Selected Solutions to Exercises from Pavel Grinfeldgs Introduction to Tensor Analysis and the Calculus of Moving Surfaces

Home , Differentiable function

Selected solutions to exercises from Pavel Grinfeld’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 to Tensor Analysis and the Calculus of Moving Surfaces, by Dr. Pavel Grinfeld.

v vi CONTENTS Part I

Part I

Chapter 1

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

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

2 2y0 = (2x0) e

2 2y0 = 4 (x0) e : 1 1 Ex. 2: Note that the above implies x0 = 2 x, y0 = 2 y. We check

@F 0 2y0 (x0; y0) = 8 (x0) e @x0 2 1 y 1 ( 2 ) = 8 x e 2 = 4xey @F (x; y) = 2xey: @x

@F 0 @F Thus, (x0; y0) = 2 (x; y) as desired. @x0 @x Ex. 3: Let a; b R, a; b = 0, and consider the "re-scaled" coordinate basis 2 6 a i = 0 0 0 j = ; 0 b where each of the above vectors is taken to be with respect to the standard 2 basis for R . 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, @F @F T = (x; y); (x; y) r @x @y 3 4 CHAPTER 1. CHAPTER 1

in standard coordinates @F @x @F @y = (x ; y ) 0 (x; y); (x ; y ) 0 (x; y) @x 0 0 @x @y 0 0 @y 0 0

(think of x0 as a function of x).

@F 1 @F 1 = (x ; y ) ; (x ; y ) @x 0 0 a @y 0 0 b 0 0 1 @F 1 @F = ; pa2 + 0 @x b2 + 0 @y 0 0 1 @F 1 @F = ; ; pi i @x0 pj0 j0 @y0 0 0 as desired.

Ex. 4: Assume

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

Then,

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

Also,

x0 a cos sin x = y0 b sin cos y 1 x cos sin x0 a = y sin cos y0 b cos sin cos2 +sin2 cos2 +sin2 x0 a = sin cos 2 2 2 2 y0 b cos +sin cos +sin cos sin x0 a = sin cos y0 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)]

cos sin 1 i = 0 sin cos 0 = cos i+ sin j cos sin 0 j = 0 sin cos 1 = sin i+ cos j

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

@F @F @F @x @F @y @F @x @F @y (x; y) i + (x; y) j = (x ; y ) 0 (x; y) + (x ; y ) 0 (x; y) i + (x ; y ) 0 (x; y) + (x ; y ) 0 (x; y) j @x @y @x 0 0 @x @y 0 0 @x @x 0 0 @y @y 0 0 @y 0 0 0 0 @F @F @F @F = (x0; y0) cos + (x0; y0) sin i+ (x0; y0) sin + (x0; y0) cos j @x @y @x @y 0 0 0 0 @F @F @F @F = (x0; y0) cos i (x0; y0) sin j + (x0; y0) sin i + (x0; y0) cos j @x0 @x0 @y0 @y0 @F @F = (x0; y0) (cos i sin j) + (x0; y0) (sin i + cos j) @x0 @y0 @F cos @F sin = (x ; y ) i j + (x ; y ) i j @x 0 0 sin @y 0 0 cos 0 0 @F @F = (x0; y0) i0 + (x0; y0) j0 @x0 @y0 [NOT SURE - I will ask about this one tomorrow]

(a; b) cooresponds to a shifted "origin," corresponds to angle for which the whole 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. 6 CHAPTER 1. CHAPTER 1 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 we are taking each directional derivative. De…ne f (h) := F (P ), i.e. parametrize along 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 A to B. Ex. 7: f (h) = F (P )2 + h2 h f 0 (h) = p F (P )2 + h2 dF (p) = fp(0) dl 0

Ex. 8: We have 1 f (h) = AP h 1 f 0 (h) = (AP h)2 1 = (AP h)2 so dF (p) = f (0) dl 0 1 = (AP )2

7 8 CHAPTER 2. CHAPTER 2

Ex. 9: Let denote the measure of angle OP P . By the Law of Sines, we have

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

From the second equation, we obtain

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

Thus,

(OP )(AP h) sin (F (P )) f (h) = arctan (OA) h + (OP )(AP h) cos (F (P )) (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 ))] 1 f 0 (h) = 2 [(OP )(AP h) sin(F (P ))]2 [(OA) h + (OP )(AP h) cos (F (P ))] 1 + [(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) = f (0) dl 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) = sin (F (P )) (OP )(AP )

Ex. 10: Clearly F (P ) = F (P ) for any choice P in such a direction. Thus, f is constant, and we have dF (p) = 0 dl

Ex. 11: Put d as the distance between P and the line from A to B. As with the previous problem, the distance from P to line AB! is also d. Thus, 1 F (P ) = (AB) d 2 1 F (P ) = (AB) d; 2

dF (p) and we have F (P ) = F (P ), so 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,

1 f (h) = (AB)(F (P ) + h) 2 1 f (h) = AB 0 2

dF (p) = f (0) dl 0 1 = AB 2

Ex. 13: (7) The gradient will point in direction !AP , and will have magnitude 1. (8) [Not sure] 10 CHAPTER 2. CHAPTER 2

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

(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 AB: 2

Ex. 14: The directional derivative in direction L would then correspond to the projection of Of onto L.

Ex. 15: [See diagram]

R ( + h) R ( ) 2 = 1 + 1 2 cos (h) k k by the Law of Cosines. So,

R ( + h) R ( ) 2 = 2 2 cos (h) k k R ( + h) R ( ) = 2 2 cos (h) k k p h = 2 2 cos 2 2 s h = 2 2 sin2 2 s h = 4 sin2 r 2 h = 2 sin : 2

Ex. 16: 2 sin h 2 1 cos h lim 2 = lim 2 2 ; h 0 h h 0 1 ! ! 11

by L’Hospital’srule, h = lim cos h 0 2 ! = 1:

Ex. 17: 2 sin h sin h lim 2 = lim 2 h 0 h h 0 h ! ! 2 sin 0 + h sin (0) = lim 2 h 0 h ! 2 = sin0 (0) = cos (0) = 1:

Ex. 18: We have R ( ) 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

R ( ) R00 ( ) cos = 1 k k k k R00 ( ) cos = 1; k k since R ( ) is of unit length. Now, let h be arbitrarily small. Since R ( ) = k k R ( + h) = R0 ( + h) = R0 ( ) = 1, we have by congruent triangles k k k k k k that R0 ( + h) R0 ( ) = R ( + h) R ( ) : Thus, k k k k R0 ( + h) R0 ( ) R00 ( ) = lim k k k k h 0 h ! R ( + h) R ( ) = lim k k h 0 h ! = R0 ( ) k k = 1: 12 CHAPTER 2. CHAPTER 2

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

cos = 1; which implies that = , or that R00 ( ) points in the opposite direction as R ( ). Chapter 3

Chapter 3

Ex. 19: We may construct a three-dimensional Cartesian coordinate system as follows: 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 coordinates

1 0 0 0 ; 1 ; 0 001 001 011 @ A @ A @ A respectively, and a vector V connecting the origin to a point with coordinates

x0 y0 0 1 z0 @ A can be expreessed by the linear combination

V =x0i+y0j+z0k:

Ex. 20: Since our space is three-dimensional, there are three continuous degrees of freedom associated with our choice of origin O. The choice of the direction 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 sphere centered at O). Finally, the "y"-axis may be chosen to lie along any line orthogonal to the x-axis; the set of all such lines lie in a plane, hence our choice of direction for the y-axis yields the sixth continuous degree of freedom (there is a bijection between the set of all such directions and points on the unit circle which lies in this plane orthogonal to the x-axis.

13 14 CHAPTER 3. CHAPTER 3

Ex. 21: Let P be an arbitrary point in a two-dimensional Euclidean space with polar coordinates r; . Assume P has cartesian coordinates (x; y). De…ne P 0 to be the point along the pole that is distance x from the origin. Note that by the orthogonality of the x; y axes, we may form a right triangle with P , the origin O, and P 0. Note that OP 0 = x and PP 0 = y; hence by the properties of right triangles, we have x = cos r y = sin ; r 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 = x2 + y2: p Also, y r sin = x r cos = tan ; so y = arctan : x

Ex. 23: De…ne the x and y coordinate of some arbitrary point P to be the Cartesian system of coordinates de…ned by applying Ex. 21 to the coordinate plane …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 - and also 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 the identity z (x; y; z) = z follows trivially from 23.

Ex. 25: Let P be a point with spherical coordinates r; ; . Let the x-axis be the polar axis, and the y-axis lie in the coordinate plane and point in the direction orthogonal to the polar axis (chosen in accordance to the right-hand rule). Finally, let hte z-axis be the longitudinal axis. Since the z-coordinate length OP 0, where P 0 is the orthogonal projection of P onto the longitudinal axis, we have by the properties of right triangles, z = r cos .

