1. 1.

Instructor’s Manual MATHEMATICAL METHODS FOR PHYSICISTS A Comprehensive Guide SEVENTH

EDITION George B. Arfken Miami University Oxford, OH Hans J. Weber University of Virginia Charlottesville, VA
Frank E. Harris University of Utah, Salt Lake City, UT; University of Florida, Gainesville, FL AMSTERDAM •
BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO •
SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier
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3. 3. Contents 1 Introduction 1 2 Errata and Revision Status 3 3 Exercise Solutions 7 1. Mathematical Preliminaries . .
. . . . . . . . . . . . . . . . . 7 2. Determinants and Matrices . . . . . . . . . . . . . . . . . . . . 27 3. Vector Analysis . . . . . . . . . . . .
. . . . . . . . . . . . . . 34 4. Tensors and Differential Forms . . . . . . . . . . . . . . . . . . 58 5. Vector Spaces . . . . . . . . . . . . .
. . . . . . . . . . . . . . 66 6. Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . 81 7. Ordinary Differential Equations . . .
. . . . . . . . . . . . . . 90 8. Sturm-Liouville Theory . . . . . . . . . . . . . . . . . . . . . . 106 9. Partial Differential Equations . . . .
. . . . . . . . . . . . . . 111 10. Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . 118 11. Complex Variable Theory . . . .
. . . . . . . . . . . . . . . . 122 12. Further Topics in Analysis . . . . . . . . . . . . . . . . . . . . 155 13. Gamma Function . . . . . . .
. . . . . . . . . . . . . . . . . . 166 14. Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 192 15. Legendre Functions . . . .
. . . . . . . . . . . . . . . . . . . . 231 16. Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . 256 17. Group Theory . . . . . . .
. . . . . . . . . . . . . . . . . . . . 268 18. More Special Functions . . . . . . . . . . . . . . . . . . . . . . 286 19. Fourier Series . . . . .
. . . . . . . . . . . . . . . . . . . . . . 323 20. Integral Transforms . . . . . . . . . . . . . . . . . . . . . . . . 332 21. Integral Equations .
. . . . . . . . . . . . . . . . . . . . . . . . 364 22. Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . 373 23. Probability and
Statistics . . . . . . . . . . . . . . . . . . . . . 387 4 Correlation, Exercise Placement 398 5 Unused Sixth Edition Exercises
425 iv
4. 4. Chapter 1 Introduction The seventh edition of Mathematical Methods for Physicists is a substantial and detailed
revision of its predecessor. The changes extend not only to the topics and their presentation, but also to the
exercises that are an important part of the student experience. The new edition contains 271 exercises that were
not in previous editions, and there has been a wide-spread reorganization of the previously existing exercises to
optimize their placement relative to the material in the text. Since many instructors who have used previous
editions of this text have favorite problems they wish to continue to use, we are providing detailed tables showing
where the old problems can be found in the new edition, and conversely, where the problems in the new edition
came from. We have included the full text of every problem from the sixth edition that was not used in the new
seventh edition. Many of these unused exercises are excellent but had to be left out to keep the book within its size
limit. Some may be useful as test questions or additional study material. Complete methods of solution have been
provided for all the problems that are new to this seventh edition. This feature is useful to teachers who want to
determine, at a glance, features of the various exercises that may not be com- pletely apparent from the problem
statement. While many of the problems from the earlier editions had full solutions, some did not, and we were
unfortunately not able to undertake the gargantuan task of generating full solutions to nearly 1400 problems. Not
part of this Instructor’s Manual but available from Elsevier’s on-line web site are three chapters that were not
included in the printed text but which may be important to some instructors. These include • A new chapter
(designated 31) on Periodic Systems, dealing with mathe- matical topics associated with lattice summations and
band theory, • A chapter (32) on Mathieu functions, built using material from two chap- ters in the sixth edition, but
expanded into a single coherent presentation, and 1
5. 5. CHAPTER 1. INTRODUCTION 2 • A chapter (33) on Chaos, modeled after Chapter 18 of the sixth edition but
carefully edited. In addition, also on-line but external to this Manual, is a chapter (designated 1) on Infinite Series
that was built by collection of suitable topics from various places in the seventh edition text. This alternate Chapter
1 contains no material not already in the seventh edition but its subject matter has been packaged into a separate
unit to meet the demands of instructors who wish to begin their course with a detailed study of Infinite Series in
place of the new Mathematical Preliminaries chapter. Because this Instructor’s Manual exists only on-line, there is
an opportunity for its continuing updating and improvement, and for communication, through it, of errors in the text
that will surely come to light as the book is used. The authors invite users of the text to call attention to errors or
ambiguities, and it is intended that corrections be listed in the chapter of this Manual entitled Errata and Revision
Status. Errata and comments may be directed to the au- thors at harris at qtp.ufl.edu or to the publisher. If users
choose to forward additional materials that are of general use to instructors who are teaching from the text, they
 will be considered for inclusion when this Manual is updated. Preparation of this Instructor’s Manual has been
greatly facilitated by the efforts of personnel at Elsevier. We particularly want to acknowledge the assis- tance of
our Editorial Project Manager, Kathryn Morrissey, whose attention to this project has been extremely valuable and
is much appreciated. It is our hope that this Instructor’s Manual will have value to those who teach from
Mathematical Methods for Physicists and thereby to their students.
6. 6. Chapter 2 Errata and Revision Status Last changed: 06 April 2012 Errata and Comments re Seventh Edition text
Page 522 Exercise 11.7.12(a) This is not a principal-value integral. Page 535 Figure 11.26 The two arrowheads in
the lower part of the circular arc should be reversed in direction. Page 539 Exercise 11.8.9 The answer is incorrect;
it should be π/2. Page 585 Exercise 12.6.7 Change the integral for which a series is sought to ∞ 0 e−xv 1 + v2 dv.
The answer is then correct. Page 610 Exercise 13.1.23 Replace (−t)ν by e−πiν tν . Page 615 Exercise 13.2.6 In
the Hint, change Eq. (13.35) to Eq. (13.44). Page 618 Eq. (13.51) Change l.h.s. to B(p + 1, q + 1). Page 624 After
Eq. (13.58) C1 can be determined by requiring consistency with the recurrence formula zΓ(z) = Γ(z + 1).
Consistency with the duplication formula then determines C2. Page 625 Exercise 13.4.3 Replace “(see Fig. 3.4)”
by “and that of the recurrence formula”. Page 660 Exercise 14.1.25 Note that α2 = ω2 /c2 , where ω is the angular
frequency, and that the height of the cavity is l. 3
7. 7. CHAPTER 2. ERRATA AND REVISION STATUS 4 Page 665 Exercise 14.2.4 Change Eq. (11.49) to Eq.
(14.44). Page 686 Exercise 14.5.5 In part (b), change l to h in the formulas for amn and bmn (denominator and
integration limit). Page 687 Exercise 14.5.14 The index n is assumed to be an integer. Page 695 Exercise 14.6.3
The index n is assumed to be an integer. Page 696 Exercise 14.6.7(b) Change N to Y (two occurrences). Page
709 Exercise 14.7.3 In the summation preceded by the cosine function, change (2z)2s to (2z)2s+1 . Page 710
Exercise 14.7.7 Replace nn(x) by yn(x). Page 723 Exercise 15.1.12 The last formula of the answer should read
P2s(0)/(2s + 2) = (−1)s (2s − 1)!!/(2s + 2)!!. Page 754 Exercise 15.4.10 Insert minus sign before P1 n(cos ζ). Page
877 Exercise 18.1.6 In both (a) and (b), change 2π to √ 2π. Page 888 Exercise 18.2.7 Change the second of the
four members of the first display equation to x + ip √ 2 ψn(x), and change the corresponding member of the second
display equation to x − ip √ 2 ψn(x). Page 888 Exercise 18.2.8 Change x + ip to x − ip. Page 909 Exercise 18.4.14
All instances of x should be primed. Page 910 Exercise 18.4.24 The text does not state that the T0 term (if
present) has an additional factor 1/2. Page 911 Exercise 18.4.26(b) The ratio approaches (πs)−1/2 , not (πs)−1 .
Page 915 Exercise 18.5.5 The hypergeometric function should read 2F1 ν 2 + 1 2 , ν 2 + 1; ν + 3 2 ; z−2 . Page
916 Exercise 18.5.10 Change (n − 1 2 )! to Γ(n + 1 2 ). Page 916 Exercise 18.5.12 Here n must be an integer.
Page 917 Eq. (18.142) In the last term change Γ(−c) to Γ(2 − c). Page 921 Exercise 18.6.9 Change b to c (two
occurrences). Page 931 Exercise 18.8.3 The arguments of K and E are m. Page 932 Exercise 18.8.6 All
arguments of K and E are k2 ; In the integrand of the hint, change k to k2 .
8. 8. CHAPTER 2. ERRATA AND REVISION STATUS 5 Page 978 Exercise 20.2.9 The formula as given assumes
that Γ > 0. Page 978 Exercise 20.2.10(a) This exercise would have been easier if the book had mentioned the
integral representation J0(x) = 2 π 1 0 cos xt √ 1 − t2 dt. Page 978 Exercise 20.2.10(b) Change the argument of
the square root to x2 − a2 . Page 978 Exercise 20.2.11 The l.h.s. quantities are the transforms of their r.h.s.
counterparts, but the r.h.s. quantities are (−1)n times the transforms of the l.h.s. expressions. Page 978 Exercise
20.2.12 The properly scaled transform of f(µ) is (2/π)1/2 in jn(ω), where ω is the transform variable. The text
assumes it to be kr. Page 980 Exercise 20.2.16 Change d3 x to d3 r and remove the limits from the first integral (it
is assumed to be over all space). Page 980 Eq. (20.54) Replace dk by d3 k (occurs three times) Page 997
Exercise 20.4.10 This exercise assumes that the units and scaling of the momentum wave function correspond to
the formula ϕ(p) = 1 (2π )3/2 ψ(r) e−ir·p/ d3 r . Page 1007 Exercise 20.6.1 The second and third orthogonality
equa- tions are incorrect. The right-hand side of the second equation should read: N, p = q = (0 or N/2); N/2, (p + q
= N) or p = q but not both; 0, otherwise. The right-hand side of the third equation should read: N/2, p = q and p + q
= (0 or N); −N/2, p = q and p + q = N; 0, otherwise. Page 1007 Exercise 20.6.2 The exponentials should be
e2πipk/N and e−2πipk/N . Page 1014 Exercise 20.7.2 This exercise is ill-defined. Disregard it. Page 1015 Exercise
20.7.6 Replace (ν − 1)! by Γ(ν) (two occurrences). Page 1015 Exercise 20.7.8 Change M(a, c; x) to M(a, c, x) (two
9. 9. CHAPTER 2. ERRATA AND REVISION STATUS 6 occurrences). Page 1028 Table 20.2 Most of the references
to equation numbers did not get updated from the 6th edition. The column of references should, in its entirety,
read: (20.126), (20.147), (20.148), Exercise 20.9.1, (20.156), (20.157), (20.166), (20.174), (20.184), (20.186),
(20.203). Page 1034 Exercise 20.8.34 Note that u(t − k) is the unit step function. Page 1159 Exercise 23.5.5 This
problem should have identified m as the mean value and M as the “random variable” describing individual student
scores. Corrections and Additions to Exercise Solutions None as of now.
10. 10. Chapter 3 Exercise Solutions 1. Mathematical Preliminaries 1.1 Infinite Series 1.1.1. (a) If un < A/np the
integral test shows n un converges for p > 1. (b) If un > A/n, n un diverges because the harmonic series diverges.
1.1.2. This is valid because a multiplicative constant does not affect the conver- gence or divergence of a series.
1.1.3. (a) The Raabe test P can be written 1 + (n + 1) ln(1 + n−1 ) ln n . This expression approaches 1 in the limit of
large n. But, applying the Cauchy integral test, dx x ln x = ln ln x, indicating divergence. (b) Here the Raabe test P
