Tensor Calc and Moving Surfaces Exercises New

7/25/2019 Tensor Calc and Moving Surfaces Exercises New

1/119

Selected solutions to exercises from Pavel

GrinfeldsIntroduction to Tensor Analysis and

the Calculus of Moving Surfaces

David Sulon

9/14/14


2/119

ii


3/119

Contents

I Part I 1

1 Chapter 1 3

2 Chapter 2 7

3 Chapter 3 13

4 Chapter 4 17

5 Chapter 5 33

6 Chapter 6 39

7 Chapter 7 47

8 Chapter 8 49

9 Chapter 9 51

II Part II 57

10 Chapter 10 59

11 Chapter 11 67

12 Chapter 12 77

III Part III 89

13 Chapter 16 101

14 Chapter 17 109

iii


4/119

iv CONTENTS


5/119

Introduction

Included in this text are solutions to various exercises from Introduction toTensor Analysis and the Calculus of Moving Surfaces, by Dr. Pavel Grinfeld.

v


6/119

vi CONTENTS


7/119

Part I

Part I

1


8/119


9/119

Chapter 1

Chapter 1

Ex. 1: We havex = 2x0, y = 2y0. Thus

F0(x0; y0) = F(2x0; 2y0)

= (2x0)2

e2y0

= 4 (x0)2

e2y0

:

Ex. 2: Note that the above impliesx0 = 12

x, y 0= 12

y. We check

@F0

@x0(x0; y0) = 8 (x0) e

2y0

= 8

1

2x

e2( 12)y

= 4xey

@F

@x(x; y) = 2xey:

Thus, @F0

@x0(x0; y0) = 2@F@x(x; y)as desired.

Ex. 3: Leta; b 2 R, a; b 6= 0, and consider the "re-scaled" coordinate basis

i0 =

a0

j0 =

0b

;

where each of the above vectors is taken to be with respect to the standardbasis for R2. Thus, given point(x; y)in standard coordinates, we have x = ax0,y = by 0, where (x0; y0) is the same point in our new coordinate system. Now,letT(x; y)be a dierentiable function. Then,

rT=

@F

@x(x; y);

@F

@y(x; y)

3


10/119

4 CHAPTER 1. CHAPTER 1

in standard coordinates

=

@F

@x0(x0; y0)

@x0

@x(x; y);

@F

@y 0(x0; y0)

@y 0

@y(x; y)

(think ofx0 as a function ofx).

=

@F

@x0(x0; y0)

1

a;@ F

@y 0(x0; y0)

1

b

=

1pa2 + 0

@F

@x0;

1

b2 + 0

@F

@y 0

= 1p

i0

i0

@F

@x0;

1pj0

j0

@F

@y 0;

as desired.

Ex. 4: Assume x0

y0

=

ab

+

cos sin sin cos

xy

:

Then,

x0 = a + (cos ) x (sin ) yy0 = b+ (sin ) x + (cos ) y

Also, x0 ay0 b

=

cos sin sin cos

xy

xy

=

cos sin sin cos

1x0 ay0 b

=

coscos2 +sin2

sincos2 +sin2

sincos2 +sin2

coscos2 +sin2

x0 ay0 b

= cos sin

sin cos

x0 ay0

b

Thus,

x = (cos ) (x0 a) + (sin ) (y0 b)y = (sin ) (x0 a) + (cos ) (y0 b) :


11/119

5

Further notice that we obtain i0;j0 from the standard basis [Note: this "basis"

would describe points be with respect to this point (a; b)]

i0 =

cos sin sin cos

10

= cos i+sin j

j0 =

cos sin sin cos

01

= sin i+cos j

Now, we have, given a function F, we compute

@F

@x(x; y) i +

@ F

@y (x; y)j =

@F

@x0(x0; y0)

@x0

@x (x; y) +

@ F

@y 0(x0; y0)

@y0

@x (x; y)

i +

@F

@x0(x0; y0)

@x0

@y (x; y) +

=

@F

@x0(x0; y0)cos +

@ F

@y 0(x0; y0)sin

i+

@F

@x0(x0; y0)sin +

@F

@y 0(x0; y0)cos

= @F

@x0(x0; y0)cos i @F

@x0(x0; y0)sin j +

@ F

@y 0(x0; y0)sin i +

@F

@y 0(x0; y0)cos j

= @F

@x0(x0; y0) (cos i sin j) + @ F

@y 0(x0; y0) (sin i + cos j)

= @F

@x0(x0; y0)

i j cos sin

+

@ F

@y 0(x0; y0)

i j sin

cos

= @F

@x0 (x0; y0) i0+

@ F

@y 0 (x0; y0)j0

[NOT SURE - I will ask about this one tomorrow]

(a; b)cooresponds to a shifted "origin," corresponds to angle for which thewhole coordinate system is rotated.

Ex. 5: We may obtain any ane orthogonal coordinate system by rotatingthe "standard" Cartesian coordinates via (1.7) and then applying a rescaling.


12/119



13/119

Chapter 2

Chapter 2

Ex. 6: See diagram.

Note: Diagrams will be added later for Ex. 7-12

Note: For Ex 7-12, let h denote the distance from P to P, where P

is a point arbitrarily close to Palong the appropriate direction for which weare taking each directional derivative. Denef(h) := F(P), i.e. parametrizealong the unit vector emanating fromPin the direction ofl(notef(0) =F(P)).Also, for points A,B, AB indicates the (unsigned) length of the vector from AtoB .

Ex. 7:

f(h) =

pF(P)2 + h2

f0(h) = hpF(P)2 +h2

dF(p)

dl = f0(0)

Ex. 8: We have

f(h) = 1

AP hf0(h) = 1

(AP h)2

= 1

(AP h)2

so

dF(p)

dl = f0(0)

= 1

(AP)2

7


14/119


Ex. 9: Let denote the measure of angleOPP. By the Law of Sines, wehave

sin(F(P))

AP h = sin( )

OA

= sin()

OAsin

OP =

sin(F(P) F(P))h

From the second equation, we obtain

sin = OPsin (F(P) F(P))h

= OP[sin (F(P))cos(F(P)) cos(F(P))sin(F(P))]

h

The, from the rst equation, we have

sin(F(P))

AP h = OP[sin (F(P))cos(F(P)) cos(F(

(OA) h

(OA) h sin(F(P)) = (OP) (AP h)[sin(F(P))cos(F(P

(OA) h tan(F(P)) = (OP) (AP h)sin(F(P)) (OP) (AP((OA) h+ (OP) (AP h)cos(F(P)))tan(F(P)) = (OP) (AP h)sin(F(P))

tan(F(P)) = (OP) (AP h)sin(F(P))

(OA) h+ (OP) (AP h)cos(F(P))F(P) = arctan

(OP) (AP h)sin(F(P

(OA) h+ (OP) (AP h)cos(

Thus,

f(h) = arctan (OP) (AP h)sin(F(P))(OA) h+ (OP) (AP h)cos(F(P))

f0(h) = (OP)sin(F(P))[(OA) h+ (OP) (AP h)cos(F(P))] (OP)sin(F(P)) (AP

[(OA) h+ (OP) (AP h)cos(F(P))]2

= (OP)sin(F(P))[(OA) h+ (OP) (AP h)cos(F(P))] (OP)sin(F(P)) (AP

[(OA) h+ (OP) (AP h)cos(F(P))]2 + [(OP) (AP h)


15/119

9

dF(p)

dl = f0(0)

= (OP)sin(F(P))[(OP) (AP)cos(F(P))] (OP)sin(F(P)) (AP) [(OA) (OP) (AP)cos(F[(OP) (AP)cos(F(P))]2 + [(OP) (AP)sin(F(P))]2

= (OP)sin(F(P)) (OP) (AP)cos(F(P)) (OP)sin(F(P)) (AP) (OA) + (OP)sin(F(P)) (A

(OP)2

(AP)2

cos2 (F(P)) + sin2 (F(P))

= (OP)sin(F(P)) (AP) (OA)

(OP)2

(AP)2

= (OA)(OP) (AP)

sin (F(P))

Ex. 10: ClearlyF(P) = F(P)for any choicePin such a direction. Thus,

f is constant, and we have dF(p)

dl = 0

Ex. 11: Put d as the distance between P and the line from A to B. Aswith the previous problem, the distance from P to line

!AB is also d. Thus,

F(P) = 1

2(AB) d

F(P) = 1

2(AB) d;

and we have F(P) = F(P), so dF(p)dl = 0 as before.

Ex. 12: Drop a perpendicular from P to!AB. Let K be this point of

intersection. Note that the lengthAK= F(P) +h. Then,

f(h) = 1

2(AB) (F(P) +h)

f0(h) = 1

2AB

dF(p)

dl = f0(0)

= 12

AB

Ex. 13: (7) The gradient will point in direction!AP, and will have magnitude

1.(8) [Not sure]


16/119


(9) The gradient will point in direction!AP(in the same direction

as was asked for the directional derivative), and thus will have magnitude

(OA)

(OP) (AP)sin (F(P))

(noteF(P)is assumed to satisfy F(P) )(10) The gradient will point in direction perpendicular to

!AB, and will have

magnitude1.(11),(12) The gradient will point in the direction perpendicular to!

AB(in the same direction as was asked for the directional derivative in Ex.12),and thus will have magnitude

1

2AB:

Ex. 14: The directional derivative in directionL would then correspond tothe projection ofOf ontoL.

Ex. 15: [See diagram]

kR (+h) R ()k2 = 1 + 1 2cos(h)

by the Law of Cosines. So,

kR (+ h)R ()k2 = 2 2cos(h)kR (+h)R ()k =

p2 2cos(h)

=

s2 2cos

2

h

2

=

s2 2

sin2

h

2

=

r4sin2

h

2

= 2 sinh

2

:

Ex. 16:

limh!0

2sin h2h

= limh!0

212

cos h2

1 ;


17/119

11

by LHospitals rule,

= limh!0

cosh

2= 1:

Ex. 17:

limh!0

2sin h2h

= limh!0

sin h2h2

= limh!0

sin

0 + h2 sin (0)h2

= sin0(0)

= cos (0)

= 1:

Ex. 18: We haveR () R0() = 0:

Dierentiating both sides, we obtain

R0() R0() +R () R00() = 0:

But, R0() is of unit length, so we have

1 +R () R00() = 0R () R00() = 1:

Now, let be the angle between R (), R00(). We thus have

kR ()k kR00()k cos = 1kR00()k cos = 1;

sinceR () is of unit length. Now, leth be arbitrarily small. SincekR ()k =kR (+ h)k =kR0(+h)k =kR0()k = 1, we have by congruent trianglesthatkR0(+h)R0()k = kR (+ h)R ()k : Thus,

kR00()k = limh!0 kR0(+h)

R0()

kh= lim

h!0

kR (+ h) R ()kh

= kR0()k= 1:


18/119


Thus, R00() is of unit length. We then have

cos = 1;

which implies that = , or that R00() points in the opposite direction asR ().