Now, project P onto the coordinate plane, and denote this point P 00. We clearly have the length OP 00 = r sin . Thus, by considering the right triangle determined 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 have

x2 + y2 + z2 = r2 sin2 cos2 + r2 sin2 sin2 + r2 cos2 q p = r sin2 cos2 + sin2 sin2 + cos2 q = r sin2 cos2 + sin2 + cos2 q = r sin2 + cos2 = r;p so r (x; y; z) = x2 + y2 + z2: p Also, z=r = cos ; so z (x; y; z) = arccos r z = arccos : x2 + y2 + z2 p 16 CHAPTER 3. CHAPTER 3

Finally, y r sin sin = x r sin cos = tan ; so y (x; y; z) = arctan : x Chapter 4

Chapter 4

Ex. 27:

x y px2+y2 px2+y2 det J = det y x " x2+y2 x2+y2 # x2 + y2 = (x2 + y2) x2 + y2 1 = p: x2 + y2 p

Ex. 28:

1 1 p12+12 p12+12 J (1; 1) = 1 1 12+12 12+12 1 1 p2 p2 = 1 1 : 2 2

Ex. 29:

cos r sin det J = det 0 sin r cos = r cos2 + sin2 = r;

2 2 Using the relationship r = x + y ; we have det J det J 0 = 1. p 17 18 CHAPTER 4. CHAPTER 4

Ex. 30: cos p2 sin J p2; = 4 4 0 p 4 sin 4 2 cos 4 p2 2 1 = p2 : " 2 1 #

Ex. 31: We evaluate the product

1 1 p2 p2 p2 2 1 JJ 0 = 1 1 p2 1 2 2 " 2 # 1 0 = ; 0 1 as desired.

Ex. 32: cos r sin J (x; y) = 0 sin r cos x y = r y x r x y px2+y2 = y ; 2 x 3 px2+y2 4 5 so x y x y px2+y2 px2+y2 px2+y2 JJ 0 = y y x 2 x 3 " x2+y2 x2+y2 # px2+y2 x2 y2 4 5 x2+y2 + x2+y2 0 = 2 x2 y " 0 x2+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) = x2 + y2 + z2 z (x; y; z) =p arccos x2 + y2 + z2 y (x; y; z) = arctan p 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 = sin cos @x @r = sin sin @y @r = cos @z

@ cos cos = @x r @ cos = @y r sin @ sin = @z r

@ sin = @x r sin @ = cos sin @y @ = 0: @z

Thus,

@r @r @r @x @y @y @ @ @ J (r; ; ) = 2 @x @y @z 3 @ @ @ 6 @x @y @z 7 4sin cos sin5 sin cos = cos cos cos sin : 2 r r sin r 3 sin cos sin 0 r sin 4 5 We then compute

@x @x @x @r @ @ @y @y @y J 0 (r; ; ) = 2 @r @ @ 3 @z @z @z 6 @r @ @ 7 4sin cos r5cos cos r sin sin = sin sin r cos sin r sin cos ; 2 cos r sin 0 3 4 5 20 CHAPTER 4. CHAPTER 4

sin cos sin sin cos sin cos r cos cos r sin sin cos cos cos sin JJ 0 = sin sin r cos sin r sin cos 2 r r sin r 3 2 3 sin cos sin 0 cos r sin 0 r sin 4 cos2 + cos2 sin2 + sin5 42 sin2 r cos sin + r cos 5cos2 sin + r cos sin sin2 0 = 1 cos sin + 1 cos sin + 1 cos cos2 sin sin2 + (cos ) cos sin + cos2 cos2 cos cos sin sin + cos2 2 r r r sin 3 1 cos sin + cos sin sin2 r cos cos sin2 (cos ) cos sin sin2 + r cos2 sin sin r sin 41 0 0 5 = 0 1 0 20 0 13 4 5

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

Ex. 34: We compute

@2f (; ) @ @F @A @F @B @F @C = + + @2 @ @a @ @b @ @c @ @ @F @A @F @ @A = + @ @a @ @a @ @ @ @F @B @F @ @B + + @ @b @ @b @ @ @ @F @C @F @ @C + + @ @c @ @c @ @ by the product rule. We continue:

@f (; ) @2F @A @2F @B @2F @C @A @F @2A = + + + @2 @a2 @ @a@b @ @a@c @ @ @a @2 @2F @A @2F @B @2F @C @B @F @2B + + + + @a@b @ @b2 @ @b@c @ @ @b @2 @2F @A @2F @B @2F @C @C @F @2C + + + + : @a@c @ @b@c @ @c2 @ @ @c @2 21

Similarly,

@2f (; ) @ @F @A @F @B @F @C = + + @@ @ @a @ @b @ @c @ @ @F @A @F @ @A = + @ @a @ @a @ @ @ @F @B @F @ @B + + @ @b @ @b @ @ @ @F @C @F @ @C + + @ @c @ @c @ @ @2F @A @2F @B @2F @C @A @F @2A = + + + @a2 @ @a@b @ @a@c @ @ @a @@ @2F @A @2F @B @2F @C @B @F @2B + + + + @a@b @ @b2 @ @b@c @ @ @b @@ @2F @A @2F @B @2F @C @C @F @2C + + + + @a@c @ @b@c @ @c2 @ @ @c @@

@f (; ) @2F @A @2F @B @2F @C @A @F @2A = + + + @2 @a2 @ @a@b @ @a@c @ @ @a @2 @2F @A @2F @B @2F @C @B @F @2B + + + + @a@b @ @b2 @ @b@c @ @ @b @2 @2F @A @2F @B @2F @C @C @F @2C + + + + : @a@c @ @b@c @ @c2 @ @ @c @2

Ex. 35: We compute

@3f (; ) @ @2F @A @2F @B @2F @C @A @F @2A = + + + @2@ @ @a2 @ @a@b @ @a@c @ @ @a @@ @ @2F @A @2F @B @2F @C @B @F @2B + + + + @ @a@b @ @b2 @ @b@c @ @ @b @@ @ @2F @A @2F @B @2F @C @C @F @2C + + + + @ @a@c @ @b@c @ @c2 @ @ @c @@ = I + J + K; 22 CHAPTER 4. CHAPTER 4 for sake of clarity,

@ @2F @A @2F @B @2F @C @A @F @2A I = + + + @ @a2 @ @a@b @ @a@c @ @ @a @@ @ @2F @A @2F @B @2F @C @A = + + @ @a2 @ @a@b @ @a@c @ @ @2F @A @2F @B @2F @C @ @A @ @F @2A @F @3A + + + + + @a2 @ @a@b @ @a@c @ @ @ @ @a @@ @a @2@ @ @2F @A @2F @B @2F @C @A = + + @ @a2 @ @a@b @ @a@c @ @ @2F @A @2F @B @2F @C @2A + + + @a2 @ @a@b @ @a@c @ @@ @2F @A @2F @B @2F @C @2A @F @3A + + + + @a2 @ @a@b @ @a@c @ @@ @a @2@ @ @2F @A @2F @B @2F @C @A = + + @ @a2 @ @a@b @ @a@c @ @ @2F @A @2F @B @2F @C @2A @F @3A +2 + + + @a2 @ @a@b @ @a@c @ @@ @a @2@ @ @2F @A @2F @ @A @ @2F @B @2F @ @B @ @2F @C @2F @ @C @A = + + + + + @ @a2 @ @a2 @ @ @ @a@b @ @a@b @ @ @ @a@c @ @a@c @ @ @ @2F @A @2F @B @2F @C @2A @F @3A +2 + + + @a2 @ @a@b @ @a@c @ @@ @a @2@ @ @2F @A @A @2F @ @A @A = + @ @a2 @ @ @a2 @ @ @ @ @2F @B @A @2F @ @B @A + + @ @a@b @ @ @a@b @ @ @ @ @2F @C @A @2F @ @C @A + + @ @a@c @ @ @a@c @ @ @ @2F @A @2F @B @2F @C @2A @F @3A +2 + + + @a2 @ @a@b @ @a@c @ @@ @a @2@ @3F @A @3F @B @3F @C @A @A @2F @2A @A = + + + @a3 @ @a2@b @ @a2@c @ @ @ @a2 @2 @ @3F @A @3F @B @3F @C @B @A @2F @2B @A + + + + @a2@b @ @a@b2 @ @a@b@c @ @ @ @a@b @2 @ @3F @A @3F @B @3F @C @C @A @2F @2C @A + + + + @a2@c @ @a@b@c @ @a@c2 @ @ @ @a@c @2 @ @2F @A @2F @B @2F @C @2A @F @3A +2 + + + @a2 @ @a@b @ @a@c @ @@ @a @2@ 23

@ @2F @A @2F @B @2F @C @B @F @2B J = + + + @ @a@b @ @b2 @ @b@c @ @ @b @@ @ @2F @A @2F @B @2F @C @B = + + @ @a@b @ @b2 @ @b@c @ @ @2F @A @2F @B @2F @C @ @B @ @F @2B @F @ @2B + + + + + @a@b @ @b2 @ @b@c @ @ @ @ @b @@ @b @ @@ @ @2F @A @2F @B @2F @C @B = + + @ @a@b @ @b2 @ @b@c @ @ @2F @A @2F @B @2F @C @2B @2F @A @2F @B @2F @C @2B @F @3B + + + + + + + @a@b @ @b2 @ @b@c @ @@ @a@b @ @b2 @ @b@c @ @@ @b @2@ @ @2F @A @2F @ @A @ @2F @B @2F @ @B @ @2F @C @2F @ @C @B = + + + + + @ @a@b @ @a@b @ @ @ @b2 @ @b2 @ @ @ @b@c @ @b@c @ @ @ @2F @A @2F @B @2F @C @2B @2F @A @2F @B @2F @C @2B @F @3B + + + + + + + @a@b @ @b2 @ @b@c @ @@ @a@b @ @b2 @ @b@c @ @@ @b @2@ @ @2F @A @B @2F @ @A @B = + @ @a@b @ @ @a@b @ @ @ @ @2F @B 2 @2F @ @B @B + + @ @b2 @ @b2 @ @ @ @ @2F @C @B @ @C @B + + @ @b@c @ @ @ @ @ @2F @A @2F @B @2F @C @2B @2F @A @2F @B @2F @C @2B @F @3B + + + + + + + @a@b @ @b2 @ @b@c @ @@ @a@b @ @b2 @ @b@c @ @@ @b @2@ @3F @A @3F @B @3F @C @A @B @2F @2A @B = + + + @a2@b @ @a@b2 @ @a@b@c @ @ @ @a@b @@ @ @3F @A @3F @B @3F @C @B 2 @2F @B @B + + + + @a@b2 @ @b3 @ @b2@c @ @ @b2 @@ @ @3F @A @3F @B @3F @C @C @B @2C @B + + + + @a@b@c @ @b2@c @ @b@c2 @ @ @ @@ @ @2F @A @2F @B @2F @C @2B @2F @A @2F @B @2F @C @2B @F @3B + + + + + + + @a@b @ @b2 @ @b@c @ @@ @a@b @ @b2 @ @b@c @ @@ @b @2@ 24 CHAPTER 4. CHAPTER 4