can be written 1 + n + 1 ln n ln 1 + 1 n + ln2 (1 + n−1 ) ln2 n , which also approaches 1 as a large-n limit. But the
Cauchy integral test yields dx x ln2 x = − 1 ln x , indicating convergence. 1.1.4. Convergent for a1 − b1 > 1.
Divergent for a1 − b1 ≤ 1. 1.1.5. (a) Divergent, comparison with harmonic series. 7
11. 11. CHAPTER 3. EXERCISE SOLUTIONS 8 (b) Divergent, by Cauchy ratio test. (c) Convergent, comparison with
δ(2). (d) Divergent, comparison with (n + 1)−1 . (e) Divergent, comparison with 1 2 (n+1) −1 or by Maclaurin
integral test. 1.1.6. (a) Convergent, comparison with δ(2). (b) Divergent, by Maclaurin integral test. (c) Convergent,
by Cauchy ratio test. (d) Divergent, by ln 1 + 1 n ∼ 1 n . (e) Divergent, majorant is 1/(n ln n). 1.1.7. The solution is
given in the text. 1.1.8. The solution is given in the text. 1.1.10. In the limit of large n, un+1/un = 1 + 1 n + O(n−2 ).
 Applying Gauss’ test, this indicates divergence. 1.1.11. Let sn be the absolute value of the nth term of the series.
(a) Because ln n increases less rapidly than n, sn+1 < sn and limn→∞ sn = 0. Therefore this series converges.
Because the sn are larger than corre- sponding terms of the harmonic series, this series is not absolutely con-
vergent. (b) Regarding this series as a new series with terms formed by combin- ing adjacent terms of the same
sign in the original series, we have an alternating series of decreasing terms that approach zero as a limit, i.e., 1
2n + 1 + 1 2n + 2 > 1 2n + 3 + 1 2n + 4 , this series converges. With all signs positive, this series is the harmonic
series, so it is not aboslutely convergent. (c) Combining adjacent terms of the same sign, the terms of the new
series satisfy 2 1 2 > 1 2 + 1 3 > 2 1 3 , 3 1 4 > 1 4 + 1 5 + 1 6 > 3 1 6 , etc. The general form of these relations is
2n n2 − n + 2 > sn > 2 n + 1 .
12. 12. CHAPTER 3. EXERCISE SOLUTIONS 9 An upper limit to the left-hand side member of this inequality is
2/(n−1). We therefore see that the terms of the new series are decreasing, with limit zero, so the original series
converges. With all signs positive, the original series becomes the harmonic series, and is therefore not absolutely
convergent. 1.1.12. The solution is given in the text. 1.1.13. Form the nth term of δ(2)−c1α1 −c2α2 and choose c1
and c2 so that when placed over the common denominator n2 (n + 1)(n + 2) the numerator will be independent of
n. The values of the ci satisfying this condition are c1 = c2 = 1, and our resulting expansion is δ(2) = α1 + α2 + ∞
n=1 2 n2(n + 1(n + 2) = 5 4 + ∞ n=1 2 n2(n + 1(n + 2) . Keeping terms through n = 10, this formula yields δ(2) ≈
1.6445; to this precision the exact value is δ(2) = 1.6449. 1.1.14. Make the observation that ∞ n=0 1 (2n + 1)3 + ∞
n=1 1 (2n)3 = δ(3) and that the second term on the left-hand side is δ(3)/8). Our summation therefore has the value
7δ(3)/8. 1.1.15. (a) Write δ(n) − 1 as ∞ p=2 p−n , so our summation is ∞ n=2 ∞ p=2 1 pn = ∞ p=2 ∞ n=2 1 pn . The
summation over n is a geometric series which evaluates to p−2 1 − p−1 = 1 p2 − p . Summing now over p, we get
∞ p=2 1 p(p − 1) = ∞ p=1 1 p(p + 1) = α1 = 1 . (b) Proceed in a fashion similar to part (a), but now the geometric
series has sum 1/(p2 + p), and the sum over p is now lacking the initial term of α1, so ∞ p=2 1 p(p + 1) = α1 − 1
(1)(2) = 1 2 .
13. 13. CHAPTER 3. EXERCISE SOLUTIONS 10 1.1.16. (a) Write δ(3) = 1 + ∞ n=2 1 n3 − ∞ n=2 1 (n − 1)n(n + 1) +
α2 = 1 + ∞ n=2 1 n3 − 1 n(n2 − 1) + 1 4 = 1 + 1 4 − ∞ n=2 1 n3(n2 − 1) . (b) Now use α2 and α4 = ∞ n=3 1 n(n2 −
1)(n2 − 4) = 1 96 : δ(3) = 1 + 1 23 + ∞ n=3 1 n3 − ∞ n=3 1 n(n2 − 1) + α2 − 1 6 − ∞ n=3 B n(n2 − 1)(n2 − 4) + Bα4
= 29 24 + B 96 + ∞ n=3 1 n3 − 1 n(n2 − 1) − B n(n2 − 1)(n2 − 4) = 29 24 − B 96 + ∞ n=3 4 − (1 + B)n2 n(n2 − 1)(n2
− 4) . The convergence of the series is optimized if we set B = −1, leading to the final result δ(3) = 29 24 − 1 96 + ∞
n=3 4 n(n2 − 1)(n2 − 4) . (c) Number of terms required for error less than 5×10−7 : δ(3) alone, 999; combined as in
part (a), 27; combined as in part (b), 11. 1.2 Series of Functions 1.2.1. (a) Applying Leibniz’ test the series
converges uniformly for ε ≤ x < ∞ no matter how small ε > 0 is. (b) The Weierstrass M and the integral tests give
uniform convergence for 1 + ε ≤ x < ∞ no matter how small ε > 0 is chosen. 1.2.2. The solution is given in the text.
1.2.3. (a) Convergent for 1 < x < ∞. (b) Uniformly convergent for 1 < s ≤ x < ∞.
14. 14. CHAPTER 3. EXERCISE SOLUTIONS 11 1.2.4. From | cos nx| ≤ 1, | sin nx| ≤ 1 absolute and uniform
convergence follow for −s < x < s for any s > 0. 1.2.5. Since | uj+2 uj | ∼ |x|2 , |x| < 1 is needed for convergence.
1.2.6. The solution is given in the text. 1.2.7. The solution is given in the text. 1.2.8. (a) For n = 0, 1, 2, . . . we find
d4n+1 sin x dx4n+1 0 = cos x|0 = 1, d4n+2 sin x dx4n+2 0 = − sin x|0 = 0, d4n+3 sin x dx4n+3 0 = − cos x|0 = −1,
d4n sin x dx4n 0 = sin x|0 = 0. Taylor’s theorem gives the absolutely convergent series sin x = ∞ n=0 (−1)n x2n+1
(2n + 1)! . (b) Similar derivatives for cos x give the absolutely convergent series cos x = ∞ n=0 (−1)n x2n (2n)! .
1.2.9. cot x = 1 x − x 3 − x3 45 − 2x5 945 − · · · , −π < x < π. 1.2.10. From coth y = ε0 = ey + e−y ey − e−y = e2y +
1 e2y − 1 we extract y = 1 2 ln ε0 + 1 ε0 − 1 . To check this we substitute this into the first relation, giving ε0 + 1
ε0 − 1 + 1 ε0 + 1 ε0 − 1 − 1 = ε0. The series coth−1 ε0 = ∞ n=0 (ε0)−2n−1 2n + 1 follows from Exercise 1.6.1.
15. 15. CHAPTER 3. EXERCISE SOLUTIONS 12 1.2.11. (a) Since d √ x dx 0 = 1 2 √ x 0 does not exist, there is no
Maclaurin expan- sion. (b) |x − x0| < x0 because the origin must be excluded. 1.2.12. lim x→x0 f(x) g(x) = f(x + (x0
− x)) g(x + (x0 − x) = lim x→x0 f(x) + (x0 − x)f (x) + · · · g(x) + (x0 − x)g (x) + · · · = lim x→x0 f (x) g (x) , where the
intermediate formal expression f(x + (x0 − x)) g(x + (x0 − x) may be dropped. 1.2.13. (a) − ln n n − 1 = ln 1 − 1 n =
− ∞ ν=1 1 νnν . Hence 1 n − ln n n − 1 = − ∞ ν=2 1 νnν < 0. (b) ln n + 1 n = ln(1 + 1 n ) = ∞ ν=2 (−1)ν−1 νnν , 1 n −
ln n + 1 n = ∞ ν=2 (−1)ν νnν > 0. Summing (a) yields 0 > n m=2 1 m − ln 2 · 3 · · · n 1 · 2 · · · (n − 1) = n m=2 1 m −
ln n → γ − 1. Thus, γ < 1. Summing (b) yields 0 < n−1 m=2 1 m − ln 2 · 3 · · · n 1 · 2 · · · (n − 1) = n−1 m=2 1 m −
ln n → γ. Hence 0 < γ < 1. 1.2.14. The solution is given in the text. 1.2.15. The solutions are given in the text.
1.2.16. If an+1 an → 1 R then (n + 2)an+1 (n + 1)an → 1 R and an+1/(n + 2) an/(n + 1) → 1 R . 1.3 Binomial
Theorem 1.3.1. P(x) = C x 3 − x3 45 + · · · . 1.3.2. Integrating termwise tan−1 1 = π 4 = ∞ n=0 (−1)n 1 0 x2n dx = ∞
n=0 (−1)n 2n + 1 .
16. 16. CHAPTER 3. EXERCISE SOLUTIONS 13 1.3.3. The solution is given in the text. Convergent for 0 ≤ x < ∞.
The upper limit x does not have to be small, but unless it is small the convergence will be slow and the expansion
relatively useless. 1.3.4. sinh−1 x = x − 1 2 x3 3 + 1 · 3 2 · 4 x5 5 − · · · , −1 ≤ x ≤ 1. 1.3.5. The expansion of the
integral has the form 1 0 dx 1 + x2 = 1 0 1 − x2 + x4 − x6 + · · · dx = 1 − 1 3 + 1 5 − 1 7 + · · · . 1.3.6. For m = 1, 2,
. . . the binomial expansion gives (1+x)−m/2 = ∞ n=0 −m/2 n xn . By mathematical induction we show that −m/2 n =
(−1)n (m + 2n − 2)!! 2n(m − 2)!!n! . 1.3.7. (a) ν = ν 1 ± ν c + ν2 c2 + · · · . (b) ν = ν 1 ± ν c . (c) ν = ν 1 ± ν c + 1 2 ν2
c2 + · · · . 1.3.8. (a) ν1 c = δ + 1/2δ2 . (b) ν2 c = δ − 3/2δ2 + · · · . (c) ν3 c = δ − 1/2δ2 + · · · . 1.3.9. w c = 1 − α2 2
− α3 2 + · · · . 1.3.10. x = 1 2 gt2 − 1 8 g3 t4 c2 + 1 16 g5 t6 c4 − · · · . 1.3.11. E = mc2 1 − γ2 2n2 − γ4 2n4 n |k| −
3 4 + · · · . 1.3.12. The solution is given in the text. 1.3.13. The two series have different, nonoverlapping
convergence intervals. 1.3.14. (a) Differentiating the geometric series ∞ n=0 xn = 1 1 − x for x = exp(−ε0/kT) yields
x (1 − x)2 = ∞ n=1 nxn . Therefore, ε = ε0x 1 − x = ε0 eε0/kT − 1 .
17. 17. CHAPTER 3. EXERCISE SOLUTIONS 14 (b) Expanding y ey − 1 = 1 + y 2 + · · · we find ε = kT(1 + ε0 2kT + ·
· · ) = kT + ε0 + · · · . 1.3.15. (a) tan−1 x = ∞ n=0 (−1)n x 0 t2n dt = ∞ n=0 (−1)n 2n + 1 x2n+1 , |x| ≤ 1. (b) Writing x
 = tan y as ix = e2iy + 1 e2iy − 1 we extract y = − i 2 ln 1 + ix 1 − ix . 1.3.16. Start by obtaining the first few terms of
the power-series expansion of the expression within the square brackets. Write 2 + 2ε 1 + 2ε = 1 + 1 1 + 2ε = 2 −
2ε + (2ε)2 − · · · , ln(1 + 2ε ε = 1 ε 2ε − (2ε)2 2 + (2ε)3 3 − · · · 2 + 2ε 1 + 2ε − ln(1 + 2ε ε = 4 3 ε2 + O(ε3 ) .
Inserting this into the complete expression for f(ε), the limit is seen to be 4/3. 1.3.17. Let x = 1/A, and write xi1 = 1
+ (1 − x)2 2x ln 1 − x 1 + x . Expanding the logarithm, ξ1 = 1 + (1 − x)2 2x −2x − 2x3 3 − · · · = 2x − 4 3 x2 + 2 3 x3
− · · · . The similar expansion of ξ2 = 2x 1 + 2x/3 yields ξ2 = 2x − 4 3 x2 + 8 9 x3 − · · · . Comparing these
expansions, we note agreement through x2 , and the x3 terms differ by (2/9)x3 , or 2/9A3 . 1.3.18. (a) Insert the
power-series expansion of arctan t and carry out the inte- gration. The series for β(2) is obtained. (b) Integrate by
parts, converting ln x into 1/x and 1/(1+x2 ) into arctan x. The integrated terms vanish, and the new integral is the
negative of that already treated in part (a).
18. 18. CHAPTER 3. EXERCISE SOLUTIONS 15 1.4 Mathematical Induction 1.4.1. Use mathematical induction. First
evaluate the claimed expression for the sum for both n − 1 and n: Sn−1 = n − 1 30 (2n − 1)n 3(n − 1)2 + 3(n − 1) −
1 = n5 5 − n4 2 + n3 3 − n 30 Sn = n 30 (2n + 1)(n + 1)(3n2 + 3n − 1) = n5 5 + n4 2 + n3 3 − n 30 Next verify that
Sn = Sn−1 + n4 . Complete the proof by verifying that S1 = 1. 1.4.2. Use mathematical induction. First, differentiate
the Leibniz formula for n − 1, getting the two terms n−1 j=0 n − 1 j d dx j+1 f(x) d dx n−1−j g(x) + n−1 j=0 n − 1 j d
dx j f(x) d dx n−j g(x) Now change the index of the first summation to (j − 1), with j ranging from 1 to n; the index
can be extended to j = 0 because the binomial coefficient n−1 −1 vanishes. The terms then combine to yield n j=0
n − 1 j − 1 + n − 1 j d dx j f(x) d dx n−j g(x) The sum of two binomial coefficients has the value n j , thereby
confirming that if the Leibnize formula is correct for n − 1, it is also correct for n. One way to verify the binomial
coefficient sum is to recognize that it is the number of ways j of n objects can be chosen: either j − 1 choices are
made from the first n − 1 objects, with the nth object the jth choice, or all j choices are made from the first n − 1
objects, with the nth object remaining unchosen. The proof is now completed by noticing that the Leibniz formula
gives a correct expression for the first derivative. 1.5 Operations on Series Expansions of Functions 1.5.1. The
partial fraction expansion is 1 1 − t2 = 1 2 1 1 + t + 1 1 − t ,
19. 19. CHAPTER 3. EXERCISE SOLUTIONS 16 with integral x −x dt 1 − t2 = 1 2 ln(1 + x) − ln(1 − x) x −x = 1 2 ln 1 +
x 1 − x x −x . The upper and lower limits give the same result, canceling the factor 1/2. 1.5.2. Start by writing the
partial-fraction expansion for p+1 using the assumed form of that for p multiplied by an additional factor 1/(n + p +
1). Thus, we want to see if we can simplify 1 p ! p j=0 p j (−1)j n + j 1 n + p + 1 to get the expected formula. Our
first step is to expand the two factors containing n into partial fractions: 1 (n + j)(n + p + 1) = 1 p + 1 − j 1 n + j − 1 n
+ p + 1 Replacing the 1/(n + j) term of our original expansion using this result and adding a new 1/(n+p+1) term
which is the summation of the above result for all j, we reach p j=0 (−1)j n + j 1 p ! p j 1 p + 1 − j + p j=0 1 p ! p j 1 p
+ 1 − j (−1)j−1 n + p + 1 Using the first formula supplied in the Hint, we replace each square bracket by the quantity
1 (p + 1)! p + 1 j , thereby identifying the first summation as all but the last term of the partial-fraction expansion for