19/119

Chapter 3

Chapter 3

Ex. 19: We may construct a three-dimensional Cartesian coordinate system asfollows: Fix an originO, then pick three points A; B; Csuch that the vectors!OA;

!OB;

!OC form an orthonormal system. Dene i =

!OA, j =

!OB, k =

!OC.

Note that in this coordinate system, A; B; Chave coordinates0@10

0

1A ;0@01

0

1A ;0@00

1

1A

respectively, and a vector Vconnecting the origin to a point with coordinates

0@x0y0z01A

can be expreessed by the linear combination

V =x0i+y0j+z0k:

Ex. 20: Since our space is three-dimensional, there are three continuousdegrees of freedom associated with our choice of origin O. The choice of thedirection of the "x"-axis yield another two continuous degrees of freedom (note

the bijection between the direction of the x-axis and a point on the unit spherecentered at O). Finally, the "y"-axis may be chosen to lie along any lineorthogonal to thex-axis; the set of all such lines lie in a plane, hence our choiceof direction for the y -axis yields the sixth continuous degree of freedom (thereis a bijection between the set of all such directions and points on the unit circlewhich lies in this plane orthogonal to the x-axis.

13


20/119


Ex. 21: LetPbe an arbitrary point in a two-dimensional Euclidean spacewith polar coordinatesr; . AssumePhas cartesian coordinates (x; y). Dene

P0 to be the point along the pole that is distance x from the origin. Note thatby the orthogonality of the x; y axes, we may form a right triangle with P, theorigin O, andP0. Note thatOP0 = x and P P0 = y; hence by the properties ofright triangles, we have

x

r = cos

y

r = sin ;

or

x = r cos

y = r sin :

Ex. 22: We see from the above that

x2 +y2 = r2 cos2 + r2 sin2

= r2;

hence we may solve for r (taken to be non-negative):

r=p

x2 + y2:

Also,

y

x =

r sin

r cos = tan ;

so= arctan

y

x:

Ex. 23: Dene thex and y coordinate of some arbitrary point Pto be theCartesian system of coordinates dened by applying Ex. 21 to the coordinateplane xed in the denition of our cylindrical coordinates. Simply dene thez(Cartesian) coordinate to be the signed distance fromPto the coordinate plane


21/119

15

(note the orthogonality of ofx,y,z by the denition of distance to a plane - andalso thatx; y do not depend on z). The equations forx; y then follow from Ex.

21, and thez (Cartesian) coordinate is equal to thez (cylindrical) by denition.

Ex. 24: The inverse relationships for r, follow from Ex. 22, and theidentity z (x; y; z) = z follows trivially from 23.

Ex. 25: LetPbe a point with spherical coordinates r; ;. Let thex-axisbe the polar axis, and the y-axis lie in the coordinate plane and point in thedirection orthogonal to the polar axis (chosen in accordance to the right-handrule). Finally, let htez-axis be the longitudinal axis. Since thez -coordinatelength OP0, where P0 is the orthogonal projection ofP onto the longitudinalaxis, we have by the properties of right triangles,

z = r cos .

Now, project P onto the coordinate plane, and denote this point P00. Weclearly have the length OP00 =r sin . Thus, by considering the right triangledetermined by the points O , P00, and the polar axis, we have

x = (OP00)cos

= r sin cos

y = (OP00)sin

= r sin sin :

Ex. 26: From Ex. 25, we havepx2 +y2 +z2 =

qr2 sin2 cos2 + r2 sin2 sin2 + r2 cos2

= r

qsin2 cos2 + sin2 sin2 + cos2

= rq

sin2

cos2 + sin2

+ cos2

= rp

sin2 + cos2

= r;

sor (x; y; z) =

px2 + y2 + z2:

Also,z =r= cos ; so

(x; y; z) = arccosz

r

= arccos zp

x2 +y2 +z2:


22/119


Finally,

y

x =

r sin sin

r sin cos

= tan ;

so (x; y; z) = arctan

y

x:


23/119

Chapter 4

Chapter 4

Ex. 27:

det J = det

" xpx2+y2

ypx2+y2

yx2+y2

xx2+y2

#

= x2 +y2

(x2 +y2)p

x2 +y2

= 1p

x2 +y2:

Ex. 28:

J(1; 1) =

1p12+12

1p12+12

112+12

112+12

=

1p2

1p2

12

12

:

Ex. 29:

det J0 = det cos r sin sin r cos

= r

cos2 + sin2

= r;

Using the relationshipr =p

x2 + y2; we havedet Jdet J0 = 1.

17


24/119


Ex. 30:

J0p2;4 = cos

4 p

2sin 4sin 4

p2cos 4

=

"p22 1p

22 1

#:

Ex. 31: We evaluate the product

JJ0 =

1p2

1p2

12

12

"p22 1p22 1

#

= 1 00 1

;as desired.

Ex. 32:

J0(x; y) =

cos r sin sin r cos

=

xr yyr

x

=

2

4

xpx2+y2

yyp

x2+y2x

3

5;

so

JJ0 =

" xpx2+y2

ypx2+y2

yx2+y2

xx2+y2

#24 xpx2+y2 yyp

x2+y2x

35

=

" x2

x2+y2 + y2

x2+y2 0

0 x2

x2+y2 + y2

x2+y2

#

= I;

similarly,J0J=I. Thus, J; J0 are inverses of each other.

Ex. 33: We use

r (x; y; z) =p

x2 + y2 +z2

(x; y; z) = arccos zp

x2 + y2 +z2

(x; y; z) = arctany

x


25/119

19

Note from our computation of the Laplacian in spherical coordinates, we have(after substituing expressions forx; y; zto obtain these results in terms ofr; ;):

@r

@x = sin cos

@r

@y = sin sin

@r

@z = cos

@

@x =

cos cos

r@

@y =

cos

r sin @

@z = sin

r

@

@x = sin

r sin @

@y = cos sin

@

@z = 0:

Thus,

J(r;;) =

264

@r@x

@r@y

@r@y

@@x

@@y

@@z

@@x

@@y

@@z

375

=

24sin cos sin sin cos cos cos

rcosr sin

sinr

sinr sin cos sin 0

35 :

We then compute

J0(r;;) =264

@x@r

@x@

@x@

@y@r

@y@

@y@

@z@r

@z@

@z@

375

=

24sin cos r cos cos r sin sin sin sin r cos sin r sin cos

cos r sin 0

35 ;


26/119


So

JJ0 =

24sin cos sin sin cos cos cos

rcosr sin sin r

sinr sin

cos sin 0

3524sin cos r cos cos r sin sin sin sin r cos sin r sin cos

cos r sin 0

35

=

24 cos2 + cos2 sin2 + sin2 sin2 r cos sin + r cos cos2 sin +1rcos sin + 1rcos sin + 1rcos cos2 sin sin2 + (cos ) cossin sin +

1r

cos sin + cos sin sin2 r cos cos sin2 (cos

=

241 0 00 1 0

0 0 1

35

[Note: Something may be o with the computation ofJ]

Ex. 34: We compute

@2f(; )

@2 =

@

@

@F

@a

@A

@ +

@ F

@b

@B

@ +

@ F

@c

@C

@

= @

@

@F

@a

@A

@ +

@ F

@a

@

@

@A

@

+ @@

@F@b

@B@

+ @ F@b

@@

@B@

+ @

@

@F

@c

@C

@ +

@ F

@c

@

@

@C

@

by the product rule. We continue:

@f(; )

@2 =

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@ +

@ F

@a

@2A

@2

+@2F

@a@b

@A

@ +

@2F

@b2

@B

@ +

@2F

@b@c

@C

@@B

@ +

@ F

@b

@2B

@2

+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@ +

@ F

@c

@2C

@2:


27/119

21

Similarly,

@2f(; )

@@ =

@

@

@F

@a

@A

@ +

@ F

@b

@B

@ +

@ F

@c

@C

@

= @

@

@F

@a

@A

@ +

@ F

@a

@

@

@A

@

+ @

@

@F

@b

@B

@ +

@ F

@b

@

@

@B

@

+ @

@

@F

@c

@C

@ +

@ F

@c

@

@

@C

@

=

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@ +

@ F

@a

@2A

@@

+@2F

@a@b

@A

@ +@2F

@b2@B

@ + @2F

@b@c

@C

@@B

@ +@ F

@b

@2B

@@

+

@2F

@a@c

@A

@+

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@ +

@ F

@c

@2C

@@

@f(; )

@2 =

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@ +

@ F

@a

@2A

@2

+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@ +

@ F

@b

@2B

@2

+@2F

@a@c

@A

@

+ @2F

@b@c

@B

@

+@2F

@c2

@C

@@C

@

+@ F

@c

@2C

@2

:

Ex. 35: We compute

@3f(; )

@2@ =

@

@

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@ +

@ F

@a

@2A

@@

+@

@

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@ +

@ F

@b

@2B

@@

+ @@

@2F@a@c

@A@

+ @2F

@b@c@B@

+ @2F

@c2@C@

@C@

+ @ F@c

@2C@@

= I+ J+ K;


