Selected solutions to exercises from Pavel Grinfeld™s …dws57/Tensor_Calc_and... · 2014-09-14 · Introduction Included in this text are solutions to various exercises from Introduction

Selected solutions to exercises from PavelGrinfeld�s Introduction to Tensor Analysis and

the Calculus of Moving Surfaces

David Sulon

9/14/14

ii

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

iv CONTENTS

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

vi CONTENTS

Part I

Part I

1

Chapter 1

Chapter 1

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

F 0(x0; y0) = F (2x0; 2y0)

= (2x0)2e2y

0

= 4 (x0)2e2y0

:

Ex. 2: Note that the above implies x0 = 12x, y

0 = 12y. We check

@F 0

@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: Let a; 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 = by0, where (x0; y0) is the same point in our new coordinate system. Now,let T (x; y) be a di¤erentiable function. Then,

rT =�@F

@x(x; y);

@F

@y(x; y)

�3

4 CHAPTER 1. CHAPTER 1

in standard coordinates

=

�@F

@x0(x0; y0)

@x0

@x(x; y);

@F

@y0(x0; y0)

@y0

@y(x; y)

�

(think of x0 as a function of x).

=

�@F

@x0(x0; y0)

1

a;@F

@y0(x0; y0)

1

b

�=

�1pa2 + 0

@F

@x0;

1

b2 + 0

@F

@y0

�=

�1pi0 � i0

@F

@x0;

1pj0 � j0

@F

@y0;

�

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

�=


��xy

��xy

�=


��1�x0 � ay0 � b

�=

� cos�cos2 �+sin2 �

sin�cos2 �+sin2 �

� sin�cos2 �+sin2 �

cos�cos2 �+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) :

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 =


��10

�= cos�i+sin�j

j0 =


��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

@y0(x0; y0)

@y0

@x(x; y)

�i+

�@F

@x0(x0; y0)

@x0

@y(x; y) +

@F

@y0(x0; y0)

@y0

@y(x; y)

�j

=

�@F

@x0(x0; y0) cos�+

@F

@y0(x0; y0) sin�

�i+

��@F@x0

(x0; y0) sin�+@F

@y0(x0; y0) cos�

�j

=@F

@x0(x0; y0) cos�i� @F

@x0(x0; y0) sin�j+

@F

@y0(x0; y0) sin�i+

@F

@y0(x0; y0) cos�j

=@F

@x0(x0; y0) (cos�i� sin�j) + @F

@y0(x0; y0) (sin�i+ cos�j)

=@F

@x0(x0; y0)

�i j

� � cos�� sin�

�+@F

@y0(x0; y0)

�i j

� �sin�cos�

�

=@F

@x0(x0; y0) i0 +

@F

@y0(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 a¢ ne orthogonal coordinate system by rotating

the "standard" Cartesian coordinates via (1.7) and then applying a rescaling.


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 P along the appropriate direction for which weare taking each directional derivative. De�ne f (h) := F (P �), i.e. parametrizealong the unit vector emanating from P in the direction of l (note f (0) = F (P )).Also, for points A,B, AB indicates the (unsigned) length of the vector from Ato B.Ex. 7:

f (h) =pF (P )2 + h2

f 0 (h) =hp

F (P )2 + h2

dF (p)

dl= f 0 (0)

Ex. 8: We have

f (h) =1

AP � h

f 0 (h) = � �1(AP � h)2

=1

(AP � h)2

so

dF (p)

dl= f 0 (0)

=1

(AP )2

7


Ex. 9: Let � denote the measure of angle OP �P . 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� =OP sin (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 (P )) sin (F (P �))]

(OA)h

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

((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 (F (P ))

�

Thus,

f (h) = arctan

�(OP ) (AP � h) sin (F (P ))

(OA)h+ (OP ) (AP � h) cos (F (P ))

�f 0 (h) =

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

1

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

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

9

dF (p)

dl= f 0 (0)

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

[(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 )) (AP ) (OP ) (AP ) cos (F (P ))

(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: Clearly F (P �) = F (P ) for any choice P � in 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 length AK = F (P ) + h. Then,

f (h) =1

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

f 0 (h) =1

2AB

dF (p)

dl= f 0 (0)

=1

2AB

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

1.(8) [Not sure]


(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 ))

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

!AB, and will have

magnitude 1.(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 direction L would then correspond tothe projection of Of onto L.

Ex. 15: [See diagram]

kR (�+ h)�R (�)k2 = 1 + 1� 2 cos (h)

by the Law of Cosines. So,

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

p2� 2 cos (h)

=

s2� 2 cos

�2h

2

�

=

s2� 2

�sin2

h

2

�=

r4 sin2

h

2

= 2 sinh

2:

Ex. 16:

limh!0

2 sin h2h

= limh!0

2�12

�cos h21

;

11

by L�Hospital�s rule,

= limh!0

cosh

2= 1:

Ex. 17:

limh!0

2 sin h2h

= limh!0

sin h2h2

= limh!0

sin�0 + h

2

�� sin (0)

h2

= sin0 (0)

= cos (0)

= 1:

Ex. 18: We haveR (�)R0 (�) = 0:

Di¤erentiating both sides, we obtain

R0 (�) �R0 (�) +R (�) �R00 (�) = 0:

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

1 +R (�) �R00 (�) = 0

R (�) �R00 (�) = �1:

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

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

since R (�) is of unit length. Now, let h be arbitrarily small. Since kR (�)k =kR (�+ h)k = kR0 (�+ h)k = kR0 (�)k = 1, we have by congruent trianglesthat kR0 (�+ h)�R0 (�)k = kR (�+ h)�R (�)k : Thus,

kR00 (�)k = limh!0

kR0 (�+ h)�R0 (�)kh

= limh!0

kR (�+ h)�R (�)kh

= kR0 (�)k= 1:


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

cos � = �1;

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

Chapter 3

Chapter 3

Ex. 19: We may construct a three-dimensional Cartesian coordinate system asfollows: Fix an origin O, then pick three points A;B;C such that the vectors�!OA;��!OB;

��!OC form an orthonormal system. De�ne i =

�!OA, j =

��!OB, k =

��!OC.

Note that in this coordinate system, A;B;C have coordinates0@100

1A ;0@010

1A ;0@001

1Arespectively, and a vector V connecting the origin to a point with coordinates0@x0y0

z0

1Acan 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 (notethe 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 the x-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


Ex. 21: Let P be an arbitrary point in a two-dimensional Euclidean spacewith polar coordinates r; �. Assume P has cartesian coordinates (x; y). De�neP 0 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, and P 0. Note that OP 0 = x and PP 0 = 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 =px2 + y2:

Also,

y

x=

r sin �

r cos �= tan �;

so� = arctan

y

x:

Ex. 23: De�ne the x and y coordinate of some arbitrary point P to be theCartesian system of coordinates de�ned by applying Ex. 21 to the coordinateplane �xed in the de�nition of our cylindrical coordinates. Simply de�ne the z(Cartesian) coordinate to be the signed distance from P to the coordinate plane

15

(note the orthogonality of of x,y,z by the de�nition of distance to a plane - andalso that x; y do not depend on z). The equations for x; y then follow from Ex.21, and the z (Cartesian) coordinate is equal to the z (cylindrical) by de�nition.

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

Ex. 25: Let P be a point with spherical coordinates r; �; �. Let the x-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 hte z-axis be the longitudinal axis. Since the z-coordinatelength OP 0, where P 0 is the orthogonal projection of P 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 P 00. Weclearly have the length OP 00 = r sin �. Thus, by considering the right triangledetermined by the points O, P 00, and the polar axis, we have

x = (OP 00) cos�

= r sin � cos�

y = (OP 00) 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 �

= rqsin2 �

�cos2 �+ sin2 �

�+ cos2 �

= rpsin2 � + cos2 �

= r;

sor (x; y; z) =

px2 + y2 + z2:

Also, z=r = cos �; so

� (x; y; z) = arccosz

r

= arccoszp

x2 + y2 + z2:


Finally,

y

x=

r sin � sin�

r sin � cos�

= tan�;

so� (x; y; z) = arctan

y

x:

Chapter 4

Chapter 4

Ex. 27:

det J = det

"xpx2+y2

ypx2+y2

�yx2+y2

xx2+y2

#

=x2 + y2

(x2 + y2)px2 + y2

=1p

x2 + y2:

Ex. 28:

J (1; 1) =

� 1p12+12

1p12+12

�112+12

112+12

�=

� 1p2

1p2

� 1212

�:

Ex. 29:

det J 0 = det

�cos � �r sin �sin � r cos �

�= r

�cos2 � + sin2 �

�= r;

Using the relationship r =px2 + y2; we have det J det J 0 = 1.

17


Ex. 30:

J 0�p2;�

4

�=

�cos �4 �

p2 sin �4

sin �4p2 cos �4

�=

"p22 �1p22 1

#:

Ex. 31: We evaluate the product

JJ 0 =

� 1p2

1p2

� 1212

�"p22 �1p22 1

#

=

�1 00 1

�;

as desired.

Ex. 32:

J 0 (x; y) =

�cos � �r sin �sin � r cos �

�=

�xr �yyr x

�

=

24 xpx2+y2

�yypx2+y2

x

35 ;so

JJ 0 =

"xpx2+y2

ypx2+y2

�yx2+y2

xx2+y2

#24 xpx2+y2

�yypx2+y2

x

35=

"x2

x2+y2 +y2

x2+y2 0

0 x2

x2+y2 +y2

x2+y2

#= I;

similarly, J 0J = I. Thus, J; J 0 are inverses of each other.

