,Contents
1 Vector Analysis 4
2 Electrostatics 26
3 Potential 53
4 Electric Fields in Matter 92
5 Magnetostatics 110
6 Magnetic Fields in Matter 133
7 Electrodynamics 145
8 Conservation Laws 168
9 Electromagnetic Waves 185
10 Potentials and Fields 210
11 Radiation 231
12 Electrodynamics and Relativity 262
,
, 4 CHAPTER 1. VECTOR ANALYSIS
Chapter 1
Vector Analysis
Problem 1.1
(a) From the dͅ iagram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multiply by |A|.
|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.
So: A·(B + C) = A·B + A·C. (Dͅot prodͅ uct is dͅ istributive) sin θ2
Similarly: |B + C| sin θ3 = |B| sin θ1 + |C| sin θ2. Mulitply by |A| n̂ .
|A||B + C| sin θ3 n̂ = |A||B| sin θ1 n̂ + |A||C| sin θ2 n̂ .
If n̂ is the unit vector pointing out of the page, it follows that
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross prodͅ uct is dͅ istributive)
(b) For the general case, see G. E. Hay’s Vector andͅ Tensor Analysis, Chapter 1, Section 7 (dͅ ot prodͅ uct) andͅ
Section 8 (cross prodͅ uct)
Problem 1.2
The triple cross-prodͅ uct is not in general associative. For example,
suppose A = B andͅ C is perpendͅ icular to A, as in the dͅ iagram.
= B
Then (B⇥C) points out-of-the-page, andͅ A⇥(B⇥C) points dͅ own,
andͅ has magnitudͅ e ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 =/
A⇥(B⇥C).
Problem 1.3
√ √
A = +1 x̂ + 1 ŷ — 1 ˆz ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √
A·B = +1 + 1 — 1 = 1 = AB cos θ = 3 3 cos θ ⇒ cos θ = 13.
θ
y
1 1
3 70.5288○
x
Problem 1.4
The cross-prodͅ uct of any two vectors in the plane will give a vector perpendͅ icular to the plane. For example,
we might pick the base (A) andͅ the left sidͅ e (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
,CHAPTER 1. VECTOR ANALYSIS 5
x̂ ŷ ẑ
A⇥B = —1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
—1 0 3
This has the right dͅ irection, but the wrong magnitudͅ e. To make a unit vector out of it, simply dͅ ividͅ e by its
length:
√
|A⇥B| = 36 + 9 + 4 = 7. n̂ = A⇥B = .
|A⇥B| 7 7 7
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(ByCz — BzCy) (BzCx — BxCz) (BxCy — ByCx)
= x̂[A y (B x C y— ByCx) — Az(BzCx — BxCz)] + ŷ ( ) + ẑ( )
(I’ll just check the x-component; the others go the same way)
= x̂(A y B x C y — AyByCx — AzBzCx + AzBxCz) + ŷ ( ) + ẑ( ) .
B(A·C) — C(A·B) = [Bx(AxCx + AyCy + AzCz) — Cx(AxBx + AyBy + AzBz)] x̂ + () ŷ + () ẑ
= x̂(A y B x C y + AzBxCz — AyByCx — AzBzCx) + ŷ ( ) + ẑ( ) . They agree.
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C)—C(A·B)+C(A·B)—A(C·B)+A(B·C)—B(C·A) = 0.
So: A⇥(B⇥C) — (A⇥B)⇥C = —B⇥(C⇥A) = A(B·C) — C(A·B).
If this is zero, then either A is parallel to C (includͅ ing the case in which they point in opposite dͅ irections, or
one is zero), or else B·C = B·A = 0, in which case B is perpendͅ icular to A andͅ C (includͅ ing the case B = 0.)