28/119


for sake of clarity,

I = @@

@2F@a2

@A@

+ @2F

@a@b@B@

+ @2F

@a@c@C@

@A@

+ @ F@a

@2A@@

= @

@

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@

+

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@

@

@A

@

+

@

@

@F

@a

@2A

@@+

@ F

@a

@3A

@2@

= @

@

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@

+

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@2A

@@

+@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@ @2A

@@+

@ F

@a

@3A

@2@

= @

@

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@

+2

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@2A

@@ +

@ F

@a

@3A

@2@

=

@

@

@2F

@a2

@A

@+

@2F

@a2@

@

@A

@

+

@

@

@2F

@a@b

@B

@ +

@2F

@a@b

@

@

@B

@

+

@

@

@2F

@a@c

@C

@

+2

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@2A

@@ +

@ F

@a

@3A

@2@

= @

@ @2F

@a2 @A

@

@A

@ +

@2F

@a2@

@ @A

@ @A

@

+ @

@

@2F

@a@b

@B

@

@A

@ +

@2F

@a@b

@

@

@B

@

@A

@

+ @

@

@2F

@a@c

@C

@

@A

@ +

@2F

@a@c

@

@

@C

@

@A

@

+2

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@2A

@@ +

@ F

@a

@3A

@2@

=

@3F

@a3@A

@+

@3F

@a2@b

@B

@ +

@3F

@a2@c

@C

@

@A

@

@A

@ +

@2F

@a2@2A

@2@A

@

+

@3F

@a2@b

@A

@ +

@3F

@a@b2@B

@ +

@3F

@a@b@c

@C

@

@B

@

@A

@ +

@2F

@a@b

@2B

@2@A

@

+

@3

F@a2@c

@A@

+ @3

F@a@b@c

@B@

+ @3

F@a@c2

@C@

@C@

@A@

+ @2

F@a@c

@2

C@2

@A@

+2

@2F

@a2@A

@+

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@2A

@@ +

@ F

@a

@3A

@2@


29/119

23

J = @

@

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@ +

@ F

@b

@2B

@@

= @

@

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@

+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@

@

@B

@

+

@

@

@F

@b

@2B

@@+

@ F

@b

@

@

@2B

@@

= @

@

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@

+

@2F@a@b

@A@

+@2F

@b2@B@

+ @2F@b@c

@C@

@2B@@

+

@2F@a@b

@A@

+@2F

@b2@B@

+ @2F@b@c

@C@

@2B@@

+@ F

@b@3B

@2@

=

@

@

@2F

@a@b

@A

@ +

@2F

@a@b

@

@

@A

@

+

@

@

@2F

@b2

@B

@ +

@2F

@b2@

@

@B

@

+

@

@

@2F

@b@c

@C

@ +

@2F

@b@c

@

@

+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@2B

@@+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@2B

@@+

@ F

@b

@3B

@2@

= @

@

@2F

@a@b

@A

@

@B

@ +

@2F

@a@b

@

@

@A

@

@B

@

+ @

@

@2F

@b2

@B

@

2+

@2F

@b2@

@

@B

@

@B

@

+ @

@ @2F

@b@c@C

@

@B

@ + @

@@C

@@B

@

+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@2B

@@+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@2B

@@+

@ F

@b

@3B

@2@

=

@3F

@a2@b

@A

@ +

@3F

@a@b2@B

@ +

@3F

@a@b@c

@C

@

@A

@

@B

@ +

@2F

@a@b

@2A

@@

@B

@

+

@3F

@a@b2@A

@ +

@3F

@b3@B

@ +

@3F

@b2@c

@C

@

@B

@

2+

@2F

@b2@B

@@

@B

@

+

@3F

@a@b@c

@A

@ +

@3F

@b2@c

@B

@ +

@3F

@b@c2@C

@

@C

@

@B

@ +

@2C

@@

@B

@

+@2F

@a@b

@A

@

+@2F

@b2

@B

@

+ @2F

@b@c

@C

@ @

2B

@@

+ @2F

@a@b

@A

@

+@2F

@b2

@B

@

+ @2F

@b@c

@C

@ @

2B

@@

+@ F

@b

@3B

@2

@


30/119


K = @

@

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@ +

@ F

@c

@2C

@@

= @

@

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@

+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@

@

@C

@

+

@

@

@F

@c

@2C

@@+

@ F

@c

@

@

@2C

@@

= @

@

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@

+

@2

F@a@c

@A@

+ @2

F@b@c

@B@

+ @2

F@c2

@C@

@

2

C@@

+

@2

F@a@c

@A@

+ @2

F@b@c

@B@

+ @2

F@c2

@C@

@

2

C@@

=

@

@

@2F

@a@c

@A

@ +

@2F

@a@c

@

@

@A

@

+

@

@

@2F

@b@c

@B

@ +

@2F

@b@c

@

@

@B

@

+

@

@

@2F

@c2

@

@

+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@2C

@@+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@2C

@@

= @

@

@2F

@a@c

@A

@

@C

@ +

@2F

@a@c

@

@

@A

@

@C

@

+ @

@

@2F

@b@c

@B

@

@C

@ +

@2F

@b@c

@

@

@B

@

@C

@

+ @

@@2F

@c2@C

@2

+@2F

@c2

@

@@C

@@C

@

+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@2C

@@+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@2C

@@

=

@3F

@a2@c

@A

@ +

@3F

@a@b@c

@B

@ +

@3F

@a@c2@C

@

@A

@

@C

@ +

@2F

@a@c

@2A

@@

@C

@

+

@3F

@a@b@c

@A

@ +

@3F

@b2@c

@B

@ +

@3F

@b@c2@C

@

@B

@

@C

@ +

@2F

@b@c

@2B

@@

@C

@

+

@3F

@a@c2@A

@+

@3F

@b@c2@B

@ +

@3F

@c3@C

@

@C

@

2+

@2F

@c2@2C

@@

@C

@

+@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@ @2C

@@+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@ @2C

@@


31/119

25

Ex. 36

@f(; )

@2 =@2F

@a2@A

@ + @2F

@a@b

@B

@ + @2F

@a@c

@C

@@A

@ +@ F

@a

@2A

@2

+

@2F

@a@b

@A

@ +

@2F

@b2@B

@ +

@2F

@b@c

@C

@

@B

@ +

@ F

@b

@2B

@2

+

@2F

@a@c

@A

@ +

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@ +

@ F

@c

@2C

@2

= @2F

@ai@aj@Aj

@

@Ai

@ +

@F

@ai@2Ai

@2

@2f(; )

@@ =

@2F

@a2@A

@ +

@2F

@a@b

@B

@ +

@2F

@a@c

@C

@

@A

@ +

@ F

@a

@2A

@@

+

@2

F@a@b @A@ + @2

F@b2 @B@ + @2

F@b@c @C@

@B@ + @ F@b @2

B@@

+

@2F

@a@c

@A

@+

@2F

@b@c

@B

@ +

@2F

@c2@C

@

@C

@ +

@ F

@c

@2C

@@

= @2F

@ai@aj@Aj

@

@Ai

@ +

@F

@ai@2Ai

@@

@f(; )

@2 =

@2F

@ai@aj@Aj

@

@Ai

@ +

@F

@ai@2Ai

@2

Ex. 37 We may generalize the above three equations, setting 1 = ,2 =, to yield

@2f(; )@@

= @2F@ai@aj

@Aj

@@Ai

@ +

@F@ai

@2Ai

@@:

This encompasses three separate identities, since we have been assuming thatwe may switch the order of partial dierentiation throughout.

Ex. 38 [Not nished]

Ex. 39 Begin withcos arccos x= x

and dierentiate both sides:

ddx

[cosarccos x] = 1

(sinarccos x) ddx

[arccos x] = 1

d

dx[arccos x] =

1(sin arccos x)


32/119


By examining triangles with unit hypotenuse, we obtain

sin arccos x= p

1 x2;

sod

dx[arccos x] = 1p

1 x2

Ex. 40: We know thatf ; g satisfy

g0(f(x)) f0(x) = 1:

Dierentiating both sides, we obtain

d

dx[g0(f(x))] f0(x) +g0(f(x))

d

dx[f0(x)] = 0

g00(f(x)) f0(x) f0(x) +g0(f(x)) f00(x) = 0

g00(f(x)) [f0(x)]2

+g0(f(x)) f00(x) = 0 (4.1)

as desired.

Ex. 41: We compute

f0(x) = ex

f00(x) = ex

g0(x) = 1

x

g00(x) = 1x2

;

So

g00(f(x)) [f0(x)]2

+g0(f(x)) f00(x) =

1

e2x

e2x + 1

ex

ex

= 1 + 1= 0;

as desired.


33/119

27

Ex. 42: We compute

f0(x) = 1p1 x2

f00(x) = 2x 12

1 x23=2

= x 1 x23=2g0(x) = sin(x)

g00(x) = cos(x) ;So

g00(f(x)) [f0(x)]2

+g0(f(x)) f00(x) = cos (arccos x) 11 x2 sin (arccos (x))

x 1 x23=2

= x

1 x2 +xp

1

x2

p1 x23

= x1 x2 +

x

1 x2= 0;

as desired.

Ex. 43: We dierentiate both sides of the second-order relationship toobtain

d

dx g00(f(x)) [f0(x)]

2+g0(f(x)) f00(x) = 0

ddx

[g00(f(x))][f0(x)]2 + g00(f(x)) ddx

h[f0(x)]2

i+ d

dx[g0(f(x))] f00(x) +g0(f(x)) d

dxf00(x) = 0

g(3) (f(x)) [f0(x)]3

+g00(f(x)) 2f0(x) f00(x) +g00(f(x)) f0(x) f00(x) +g0(f(x)) f(3) (x) = 0g(3) (f(x)) [f0(x)]

3+ 3g00(f(x)) f0(x) f00(x) +g0(f(x)) f(3) (x) = 0

Ex. 44:Ex. 45:Ex. 46:Ex. 47:

Ex. 48: We begin with the identity (note the top indices should be consid-ered "rst")

Jii0Ji0

j =ij ;

and write out the dependences on unprimed coordinates:

Jii0(Z0(Z)) Ji

0

j (Z) = ij(Z) (4.2)


34/119


(note, however, that the Krnicker delta is constant with respect to the un-

primed coordinates Z). We dierentiate both sides of (4.2) with respect toZk:

@

@Zk

hJii0(Z

0(Z)) Ji0

j (Z)i

= @

@Zk

ij(Z)

@

@Zk

hJii0(Z

0(Z)) Ji0

j (Z)i

= 0

@

@Zk

Jii0(Z0(Z))

Ji

0

j (Z) +Jii0(Z

0(Z)) @

@Zk

hJi

0

j (Z)i

= 0;

since dierentiation passes through the implied summation overi0. Then, usingthe denition of the Jacobian,

@@Zk

@Zi@Zi0

(Z0(Z))

@Zi0@Zj

(Z) + @Zi@Zi0

(Z0(Z)) @@Zk

@Zi0@Zj

(Z)

= 0(4.3)

@2Zi

@Zk0@Zi0(Z0(Z))

@Zk0

@Zk (Z)

@Zi0

@Zj (Z) +

@Zi

@Zi0 (Z0(Z))

@2Zi0

@Zk@Zj(Z) = 0;(4.4)

applying the chain rule to the rst term, and implying summation over newindexk 0. Then, if we dene the "Hessian" object

Jik0;i0 := @2Zi

@Zk0@Zi0(Z0)

with an analogous denition for Ji0

k;i, we write (4.3) concisely:

Jik0;i0Jk0

k Ji0

j +Jii0J

i0

k;j = 0;

or, using a renaming of dummy indicex k 0 to j 0 and a reversing of the order ofpartial derivatives,

Jii0j0Ji0

j Jj0

k +Jii0J

i0

jk = 0:

This above tensor relationship represents n3 identities.Ex. 49: Since each Jii0 is constant for a transformation from one ane

coordinate system to another, each second derivative vanishes, and hence eachJii0j0 = 0.