@ @2F @A @2F @B @2F @C @C @F @2C K = + + + @ @a@c @ @b@c @ @c2 @ @ @c @@ @ @2F @A @2F @B @2F @C @C = + + @ @a@c @ @b@c @ @c2 @ @ @2F @A @2F @B @2F @C @ @C @ @F @2C @F @ @2C + + + + + @a@c @ @b@c @ @c2 @ @ @ @ @c @@ @c @ @@ @ @2F @A @2F @B @2F @C @C = + + @ @a@c @ @b@c @ @c2 @ @ @2F @A @2F @B @2F @C @2C @2F @A @2F @B @2F @C @2C @F @3C + + + + + + + @a@c @ @b@c @ @c2 @ @@ @a@c @ @b@c @ @c2 @ @@ @c @2@ @ @2F @A @2F @ @A @ @2F @B @2F @ @B @ @2F @C @2F @ @C @C = + + + + + @ @a@c @ @a@c @ @ @ @b@c @ @b@c @ @ @ @c2 @ @c2 @ @ @ @2F @A @2F @B @2F @C @2C @2F @A @2F @B @2F @C @2C @F @3C + + + + + + + @a@c @ @b@c @ @c2 @ @@ @a@c @ @b@c @ @c2 @ @@ @c @2@ @ @2F @A @C @2F @ @A @C = + @ @a@c @ @ @a@c @ @ @ @ @2F @B @C @2F @ @B @C + + @ @b@c @ @ @b@c @ @ @ @ @2F @C 2 @2F @ @C @C + + @ @c2 @ @c2 @ @ @ @2F @A @2F @B @2F @C @2C @2F @A @2F @B @2F @C @2C @F @3C + + + + + + + @a@c @ @b@c @ @c2 @ @@ @a@c @ @b@c @ @c2 @ @@ @c @2@ @3F @A @3F @B @3F @C @A @C @2F @2A @C = + + + @a2@c @ @a@b@c @ @a@c2 @ @ @ @a@c @@ @ @3F @A @3F @B @3F @C @B @C @2F @2B @C + + + + @a@b@c @ @b2@c @ @b@c2 @ @ @ @b@c @@ @ @3F @A @3F @B @3F @C @C 2 @2F @2C @C + + + + @a@c2 @ @b@c2 @ @c3 @ @ @c2 @@ @ @2F @A @2F @B @2F @C @2C @2F @A @2F @B @2F @C @2C @F @3C + + + + + + + @a@c @ @b@c @ @c2 @ @@ @a@c @ @b@c @ @c2 @ @@ @c @2@ 25

Ex. 36 @f (; ) @2F @A @2F @B @2F @C @A @F @2A = + + + @2 @a2 @ @a@b @ @a@c @ @ @a @2 @2F @A @2F @B @2F @C @B @F @2B + + + + @a@b @ @b2 @ @b@c @ @ @b @2 @2F @A @2F @B @2F @C @C @F @2C + + + + @a@c @ @b@c @ @c2 @ @ @c @2 @2F @Aj @Ai @F @2Ai = + @ai@aj @ @ @ai @2

@2f (; ) @2F @A @2F @B @2F @C @A @F @2A = + + + @@ @a2 @ @a@b @ @a@c @ @ @a @@ @2F @A @2F @B @2F @C @B @F @2B + + + + @a@b @ @b2 @ @b@c @ @ @b @@ @2F @A @2F @B @2F @C @C @F @2C + + + + @a@c @ @b@c @ @c2 @ @ @c @@ @2F @Aj @Ai @F @2Ai = + @ai@aj @ @ @ai @@

@f (; ) @2F @Aj @Ai @F @2Ai = + @2 @ai@aj @ @ @ai @2 Ex. 37 We may generalize the above three equations, setting 1 = , 2 = , to yield @2f (; ) @2F @Aj @Ai @F @2Ai = + : @ @ @ai@aj @ @ @ai @ @

This encompasses three separate identities, since we have been assuming that we may switch the order of partial di¤erentiation throughout.

Ex. 38 [Not …nished]

Ex. 39 Begin with cos arccos x = x and di¤erentiate both sides: d [cos arccos x] = 1 dx d (sin arccos x) [arccos x] = 1 dx d 1 [arccos x] = dx (sin arccos x) 26 CHAPTER 4. CHAPTER 4

By examining triangles with unit hypotenuse, we obtain

sin arccos x = 1 x2; p so d 1 [arccos x] = dx p1 x2

Ex. 40: We know that f; g satisfy

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

Di¤erentiating both sides, we obtain

d d [g (f (x))] f (x) + g (f (x)) [f (x)] = 0 dx 0 0 0 dx 0 g00 (f (x)) f 0 (x) f 0 (x) + g0 (f (x)) f 00 (x) = 0

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

Ex. 41: We compute

x f 0 (x) = e x f 00 (x) = e 1 g (x) = 0 x 1 g00 (x) = ; x2

2 1 2x 1 x g00 (f (x)) [f 0 (x)] + g0 (f (x)) f 00 (x) = e + e e2x ex = 1 + 1 = 0; as desired. 27

Ex. 42: We compute 1 f 0 (x) = p1 x2 1 2 3=2 f 00 (x) = 2x 1 x 2 3=2 = x 1 x2 g0 (x) = sin (x) g00 (x) = cos (x) ; So

2 1 2 3=2 g00 (f (x)) [f 0 (x)] + g0 (f (x)) f 00 (x) = cos (arccos x) sin (arccos (x)) x 1 x 1 x2 x xp1 x2 = + 3 1 x2 p 2 1 x x x = + 1 x2 1 x2 = 0; as desired.

Ex. 43: We di¤erentiate both sides of the second-order relationship to obtain d g (f (x)) [f (x)]2 + g (f (x)) f (x) = 0 dx 00 0 0 00 d d d d [g (f (x))] [f (x)]2 + g (f (x)) [f (x)]2 + [g (f (x))] f (x) + g (f (x)) f (x) = 0 dx 00 0 00 dx 0 dx 0 00 0 dx 00 (3) 3 h i (3) g (f (x)) [f 0 (x)] + g00 (f (x)) 2f 0 (x) f 00 (x) + g00 (f (x)) f 0 (x) f 00 (x) + g0 (f (x)) f (x) = 0 (3) 3 (3) g (f (x)) [f 0 (x)] + 3g00 (f (x)) f 0 (x) f 00 (x) + g0 (f (x)) f (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 i J i0 = i ; i0 j j and write out the dependences on unprimed coordinates:

i i0 i J (Z0 (Z)) J (Z) = (Z) (4.2) i0 j j 28 CHAPTER 4. CHAPTER 4

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

@ i i0 @ i J (Z0 (Z)) J (Z) = (Z) @Zk i0 j @Zk j h i @ i i0 J (Z0 (Z)) J (Z) = 0 @Zk i0 j h i @ i i0 i @ i0 J (Z0 (Z)) J (Z) + J (Z0 (Z)) J (Z) = 0; @Zk i0 j i0 @Zk j h i since di¤erentiation passes through the implied summation over i0. Then, using the de…nition of the Jacobian,

i i i i @ @Z @Z 0 @Z @ @Z 0 (Z0 (Z)) (Z) + (Z0 (Z)) (Z) =(4.3) 0 @Zk @Zi0 @Zj @Zi0 @Zk @Zj 2 i k0 i i 2 i0 @ Z @Z @Z 0 @Z @ Z (Z0 (Z)) (Z) (Z) + (Z0 (Z)) (Z) =(4.4) 0; @Zk0 @Zi0 @Zk @Zj @Zi0 @Zk@Zj applying the chain rule to the …rst term, and implying summation over new index k0. Then, if we de…ne the "Hessian" object

2 i i @ Z Jk ;i := (Z0) 0 0 @Zk0 @Zi0

i0 with an analogous de…nition for Jk;i, we write (4.3) concisely:

J i J k0 J i0 + J i J i0 = 0; k0;i0 k j i0 k;j

or, using a renaming of dummy indicex k0 to j0 and a reversing of the order of partial derivatives, J i J i0 J j0 + J i J i0 = 0: i0j0 j k i0 jk

This above tensor relationship represents n3 identities. Ex. 49: Since each J i is constant for a transformation from one a¢ ne i0 coordinate system to another, each second derivative vanishes, and hence each J i = 0. i0j0

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

J i J i0 J j0 + J i J i0 = 0; i0j0 j k i0 jk 29

k Then, letting k0 be arbitrary, we multiply both sides by J , implying summation k0 over k:

J i J i0 J j0 + J i J i0 J k = 0 i0j0 j k i0 jk k0 Jhi J i0 J j0 J k + J i J ii0 J k = 0 i0j0 j k k0 i0 jk k0 but, J j0 J k = j0 ; so k k0 k0 J i J i0 j0 + J i J i0 J k = 0: i0j0 j k0 i0 jk k0

j0 Note that we have = 1 if and only if j0 = k0, so the …rst term is equal to k0 i i0 J J . After re-naming k0 = j0, we obtain i0k0 j

J i J i0 + J i J i0 J k = 0: (4.5) i0j0 j i0 jk j0

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

J i J i0 + J i J i0 J k J j = 0 i0j0 j i0 jk j0 k0 Jhi J i0 J j + J i J i0 Jik J j = 0 i0j0 j k0 i0 jk j0 k0 J i i0 + J i J i0 J k J j = 0 i0j0 k0 i0 jk j0 k0 J i i0 + J i0 J i J k J j = 0: i0j0 k0 jk i0 j0 k0

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

J i i0 + J h0 J i J k J j = 0: i0j0 k0 jk h0 j0 k0

Noting that the …rst term is zero for all i0 = k0, and setting k0 = i0: 6 J i + J h0 J i J k J j = 0: i0j0 jk h0 j0 i0

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

J i + J k0 J i J k J j = 0: i0j0 jk k0 j0 i0

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

J i + J k0 J i J j J k = 0 i0j0 kj k0 j0 i0 30 CHAPTER 4. CHAPTER 4

Ex. 52: Return to

J i J i0 J j0 + J i J i0 = 0; i0j0 j k i0 jk and write out the dependences

i i0 j0 i i0 J (Z0 (Z)) J (Z) J (Z) + J (Z0 (Z)) J (Z) = 0 i0j0 j k i0 jk

2 i i0 j i 2 i0 @ Z @Z @Z 0 @Z @ Z (Z0 (Z)) (Z) (Z) + (Z0 (Z)) (Z) = 0 @Zi0 @Zj0 @Zj @Zk @Zi0 @Zj@Zk

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

2 i i0 j i 2 i0 @ @ Z @Z @Z 0 @Z @ Z 0 = m i j (Z0 (Z)) j (Z) k (Z) + i (Z0 (Z)) j k (Z) @Z "@Z 0 @Z 0 @Z @Z @Z 0 @Z @Z #

2 i i0 j 2 i i0 j 2 i i0 j @ @ Z @Z @Z 0 @ Z @ @Z @Z 0 @ Z @Z @ @Z 0 = (Z0 (Z)) (Z) (Z) + (Z0 (Z)) (Z) (Z) + (Z0 (Z)) (Z) (Z) @Zm @Zi0 @Zj0 @Zj @Zk @Zi0 @Zj0 @Zm @Zj @Zk @Zi0 @Zj0 @Zj @Zm @Zk " # @ @Zi @2Zi0 @Zi @ @2Zi0 + (Z0 (Z)) (Z) + (Z0 (Z)) (Z) @Zm @Zi0 @Zj@Zk @Zi0 @Zm @Zj@Zk " # 3 i m0 i0 j 2 i 2 i0 j 2 i i0 2 j @ Z @Z @Z @Z 0 @ Z @ Z @Z 0 @ Z @Z @ Z 0 = i j m (Z0 (Z)) m (Z) j (Z) k (Z) + i j (Z0 (Z)) m j (Z) k (Z) + i j (Z0 (Z)) j (Z) m k (Z) "@Z 0 @Z 0 @Z 0 @Z # @Z @Z @Z 0 @Z 0 @Z @Z @Z @Z 0 @Z 0 @Z @Z @Z @2Zi @Zm0 @2Zi0 @Zi @3Zi0 + i m (Z0 (Z)) m (Z) j k (Z) + i (Z0 (Z)) j k m (Z) "@Z 0 @Z 0 @Z # @Z @Z @Z 0 @Z @Z @Z = J i J m0 J i0 J j0 + J i J i0 J j0 + J i J i0 J j0 + J i J m0 J i0 + J i J i0 ; i0j0m0 m j k i0j0 jm k i0j0 j km i0m0 m jk i0 jkm

so, setting k0 = m0 as a dummy index:

J i J i0 J j0 J k0 + J i J j0 J i0 + J i J i0 J j0 + J i J i0 J k0 + J i J i0 = 0: (4.6) i0j0k0 j k m i0j0 k jm i0j0 j km i0k0 jk m i0 jkm

Then, multiply both sides by J m , implying summation over m: m0

J i J i0 J j0 J k0 J m + J i J j0 J i0 J m + J i J i0 J j0 J m + J i J i0 J k0 J m + J i J i0 J m = 0 i0j0k0 j k m m0 i0j0 k jm m0 i0j0 j km m0 i0k0 jk m m0 i0 jkm m0 J i J i0 J j0 k0 + J i J j0 J i0 J m + J i J i0 J j0 J m + J i J i0 k0 + J i J i0 J m = 0; i0j0k0 j k m0 i0j0 k jm m0 i0j0 j km m0 i0k0 jk m0 i0 jkm m0

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

J i J i0 J j0 +J i J j0 J i0 J m +J i J i0 J j0 J m +J i J i0 +J i J i0 J m = 0 (4.7) i0j0k0 j k i0j0 k jm k0 i0j0 j km k0 i0k0 jk i0 jkm k0 31

Next, in an analogous manner, multiply both sides by J k for arbitrary m : m0 0

J i J i0 J j0 J k + J i J j0 J i0 J mJ k + J i J i0 J j0 J mJ k + J i J i0 J k + J i J i0 J mJ k = 0 i0j0k0 j k m0 i0j0 k jm k0 m0 i0j0 j km k0 m0 i0k0 jk m0 i0 jkm k0 m0 J i J i0 j0 + J i J j0 J i0 J mJ k + J i J i0 J j0 J mJ k + J i J i0 J k + J i J i0 J mJ k = 0; i0j0k0 j m0 i0j0 k jm k0 m0 i0j0 j km k0 m0 i0k0 jk m0 i0 jkm k0 m0

rename the dummy index h0 = j0 in all but the …rst term:

J i J i0 j0 +J i J h0 J i0 J mJ k +J i J i0 J h0 J mJ k +J i J i0 J k +J i J i0 J mJ k = 0; i0j0k0 j m0 i0h0 k jm k0 m0 i0h0 j km k0 m0 i0k0 jk m0 i0 jkm k0 m0

then, as in the previous exercises, set m0 = j0:

J i J i0 +J i J h0 J i0 J mJ k +J i J i0 J h0 J mJ k +J i J i0 J k +J i J i0 J mJ k = 0; i0j0k0 j i0h0 k jm k0 j0 i0h0 j km k0 j0 i0k0 jk j0 i0 jkm k0 j0 (4.8)

Finally, multiply both sides by J j : m0

J i J i0 J j + J i J h0 J i0 J mJ k J j + J i J i0 J h0 J mJ k J j + J i J i0 J k J j + J i J i0 J mJ k J j = 0 i0j0k0 j m0 i0h0 k jm k0 j0 m0 i0h0 j km k0 j0 m0 i0k0 jk j0 m0 i0 jkm k0 j0 m0 J i i0 + J i J h0 J i0 J mJ k J j + J i J i0 J h0 J mJ k J j + J i J i0 J k J j + J i J i0 J mJ k J j = 0; i0j0k0 m0 i0h0 k jm k0 j0 m0 i0h0 j km k0 j0 m0 i0k0 jk j0 m0 i0 jkm k0 j0 m0

rename the dummy index i0 to g0 in all but the …rst term:

J i i0 +J i J h0 J g0 J mJ k J j +J i J g0 J h0 J mJ k J j +J i J g0 J k J j +J i J g0 J mJ k J j = 0: i0j0k0 m0 g0h0 k jm k0 j0 m0 g0h0 j km k0 j0 m0 g0k0 jk j0 m0 g0 jkm k0 j0 m0

Then, set m0 = i0:

J i +J i J h0 J g0 J mJ k J j +J i J g0 J h0 J mJ k J j +J i J g0 J k J j +J i J g0 J mJ k J j = 0: i0j0k0 g0h0 k jm k0 j0 i0 g0h0 j km k0 j0 i0 g0k0 jk j0 i0 g0 jkm k0 j0 i0 (4.9)

Ex. 53: We have the following relationship

00 Z Z0 Z (Z) = Z; or for each i,

i 00 i Z Z0 Z (Z) = Z . Di¤erentiating both sides with respect to Zj, we obtain

@ Zi (Z (Z (Z))) = i : @Zj 0 00 j 32 CHAPTER 4. CHAPTER 4

Then, we apply the chain rule twice:

i @Z @ i [Z0 (Z00 (Z))] = j @Zi0 @Zj i i0 @Z @Z @ i00 i Z (Z) = j @Zi0 @Zi00 @Zj i h i0 i00i @Z @Z @Z i = j; @Zi0 @Zi00 @Zj or J i J i0 J i00 = i : i0 i00 j j Chapter 5

Chapter 5

i Ex. 61: j. i Ex. 62: Assume U is an arbitrary nontrivial linear combination U Zi of coordinate bases Zi. Since U U > 0 and i j U U = ZijU U ; or in matrix notation U U = U T ZU, this condition implies Z = Zij is positive de…nite.

Ex. 63:

V = pV V k k i j = ZijV V q

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

T 1 T y = Z x T 1 T = x Z T 1 = x Z ;

33 34 CHAPTER 5. CHAPTER 5

1 since Z is symmetric. Then, since Z is positive de…nite, note 0 < yT Zy T 1 1 = x Z ZZ x T 1 = x Z x.