second formula supplied in the Hint, we now identify the quan- tity within square brackets as p+1 j=1 (−1)j−1 p + 1 j
− (−1)p p + 1 p + 1 + (−1)−1 p + 1 0 = 1 + (−1)p+1 − 1 = (−1)p+1 ,
20. 20. CHAPTER 3. EXERCISE SOLUTIONS 17 so the second summation reduces to (−1)p+1 (p + 1)! 1 n + p + 1 ,
as required. Our proof by mathematical induction is now completed by observing that the partial-fraction formula is
correct for the case p = 0. 1.5.3. The formula for un(p) follows directly by inserting the partial fraction
decomposition. If this formula is summed for n from 1 to infinity, all terms cancel except that containing u1, giving
the result ∞ n=1 un(p) = u1(p − 1) p . The proof is then completed by inserting the value of u1(p − 1). 1.5.4. After
inserting Eq. (1.88) into Eq. (1.87), make a change of summation variable from n to p = n − j, with the ranges of j
and p both from zero to infinity. Placing the p summation outside, and moving quantities not dependent upon j
outside the j summation, reach f(x) = ∞ p=0 (−1)p cp xp (1 + x)p+1 ∞ j=0 p + j j x 1 + x j . Using now Eq. (1.71), we
identify the binomial coefficient in the above equation as p + j j = (−1)j −p − 1 j , so the j summation reduces to ∞
j=0 −p − 1 j − x 1 + x j = 1 − x 1 + x −p−1 = (1 + x)p+1 . Insertion of this expression leads to the recovery of Eq.
(1.86). 1.5.5. Applying Eq. (1.88) to the coefficients in the power-series expansion of arctan(x), the first 18 an (a0
through a17) are: 0, −1, 2, −8/3, 8/3, −28/15, 8/15, 64/105, −64/105,−368/15, 1376/315, 1376/315, −25216/3465,
25216/3465, −106048/45045, −305792/45045, 690176/45045, −690176/45045, 201472/765765. Using these in
Eq. (1.87) for x = 1, the terms through a17 yield the approximate value arctan(1) ≈ 0.785286, fairly close to the
exact value at this precision, 0.785398. For this value of x, the 18th nonzero term in the power series is −1/35,
showing that a power series for x = 1 cut off after 18 terms would barely give a result good to two significant
figures. The 18-term Euler expansion yields arctan(1/ √ 3) ≈ 0.523598, while the exact value at this precision is
0.523599.
21. 21. CHAPTER 3. EXERCISE SOLUTIONS 18 1.6 Some Important Series 1.6.1. For |x| < 1, ln 1 + x 1 − x = ∞ ν=0
xν ν (−1)ν−1 + 1 = 2 ∞ n=0 x2n+1 2n + 1 . 1.7 Vectors 1.7.1. Ax = Ay = Az = 1. 1.7.2. The triangle sides are given
by AB = B − A, BC = C − B, CA = A − C with AB + BC + CA = (B − A) + (C − B) + (A − C) = 0. 1.7.3. The solution is
given in the text. 1.7.4. If vi = vi − v1, ri = ri − r1, are the velocities and distances, respectively, from the galaxy at
r1, then vi = H0(ri − r1) = H0ri holds, i.e., the same Hubble law. 1.7.5. With one corner of the cube at the origin, the
space diagonals of length√ 3 are: (1, 0, 1) − (0, 1, 0) = (1, −1, 1), (1, 1, 1) − (0, 0, 0) = (1, 1, 1), (0, 0, 1) − (1, 1, 0)
= (−1, −1, 1), (1, 0, 0) − (0, 1, 1) = (1, −1, −1). The face diagonals of length √ 2 are: (1, 0, 1) − (0, 0, 0) = (1, 0, 1),
(1, 0, 0) − (0, 0, 1) = (1, 0, −1); (1, 0, 0) − (0, 1, 0) = (1, −1, 0), (1, 1, 0) − (0, 0, 0) = (1, 1, 0); (0, 1, 0) − (0, 0, 1) =
(0, 1, −1), (0, 1, 1) − (0, 0, 0) = (0, 1, 1). 1.7.6. (a) The surface is a plane passing through the tip of a and
perpendicular to a. (b) The surface is a sphere having a as a diameter: (r − a) · r = (r − a/2)2 − a2 /4 = 0. 1.7.7. The
solution is given in the text. 1.7.8. The path of the rocket is the straight line r = r1 + tv, or in Cartesian coordinates
x(t) = 1 + t, y(t) = 1 + 2t, z(t) = 1 + 3t.
22. 22. CHAPTER 3. EXERCISE SOLUTIONS 19 We now minimize the distance |r − r0| of the observer at the point r0
= (2, 1, 3) from r(t), or equivalently (r−r0)2 =min. Differentiating the rocket path with respect to t yields ˙r = ( ˙x, ˙y,
˙z) = v. Setting d dt (r − r0)2 = 0 we obtain the condition 2(r − r0) · ˙r = 2[r1 − r0 + tv] · v = 0. Because ˙r = v is the
tangent vector of the line, the geometric meaning of this condition is that the shortest distance vector through r0 is
perpendicular to the line, or the velocity of the rocket. Now solving for t yields the ratio of scalar products t = − (r1 −
r0) · v v2 = − (−1, 0, −2) · (1, 2, 3) (1, 2, 3) · (1, 2, 3) = 1 + 0 + 6 1 + 4 + 9 = 1 2 . Substituting this parameter value
into the rocket path gives the point rs = (3/2, 2, 5/2) on the line that is closest to r0. The shortest distance is d = |r0
− rs| = |(−1/2, 1, −1/2)| = 2/4 + 1 = 3/2. 1.7.9. Consider each corner of the triangle to have a unit of mass and be
located at ai from the origin where, for example, a1 = (2, 0, 0), a2 = (4, 1, 1), a3 = (3, 3, 2). Then the center of
mass of the triangle is 1 3 (a1 + a2 + a3) = acm = 1 3 (2 + 4 + 3, 1 + 3, 1 + 2) = 3, 4 3 , 1 . The three midpoints are
located at the point of the vectors 1 2 (a1 + a2) = 1 2 (2 + 4, 1, 1) = 3, 1 2 , 1 2 1 2 (a2 + a3) = 1 2 (4 + 3, 1 + 3, 1 +
2) = 7 2 , 2, 3 2 1 2 (a3 + a1) = 1 2 (2 + 3, 3, 2) = 5 2 , 3 2 , 1 .
23. 23. CHAPTER 3. EXERCISE SOLUTIONS 20 We start from each corner and end up in the center as follows (2, 0,
0) + 2 3 7 2 , 2, 3 2 − (2, 0, 0) = 3, 4 3 , 1 a1 + 2 3 1 2 (a2 + a3) − a1 = 1 3 (a1 + a2 + a3), (4, 1, 1) + 2 3 5 2 , 3 2 ,
1 − (4, 1, 1) = 3, 4 3 , 1 a2 + 2 3 1 2 (a1 + a3) − a2 = 1 3 (a1 + a2 + a3), (3, 3, 2) + 2 3 3, 1 2 , 1 2 − (3, 3, 2) = 3, 4
3 , 1 a3 + 2 3 1 2 (a1 + a2) − a3 = 1 3 (a1 + a2 + a3). 1.7.10. A2 = A2 = (B − C)2 = B2 + C2 − 2BC cos ζ with ζ the
angle between ˆB and ˆC. 1.7.11. P and Q are antiparallel; R is perpendicular to both P and Q. 1.8 Complex
Numbers and Functions 1.8.1. (a) (x + iy)−1 = x − iy x2 + y2 . (b) x + iy = reiζ gives (x + iy)−1 = e−iζ r = 1 r (cos ζ −
i sin ζ) = x − iy r2 = x − iy x2 + y2 . 1.8.2. If z = reiζ , √ z = √ reiζ/2 = √ r(cos ζ/2 + i sin ζ/2). In particular, √ i = eiπ/4
= 1 + i √ 2 or √ i = e−i3π/4 . 1.8.3. einζ = cos nζ+i sin nζ = (eiζ )n = (cos ζ+i sin ζ)n = n ν=0 n ν cosn−ν ζ(i sin ζ)ν .
Separating real and imaginary parts we have cos nζ = [n/2] ν=0 (−1)ν n 2ν cosn−2ν ζ sin2ν ζ, sin nζ = [n/2] ν=0
(−1)ν n 2ν + 1 cosn−2ν−1 ζ sin2ν+1 ζ.
24. 24. CHAPTER 3. EXERCISE SOLUTIONS 21 1.8.4. N−1 n=0 (eix )n = 1 − eiNx 1 − eix = eiNx/2 eix/2 eiNx/2 −
e−iNx/2 eix/2 − e−ix/2 = ei(N−1)x/2 sin(Nx/2)/ sin(x/2). Now take real and imaginary parts to get the result. 1.8.5.
(a) sinh(iz) = ∞ n=0 (iz)2n+1 (2n + 1)! = i ∞ n=0 (−1)n z2n+1 (2n + 1)! = i sin z. All other identities are shown
similarly. (b) ei(z1+z2) = cos(z1 + z2) + i sin(z1 + z2) = eiz1 eiz2 = (cos z1 + i sin z1)(cos z2 + i sin z2) = cos z1
cos z2 − sin z1 sin z2 + i(sin z1 cos z2 + sin z2 cos z1). Separating this into real and imaginary parts for real z1, z2
proves the addition theorems for real arguments. Analytic continuation extends them to the complex plane. 1.8.6.
(a) Using cos iy = cosh y, sin iy = i sinh y, etc. and the addition theorem we obtain sin(x + iy) = sin x cosh y + i cos
x sinh y, etc. (b) | sin z|2 = sin(x + iy) sin(x − iy) = sin2 x cosh2 y + cos2 x sinh2 y = sin2 x(cosh2 y − sinh2 y) +
sinh2 y = sin2 x + sinh2 y, etc. 1.8.7. (a) Using cos iy = cosh y, sin iy = i sinh y, etc. and the addition theorem we
obtain sinh(x + iy) = sinh x cos y + i cosh x sin y, etc. (b) | cosh(x+iy)|2 = cosh(x+iy) cosh(x−iy) = cosh2 x cos2
y+sinh2 x sin2 y = sinh2 x + cos2 y, etc. 1.8.8. (a) Using Exercise 1.8.7(a) and rationalizing we get tanh(x + iy) =
sinh x cos y + i cosh x sin y cosh x cos y + i sinh x sin y = 1 2 sinh 2x(cos2 y + sin2 y) + i 2 sin 2y(cosh2 x − sinh2
x) cosh2 x cos2 y + sinh2 x sin2 y = 1 2 sinh 2x + i sin 2y cos2 y + sinh2 x = sinh 2x + sin 2y cos 2y + cosh 2x . (b)
Starting from cosh(x + iy) sinh(x + iy) this is similarly proved.
25. 25. CHAPTER 3. EXERCISE SOLUTIONS 22 1.8.9. The expansions relevant to this exercise are tan−1 x = x − x3
3 + x5 5 − · · · ln(1 − ix) = −ix + x2 2 + ix3 3 − · · · ln(1 + ix) = ix + x2 2 − ix3 3 − · · · The desired identity follows
directly by comparing the expansion of tan−1 x with i/2 times the difference of the other two expansions. 1.8.10. (a)
The cube roots of −1 are −1, eπi/3 = 1/2 + i √ 3/2, and e−πi/3 = 1/2 − i √ 3/2, so our answers are −2, 1 + i √ 3, and
1 − i √ 3. (b) Write i as eπi/2 ; its 1/4 power has values e(πi/2+2nπ)/4 for all integer n; there are four distinct values:
eiπ/8 = cos π/8 + i sin π/8, e5iπ/8 = cos 5π/8 + i sin 5π/8, e9iπ/8 = −eiπ/8 , and e13iπ/8 = −e5iπ/8 . (c) eiπ/4 has
the unique value cos π/4 + i sin π/4 = (1 + i)/ √ 2. 1.8.11. (a) (1 + i)3 has a unique value. Since 1 + i has magnitude
√ 2 and is at an angle of 45◦ = π/4, (1 + i)3 will have magnitude 23/2 and argument 3π/4, so its polar form is 23/2
e3iπ/4 . (b) Since −1 = eπi , its 1/5 power will have values e(2n+1)πi for all integer n. There will be five distinct
values: ekπi/5 with k = 1, 3, 5, 7, and 9. 1.9 Derivatives and Extrema 1.9.1. Expand first as a power series in x,
with y kept at its actual value. Then expand each term of the x expansion as a power series in y, regarding x as
fixed. The nth term of the x expansion will be xn n! ∂ ∂x n f(x, y) x=0,y=0 The mth term in the y expansion of the xn
term is therefore xn n! ym m! ∂ ∂y m ∂ ∂x n f(x, y) x=0,y=0 The coefficient in the above equation can be written 1
(m + n)! (m + n)! m!n! = 1 (m + n)! m + n n . Using the right-hand side of the above equation and collecting together
all terms with the same value of m + n, we reach the form given in the exercise.
26. 26. CHAPTER 3. EXERCISE SOLUTIONS 23 1.9.2. The quantities αi are regarded as independent of the xi when
the differ- entiations are applied. Then, expansion of the differential operator raised to the power n will, when
combined with tn , produce terms with a total of n derivatives applied to f, with each term containing a power of
each xi equal to the number of times xi was differentiated. The coefficient of each distinct term in the expansion of
this nth order derivative will be the number of ways that derivative combination occurs in the expansion; the term in
which each xj derivative is applied nj times occurs in the following number of ways: n! n1!n2! · · · , with the sum of
the ni equal to n. Inserting this formula, we obtain the same result that would be obtained if we expanded, first in
x1, then in x2, etc. 1.10 Evaluation of Integrals 1.10.1. Apply an integration by parts to the integral in Table 1.2
defining the gamma function, for integer n > 0: Γ(n) = ∞ 0 tn−1 e−t dt = tn n e−t ∞ 0 + ∞ 0 tn n e−x dx = Γ(n + 1) n .
Rearranging to Γ(n + 1) = nΓ(n), we apply mathematical induction, not- ing that if Γ(n) = (n−1)!, then also Γ(n+1) =