Ex. 50: Begin with the identity derived in Ex. 48:

Jii0j0Ji0

j Jj0

k +Jii0J

i0

jk = 0;


35/119

29

Then, lettingk0be arbitrary, we multiply both sides byJkk0, implying summation

over k : hJii0j0J

i0

j Jj0

k + Jii0J

i0

jk

iJkk0 = 0

Jii0j0Ji0

j Jj0

k Jkk0+ J

ii0J

i0

jkJkk0 = 0

but,Jj0

k Jkk0 =

j0

k0 ; so

Jii0j0Ji0

j j0

k0+ Jii0J

i0

jkJkk0 = 0:

Note that we have j0

k0 = 1 if and only if j0 = k0, so the rst term is equal to

Jii0k0Ji0

j . After re-namingk0 = j 0, we obtain

Jii0j0Ji0

j +Jii0J

i0

jkJkj0 = 0: (4.5)

Ex. 51: Let k0 be arbitrary, and multiply both sides of (4.5) by Jjk0 ,implying summation over j :h

Jii0j0Ji0

j +Jii0J

i0

jkJkj0

iJjk0 = 0

Jii0j0Ji0

j Jjk0+ J

ii0J

i0

jkJkj0J

jk0 = 0

Jii0j0i0

k0+ Jii0J

i0

jkJkj0J

jk0 = 0

Jii0j0i0

k0+ Ji0

jkJii0J

kj0J

jk0 = 0:

Rename the dummy index in the second term i0 = h0. Then,

Jii0j0i0

k0+ Jh0

jkJih0J

kj0J

jk0 = 0:

Noting that the rst term is zero for all i06=k0, and settingk 0 = i0:Jii0j0+ J

h0

jkJih0J

kj0J

ji0 = 0:

We then may re-introduce k 0 as a dummy index:

Jii0j0+ Jk0

jkJik0Jkj0Jji0 = 0:

Then, switch the roles ofj; k as dummy indices:

Jii0j0+ Jk0

kjJik0J

jj0J

ki0 = 0


36/119


Ex. 52: Return to

Jii0j0Ji0

j Jj0

k + Jii0J

i0

jk = 0;

and write out the dependences

Jii0j0(Z0(Z)) Ji

0

j (Z) Jj0

k (Z) +Jii0(Z

0(Z)) Ji0

jk(Z) = 0

@2Zi

@Zi0@Zj0(Z0(Z))

@Zi0

@Zj (Z)

@Zj0

@Zk (Z) +

@Zi

@Zi0 (Z0(Z))

@2Zi0

@Zj@Zk(Z) = 0

Then, dierentiate both sides with respect to Zm:

0 = @

@Zm

" @2Zi

@Zi0@Zj0 (Z0(Z))

@Zi0

@Zj (Z)

@Zj0

@Zk (Z) +

@Zi

@Zi0(Z0(Z))

@2Zi0

@Zj@Zk(Z)

#

= @

@Zm

@2Zi

@Zi0@Zj0 (Z0(Z))

@Zi

0

@Zj (Z)

@Zj0

@Zk (Z) +

@2Zi

@Zi0@Zj0 (Z0(Z))

@

@Zm

"@Zi

0

@Zj (Z)

#@Zj

@Z

+ @

@Zm

@Zi

@Zi0 (Z0(Z))

@2Zi

0

@Zj@Zk(Z) +

@Zi

@Zi0 (Z0(Z))

@

@Zm

" @2Zi

0

@Zj@Zk(Z)

#

=

" @3Zi

@Zi0@Zj0@Zm0 (Z0(Z))

@Zm0

@Zm (Z)

#@Zi

0

@Zj (Z)

@Zj0

@Zk (Z) +

@2Zi

@Zi0@Zj0 (Z0(Z))

@2Zi0

@Zm@Zj(Z

+"

@

2

Z

i

@Zi0@Zm0 (Z0(Z))@Z

m0

@Zm (Z)#

@

2

Z

i0

@Zj@Zk (Z) + @Z

i

@Zi0 (Z0(Z)) @

3

Z

i0

@Zj@Zk@Zm(Z)

= Jii0j0m0Jm0

m Ji0

j Jj0

k +Jii0j0J

i0

jmJj0

k +Jii0j0J

i0

j Jj0

km+Jii0m0J

m0

m Ji0

jk +Jii0J

i0

jkm;

so, settingk 0= m0 as a dummy index:

Jii0j0k0Ji0

j Jj0

k Jk0

m +Jii0j0J

j0

k Ji0

jm+Jii0j0J

i0

j Jj0

km+Jii0k0J

i0

jkJk0

m +Jii0J

i0

jkm = 0: (4.6)

Then, multiply both sides by Jmm0 , implying summation over m:

Jii0j0k0Ji0

j Jj0

k Jk0

mJmm0+ J

ii0j0J

j0

k Ji0

jmJmm0+ J

ii0j0J

i0

j Jj0

kmJmm0+ J

ii0k0J

i0

jkJk0

mJmm0+ J

ii0J

i0

jkmJmm0 = 0

Jii0j0k0Ji0

j Jj0

k k0

m0+ Jii0j0Jj0

k Ji0

jmJmm0+ Jii0j0Ji0

j Jj0

kmJmm0+ Jii0k0Ji0

jkk0

m0+ Jii0Ji0

jkmJmm0 = 0;

This holds for all m0, so specically for m0= k0, the above identity reads

Jii0j0k0Ji0

j Jj0

k + Jii0j0J

j0

k Ji0

jmJmk0+ J

ii0j0J

i0

j Jj0

kmJmk0+ J

ii0k0J

i0

jk + Jii0J

i0

jkmJmk0 = 0 (4.7)


37/119

31

Next, in an analogous manner, multiply both sides by Jkm0 for arbitrarym0:

Jii0j0k0Ji0

j Jj0

k Jkm0+ Jii0j0Jj0

k Ji0

jmJmk0 Jkm0+ Jii0j0Ji0

j Jj0

kmJmk0 Jkm0+ Jii0k0Ji0

jkJkm0+ Jii0Ji0

jkmJmk0 Jkm0 = 0

Jii0j0k0Ji0

j j0

m0+ Jii0j0J

j0

k Ji0

jmJmk0 J

km0+ J

ii0j0J

i0

j Jj0

kmJmk0 J

km0+ J

ii0k0J

i0

jkJkm0+ J

ii0J

i0

jkmJmk0 J

km0 = 0;

rename the dummy index h0= j 0 in all but the rst term:

Jii0j0k0Ji0

j j0

m0+Jii0h0J

h0

k Ji0

jmJmk0 J

km0+J

ii0h0J

i0

j Jh0

kmJmk0 J

km0+J

ii0k0J

i0

jkJkm0+J

ii0J

i0

jkmJmk0 J

km0 = 0;

then, as in the previous exercises, set m0 = j 0:

Jii0j0k0Ji0

j +Jii0h0J

h0

k Ji0

jmJmk0 J

kj0+J

ii0h0J

i0

j Jh0

kmJmk0 J

kj0+J

ii0k0J

i0

jkJkj0+J

ii0J

i0

jkmJmk0 J

kj0 = 0;

(4.8)

Finally, multiply both sides by Jjm0 :

Jii0j0k0Ji0

j Jjm0+ J

ii0h0J

h0

k Ji0

jmJmk0 J

kj0J

jm0+ J

ii0h0J

i0

j Jh0

kmJmk0 J

kj0J

jm0+ J

ii0k0J

i0

jkJkj0J

jm0+ J

ii0J

i0

jkmJmk0 J

kj0J

jm0 = 0

Jii0j0k0i0

m0+ Jii0h0J

h0

k Ji0

jmJmk0 J

kj0J

jm0+ J

ii0h0J

i0

j Jh0

kmJmk0 J

kj0J

jm0+ J

ii0k0J

i0

jkJkj0J

jm0+ J

ii0J

i0

jkmJmk0 J

kj0J

jm0 = 0;

rename the dummy index i0 tog 0 in all but the rst term:

Jii0j0k0i0

m0+Jig0h0J

h0

k Jg0

jmJmk0 J

kj0J

jm0+J

ig0h0J

g0

j Jh0

kmJmk0 J

kj0J

jm0+J

ig0k0J

g0

jkJkj0J

jm0+J

ig0J

g0

jkmJmk0 J

kj0J

jm0 = 0:

Then, setm0 = i0:

Jii0j0k0+Jig0h0J

h0

k Jg0

jmJmk0 J

kj0J

ji0+J

ig0h0J

g0

j Jh0

kmJmk0 J

kj0J

ji0+J

ig0k0J

g0

jkJkj0J

ji0+J

ig0J

g0

jkmJmk0 J

kj0J

ji0 = 0:

(4.9)

Ex. 53: We have the following relationship

Z

Z0

Z00

(Z)

= Z;

or for each i,

Zi

Z0

Z

00

(Z)

= Zi

.

Dierentiating both sides with respect to Zj , we obtain

@

@Zj

Zi (Z0(Z00(Z)))

= ij :


38/119


Then, we apply the chain rule twice:

@Zi

@Zi0@

@Zj [Z0(Z00(Z))] = ij

@Zi

@Zi0@Zi

0

@Zi00@

@Zj

hZi

00

(Z)i

= ij

@Zi

@Zi0@Zi

0

@Zi00@Zi

00

@Zj = ij ;

orJii0J

i0

i00Ji00

j =ij :


39/119

Chapter 5

Chapter 5

Ex. 61: ij.

Ex. 62: Assume U is an arbitrary nontrivial linear combination UiZi ofcoordinate bases Zi. Since

U U >0

and

U U=ZijUiUj ;

or in matrix notation

U U=UTZU,

this condition implies Z= Zij is positive denite.

Ex. 63:

kVk =p

V V=q

ZijViVj

Ex. 64: Put Z = Zij . Thus, Z1 = Zij by denition. Let x be an

arbitrary nontrivial vector. Then, deney = Z1

x. We have

yT =

Z1xT

= xT

Z1T

= xTZ1;

33


40/119


sinceZ1 is symmetric. Then, sinceZ is positive denite, note

0 < yTZy

= xTZ1ZZ1x

= xTZ1x.

Sincex was arbitrary, this implies thatZ=Zij >0.

Ex. 65:

Zi Zj = ZikZk Zj= ZikZkj

by denition. But,ZikZkj =

ij ;

so we haveZi Zj =ij

Ex. 66, 67: [Not sure - which coordinate system are we in (if any?)]