Problem 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) =2 —2
√
= 4+4+1= 3
2
3
—23 3
Problem 1.8
(a) A¯y B̄ y + A¯z B̄ z = (cos Ay + sin Az)(cos By + sin Bz) + (— sin Ay + cos Az)( —sin By + cos Bz)
= cos2 AyBy + sin cos (AyBz + AzBy) + sin2 AzBz + sin2 AyBy — sin cos (AyBz + AzBy) +
2
cos AzBz
= (cos2 + sin2 )AyBy + (sin2 + cos2 )AzBz = AyBy + AzBz. X
(b) (Ax)2 + (Ay)2 + (Az)2 = C3 AiAi = C3 C3j=1 RijAj C3 RikAk = Cj,k (CiRij Rik) Aj Ak.
i=1 i=1 k=1
⇢
1 if j = k
This equals A2x + A2y + A2zprovidͅ edͅ C3i=1RijRik =
0 if j /= k
Moreover, if R is to preserve lengths for all vectors A, then this condͅ ition is not only sufficient but also
necessary.
2
For
2
suppose
2
A = (1, 0, 0). Then Cj,k (Ci RijRik) AjAk = Ci Ri1Ri1, andͅ this must equal 1 (since we
wantAx +Ay +Az = 1). Likewise, Ci=1 3 R R 3
i2 i2 = C i=1 R i3 R i3 = 1. To check the case j /= k, choose A = (1, 1, 0).
Then we want 2 = Cj,k (Ci RijRik) AjAk = Ci Ri1Ri1 + Ci Ri2Ri2 + Ci Ri1Ri2 + Ci Ri2Ri1. But we alreadͅ y
know that the first two sums are both 1; the thirdͅ andͅ fourth are equal, so Ci Ri1Ri2 = Ci Ri2Ri1 = 0, andͅ so
on for other unequal combinations of j, k. X In matrix notation: R̃ R = 1, where R̃ is the transpose of R.
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
, 6 CHAPTER 1. VECTOR ANALYSIS
Problem 1.9
Looking dͅ own the axis:
✒
A 120○ rotation carries the z axis into the y (= z) axis, y into x (= y), andͅ x into z (= x). So Ax = Az,
Ay = Ax, 0Az = Ay1
.
001
R =@ 1 0 0A
010
Problem 1.10
(a) No change. (Ax = Ax, Ay = Ay, Az = Az)
(b) —→ — in the sense (Ax = —Ax, Ay = —Ay, Az = —Az)
(c) (A⇥B) —→ (—A)⇥( —B) = (A⇥B). That is, if C = A⇥B, . No minus sign, in contrast to
behavior of an “ordͅ inary” vector, as given by (b). If A andͅ B are p seudͅ ovectors, then (A⇥B) —→ (A)⇥(B) =
(A⇥B). So the cross-prodͅ uct of two pseudͅ ovectors is again a pseudͅ ovector. In the cross-prodͅ uct of a vector
andͅ a pseudͅ ovector, one changes sign, the other dͅ oesn’t, andͅ therefore the cross-prodͅ uct is itself a vector.
Angular momentum (L = r⇥p) andͅ torque (N = r⇥F) are pseudͅ ovectors.
(d) A·(B⇥C) —→ (—A)·((—B)⇥(—C)) = —A·(B⇥C). So, if a = A·(B⇥C), then a —→ —a; a pseudͅ oscalar
changes sign undͅ er inversion of coordͅ inates.
Problem 1.11
(a)rf = 2x x̂ + 3y2 ŷ + 4z3 ẑ
(b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ
(c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ
Problem 1.12
(a) rh = 10[(2y — 6x — 18) x̂ + (2x — 8y + 2 8) ŷ ] . rh = 0 at summit, so
2y — 6x — 18 = 0
2y — 18 — 24y + 84 = 0.
2x — 8y + 28 = 0 =⇒ 6x — 24y + 84 = 0
22y = 66 =⇒ y = 3 =⇒ 2x — 24 + 28 = 0 =⇒ x = —2.
Top is 3 miles north, 2 miles west, of South Hadͅ ley.
(b) Putting in x = —2, y = 3:
h = 10(—12 — 12 — 36 + 36 + 84 + 12) = 720 ft.
(c) Putting in x = 1, y = 1: rh = 10[(2 — 6 — 18) x̂ + (2 — 8 + 28) ŷ ] = 10(—22 x̂ + 22 ŷ ) = 220(— x̂ + ŷ ) .
√
|rh| = 220 2 ≈ 311 ft/mile; dͅ irection: northwest.
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
, CHAPTER 1. VECTOR ANALYSIS 7
Problem 1.13
p
= (x — x′) x̂ + (y — y′) ŷ + (z — z′) ẑ ; = (x — x′)2 + (y — y′)2 + (z — z′)2.