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

Ex. 65:

i ik Z Zj = Z Zk Zj ik = Z Zkj by de…nition. But, ik i Z Zkj = j; so we have i i Z Zj = j

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

Ex. 68: Use the de…nition

i ij Z = Z Zj i ij ZikZ = ZikZ Zj j = kZj = Zk:

Thus,

i Zk = ZikZ i = ZkiZ ;

since Zik is symmetric.

Ex. 69: Examine

i j in j ZkiZ Z = ZkiZ Zn Z n j = k n j = k; 35

n j i j since k n = 1 i¤ k = n and n = j, or by transitivity, i¤ k = j. Thus, Z Z ij determines the matrix inverse of Zki, which must be Z by uniqueness of matrix inverse. Ex. 70: Because the inverse of a matrix is uniquely determined, we have that Zi are uniquely determined.

Ex. 71:

ij i j Z Zjk = Z Z Zjk i = Z Zk; from 5.17 i = k from 5.16.

Ex. 72: We compute

1 1 1 Z Z2 = i j j 3 3 1 1 = i j j j 3 3 1 1 = i j cos j 2 3 k k k k 3 3 k k 1 1 1 = (2) (1) 12 3 2 3 = 0;

1 thus, Z ; Z2 are orthogonal. We further compute

2 1 4 Z Z1 = i+ j i 3 3 1 4 = i 2 + i j cos 3 k k 3 k k k k 3 1 4 1 = (4) + (2) (1) 3 3 2 = 0;

2 so Z ; Z1 are orthogonal. 36 CHAPTER 5. CHAPTER 5

Ex. 73:

1 1 1 Z Z1 = i j i 3 3 1 1 = i 2 i j cos 3 k k 3 k k k k 3 4 2 1 = 3 3 2 = 1: 2 1 4 Z Z2 = i+ j j 3 3 1 4 = i j cos + j 2 3 k k k k 3 3 k k 2 1 4 = + 3 2 3 = 1:

Ex. 74: 1 1 1 4 Z1 Z2 = i j i+ j 3 3 3 3 1 4 1 4 = i 2 + i j cos + i j cos j 2 9 k k 9 k k k k 3 9 k k k k 3 9 k k 4 4 1 4 = + + 9 9 9 9 3 = 9 1 = : 3

Ex. 75: Let R denote the position vector. We compute

@R (Z) Z3 = Z3 @R (r; ; z) = @z for cylindrical coordinates

R (r; ; z + h) R (r; ; z) = lim : h 0 h ! 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 vector pointing in the z direction.

Ex. 76: The computations of the diagonal elements Z11 and Z22 are the same as for polar coordinates; moreover the zero o¤-diagonal entries Z12, Z21 follow 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 are zero. Morover, since Z3 is of unit length; we have Z33 = Z3 Z3 = 1. Thus, we have 1 0 0 2 Zij = 0 r 0 . 20 0 13 4 5 ij Now, since Z is de…ned to be the inverse of Zij, we may easily compute

1 0 0 ij 2 Z = 0 r 0 ; 20 0 13 4 5 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

3 3j Z = Z Zj

= 0Z1 + 0Z2 + 1Z3

= Z3 38 CHAPTER 5. CHAPTER 5 Chapter 6

Chapter 6

Ex. 87: Look at

ij i0 j0 ij i0 j0 j k Z J J Zj k = Z J J ZjkJ J i j 0 0 i j j0 k0

by the tensor property of Zjk

= ZijZ J i0 J k jk i k0 = i J i0 J k k i k0 = J i0 J k k k0 = i0 ; k0

ij i0 j0 so, in linear algebra terms, we have that Z Ji Jj is the matrix inverse of Zj0k0 . ij i0 j0 i0j0 By uniqueness of matrix inverses, this forces Z Ji Jj = Z , as desired.

Ex. 88: Let Z,Z0 be two coordinate systems. Write the unprimed coordinates in terms of the primed coordinates

Z = Z (Z0) :

Then, @F (Z) @F (Z (Z )) = 0 @Zi0 @Zi0 @F @Zi = @Zi @Zi0 @F = J i ; @Zi i0

@F so @Zi is a covariant tensor.

39 40 CHAPTER 6. CHAPTER 6

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

@T S = i ij Zj

@Ti0 Si j = 0 0 Zj0 so

@Ti0 Si j = 0 0 @Zj0 @ i = TiJi ; @Zj0 0 since T is a covariant tensor,

@ i = Ti (Z0 (Z)) Ji (Z0) Zj0 0 i j @T @Z i i = Ji + TiJi j @Zj @Zj0 0 0 0 j i i = SijJ J + TiJ j0 i0 i0j0 j i = SijJ Ji 6 j0 0

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

@Ti Zj is not a covariant tensor.

Ex. 90: Compute

@Ti0 @Tj0 Si j = 0 0 @Zj0 @Zi0 @Ti j i i @Tj i j i = J Ji + TiJi j Ji J + TiJj i @Zj j0 0 0 0 @Zi 0 j0 0 0 by the above, interchanging the rolls of i0; j0 for the second term:

@Ti j i i @Tj j i i = J Ji + TiJi j J Ji + TiJi j ; @Zj j0 0 0 0 @Zi j0 0 0 0 41

since J i = J i , j0i0 i0j0

@Ti j i i @Tj j i i = J Ji + TiJi j J Ji TiJi j @Zj j0 0 0 0 @Zi j0 0 0 0 @Ti @Tj j i = J Ji @Zj @Zi j0 0 i j = SijJ J ; i0 j0

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

Ex. 91: Put @T i Sij = ; @Zj so

@T i0 Si0j0 = @Zj0

@ i i0 = T Ji ; @Zj0 h i since T is a contravariant tensor,

@ i i0 i @ i0 = T (Z0 (Z)) Ji + T Ji (Z (Z0)) @Zj0 @Zj0 i i0 j h i @T j0 i0 i @Ji @Z = Jj Ji + T @Zj @Zj @Zj0 = SijJ i0 J j0 + T iJ i0 J j i j ij j0 = SijJ i0 J j0 6 i j except in the trivial case.

Ex. 92: We have @Z k = Zk i ; ij @Zj so in primed coordinates,

k0 k0 @Zi0 i j = Z 0 0 @Zj0

k0 @Zi i j i = Z Ji J + ZiJi j @Zj 0 j0 0 0 42 CHAPTER 6. CHAPTER 6

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

k k0 @Zi i j i Z Jk Ji J + ZiJi j @Zj 0 j0 0 0 since Zk is a contravariant tensor,

k @Zi k0 i j k k0 i = Z Jk Ji J + Z Zi Jk Ji j @Zj 0 j0 0 0 k @Zi k0 i j k k0 i = Z Jk Ji J + i Jk Ji j @Zj 0 j0 0 0

k @Zi k0 i j k0 i = Z Jk Ji J + Ji Ji j @Zj 0 j0 0 0

k @Zi k0 i j k k0 i j = Z Jk Ji J = ijJk Ji J 6 @Zj 0 j0 0 j0 except in the trivial case.

Ex. 93: Compute

@Ti0j0 @ i j = TijJi Jj @Zk0 @Zk0 0 0 @ h i ji @ i j i @ j = [Tij] Ji Jj + Tij Ji Jj + TijJi Jj @Zk0 0 0 @Zk0 0 0 0 @Zk0 0 @ i j i j i j h i = [Tij] Ji Jj + TijJi k Jj + TijJi Jj k @Zk0 0 0 0 0 0 0 0 0 @ i j i j i j = [Tij (Z (Z0))] Ji Jj + TijJi k Jj + TijJi Jj k @Zk0 0 0 0 0 0 0 0 0 k @Tij @Z i j i j i j = Ji Jj + TijJi k Jj + TijJi Jj k @Zk @Zk0 0 0 0 0 0 0 0 0 @Tij k i j i j i j = Jk Ji J + TijJi k J + TijJi J : @Zk 0 0 j0 0 0 j0 0 j0k0 43

Thus, from 5.66,

k0 1 k0m0 @Zm0i0 @Zm0j0 @Zi0j0 i j = Z + 0 0 2 @Zj0 @Zi0 @Zm0

1 km k0 m0 @Zm0i0 @Zm0j0 @Zi0j0 = Z Jk Jm + 2 @Zj0 @Zi0 @Zm0 1 km k0 m0 @Zmi j m i m i m i = Z Jk Jm ( J Jm Ji + ZmiJm j Ji + ZmiJm Ji j 2 @Zj j0 0 0 0 0 0 0 0 0 @Zmj i m j m j m j + Ji Jm J + ZmjJm i J + ZmjJm J @Zi 0 0 j0 0 0 j0 0 i0j0 @Zij m i j i j i j Jm Ji J ZijJi m J ZijJi J ) @Zm 0 0 j0 0 0 j0 0 j0m0

1 km k0 m0 @Zmi j m i @Zmj i m j @Zij m i j = Z Jk Jm J Jm Ji + Ji Jm J Jm Ji J 2 @Zj j0 0 0 @Zi 0 0 j0 @Zm 0 0 j0 1 km k0 m0 m i m i + Z J J ZmiJ J + ZmiJ J 2 k m m0j0 i0 m0 i0j0

1 km k0 m0 m j m j + Z Jk Jm ZmjJm i J + ZmjJm J 2 0 0 j0 0 i0j0 1 km k0 m0 i j i j Z Jk Jm ZijJi m J + ZijJi J 2 0 0 j0 0 j0m0 1 km k0 j i @Zmi @Zmj @Zij = Z Jk J Ji + 2 j0 0 @Zj @Zi @Zm 1 km k0 m0 m i m i + Z J J ZmiJ J + ZmiJ J 2 k m m0j0 i0 m0 i0j0