n!. To complete the proof, we directly evaluate the integral Γ(1) = ∞ 0 e−x dx = 1, which is 0!. 1.10.2. This integral
can also be evaluated using contour integration (see Exam- ple 11.8.5). A method motivated by the discussion of
this section starts by multiplying the integrand by e−αx and considering the value of this integral when α = 0. We
can start by differentiating the integral by the parameter α, corresponding to I(α) = ∞ 0 sin xe−αx x dx, I (α) = − ∞ 0
e−αx sin x dx = − 1 α2 + 1 , where the integral for I is ientified as having the value found in Example 1.10.4. We
 now integrate the expression for I , writing it as the indefinite integral I(α) = − tan−1 α + C . The value of C is now
determined from the value of I(∞), which from the form of I must be zero. Thus, C = tan−1 ∞ = π/2, and, since
tan−1 0 = 0, we find I(0) = π/2. 1.10.3. Write the integrand as 1 cosh x = 2 ex + e−x = 2e−x 1 + e−2x = 2(e−x −
e−3x + e−5x − · · · ).
27. 27. CHAPTER 3. EXERCISE SOLUTIONS 24 Now integrate term by term; each integrand is a simple exponential.
The result is 2 1 − 1 3 + 1 5 − 1 7 + · · · . The series in parentheses is that discussed in Exercise 1.3.2, with value
π/4. Our integral therefore has value π/2. 1.10.4. Expand the integrand as a power series in e−ax and integrate
term by term: ∞ 0 dx eax + 1 = ∞ 0 e−ax − e−2ax + e−3ax − · · · = 1 a − 1 2a + 1 3a − · · · After factoring out (1/a),
the series that remains is that identified in Eq. (1.53) as ln 2, so our integral has value ln(2)/a. 1.10.5. Integrate by
parts, to raise the power of x in the integrand: ∞ π sin x x2 dx = ∞ π cos x x dx . Note that the integrated terms
vanish. The integral can now be recognized (see Table 1.2) as −Ci(π). 1.10.6. This is a case of the integral I(α)
defined in the solution of Exercise 1.10.2, with α = 1. We therefore have I(α) = π 2 − tan−1 α; I(1) = π 2 − π 4 = π 4
. 1.10.7. Write erf as an integral and interchange the order of integration. We get x 0 erf(t) dt = 2 √ π x 0 dx t 0
e−u2 du = 2 √ π x 0 e−u2 du x u dt = 2 √ π x 0 e−u2 (x − u)du = x erf(x) − 1 √ π x 0 2ue−u2 du = x erf(x) + 1 √ π
e−x2 − 1 . 1.10.8. Write E1 as an integral and interchange the order of integration. Now the outer (u) integration
must be broken into two pieces: x 1 E1(t)dt = x 1 dt ∞ t e−u u du = x 1 e−u u du u 1 dt + ∞ x e−u u du x 1 dt = x 1
e−u u (u − 1) du + ∞ x e−u u (x − 1) du = e−1 − e−x − E1(1) + E1(x) + (x − 1)E1(x) = e−1 − e−x − E1(1) + xE1(x).
28. 28. CHAPTER 3. EXERCISE SOLUTIONS 25 1.10.9. Change the variable of integration to y = x + 1, leading to ∞
0 e−x x + 1 dx = ∞ 1 e−y+1 y dy = e E1(1). 1.10.10. After the integration by parts suggested in the text, with [tan−1
x]2 dif- ferentiated and dx/x2 integrated, the result is I(1), where I(a) = ∞ 0 2 tan−1 ax x(x2 + 1) dx We now
differentiate I(a) with respect to the parameter a, reaching after a partial-fraction decomposition I (a) = 2 ∞ 0 dx (x2
+ 1)(a2x2 + 1) = 2 1 − a2 ∞ 0 1 x2 + 1 − a2 a2x2 + 1 dx = 2 1 − a2 π 2 − a2 π 2a = π 1 + a . Integrating with
respect to a, we get I(a) = π ln(1 + a) + C, with C set to zero to obtain the correct result I(0) = 0. Then, setting a =
1, we find I(1) = π ln 2, as required. 1.10.11. Integrating over one quadrant and multiplying by four, the range of x
is (0, a) and, for given x, the range of y is from 0 to the positive y satisfying the equation for the ellipse. Thus, A = 4
a 0 dx b √ a2−x2/a 0 dy = 4b a a 0 a2 − x2 dx = 4b a a2 π 4 = πab. 1.10.12. Draw the dividing line at y = 1/2. Then
the contribution to the area for each y between 1/2 and 1 is 2 1 − y2, so A = 2 1 1/2 1 − y2 dy = π 3 − √ 3 4 . A
simple explanation of these two terms is that π/3 is the area of the sector that includes the piece in question, while
√ 3/4 is the area of the triangle that is the part of the sector not included in that piece. 1.11 Dirac Delta Function
1.11.1. The mean value theorem gives lim n→∞ f(x)δn(x)dx = lim n→∞ n 1/2n −1/2n f(x)dx = lim n→∞ n n f(ξn) =
f(0), as − 1 2n ≤ ξn ≤ 1 2n .
29. 29. CHAPTER 3. EXERCISE SOLUTIONS 26 1.11.2. Use the elementary integral dx 1 + x2 = arctan z, thus
reaching ∞ −∞ dx 1 + n2x2 = π n . 1.11.4. ∞ −∞ f(x)δ(a(x − x1))dx = 1 a ∞ −∞ f((y + y1)/a)δ(y)dy = 1 a f y1 a = 1 a
f(x1) = ∞ −∞ f(x)δ(x − x1) dx a . 1.11.5. The left-hand side of this equation is only nonzero in the neighborhood of x
= x1, where it is a case of Exercise 1.11.4, and in the neighborhood of x = x2, where it is also a case of Exercise
1.11.4. In both cases, the quantity playing the role of a is |x1 − x2|. 1.11.7. Integrating by parts we find ∞ −∞ δ
(x)f(x) dx = − ∞ −∞ f (x)δ(x)dx = −f (0). 1.11.9. (a) Inserting the given form for δn(x) and changing the variable of
inte- gration to nx, we obtain a result that is independent of n. The indefinite integral of 1/ cosh2 x is tanh(x), which
approaches +1 as x → +∞ and −1 as x → −∞, thus confirming the normalization claimed for δn. (b) The behavior of
tanh(x) causes the right-hand side of this equation to approach zero for large n and negative x, but to approach +1
for large n and positive x.
30. 30. CHAPTER 3. EXERCISE SOLUTIONS 27 2. Determinants and Matrices 2.1 Determinants 2.1.1. (a) −1. (b)
−11. (c) 9/ √ 2. 2.1.2. The determinant of the coefficients is equal to 2. Therefore no nontrivial solution exists. 2.1.3.
Given the pair of equations x + 2y = 3, 2x + 4y = 6. (a) Since the coefficients of the second equation differ from
those of the first one just by a factor 2, the determinant of (lhs) coefficients is zero. (b) Since the inhomogeneous
terms on the right-hand side differ by the same factor 2, both numerator determinants also vanish. (c) It suffices to
solve x + 2y = 3. Given x, y = (3 − x)/2. This is the general solution for arbitrary values of x. 2.1.4. (a) Cij is the
quantity that multiplies aij in the expansion of the deter- minant. The sum over i collects the quantities that multiply
all the aij in column j of the determinant. (b) These summations form determinants in which the same column (or
row) appears twice; the determinant is therefore zero, 2.1.5. The solution is given in the text. 2.1.6. If a set of forms
is linearly dependent, one of them must be a linear combination of others. Form the determinant of their
coefficients (with each row describing one of the forms) and subtract from one row the linear combination of other
rows that reduces that row to zero. The determinant (whose value is not changed by the operation) will be seen to
be zero. 2.1.7. The Gauss elimination yields 10x1 + 9x2 + 8x3 + 4x4 + x5 = 10, x2 + 2x3 + 3x4 + 5x5 + 10x6 = 5,
10x3 + 23x4 + 44x5 − 60x6 = −5, 16x4 + 48x5 − 30x6 = 15, 48x5 + 498x6 = 215, −11316x6 = −4438, so x6 =
2219/5658, x5 = (215 − 498x6)/48, x4 = (15 + 30x6 − 48x5)/16,
31. 31. CHAPTER 3. EXERCISE SOLUTIONS 28 x3 = (−5 + 60x6 − 44x5 − 23x4)/10, x2 = 5 − 10x6 − 5x5 − 3x4 −
2x3, x1 = (10 − x5 − 4x4 − 8x3 − 9x2)/10. 2.1.8. (a) δii = 1 (not summed) for each i = 1, 2, 3. (b) δijεijk = 0 because
δij is symmetric in i, j while εijk is antisymmetric in i, j. (c) For each ε in εipqεjpq to be non-zero, leaves only one
value for i and j, so that i = j. Interchanging p and q gives two terms, hence the factor 2. (d) There are 6
permutations i, j, k of 1, 2, 3 in εijkεijk = 6. 2.1.9. Given k implies p = q for εpqk = 0. For εijk = 0 requires either i = p
and so j = q, or i = q and then j = p. Hence εijkεpqk = δipδjp − δiqδjp. 2.2 Matrices 2.2.1. Writing the product
matrices in term of their elements, AB = ( m aimbmk), BC = ( n bincnk), (AB)C = n m aimbmn cnk = mn aimbmncnk
= A(BC) = m aim n bmncnk , because products of real and complex numbers are associative the paren- theses
can be dropped for all matrix elements. 2.2.2. Multiplying out (A + B)(A − B) = A2 + BA − AB − B2 = A2 − B2 + [B,
A]. 2.2.3. (a) (a1 + ib1) − (a2 + ib2) = a1 − a2 + i(b1 − b2) corresponds to a1 b1 −b1 a1 − a2 b2 −b2 a2 = a1 − a2
 b1 − b2 −(b1 − b2) a1 − a2 , i.e., the correspondence holds for addition and subtraction. Similarly, it holds for
multiplication because first (a1 + ib1)(a2 + ib2) = (a1a2 − b1b2) + i(a1b2 + a2b1) and matrix multiplication yields a1
b1 −b1 a1 a2 b2 −b2 a2 = a1a2 − b1b2 a1b2 + a2b1 −(a1b2 + a2b1) a1a2 − b1b2 .
32. 32. CHAPTER 3. EXERCISE SOLUTIONS 29 (b) (a + ib)−1 ←→ a/(a2 + b2 ) −b/(a2 + b2 ) b/(a2 + b2 ) a/(a2 + b2
) . 2.2.4. A factor (−1) can be pulled out of each row giving the (−1)n overall. 2.2.5. (a) First we check that ab b2
−a2 −ab ab b2 −a2 −ab = a2 b2 − a2 b2 ab3 − ab3 −a3 b + a3 b −a2 b2 + a2 b2 = 0. Second, to find the
constraints we write the general matrix as A B C D A B C D = A2 + BC B(A + D) C(A + D) BC + D2 = 0 giving D =
−A, D2 = −BC = A2 . This implies, if we set B = b2 , C = −a2 without loss of generality, that A = ab = −D. 2.2.6. n =
6. 2.2.7. Expanding the commutators we find [A, [B, C]] = A[B, C] − [B, C]A = ABC − ACB − BCA + CBA, [B, [A, C]]
= BAC − BCA − ACB + CAB, [C, [A, B]] = CAB − CBA − ABC + BAC, and subtracting the last double commutator
from the second yields the first one, since the BAC and CAB terms cancel. 2.2.8. By direct multiplication of the
matrices we find [A, B] = AB = C, BA = 0, etc. 2.2.9. These results can all be verified by carrying out the indicated
matrix multiplications. 2.2.10. If aik = 0 = bik for i > k, then also m aimbmk = i≤m≤k aimbmk = 0, as the sum is
empty for i > k. 2.2.11. By direct matrix multiplications and additions. 2.2.12. By direct matrix multiplication we
verify all claims. 2.2.13. By direct matrix multiplication we verify all claims.
33. 33. CHAPTER 3. EXERCISE SOLUTIONS 30 2.2.14. For i = k and aii = akk we get for the product elements (AB)ik
= ( n ainbnk) = (aiibik) = (BA)ik = ( n binank) = (bikakk). Hence bik = 0 for i = k. 2.2.15. m aimbmk = aiibiiδik = m
bimamk. 2.2.16. Since trace ABC = trace BCA, choose one of the foregoing in which two commuting matrices
appear adjacent to each other and interchange their order. Then make a cyclic permutation if needed to reach
CBA. 2.2.17. Taking the trace, we find from [Mi, Mj] = iMk that i trace(Mk) = trace(MiMj − MjMi) = trace(MiMj) −
trace(MiMj) = 0. 2.2.18. Taking the trace of A(BA) = −A2 B = −B yields −tr(B) = tr(A(BA)) = tr(A2 B) = tr(B). 2.2.19.
(a) Starting from AB = −BA, multiply on the left by B−1 and take the trace. After simplification, we get trace B =
−trace B, so trace B = 0. 2.2.20. This is proved in the text. 2.2.21. (a) A unit matrix except that Mii = k, (b) A unit
matrix except that Mim = −K, (c) A unit matrix except that Mii = Mmm = 0 and Mmi − Mim = 1. 2.2.22. Same