Ex. 68: Use the denition

Zi = ZijZj

ZikZi = ZikZ

ijZj

= jkZj

= Zk:

Thus,

Zk = ZikZi

= ZkiZi;

sinceZik is symmetric.

Ex. 69: Examine

ZkiZi Zj = ZkiZinZn Zj

= nkjn

= jk;


41/119

35

sincenkjn = 1 ik = n andn= j , or by transitivity, ik = j . Thus, Z

i

Zj

determines the matrix inverse ofZki, which must beZij by uniqueness of matrixinverse.

Ex. 70: Because the inverse of a matrix is uniquely determined, we havethat Zi are uniquely determined.

Ex. 71:

ZijZjk = Zi ZjZjk

= Zi Zk;

from 5.17=i

k

from 5.16.

Ex. 72: We compute

Z1 Z2 =

1

3i1

3j

j

= 1

3i j1

3j j

= 1

3kik kjk cos

31

3kjk2

= 1

3(2)(1)

1

2

1

312

= 0;

thus, Z1; Z2 are orthogonal. We further compute

Z2 Z1 =1

3i+

4

3j

i

= 13kik2 +4

3kik kjk cos

3

=

1

3

(4) +4

3

(2)(1)12

= 0;

so Z2; Z1 are orthogonal.


42/119


Ex. 73:

Z1 Z1 = 13

i13

j i

= 1

3kik2 1

3kik kjk cos

3

= 4

3 2

3

1

2

= 1:

Z2 Z2 =1

3i+

4

3j

j

= 13kik kjk cos

3+

4

3kjk2

=

2

31

2+4

3= 1:

Ex. 74:

Z1 Z2 =

1

3i1

3j

1

3i+

4

3j

= 19kik2 +4

9kik kjk cos

3+

1

9kik kjk cos

3 4

9kjk2

= 49

+4

9+

1

9 4

9

= 39

= 13

:

Ex. 75: Let Rdenote the position vector. We compute

Z3 = @R (Z)

Z3

= @R (r;;z)

@z

for cylindrical coordinates

= limh!0

R (r;;z+ h) R (r;;z)h

:


43/119

37

But, R (r;;z+ h)

R (r;;z) is clearly a vector of length h pointing in the z

direction; thus, for any h,

R (r;;z+ h)R (r;;z)h

is the unit vector pointing in the z direction. This implies Z3 is the unit vectorpointing in thez direction.

Ex. 76: The computations of the diagonal elementsZ11 and Z22 are thesame as for polar coordinates; moreover the zero o-diagonal entries Z12, Z21follow from the orthogonality ofZ1, Z2. By denition of cylindrical coordinates,thez axis is perpendicular to the coordinate plane (upon which Z1, Z2lie); thus,since Z3 points in the z direction, we have that Z3 is perpendicular to both Z1,

Z2. This implies that the o-diagonal entries in row3 and column 3 ofZij arezero. Morover, sinceZ3 is of unit length; we have Z33 = Z3 Z3 = 1. Thus,we have

Zij =

241 0 00 r2 0

0 0 1

35 .

Now, since Zij is dened to be the inverse ofZij, we may easily compute

Zij =

241 0 00 r2 0

0 0 1

35 ;

since the inverse of a diagonal matrix (with non-zero diagonal entries, of course)is the diagonal matrix with corresponding reciprocal diagonal entries.

Ex. 77: We have

Z3 = Z3jZj

= 0Z1+ 0Z2+ 1Z3

= Z3


44/119



45/119

Chapter 6

Chapter 6

Ex. 87: Look at

ZijJi0

i Jj0

j Zj0k0 =ZijJi

0

i Jj0

j ZjkJjj0J

kk0

by the tensor property ofZjk

= ZijZjkJi0

i Jkk0

= ikJi0

i Jkk0

= Ji0

kJkk0

= i0

k0;

so, in linear algebra terms, we have that ZijJi0

i Jj0

j is the matrix inverse ofZj0k0 .

By uniqueness of matrix inverses, this forces ZijJi0

i Jj0

j =Zi0j0, as desired.

Ex. 88: LetZ,Z0 be two coordinate systems. Write the unprimed coordi-nates in terms of the primed coordinates

Z= Z(Z0) :

Then,

@F(Z)

@Zi0 =

@F(Z(Z0))

@Zi0

= @F@Zi @Z

i

@Zi0

= @F

@ZiJii0;

so @F@Zi is a covariant tensor.

39


46/119


Ex. 89: We show the general case (since by the previous exercise, we knowthat the collection of rst partial derivatives is a covariant tensor). Dene,

given a covariant tensor eld Ti

Sij = @Ti

Zj

Si0j0 = @Ti0

Zj0

so

Si0j0 = @Ti0

@Zj0

= @

@Zj0 TiJii0 ;sinceT is a covariant tensor,

= @

Zj0

Ti(Z0(Z)) Jii0(Z

0)

= @Ti

@Zj@Zj

@Zj0Jii0+ TiJ

ii0j0

= SijJjj0J

ii0+ TiJ

ii0j0

6= SijJjj0Jii0

(except in the trivial case where Ti = 0). Thus, in general, the collection

@TiZj

is not a covariant tensor.

Ex. 90: Compute

Si0j0 = @Ti0

@Zj0 @Tj0

@Zi0

= @Ti

@Zj

Jjj0Jii0+ TiJ

ii0j0

@Tj

@Zi

Jii0Jjj0+ TiJ

ij0i0

by the above, interchanging the rolls ofi0; j0 for the second term:

= @Ti@Zj

Jjj0Jii0+ TiJ

ii0j0

@Tj@Zi

Jjj0Jii0+ TiJ

ii0j0

;


47/119

41

sinceJij0i0 =Jii0j0,

= @Ti

@ZjJjj0J

ii0+ TiJ

ii0j0

@Tj@Zi

Jjj0Jii0 TiJii0j0

=

@Ti@Zj

@Tj@Zi

Jjj0J

ii0

= SijJii0J

jj0;

so this skew-symmetric part Sij is indeed a covariant tensor.

Ex. 91: Put

Sij = @Ti

@Zj

;

so

Si0j0 =

@Ti0

@Zj0

= @

@Zj0

hTiJi

0

i

i;

sinceTis a contravariant tensor,

= @

@Zj0 Ti (Z0(Z))

Ji

0

i +Ti @

@Zj0 hJi

0

i (Z(Z0))

i= @T

i

@ZjJj

0

j Ji0

i + Ti @Ji

0

i

@Zj@Zj

@Zj0

= SijJi0

i Jj0

j +TiJi

0

ijJjj0

6= SijJi0i Jj0

j

except in the trivial case.

Ex. 92: We have

kij = Zk @Zi

@Zj;

so in primed coordinates,

k0

i0j0 = Zk0 @Zi0

@Zj0

= Zk0

@Zi@Zj

Jii0Jjj0+ ZiJ

ii0j0


48/119


by our work done earlier (note that Zi is a covariant tensor)

ZkJk

0

k

@Zi@Zj

Jii0Jjj0+ ZiJ

ii0j0

sinceZk is a contravariant tensor,

=

Zk @Zi

@Zj

Jk

0

k Jii0J

jj0+

Zk Zi

Jk0

k Jii0j0

= Zk @Zi

@ZjJk0

k

Ji

i0J

j

j0+

k

i

Jk0

k

Ji

i0

j0

=

Zk @Zi

@Zj

Jk

0

k Jii0J

jj0+ J

k0

i Jii0j0

6=

Zk @Zi@Zj

Jk

0

k Jii0J

jj0 =

kijJ

k0

k Jii0J

jj0

except in the trivial case.

Ex. 93: Compute

@Ti0j0

@Zk0 =

@

@Zk0

hTijJ

ii0J

jj0

i=

@

@Zk0 [Tij ] J

ii0J

jj0+ Tij

@

@Zk0

Jii0

Jjj0+ TijJii0

@

@Zk0

hJjj0i

= @

@Zk0 [Tij ] J

ii0J

jj0+ TijJ

ii0k0J

jj0+ TijJ

ii0J

jj0k0

= @

@Zk0 [Tij(Z(Z

0))] Jii0Jjj0+ TijJ

ii0k0J

jj0+ TijJ

ii0J

jj0k0

=

@Tij@Zk

@Zk

@Zk0 Jii0J

j

j0+ TijJii0k0J

j

j0+ TijJii0J

j

j0k0

= @Tij

@ZkJkk0J

ii0J

jj0+ TijJ

ii0k0J

jj0+ TijJ

ii0J

jj0k0:


49/119

43

Thus, from 5.66,

k0

i0j0 = 1

2 Zk0

m

0 @Zm0i0@Zj0 +

@ Zm0j0

@Zi0 @ Zi0j0

@Zm0

= 1

2ZkmJk

0

k Jm0

m

@Zm0i0

@Zj0 +

@ Zm0j0

@Zi0 @ Zi0j0

@Zm0

= 1

2ZkmJk

0

k Jm0

m (@Zmi@Zj

Jjj0Jmm0J

ii0+ ZmiJ

mm0j0J

ii0+ ZmiJ

mm0J

ii0j0

+@Zmj@Zi

Jii0Jmm0J

jj0+ ZmjJ

mm0i0J

jj0+ ZmjJ

mm0J

ji0j0

@Zij@Zm

Jmm0Jii0J

jj0ZijJii0m0Jjj0ZijJii0Jjj0m0)