Ex. 33: We use

r (x; y; z) =px2 + y2 + z2

� (x; y; z) = arccoszp

x2 + y2 + z2

� (x; y; z) = arctany

x

19

Note from our computation of the Laplacian in spherical coordinates, we have(after substituing expressions for x; y; z to obtain these results in terms of r; �; �):

@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�

rcos�r sin � � sin �r

� sin�r sin � cos� sin� 0

35 :We then compute

J 0 (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 ;


So

JJ 0 =

24sin � cos� sin � sin� cos �cos � cos�

rcos�r sin � � sin �r

� sin�r 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 � + r cos � sin � sin2 � 0

� 1r cos � sin � +1r cos� sin�+

1r cos � cos

2 � sin � sin2 � + (cos �) cos�sin � sin�+ cos2 � cos2 � � cos � cos� sin � sin�+ cos2 �

� 1r cos� sin�+ cos� sin � sin2 � r cos � cos� sin2 �� (cos �) cos�sin � sin� sin2 �+ r cos2 � sin � sin�

35=

241 0 00 1 00 0 1

35

[Note: Something may be o¤ with the computation of J ]

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:

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;


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

@

@�

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@�

�+@

@�

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@a@c

�@C

@�+@2F

@a@c

@

@�

�@C

@�

��@A

@�

+2

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@�+@2F

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@B

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@C

@�

�@2A

@�@�+@F

@a

@3A

@2�@�

=@

@�

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@a2

�@A

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@�+@2F

@a2@

@�

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@�

+@

@�

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�@B

@�

@A

@�+@2F

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@

@�

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@�

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@�

+@

@�

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@

@�

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@�

+2

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@B

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@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

@�

+

�@3F

@a2@c

@A

@�+

@3F

@a@b@c

@B

@�+

@3F

@a@c2@C

@�

�@C

@�

@A

@�+@2F

@a@c

@2C

@�2@A

@�

+2

�@2F

@a2@A

@�+@2F

@a@b

@B

@�+@2F

@a@c

@C

@�

�@2A

@�@�+@F

@a

@3A

@2�@�

23

J =@

@�

��@2F

@a@b

@A

@�+@2F

@b2@B

@�+@2F

@b@c

@C

@�

�@B

@�+@F

@b

@2B

@�@�

�=

@

@�

�@2F

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@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

@

@�

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�=

@

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�@2F

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@A

@�+@2F

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@�+@2F

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+

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@�

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@�@�+

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@�@�+@F

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@3B

@�2@�

=

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@

@�

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@

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+

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@�@�+

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@�@�+@F

@b

@3B

@�2@�

=@

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@

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+@

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��@B

@�

�2+@2F

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@�

�@B

@�

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+@

@�

�@2F

@b@c

�@C

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@B

@�+@

@�

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@�

+

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@A

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@�@�+

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@�+@2F

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@�+@2F

@b@c

@C

@�

�@2B

@�@�+@F

@b

@3B

@�2@�

=

�@3F

@a2@b

@A

@�+

@3F

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@�+

@3F

@a@b@c

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@�

�@A

@�

@B

@�+@2F

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@2A

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+

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@�+@3F

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@�+@3F

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@�

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@�

�2+@2F

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+

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+

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@�@�+@F

@b

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@�2@�


K =@

@�

��@2F

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@�+@2F

@b@c

@B

@�+@2F

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@�

�@C

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+

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+

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=

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+

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=

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+

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@2B

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+

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@�

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@�

+

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@�

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@b@c

@B

@�+@2F

@c2@C

@�

�@2C

@�@�+@F

@c

@3C

@�2@�

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

@�@�

+

�@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

@�@�

=@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 di¤erentiation throughout.

Ex. 38 [Not �nished]

Ex. 39 Begin withcos arccosx = x

and di¤erentiate both sides:

d

dx[cos arccosx] = 1

� (sin arccosx) ddx[arccosx] = 1

d

dx[arccosx] =

�1(sin arccosx)


By examining triangles with unit hypotenuse, we obtain

sin arccosx = �p1� x2;

sod

dx[arccosx] = � 1p

1� x2

Ex. 40: We know that f; g satisfy

g0 (f (x)) f 0 (x) = 1:

Di¤erentiating both sides, we obtain

d

dx[g0 (f (x))] f 0 (x) + g0 (f (x))

d

dx[f 0 (x)] = 0

g00 (f (x)) f 0 (x) f 0 (x) + g0 (f (x)) f 00 (x) = 0

g00 (f (x)) [f 0 (x)]2+ g0 (f (x)) f 00 (x) = 0 (4.1)

as desired.

Ex. 41: We compute

f 0 (x) = ex

f 00 (x) = ex

g0 (x) =1

x

g00 (x) = � 1x2;

So

g00 (f (x)) [f 0 (x)]2+ g0 (f (x)) f 00 (x) = � 1

e2xe2x +

1

exex

= �1 + 1= 0;

as desired.

27

Ex. 42: We compute

f 0 (x) =1p1� x2

f 00 (x) = �2x12

�1� x2

��3=2= �x

�1� x2

��3=2g0 (x) = � sin (x)g00 (x) = � cos (x) ;

So

g00 (f (x)) [f 0 (x)]2+ g0 (f (x)) f 00 (x) = � cos (arccosx) 1

1� x2 � sin (arccos (x))��x�1� x2

��3=2�= � x

1� x2 +xp1� x2

p1� x23

= � x

1� x2 +x

1� x2= 0;

as desired.

Ex. 43: We di¤erentiate both sides of the second-order relationship toobtain

d

dx

�g00 (f (x)) [f 0 (x)]

2+ g0 (f (x)) f 00 (x)

�= 0

d

dx[g00 (f (x))] [f 0 (x)]

2+ g00 (f (x))

d

dx

h[f 0 (x)]

2i+d

dx[g0 (f (x))] f 00 (x) + g0 (f (x))

d

dxf 00 (x) = 0

g(3) (f (x)) [f 0 (x)]3+ g00 (f (x)) � 2f 0 (x) f 00 (x) + g00 (f (x)) f 0 (x) f 00 (x) + g0 (f (x)) f (3) (x) = 0

g(3) (f (x)) [f 0 (x)]3+ 3g00 (f (x)) f 0 (x) f 00 (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")

J ii0Ji0

j = �ij ;

and write out the dependences on unprimed coordinates:

J ii0 (Z0 (Z)) J i

0

j (Z) = �ij (Z) (4.2)


(note, however, that the Krönicker delta is constant with respect to the un-primed coordinates Z). We di¤erentiate both sides of (4.2) with respect toZk:

@

@Zk

hJ ii0 (Z

0 (Z)) J i0

j (Z)i=

@

@Zk��ij (Z)

�@

@Zk

hJ ii0 (Z

0 (Z)) J i0

j (Z)i= 0

@

@Zk�J ii0 (Z

0 (Z))�J i

0

j (Z) + Jii0 (Z

0 (Z))@

@Zk

hJ i

0

j (Z)i= 0;

since di¤erentiation passes through the implied summation over i0. Then, usingthe de�nition of the Jacobian,

@

@Zk

�@Zi

@Zi0(Z 0 (Z))

�@Zi0

@Zj(Z) +

@Zi

@Zi0(Z 0 (Z))

@

@Zk

�@Zi0

@Zj(Z)

�= 0(4.3)

@2Zi

@Zk0@Zi0(Z 0 (Z))

@Zk0

@Zk(Z)

@Zi0

@Zj(Z) +

@Zi

@Zi0(Z 0 (Z))

@2Zi0

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

applying the chain rule to the �rst term, and implying summation over newindex k0. Then, if we de�ne the "Hessian" object

J ik0;i0 :=@2Zi

@Zk0@Zi0(Z 0)

with an analogous de�nition for J i0

k;i, we write (4.3) concisely:

J ik0;i0Jk0

k Ji0

j + Jii0J

i0

k;j = 0;

or, using a renaming of dummy indicex k0 to j0 and a reversing of the order ofpartial derivatives,

J ii0j0Ji0

j Jj0

k + Jii0J

i0

jk = 0:

This above tensor relationship represents n3 identities.Ex. 49: Since each J ii0 is constant for a transformation from one a¢ ne

coordinate system to another, each second derivative vanishes, and hence eachJ ii0j0 = 0.

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

J ii0j0Ji0

j Jj0

k + Jii0J

i0

jk = 0;

29

Then, letting k0 be arbitrary, we multiply both sides by Jkk0 , implying summationover k: h

J ii0j0Ji0

j Jj0

k + Jii0J

i0

jk

iJkk0 = 0

J ii0j0Ji0

j Jj0

k Jkk0 + J

ii0J

i0

jkJkk0 = 0

but, Jj0

k Jkk0 = �

j0

k0 ; so

J ii0j0Ji0

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 toJ ii0k0J

i0

j . After re-naming k0 = j0, we obtain

J ii0j0Ji0

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

J ii0j0Ji0

j + Jii0J

i0

jkJkj0

iJjk0 = 0

J ii0j0Ji0

j Jjk0 + J

ii0J

i0

jkJkj0J

jk0 = 0

J ii0j0�i0

k0 + Jii0J

i0

jkJkj0J

jk0 = 0

J ii0j0�i0

k0 + Ji0

jkJii0J

kj0J

jk0 = 0:

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

J ii0j0�i0

k0 + Jh0

jkJih0J

kj0J

jk0 = 0:

Noting that the �rst term is zero for all i0 6= k0, and setting k0 = i0:

J ii0j0 + Jh0

jkJih0J

kj0J

ji0 = 0:

We then may re-introduce k0 as a dummy index:

J ii0j0 + Jk0

jkJik0J

kj0J

ji0 = 0:

Then, switch the roles of j; k as dummy indices:

J ii0j0 + Jk0

kjJik0J

jj0J

ki0 = 0


Ex. 52: Return to

J ii0j0Ji0

j Jj0

k + Jii0J

i0

jk = 0;

and write out the dependences

J ii0j0 (Z0 (Z)) J i

0

j (Z) Jj0

k (Z) + Jii0 (Z

0 (Z)) J i0

jk (Z) = 0

@2Zi

@Zi0@Zj0(Z 0 (Z))

@Zi0

@Zj(Z)

@Zj0

@Zk(Z) +

@Zi

@Zi0(Z 0 (Z))

@2Zi0

@Zj@Zk(Z) = 0

Then, di¤erentiate both sides with respect to Zm:

0 =@

@Zm

"@2Zi

@Zi0@Zj0(Z 0 (Z))

@Zi0

@Zj(Z)

@Zj0

@Zk(Z) +

@Zi

@Zi0(Z 0 (Z))

@2Zi0

@Zj@Zk(Z)

#

=@

@Zm

�@2Zi

@Zi0@Zj0(Z 0 (Z))

�@Zi

0

@Zj(Z)

@Zj0

@Zk(Z) +

@2Zi

@Zi0@Zj0(Z 0 (Z))

@

@Zm

"@Zi

0

@Zj(Z)

#@Zj0

@Zk(Z) +

@2Zi

@Zi0@Zj0(Z 0 (Z))

@Zi0

@Zj(Z)

@

@Zm

�@Zj0

@Zk(Z)

�

+@

@Zm

�@Zi

@Zi0(Z 0 (Z))

�@2Zi

0

@Zj@Zk(Z) +

@Zi

@Zi0(Z 0 (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(Z 0 (Z))

@2Zi0

@Zm@Zj(Z)

@Zj0

@Zk(Z) +

@2Zi

@Zi0@Zj0(Z 0 (Z))

@Zi0

@Zj(Z)

@2Zj0

@Zm@Zk(Z)

+

"@2Zi

@Zi0@Zm0 (Z0 (Z))

@Zm0

@Zm(Z)

#@2Zi

0

@Zj@Zk(Z) +

@Zi

@Zi0(Z 0 (Z))

@3Zi0

@Zj@Zk@Zm(Z)