(a) states that T moves people from area j but does not change their total number. (b) Write the component
equation j TijPj = Qi and sum over i. This summation replaces Tij by unity, leaving that the sum over Pj equals the
sum over Qi, hence conserving people. 2.2.25. The answer is given in the text. 2.2.26. If O−1 i = ˜Oi, i = 1, 2, then
(O1O2)−1 = O−1 2 O−1 1 = ˜O2 ˜O1 = O1O2. 2.2.27. Taking the determinant of ˜AA = 1 and using the product
theorem yields det(˜A) det(A) = 1 = det2 (A) implying det(A) = ±1. 2.2.28. If ˜A = −A, ˜S = S, then trace(SA) =
trace(SA) = trace(˜A˜S) = −trace(AS).
34. 34. CHAPTER 3. EXERCISE SOLUTIONS 31 2.2.29. From ˜A = A−1 and det(A) = 1 we have A−1 = a22 −a12
−a21 a11 = ˜A = a11 a21 a12 a22 . This gives det(A) = a2 11 + a2 12 = 1, hence a11 = cos ζ = a22, a12 = sin ζ =
−a21, the standard 2 × 2 rotation matrix. 2.2.30. Because ε is real, det(A∗ ) = ik εi1i2...in a∗ 1i1 a∗ 2i2 · · · a∗ nin =
ik εi1i2...in a1i1 a2i2 · · · anin ∗ = (det A)∗ . Because, for any A, det(A) = det(˜A), det(A∗ ) = det(A† ). 2.2.31. If Jx
and Jy are real, so also must be their commutator, so the commuta- tion rule requires that Jz be pure imaginary.
2.2.32. (AB)† = A∗B∗ = ˜B∗ ˜A∗ = B† A† . 2.2.33. As Cjk = n S∗ njSnk, trace (C) = nj |Snj|2 . 2.2.34. If A† = A, B† =
B, then (AB + BA)† = B† A† + A† B† = AB + BA, −i(B† A† − A† B† ) = i(AB − BA). 2.2.35. If C† = C, then (iC−)† ≡
(C† − C)† = C − C† = −iC† −, i.e. (C−)† = C−. Similarly C† + = C+ = C + C† . 2.2.36. −iC† = (AB − BA)† = B† A† −
A† B† = BA − AB = −iC. 2.2.37. (AB)† = B† A† = BA = AB yields [A, B] = 0 as the condition, that is, the answer in
the text. 2.2.38. (U† )† = U = (U−1 )† . 2.2.39. (U1U2)† = U† 2U† 1 = U−1 2 U−1 1 = (U1U2)−1 . 2.2.40. Start by
noting the relationships ζiζj + ζjζi = 0 if i = j, and ζ2 i = 12; see Eq. (2.59); for proof add Eqs. (2.29) and (2.30).
Then, (p · ζ)2 = (pxζ1 + pyζ2 + pzζ3)2 expands to p2 xζ2 1 + p2 yζ2 2 + p2 zζ2 3 + pxpy(ζ1ζ2 + ζ2ζ1) +
pxpz(ζ1ζ3 + ζ3ζ1) + pypz(ζ1ζ2 + ζ2ζ1) = p2 x + p2 y + p2 z = p2 .
35. 35. CHAPTER 3. EXERCISE SOLUTIONS 32 2.2.41. Writing γ0 = ζ3 ⊗ 1 and γi = γ ⊗ ζi (i = 1, 2, 3), where γ = 0
1 −1 0 , and noting fron Eq. (2.57) that if C = A ⊗ B and C = A ⊗ B then CC = AA ⊗ BB , (γ0 )2 = ζ2 3 ⊗ 12 2 =
12 ⊗ 12 = 14, (γi )2 = γ2 ⊗ ζ2 i = (−12) ⊗ 12 = −14 γ0 γi = ζ3γ ⊗ 12ζi = ζ1 ⊗ ζi, γi γ0 = γζ3 ⊗ ζi12 = (−ζ1) ⊗
ζi γi γj = γ2 ⊗ ζiζj γj γi = γ2 ⊗ ζjζi It is obvious from the second line of the above equation set that γ0 γi + γi γ0 =
0; from the third line of the equation set we find γi γj + γj γi is zero if j = i because then ζjζi = −ζiζj. 2.2.42. The
anticommutation can be demonstrated by matrix multiplication. 2.2.43. These results can be confirmed by carrying
out the indicated matrix op- erations. 2.2.44. Since γ2 5 = 14, 1 4 (14 + γ5)2 = 1 4 (14 + 2γ5 + 14) = 1 2 (14 + γ5).
2.2.47. Since ˜C = −C = C−1 , and Cγ0 C−1 = −γ0 = −˜γ0 , Cγ2 C−1 = −γ2 = −˜γ2 , Cγ1 C−1 = γ1 = −˜γ1 , Cγ3 C−1
= γ3 = −˜γ3 , we have CγµC−1 = ˜C−1 ˜γµ ˜C = C˜γµ C−1 = −˜γµ . 2.2.48. (a) Written as 2 × 2 blocks, the matrices
αi and the wave function Ψ are αi = 0 ζi ζi 0 and Ψ = ΨL ΨS . In block form, Eq. (2.73) becomes mc2 0 0 −mc2 +
0 ζ1px ζ1px 0 + 0 ζ2py ζ2py 0 + 0 ζ3pz ζ3pz 0 ΨL ΨS = E ΨL ΨS
36. 36. CHAPTER 3. EXERCISE SOLUTIONS 33 The solution is completed by moving the right-hand side of the
above equation to the left, written in the form −E 0 0 −E and combining all the terms by matrix addition. 2.2.49. The
requirements the gamma matrices must satisfy are Eqs. (2.74) and (2.75). Use the same process that was
illustrated in the solution to Exer- cise 2.2.41, but now with γ0 = ζ1 ⊗ 12. 2.2.50. In the Weyl representation, the
matrices αi and the wave function Ψ, written as 2 × 2 blocks, take the forms αi = −ζi 0 0 ζi and Ψ1 Ψ2 . Then
proceed as in the solution to Exercise 2.2.48, obtaining the matrix equation 0 mc2 mc2 0 + −ζ · p 0 0 ζ · p Ψ1 Ψ2
= E Ψ1 Ψ2 . Here we wrote ζ · p for ζ1px + ζ2py + ζ3pz. If m is negligible, this matrix equation becomes two
independent equa- tions, one for Ψ1, and one for Ψ2. In this limit, one set of solutions will be with Ψ2 = 0 and Ψ1 a
solution to −ζ · pΨ1 = EΨ1; a second set of solutions will have zero Ψ1 and a set of Ψ2 identical to the previously
found set of Ψ1 but with values of E of the opposite sign. 2.2.51. (a) Form r † r = (Ur)† Ur = r† U† Ur = r† r. (b) If
for all r, r † r = r† U† Ur, then we must have U† U = 1.
37. 37. CHAPTER 3. EXERCISE SOLUTIONS 34 3. Vector Analysis 3.1 Review of Basic Properties (no exercises) 3.2
Vectors in 3-D Space 3.2.1. P × Q = (PxQy − PyQx)ˆx × ˆy = (PxQy − PyQx)ˆz. 3.2.2. (A × B)2 = A2 B2 sin2 ζ = A2
B2 (1 − cos2 ζ) = A2 B2 − (A · B)2 with ζ the angle between ˆA and ˆB. 3.2.3. The vector P is at an angle ζ (in the
positive direction) from the x axis, while Q is at an angle −ϕ. The angle between these vectors is therefore ζ + ϕ.
Both vectors are of unit length. Therefore P · Q = cos(ζ + ϕ) and the z component of Q × P is sin(ζ + ϕ). 3.2.4. A =
U × V = −3ˆy − 3ˆz, A/A = −(ˆy + ˆz)/ √ 2. 3.2.5. If a and b both lie in the xy-plane their cross product is in the z-
direction. The same is valid for c × d ∼ ˆz. The cross product of two parallel vectors is zero. Hence (a × b) × (c × d)
= 0. 3.2.6. Cross A−B−C = 0 into A to get −A×C = A×B, or C sin β = B sin γ, etc. 3.2.7. B = ˆx + 2ˆy + 4ˆz. 3.2.8. (a)
A · B × C = 0, A is the plane of B and C. The parallelpiped has zero height above the BC plane and therefore zero
volume. (b) A × (B × C) = −ˆx + ˆy + 2ˆz. 3.2.9. Applying the BAC-CAB rule we obtain [a · cb − a · bc] + [b · ac − b ·
ca] + [c · ba − c · ab] = 0. 3.2.10. (a) ˆr · Ar = A · ˆr. (b) ˆr · At = −ˆr · [ˆr × (ˆr × A)] = 0. 3.2.11. The scalar triple
product A · B × C is the volume spanned by the vectors. 3.2.12. A · B × C = −120, A × (B × C) = −60ˆx − 40ˆy +
50ˆz, C × (A × B) = 24ˆx + 88ˆy − 62ˆz, B × (C × A) = 36ˆx − 48ˆy + 12ˆz. 3.2.13. (A × B) · (C × D) = [(A × B) × C] ·
D = [(A · C)B − (B · C)A] · D = (A · C)(B · D) − (A · D)(B · C).
38. 38. CHAPTER 3. EXERCISE SOLUTIONS 35 3.2.14. Using the BAC-CAB rule with A × B as the first vector we
obtain (A × B) × (C × D) = (A × B) · DC − (A × B) · CD. 3.2.15. The answer is given in the text. 3.3 Coordinate