= 1

2ZkmJk

0

k Jm0

m

@Zmi@Zj

Jjj0Jmm0J

ii0+

@ Zmj@Zi

Jii0Jmm0J

jj0

@Zij@Zm

Jmm0Jii0J

jj0

+ 12

ZkmJk0

k Jm0

m

ZmiJmm0j0Jii0+ ZmiJmm0Jii0j0

+1

2ZkmJk

0

k Jm0

m

ZmjJ

mm0i0J

jj0+ ZmjJ

mm0J

ji0j0

1

2ZkmJk

0

k Jm0

m

ZijJ

ii0m0J

jj0+ ZijJ

ii0J

jj0m0

=

1

2ZkmJk

0

k Jjj0J

ii0

@Zmi@Zj

+@ Zmj

@Zi @Zij

@Zm

+1

2ZkmJk

0

k Jm0

m

ZmiJ

mm0j0J

ii0+ ZmiJ

mm0J

ii0j0

+1

2ZkmJk

0

k Jm0

m

ZmjJ

mm0i0J

jj0+ ZmjJ

mm0J

ji0j0

1

2 Zkm

Jk0

k Jm0

m

ZijJii0m0J

jj0+ ZijJ

ii0J

jj0m0

= kijJk0

k Jjj0J

ii0

+1

2ZkmZmiJ

k0

k Jm0

m

Jmm0j0J

ii0+ J

mm0J

ii0j0

+1

2ZkmZmjJ

k0

k Jm0

m

Jmm0i0J

jj0+ J

mm0J

ji0j0

1

2ZkmZijJ

k0

k Jm0

m

Jii0m0J

jj0+ J

ii0J

jj0m0

= kijJ

k0

k Jjj0J

ii0

+1

2ki J

k0

k Jm0

m

Jmm0j0J

ii0+ J

mm0J

ii0j0

+ 12

kj Jk0

k Jm0

m

Jmm0i0Jjj0+ Jmm0Jji0j0

12

ZkmZijJk0

k Jm0

m

Jii0m0J

jj0+ J

ii0J

jj0m0

= kijJ

k0

k Jjj0J

ii0

+1

2Jk

0

i Jm0

m

Jmm0j0J

ii0+ J

mm0J

ii0j0

+1

2Jk

0

j Jm0

m

Jmm0i0J

jj0+ J

mm0J

ji0j0

1

2ZkmZijJ

k0

k Jm0

m

Jii0m0J

jj0+ J

ii0J

jj0m0

= kijJ

k0

k Jjj0J

ii0

+ 12

Jk0

i Jm0m J

mm0j0J

ii0+ 12

Jk0

i Jm0m J

mm0J

ii0j0

1 k0 m0 m 1 k0 m0 m


50/119


= kijJk0

k Jjj0J

ii0

+1

2Jk

0

i Jii0J

m0

m Jmm0j0+

1

2Jk

0

i Jii0j0

+1

2Jk

0

j Jjj0J

m0

m Jmm0i0+

1

2Jk

0

j Jji0j0

12

ZkmZijJk0

k Jm0

m Jii0m0J

jj0

1

2ZkmZijJ

k0

k Jm0

m Jii0J

jj0m0

= kijJk0

k Jjj0J

ii0+ J

k0

i Jii0j0+

1

2k

0

i0Jm0

m Jmm0j0+

1

2k

0

j0 Jm0

m Jmm0i0

1

2ZkmZijJ

jj0J

k0

k Jm0

m Jii0m0

1

2Zk

= kijJk0

k Jjj0J

ii0+ J

k0

i Jii0j0+ 0:

Thus, we havek

0

i0j0 = kijJ

k0

k Jjj0J

ii0+ J

k0

i Jii0j0:

Ex. 94: We show the result for degree-one covariant tensors. The general-ization to other tensors is then evident.

AssumeTi= 0. Then, sinceTi0 =TiJii0, we have Ti0 = 0.

Ex. 95: Note that in such a coordinate change,

Ji0

i =Ai0

i;

so the "Hessian" objectJi

0

ij = 0;

sinceAi0

i is assumed to be constant with respect toZ. This means that all but

the rst terms in the above computations of @Ti0Zj0

, @Ti0

@Zj0, @Ti0j0

@Zk0, and evenk

0

i0j0 arezero, and hence we have that each of these objects have this "tensor property"

with respect to coordinate changes that are linear transformations.

Ex. 96: Since the sum of two tensors is a tensor, we may inductively showthat the sum of nitely many tensors is a tensor. We must show that for anyconstant c,

cAijk


51/119

45

is a tensor. Compute

cAi0

j0k0 =cAijkJi0

i Jjj0Jkk0;

sinceAijk is a tensor. Thus, by the above, we have if eachA (n)ijk is a tensor,

then the sumNX

n=1

cnA (n)ijk

is a tensor. Thus, linear combinations of tensors are tensors.

Ex. 97: We have that

SiTij = Si

ikT

kj

=

i

kSiTkj

;

which is a tensor, since both ik and SiTkj are tensors by the previous section

and by the fact that the product of two tensors is a tensor.

Ex. 98: ii = n by the summation convention. Thus, ii returns the

dimension of the ambient space.

Ex. 99: We haveVij =V

kijZk;

so

VijZm = VkijZk Zm= Vkij

mk

= Vmij

so substitutingm = k, we have an expression for the components

Vkij = VijZk

So,

Vk0

i0j0 = Vi0j0Zk0

= VijJii0Jjj0ZkJk0

k ;

since both Vij ; Zk are tensors

= VijZkJii0Jjj0Jk0

k


52/119


by linearity

=VkijJii0Jjj0Jk0

k

as desired.

Ex. 100: Fix a coordinate systemZi

Tijk =Tij

k Ji

iJjj

Jkk ;

so

Tijk Ji0

i Jj0

j Jkk0 = T

ij

kJi

iJjj

Jkk Ji0

i Jj0

j Jkk0

= Tijk

i0

ij0

jkk0

= Ti0j0

k0 ;

as desired.


53/119

Chapter 7

Chapter 7

47


54/119



55/119

Chapter 8

Chapter 8

49


56/119



57/119

Chapter 9

Chapter 9

Ex. 183: Assumen = 3. Given aij , put A as the determinant. We dene

A= eijkai1aj2ak3:

Note that switching the roles of1; 2 in the above equation yields

eijkai2aj1ak3 = eijkaj1ai2ak3

= ejikai1aj2ak3

= eijkai1aj2ak3= A

Generalizing, we let (r;s;t)be a permutation of(1; 2; 3). We may then seethat

A= eijkerstairajsakt

(note that the summation convention is not implied in the above line). Then,since there are 3! permutations of(1; 2; 3), we may write

3!A=X

permutations(r;s;t)

eijkerstairajsakt

But, erst = 0 for (r;s;t) that is not a permutation; hence we may sum overall0

r;s;t

3, and apply the Einstein summation convention:

3!A= eijkerstairajsakt;

or

A= 1

3!eijkerstairajsakt

51


58/119


We may similarly show that for aij , we have

A= 13!

eijkai1aj2ak3:

Ex. 184: We have

123rst ar2a

s2a

t3 =

123srt a

s2a

r2a

t3

after index renaming

= 123srt ar2a

s2a

t3

=

123rst a

r2a

s2a

t3:

Since123rst a

r2a

s2a

t3 = 123rst ar2as2at3;

we need123rst a

r2a

s2a

t3 = 0:

The result for 132srt as2a

r3a

t2 follows similarly.

Ex. 185: Note

123rst =e123erst = 1 erst = erst;

so123rst a

r1a

s2a

t3= ersta

r1a

s2a

t3 = A:

Also,

132rst ar2a

s3a

t1 =

132rst a

t1a

r2a

s3

= 132str ar1a

s2a

t3

=

123

strar

1as

2at

3= estrar1as2at3= A:


59/119

53

[Note: Is there an error somewhere - should this beA?]

Ex. 186: DeneAir =

1

2!eijkerstajsatk:

We check that@A

@air=Air:

Check

@A

@alu=

1

3!eijkerst

@(airajsakt)

@alu

= 1

3!

eijkerst @air

@alu

ajsakt+air@ajs

@alu

akt+airajs@akt

@alu

= 1

3!eijkerst

hli

urajsakt+air

lj

usakt+airajs

lk

ut

i=

1

3!

heijklie

rsturajsakt+aireijklje

rstusakt+airajseijklke

rstut

i=

1

3!

eljkeustajsakt+ aire

ilkerutakt+airajseijlersu

=

1

3!

eljkeustajsakt+ e

ilkerutairakt+eijlersuairajs

=

1

3!

3eljkeustajsakt

;

after an index renaming,

= 1

2!eljkeustajsakt

= Alu

as desired. Similarly, if we dene

Air = 1

2!eijkersta

jsatk;

we have@A

@air =Air

by a similar argument.Ex. 187: In cartesian coordinates,

Zij =ij;


60/119


so

Z = jZj= jIj= 1:

Thus, pZ= 1:

In polar coordinates,

[Zij ] = 1 00 r2

;so

Z =

1 00 r2

= r2;

hence pZ= r:

In spherical coordinates,

[Zij ] =

241 0 00 r2 0

0 0 r2 sin2

35 ;

so

Z =

1 0 00 r2 00 0 r2 sin2

= r4 sin2 ;

thus, pZ= r2 sin :


61/119

55

Ex. 188: We compute, using the Voss-Weyl formula,

ririF = 1pZ

@@ZipZZij @F

@Zj

= 1

r2 sin

@

@r

r2 sin (1)

@F

@r

+

@

@

r2 sin r2

@F

@

+

@

@

r2 sin r2 sin2

@F

@

= 1

r2 sin

@

@r

r2 sin

@F

@r

+

@

@

r4 sin

@F

@

+

@

@

r4 sin3

@F

@

= 1

r2 sin

sin

@

@r

r2

@F

@r

+ r4

@

@

sin

@F

@

+r4 sin3

@

@

@F

@

= 1

r2 sin

sin

2r

@F

@r +r2

@2F

@r2

+r4

cos

@F

@+ sin

@2F

@2

+r4 sin3

@2F

@2

= 1

r22r @F

@r +r2

@2F

@r2+ r2

sin cos @F

@+ sin

@2F

@2+r2 sin2 @2F

@2

= 2

r

@F

@r +

@2F

@r2 + r2 cot

@F

@ + r2

@2F

@2 +r2 sin2

@2F

@2

= @2F

@r2 +

2

r

@F

@r + r2

@2F

@2 +r2 cot

@F

@ +r2 sin2

@2F

@2

Ex. 189: We compute, for cylindrical coordinates

ririF = 1pZ

@

@Zi p

ZZij@F

@Zj=

1

r

@

@r

r (1)

@F

@r

+

@

@

r r2 @F

@

+

@

@z

r (1)

@F

@z

= 1

r

@

@r

r

@F

@r

+ r3

@2F

@2 +r

@2F

@z2

= 1

r

@F

@r +r

@2F

@r2 +r3

@2F

@2 +r

@2F

@z2

= @2F

@r2 +

1

r

@F

@r + r2

@2F

@2 +

@2F

@z2:


62/119



63/119

Part II

Part II

57


64/119


65/119

Chapter 10

Chapter 10

Ex. 213: Note that

ZiZj =

S Zi

(S Zj)

=

S(S Zj) Zi

= (S Zj) S Zi6= ij ;

since(S Zj) S

is merely the projection ofZj onto the tangent space. [ASK]

EARLIER ATTEMPT:

ZiZj =

ij

S Zi

ZjS = ij

SS ZiZj = ijSZj Zi = ijSZj Zi = ZjZi;

which forcesSZj = Zj

??? [Not sure - maybe a dimensional argument?]