= J ii0j0m0Jm0

m J i0

j Jj0

k + Jii0j0J

i0

jmJj0

k + Jii0j0J

i0

j Jj0

km + Jii0m0Jm

0

m J i0

jk + Jii0J

i0

jkm;

so, setting k0 = m0 as a dummy index:

J ii0j0k0Ji0

j Jj0

k Jk0

m + J ii0j0Jj0

k Ji0

jm + Jii0j0J

i0

j Jj0

km + Jii0k0J

i0

jkJk0

m + J ii0Ji0

jkm = 0: (4.6)

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

J ii0j0k0Ji0

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

J ii0j0k0Ji0

j Jj0

k �k0

m0 + J ii0j0Jj0

k Ji0

jmJmm0 + J ii0j0J

i0

j Jj0

kmJmm0 + J ii0k0J

i0

jk�k0

m0 + J ii0Ji0

jkmJmm0 = 0;

This holds for all m0, so speci�cally for m0 = k0, the above identity reads

J ii0j0k0Ji0

j Jj0

k +Jii0j0J

j0

k Ji0

jmJmk0 +J

ii0j0J

i0

j Jj0

kmJmk0 +J

ii0k0J

i0

jk+Jii0J

i0

jkmJmk0 = 0 (4.7)

31

Next, in an analogous manner, multiply both sides by Jkm0 for arbitrary m0:

J ii0j0k0Ji0

j Jj0

k Jkm0 + J ii0j0J

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

J ii0j0k0Ji0

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 = j0 in all but the �rst term:

J ii0j0k0Ji0

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 = j0:

J ii0j0k0Ji0

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 :

J ii0j0k0Ji0

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

J ii0j0k0�i0

m0 + J ii0h0Jh0

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 to g0 in all but the �rst term:

J ii0j0k0�i0

m0+J ig0h0Jh0

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, set m0 = i0:

J ii0j0k0+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�Z 0�Z

00(Z)��= Z;

or for each i,

Zi�Z 0�Z

00(Z)��= Zi.

Di¤erentiating both sides with respect to Zj , we obtain

@

@Zj�Zi (Z 0 (Z 00 (Z)))

�= �ij :


Then, we apply the chain rule twice:

@Zi

@Zi0@

@Zj[Z 0 (Z 00 (Z))] = �ij

@Zi

@Zi0@Zi

0

@Zi00@

@Zj

hZi

00(Z)i= �ij

@Zi

@Zi0@Zi

0

@Zi00@Zi

00

@Zj= �ij ;

orJ ii0J

i0

i00Ji00

j = �ij :

Chapter 5

Chapter 5

Ex. 61: �ij .Ex. 62: Assume U is an arbitrary nontrivial linear combination U iZi of

coordinate bases Zi. SinceU � U > 0

andU � U = ZijU iU j ;

or in matrix notationU � U = UTZU ,

this condition implies Z = Zij is positive de�nite.

Ex. 63:

kV k =pV � V

=qZijV iV j

Ex. 64: Put Z = Zij . Thus, Z�1 = Zij by de�nition. Let x be anarbitrary nontrivial vector. Then, de�ne y = Z�1x. We have

yT =�Z�1x

�T= xT

�Z�1

�T= xTZ�1;

33


since Z�1 is symmetric. Then, since Z is positive de�nite, note

0 < yTZy

= xTZ�1ZZ�1x

= xTZ�1x.

Since x was arbitrary, this implies that Z = Zij > 0.

Ex. 65:

Zi � Zj = ZikZk � Zj= ZikZkj

by de�nition. 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 de�nition

Zi = ZijZj

ZikZi = ZikZ

ijZj

= �jkZj

= Zk:

Thus,

Zk = ZikZi

= ZkiZi;

since Zik is symmetric.

Ex. 69: Examine

ZkiZi � Zj = ZkiZ

inZn � Zj

= �nk�jn

= �jk;

35

since �nk�jn = 1 i¤ k = n and n = j, or by transitivity, i¤ k = j. Thus, Zi � Zj

determines the matrix inverse of Zki, which must be Zij by uniqueness of matrixinverse.Ex. 70: Because the inverse of a matrix is uniquely determined, we have

that Zi are uniquely determined.

Ex. 71:

ZijZjk = Zi � ZjZjk= Zi � Zk;

from 5.17= �ik

from 5.16.

Ex. 72: We compute

Z1 � Z2 =

�1

3i�13j

�� j

=1

3i � j�1

3j � j

=1

3kik kjk cos �

3�13kjk2

=1

3(2) (1)

�1

2

��1312

= 0;

thus, Z1;Z2 are orthogonal. We further compute

Z2 � Z1 =

��13i+4

3j

�� i

= �13kik2 + 4

3kik kjk cos �

3

= �13(4) +

4

3(2) (1)

�1

2

�= 0;

so Z2;Z1 are orthogonal.


Ex. 73:

Z1 � Z1 =

�1

3i�13j

�� i

=1

3kik2 � 1

3kik kjk cos �

3

=4

3� 23

�1

2

�= 1:

Z2 � Z2 =

��13i+4

3j

�� j

= �13kik kjk cos �

3+4

3kjk2

= �23

�1

2

�+4

3

= 1:

Ex. 74:

Z1 � Z2 =

�1

3i�13j

��13i+4

3j

�= �1

9kik2 + 4

9kik kjk cos �

3+1

9kik kjk cos �

3� 49kjk2

= �49+4

9+1

9� 49

= �39

= �13:

Ex. 75: Let R denote 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

:

37

But, R (r; �; z + h)�R (r; �; z) is clearly a vector of length h pointing in the zdirection; 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 the z direction.

Ex. 76: The computations of the diagonal elements Z11 and Z22 are thesame as for polar coordinates; moreover the zero o¤-diagonal entries Z12, Z21follow from the orthogonality of Z1, Z2. By de�nition of cylindrical coordinates,the z axis is perpendicular to the coordinate plane (upon which Z1, Z2 lie); 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 row 3 and column 3 of Zij arezero. Morover, since Z3 is of unit length; we have Z33 = Z3 � Z3 = 1. Thus,we have

Zij =

241 0 00 r2 00 0 1

35 .Now, since Zij is de�ned to be the inverse of Zij , we may easily compute

Zij =

241 0 00 r�2 00 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


Chapter 6

Chapter 6

Ex. 87: Look at

ZijJ i0

i Jj0

j Zj0k0 = ZijJ i

0

i Jj0

j ZjkJjj0J

kk0

by the tensor property of Zjk

= ZijZjkJi0

i Jkk0

= �ikJi0

i Jkk0

= J i0

k Jkk0

= �i0

k0 ;

so, in linear algebra terms, we have that ZijJ i0

i Jj0

j is the matrix inverse of Zj0k0 .

By uniqueness of matrix inverses, this forces ZijJ i0

i Jj0

j = Zi0j0 , as desired.

Ex. 88: Let Z,Z 0 be two coordinate systems. Write the unprimed coordi-nates in terms of the primed coordinates

Z = Z (Z 0) :

Then,

@F (Z)

@Zi0=

@F (Z (Z 0))

@Zi0

=@F

@Zi@Zi

@Zi0

=@F

@ZiJ ii0 ;

so @F@Zi is a covariant tensor.

39


Ex. 89: We show the general case (since by the previous exercise, we knowthat the collection of �rst partial derivatives is a covariant tensor). De�ne,given a covariant tensor �eld Ti

Sij =@TiZj

Si0j0 =@Ti0

Zj0

so

Si0j0 =@Ti0

@Zj0

=@

@Zj0�TiJ

ii0�;

since T is a covariant tensor,

=@

Zj0�Ti (Z

0 (Z)) J ii0 (Z0)�

=@T i

@Zj@Zj

@Zj0J ii0 + TiJ

ii0j0

= SijJjj0J

ii0 + TiJ

ii0j0

6= SijJjj0J

ii0

(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� @Tj

0

@Zi0

=@Ti@Zj

Jjj0Jii0 + TiJ

ii0j0 �

�@Tj@Zi

J ii0Jjj0 + TiJ

ij0i0

�

by the above, interchanging the rolls of i0; j0 for the second term:

=@Ti@Zj

Jjj0Jii0 + TiJ

ii0j0 �

�@Tj@Zi

Jjj0Jii0 + TiJ

ii0j0

�;

41

since J ij0i0 = Jii0j0 ,

=@Ti@Zj

Jjj0Jii0 + TiJ

ii0j0 �

@Tj@Zi

Jjj0Jii0 � TiJ ii0j0

=

�@Ti@Zj

� @Tj@Zi

�Jjj0J

ii0

= SijJii0J

jj0 ;

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

Ex. 91: Put

Sij =@T i

@Zj;

so

Si0j0 =

@T i0

@Zj0

=@

@Zj0

hT iJ i

0

i

i;

since T is a contravariant tensor,

=@

@Zj0�T i (Z 0 (Z))

�J i

0

i + Ti @

@Zj0

hJ i

0

i (Z (Z0))i

=@T i

@ZjJj

0

j Ji0

i + Ti @J

i0

i

@Zj@Zj

@Zj0

= SijJ i0

i Jj0

j + TiJ i

0

ijJjj0

6= SijJ i0

i Jj0

j

except in the trivial case.