1 km k0 m0 m j m j + Z Jk Jm ZmjJm i J + ZmjJm J 2 0 0 j0 0 i0j0 1 km k0 m0 i j i j Z Jk Jm ZijJi m J + ZijJi J 2 0 0 j0 0 j0m0 = k J k0 J j J i ij k j0 i0

1 km k0 m0 m i m i + Z ZmiJ J J J + J J 2 k m m0j0 i0 m0 i0j0

1 km k0 m0 m j m j + Z ZmjJk Jm Jm i J + Jm J 2 0 0 j0 0 i0j0 1 km k0 m0 i j i j Z ZijJk Jm Ji m J + Ji J 2 0 0 j0 0 j0m0 = k J k0 J j J i ij k j0 i0 1 + kJ k0 J m0 J m J i + J m J i 2 i k m m0j0 i0 m0 i0j0

1 k k0 m0 m j m j + j Jk Jm Jm i J + Jm J 2 0 0 j0 0 i0j0 1 km k0 m0 i j i j Z ZijJk Jm Ji m J + Ji J 2 0 0 j0 0 j0m0 = k J k0 J j J i ij k j0 i0 1 + J k0 J m0 J m J i + J m J i 2 i m m0j0 i0 m0 i0j0

1 k0 m0 m j m j + Jj Jm Jm i J + Jm J 2 0 0 j0 0 i0j0 1 km k0 m0 i j i j Z ZijJk Jm Ji m J + Ji J 2 0 0 j0 0 j0m0 = k J k0 J j J i ij k j0 i0 1 1 + J k0 J m0 J m J i + J k0 J m0 J m J i 2 i m m0j0 i0 2 i m m0 i0j0

1 k0 m0 m j 1 k0 m0 m j + Jj Jm Jm i J + Jj Jm Jm J 2 0 0 j0 2 0 i0j0

1 km k0 m0 i j 1 km k0 m0 i j Z ZijJk Jm Ji m J Z ZijJk Jm Ji J 2 0 0 j0 2 0 j0m0 44 CHAPTER 6. CHAPTER 6

= k J k0 J j J i ij k j0 i0 1 1 + J k0 J i J m0 J m + J k0 J i 2 i i0 m m0j0 2 i i0j0

1 k0 j m0 m 1 k0 j + Jj J Jm Jm i + Jj J 2 j0 0 0 2 i0j0

1 km k0 m0 i j 1 km k0 m0 i j Z ZijJk Jm Ji m J Z ZijJk Jm Ji J 2 0 0 j0 2 0 j0m0

k k0 j i k0 i 1 k0 m0 m 1 k0 m0 m 1 km j k0 m0 i 1 km i k0 m0 j = ijJk J Ji + Ji Ji j + i Jm Jm j + j Jm Jm i Z ZijJ Jk Jm Ji m Z ZijJi Jk Jm J j0 0 0 0 2 0 0 0 2 0 0 0 2 j0 0 0 2 0 j0m0 = k J k0 J j J i + J k0 J i + 0: ij k j0 i0 i i0j0

Thus, we have k0 = k J k0 J j J i + J k0 J i : i0j0 ij k j0 i0 i i0j0

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

i Assume Ti = 0. Then, since Ti = TiJ , we have Ti = 0. 0 i0 0

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

i0 i0 Ji = Ai ; so the "Hessian" object i0 Jij = 0;

i0 since Ai is assumed to be constant with respect to Z. This means that all but i @T @T 0 @Ti j k the …rst terms in the above computations of i0 , , 0 0 , and even 0 are Zj0 @Zj0 @Zk0 i0j0 zero, 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 show that the sum of …nitely many tensors is a tensor. We must show that for any constant c, i cAjk 45

is a tensor. Compute cAi0 = cAi J i0 J j J k ; j0k0 jk i j0 k0

i i since Ajk is a tensor. Thus, by the above, we have if each A (n)jk is a tensor, then the sum N i cnA (n)jk n=1 X is a tensor. Thus, linear combinations of tensors are tensors.

Ex. 97: We have that

ij i kj SiT = SikT i kj = kSiT ;

i kj which is a tensor, since both k and SiT are tensors by the previous section and by the fact that the product of two tensors is a tensor.

i i Ex. 98: i = n by the summation convention. Thus, i returns the dimension of the ambient space.

Ex. 99: We have k Vij = Vij Zk; so

m k m Vij Z = V Zk Z ij k m = Vij k m = Vij so substituting m = k, we have an expression for the components

k k V = Vij Z ij

So,

k0 k0 V = Vi j Z i0j0 0 0 i j k k0 = VijJi J Z Jk ; 0 j0

k since both Vij; Z are tensors

k i j k0 = Vij Z Ji J Jk 0 j0 46 CHAPTER 6. CHAPTER 6

by linearity = V kJ i J j J k0 ij i0 j0 k as desired.

Ex. 100: Fix a coordinate system Zi

T ij = T ijJ iJ jJ k; k k i j k so

T ijJ i0 J j0 J k = T ijJ iJ jJ kJ i0 J j0 J k k i j k0 k i j k i j k0 = T iji0 j0 k k i j k0 = T i0j0 ; k0 as desired. Chapter 7

Chapter 7

47 48 CHAPTER 7. CHAPTER 7 Chapter 8

Chapter 8

49 50 CHAPTER 8. CHAPTER 8 Chapter 9

Chapter 9

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

ijk A = e ai1aj2ak3:

Note that switching the roles of 1; 2 in the above equation yields

ijk ijk e ai2aj1ak3 = e aj1ai2ak3 jik = e ai1aj2ak3 ijk = e ai1aj2ak3 = A

Generalizing, we let (r; s; t) be a permutation of (1; 2; 3). We may then see that ijk rst A = e e airajsakt

(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

ijk 3!A = e erstairajsakt permutations (r;s;tX)

But, erst = 0 for (r; s; t) that is not a permutation; hence we may sum over all 0 r; s; t 3, and apply the Einstein summation convention: ijk 3!A = e erstairajsakt; or 1 A = eijke a a a 3! rst ir js kt

51 52 CHAPTER 9. CHAPTER 9

We may similarly show that for aij, we have 1 A = e ai1aj2ak3: 3! ijk

Ex. 184: We have

123 r s t 123 s r t rst a2a2a3 = srt a2a2a3 after index renaming

123 r s t = srt a2a2a3 = 123arasat : rst 2 2 3

Since 123arasat = 123arasat ; rst 2 2 3 rst 2 2 3 we need 123 r s t rst a2a2a3 = 0:

132 s r t The result for srt a2a3a2 follows similarly.

Ex. 185: Note

123 123 = e erst = 1 erst = erst; rst so 123 r s t r s t rst a1a2a3 = ersta1a2a3 = A:

Also,

132 r s t 132 t r s rst a2a3a1 = rst a1a2a3 132 r s t = str a1a2a3 123 r s t = str a1a2a3 r s t = estra a a 1 2 3 = A: 53

[Note: Is there an error somewhere - should this be A?] Ex. 186: De…ne 1 Air = eijkersta a : 2! js tk

We check that @A = Air: @air

Check @A 1 @ (a a a ) = eijkerst ir js kt @alu 3! @alu 1 @a @a @a = eijkerst ir a a + a js a + a a kt 3! @a js kt ir @a kt ir js @a lu lu lu 1 = eijkerst lua a + a l ua + a a l u 3! i r js kt ir j s kt ir js k t 1 h i = eijklerstua a + a eijkl erstua + a a eijkl erstu 3! i r js kt ir j s kt ir js k t 1 h i = eljkeusta a + a eilkeruta + a a eijlersu 3! js kt ir kt ir js 1 = eljkeusta a + eilkeruta a + eijlersua a 3! js kt ir kt ir js 1 = 3eljkeusta a ; 3! js kt after an index renaming, 1 = eljkeusta a 2! js kt = Alu as desired. Similarly, if we de…ne 1 A = e e ajsatk; ir 2! ijk rst we have @A = A @air ir

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

i Zij = j; 54 CHAPTER 9. CHAPTER 9

Z = Z j j = I j j = 1:

Thus, pZ = 1:

In polar coordinates, 1 0 [Z ] = ; ij 0 r2 so 1 0 Z = 0 r2

2 = r ;

hence pZ = r:

In spherical coordinates,

1 0 0 2 [Zij] = 0 r 0 ; 20 0 r2 sin2 3 4 5 so 1 0 0 Z = 0 r2 0

0 0 r2 sin2

4 2 = r sin ;

thus, pZ = r2 sin : 55

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

i 1 @ ij @F i F = pZZ r r pZ @Zi @Zj 1 @ @F @ @F @ @F = r2 sin (1) + r2 sin r2 + r2 sin r2 sin2 r2 sin @r @r @ @ @ @ 1 @ @F @ @F @ @F = r2 sin + r4 sin + r4 sin3 r2 sin @r @r @ @ @ @ 1 @ @F @ @F @ @F = sin r2 + r4 sin + r4 sin3 r2 sin @r @r @ @ @ @ 1 @F @2F @F @2F @2F = sin 2r + r2 + r4 cos + sin + r4 sin3 r2 sin @r @r2 @ @2 @2 1 @F @2F r2 @F @2F @2F = 2r + r2 + cos + sin + r2 sin2 r2 @r @r2 sin @ @2 @2 2 @F @2F @F @2F @2F = + + r2 cot + r2 + r2 sin2 r @r @r2 @ @2 @2 @2F 2 @F @2F @F @2F = + + r2 + r2 cot + r2 sin2 @r2 r @r @2 @ @2

Ex. 189: We compute, for cylindrical coordinates

i 1 @ ij @F i F = pZZ r r pZ @Zi @Zj 1 @ @F @ @F @ @F = r (1) + r r2 + r (1) r @r @r @ @ @z @z 1 @ @F @2F @2F = r + r3 + r r @r @r @2 @z2 1 @F @2F @2F @2F = + r + r3 + r r @r @r2 @2 @z2 @2F 1 @F @2F @2F = + + r2 + : @r2 r @r @2 @z2 56 CHAPTER 9. CHAPTER 9 Part II

Part II

Chapter 10

Ex. 213: Note that

i i Z Z = S Z (S Zj) j i = S (S Zj) Z i = (S Zj) S Z = i ; 6 j since

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

EARLIER ATTEMPT:

i i Z Zj = j i i S Z Zj S = j i i S S Z Zj = j i i S Zj Z = j i i S Zj Z = Zj Z ; which forces

S Zj = Zj

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

59 60 CHAPTER 10. CHAPTER 10

Ex. 214: We have i i T = T Z :

Now,

i i T Zi = T Z Zi

= 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; since N N =1.