(a) r( 2) = ∂x[(x—x )
∂ ′ 2 +(y —y′)2 +(z —z′)2] x̂ + ∂y () ŷ
∂ + ∂z () ẑ
∂ = 2(x—x′) x̂ + 2(y —y′) ŷ + 2(z —z′) ẑ = 2 .
1 1 1
(b) r( 1 ) = ∂ ∂x
[(x — x′)2 + (y — y′)2 + (z — z′)2]— 2 x̂ + ∂∂y()— 2 ŷ + ∂z
∂ ()— 2 ẑ
1 — 32 1 — 23 1 —23
= — 2 () 2(x — x ) x̂ — 2 () 2(y — y ) ŷ — 2 () 2(z — z ) ẑ
′ ′ ′
3
= —()— 2 [(x — x′) x̂ + (y — y′) ŷ + (z — z′) ẑ ] = —(1/ 3) = —(1/ 2) ˆ .
n—1 ∂ n—1( 1 1 2
2
Problem 1.14
y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 .
z = —y sin + z cos ; multiply by cos : z cos = —y sin cos + z cos2 .
Adͅ dͅ : y sin + z cos = z(sin2 + cos2 ) = z. Likewise, y cos — z sin = y.
So ∂y∂y = cos ; ∂z = — sin ; ∂y = sin ; ∂z = cos . Therefore
∂y ∂z ∂z
)
(rf) y = ∂f
∂y = ∂f ∂y + ∂f
∂y ∂y ∂y =
∂z ∂z + cos (rf )y + sin (rf )z So rf transforms as a vector. qedͅ
(rf ) z = ∂f =
∂z
∂f ∂y + ∂f ∂z = — sin (rf ) + cos (rf )
∂y ∂z ∂z ∂z y z
Problem 1.15
(a) 2 2
Problem 1.16
h 3
i
r·v = ∂ ( x ) + ∂ ( y ) + ∂ ( z ) = ∂ x(x2 + y2 + z2)— 2
h ∂x r3 ∂y r3
3
i ∂z r3h ∂x
3
i
∂
+ ∂y y(x + y + z ) 2 + ∂z z(x + y2 + z2)— 2
2 2 2 — ∂ 2
3 5 3 5 3
= ()— 2 + x(—3/2)()— 2 2x + ()— 2 + y(—3/2)()— 2 2y + ()— 2
5
+ z(—3/2)()— 2 2z = 3r—3 — 3r—5(x2 + y2 + z 2 ) = 3r—3 — 3r—3 = 0.
This conclusion is surprising, because, from the dͅ iagram, this vector fieldͅ is obviously dͅ iverging away from the
origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the
origin our calculation is no goodͅ , since r = 0, andͅ the expression for v blows up. In fact, r·v is infinite at
that one point, andͅ zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
vy = cos vy + sin vz; vz = — sin vy + cos vz.
⇣ ⌘ ⇣ ⌘
∂vy ∂vy ∂y
= ∂vy cos + ∂vz sin = + ∂vy ∂z cos + ∂vz ∂y + ∂vz ∂z sin . Use result in Prob. 1.14:
∂y ∂y ∂y ⌘ ∂y ∂y ⇣ ∂z ∂y ⌘∂y ∂y ∂z ∂y
⇣ ∂vy ∂vy sin
∂vz
cos + cos + cos + ∂vz sin sin .
= ∂y ∂z ⇣ ∂y ⌘ ∂z ⇣ ⌘
∂v ∂v
∂vz
= — y sin + ∂vz cos = — y ∂y + ∂vy ∂z sin + ∂vz ∂y + ∂vz ∂z cos
∂z ⇣∂z ∂z ⌘ ∂y ∂z
⇣ ∂z ∂z ∂y ∂z
⌘ ∂z ∂z
∂v
= — — y sin + ∂vy cos sin + — ∂vz sin + ∂vz cos cos . So
∂y ∂z ∂y ∂z
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
1 Vector Analysis 4
2 Electrostatics 26
3 Potential 53
4 Electric Fields in Matter 92
5 Magnetostatics 110
6 Magnetic Fields in Matter 133
7 Electrodynamics 145
8 Conservation Laws 168
9 Electromagnetic Waves 185
10 Potentials and Fields 210
11 Radiation 231
12 Electrodynamics and Relativity 262
,
, 4 CHAPTER 1. VECTOR ANALYSIS
Chapter 1
Vector Analysis
Problem 1.1
(a) From the dͅ iagram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multiply by |A|.
|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.