Ex. 92: We have

�kij = Zk � @Zi@Zj

;

so in primed coordinates,

�k0

i0j0 = Zk0� @Zi

0

@Zj0

= Zk0��@Zi@Zj

J ii0Jjj0 + ZiJ

ii0j0

�


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

�ZkJk

0

k

��@Zi@Zj

J ii0Jjj0 + ZiJ

ii0j0

�

since Zk is a contravariant tensor,

=

�Zk � @Zi

@Zj

�Jk

0

k Jii0J

jj0 +

�Zk � Zi

�Jk

0

k Jii0j0

=

�Zk � @Zi

@Zj

�Jk

0

k Jii0J

jj0 + �

ki J

k0

k Jii0j0

=

�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�J ii0�Jjj0 + TijJ

ii0@

@Zk0

hJjj0i

=@

@Zk0[Tij ] J

ii0J

jj0 + TijJ

ii0k0J

jj0 + TijJ

ii0J

jj0k0

=@

@Zk0[Tij (Z (Z

0))]J ii0Jjj0 + TijJ

ii0k0J

jj0 + TijJ

ii0J

jj0k0

=@Tij@Zk

@Zk

@Zk0J ii0J

jj0 + TijJ

ii0k0J

jj0 + TijJ

ii0J

jj0k0

=@Tij@Zk

Jkk0Jii0J

jj0 + TijJ

ii0k0J

jj0 + TijJ

ii0J

jj0k0 :

43

Thus, from 5.66,

�k0

i0j0 =1

2Zk

0m0�@Zm0i0

@Zj0+@Zm0j0

@Zi0� @Zi

0j0

@Zm0

�=

1

2ZkmJk

0

k Jm0

m

�@Zm0i0

@Zj0+@Zm0j0

@Zi0� @Zi

0j0

@Zm0

�=

1

2ZkmJk

0

k Jm0

m (@Zmi@Zj

Jjj0Jmm0J ii0 + ZmiJ

mm0j0J

ii0 + ZmiJ

mm0J ii0j0

+@Zmj@Zi

J ii0Jmm0J

jj0 + ZmjJ

mm0i0J

jj0 + ZmjJ

mm0J

ji0j0

� @Zij@Zm

Jmm0J ii0Jjj0 � ZijJ

ii0m0J

jj0 � ZijJ

ii0J

jj0m0)

=1

2ZkmJk

0

k Jm0

m

�@Zmi@Zj

Jjj0Jmm0J ii0 +

@Zmj@Zi

J ii0Jmm0J

jj0 �

@Zij@Zm

Jmm0J ii0Jjj0

�+1

2ZkmJk

0

k Jm0

m

�ZmiJ

mm0j0J

ii0 + ZmiJ

mm0J ii0j0

�+1

2ZkmJk

0

k Jm0

m

�ZmjJ

mm0i0J

jj0 + ZmjJ

mm0J

ji0j0

��12ZkmJk

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

��12ZkmJk

0

k Jm0

m

�ZijJ

ii0m0J

jj0 + ZijJ

ii0J

jj0m0

�= �kijJ

k0

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

��12ZkmZijJ

k0

k Jm0

m

�J ii0m0J

jj0 + J

ii0J

jj0m0

�= �kijJ

k0

k Jjj0J

ii0

+1

2�ki J

k0

k Jm0

m

�Jmm0j0J

ii0 + J

mm0J ii0j0

�+1

2�kjJ

k0

k Jm0

m

�Jmm0i0J

jj0 + J

mm0J

ji0j0

��12ZkmZijJ

k0

k Jm0

m

�J ii0m0J

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

��12ZkmZijJ

k0

k Jm0

m

�J ii0m0J

jj0 + J

ii0J

jj0m0

�= �kijJ

k0

k Jjj0J

ii0

+1

2Jk

0

i Jm0

m Jmm0j0Jii0 +

1

2Jk

0

i Jm0

m Jmm0J ii0j0

+1

2Jk

0

j Jm0

m Jmm0i0Jjj0 +

1

2Jk

0

j Jm0

m Jmm0Jji0j0

�12ZkmZijJ

k0

k Jm0

m J ii0m0Jjj0 �

1

2ZkmZijJ

k0

k Jm0

m J ii0Jjj0m0


= �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

�12ZkmZijJ

k0

k Jm0

m J ii0m0Jjj0 �

1

2ZkmZijJ

k0

k Jm0

m J ii0Jjj0m0

= �kijJk0

k Jjj0J

ii0 + J

k0

i Jii0j0 +

1

2�k

0

i0 Jm0

m Jmm0j0 +1

2�k

0

j0 Jm0

m Jmm0i0 �1

2ZkmZijJ

jj0J

k0

k Jm0

m J ii0m0 �1

2ZkmZijJ

ii0J

k0

k Jm0

m Jjj0m0

= �kijJk0

k Jjj0J

ii0 + J

k0

i Jii0j0 + 0:

Thus, we have�k

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.

Assume Ti = 0. Then, since Ti0 = TiJ ii0 , we have Ti0 = 0.

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

J i0

i = Ai0

i ;

so the "Hessian" objectJ i

0

ij = 0;

since Ai0

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

the �rst terms in the above computations of @Ti0Zj0

, @Ti0

@Zj0,@Ti0j0

@Zk0, and even �k

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

45

is a tensor. ComputecAi

0

j0k0 = cAijkJ

i0

i Jjj0J

kk0 ;

since Aijk is a tensor. Thus, by the above, we have if each A (n)ijk is a tensor,

then the sumNXn=1

cnA (n)ijk

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

Ex. 97: We have that

SiTij = Si�

ikT

kj

= �ikSiTkj ;

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 thedimension of the ambient space.

Ex. 99: We haveVij = V

kijZk;

so

Vij � Zm = V kijZk � Zm

= V kij�mk

= V mij

so substituting m = k, we have an expression for the components

V kij = Vij � Zk

So,

V k0

i0j0 = Vi0j0 � Zk0

= VijJii0J

jj0 � Z

kJk0

k ;

since both Vij ;Zk are tensors

= Vij � ZkJ ii0Jjj0J

k0

k


by linearity= V kijJ

ii0J

jj0J

k0

k

as desired.

Ex. 100: Fix a coordinate system Zi

T ijk = T ijkJ iiJjjJkk ;

so

T ijk Ji0

i Jj0

j Jkk0 = T ij

kJ iiJjjJkkJ

i0

i Jj0

j Jkk0

= T ijk�i

0

i �j0

j�kk0

= T i0j0

k0 ;

as desired.

Chapter 7

Chapter 7

47


Chapter 8

Chapter 8

49


Chapter 9

Chapter 9

Ex. 183: Assume n = 3. Given aij , put A as the determinant. We de�ne

A = eijkai1aj2ak3:

Note that switching the roles of 1; 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 overall 0 � r; s; t � 3, and apply the Einstein summation convention:

3!A = eijkerstairajsakt;

orA =

1

3!eijkerstairajsakt

51


We may similarly show that for aij , we have

A =1

3!eijka

i1aj2ak3:

Ex. 184: We have

�123rst ar2as2at3 = �

123srt a

s2ar2at3

after index renaming

= �123srt ar2as2at3

= ��123rst ar2as2at3:

Since�123rst a

r2as2at3 = ��123rst a

r2as2at3;

we need�123rst a

r2as2at3 = 0:

The result for �132srt as2ar3at2 follows similarly.

Ex. 185: Note

�123rst = e123erst = 1 � erst = erst;

so�123rst a

r1as2at3 = ersta

r1as2at3 = A:

Also,

�132rst ar2as3at1 = �132rst a

t1ar2as3

= �132str ar1as2at3

= ��123str ar1as2at3

= �estrar1as2at3= A:

53

[Note: Is there an error somewhere - should this be �A?]

Ex. 186: De�neAir =

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

h�li�

urajsakt + air�

lj�usakt + airajs�

lk�ut

i=

1

3!

heijk�lie

rst�urajsakt + aireijk�lje

rst�usakt + airajseijk�lke

rst�ut

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 de�ne

Air =1

2!eijkersta

jsatk;

we have@A

@air= Air

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

Zij = �ij;


so

Z = jZ��j= 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 00 0 r2 sin2 �

35 ;so

Z =

��1 0 00 r2 00 0 r2 sin2 �

��= r4 sin2 �;

thus, pZ = r2 sin �:

55

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

ririF =1pZ

@

@Zi

�pZZij

@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

r2

�2r@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

�pZZij

@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:


Part II

Part II

57

Chapter 10

Chapter 10

Ex. 213: Note that

Zi�Z�j =

�S� � Zi

�(S� � Zj)

=�S� (S

� � Zj) � Zi�

= (S� � Zj)S� � Zi

6= �ij ;

since(S� � Zj)S�

is merely the projection of Zj onto the tangent space. [ASK]

EARLIER ATTEMPT:

Zi�Z�j = �ij�

S� � Zi�Zj�S

�� = �ij

S�S�� ZiZj� = �ij

S�Zj� � Zi = �ij

S�Zj� � Zi = Zj � Zi;

which forcesS�Zj� = Zj

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

59


Ex. 214: We haveT i = T�Zi�:

Now,

T iZ�i = T �Zi�Z�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 �N�V �N= 0;

sinceN �N =1.

Ex. 216: We compute

P ijPjk = N iNjN

jNk

= N i (1)Nk

= P ik;

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:

61

Ex. 218: Similarly to 216, we have, given de�nition T ij = NiNj

T ijTjk = N iNjN

jNk

= N iNk

= T ik:

[Note: This seems like the exact same problem - do we mean to de�neT ij = T

iTj?]

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

N iNj + Zi�Z

�j = �

ij :

Contract both sides with Ni :

N iNjNi + Zi�Z

�j Ni = �ijNi

NiNiNj +NiZ

i�Z

�j = Nj

NiNiNj + 0 = Nj

NiNiNj = Nj ;

where the third line follows fromNiZi� = 0. Now, this holds for allNj , for whichat least one is nonzero (we cannot have the normal vector be zero). Hence, wehave

NiNi = 1;

as desired.

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

�14�ijkrstT

tjT

sk = �1

4�ijkrtsT

sj T

tk

=1

4�ijkrstT

sj T

tk;

so

N iNr =1

4�ijkrstT

sj T

tk �

1

4�ijkrstT

tjT

sk

= 2

�1

4�ijkrstT

sj T

tk

�=

1

2�ijkrstT

sj T

tk:


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

J�0

� =@S�

0

@S�:

Ex. 222: From before, we have

@Zij@Zk

= Zli�ljk + Zlj�

lik:

From the analogous de�nitions of S�� ; 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� ��!

�=

1

2

��

�� + S

�!S�� + �

��

�� + S

�!S� ��!� � S�!S�� ! � S�!S� ��!

�=

1

2

�2��

��

�= �� ;

as desired.

Ex. 223: Assume the ambient space is re¤ered to a¢ ne coordiantes. Wehave

�� = Z�i@Zi�@S

+ �ijkZ�� Z

j�Z

k�

= Z�i@Zi�@S

+ 0;

since �ijk = 0 in a¢ ne coordinates.

63

Ex. 224 [Still Working]

Ex. 225 We compute, given

Z1 (�; �) = R

Z2 (�; �) = �

Z3 (�; �) = �

Zi� =@Zi

@S�

=

264@Z1

@S1@Z1

@S2@Z2

@S1@Z2

@S2@Z3

@S1@Z3

@S2

375=

240 01 00 1

35 ;then, note that since �

0 1 00 0 1

�240 01 00 1

35 = �1 00 1

�;

we have

Z�i =

�0 1 00 0 1

�

N i =

24010

35�24001

35=

��i j k0 1 00 0 1

��= i

=

24100

35S1 = Zi1Zi

= Z21Z2

= R cos � cos�i+R cos � sin�j�R sin �kS2 = Zi2Zi

= Z32Z3

= �R sin � sin�i+R sin � cos�j


S�� =

�R2 cos2 � cos2 �+R2 cos2 � sin2 �+R2 sin2 � �R2 cos � cos� sin � sin�+R2 cos � sin� sin � cos�

�R2 cos � cos� sin � sin�+R2 cos � sin� sin � cos� R2 sin2 � sin2 �+R2 sin2 � cos2 �

�=

�R2 00 R2 sin2 �

�[ASK - should this be the same as when the ambient coordinates are Cart.?]