e xy, xz, and yz planes. If an incoming ray

strikes the xy plane, the z component of its direction of propagation is reversed. A strike on the xz plane reverses
its y component, and a strike on the yz plane reverses its x component. These properties apply for an arbitrary
direction of incidence, and together the reverse the propagation direction to the opposite of its incidence
orientation. 3.3.3. Because S is orthogonal, its transpose is also its inverse. Therefore (x )T = (Sx)T = xT ST = xT
S−1 . Then (x

39. 39.

−0.48 −0.

. (c) (a × b) · c

−0.40 −8.84 7.12

40. 40. CHAPTER 3. EXERCISE SOLUTIONS 37 (e) Note that S is an improper rotation. The fact that S(a × b) has
components of opposite sign to a ×b shows that a×b is a pseudovector. The difference in sign between (a × b) · c
and (a × b ) · c shows that (a × b) · c is a pseudoscalar. The equality of the vectors S(a × (b × c)) and a × (b × c )
shows that a × (b × c) is a vector. 3.4 Rotations in IR3 3.4.1. The Euler rotations defined here differ from those in
the text in that the inclination of the polar axis (in amount β, in now about the x1 axis rather than the x2 axis.
Therefore, to achieve the same polar orientation, we must place the x1 axis where the x2 axis was using the text
rotation. This requires an additional first rotation of π/2. After inclining the polar axis, the rotational position is now
π/2 greater (counterclockwise) than from the text rotation, so the third Euler angle must be π/2 less than its original
value. 3.4.2. (a) α = 70◦ , β = 60◦ , γ = −80◦ . (b) The answer is in the text. 3.4.3. The angle changes lead to cos α
→ − cos α, sin α → − sin α; cos β → cos β, sin β → − sin β; sin γ → − sin γ, cos γ → − cos γ. From these we verify
that each matrix element of Eq. (3.37) stays the same. 3.4.4. (a) Each of the three Euler rotations is an orthogonal
matrix, so their matrix product must also be orthogonal. Therefore its transpose, ˜S, must equal its inverse, S−1 .
(b) This equation simply carries out the three Euler rotations in reverse order, each in the opposite direction. 3.4.5.
(a) The projection of r on the rotation axis is not changed by the rotation; it is (r · ˆn)ˆn. The portion of r
perpendicular to the rotation axis can be written r − (r · ˆn)ˆn. Upon rotation through an angle Φ, this vector
perpendicular to the rotation axis will consist of a vector in its original direction (r − (r · ˆn)ˆn) cos Φ plus a vector
perpendicular both to it and to ˆn given by (r − (r · ˆn)ˆn) sin Φ × ˆn; this reduces to r × ˆn sin Φ. Adding these
contributions, we get the required result. (b) If ˆn = ˆez, the formula yields r = x cos Φˆex+y cos Φˆey+z cos Φˆez+y
sin Φˆex−x sin Φˆey+z(1−cos Φ)ˆez . Simplifying, this reduces to r = (x cos Φ + y sin Φ)ˆex + (−x sin Φ + y cos
Φ)ˆey + zˆez . This corresponds to the rotational transformation given in Eq. (3.35).
41. 41. CHAPTER 3. EXERCISE SOLUTIONS 38 (c) Expanding r 2 , recognizing that the second term of r is
orthogonal to the first and third terms, r 2 = r2 cos2 Φ + (r × ˆn) · (r × ˆn) sin2 Φ + (ˆn · r)2 (1 − cos Φ)2 + 2(ˆn · r)2
cos Φ(1 − cos Φ) . Using an identity to make the simplification (r × ˆn) · (r × ˆn) = (r · r)(ˆn · n) − (r · ˆn)2 = r2 − (r ·
ˆn)2 , we get r 2 = r2 + (r · ˆn)2 (− sin2 Φ + 1 + cos2 Φ − 2 cos2 Φ) = r2 . 3.5 Differential Vector Operators 3.5.1.
(a) −3(14)−5/2 (ˆx + 2ˆy + 3ˆz). (b) 3/196. (c) −1/(14)1/2 , −2/(14)1/2 , −3/(14)1/2 . 3.5.2. The solution is given in the
text. 3.5.3. From r12 = (x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 we obtain 1r12 = r1 − r2 r12 = ˆr12 by differentiating
componentwise. 3.5.4. dF = F(r + dr, t + dt) − F(r,t) = F(r + dr, t + dt) − F(r,t + dt) + F(r,t + dt) − F(r,t) = (dr · )F + ∂F
∂t dt. 3.5.5. (uv) = v u + u v follows from the product rule of differentiation. (a) Since f = ∂f ∂u u + ∂f ∂v v = 0, u and
 v are parallel so that ( u) × ( v) = 0, and vice versa. (b) If ( u)×( v) = 0, the two-dimensional volume spanned by u
and v, also given by the Jacobian J u, v x, y = ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y , vanishes.
42. 42. CHAPTER 3. EXERCISE SOLUTIONS 39 3.5.6. (a) From ˙r = ωr(−ˆx sin ωt + ˆy cos ωt), we get r × ˙r = ˆzωr2
(cos2 ωt + sin2 ωt) = ˆzωr2 . (b) Differentiating ˙r above we get ¨r = −ω2 r(ˆx cos ωt + ˆy sin ωt) = −ω2 r. 3.5.7. The
time derivative commutes with the transformation because the coef- ficients aij are constants. Therefore dVj/dt
satisfies the same transforma- tion law as Vj. 3.5.8. The product rule directly implies (a) and (b). 3.5.9. The product
rule of differentiation in conjunction with (a × b) · c = a · (b × c), etc. gives · (a × b) = b · ( × a) − a · ( × b). 3.5.10. If
L = −ir × , then the determinant form of the cross product gives Lz = −i x ∂ ∂y − y ∂ ∂x , (in units of ), etc. 3.5.11.
Carry out the indicated operations, remembering that derivatives operate on everything to their right in the current
expression as well as on the function to which the operator is applied. Therefore, LxLy = − y ∂ ∂z − z ∂ ∂y z ∂ ∂x −
x, ∂ ∂z = − y ∂ ∂x + yz ∂2 ∂z∂x − z2 ∂2 ∂y∂x − xy ∂2 ∂z2 + zx ∂2 ∂y∂z . LyLx = − z ∂ ∂x − x, ∂ ∂z y ∂ ∂z − z ∂ ∂y = −
zy ∂2 ∂x∂z − xy ∂2 ∂z2 − z2 ∂2 ∂x∂y + xz ∂2 ∂z∂y + x ∂ ∂y . Combining the above, LxLy − LyLx = x ∂ ∂y − u ∂ ∂x =
iLz. 3.5.12. [a · L, b · L] = aj[Lj, Lk]bk = iεjklajbkLl = i(a × b) · L. 3.5.13. The stream lines of b are solutions of the
differential equation dy dx = by bx = x −y . Writing this differential equation as xdx + ydy = 0, we see that it can be
integrated to yield x2 /2 + y2 /2 =constant, equivalent to x2 + y2 = C2 ,
43. 43. CHAPTER 3. EXERCISE SOLUTIONS 40 the equation for a family of circles centered at the coordinate origin.
To determine the direction of the stream lines, pick a convenient point on a circle, e.g., the point (+1, 0). Here bx =
0, by = +1, which corresponds to counterclockwise travel. 3.6 Differential Vector Operators: Further Properties
3.6.1. By definition, u × v is solenoidal if · (u × v) = 0. But we have the identity · (u × v) = v · ( × u) − u · ( × v) . If a
vector w is irrotational, ×w = 0, so if u and v are both irrotational, the right-hand side of the above equation is zero,
proving that u × v is solenoidal. 3.6.2. If × A = 0, then · (A × r) = r · × A − A · ( × r) = 0 − 0 = 0. 3.6.3. From v = ω × r
we get · (ω × r) = −ω · ( × r) = 0. 3.6.4. Forming the scalar product of f with the identity × (gf) = g × f + ( g) × f ≡ 0
we obtain the result, because the second term of the identity is perpen- dicular to f. 3.6.5. Applying the BAC-CAB
rule naively we obtain ( · B)A − ( · A)B, where still acts on A and B. Thus, the product rule of differentiation
generates two terms out of each which are ordered so that acts only on what comes after the operator. That is, (
·B)A → A( ·B)+(B· )A, and similarly for the second term. Hence the four terms. 3.6.6. Write the x components of all
the terms on the right-hand side of this equation. We get [(A × ) × B]x = Az ∂Bz ∂x − Ax ∂Bz ∂z − Ax ∂By ∂y + Ay
∂By ∂x , [(B × ) × A]x = Bz ∂Az ∂x − Bx ∂Az ∂z − Bx ∂Ay ∂y + By ∂Ay ∂x , [A( · B)]x = Ax ∂Bx ∂x + Ax ∂By ∂y + Ax
∂Bz ∂z , [B( · A)]x = Bx ∂Ax ∂x + Bx ∂Ay ∂y + Bx ∂Az ∂z . All terms cancel except those corresponding to the x
component of the left-hand side of the equation.
44. 44. CHAPTER 3. EXERCISE SOLUTIONS 41 3.6.7. Apply the BAC-CAB rule to get A × ( × A) = 1 2 (A2 ) − (A ·
)A. The factor 1/2 occurs because operates only on one A. 3.6.8. (A · B × r) = (r · A × B) = ˆex(A × B)x + ˆey(A ×
B)y + ˆez(A × B)z = A × B. 3.6.9. It suffices to check one Cartesian component; we take x. The x component of the
left-hand side of Eq. (3.70) is ∂ ∂y ( × V)z − ∂ ∂z ( × V)y = ∂2 Vy ∂y∂x − ∂2 Vx ∂y2 − ∂2 Vx ∂z2 + ∂2 Vz ∂z∂x . The x
component of the right-hand side is ∂ ∂x ∂Vx ∂x + ∂Vy ∂y + ∂Vz ∂z − ∂2 Vx ∂x2 + ∂2 Vx ∂y2 + ∂2 Vx ∂z2 . After
canceling the two right-hand-side occurrences of ∂2 Vx/∂x2 these two expressions contain identical terms. 3.6.10.
× (ϕ ϕ) = ϕ × ϕ + ϕ × ( ϕ) = 0 + 0 = 0. 3.6.11. (a) If F or G contain an additive constant, it will vanish on application
of any component of . (b) If either vector contains a term f, it will not affect the curl because × ( f) = 0. 3.6.12. Use
the identity v × ( × v) = (v · v) − (v · )v . Taking the curl and noting that the first term on the right-hand side then
vanishes, we obtain the desired relation. 3.6.13. Using Exercise 3.5.9, · ( u × v) = ( v) · ( × u) − ( u) · ( × v) = 0 − 0
= 0. 3.6.14. 2 ϕ = · ϕ = 0, and × ϕ = 0. 3.6.15. From Eq. (3.70), × ( × A) = − 2 A if · A = 0. 3.6.16. Use the identity 2
(fg) = f 2 g +g 2 f +2( f)·( g) with f = g = Φ. Then we find 2 Ψ = k 2 2Φ 2 Φ + 2( Φ) · ( Φ) , which satisfies the heat
conduction equation because 2 Φ = 0.
45. 45. CHAPTER 3. EXERCISE SOLUTIONS 42 3.6.17. Start by forming the matrix