So: A·(B + C) = A·B + A·C. (Dͅot prodͅ uct is dͅ istributive) sin θ2
Similarly: |B + C| sin θ3 = |B| sin θ1 + |C| sin θ2. Mulitply by |A| n̂ .
|A||B + C| sin θ3 n̂ = |A||B| sin θ1 n̂ + |A||C| sin θ2 n̂ .
If n̂ is the unit vector pointing out of the page, it follows that
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross prodͅ uct is dͅ istributive)
(b) For the general case, see G. E. Hay’s Vector andͅ Tensor Analysis, Chapter 1, Section 7 (dͅ ot prodͅ uct) andͅ
Section 8 (cross prodͅ uct)
Problem 1.2
The triple cross-prodͅ uct is not in general associative. For example,
suppose A = B andͅ C is perpendͅ icular to A, as in the dͅ iagram.
= B
Then (B⇥C) points out-of-the-page, andͅ A⇥(B⇥C) points dͅ own,
andͅ has magnitudͅ e ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 =/
A⇥(B⇥C).
Problem 1.3
√ √
A = +1 x̂ + 1 ŷ — 1 ˆz ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √
A·B = +1 + 1 — 1 = 1 = AB cos θ = 3 3 cos θ ⇒ cos θ = 13.
θ
y
1 1
3 70.5288○
x
Problem 1.4
The cross-prodͅ uct of any two vectors in the plane will give a vector perpendͅ icular to the plane. For example,
we might pick the base (A) andͅ the left sidͅ e (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
,CHAPTER 1. VECTOR ANALYSIS 5
x̂ ŷ ẑ
A⇥B = —1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
—1 0 3
This has the right dͅ irection, but the wrong magnitudͅ e. To make a unit vector out of it, simply dͅ ividͅ e by its
length:
√
|A⇥B| = 36 + 9 + 4 = 7. n̂ = A⇥B = .
|A⇥B| 7 7 7
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(ByCz — BzCy) (BzCx — BxCz) (BxCy — ByCx)
= x̂[A y (B x C y— ByCx) — Az(BzCx — BxCz)] + ŷ ( ) + ẑ( )
(I’ll just check the x-component; the others go the same way)
= x̂(A y B x C y — AyByCx — AzBzCx + AzBxCz) + ŷ ( ) + ẑ( ) .
B(A·C) — C(A·B) = [Bx(AxCx + AyCy + AzCz) — Cx(AxBx + AyBy + AzBz)] x̂ + () ŷ + () ẑ
= x̂(A y B x C y + AzBxCz — AyByCx — AzBzCx) + ŷ ( ) + ẑ( ) . They agree.
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C)—C(A·B)+C(A·B)—A(C·B)+A(B·C)—B(C·A) = 0.
So: A⇥(B⇥C) — (A⇥B)⇥C = —B⇥(C⇥A) = A(B·C) — C(A·B).
If this is zero, then either A is parallel to C (includͅ ing the case in which they point in opposite dͅ irections, or
one is zero), or else B·C = B·A = 0, in which case B is perpendͅ icular to A andͅ C (includͅ ing the case B = 0.)
Problem 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) =2 —2
√
= 4+4+1= 3
2
3
—23 3
Problem 1.8
(a) A¯y B̄ y + A¯z B̄ z = (cos Ay + sin Az)(cos By + sin Bz) + (— sin Ay + cos Az)( —sin By + cos Bz)
= cos2 AyBy + sin cos (AyBz + AzBy) + sin2 AzBz + sin2 AyBy — sin cos (AyBz + AzBy) +
2
cos AzBz
= (cos2 + sin2 )AyBy + (sin2 + cos2 )AzBz = AyBy + AzBz. X
(b) (Ax)2 + (Ay)2 + (Az)2 = C3 AiAi = C3 C3j=1 RijAj C3 RikAk = Cj,k (CiRij Rik) Aj Ak.
i=1 i=1 k=1
⇢
1 if j = k
This equals A2x + A2y + A2zprovidͅ edͅ C3i=1RijRik =
0 if j /= k
Moreover, if R is to preserve lengths for all vectors A, then this condͅ ition is not only sufficient but also
necessary.
