.8
A Current Density Which Varies Linearly in Time
Let z be the symmetry axis of the solenoid. The simplest guess for the exterior magnetic
field is that the magnetostatic field does not change; namely,
B(ρ > b, t) = 0.
Now consider the integral form of the Ampère-Maxwell law:
17
Chapter 2
The Maxwell Equations
d · B = µ0 Ienc +
1 d
c2 dt
C
dS · E.
S
Symmetry suggests that B(r, t) = B(ρ, t)ẑ and E(r, t) = E(ρ, t)φ̂. Therefore, we choose a
rectangular Ampèrian circuit C of the same kind used to solve the magnetostatic problem.
One leg of length points along −ẑ and lies outside the solenoid. The other leg points along
+ẑ, has length , and lies at a radius ρ < b. The other two legs are aligned with ρ̂. In that
case, the foregoing gives
1 d
Bz = µ0 K0 (t/τ ) + 2
c dt
dS · E.
S
If we guess that the electric field does not depend on time, a solution of this equation is
B(ρ < b, t) = µ0 K0 (t/τ )ẑ.
This magnetic field satisfies ∇ · B = 0 everywhere because the field lines end at infinity only.
We turn next to Faraday’s law and choose a circular loop C lying in the x-y plane with
radius ρ:
d
d · E = −
dS · B.
dt
C
S
Evaluating this for ρ < b and ρ > b gives
⎧
1
⎪
⎪
µ K ρφ̂
⎪
⎨ 2τ 0 0
E(ρ, t) =
⎪
⎪
b2
1
⎪
⎩
µ0 K0 φ̂
2τ
ρ
ρ < b,
ρ > b.
This electric field satisfies ∇ · E = 0 everywhere because the electric field lines close on
themselves.
2.9
A Charge Density Which Varies Linearly in Time
It is clear that there is a positive inward flux of charge at the origin, and so the charge
density at the origin must be increasing. Away from the origin, every volume element has
both and inward and outward flux. However, these are not equal, and so charge is actually
building up in every volume element (at, it turns out, the same rate). To see this, consider
a volume dV = r2 drdΩ centered on an arbitary point x. Charge flows radially inward into
the surface area element at r + dr with area (r + dr)2 dΩ; the current is entirely in the
radial direction with jr = −(α/3)(r + dr). Thus the flux through this surface (and into the
volume) is (α/3)(r + dr)3 dΩ = (α/3)(r3 + 3r2 dr)dΩ. Similarly, the flux through the area
element (and out of the volume) at r with area r2 dΩ where the current is jr = −(α/3)r
is (α/3)r3 dΩ. The difference is αr2 drdΩ = αdV , and we see that this does not depend on
the location. Also, as required, this matches the rate of change of the total charge in the
volume, (∂ρ/∂t)dV = αdV.
18
Chapter 2
The Maxwell Equations
Source: Prof. M. Srednicki, University of California, Santa Barbara (private communication).
2.10
Coulomb Repulsion in One Dimension
Newton’s equation of motion for the released particle is
mẍ =
q2 1
.
4π0 x2
If we let A = q 2 /4π0 m, the equation of motion is
A
.
x2
ẍ =
To integrate this, we multiply by ẋ to get
A
ẋẍ = 2 ẋ
x
or
d
dt
1 2
ẋ
2
d
=−
dt
A
.
x
Using the initial conditions, this integrates to
1 2
A A
ẋ = − .
2
d
x
Therefore, as x → ∞ the speed approaches υ =
2.11
2A/d.
Ampère-Maxwell Matching Conditions
(a) The fields for this problem are
B(r, t) = Θ(z)B1 (r, t) + Θ(−z)B2 (r, t)
E(r, t) = Θ(z)E1 (r, t) + Θ(−z)E2 (r, t)
j(r, t) = Θ(z)j1 (r, t) + Θ(−z)j 2 (r, t) + K(rS , t)δ(z).
Then,
∇ × B = Θ(z)∇ × B1 − B1 × ∇Θ(z) + Θ(−z)∇ × B2 − B2 × ∇Θ(−z)
and
∂E1 (r, t)
∂E2 (r, t)
∂E
= Θ(z)
+ Θ(−z)
.
