Solutions Manual
Modern Electrodynamics
By
Andrew Zangwill
( All Chapters Included - 100% Verified Solutions )
1
,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).
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,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
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, Chapter 1 Mathematical Preliminaries
(c)
[∇ × (g × r)]i = ǫij k ∂j ǫk ℓm gℓ rm
= ǫk ij ǫk ℓm ∂j (gℓ rm )
= (δiℓ δj m − δim δj ℓ )∂j (gℓ rm )
= ∂j (gi rj ) − ∂j (gj ri )
= rj gj gi + gi (∇ · r) − ri (∇ · g) − δij gj
= (r · ∇)gi + 3gi − ri (∇ · g) − gi
Therefore,
∂g
∇ × (g × r) = 2g + r − r(∇ · g).
∂r
1.5 Delta Function Identities
(a) Let f (x) be an arbitrary function. Then, if a > 0, a change of variable to y = ax gives
∞ ∞
1 1
dxf (x)δ(ax) = dyf (y/a)δ(y) = f (0).
a a
−∞ −∞
However, if a < 0,
∞ −∞
∞
1 1 1
dxf (x)δ(ax) = dyf (y/a)δ(y) = − dyf (y/a)δ(y) = − f (0).
a a a
−∞ ∞ −∞
1
These two results are summarized by δ(ax) = 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),
∞ ∞
1
dxf (x)δ[g(x)] ≈ dxf (x)g[(x − x0 )g ′ (x0 )] = δ(x − x0 ).
|g ′ (x0 )|
−∞ −∞
A similar contribution comes from each distinct zero xm . Adding these together gives
the advertised result.
3
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Modern Electrodynamics
By
Andrew Zangwill
( All Chapters Included - 100% Verified Solutions )
1
,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
2
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,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
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, Chapter 1 Mathematical Preliminaries
(c)
[∇ × (g × r)]i = ǫij k ∂j ǫk ℓm gℓ rm
= ǫk ij ǫk ℓm ∂j (gℓ rm )
= (δiℓ δj m − δim δj ℓ )∂j (gℓ rm )
= ∂j (gi rj ) − ∂j (gj ri )
= rj gj gi + gi (∇ · r) − ri (∇ · g) − δij gj
= (r · ∇)gi + 3gi − ri (∇ · g) − gi
Therefore,
∂g
∇ × (g × r) = 2g + r − r(∇ · g).
∂r
1.5 Delta Function Identities
(a) Let f (x) be an arbitrary function. Then, if a > 0, a change of variable to y = ax gives
∞ ∞
1 1
dxf (x)δ(ax) = dyf (y/a)δ(y) = f (0).
a a
−∞ −∞
However, if a < 0,
∞ −∞
∞
1 1 1
dxf (x)δ(ax) = dyf (y/a)δ(y) = − dyf (y/a)δ(y) = − f (0).
a a a
−∞ ∞ −∞
1
These two results are summarized by δ(ax) = 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),
∞ ∞
1
dxf (x)δ[g(x)] ≈ dxf (x)g[(x − x0 )g ′ (x0 )] = δ(x − x0 ).
|g ′ (x0 )|
−∞ −∞
A similar contribution comes from each distinct zero xm . Adding these together gives
the advertised result.
3
4
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