S�� =

�R�2 00 R�2 sin�2 �

�pS =

s��R2 00 R2 sin2 �

��= R2 sin �:

Now, recall the Christo¤el symbols for the ambient space (in spherical coords):

�122 = �r�133 = �r sin2 �

�212 = �221 =1

r

�233 = � sin � cos �

�313 = �331 =1

r

�223 = �232 = cot �:

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

�� = Z�i

@Zi�@S

+ �ijkZ�� Z

j�Z

k� ;

65

we compute

�111 = Z1i@Zi1@S1

+ �ijkZ1� Z

j1Z

k1

= Z12@Z21@S1

+ �2jkZ12Z

j1Z

k1

= Z12@Z21@S1

+ �222Z12Z

21Z

21

=@Z21@S1

= 0

�121 = �112 = 0 + �ijkZ

1� Z

j2Z

k1

= �2jkZ12Z

j2Z

k1

= �232Z12Z

32Z

21

= cot � (1) (1) (1)

= cot �

�122 = �ijkZ1� Z

j2Z

k2

= �2jkZ12Z

j2Z

k2

= �233Z12Z

32Z

32

= � sin � cos �

�211 = �ijkZ2� Z

j1Z

k1

= �3jkZ23Z

j1Z

k1

= �322Z23Z

21Z

21

= 0

�221 = �212 = �ijkZ

2� Z

j2Z

k1

= �3jkZ23Z

j2Z

k1

= �332Z23Z

32Z

21

= 0

�222 = �ijkZ2� Z

j2Z

k2

= �333Z23Z

32Z

32

= 0;

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


Ex. 226: We have q(x0 (s))

2+ (y0 (s))

2= 1;

since this is an arc-length parametrization. Thus,

N i =

�y0 (s)�x0 (s)

�S�� = (x0 (s))

2+ (y0 (s))

2

= 1

S�� = 1pS = 1

[ASK why x0x00 + y0y00 = 0].

Ex. 227: Simply denote t = 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-substitute x = 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.

Chapter 11

Chapter 11

Ex. 229: This follows similarly as with the ambient covariant derviative, us-ing the tensor properties of T �� (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!

�

r S�� =@S��@S

� �� S�� S��

=@S��@S

� �� S�� S��

=@S��@S

� 12S�!

�@S!�@S

+@S! @S�

� @S� @S!

�S��

1

2S�!

�@S!�@S

+@S! @S�

� @S� @S!

�S��

=@S��@S

� 12�!�

�@S!�@S

+@S! @S�

� @S� @S!

�� 12�!�

�@S!�@S

+@S! @S�

� @S� @S!

�=

@S��@S

� 12

�@S��@S

+@S� @S�

� @S� @S�

�� 12

�@S��@S

+@S� @S�

� @S� @S�

�=

@S��@S

� 12

@S��@S

� 12

@S��@S

= 0:

Similarly, we may show that in the contravariant case,

r S�� = 0:

67


For the Levy-Civita symbols, note

"�� =pSe��

"�� =1pSe��

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 r S�� =r "�� = r "�� = 0:

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

Ex. 233: We compute, using

S�� =

�R�2 00 R�2 sin�2 �

�;

the following:

r�r�F =1pS

@

@S�

�pSS��

@F

@S�

�=

1

R2 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 �R�2

@F

@�

�+

1

R2 sin �

@

@�

�R2 sin �R�2 sin�2 �

@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�� =

�R�2 00 1

�pS = R;

69

so

r�r�F =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�)

�20

0 r�2

�pS = r (R+ r cos�) ;

and compute

r�r�F =1pS

@

@S�

�pSS��

@F

@S�

�=

1

r (R+ r cos�)

@

@�

�r (R+ r cos�)S1�

@F

@S�

�+

1

r (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�) r�2

@F

@�

�=

1

(R+ r cos�)2

@2F

@�2+

1

r2 (R+ r cos�)

@

@�

�(R+ r cos�)

@F

@�

�:

Ex. 236: We have

S�� =

"r (z)

�20

0 11+r0(z)2

#pS = r (z)

q1 + r0 (z)

2;


Thus,

r�r�F =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)

q1 + 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+ 1

r (z)2

@2F

@�2:

Ex. 237: These were computed earlier.

Ex. 238: We compute:

r Zi =@Zi@S

� Zj �kijZk

=@Zi@Zm

@Zm

@S � Zj �kijZk

=@Zi@Zm

Zm � Zj �kijZk

= �kimZkZm � Zj �kijZk

=�Zm �

kim � Zj �kij

�Zk

= 0;

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

r Zij = 0

by the product rule. Similarly,

r Zij = 0:

[Levy-Civita Symbols to come]

71

Ex. 239: Begin with equation 10.41:

NiNi = 1;

and compute take the surface covariant derivative of both sides:

0 = r��NiN

i�

= r�NiN i +Nir�N i

= r��N jZij

�NkZ

ik +Nir�N i

=�r�N jZij

�NkZ

ik +Nir�N i;

by the metrinilic property,

= r�N jNk�kj +Nir�N i

= r�NkNk +Nir�N i

= Nkr�Nk +Nir�N i

= 2Nir�N i;

after index renaming. Thus,

Nir�N i = 0:


Ex. 240: We compute

"ijk"��Z�j Nk = "ijk"��Z

�j

�1

2"kmn"

�Zm Zn�

�=

1

2"ijk"kmn"

�"��Z�j Z

m Z

n�

=1

2�ijkkmn�

��Z

�j Z

m Z

n�

= �12�ijkmkn�

��Z

�j Z

m Z

n�

=1

2�ijkmnk�

��Z

�j Z

m Z

n�

=1

22�ijmn�

��Z

�j Z

m Z

n�

= �ijmn� ��Z

�j Z

m Z

n�

=��im�

jn � �jm�in

� ��

��

�

�Z�j Z

m Z

n�

= �im�jn� ��

��Z

�j Z

m Z

n� � �jm�in� ��Z

�j Z

m Z

n� � �im�jn��

�Z

�j Z

m Z

n� + �

jm�

in��

�Z

�j Z

m Z

n�

= � ��Z

�j Z

i Z

j� � �

��

��Z

�j Z

j Z

i� � ��

�Z

�j Z

i Z

j� + �

��

�Z

�j Z

j Z

i�

= � ��Z

i � � �� Zi� � ��

��Z

i + �

��

�� Z

i�

= Zi� � Zi� � Zi� + Zi�

[Some factors of 2 needed?]

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

r�T ij = Zk�rkT ij :

Thus,r�u = Zk�rku

andr�u = Z�krku:

Thus,

r r�u = r �Z�krku

�= r Z�krku+ Z�kr rku= B� N

krku+ Z�kZm rmrku:= B� N

krku+ Z�kZn Zmnrmrku= B� N

krku+ Z�kZ�nS� Zmnrmrku

73

Now, set = � and contract:

r�r�u = B��Nkrku+ Z�kZ�nS��Zmnrmrku

= B��Nkrku+ ��Z

k�Z

�nZ

mnrmrku= B��N

krku+ Zk�Z�nZmnrmrku:

Now,NkNn + Z

k�Z

�n = �

kn;

soZk�Z

�n = �

kn �NkNn:

We substitute in the above:

r�r�u = B��Nkrku+

��kn �NkNn

�Zmnrmrku

= B��Nkrku+ Zkmrmrku�NkNnZ

mnrmrku= B��N

krku+rmrmu�NmNkrmrku;

or after renaming dummy indices,

r�r�u = B��N iriu+ririu�N iN jrirju;

orN iN jrirju = ririu�r�r�u+B��N iriu

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

= N i:

also, compute

d

ds

�dZi

dsZi

�=

d2Zi

ds2Zi +

dZi

ds

dZids

(Z (s))

=d2Zi

ds2Zi +

dZi

ds

@Zi@Zk

dZk

ds

=d2Zi

ds2Zi +

dZi

ds�nikZn

dZk

ds;


so at s = 0;

d

ds

�dZi

dsZi

�js=0 =

d2Zi

ds2Zi +N

i�nikZnNk

=d2Zi

ds2Zi +N

iNk�nikZn

=d2Zi

ds2Zi +N

jNk�njkZi

=

�d2Zi

ds2+N jNk�ijk

�Zi

Now, examine the LHS of the above. Since

d

ds

�dZi

dsZi

�

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

d2Zi

ds2= �N jNk�ijk;

Then, de�neF (s) = u (Z (s)) ;

so

F 0 (s) =@u

@Zi(Z (s))

dZi

ds(s)

F 00 (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) :

at s = 0 :

F 00 (0) =@2u

@Zi@Zj(Z (0))N iN j +

@u

@Zi(Z (0))

d2Zi

ds2(0)

=@

@Zj[riu]N iN j +riu

d2Zi

ds2(0)

now, note

75

rjriu =@riu@Zj

� �kijrku

@riu@Zj

= rjriu+ �kijrku;

so

F 00 (0) =�rjriu+ �kijrku

�N iN j +riu

d2Zi

ds2(0)

= rjriuN iN j + �kijrkuN iN j +riud2Zi

ds2(0)

= N iN jrjriu+N iN j�kijrku�N jNk�ijkriu= N iN jrjriu;

thus@2u

@n2= F 00 (0) = N iN jrirju

after renaming indices.


Chapter 12

Chapter 12

Ex. 243: This follows from the de�nition 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


Ex. 247: Examine

R�� = R ��

= S� R��

= S� R ��

= S �R ��

= R��

= R �� = R��:

Ex. 248: We may easily see the symmetry of the Einstein tensor from thefact that both R�� and S�� are symmetric.

Ex. 249: Note

G�� = R� S � � 1

2RS� S

�

= R� S � � 1

2R��;

so

G�� = R� S � �R

= R�� R= R�R= 0;

since R�� = R by de�nition.

Ex. 250: We compute

(r�r� �r�r�)T = (r�r� �r�r�)T �S� = R��"��T

"S�

= R��"��S� T"

= R "��T"

= R "��T!S"!

= �R" ��T!S"!

= �R" ��S"!T!= �R!� ��T!= �R�� T�;

79

with index renaming at the last step.

Ex. 251: The invariant case follows from the commutativity of partialderivatives. Now, we consider the covariant case:

(r�r� �r�r�)T i = r�r�T i �r�r�T i

=@�r�T i

�@S�

� � ��r Ti � @�r�T i

�@S�

� � ��r Ti

!