Ex. 216: We compute

i j i j Pj Pk = N NjN Nk i = N (1) Nk i = Pk; as desired.

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

(V T) S = (V (V S ) S ) S = V S (V S ) S S = V S (V S ) = V S V S = 0: 61

i i Ex. 218: Similarly to 216, we have, given de…nition Tj = N Nj

i j i j Tj Tk = N NjN Nk i = N Nk i = Tk:

[Note: This seems like the exact same problem - do we mean to de…ne i i Tj = T Tj?] Ex. 219: We have [Note that this implies 213 additionally]

i i i N Nj + Z Zj = j:

Contract both sides with Ni : i i i N NjNi + Z Zj Ni = jNi i i NiN Nj + NiZ Zj = Nj i NiN Nj + 0 = Nj i NiN Nj = Nj;

i where the third line follows from NiZ = 0. Now, this holds for all Nj, for which at least one is nonzero (we cannot have the normal vector be zero). Hence, we have i NiN = 1; as desired.

Ex. 220: Using similar manipulations of indices to the earlier discussion of the Levy-Civita symbols, we derive 1 1 ijkT tT s = ijkT sT t 4 rst j k 4 rts j k 1 = ijkT sT t; 4 rst j k so

i 1 ijk s t 1 ijk t s N Nr = T T T T 4 rst j k 4 rst j k 1 = 2 ijkT sT t 4 rst j k 1 = ijkT sT t: 2 rst j k 62 CHAPTER 10. CHAPTER 10

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

@S 0 J 0 = : @S

Ex. 222: From before, we have

@Z ij = Z l + Z l : @Zk li jk lj ik

From the analogous de…nitions of S ; we have

@S = S + S @S compute

1 @S @S @S S ! ! + ! 2 @S @S @S! 1 ! = S S! + S + S! + S S + S 2 ! ! ! ! 1 ! = S S! + S + S! + S S S 2 ! ! ! ! 1 ! ! ! ! = + S S + + S S S S S S 2 ! ! ! 1 = 2 2

= ; as desired.

Ex. 223: Assume the ambient space is re¤ered to a¢ ne coordiantes. We have @Zi = Z + i Z Zj Zk i @S jk @Zi = Z + 0; i @S

i since jk = 0 in a¢ ne coordinates. 63

Ex. 224 [Still Working]

Ex. 225 We compute, given Z1 (; ) = R Z2 (; ) = Z3 (; ) =

@Zi Zi = @S @Z1 @Z1 @S1 @S2 = @Z2 @Z2 2 @S1 @S2 3 @Z3 @Z3 6 @S1 @S2 7 40 0 5 = 1 0 ; 20 13 4 5 then, note that since 0 0 0 1 0 1 0 1 0 = ; 0 0 1 2 3 0 1 0 1 4 5 we have 0 1 0 Z = i 0 0 1 0 0 N i = 1 0 203 213 4i5 j4 k 5 = 0 1 0

0 0 1

= i 1 = 0 203 4 5 i S1 = Z1Zi 2 = Z1 Z2 = R cos cos i + R cos sin j R sin k i S2 = Z2Zi 3 = Z2 Z3 = R sin sin i + R sin cos j 64 CHAPTER 10. CHAPTER 10

R2 cos2 cos2 + R2 cos2 sin2 + R2 sin2 R2 cos cos sin sin + R2 cos sin sin cos S = R2 cos cos sin sin + R2 cos sin sin cos R2 sin2 sin2 + R2 sin2 cos2 R2 0 = [ASK - should this be the same as when the ambient coordinates are Cart.?] 0 R2 sin2 2 R 0 S = 2 2 0 R sin R2 0 pS = 0 R2 sin2 s

= R2 sin :

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

1 = r 22 1 = r sin2 33 1 2 = 2 = 12 21 r 2 = sin cos 33 1 3 = 3 = 13 31 r 2 2 23 = 32 = cot :

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

@Zi = Z + i Z Zj Zk; i @S jk 65 we compute

@Zi 1 = Z1 1 + i Z1ZjZk 11 i @S1 jk 1 1 @Z2 = Z1 1 + 2 Z1ZjZk 2 @S1 jk 2 1 1 @Z2 = Z1 1 + 2 Z1Z2Z2 2 @S1 22 2 1 1 @Z2 = 1 @S1 = 0 1 1 i 1 j k 21 = 12 = 0 + jkZ Z2 Z1 2 1 j k = jkZ2 Z2 Z1 2 1 3 2 = 32Z2 Z2 Z1 = cot (1) (1) (1) = cot 1 i 1 j k 22 = jkZ Z2 Z2 2 1 j k = jkZ2 Z2 Z2 2 1 3 3 = 33Z2 Z2 Z2 = sin cos

2 i 2 j k 11 = jkZ Z1 Z1 3 2 j k = jkZ3 Z1 Z1 3 2 2 2 = 22Z3 Z1 Z1 = 0 2 2 i 2 j k 21 = 12 = jkZ Z2 Z1 3 2 j k = jkZ3 Z2 Z1 3 2 3 2 = 32Z3 Z2 Z1 = 0 2 i 2 j k 22 = jkZ Z2 Z2 3 2 3 3 = 33Z3 Z2 Z2 = 0;

i @Z (note @S vanishes in each computation). 66 CHAPTER 10. CHAPTER 10

Ex. 226: We have

2 2 (x0 (s)) + (y0 (s)) = 1; q since this is an arc-length parametrization. Thus,

y (s) N i = 0 x0 (s) 2 2 S = (x0 (s)) + (y0 (s)) = 1 S = 1 pS = 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)

2 2 2 r0 (s) + r (s) 0 (s) = 1, q so the results follow similarly to the above cases. Chapter 11

Chapter 11

Ex. 229: This follows similarly as with the ambient covariant derviative, using the tensor properties of T (S) given in surface coordinates, and using the 0 analogous Jacobians J .

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 the ambient case.

Ex. 231: We compute, using 1 @S @S @S = S ! ! + ! 2 @S @S @S!

@S S = S S r @S @S = S S @S @S 1 ! @S! @S! @S 1 ! @S! @S! @S = S + S S + S @S 2 @S @S @S! 2 @S @S @S! @S 1 @S @S @S 1 @S @S @S = ! ! + ! ! ! + ! @S 2 @S @S @S! 2 @S @S @S! @S 1 @S @S @S 1 @S @S @S = + + @S 2 @S @S @S 2 @S @S @S @S 1 @S 1 @S = @S 2 @S 2 @S = 0:

Similarly, we may show that in the contravariant case,

S = 0: r 67 68 CHAPTER 11. CHAPTER 11

For the Levy-Civita symbols, note

" = pSe 1 " = e pS

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 S = r " = " = 0: r r Ex. 232: Commutativity with contraction follows exactly as in the ambient case.

Ex. 233: We compute, using

2 R 0 S = 2 2 ; 0 R sin the following:

1 @ @F F = pSS r r pS @S @S 1 @ @F 1 @ @F = R2 sin S1 + R2 sin S2 R2 sin @ @S R2 sin @ @S 1 @ @F 1 @ @F = R2 sin S11 + R2 sin S22 R2 sin @ @ R2 sin @ @ 1 @ 2 2 @F 1 @ 2 2 2 @F = R sin R + R sin R sin R2 sin @ @ R2 sin @ @ 1 @ @F 1 @ 1 @F = sin + R2 sin @ @ R2 sin @ sin @ 1 @ @F 1 @2F = sin + : R2 sin @ @ R2 sin2 @2

Ex. 234: For the surface of a cylinder, note R 2 0 S = 0 1 pS = R; 69

1 @ @F F = pSS r r pS @S @S 1 @ @F 1 @ @F = RS11 + RS22 R @ @ R @z @z 1 @2F @2F = + : R2 @2 @z2

Ex. 235: Note

2 (R + r cos ) 0 S = 0 r 2 pS = r (R + r cos ) ; and compute

1 @ @F F = pSS r r pS @S @S 1 @ @F = r (R + r cos ) S1 r (R + r cos ) @ @S 1 @ @F + r (R + r cos ) S2 r (R + r cos ) @ @S 1 @ 2 @F = r (R + r cos )(R + r cos ) r (R + r cos ) @ @ 1 @ @F + r (R + r cos ) r 2 r (R + r cos ) @ @ 1 @2F 1 @ @F = + (R + r cos ) : R r 2 @2 r2 (R + r cos ) @ @ ( + cos )

Ex. 236: We have

2 r (z) 0 S = 1 0 2 " 1+r0(z) #

2 pS = r (z) 1 + r0 (z) ; q 70 CHAPTER 11. CHAPTER 11

Thus,

1 @ @F F = pSS r r pS @S @S 1 @ @F 1 @ @F = pSS11 + pSS22 2 @ @ 2 @z @z r (z) 1 + r0 (z) r (z) 1 + r0 (z)

q 1 @ 2 q 2 @F = r (z) 1 + r0 (z) r (z) 2 @ @ r (z) 1 + r0 (z) q q 1 @ 2 1 @F + r (z) 1 + r0 (z) 2 2 @z 1 + r0 (z) @z ! r (z) 1 + r0 (z) q q 1 @ r (z) @F 1 @2F = + 2 2 : 2 @z 0 2 @z 1 r (z) @ r (z) 1 + r0 (z) 1 + r0 (z) q @q A

Ex. 237: These were computed earlier.