∂t
∂t
∂t
However, B1 , E1 , and j1 satisfy the Ampère-Maxwell law, as do B2 , E2 , and j2 .
Therefore, the time derivative disappears when we write out this law using the previous
two equations to get
−B1 × ∇Θ(z) − B2 × ∇Θ(−z) = µ0 Kδ(z).
19
Chapter 2
The Maxwell Equations
This simplifies because ∇Θ(±z) = ±ẑδ(z). Using this information gives
[B2 − B1 ]δ(z) × ẑ = µ0 Kδ(z).
If we use the square brackets to enforce the delta function evaluation of B1 and B2
infinitesimally near to (but on opposite sides of) z = 0, we get the matching condition,
ẑ × [B1 − B2 ] = µ0 K.
(b) We can apply this result to an arbitrary point rS on a non-flat interface because the
fields involved in the matching condition are evaluated infinitesimally close to rS . From
that distance, the interface looks flat and the result proved in part (a) is applicable.
Using our usual convention that n̂2 is the outward normal from region 2, the matching
condition of part (a) generalizes to
n̂2 × [B1 − B2 ] = µ0 K(rS ).
2.12
A Variation of Gauss’ Law
(a) Because ∇ × E = 0 , we can still define a scalar potential E = −∇ϕ. Substituting this
into the Podolsky-Gauss equation gives
0 (1 − a2 ∇2 )∇2 ϕ(r) = −qδ(r).
Integrating this over a small spherical volume of radius R centered on the origin gives
−q = 0
d r ∇ · ∇ϕ − a ∇∇ ϕ = 0
3
2
2
dS · ∇ϕ − a2 ∇∇2 ϕ .
By symmetry, we may assume that ϕ(r) = ϕ(r). Therefore, writing out the gradient
and Laplacian operators and doing the integral over r = R gives
−q = 4π0 R
or
2
dϕ
1 d
2 d
2 dϕ
−a
R
dR
dR R2 dR
dR
1 d
q
= ϕ − a2 2
4π0 R
R dR
dϕ
R
dR
2
.
Using the suggested ansatz, ϕ(r) = qu(r)/4π0 R, simplifies this equation to
a2
d2 u
= u − 1.
dr2
Since u(r) cannot diverge at infinity, this integrates immediately to u(r) = 1 +
B exp(−r/a). Therefore,
ϕ(r) =
q
{1 + B exp(−ar)} .
4π0 r
20
Chapter 2
The Maxwell Equations
The entire point of this exercise was to eliminate the divergence of the field at the
origin. The corresponding divergence of the potential disappears only if B = −1.
Therefore, we conclude that
ϕ(r) =
so
E(r) = −r̂
q
{1 − exp(−r/a)}
4π0 r
∂ϕ
q r̂
=
[1 − (1 + r/a) exp(−r/a)] .
∂r
4π0 r2
(b) The parameter a has dimensions of length so, by analogy with meson theory, we
may regard it as the de Broglie wavelength of a particle which mediates the modified
Coulomb interaction.
Source: B. Podolsky, Physical Review 62, 68 (1942).
2.13
If the Photon Had Mass . . .
r2
r1
Φ
Φ
(a) The equation to be solved is ∇2 ϕ = ϕ/L2 because ρ = 0 between the shells. By
spherical symmetry,
1 d
ϕ
2 dϕ
r
= 2.
r2 dr
dr
L
The substitution ϕ = u/r simplifies this equation to d2 u/dr2 = u/L2 , which is solved
by real exponentials. Therefore,
ϕ(r) =
1
aer /L + be−r /L .
r
This has the proposed form. The constants are determined by the boundary conditions
ϕ(r1 ) = ϕ(2) = Φ. After a bit of algebra, we find
ϕ(r) = Φ
r2 sinh[(r − r1 )/L]
r1 sinh[(r2 − r)/L]
+
r sinh[(r2 − r1 )/L]
r sinh[(r2 − r1 )/L]
.
If ∆ = (r2 − r1 )/L, the associated electric field E = −∇ϕ is
E(r)
=
sinh[(r − r1 )/L] cosh[(r − r1 )/L]
Φr̂
−
r2
sinh ∆
r2
rL
sinh[(r2 − r)/L] cosh[(r2 − r)/L]
+ r1
+
r2
rL
21