=@�r�T i

�@S�

�@�r�T i

�@S�

=@

@S�

�@T i

@S�+ Zk��

ikmT

m

�� @

@S�

�@T i

@S�+ Zk��

ikmT

m

�=

@2T i

@S�@S�+

@

@S��Zk��

ikmT

m�� @2T i

@S�@S�� @

@S��Zk��

ikmT

m�

=@

@S��Zk��

ikmT

m�� @

@S��Zk��

ikmT

m�

=@Zk�@S�

�ikmTm + Zk�

@�ikm@S�

+ Zk��ikm

@Tm

@S�

�@Zk�

@S��ikmT

m � Zk�@�ikm@S�

Tm � Zk��ikm@Tm

@S�

= 0 [not sure yet]


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.

81

Ex. 253: Compute

r"R�� +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�� ! + �! �R��!"

=@R�� @S"

� �!�"R!� � � �!�"R�! �

+@R��"@S

� �!� R!��" � �!� R�!�"

+@R��" @S�

� �!��R!�" � �!��R�!"

=@R�� @S"

� �!�"�@�!;��@S

� @�!; �@S�

+ ��;!�� ;��

� !

�� !�"

�@��;�!@S

� @��; !@S�

+ ��;�� ! � ��;!��

�+:::

Ex. 254: Compute

1

4" �"��R �� =

1

4" �"��

R1212S

" �"��

=1

4(2) (2)

R1212S

=R1212S

= K;

as desired.


Ex. 255: Compute

K (S� S�� S��S� ) =1

4" �"��R �� (S� S�� S��S� )

=1

4" �"��S� S��R ��

1

4" �"��S��S�

=1

4" �"��S� S��R �� +

1

4" �"��S��S�

=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

2K��

��

= K

Ex. 257: This follows since

r�S� = r�S�;

henceNB�� = NB��;

orB�� = B��:

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 �� = B�2�

then,

B��B�� = tr B�2�

= �1 + �2;

where �1; �2 are the eigenvalues of B�2� . But, since �1 = �

2 and �2 = (��)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 � ba

�= a

�e(z�b)=a + e(b�z)=a

2

�=

a

2e(z�b)=a +

a

2e(b�z)=a

r0 (z) =1

2e(z�b)=a � 1

2e(b�z)=a

r00 (z) =1

2ae(z�b)=a +

1

2ae(b�z)=a:


Compute

r00 (z) r (z)� r0 (z)2 =

�1

2ae(z�b)=a +

1

2ae(b�z)=a

��a2e(z�b)=a +

a

2e(b�z)=a

��1

2e(z�b)=a � 1

2e(b�z)=a

�2=

1

4e2(z�b)=a � 1

4e2(b�z)=a �

�1

4e2(z�b)=a � 1 + 1

4e2(b�z)=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

@S�dS�

dt= S�V

�

= V �S�

as desired.

85

Ex. 261: Compute

A =dV

dt

=d

dt[V �S�]

=dV �

dtS� + V

� dS�dt

(S (t))

=dV �

dtS� + V

� @S�@S�

dS�

dt

=dV �

dtS� + V

�V �@S�@S�

=dV �

dtS� + V

�V ��r�S� + � ��S

�=

dV �

dtS� + V

�V �� S + V�V �r�S�

=dV �

dtS� + V

�V �� S� + V�V �r�S�

=�V �

�tS� + V

�V �r�S�

=�V �

�tS� +NV

�V �B��

=�V �

�tS� +NB��V

�V � ;

as desired.

Ex. 262: We de�ne

�T��t

=dT��dt

+ V �� !T!� � V �! �T�!:

Ex. 263: Look at

�T�0

�0

�t=

dT�0

�0

dt+ V ��

0

!T!�0 � V �! �0T�

0

!

=d

dt

�T�� J

�0

� (S (t)) J��0 (S0 (t))

�+ V ��

0

!T!� J

��0 � V

�! �0T�! J

�0

�

=dT��dtJ�

0

� J��0 + T

��

@J�0

�

@S dS

dtJ��0 + T

��

@J��0

@S 0J�

0

�

dS 0

dt+ V ��

0

!T!� J

��0 � V

�! �0T�! J

�0

�

=dT��dtJ�

0

� J��0 + T

�� J

�0

�V J��0 + T

�� J

� 0�0V

0J�0

� + V ��0

!T!� J

��0 � V

�! �0T�! J

�0

�

= ::: ???

=�T��tJ�

0

� J��0


Ex. 264: These follow from the properties of the standard derivative.

Ex. 265: This also follows from the properties of the standard derivative.

Ex. 266: Not sure - are we considering the surface metrics as functions oftime? In that case, would this be a moving surface, and then we would requirethe derivative in Part III?

Ex. 267: Compute

�S��t

=dS� (S (t))

dt� V �! �S!

=@S�@S�

dS�

dt� V �! �S!

=�r�S� + �!��S!

�V � � V �! �S!

= r�S�V � + V ��!��S! � V �! �S!= r�S�V �

= NB��V�

= NV �B�� ;

as desired.

Ex. 268: This follows from the sum and product rules.

Ex. 269: [Not �nished]

Ex. 270: For a cylinder, we have

B�� =

�� 1R 00 0

�;

so

K = jB�� j

= 0

�� 1R

�= 0:

87

Ex. 271: For a cone, we have

B�� =

�� cot�r 00 0

�;

which has determinant zero. Thus,

K = 0:

Ex. 272: For a sphere, we have

B�� =

�� 1R 00 � 1

R

�;

so

K = jB�� j =�� 1R

�� 1R

�=

1

R2

Ex. 273: For a torus, we have

B�� =

�� cos�R+r cos� 0

0 � 1r

�;

so

K =cos�

r (R+ r cos�)

Ex. 274: For a surface of revolution, we have

B�� =

24� 1

r(z)p1+r0(z)2

0

0 r00(z)

(1+r0(z)2)3=2

35 ;so

K = � 1

r (z)

q1 + r0 (z)

2

r00 (z)�1 + r0 (z)

2�3=2

= � r00 (z)

r (z)�1 + r0 (z)

2�2


Ex. 275: We integrate ZS

KdS =

ZS

1

R2dS

=1

R2

ZS

dS

=1

R24�R2

= 4�

Ex. 276: We integrateZS

cos�

r (R+ r cos�)dS =

Z 2�

0

Z 2�

0

cos�

r (R+ r cos�)

pSd�d�

=

Z 2�

0

Z 2�

0

cos�

r (R+ r cos�)r (R+ r cos�) d�d�

=

Z 2�

0

Z 2�

0

cos�d�d�

= 2� (sin (2�)� sin (0))= 0:

Part III

Part III

89

91

Ex. 291: We have

� (�) = arccot (At cot�) ;

so

� = arccot (At cot�)

cot � = At cot�1

Atcot � = cot�

� (�) = arccot

�1

Atcot �

�

J�t =@S� (t; S0)

@t

=@

@tarccot

�1

Atcot �

�= � 1

1 + 1A2t2 cot

2 �� cot �

At2

=cot �

At2 + 1A cot

2 �

=A cot �

A2t2 + cot2 �

J�0

t =@

@tarccot (At cot�)

= � A cot�

1 +A2t2 cot2 �

Ex. 292:

V � =@Zi

@t=

�A cos�0

�

92

Ex. 293:

V i =@Zi

@t=

2664@@t

�A cos �p

cos2 �+A2t2 sin2 �

�@@t

�A sin �p


�3775

=

24� 2A2t sin2 ��A cos �2pcos2 �+A2t2 sin2 �

3

� 2A2t sin2 ��A sin �2pcos2 �+A2t2 sin2 �

3

35=

24� A3t sin2 � cos �pcos2 �+A2t2 sin2 �

3

� A3t sin3 �pcos2 �+A2t2 sin2 �

3

35 :

Ex. 294: Clearly, the above expressions do not show the tensor propertywith respect to changes in surface coordinates.

Ex. 295: First, note our parametrization:

Zi (�) =

�At cos�sin�

�Then, compute the shift tensors:

Zi� =@Zi

@S�

=

��At sin�cos�

�;

soZ�i =

�� 1At sin� cos�

�;

since we needZ�i Z

i� = �

�� :

Thus,

V � = V iZ�i

=�� 1At sin� cos�

� �A cos�0

�=

�� 1t sin�

�:

93

Ex. 296: Recall

Zi (�) =

24 A cos �pcos2 �+A2t2 sin2 �

A sin �pcos2 �+A2t2 sin2 �

35 ;and note that

@

@�

pcos2 � +A2t2 sin2 � =

�2 cos � sin � + 2A2t2 sin � cos �2pcos2 � +A2t2 sin2 �

=A2t2 sin � cos � � cos � sin �p

cos2 � +A2t2 sin2 �

so

Zi� =@Zi

@S�

=1

cos2 � +A2t2 sin2 �

24pcos2 � +A2t2 sin2 � (�A sin �)�A cos �A2t2 sin � cos ��cos � sin �p

cos2 �+A2t2 sin2 �pcos2 � +A2t2 sin2 � (A cos �)�A sin �A2t2 sin � cos ��cos � sin �p


35 ;

The result is V � = 0, since the motion of a particle on the surface withconstant � is normal to tangent space; hence the projection V � = V iZ�i = 0.

Ex. 297: We see that V � is not a tensor with respect to changes in ambientcoordinates.

Ex. 298: We haveV = V iZi;

soV � Zj = V j ;

con�rming that V i is a tensor (V is an invariant and Zj is a tensor).

Ex. 299: Both V i and Z�i are tensors with regard to ambient coordinatechanges; hence the contraction V � = V iZ�i is a tensor with regard to ambientcoordinate changes.

Ex. 300: This was done before.

Ex. 301: Both parametrizations use Cartesian ambient coordinates. Thus,the Jacobians J i

0

i are the identity. We compute

V iJ i0

i + Zi�J

i0

i J�t = V i + Zi�J

�t

=

�A cos�0

�+

A cot �

A2t2 + cot2 �

�� 1At sin�cos�

�:

94

[To be continued]

Ex. 302: Write

V �0= V i

0Z�

0

i0

=�V iJ i

0

i + Zi�J

i0

i J�t

�Z�j J

ji0J

�0

�

= V iJ i0

i Z�j J

ji0J

�0

� + Zi�Ji0

i J�t Z

�j J

ji0J

�0

�

= V iZ�j J�0

� �ji + Z

i�J

�t Z

�j J

�0

� �ji

= V jZ�j J�0

� + Zj�J�t Z

�j J

�0

�

= V �J�0

� + ��J�t J

�0

�

= V �J�0

� + J�0

� J�t ;

as desired.