∂y + i − 1 c2
∂Ex ∂t − ∂By ∂z + ∂Bz ∂y ∂By ∂t − ∂Ez ∂x + ∂Ex ∂z + i − 1 c2 ∂Ey ∂t − ∂Bz ∂x + ∂Bx ∂z ∂Bz ∂t − ∂Ex ∂y + ∂Ey ∂x + i

components of the above vector, we recover two Maxwell equations. 3.6.18. By direct matrix multiplication we
verify this equation. 3.7 Vector Integration 3.7.1. A triangle ABC has area 1 2 |B − A| |C − A| sin ζ, where ζ is the
angle between B−A and C−A. This area can be written |(B−A)×(C−A)|/2. Expanding, Area ABC = |A × B + B × C +
C × A|/2 . Applying this formula to OAB, we get just |A × B|/2. Continuing to the other three faces, the total area is
Area = |A × B| + |B × C| + |C × A| + |A × B + B × C + C × A| 2 . 3.7.2. Let us parameterize the circle C as x = cos ϕ,
y = sin ϕ with the polar angle ϕ so that dx = − sin ϕ dϕ, dy = cos ϕ dϕ. Then the force can be written as F = −ˆx sin
ϕ + ˆy cos ϕ. The work becomes − C xdy − ydx x2 + y2 = −π 0 (− sin2 ϕ − cos2 ϕ) dϕ = π. Here we spend energy.
If we integrate counterclockwise from ϕ = 0 to π we find the value −π, because we are riding with the force. The
work is path dependent which is consistent with the physical interpretation that
46. 46. CHAPTER 3. EXERCISE SOLUTIONS 43 F · dr ∼ xdy − ydx = Lz is proportional to the z-component of orbital
angular momentum (involving circulation, as discussed in Section 3.5). If we integrate along the square through the
points (±1, 0), (0, −1) sur- rounding the circle we find for the clockwise lower half square path − F · dr = − −1 0
Fydy|x=1 − −1 1 Fxdx|y=−1 − 0 −1 Fydy|x=−1 = 1 0 dy 1 + y2 + 1 −1 dx x2 + (−1)2 + 0 −1 dy (−1)2 + y2 = arctan(1)
+ arctan(1) − arctan(−1) − arctan(−1) = 4 · π 4 = π, which is consistent with the circular path. 3.7.3. The answer
depends upon the path that is chosen. A simple possibility is to move in the x direction from (1,1) to (3,1) and then
in the y direction from (3,1) to (3,3). The work is the integral of F·ds. For the first segment of the path the work is Fx
dx; for the second segment it is Fy dy. These correspond to the specific integrals w1 = 3 1 (x−1) dx = x2 2 − x 3 1
= 2, w2 = 3 1 (3+y) dy = 3y + y2 2 3 1 = 10. 3.7.4. Zero. 3.7.5. 1 3 r · dζ = x 3 dydz + y 3 dzdx + z 3 dxdy = 1 3 1 0
 dy 1 0 dz + · · · = 3 3 = 1. Here the factor x in the first term is constant and therefore outside the integral; it is 0 for
one face of the cube and unity for the opposite one. Similar remarks apply to the factors y, z in the other two terms
which contribute equally. 3.8 Integral Theorems 3.8.1. For a constant vector a, its divergence is zero. Using Gauss’
theorem we have 0 = V · adη = a · S dζ, where S is the closed surface of the finite volume V . As a = 0 is arbitrary,
S dζ = 0 follows.
47. 47. CHAPTER 3. EXERCISE SOLUTIONS 44 3.8.2. From · r = 3 in Gauss’ theorem we have V · rdη = 3 V dη = 3V
= S r · dζ, where V is the volume enclosed by the closed surface S. 3.8.3. Cover the closed surface by small (in
general curved) adjacent rectangles Si whose circumference are formed by four lines Li each. Then Stokes’
theorem gives S ( × A) · dζ = i Si ( × A) · dζ = i Li A · dl = 0 because all line integrals cancel each other. 3.8.4.
Apply Gauss’ theorem to · (ϕE) = ϕ · E + ϕ · E = −E2 + ε−1 0 ϕρ, where S→∞ ϕE · dζ = 0. 3.8.5. First, show that Ji
= · (xJ) by writing · (xJ) = x · J + ( x) · J = 0 + ˆex · J = Jx . Since J is zero on the boundary, so is xJ, so by Gauss’
theorem we have · (xJ)dη = 0, equivalent to Jx dη = 0. 3.8.6. By direct calculation we can find that × t = 2ez. Then,
by Stokes’ theorem, the line integral has the value 2A. 3.8.7. (a) As r × dr/2 is the area of the infinitesimal triangle, r
× dr is twice the area of the loop. (b) From dr = (−ˆxa sin ζ + ˆyb cos ζ)dζ and ˆx × ˆy = ˆz we obtain r × dr =
ˆzab(cos2 ζ + sin2 ζ) and r × dr = ˆzab 2π 0 dζ = ˆz2abπ. 3.8.8. We evaluate the surface integral with P = r. Note
that dζ = ˆez dA, and that, evaluating components, dζ × = −ˆex ∂ ∂y + ˆey ∂ ∂x . Then form (dζ × ) × r. The x and y
components of this expression vanish; the z component is − ∂ ∂y y − ∂ ∂x x = −2 . The surface integral then has the
value −2A, where A is the area of the loop. Note that the alternate form of Stokes’ theorem equates this surface
integral to − r × dr.
48. 48. CHAPTER 3. EXERCISE SOLUTIONS 45 3.8.9. This follows from integration by parts shifting from v to u. The
inte- grated term cancels for a closed loop. 3.8.10. Use the identity of Exercise 3.8.9, i.e. (uv)·dλ = 0, and apply
Stokes’ theorem to 2 S u v · dζ = (u v − v u) · dλ = S × (u v − v u) · dζ = 2 S ( u × v) · dζ. 3.8.11. Starting with
Gauss’ theorem written as ∂V B · dζ = V · B dη, substitute B = a × P, where a is a constant vector and P is
arbitrary. The left-hand integrand then becomes (a × P) · dζ = (P × dζ) · a. The right-hand integrand expands into
P·( ×a)−a·( ×P), the first term of which vanishes because a is a constant vector. Our Gauss’ theorem equation can
then be written a · ∂V P × dζ = −a · V × P dη . Rearranging to a · ∂V P × ζ + V × P dη = 0, we note that because
the constant direction of a is arbitrary the quantity in square brackets must vanish; its vanishing is equivalent to the
relation to be proved. 3.8.12. Start from Stokes’ theorem, S ( × B · dζ = ∂S B · dr and substitute B = ϕ a, where a is
a constant vector and ϕ is an arbitrary scalar function. Because a is constant, the quantity × ϕ a reduces to ( ϕ) × a,
and the left-side integrand is manipulated as follows: ( ϕ) × a · dζ = (dζ × ϕ) · a. The Stokes’ theorem formula can
then be written a · S dζ × ϕ = a · ∂S ϕ dr. Because a is arbitrary in direction, the integrals on the two sides of this
equation must be equal, proving the desired relation.
49. 49. CHAPTER 3. EXERCISE SOLUTIONS 46 3.8.13. Starting from Stokes’ theorem as written in the solution to
Exercise 3.8.12, set B = a × P. This substitution yields S ( × (a × P)) · dζ = ∂S (a × P) · dr Applying vector identities
and remembering that a is a constant vector, the left- and right-side integrands can be manipulated so that this
equation becomes − S a · ((dζ × ) × P) = ∂S (P × dr) · a. Bringing a outside the integrals and rearranging, we reach
a · S (dζ × ) × P − ∂S dr × P = 0 . Since the direction of a is arbitrary, the quantity within the square brackets
vanishes, thereby confirming the desired relation. 3.9 Potential Theory 3.9.1. The solution is given in the text.
3.9.2. ϕ(r) = Q 4π ε0r , a ≤ r < ∞, ϕ(r) = Q 4π ε0a 3 2 − 1 2 r2 a2 , 0 ≤ r ≤ a. 3.9.3. The gravitational acceleration in
the z-direction relative to the Earth’s surface is − GM (R + z)2 + GM R2 ∼ 2z GM R3 for 0 ≤ z R. Thus, Fz = 2z
GmM R3 , and Fx = −x GmM (R + x)3 ∼ −x GmM R3 , Fy = −y GmM (R + x)3 ∼ −y GmM R3 . Integrating F = − V
yields the potential V = GmM R3 z2 − 1 2 x2 − 1 2 y2 = GmMr2 2R3 (3z2 −r2 ) = GmMr2 R3 P2(cos ζ). 3.9.4. The
answer is given in the text. 3.9.5. The answer is given in the text.
50. 50. CHAPTER 3. EXERCISE SOLUTIONS 47 3.9.6. A = 1 2 (B × r) for constant B implies B = × A = 1 2 B · r − 1 2
B · r = 3 2 − 1 2 B. 3.9.7. (a) This is proved in Exercise 3.6.14. (b) 2 × A = × (u v − v u) = u × v − v × u = 2 u × v.
3.9.8. If A = A + Λ, then B = × A = × A + × Λ = B because × Λ = 0, and A · dr = A · dr + Λ · dr = A · dr because b a
Λ · dr = Λ|b a = 0 for b = a in a closed loop. 3.9.9. Using Green’s theorem as suggested in the problem and the
formula for the Laplacian of 1/r (where r is the distance from P), the volume integral of Green’s theorem reduces to
V (−ϕ) 2 1 r dη = V (−ϕ) [−4πδ(r)] dη = 4πϕ(P) . The surface integrals, for a sphere of radius a centered at P, are S
1 a ϕ − ϕ 1 r dζ . Using (1/r) = −ˆer/r2 , the second term of the surface integral yields 4π times ϕ , the average of ϕ
on the sphere. The first surface-integral term vanishes by Gauss’ theorem because · ϕ vanishes everywhere within
the sphere. We thus have the final result 4πϕ0 = 4π ϕ . 3.9.10. Use × A = B = µH, D = εE with ∂E ∂t = 0 in × H = ∂D
∂t + J = × ( × A)/µ = ( · A − 2 A)/µ = J so that − 2 A = µJ follows. 3.9.11. Start from Maxwell’s equation for × B and
substitute for the fields B and E in terms of the potentials A and ϕ. The relevant equations are × B = 1 c2 ∂E ∂t +
µ0J, B = × A, E = − ϕ − ∂A ∂t × ( × A) = − 1 c2 ∂ϕ ∂t − 1 c2 ∂2 A ∂t2 + µ0J
51. 51. CHAPTER 3. EXERCISE SOLUTIONS 48 Next manipulate the left-hand side using Eqs. (3.70) and (3.109): × (
× A) = − 2 A + ( · A = − 2 A − 1 c2 ∂ϕ ∂t . Inserting this result for × ( × A), the terms in ∂ϕ/∂t cancel and the desired
formula is obtained. 3.9.12. Evaluate the components of × A. ( × A)x = ∂Az ∂y − ∂Ay ∂z = ∂Az ∂y = − ∂ ∂y x x0 By(x,
y0, z) dx − y y0 Bx(x, y, z) dy = 0 + Bx(x, y, z) , ( × A)y = ∂Ax ∂z − ∂Az ∂x = − ∂ ∂z y y0 Bz(x, y, z) dy + ∂ ∂x x x0
By(x, y0, z) dx − y y0 Bx(x, y, z) dy = − y y0 ∂Bz ∂z dy + By(x, y0, z) − y y0 ∂Bx ∂x dy . The evaluation of ( × A)y is
now completed by using the fact that · B = 0, so we continue to ( × A)y = y y0 ∂By ∂y dy + By(x, y0, z) = By(x, y, z),
( × A)z = ∂Ay ∂x − ∂Ax ∂y = − ∂Ax ∂y = ∂ ∂y y y0 Bz(x, y, z) dy = Bz(x, y, z). 3.10 Curvilinear Coordinates 3.10.1. (a)
In the xy-plane different u, v values describe a family of hyperbolas in the first and third quadrants with foci along
the diagonal x = y and asymptotes given by xy = u = 0, i.e. the x- and y-axes, and orthogonal hyperbolas with foci
along the x-axis with asymptotes given by v = 0, i.e. the lines x±y. The values z =constant describe a family of
planes parallel to the xy-plane. (c) For u =const. and v =const. we get from x2 − y2 = v, xdx − ydy = 0, or dx/dy =
 y/x, dy/dx = x/y. Thus, on the x-axis these hyperbolas have a vertical tangent. Similarly xy = u =const. gives xdy +
ydx = 0, or dy/dx = −y/x. The product of these slopes is equal to −1, which proves
52. 52. CHAPTER 3. EXERCISE SOLUTIONS 49 orthogonality. Alternately, from ydx+xdy = du, 2xdx+2ydy = dv we
get by squaring and adding that (x2 + y2 )(dx2 + dy2 ) = du2 + dv2 /4. Here, the mixed terms dudv, dxdy drop out,
proving again orthogonality. (d) The uvz-system is left-handed. This follows from the negative Jaco- bian ∂(x, y)
∂(u, v) = − 1 x2 + y2 . To prove this, we differentiate the hyperbolas with respect to u and v giving, respectively, y
∂x ∂u + x ∂y ∂u = 1, y ∂x ∂v + x ∂y ∂v = 0, x ∂x ∂u − y ∂y ∂u = 0, x ∂x ∂v − y ∂y ∂v = 1 2 . Solving for the partials we
obtain ∂x ∂u = y x2 + y2 = y x ∂y ∂u , ∂x ∂v = x 2(x2 + y2) = − x y ∂y ∂u . From these we find the Jacobian given
above. The coordinate vectors are ∂r ∂u = ∂x ∂u , ∂y ∂u = ∂x ∂u 1, x y , ∂r ∂v = ∂x ∂v , ∂y ∂v = ∂x ∂v 1, − y x . 3.10.2.
These elliptical cylinder coordinates can be parameterized as x = c cosh u cos v, y = c sinh u sin v, z = z, (using c
instead of a). As we shall see shortly, the parameter 2c > 0 is the distance between the foci of ellipses centered at
the origin of the x, y-plane and described by different values of u =const. Their major and minor half-axes are
respectively a = c cosh u and b = c sinh u. Since b a = tanh u = 1 − 1 cosh2 u = 1 − ε2, the eccentricity ε = 1/ cosh
u, and the distance between the foci 2aε = 2c, proving the statement above. As u → ∞, ε → 0 so that the ellipses
become circles. As u → 0, the ellipses become more elongated until, at u = 0, they shrink to the line segment
between the foci. Different values of v =const. describe a family of hyperbolas. To show orthogonality of the
ellipses and hyperbolas we square and add the coordinate differentials dx = c sinh u cos vdu − c cosh u sin vdv, dy
= c cosh u sin vdu + c sinh u cos vdv,
53. 53. CHAPTER 3. EXERCISE SOLUTIONS 50 to obtain dx2 + dy2 = c2 (sinh2 u cos2 v + cosh2 u sin2 v)(du2 +
dv2 ) = c2 (cosh2 u − cos2 v)(du2 + dv2 ). Since there is no cross term dudv, these coordinates are locally
orthogonal. Differentiating the ellipse and hyperbola equations with respect to u and v we can determine ∂x/∂u, . . .
, just as in Exercise 3.10.1, and obtain the coordinate vectors ∂r/∂u and ∂r/∂v. 3.10.3. From the component
definition (projection) a = i ˆqia · ˆqi ≡ i aqi ˆqi and a similar expression for b, get a · b = ij ˆqi · ˆqja · ˆqib · ˆqj = i a ·
ˆqib · ˆqi = i aqi bqi using orthogonality, i.e. ˆqi · ˆqj = δij. 3.10.4. (a) From Eq. (3.141) with ˆe1 = ˆq1 and (ˆe1)1 = 1,
(ˆe1)2 = (ˆe1)3 = 0, we get · ˆe1 = 1 h1h2h3 ∂(h2h3) ∂q1 . (b) From Eq. (3.143) with h2V2 → 0, h3V3 → 0, we get
× ˆe1 = 1 h1 ˆe2 1 h3 ∂h1 ∂q3 − ˆe3 1 h2 ∂h1 ∂q2 . 3.10.5. This problem assumes that the unit vectors ˆqi are
orthogonal. From dr = ∂r ∂qi dqi we see that the ∂r ∂qi are tangent vectors in the directions ˆei = ˆqi with lengths hi.
This establishes the first equation of this problem. Writing (for any i) ˆei · ˆei = 1 h2 i ∂r ∂qi · ∂r ∂qi = 1 h2 i ∂x ∂qi 2
+ ∂y ∂qi 2 + ∂z ∂qi 2 = 1 , we confirm the formula for hi. If we now differentiate hiˆei = ∂r/∂qi with respect to qj (with j
= i) and note that the result is symmetric in i and j, we get ∂(hiˆei) ∂qj = ∂2 r ∂qi∂qj = ∂(hjˆej) ∂qi . Expanding the
differentiations of the left and right members of this equa- tion and equating the results, ∂hi ∂qj ˆei + hi ∂ˆei ∂qj = ∂hj
∂qi ˆej + hj ∂ˆej ∂qi .
54. 54. CHAPTER 3. EXERCISE SOLUTIONS 51 Since ∂ˆei/∂qj must be a vector in the ˆej direction, we are able to
establish the second equation of the exercise. To prove the last relation, we differentiate ˆei · ˆei = 1 and ˆei · ˆej =
0 with respect to qi. We find ˆei · ∂ˆei ∂qi = 0, ∂ˆei ∂qi · ˆej = −ˆei · ∂ˆej ∂qi . These equations show that ∂ˆei/∂qi has
no component in the ˆei direction and that its components in the ˆej directions are −ˆei · ∂ˆej/∂qi. Using the second
formula to write these derivatives in terms of the hi, we reach the final equation of this exercise. 3.10.6. The
solution is given in the text. 3.10.7. The solution is given in the text. 3.10.8. Using the formulas from Exercise
3.10.5, with hρ = hz = 1 and hϕ = ρ, nonzero terms only result if the hi being differentiated is hϕ, and then only if
differentiated with respect to ρ. These conditions cause all the first derivatives of the unit vectors to vanish except
for the two cases listed in the exercise; those cases are straightforward applications of the formulas. 3.10.9. The
formula given in the exercise is incorrect because it neglects the ϕ- dependence of ˆeρ. When this is properly
included, instead of ∂Vρ/∂ρ we get ρ−1 ∂(ρVρ)/∂ρ. 3.10.10. (a) r = (x, y, z) = (x, y) + zˆz = ρˆρ + zˆz. (b) From Eq.
(3.148) we have · r = 1 ρ ∂ρ2 ∂ρ + ∂z ∂z = 2 + 1 = 3. From Eq. (3.150) with Vρ = ρ, Vϕ = 0, Vz = z we get × r = 0.
3.10.11. (a) The points x, y, z and −x, −y, −z have the same value of ρ, values of z of opposite sign, and if x = ρ
cos ϕ, y = ρ sin ϕ, then −x and −y must have a value of ϕ displaced from the original ϕ value by π. (b) A unit vector
ˆez will always be in the same (the +z) direction, but the change by π in ϕ will cause the ˆeρ unit vector to change
sign under inversion. The same is true of ˆeϕ. 3.10.12. The solution is given in the text. 3.10.13. The solution is
given in the text.
55. 55. CHAPTER 3. EXERCISE SOLUTIONS 52 3.10.14. Using Vz ≡ 0 we obtain × V|ρ = 1 ρ ∂(ρVϕ(ρ, ϕ)) ∂z = 0, ×
V|ϕ = 1 ρ ∂(Vρ(ρ, ϕ)) ∂z = 0, × V|z = 1 ρ ∂(ρVϕ(ρ, ϕ)) ∂ρ − ∂Vρ(ρ, ϕ) ∂ϕ . 3.10.15. The solution is given in the text.
3.10.16. (a) F = ˆϕ1 ρ . (b) × F = 0, ρ = 0. (c) 2π 0 F · ˆϕρdϕ = 2π. (d) × F is not defined at the origin. A cut line from
the origin out to infinity (in any direction) is needed to prevent one from encircling the origin. The scalar potential ψ
= ϕ is not single-valued. 3.10.17. The solution is given in the text. 3.10.18. The solution is given in the text. 3.10.19.
Resolving the unit vectors of spherical polar coordinates into Cartesian components was accomplished in Exercise
3.10.18 involving an orthogonal matrix. The inverse is the transpose matrix, i.e. ˆx = ˆr sin ζ cos ϕ + ˆζ cos ζ cos ϕ
− ˆϕ sin ϕ, ˆy = ˆr sin ζ sin ϕ + ˆζ cos ζ sin ϕ + ˆϕ cos ϕ, ˆz = ˆr cos ζ − ˆζ sin ζ. 3.10.20. (a) The transformation
between Cartesian and spherical polar coordinates is not represented by a constant matrix, but by a matrix whose
compo- nents depend upon the value of r. A matrix equation of the indicated type has no useful meaning because
the components of B depend upon both r and r . (b) Using the fact that both the Cartesian and spherical polar
coordinate systems are orthogonal, the transformation matrix between a Cartesian- component vector A and its
spherical-
ˆey ˆeζ · ˆez ˆeϕ · ˆex ˆe
56. 56.

the same point, whose angular coordinates are (ζ, ϕ). To check orthogonality, transpose and check the product UT
 U. We find UT U = 1. 3.10.21. One way to proceed is to first obtain the transformation of a vector A to its
representation A in cylindrical coordinates. Letting V be the trans-

are associated with the same point, which has angular coordinate ϕ. We now convert from spherical polar to
cylindrical coordi- nates in two steps, of which the first is from spherical polar to Cartesian coordinates,
accomplished by the transformation UT , the inverse of the transformation U of Exercise 3.10.20(b). We then apply
transformation V to convert to cylindrical coordinates. The overall transformation matrix W is then the matrix
product VUT . Th

this transformation is represented by the transpose of W.