59


66/119


Ex. 214: We haveTi =TZi:

Now,

TiZi = TZiZ

i

= T= T;

as desired.

Ex. 215: We show that (V P) N = 0. Compute(V

P)

N = (V

(V

N) N)

N

= V N (V N) (N N)= V NV N= 0;

sinceN N =1.

Ex. 216: We compute

PijPjk = NiNjNjNk

= Ni (1) Nk

= Pik;

as desired.

Ex. 217: We show that V T is orthogonal to the tangent plane. Wecompute

(V T) S = (V (V S) S) S= V S (V S) S S

= V S

(V S

)

= V S V S= 0:


67/119

61

Ex. 218: Similarly to 216, we have, given denitionTij =NiNj

TijTjk = NiNjNjNk

= NiNk

= Tik:

[Note: This seems like the exact same problem - do we mean to deneTij =T

iTj?]

Ex. 219: We have [Note that this implies 213 additionally]

NiNj+ ZiZ

j =

ij :

Contract both sides withNi:

NiNjNi+ ZiZ

jNi =

ijNi

NiNiNj+ NiZ

iZ

j = Nj

NiNiNj+ 0 = Nj

NiNiNj = Nj ;

where the third line follows fromNiZi= 0. Now, this holds for allNj , for which

at least one is nonzero (we cannot have the normal vector be zero). Hence, wehave

NiN

i

= 1;

as desired.

Ex. 220: Using similar manipulations of indices to the earlier discussion ofthe Levy-Civita symbols, we derive

14

ijkrstTtj T

sk =

1

4ijkrtsT

sjT

tk

= 1

4ijkrstT

sjT

tk;

so

NiNr = 14

ijkrstTsjTtk 14 ijkrstTtj Tsk

= 2

1

4ijkrstT

sjT

tk

= 1

2ijkrstT

sjT

tk:


68/119


Ex. 221: This result follows exactly as was done earlier, except we use thenew denition of the Jacobian for surface coordinates

J0

=@S

0

@S:

Ex. 222: From before, we have

@Zij@Zk

=Zliljk +Zlj

lik:

From the analogous denitions ofS; we have

@S@S

=S+ S

compute

1

2S!

@S!@S

+@ S!

@S @ S

@S!

= 1

2S!

S!+S

!+ S!

+ S

!

S

! +S

!

=

1

2S! S!

+S

!+ S!

+ S

! S! S!

= 12

+ S

!S+

+ S

!S! S!S! S!S!

=

1

2

2

= ;

as desired.

Ex. 223: Assume the ambient space is reered to ane coordiantes. Wehave

= Z

i

@Zi

@S + i

jkZ

Z

j

Zk

= Zi@Zi@S

+ 0;

sinceijk = 0 in ane coordinates.


69/119

63

Ex. 224 [Still Working]

Ex. 225 We compute, givenZ1 (; ) = R

Z2 (; ) =

Z3 (; ) =

Zi = @Zi

@S

=

264

@Z1

@S1@Z1

@S2@Z2

@S1@Z2

@S2@Z3

@S1@Z3

@S2

375

=240 0

1 00 1

35;

then, note that since

0 1 00 0 1

240 01 00 1

35= 1 0

0 1

;

we have

Zi =

0 1 00 0 1

Ni =2401

0

35 24001

35

=

i j k

0 1 00 0 1

= i

=

2410

0

35

S1 = Zi1Zi

= Z21Z2

= R cos cos i +R cos sin j R sin kS2 = Z

i2Zi

= Z32Z3

= R sin sin i +R sin cos j


70/119


S = R2 cos2 cos2 + R2 cos2 sin2 + R2 sin2 R2 cos cos sin sin +R2 c

R2 cos cos sin sin +R2 cos sin sin cos R2 sin2 sin2 +R2 sin

=

R2 00 R2 sin2

[ASK - should this be the same as when the ambient coordinates a

S =

R2 0

0 R2 sin2

pS =

sR2 00 R2 sin2

= R2 sin :

Now, recall the Christoel symbols for the ambient space (in spherical coords):

122 = r133 = r sin2 212 =

221 =

1

r233 = sin cos 313

= 331

=1

r223 =

232 = cot :

Now, setting as coord. 1 and as coord. 2, and using

=Z

i

@Zi@S +

ijkZ

Z

jZ

k ;


71/119

65

we compute

111 = Z1i

@Zi1@S1

+ ijkZ1Z

j1Z

k1

= Z12@Z21@S1

+ 2jkZ12Z

j1Z

k1

= Z12@Z21@S1

+ 222Z12Z

21Z

21

= @Z21

@S1

= 0

121 = 112= 0 +

ijkZ

1Z

j2Z

k1

= 2jkZ12Z

j2Z

k1

= 232Z

12Z

32Z

21

= cot (1) (1) (1)

= cot

122 = ijkZ

1Z

j2Z

k2

= 2jkZ12Z

j2Z

k2

= 233Z12Z

32Z

32

= sin cos

211 = ijkZ2Zj1Zk1

= 3jkZ23Z

j1Z

k1

= 322Z23Z

21Z

21

= 0

221 = 212=

ijkZ

2Z

j2Z

k1

= 3jkZ23Z

j2Z

k1

= 332Z23Z

32Z

21

= 0

222 = ijkZ

2Z

j2Z

k2

= 333Z23Z

32Z

32

= 0;

(note @Zi@S vanishes in each computation).


72/119


Ex. 226: We have

q(x0(s))2 + (y0(s))2 = 1;

since this is an arc-length parametrization. Thus,

Ni =

y0(s)x0(s)

S = (x0(s))

2+ (y0(s))

2

= 1

S = 1pS = 1

[ASK why x0x00+ y0y00= 0].

Ex. 227: Simply denotet = x, and then we have the parametrization

x (t) = t

y (t) = y (t) ;

and ccompute these objects in the preceding section, noting that x0(t) = 1.The, re-substitutex = t.

Ex. 228: Again, we have (in polar coordinates)qr0(s)2 +r (s)2 0(s)2 = 1,

so the results follow similarly to the above cases.


73/119

Chapter 11

Chapter 11

Ex. 229: This follows similarly as with the ambient covariant derviative, us-ing the tensor properties ofT(S) given in surface coordinates, and using theanalogous Jacobians J

0

.

Ex. 230: The sum rule is clear from the sum rule of the partial derivative,and the properties of contraction. Also, the product rule follows as with theambient case.

Ex. 231: We compute, using

=1

2S!

@S!@S

+@ S!

@S @ S

@S!

rS = @S@S

S S

= @S

@S S S

= @S

@S 1

2S!

@S!@S

+@ S!

@S @ S

@S!

S 1

2S!

@S!@S

+@ S!

@S @ S

@S!

S

= @S

@S 1

2!

@S!@S

+@ S!

@S @ S

@S!

1

2!

@S!@S

+@ S!

@S @ S

@S!

= @S

@S 1

2

@S@S

+@ S

@S @ S

@S

1

2

@S@S

+@ S

@S @ S

@S

= @S

@S

1

2

@S

@S

1

2

@S

@S

= 0:

Similarly, we may show that in the contravariant case,

rS = 0:

67


74/119


For the Levy-Civita symbols, note

" =p

Se

" = 1p

Se

The result follows similarly to the ambient case, carefully noting that

= S @S

@S:

The delta systems follow from the product rule and the fact that rS =r"= r"

= 0:

Ex. 232: Commutativity with contraction follows exactly as in the ambientcase.

Ex. 233: We compute, using

S =

R2 0

0 R2 sin2

;

the following:

rrF = 1pS

@

@S

pSS

@F

@S

= 1R2 sin

@@

R2 sin S1 @F

@S

+ 1

R2 sin @

@

R2 sin S2 @F

@S

= 1

R2 sin

@

@

R2 sin S11

@F

@

+

1

R2 sin

@

@

R2 sin S22

@F

@

= 1

R2 sin

@

@

R2 sin R2

@F

@

+

1

R2 sin

@

@

R2 sin R2 sin2

@F

@

= 1

R2 sin

@

@

sin

@F

@

+

1

R2 sin

@

@

1

sin

@F

@

= 1

R2 sin

@

@

sin

@F

@

+

1

R2 sin2

@2F

@2:

Ex. 234: For the surface of a cylinder, note

S =

R2 0

0 1

p

S = R;


75/119

69

so

rrF = 1pS

@

@S

pSS

@F

@S

= 1

R

@

@

RS11

@F

@

+

1

R

@

@z

RS22

@F

@z

= 1

R2@2F

@2 +

@2F

@z2:

Ex. 235: Note

S =

(R+r cos )2 0

0 r2

p

S = r (R+r cos ) ;

and compute

rrF = 1pS

@

@S

pSS

@F

@S

= 1

r (R+ r cos )

@

@

r (R+r cos ) S1

@F

@S

+ 1r (R+ r cos ) @@

r (R+ r cos ) S2 @F@S

= 1

r (R+ r cos )

@

@

r (R+r cos ) (R+ r cos )2

@ F

@

+ 1

r (R+ r cos )

@

@

r (R+ r cos ) r2

@F

@

= 1

(R+r cos )2@2F

@2 +

1

r2 (R+r cos )

@

@

(R+ r cos )

@F

@

:

Ex. 236: We have

S =

"r (z)2 0

0 11+r0(z)2

#

pS = r (z)

q1 +r0(z)2;


76/119


Thus,

rrF = 1pS

@

@S

pSS

@F

@S

= 1

r (z)

q1 +r0(z)2

@

@

pSS11

@F

@

+

1

r (z)

q1 +r0(z)2

@

@z

pSS22

@F

@z

= 1

r (z)

q1 +r0(z)2

@

@

r (z)

q1 +r0(z)2r (z)2

@ F

@

+ 1

r (z)q

1 +r0(z)2

@

@z

r (z)

q1 +r0(z)2

1

1 +r0(z)2@F

@z

!

= 1

r (z)

q1 +r0(z)2

@@z

0@ r (z)q1 +r0(z)2

@F@z

1A+ 1r (z)2

@2F@2

:

Ex. 237: These were computed earlier.