2
For
2
suppose
2
A = (1, 0, 0). Then Cj,k (Ci RijRik) AjAk = Ci Ri1Ri1, andͅ this must equal 1 (since we
wantAx +Ay +Az = 1). Likewise, Ci=1 3 R R 3
i2 i2 = C i=1 R i3 R i3 = 1. To check the case j /= k, choose A = (1, 1, 0).
Then we want 2 = Cj,k (Ci RijRik) AjAk = Ci Ri1Ri1 + Ci Ri2Ri2 + Ci Ri1Ri2 + Ci Ri2Ri1. But we alreadͅ y
know that the first two sums are both 1; the thirdͅ andͅ fourth are equal, so Ci Ri1Ri2 = Ci Ri2Ri1 = 0, andͅ so
on for other unequal combinations of j, k. X In matrix notation: R̃ R = 1, where R̃ is the transpose of R.
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
, 6 CHAPTER 1. VECTOR ANALYSIS
Problem 1.9
Looking dͅ own the axis:
✒
A 120○ rotation carries the z axis into the y (= z) axis, y into x (= y), andͅ x into z (= x). So Ax = Az,
Ay = Ax, 0Az = Ay1
.
001
R =@ 1 0 0A
010
Problem 1.10
(a) No change. (Ax = Ax, Ay = Ay, Az = Az)
(b) —→ — in the sense (Ax = —Ax, Ay = —Ay, Az = —Az)
(c) (A⇥B) —→ (—A)⇥( —B) = (A⇥B). That is, if C = A⇥B, . No minus sign, in contrast to
behavior of an “ordͅ inary” vector, as given by (b). If A andͅ B are p seudͅ ovectors, then (A⇥B) —→ (A)⇥(B) =
(A⇥B). So the cross-prodͅ uct of two pseudͅ ovectors is again a pseudͅ ovector. In the cross-prodͅ uct of a vector
andͅ a pseudͅ ovector, one changes sign, the other dͅ oesn’t, andͅ therefore the cross-prodͅ uct is itself a vector.
Angular momentum (L = r⇥p) andͅ torque (N = r⇥F) are pseudͅ ovectors.
(d) A·(B⇥C) —→ (—A)·((—B)⇥(—C)) = —A·(B⇥C). So, if a = A·(B⇥C), then a —→ —a; a pseudͅ oscalar
changes sign undͅ er inversion of coordͅ inates.
Problem 1.11
(a)rf = 2x x̂ + 3y2 ŷ + 4z3 ẑ
(b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ
(c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ
Problem 1.12
(a) rh = 10[(2y — 6x — 18) x̂ + (2x — 8y + 2 8) ŷ ] . rh = 0 at summit, so
2y — 6x — 18 = 0
2y — 18 — 24y + 84 = 0.
2x — 8y + 28 = 0 =⇒ 6x — 24y + 84 = 0
22y = 66 =⇒ y = 3 =⇒ 2x — 24 + 28 = 0 =⇒ x = —2.
Top is 3 miles north, 2 miles west, of South Hadͅ ley.
(b) Putting in x = —2, y = 3:
h = 10(—12 — 12 — 36 + 36 + 84 + 12) = 720 ft.
(c) Putting in x = 1, y = 1: rh = 10[(2 — 6 — 18) x̂ + (2 — 8 + 28) ŷ ] = 10(—22 x̂ + 22 ŷ ) = 220(— x̂ + ŷ ) .
√
|rh| = 220 2 ≈ 311 ft/mile; dͅ irection: northwest.
O
c 2012 Pearson Edͅ ucation, Inc., Upper Sadͅ dͅ le River, NJ. All rights reservedͅ . This material is
protectedͅ undͅ er all copyright laws as they currently exist. No portion of this material may be
reprodͅ ucedͅ , in any form or by any means, without permission in writing from the publisher.