57. 57. CHAPTER 3. EXERCISE SOLUTIONS 54 3.10.22. (a) Differentiating ˆr2 = 1 we get ∂r ∂r = (sin ζ cos ϕ, sin ζ
sin ϕ, cos ζ) = ˆr, ∂r ∂ζ = r(cos ζ cos ϕ, cos ζ sin ϕ, − sin ζ) = rˆζ, ∂r ∂ϕ = r(− sin ζ sin ϕ, sin ζ cos ϕ, 0) = r sin ζ ˆϕ.
(b) With given by ˆr ∂ ∂r + ˆζ 1 r ∂ ∂ζ + ˆϕ 1 r sin ζ ∂ ∂ϕ , the alternate derivation of the Laplacian is given by dotting
this into itself. In conjunction with the derivatives of the unit vectors above this gives · = ˆr · ∂ ∂r ˆr ∂ ∂r + ˆζ · 1 r ∂ˆr
∂ζ ∂ ∂r + ˆϕ · 1 r sin ζ ∂ˆr ∂ϕ ∂ ∂r + ˆϕ · 1 r sin ζ ∂ ˆζ ∂ϕ 1 r ∂ ∂ζ + ˆϕ · 1 r sin ζ ∂ ∂ϕ ˆϕ 1 r sin ζ ∂ ∂ϕ = ∂2 ∂r2 + 1 r2
∂2 ∂ζ2 + 2 r ∂ ∂r + tan ζ r2 ∂ ∂ζ + 1 r2 sin2 ζ ∂2 ∂ϕ2 . Note that, with 1 r2 sin2 ζ ∂ ∂ζ sin ζ ∂ ∂ζ = tan ζ r2 ∂ ∂ζ + 1
r2 ∂2 ∂ζ2 , we get the standard result using Exercise 3.10.34 for the radial part. 3.10.23. The solution is given in
the text. 3.10.24. Vζ, Vϕ ∼ 1/r. 3.10.25. (a) Since r = x2 + y2 + z2, changes of sign in x, y, and z leave r un-
changed. Since z → −z, cos ζ changes sign, converting ζ into π − ζ. Sign changes in x and y require that both sin
ϕ and cos ϕ change sign; this requires that ϕ change to ϕ ± π. (b) Since the coordinate point is after inversion on
the opposite side of the polar axis, increases in r or ϕ correspond to displacements in directions opposite to their
effect before inversion. Both before and after inversion, an increase in ζ is in a direction tangent to the same circle
of radius r that
58. 58. CHAPTER 3. EXERCISE SOLUTIONS 55 passes through both the north and south poles of the coordinate
system. The two tangent directions are parallel because they are at opposite points of the circle, and both are in
the southerly tangent direction. They are therefore in the same direction. 3.10.26. (a) A · r = Ax ∂r ∂x + Ay ∂r ∂y +
Az ∂r ∂z = A because ∂r ∂x = ˆx, ∂r ∂y = ˆy, ∂r ∂z = ˆz. (b) Using ∂ˆr ∂ζ = ˆζ, ∂ˆr ∂ϕ = sin ζ ˆϕ and in polar
coordinates from Exer- cise 3.10.22 we get A · r = A · ˆr ∂r ∂r + A · ˆζ ∂ˆr ∂ζ + A · ˆϕ sin ζ ∂ˆr ∂ϕ = Arˆr + Aζ ˆζ + Aϕ
ˆϕ = A. 3.10.27. The solution is given in the text. 3.10.28. The solution is given in the text. 3.10.29. From Exercise
3.10.32 and using the Cartesian decomposition in Exercise 3.10.18 ˆζz = − sin ζ we get Lz = −i sin ζ sin ζ ∂ ∂ϕ .
3.10.30. Use Exercise 3.10.32 to get this result. 3.10.31. Solving this problem directly in spherical coordinates is
somewhat chal- lenging. From the definitions of the unit vectors, one can establish ∂ˆeζ ∂ϕ = cos ζˆeϕ , ∂ˆeϕ ∂ϕ = −
sin ζ ˆer − cos ζˆeζ , ∂ˆeζ ∂ζ = −ˆer, ∂ˆeϕ ∂ζ = 0 , ˆer × ˆeζ = ˆeϕ, ˆeζ × ˆeϕ = ˆer, ˆeϕ × ˆer = ˆeζ. We now write L ×
L and expand it into its four terms, which we process individually. When a unit vector is to be differentiated, the
differentiation should be carried out before evaluating the cross product. This first term only has a contribution
when the second ˆeζ is differentiated: − ˆeζ sin ζ ∂ ∂ϕ ˆeζ sin ζ ∂ ∂ϕ = − ˆeζ sin ζ × ∂ˆeζ ∂ϕ 1 sin ζ ∂ ∂ϕ = −(ˆeζ ×
ˆeϕ) cos ζ sin2 ζ ∂ ∂ϕ .
59. 59. CHAPTER 3. EXERCISE SOLUTIONS 56 Next we process − ˆeϕ ∂ ∂ζ ˆeϕ ∂ ∂ζ = −ˆeϕ × ∂ˆeϕ ∂ζ ∂ ∂ζ − (ˆeϕ ×
ˆeϕ) ∂2 ∂ζ2 = 0 . Then ˆeζ sin ζ ∂ ∂ϕ ˆeϕ ∂ ∂ζ = ˆeζ × ˆeϕ sin ζ ∂2 ∂ϕ∂ζ ˆeζ sin ζ × (−ˆer sin ζ − ˆeζ cos ζ) ∂ ∂ζ =
ˆer sin ζ ∂2 ∂ϕ∂ζ + ˆeϕ ∂ ∂ζ . Finally, ˆeϕ ∂ ∂ζ ˆeζ sin ζ ∂ ∂ϕ = −ˆer sin ζ ∂2 ∂ζ∂ϕ − (ˆeϕ × ˆeζ) − cos ζ sin2 ζ ∂ ∂ϕ +
(ˆeϕ × (−ˆer) 1 sin ζ ∂ ∂ϕ . Several of the terms in the above expressions cancel. The remaining terms correspond
to iL. 3.10.32. (a) Using = ˆr ∂ ∂r + ˆζ 1 r ∂ ∂ζ + ˆϕ 1 r sin ζ ∂ ∂ϕ and r = rˆr, ˆr × ˆζ = ˆϕ, ˆr × ˆϕ = −ˆζ, we find L = −i(r
× ) = −i ˆϕ ∂ ∂ζ − ˆζ 1 sin ζ ∂ ∂ϕ . (b) Using Eq. (2.44), ˆζz = − sin ζ we find Lz = −i ∂ ∂ϕ , and from ˆζx = cos ζ cos
ϕ, ˆϕx = − sin ϕ we get Lx = i sin ϕ ∂ ∂ζ + i cot ζ cos ϕ ∂ ∂ϕ ; from ˆζy = cos ζ sin ϕ, ˆϕy = cos ϕ we get Ly = −i cos ϕ
∂ ∂ζ + i cot ζ sin ϕ ∂ ∂ϕ . (c) Squaring and adding gives the result.
60. 60. CHAPTER 3. EXERCISE SOLUTIONS 57 3.10.33. (a) Using ˆr × r = 0 and = ˆr ∂ ∂r − i r × L r2 and the BAC-
CAB rule we get −ir × = − 1 r2 r × (r × L) = − 1 r2 (r · Lr − r2 L) = L because L · r = 0. (b) It suffices to verify the x-
component of this equation. Substituting the formula for L, the result to be proved is × (r × ) = r 2 − (1 + r · ) . The
x-component of the left-hand side expands into × (r × ) x = ∂ ∂y x ∂ ∂y − y ∂ ∂x − ∂ ∂z z ∂ ∂x − x ∂ ∂z = x ∂2 ∂y2 − ∂
∂x − y ∂2 ∂y∂x − ∂ ∂x − z ∂2 ∂z∂x + x ∂2 ∂z2 . The x-component of the right-hand side is x ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2
− ∂ ∂x − ∂ ∂x x ∂ ∂x + y ∂ ∂y + z ∂ ∂z . The left- and right-hand sides simplify to identical expressions. 3.10.34. From
(a) 1 r2 d dr r2 = d dr + 2 r we get (c), and vice versa. From the inner d dr r = r d dr + 1 in (b) we get 1 r d2 dr2 r = 1
r d dr + d2 dr2 + d dr , hence (c), and vice versa. 3.10.35. (a) × F = 0, r ≥ P/2. (b) F · dλ = 0. This suggests (but
does not prove) that the force is conservative. (c) Potential = P cos ζ/r2 , dipole potential. 3.10.36. Solutions are
given in the text. 3.10.37. E(r) = 3ˆr(p · ˆr) − p 4π ε0r3 .
61. 61. CHAPTER 3. EXERCISE SOLUTIONS 58 4. Tensors and Differential Forms 4.1 Tensor Analysis 4.1.1. This is
a special case of Exercise 4.1.2 with B0 ij = 0. 4.1.2. If A0 ij = B0 ij in one frame of reference, then define a
coordinate transfor- mation from that frame to an arbitrary one: xi = xi(x0 j ), so that Aij = ∂xi ∂x0 α ∂xj ∂x0 β A0 αβ
= ∂xi ∂x0 α ∂xj ∂x0 β B0 αβ = Bij. 4.1.3. Make a boost in the z-direction. If Az = Az = A0 = 0, then A 0 = 0 in the
boosted frame by the Lorentz transformation, etc. 4.1.4. Since T12 = ∂xi ∂x1 ∂xk ∂x2 Tik = cos ζ sin ζ T11 + cos2 ζ
T12 sin2 ζ T21 − sin ζ cos ζ T22 we find T12 = T12 for a rotation by π, but T12 = −T21 for a rotation by π/2.
Isotropy demands T21 = 0 = T12. Similarly all other off-diagonal components must vanish, and the diagonal ones
are equal. 4.1.5. The four-dimensional fourth-rank Riemann–Christoffel curvature tensor of general relativity, Riklm
has 44 = 256 components. The antisymmetry of the first and second pair of indices, Riklm = −Rikml = −Rkilm,
reduces these pairs to 6 values each, i.e. 62 = 36 components. They can be thought of as a 6 × 6 matrix. The
 symmetry under exchange of pair indices, Riklm = Rlmik, reduces this matrix to 6 · 7/2 = 21 components. The
Bianchi identity, Riklm +Rilmk +Rimkl = 0, reduces the independent components to 20 because it represents one
constraint. Note that, upon using the permutation symmetries one can always make the first index equal to zero
followed by the other indices which are all different from each other. 4.1.6. Each component has at least one
repeated index and is therefore zero. 4.1.7. As the gradient transforms like a vector, it is clear that the gradient of a
tensor field of rank n is a tensor of rank n + 1. 4.1.8. The contraction of two indices removes two indices, while the
derivative adds one, so (n + 1) − 2 = n − 1. 4.1.9. The scalar product of the four-vectors ∂µ = 1 c ∂ ∂t , − and ∂µ = 1
c ∂ ∂t , is the scalar ∂2 = 1 c2 ∂2 ∂t2 − 2 .
62. 62. CHAPTER 3. EXERCISE SOLUTIONS 59 4.1.10. The double summation KijAiBj is a scalar. That Kij is a
second-rank tensor follows from the quotient theorem. 4.1.11. Since KijAjk = Bik is a second-rank tensor the
quotient theorem tells us that Kij is a second-rank tensor. 4.2 Pseudotensors, Dual Tensors 4.2.1. The direct
product εijkClm is a tensor of rank 5. Contracting 4 indices leaves a tensor of rank 1, a vector. Inverting gives Cjk =
εjkiCi, a tensor of rank 2. 4.2.2. The generalization of the totally antisymmetric εijk from three to n di- mensions has
n indices. Hence the generalized cross product εijk...AiBj is an antisymmetric tensor of rank n − 2 = 1 for n = 3.
4.2.3. The solution is given in the text. 4.2.4. (a) As each δij is isotropic, their direct product must be isotropic as
well. This is valid for any order of the indices. The last statement implies (b) and (c). 4.2.5. The argument relating
to Eq. (4.29) holds in two dimensions, too, with δij = det(a)aipajqδpq. No contradiction arises because εij is
antisymmetric while δij is symmetric. 4.2.6. ij = 0 1 −1 0 . If R = cos ϕ sin ϕ − sin ϕ cos ϕ is a rotation, then cos ϕ sin
ϕ − sin ϕ cos ϕ 0 1 −1 0 cos ϕ − sin ϕ sin ϕ cos ϕ = 0 1 −1 0 . 4.2.7. If Ak = 1 2 εijkBij with Bij = −Bji, then 2εmnkAk
= εmnkεijk = (δmiδnj − δmjδni)Bij = Bmn − Bnm = 2Bmn. 4.3 Tensors in General Coordinates 4.3.1. The vector εi is
completely specified by its projections onto the three linearly independent εk, i.e., by the requirements that εi ·εj =
δi j. Taking the form given in the exercise, we form εi · εi = (εj × εk) · εi (εj × εk) · εi = 1 , εi · εj = (εj × εk) · εj (εj × εk)
· εi = 0 , the zero occurring because the three vectors in the scalar triple product are not linearly independent. The
above equations confirm that εi is the contravariant version of εi.
63. 63. CHAPTER 3. EXERCISE SOLUTIONS 60 4.3.2. (a) From the defining formula, Eq. (4.40), the orthogonality of
the εi im- plies that gij = 0 when i = j. (b) See the answer to part (c). (c) From Eq. (4.46) with the εi orthogonal, the
εi must also be orthog- onal and have magnitudes that are the reciprocals of the εi. Then, from Eq. (4.47), the gii
must be the reciprocals of the gii. 4.3.3. This exercise assumes use of the Einstein summation convention.
Inserting the definitions of the εi and εi and evaluating the scalar products, we reach (εi · εj )(εj · εk) = ∂qi ∂x ∂qj ∂x
+ ∂qi ∂y ∂qj ∂y + ∂qi ∂z ∂qj ∂z ∂x ∂qj ∂x ∂qk ∂qi ∂x ∂qj ∂x + ∂y ∂qj ∂y ∂qk + ∂z ∂qj ∂z ∂qk The term of the product
arising from the first term of each factor has the form ∂qi ∂x ∂qj ∂x ∂x ∂qj ∂x ∂qk = ∂qi ∂x ∂x ∂qk j ∂x ∂qj ∂qj ∂x = ∂qi
∂x ∂x ∂qk , where we have noted that the j summation is the chain-rule expansion for ∂x/∂x, which is unity. The
products arising from the second terms and third terms of both factors have analogous forms, and the sum of
these “diagonal” terms is also a chain-rule expansion: ∂qi ∂x ∂x ∂qk + ∂qi ∂y ∂y ∂qk + ∂qi ∂z ∂z ∂qk = ∂qi ∂qk = δi k .
The remaining terms of the original product expression all reduce to zero; we illustrate with ∂qi ∂x ∂qj ∂x ∂y ∂qj ∂y
∂qk = ∂qi ∂x ∂y ∂qk j ∂y ∂qj ∂qj ∂x . Here the j summation is the chain-rule expansion of ∂y/∂x and therefore
vanishes. 4.3.4. Starting from Eq. (4.54), Γm jk = εm · (∂εk/∂qj ), we see that a proof that ∂εk/∂qj = ∂εj/∂qk would
also demonstrate that Γm jk = Γm kj. From the definition of εk, we differentiate with respect to qj , reaching ∂εk ∂qj
= ∂2 x ∂qj∂qk ˆex + ∂2 y ∂qj∂qk ˆey + ∂2 z ∂qj∂qk ˆez . Because the coordinates are differentiable functions the
right-hand side of this equation is symmetric in j and k, indicating that j and k can be interchanged without
changing the value of the left-hand side of the equation.