Ex. 303: Let the unprimed coordinates denote the �rst parametrization.Then, note

J�0

� =d

d�(arccot (At cot�))

= � 1

1 +A2t2 cot2 �

��At csc2 �

�=

At csc2 �

1 +A2t2 cot2 �

V �J�0

� + J�0

� J�t = �1

tsin�

At csc2 �

1 +A2t2 cot2 �+

A cot �

A2t2 + cot2 �

At csc2 �

1 +A2t2 cot2 �=

[perhaps I am using the wrong Jacobians]

Ex. 304: We have

Zi =

�t cos�sin�

�;

so

S� =d

d�(t cos�) i+

d

d�(sin�) j;

since our ambient space is in Cartesian coordinates. So,

S� = �t sin�i+ cos�j;

95

and choose the outward normal

N = cos�i+ t sin�j;

and thus

N i =

�cos�t sin�

�;

andNi =

�cos� t sin�

�;

since our ambient space is in Cartesian coordinates. Thus,

C = V iNi

=�cos� t sin�

� �cos�0

�= cos2 �:

Ex. 305: As before, compute

S� =d

d�

cos �p

cos2 � + t2 sin2 �

!i+

d

d�

sin �p

cos2 � + t2 sin2 �

!j

=1

cos2 � + t2 sin2 �

pcos2 � + t2 sin2 � (� sin �)� cos � t

2 sin � cos � � cos � sin �pcos2 � + t2 sin2 �

!i+pcos2 � + t2 sin2 � (cos �)� sin � t

2 sin � cos � � cos � sin �pcos2 � + t2 sin2 �

j

!

=

� sin �p

cos2 � + t2 sin2 �� t

2 sin � cos2 � � cos2 � sin �pcos2 � + t2 sin2 �

3

!i+

cos �p

cos2 � + t2 sin2 �� t

2 sin2 � cos � � cos � sin2 �pcos2 � + t2 sin2 �

3

!j;

so

Ni =h

cos �pcos2 �+t2 sin2 �

� � cos � sin2 �+t2 cos � sin2 �(cos2 �+t2 sin2 �)

32

sin �pcos2 �+t2 sin2 �

+ � cos2 � sin �+t2 cos2 � sin �(cos2 �+t2 sin2 �)

32

i;

and

C = V iNi

=h

cos �pcos2 �+t2 sin2 �

� � cos � sin2 �+t2 cos � sin2 �(cos2 �+t2 sin2 �)

32

sin �pcos2 �+t2 sin2 �

+ � cos2 � sin �+t2 cos2 � sin �(cos2 �+t2 sin2 �)

32

i24� t sin2 � cos �pcos2 �+A2t2 sin2 �

3

� t sin3 �pcos2 �+A2t2 sin2 �

3

35=

96

[Note: Shouldn�t both be 1 by geometric considerations?]

Ex. 306: We have

V �0= V �J�

0

� + J�0

� J�t

@T (t; S0)

@t=

@T (t; S)

@t+ J�t r�t;

so compute

_rT (t; S0) =@T (t; S0)

@t� V �

0(t; S0)r�0T

=@T (t; S)

@t+ J�t r�t� V �

0r�TJ�

0

� ;

since r�0T has the tensor property. But,

V �0= V J�

0

+ J�0

J t ;

so

_rT (t; S0) =@T (t; S)

@t+ J�t r�t�

�V J�

0

+ J�0

J t

�r�TJ�a0

=@T (t; S)

@t+ J�t r�t�

�V �� + J

t �

�

�r�T

=@T (t; S)

@t+ J�t r�t� V �r�T � J

�t r�T

=@T (t; S)

@t� V �r�T

=@T (t; S)

@t� V �r�T;

so _rT does not depend on changes in suface coordinates.

Ex. 307: This follows from the sum rule for partial derivatives and thecovariant derivative

Ex. 308: This follows from the product rule for partial derivatives and thecovariant derivative.

Ex. 309: Same as above

Ex. 310: This follows because the numerator of 15.33 would be zero.

Ex. 311: This follows from the de�nition of C, since we take our R (St + h)in the normal direction.

97

Ex. 312: We haveS� = Z

i�Zi:

Ex. 313: Begin with

_rR =�V i � V �Zi�

�Zi

andV � = V jZ�j ;

so

_rR =�V i � V jZ�j Zi�

�Zi

=�V i � V j

��ij �N iNj

��Zi

=�V i � V i + V jN iNj

�Zi

= V jN iNjZi;

as desired.

Ex. 314: Compute

d

dt

ZS

NB��dS =d

dt

ZS

r�S�dS

=

ZS

_r (r�S�) dS �ZS

CB��r�S�dS

=

ZS

@ (r�S�)@t

� V �r�r�S�dS �ZS

CB��r�S�dS

=

ZS

@ (r�S�)@t

� V iZ�i r�r�S�dS �

ZS

CB��r�S�dS

[ask about integral problems]

Ex. 315: By the above, ZS

NB��dS

is constant. Since our surface is of genus zero, we may smoothly append oursurface evolution so that for all t � T for some T , S is a sphere of constant

98

radius 1. Since the above quantity is constant for all t, then it must be equalto

�2ZS

NdS

for all t, since for a sphere,

B�� =�2R:

But,RSNdS = 0 (our surface is closed), so we have that

RSNB��dS vanishes.

Ex. 316: Need to show:

d

dt

Z

Fd =

Z

@F

@td+

ZS

CFdS;

i.e.d

dt

Z A2

A1

Z b(t;x)

0

F (x; y) dydx =

ZS

CFdS;

since @F@t = 0.

Clearly, C is zero on all of S except for the portion given by the graph of b. LetB denote the surface given by this graph. Then,:Z

S

CFdS =

ZB

CFdB

Clearly, S� is the vector (given relative to the ambient Cartesian basis)

S� = i+ bxj,

so

N =1p1 + b2x

(�bxi+ j) ;

and

V i =

�0bt

�;

so

C = V iNi

=btp1 + b2x

:

99

Now, at each t, our surface has line elementp1 + b2x, soZ

B

CFdB =

Z A2

A1

btF (x; b (t; x)) dx

=

Z A2

A1

d

dt

Z b(t;x)

0

F (x; y) dydx;

by FTC,

=d

dt

Z A2

A1

Z b(t;x)

0

F (x; y) dydx

Ex. 317: We have

U i0=

@T i0(t; S0)

@t

=@

@t

�T i (t; S) J i

0

i (Z (t; S))�

=@

@tT i (t; S (t; S0)) J i

0

i + Ti @

@t

�J i

0

i (Z (t; S))�

=

�@T i

@t+@T i

@S�@

@tS��J i

0

i + Ti @J

i0

i

@Zj@

@tZ (t; S)

=

�@T i

@t+@T i

@S�@S�

@t

�J i

0

i + TiJ i

0

ji

�@Zj

@t+@Zj

@S�@S�

@t

�=

�U i +

@T i

@S�J�t

�J i

0

i + TiJ i

0

ji

�V j + Zj�J

�t

�= U iJ i

0

i +@T i

@S�J�t J

i0

i + TiJ i

0

jiVj + T iJ i

0

jiZj�J

�t ;

which is the desired result.

Ex. 318: This follows similarly to the above.

Ex. 319: Write

U�0=

@T�0(t; S0)

@t

=@

@t

�T� (t; S) J�

0

� (t; S)�

=@

@tT� (t; S) J�

0

� + T�@

@tJ�

0

� (t; S)

=

�@T�

@t+@T�

@S�@S�

@t

�J�

0

� + T�

@J�

0

�

@t+@J�

0

�

@S�@S�

@t

!

= U�J�0

� +@T�

@S�J�t J

�0

� + T�J�0

�t + T�J�

0

��J�t :

100

Ex. 320: The covariant case is analogous to the above.

Ex. 321: Note

V �r�T i0= V �r�T iJ i

0

i

= V ��@T i

@S�(t; S) + �ijkT

jZj�

�J i

0

i

= V j0Z�j0

�@T i

@S�(t; S) + �ijkT

kZj�

�J i

0

i

=�V jJj

0

j + Zj�J

j0

j J�t

�� @T i@S�

(t; S) + �ijkTkZj�

�J i

0

i

=

�V jJj

0

j

@T i

@S�+ V jJj

0

j �ijkT

kZj� + Zj�J

j0

j J�t

@T i

@S�+ Zj�J

j0

j J�t �

ijkT

kZj�

�J i

0

i

= V jJj0

j

@T i

@S�J i

0

i + VjJj

0

j �ijkT

kZj�Ji0

i + Zj�J

j0

j J�t

@T i

@S�J i

0

i + Zj�J

j0

j J�t �

ijkT

kZj�Ji0

i ;

so given our work above,

@T i0

@t� V �r�T i

0=

@T i

@tJ i

0

i +@T i

@S�J�t J

i0

i + TiJ i

0

jiVj + T iJ i

0

jiZj�J

�t

��V jJj

0

j

@T i

@S�J i

0

i + VjJj

0

j �ijkT

kZj�Ji0

i + Zj�J

j0

j J�t

@T i

@S�J i

0

i + Zj�J

j0

j J�t �

ijkT

kZj�Ji0

i

�

[not �nished]

Chapter 13

Chapter 16

Ex. 325: Assume the sum, product rules hold, in addition to commutativitywith contraction and the metrinilic property with respect to the ambient basis.Then, compute

_rT = _r�T iZi

�= _rT iZi + T i _rZi;

by commutativity with contraction and the product rule,

= T iZi;

since the second term would be zero by the metrinilic property.

Ex. 326: Compute

@Zi@t

=@Zi (Z (t))

@t

=@Zi@Zj

@Zj

@t

= �kijZkVj

= V j�kijZk;

as desired.

Ex. 327: WriteT = TiZ

i;

so_rTiZi = _rT =@Ti

@tZi + Ti

@Zi

@t� V r TiZi;

101


but

@Zi

@t=

@Zi

@Zj@Zj

@t

= ��ijkZkV j

= ��ijkV jZk;

so we have_rTi =

@Ti@t� V r Ti � V j�kijTk;

after index renaming in the second term of the above expression.

Ex. 328: We know _rT is invariant, and since _rT = _rT iZi and Zi is atensor, by the quotient law, _rT i must be a tensor. An argument using changesin coordinates would follow similarly to the covariant derivative computations.

Ex. 329: This also follows from the fact that _rT = _rTiZi and by thequotient law. An argument using changes in coordinates would also followsimilarly.

Ex. 330: PutSj = T

ijZi:

then, clearly,_rSj = _rT ijZi:

Dot both sides with Zk:

_rSj � Zk = _rT ijZi � Zk

_rSj � Zk = _rT kj :

Since the LHS is clearly a tensor, the RHS is as well.

Ex. 331: We may use induction with a process similar to 330 to extend thisresult to arbitrary indices.

Ex. 332: Assume Si; T ji are arbitrary tensors. Put

U j = SiT ji :

103

Now, compute

_r�SiT ji

�=

@�SiT ji

�@t

� V r �SiT ji

�+ V n�jnkS

iT ji

=@U j

@t� V r U j + V n�jnkS

iT ji ;

since both the partial derivatives and the covariant surface derivatives commutewith contraction.