, CHAPTER 1. VECTOR ANALYSIS 7
Problem 1.13
p
= (x — x′) x̂ + (y — y′) ŷ + (z — z′) ẑ ; = (x — x′)2 + (y — y′)2 + (z — z′)2.
(a) r( 2) = ∂x[(x—x )
∂ ′ 2 +(y —y′)2 +(z —z′)2] x̂ + ∂y () ŷ
∂ + ∂z () ẑ
∂ = 2(x—x′) x̂ + 2(y —y′) ŷ + 2(z —z′) ẑ = 2 .
1 1 1
(b) r( 1 ) = ∂ ∂x
[(x — x′)2 + (y — y′)2 + (z — z′)2]— 2 x̂ + ∂∂y()— 2 ŷ + ∂z
∂ ()— 2 ẑ
1 — 32 1 — 23 1 —23
= — 2 () 2(x — x ) x̂ — 2 () 2(y — y ) ŷ — 2 () 2(z — z ) ẑ
′ ′ ′
3
= —()— 2 [(x — x′) x̂ + (y — y′) ŷ + (z — z′) ẑ ] = —(1/ 3) = —(1/ 2) ˆ .
n—1 ∂ n—1( 1 1 2
2
Problem 1.14
y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 .
z = —y sin + z cos ; multiply by cos : z cos = —y sin cos + z cos2 .
Adͅ dͅ : y sin + z cos = z(sin2 + cos2 ) = z. Likewise, y cos — z sin = y.
So ∂y∂y = cos ; ∂z = — sin ; ∂y = sin ; ∂z = cos . Therefore
∂y ∂z ∂z
)
(rf) y = ∂f
∂y = ∂f ∂y + ∂f
∂y ∂y ∂y =
∂z ∂z + cos (rf )y + sin (rf )z So rf transforms as a vector. qedͅ
(rf ) z = ∂f =
∂z
∂f ∂y + ∂f ∂z = — sin (rf ) + cos (rf )
∂y ∂z ∂z ∂z y z
Problem 1.15
(a) 2 2
Problem 1.16
h 3
i
r·v = ∂ ( x ) + ∂ ( y ) + ∂ ( z ) = ∂ x(x2 + y2 + z2)— 2
h ∂x r3 ∂y r3
3
i ∂z r3h ∂x
3
i
∂
+ ∂y y(x + y + z ) 2 + ∂z z(x + y2 + z2)— 2
2 2 2 — ∂ 2
3 5 3 5 3
= ()— 2 + x(—3/2)()— 2 2x + ()— 2 + y(—3/2)()— 2 2y + ()— 2
5
+ z(—3/2)()— 2 2z = 3r—3 — 3r—5(x2 + y2 + z 2 ) = 3r—3 — 3r—3 = 0.
This conclusion is surprising, because, from the dͅ iagram, this vector fieldͅ is obviously dͅ iverging away from the
origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the
origin our calculation is no goodͅ , since r = 0, andͅ the expression for v blows up. In fact, r·v is infinite at
that one point, andͅ zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
vy = cos vy + sin vz; vz = — sin vy + cos vz.
⇣ ⌘ ⇣ ⌘
∂vy ∂vy ∂y
= ∂vy cos + ∂vz sin = + ∂vy ∂z cos + ∂vz ∂y + ∂vz ∂z sin . Use result in Prob. 1.14:
∂y ∂y ∂y ⌘ ∂y ∂y ⇣ ∂z ∂y ⌘∂y ∂y ∂z ∂y
⇣ ∂vy ∂vy sin
∂vz
cos + cos + cos + ∂vz sin sin .
= ∂y ∂z ⇣ ∂y ⌘ ∂z ⇣ ⌘
∂v ∂v
∂vz
= — y sin + ∂vz cos = — y ∂y + ∂vy ∂z sin + ∂vz ∂y + ∂vz ∂z cos
∂z ⇣∂z ∂z ⌘ ∂y ∂z
⇣ ∂z ∂z ∂y ∂z
⌘ ∂z ∂z
∂v
= — — y sin + ∂vy cos sin + — ∂vz sin + ∂vz cos cos . So
∂y ∂z ∂y ∂z
O
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