Solution Manual
for
Modern Electrodynamics
Andrew Zangwill
School of Physics
Georgia Institute of Technology
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
,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
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
,Chapter 1 Mathematical Preliminaries
Chapter 1: Mathematical Preliminaries
1.1 Levi-Cività Practice I
(a) ǫ123 = ê1 · (ê2 × ê3 ) = ê1 · ê1 = 1. The cyclic property of the triple scalar product
guarantees that ǫ231 = ǫ312 = 1 also. Similarly, ǫ132 = ê1 · (ê3 × ê2 ) = −ê1 · ê1 = −1
with ǫ321 = ǫ213 = −1 also. Finally, ǫ122 = 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 = ǫ123 ê1 a2 b3 + ǫ132 ê1 a3 b2
+ ǫ213 ê2 a1 b3 + ǫ231 ê2 a3 b1
+ ǫ312 ê3 a1 b2 + ǫ321 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 ǫ 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 ]
= ih̄ǫm pi L̂i am bp
= ih̄L̂i ǫim p am bp
= ih̄L̂ · (a × b).
1
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
, Chapter 1 Mathematical Preliminaries
1.2 Levi-Civitá Practice II
(a) δii = 1 + 1 + 1 = 3
(b) δij ǫij k = ǫiik = 0
(c) ǫij k ǫℓj k = ǫj k i ǫj k ℓ = δk k δiℓ − δk ℓ δik = 3δiℓ − δiℓ = 2δiℓ
(d) ǫij k ǫij k = δj j δk k − δj k δk j = 9 − δk k = 6
1.3 Vector Identities
(a) (A × B) · (C × D) = ǫij k Aj Bk ǫim p Cm Dp = ǫij k ǫim 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 ǫij k fj gk = ǫij k fj ∂i gk + ǫij k gk ∂i fj = fj ǫj k i ∂i gk + gk ǫk ij ∂i fj
= gk ǫk ij ∂i fj − fj ǫj ik ∂i gk = g · (∇ × f ) − f · (∇ × g)
(c) [(A × B) × (C × D)]i = ǫij k {A × B}j {C × D}k = ǫij k ǫj m p ǫk st Am Bp Cs Dt
= ǫj k i ǫj m p ǫk st Am Bp Cs Dt = (δk m δip − δk p δim )ǫk st Am Bp Cs Dt
= ǫk st Ak Bi Cs Dt − ǫk st Ai Bk Cs Dt = Ak ǫk st Cs Dt Bi − Bk ǫk 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 +iǫij k σk )ai bj = ai bi +iǫk 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 = ǫij k ∂j (f gk ) = f ǫij k ∂j gk + ǫij k (∂j f )gk = f [∇ × g]i + [∇f × g]i
2
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
for
Modern Electrodynamics
Andrew Zangwill
School of Physics
Georgia Institute of Technology
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
,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
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
,Chapter 1 Mathematical Preliminaries
Chapter 1: Mathematical Preliminaries
1.1 Levi-Cività Practice I
(a) ǫ123 = ê1 · (ê2 × ê3 ) = ê1 · ê1 = 1. The cyclic property of the triple scalar product
guarantees that ǫ231 = ǫ312 = 1 also. Similarly, ǫ132 = ê1 · (ê3 × ê2 ) = −ê1 · ê1 = −1
with ǫ321 = ǫ213 = −1 also. Finally, ǫ122 = 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 = ǫ123 ê1 a2 b3 + ǫ132 ê1 a3 b2
+ ǫ213 ê2 a1 b3 + ǫ231 ê2 a3 b1
+ ǫ312 ê3 a1 b2 + ǫ321 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 ǫ 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 ]
= ih̄ǫm pi L̂i am bp
= ih̄L̂i ǫim p am bp
= ih̄L̂ · (a × b).
1
FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE
, Chapter 1 Mathematical Preliminaries
1.2 Levi-Civitá Practice II
(a) δii = 1 + 1 + 1 = 3
(b) δij ǫij k = ǫiik = 0
(c) ǫij k ǫℓj k = ǫj k i ǫj k ℓ = δk k δiℓ − δk ℓ δik = 3δiℓ − δiℓ = 2δiℓ
(d) ǫij k ǫij k = δj j δk k − δj k δk j = 9 − δk k = 6
1.3 Vector Identities
(a) (A × B) · (C × D) = ǫij k Aj Bk ǫim p Cm Dp = ǫij k ǫim 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 ǫij k fj gk = ǫij k fj ∂i gk + ǫij k gk ∂i fj = fj ǫj k i ∂i gk + gk ǫk ij ∂i fj
= gk ǫk ij ∂i fj − fj ǫj ik ∂i gk = g · (∇ × f ) − f · (∇ × g)
(c) [(A × B) × (C × D)]i = ǫij k {A × B}j {C × D}k = ǫij k ǫj m p ǫk st Am Bp Cs Dt
= ǫj k i ǫj m p ǫk st Am Bp Cs Dt = (δk m δip − δk p δim )ǫk st Am Bp Cs Dt
= ǫk st Ak Bi Cs Dt − ǫk st Ai Bk Cs Dt = Ak ǫk st Cs Dt Bi − Bk ǫk 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 +iǫij k σk )ai bj = ai bi +iǫk 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 = ǫij k ∂j (f gk ) = f ǫij k ∂j gk + ǫij k (∂j f )gk = f [∇ × g]i + [∇f × g]i
2
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