Ex. 238: We compute:

rZi = @Zi@S

ZjkijZk

= @Zi

@Zm

@Zm

@SZj

kijZk

= @Zi

@ZmZm ZjkijZk

= kimZkZm ZjkijZk

=

Zm kim Zjkij

Zk

= 0;

after index renaming. The contravariant case follows similarly. Also, sinceZij = Zi Zj ; we have

rZij = 0

by the product rule. Similarly,

rZij = 0:

[Levy-Civita Symbols to come]


77/119

71

Ex. 239: Begin with equation 10.41:

NiNi = 1;

and compute take the surface covariant derivative of both sides:

0 = r NiNi

= rNiNi +NirNi= r

NjZij

NkZ

ik +NirNi=rNjZijNkZik + NirNi;

by the metrinilic property,

=

rN

jNkkj + Ni

rN

i

= rNkNk+NirNi= NkrNk +NirNi= 2NirNi;

after index renaming. Thus,

NirNi

= 0:


78/119


Ex. 240: We compute

"ijk"ZjNk = "ijk"Zj

12 "kmn"ZmZn

=

1

2"ijk"kmn"

"ZjZ

mZ

n

= 1

2ijkkmn

Z

jZ

mZ

n

= 12

ijkmknZ

jZ

m Z

n

= 1

2ijkmnk

Z

jZ

mZ

n

= 1

22ijmn

Z

jZ

m Z

n

= ijmn

ZjZm Z

n

=

imjn jmin

ZjZ

m Z

n

= imjn

Z

jZ

m Z

n jminZjZm Zn imjnZjZm Zn +jmin

= Z

jZ

iZ

j ZjZjZi ZjZiZj +ZjZjZi

=

Z

i Zi Zi+Zi

= Zi Zi Zi+Zi

[Some factors of 2 needed?]

Ex. 241: Note that for a general covariant-contravariant tensor, we have

rTij =ZkrkTij :

Thus,ru= Zkrku

andru= Zkrku:

Thus,

r

ru =

rZkrku= rZkrku+ Zkrrku

= BNkrku+ ZkZmrmrku:

= BNkrku+ ZkZnZmnrmrku

= BNkrku+ ZkZnSZmnrmrku


79/119

73

Now, set = and contract:

rru = BNkrku+ZkZnSZmnrmrku= BN

krku+ ZkZnZmnrmrku= BN

krku+ZkZnZmnrmrku:

Now,NkNn+ Z

kZ

n =

kn;

soZkZ

n =

kn NkNn:

We substitute in the above:

rru = BNkrku+

kn NkNn

Zmnrmrku= BN

krku+Zkmrmrku NkNnZmnrmrku= BN

krku+ rmrmu NmNkrmrku;

or after renaming dummy indices,

rru= BNiriu+ ririu NiNjrirju;

orNiNjrirju= ririurru+BNiriu

Ex. 242: LetZi (s)

be the parametrization of the line normal to the surface, emanating from pointZi0. Note that we have

dZi

ds (0) = lim

h!0

Zi (h) Zi0h

=Ni:

also, compute

d

dsdZi

ds Zi

= d2Zi

ds2 Zi+

dZi

ds

dZi

ds (Z(s))

= d2Zi

ds2 Zi+

dZi

ds

@Zi@Zk

dZk

ds

= d2Zi

ds2 Zi+

dZi

dsnikZn

dZk

ds ;


80/119


so at s = 0;

d

ds

dZi

ds Zi

js=0 = d

2Zi

ds2 Zi+N

inikZnNk

= d2Zi

ds2 Zi+N

iNknikZn

= d2Zi

ds2 Zi+N

jNknjkZi

=

d2Zi

ds2 + NjNkijk

Zi

Now, examine the LHS of the above. Since

d

ds

dZi

ds Zi

represents the second derivative of a line, the LHS vanishes. Thus,

d2Zi

ds2 = NjNkijk ;

Then, deneF(s) = u (Z(s)) ;

so

F0(s) = @u

@Zi(Z(s))

dZi

ds (s)

F00(s) = d

ds

@u

@Zi(Z(s))

dZi

ds (s) +

@u

@Zi(Z(s))

d

ds

dZi

ds (s)

= @2u

@Zi@Zj(Z(s))

dZi

ds (s)

dZj

ds (s) +

@u

@Zi(Z(s))

d2Zi

ds2 (s) :

ats = 0 :

F00(0) = @2u

@Zi@Zj (Z(0)) Ni

Nj

+

@u

@Zi (Z(0))

d2Zi

ds2 (0)

= @

@Zj [riu] NiNj +riu d

2Zi

ds2 (0)

now, note


81/119

75

rjriu = @riu@Zj kijrku@riu@Zj

= rjriu+ kijrku;

so

F00(0) =rjriu+ kijrkuNiNj + riu d2Zids2 (0)

= rjriuNiNj + kijrkuNiNj +riud2Zi

ds2 (0)

= NiNjrjriu+ NiNjkijrkuNjNkijkriu

= Ni

Nj

rjriu;

thus@2u

@n2 =F00(0) =NiNjrirju

after renaming indices.


82/119



83/119

Chapter 12

Chapter 12

Ex. 243: This follows from the denition and from lowering the index.

Ex. 244: We have

R =@@S

@

@S + !

! !!;

so

R =@@S

@

@S + !

! !!

=@

@S@

@S + !

!

!

!

=@@S

@

@S

= 0:

Ex. 245: This was done for the nal exam.

Ex. 246: Compute

R = R (12.5)

= R (12.3)= R (12.5).

77


84/119


Ex. 247: Examine

R = R

= SR

= SR

= SR

= R= R= R:

Ex. 248: We may easily see the symmetry of the Einstein tensor from the

fact that both R andS are symmetric.

Ex. 249: Note

G = RS 1

2RSS

= RS 1

2R;

so

G = RS R

= R R= R R= 0;

sinceR= R by denition.

Ex. 250: We compute

(rr rr) T = (rr rr) TS= R"T

"S

= R"ST"

= R"T"

= R"T!S"!

= R"T!S"!= R"S"!T!= R!T!= RT;


85/119

79

with index renaming at the last step.

Ex. 251: The invariant case follows from the commutativity of partial

derivatives. Now, we consider the covariant case:

(rr rr) Ti = rrTi rrTi

= @rTi@S

rTi

@rTi

@S rTi

!

= @rTi@S

@rTi

@S

= @@S

@T

i

@S +ZkikmTm

@@S @Ti

@S +ZkikmTm

=

@2Ti

@S@S +

@

@S

ZkikmT

m @2Ti

@S@S @

@S

ZkikmT

m

= @

@S

ZkikmT

m @

@S

ZkikmT

m

=@Zk@S

ikmTm +Zk

@ikm@S

+Zkikm

@Tm

@S

@Zk

@SikmT

m Zk@ikm@S

Tm Zkikm@Tm

@S

= 0 [not sure yet]


86/119


Ex. 252: Look at

R+R+ R = @;

@S @;

@S + !;

! !;!

+@;

@S @;

@S + !;

! !;!

+@;

@S @;

@S + !;

! !;!

= @;@S

@;@S

+ !;! !;!

+@;

@S @;

@S + !;

! !;!

+@;

@S @;

@S + !;

! !;!

= !;!!;!

+!;! !;!

+!;! !;!

= S!""

! S!""!

+S!""

! S!""!

+S!""

!

S!"

"

!

= S!""

! S"!!"

+S!""

! S"!!"

+S!""

! S"!!"

= S!""

! S!""!

+S!""

! S!""!

+S!""

! S!""!

= 0;

as desired.


87/119

81

Ex. 253: Compute

r"R + rR"+ rR" = @R@S"

!"R! !"R! !"R! !"R!

+@R"

@S !R!" !R!" !R!" !"R!

+@R"

@S !R!"!R!" !"R! !R"!

= @R

@S" !"R! !"R! !"R! !"R!

+@R"

@S !R!" !R!" !R!"+ !"R!

+@R"

@S

!R!"

!R!"+

!"R!+

!R!"

= @R

@S" !"R! !"R!

+@R"

@S !R!" !R!"

+@R"

@S !R!"!R!"

= @R

@S" !"

@!;

@S @!;

@S + ;!

;!

!"

@

+:::

Ex. 254: Compute

1

4""R =

1

4""

R1212S

""

= 1

4(2)(2)

R1212S

= R1212

S= K;

as desired.


88/119


Ex. 255: Compute

K(SS SS) = 14

""R(SS SS)

= 1

4""SSR 1

4""SS

= 1

4""SSR+

1

4""SS

= 1

4

R+

1

4

R

= 1

4(2) R+

1

4(2) R

= R :

Ex. 256: Note1

2R =

1

2R!S

!S ;

so

1

2R =

1

2R!S

!S

= 1

2K"!"S

!S

= 1

2

K

= K

Ex. 257: This follows since

rS = rS;

hence

NB = NB;

or

B = B:


89/119

83

Ex. 258: Note that since B = 0, we have that both eigenvalues of B

are equal in absolute value and are negatives of each other; denote them ;

.

Thus,jB j = 2:

Now,

B B :=C

:

In linear algebra terms, we have

C =B2

then,

B B = tr B

2

= 1+2;

where1; 2 are the eigenvalues ofB2 . But, since1 =

2 and2 = ()2 =2 by the properties of eigenvalues, we have

B B = 2

2

= 2 jB j

by the above.

Ex. 259: We compute, given

r (z) = a cosh

z b

a

= a

e(zb)=a +e(bz)=a

2

= a

2e(zb)=a +

a

2e(bz)=a

r0(z) =1

2e(zb)=a 1

2e(bz)=a

r00(z) = 1

2ae(zb)=a +

1

2ae(bz)=a:


90/119


Compute

r00(z) r (z) r0(z)2 =

1

2ae(zb)=a +

1

2ae(bz)=a

a2

e(zb)=a +a

2e(bz)=a

1

2e(zb)=a

= 1

4e2(zb)=a 1

4e2(bz)=a

1

4e2(zb)=a 1 +1

4e2(bz)=a

= 1

Thus,

r00(z) r0(z) r0(z)2

1 = 0

B = r00(z) r0(z) r0(z)2 1

r (z)

q1 +r0(z)2

= 0;

as desired.

Ex. 260: We have

V = dR

dt (S(t))

= @R

@SdS

dt= SV

= VS

as desired.


91/119

85

Ex. 261: Compute

A = dV

dt

= d

dt[VS]

= dV

dt S+V

dSdt

(S(t))

= dV

dt S+V

@S@S

dS

dt

= dV

dt S+V

V@S@S

= dV

dt S+V

VrS+ S

=

dV

dt S+V

V

S+ V

V

rS=

dV

dt S+V

VS+VVrS

= V

t S+V

VrS

= V

t S+ NV

VB

= V

t S+ NBV

V ;

as desired.

Ex. 262: We dene

Tt

=dT

dt + V!T

! V!T!:

Ex. 263: Look at

T0

0

t =

dT0

0

dt

Tensor Calc and Moving Surfaces Exercises New

Documents

Transcript of Tensor Calc and Moving Surfaces Exercises New