Ex. 333: This follows from the sum and product rules for the partialderivatives and the covariant surface derivative.

Ex. 334: Note

r T ij = Zk rkT ij ;

and compute

_rT ij (t; S) =@T ij (t; Z (t; S))

@t� V r T ij (t; Z (t; S)) + V k�ikmTmj (t; Z (t; S))� V k�mkjT im (t; Z (t; S))

=@T ij@t

+@T ij@Zk

@Zk

@t� V Zk rkT ij + V k�ikmTmj � V k�mkjT im

=@T ij@t

+@T ij@Zk

V k � V Zk rkT ij + V k�ikmTmj � V k�mkjT im

=@T ij@t

+

@T ij@Zk

+ �ikmTmj � �mkjT im

!V k � V Zk rkT ij

=@T ij@t

+rkT ijV k � V Zk rkT ij

=@T ij@t

+�V k � V Zk

�rkT ij

=@T ij@t

+ CNkrkT ij ;

where the last step follows from the observation that V k � V Zk is the normalcomponent.

Ex. 335: Note that @Zi@t = 0; and the second term of the above is also zero

by the metrinilic property for covariant derivatives. Hence, _rZi = 0: the otherresults follow similarly.


Ex. 336: Compute

@S� (Z (t; S))

@t=

@S�@Zi

@Zi

@t

=@S�@Zi

V i

=@S�@S�

@S�

@ZiV i

=@2R

@S�@S�Z�i V

i

=@S�@S�

Z�i Vi

= [not sure]

Ex. 337: Simply decompose V into its tangential and normal coordinates,to obtain the substitution used for the RHS.

Ex. 338: Compute

r� (V �S� + CN) = r�V �S� + V �r�S� +r�CN+Cr�N= r�V �S� + V �NB�� +Nr�C + C

��B��S�

�= r�V �S� + V �NB�� +Nr�C � CB��S�:

Ex. 339: Use

_rT =@T�

@tS� + T

� @S�@t� V �r�T�S� � V �T�NB��

=@T�

@tS� + T

��r�V �S� + V �NB�� +Nr�C � CB��S�

�� V �r�T�S� � V �T�NB��

=@T�

@tS� +r�V �T �S� + V �T �NB�� + T �Nr�C � CB��T �S� � V �r�T�S� � V �T�NB��

=@T�

@tS� + T

�r�V �S� + T �r�CN� T �CB��S� � V �r�T�S�

=@T�

@tS� + T

�r�V �S� + T�r�CN� T �CB��S� � V �r�T�S�:

Ex. 340: Simply decompose T with respect to the contravariant basis S�,and use the similar decomposition

V = V �S� + CN:

105

Ex. 341: Note

0 = _rS��

=@S��@t� V !r!S�� (r�V ! � CB!�)S!� �

�r�V ! � CB!�

�S�!

=@S��@t�r�V !S!� + CB!�S!� �r�V !S�! + CB!�S�!

=@S��@t�r�V� �r�V� + CB�� + CB��

=@S��@t�r�V� �r�V� + 2CB�� ;

so@S��@t

= r�V� +r�V� � 2CB�� :

The contravariant case follows similarly.

Ex. 342: First, examine

@S

@S��= SS�� ;

using the properties of the cofactor matrix. Then, compute

@S

@t=

@S

@S��@S��

@t

= SS�� (r�V� +r�V� � 2CB��)= S

�r�V � +r�V � � 2CB��

�= 2S (r�V � � CB��) :

Then,

@pS

@t=

1

2pS

@S

@t

=pS (r�V � � CB��) ;


and

@�pS��1

@t= � 1�p

S�2 @pS@t

= � 1S

pS (r�V � � CB��)

= � 1pS(r�V � � CB��) :

Ex. 343: Note"�� =

pSe�� ;

so

@"��@t

=@pS

@te�� +

pS@e��@t

=@pS

@te�� ;

since e�� does not depend on t.

=pS�r V � CB

�e��

= "��r V � CB

�:

Ex. 344: Note

"�� =�pS��1

e�� ;

so

@"��

@t=

@�pS��1

@te�� +

pS@e��

@t

=@�pS��1

@te��

= � 1pS

�r V � CB

�e��

= �"��r V � CB

�:

107

Ex. 345: Simply write

_r"�� = _r�" �S

�S��

= _r" �S �S�� ;

by the metrinilic property,= 0;

by the above.

Ex. 346: Use Gauss�Theorema Egregium:

K = jB�� j

=1

2"��"��B

��B

�� ;

so

2 _rK = _r"��"��B��B�� + "�� _r"��B��B�� + "��"�� _rB��B�� + "��"��B�� _rB��= "��"�� _rB��B�� + "��"��B�� _rB�� ;

since the �rst two terms vanish,

= "��"��r�r�C + CB� B �

�B�� + "

��"��B��

�r�r�C + CB� B �

�= "��"��B

��r�r�C + C"��"��B��B� B � + "��"��B��r�r�C + C"��"��B��B� B �

= ��

�B��r�r�C +B��r�r�C +B��B� B �C +B��B� B �C

�=

��

��

��

�B��r�r�C +B��r�r�C +B��B� B �C +B��B� B �C

= ��B

��r�r�C + ��B��r�r�C + ��B��B� B �C + ��B��B� B �C

��

��B

��r�r�C + �

��B

��r�r�C + �

��B

��B

� B

�C + �

��B

��B

� B

�C�

= B��r�r�C +B��r�r�C +B��B� B �C +B��B� B �C��B��r�r�C +B��r�r�C +B��B� B �C +B��B� B �C

�= 2B��r�r�C � 2B��r�r�C + 2B��B� B

�C � 2B

��B

� B

�C

= 2�B��r�r�C �B��r�r�C +

�B��B

� B

� �B

��B

� B

�

�C�

= 2�B��r�r�C �B��r�r�C +B��KC

�;


which gives us the desired result [Note: I believe the last equality follows fromGauss�Theorema Egregium].

Chapter 14

Chapter 17

Ex. 347: Note that u is an invariant with respect to ambient indices, and thusriu = @u

@Zi . Write

@

@triu =

@

@t

@u

@Zi

=@

@Zi@u

@t

= ri@

@tu;

under smoothness assumptions (hence the partials commute), and since @u@t is

an invariant.

Ex. 348: First, compute, given polar coordinates

riu =@u

@Zi

=

24 @@r

�J0(�r)p�J1(�)

�@@�

�J0(�r)p�J1(�)

�35=

"��J1(�r)p�J1(�)

0

#

Now, since for polar coordinates,

Zij =

�1 00 r�2

�;

109


so

riu = Zijrju

=

�1 00 r�2

�"��J1(�r)p�J1(�)

0

#

=

"��J1(�r)p�J1(�)

0

#:

thus,

riuriu =��2J1 (�r)2

�J1 (�)2 :

Now, at t = 0, our surface yields r = 1, so

riuriu =��2�

Next, compute C for our surface evolution. Consider, in Cartesian coordinates,

V i =

�@x@t@y@t

�=

�a cos�b sin�

�

With respect to the Cartesian basis,

S� = � (1 + at) sin�i+ (1 + bt) cos�j

Ni =1q

(1 + at)2cos2 �+ (1 + bt)

2sin2 �

�(1 + bt) cos�(1 + at) sin�

�;

at t = 0,

Ni =

�cos�sin�

�

C = V iNi

= a cos2 �+ b sin2 �;

111

thus, by the Hadamard formula,

�1 =

Z 2�

0

�a cos2 �+ b sin2 �

��2�

�d�

=��2�

Z 2�

0

a cos2 �+ b sin2 �d�

=��2�

Z 2�

0

a�cos2 �

�+ b

�1� cos2 �

�d�

=��2�

Z 2�

0

a cos2 �+ b cos2 �d�� 2b�2

=��2�(a+ b)

Z 2�

0

cos2 �d�� 2b�2

=��2�(a+ b)

��

2+1

4sin 2�

�2�0

� 2b�2

=��2�(a+ b)� � 2b�2

= ��2 (a+ b)� 2b�2

= ��2 (a+ b) :

Since � = �2, we have the desired result.

Ex. 349: [Not sure]

Ex. 350: Want to show:

�1 =

ZS

�u1Niriu�Niriu1

�dS:

Dirichlet:

�1 = �ZS

CriuriudS

Neumann:

�1 =

Ex. 354: We have

��_rC + 2V �r�C +B��V �V �

�= �B��

_rV � + V �r�V � � Cr�C � CV �B�� = 0:


Write

V � = V iZ�i

and

C = V iNi;

Begin with the second, and contract with S�:

_rV �S� + V �r�V �S� � Cr�CS� � CV �B��S� = 0

_r (V �S�)� V � _rS� + V �r� (V �S�)� V �V �r�S� � Cr�CS� � CV �B��S� = 0

_r (V �S�)� V �Nr�C + V �r� (V �S�)� V �V �B��N� Cr�CS� � CV �B��S� = 0:

Then, manipulate the �rst:

_rC + 2V �r�C +B��V �V � =�

�B�� ;

and multiply by N:

_rCN+ 2V �r�CN+B��V �V �N =�

�B��N

_r (CN)� C _rN+ 2V �r�CN+B��V �V �N =�

�B��N

_r (CN)� C (�S�r�C) + 2V �r�CN+B��V �V �N =�

�B��N

_r (CN) + CS�r�C + 2V �r�CN+B��V �V �N =�

�B��N

_r (CN) + 2V �r�CN+ V �V �B��N+Cr�CS� =�

�B��N:

113

Now, add the results of both manipulations to each other:

_r (V �S�) + _r (CN) + V �r�CN+ V �r� (V �S�)� CV �B��S� =�

�B��N

_rV + V �r�CN+ V �r� (V �S�)� CV �B��S� =�

�B��N

_rV + V �r� (CN)� V �Cr�N+ V �r� (V �S�)� CV �B��S� =�

�B��N

_rV+V �r�V � V �Cr�N� CV �B��S� =�

�B��N

_rV+V �r�V � V �Cr��N iZi

�� CV �B��S� =

�

�B��N

_rV+V �r�V � V �Cr�N iZi � CV �B��S� =�

�B��N [note the metrinilic property]

_rV+V �r�V + V �CZi�B��Zi � CV �B��S� =�

�B��N

_rV+V �r�V + V �CB��Zi�Zi � CV �B��S� =�

�B��N

_rV+V �r�V + CV �B��S� � CV �B��S� =�

�B��N

_rV+V �r�V =�

�B��N

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Selected solutions to exercises from Pavel Grinfeld™s …dws57/Tensor_Calc_and... · 2014-09-14 · Introduction Included in this text are solutions to various exercises from Introduction