Ex. 238: We compute:

@Zi j k Zi = Z Zk r @S ij m @Zi @Z j k = Z Zk @Zm @S ij @Zi m j k = Z Z Zk @Zm ij k m j k = ZkZ Z Zk im ij m k j k = Z Z Zk im ij = 0; after index renaming. The contravariant case follows similarly. Also, since Zij = Zi Zj; we have Zij = 0 r by the product rule. Similarly,

ij Z = 0: r

[Levy-Civita Symbols to come] 71

Ex. 239: Begin with equation 10.41:

i NiN = 1;

and compute take the surface covariant derivative of both sides:

i 0 = NiN r i i = N iN + Ni N r j r ik i = N Zij NkZ + Ni N r r j ik i = N Zij NkZ + Ni N ; r r

by the metrinilic property,

j k i = N Nk + Ni N r j r k i = N Nk + Ni N r k r i = Nk N + Ni N r i r = 2Ni N ; r

after index renaming. Thus,

i Ni N = 0: r 72 CHAPTER 11. CHAPTER 11

Ex. 240: We compute 1 "ijk" Z N = "ijk" Z " " ZmZn j k j 2 kmn 1 = "ijk" " " Z ZmZn 2 kmn j 1 = ijk Z ZmZn 2 kmn j 1 = ijk Z ZmZn 2 mkn j 1 = ijk Z ZmZn 2 mnk j 1 = 2ij Z ZmZn 2 mn j ij m n = mn Zj Z Z = i j j i Z ZmZn m n m n j = i j Z ZmZn j i Z ZmZn i j Z ZmZn + j i Z ZmZn m n j m n j m n j m n j = Z Zi Zj Z Zj Zi Z Zi Zj + Z Zj Zi j j j j = Zi Zi Zi + Zi = Zi Zi Zi + Zi

[Some factors of 2 needed?]

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

i k i T = Z kT : r j r j

Thus, k u = Z ku r r and k u = Z ku: r r

Thus,

k u = Z ku r r r r k k = Z ku + Z ku r k r k rm r = B N ku + Z Z m ku: r r r k k mn = B N ku + Z Zn Z m ku r r r k k mn = B N ku + Z Z S Z m ku r n r r 73

Now, set = and contract:

k k mn u = B N ku + Z Z S Z m ku r r r n r r k k mn = B N ku + Z Z Z m ku r n r r k k mn = B N ku + Z Z Z m ku: r n r r

Now, k k k N Nn + Z Zn = n; so k k k Z Z = N Nn: n n

We substitute in the above:

k k k mn u = B N ku + N Nn Z m ku r r r n r r k km k mn = B N ku + Z m ku N NnZ m ku kr rm r m k r r = B N ku + m u N N m ku; r r r r r or after renaming dummy indices,

i i i j u = B N iu + i u N N i ju; r r r r r r r or i j i i N N i ju = i u u + B N iu r r r r r r r Ex. 242: Let Zi (s) be the parametrization of the line normal to the surface, emanating from point i Z0. Note that we have dZi Zi (h) Zi (0) = lim 0 = N i: ds h 0 h ! also, compute d dZi d2Zi dZi dZ Z = Z + i (Z (s)) ds ds i ds2 i ds ds d2Zi dZi @Z dZk = Z + i ds2 i ds @Zk ds d2Zi dZi dZk = Z + n Z ; ds2 i ds ik n ds 74 CHAPTER 11. CHAPTER 11

so at s = 0;

i 2 i d dZ d Z i n k Zi s=0 = Zi + N ZnN ds ds j ds2 ik d2Zi = Z + N iN kn Z ds2 i ik n d2Zi = Z + N jN kn Z ds2 i jk i d2Zi = + N jN ki Z ds2 jk i

Now, examine the LHS of the above. Since

d dZi Z ds ds i represents the second derivative of a line, the LHS vanishes. Thus,

d2Zi = N jN ki ; ds2 jk

Then, de…ne F (s) = u (Z (s)) ; so @u dZi F (s) = (Z (s)) (s) 0 @Zi ds d @u dZi @u d dZi F (s) = (Z (s)) (s) + (Z (s)) (s) 00 ds @Zi ds @Zi ds ds @2u dZi dZj @u d2Zi = (Z (s)) (s) (s) + (Z (s)) (s) : @Zi@Zj ds ds @Zi ds2 at s = 0 : @2u @u d2Zi F (0) = (Z (0)) N iN j + (Z (0)) (0) 00 @Zi@Zj @Zi ds2 2 i @ i j d Z = [ iu] N N + iu (0) @Zj r r ds2 now, note 75

@ iu k j iu = r ku r r @Zj ijr @ iu k r = j iu + ku; @Zj r r ijr so

2 i k i j d Z F 00 (0) = j iu + ku N N + iu (0) r r ijr r ds2 2 i i j k i j d Z = j iuN N + kuN N + iu (0) r r ijr r ds2 i j i j k j k i = N N j iu + N N ku N N iu r r ijr jkr i j = N N j iu; r r thus 2 @ u i j = F 00 (0) = N N i ju @n2 r r after renaming indices. 76 CHAPTER 11. CHAPTER 11 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 78 CHAPTER 12. CHAPTER 12

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 the fact that both R and S are symmetric.

Ex. 249: Note

1 G = R S RS S 2 1 = R S R ; 2 so

G = R S R = R R = R R = 0;

since R = R by de…nition.

Ex. 250: We compute

( ) T = ( ) T S r r r r r r " r r = 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 partial derivatives. Now, we consider the covariant case:

i i i ( ) T = T T r r r r r r r r i i @ T @ T = r T i r T i @S @S r r ! i i @ T @ T = r r @S @S @ @T i @ @T i = + Zki T m + Zki T m @S @S km @S @S km @2T i @ @2T i @ = + Zki T m Zki T m @S @S @S km @S @S @S km @ @ = Zki T m Zki T m @S km @S km @Zk @i @Tm = i T m + Zk km + Zki @S km @S km @S @Zk @i @T m i T m Zk km T m Zki @S km @S km @S = 0 [not sure yet] 80 CHAPTER 12. CHAPTER 12

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 ! 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 @R ! @!; @!; ! @ ;! @ ; ! = + ;! ; + ; ;! @S" " @S @S ! " @S @S ! +:::

Ex. 254: Compute

1 1 R " " R = " " 1212 " " 4 4 S 1 R = (2) (2) 1212 4 S R = 1212 S = K;

as desired. 82 CHAPTER 12. CHAPTER 12

Ex. 255: Compute

1 K (S S S S ) = " " R (S S S S ) 4 1 1 = " " S S R " " S S 4 4 1 1 = " " S S R + " " S S 4 4 1 1 = R + R 4 4 1 1 = (2) R + (2) R 4 4 = R :

Ex. 256: Note 1 1 ! R = R! S S ; 2 2 so

1 1 ! R = R! S S 2 2 1 = K" " S! S 2 ! 1 = K 2 = K

Ex. 257: This follows since

S = S ; r r hence NB = NB ; or B = 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, 2 B = : j j

Now,

B B := C :

In linear algebra terms, we have

2 C = B then,

2 B B = tr B = 1 + 2;

2 2 2 where 1; 2 are the eigenvalues of B . But, since 1 = and 2 = ( ) = 2 by the properties of eigenvalues, we have

2 B B = 2 = 2 B j j by the above.

Ex. 259: We compute, given

z b r (z) = a cosh a e(z b)=a + e(b z)=a = a 2 a a = e(z b)=a + e(b z)=a 2 2

1 (z b)=a 1 (b z)=a r0 (z) = e e 2 2

1 1 r (z) = e(z b)=a + e(b z)=a: 00 2a 2a 84 CHAPTER 12. CHAPTER 12

Compute

2 2 1 (z b)=a 1 (b z)=a a (z b)=a a (b z)=a 1 (z b)=a 1 (b z)=a r00 (z) r (z) r0 (z) = e + e e + e e e 2a 2a 2 2 2 2 1 2(z b)=a 1 2(b z)=a 1 2(z b)=a 1 2(b z)=a = e e e 1 + e 4 4 4 4 = 1

Thus,

2 r00 (z) r0 (z) r0 (z) 1 = 0

2 r00 (z) r0 (z) r0 (z) 1 B = 2 r (z) 1 + r0 (z) = 0; q

as desired.

Ex. 260: We have

dR V = (S (t)) dt @R dS = @S dt

= S V

= V S

as desired. 85

Ex. 261: Compute dV A = dt d = [V S ] dt dV dS = S + V (S (t)) dt dt dV @S dS = S + V dt @S dt dV @S = S + V V dt @S

dV = S + V V S + S dt r

dV = S + V V S + V V S dt r

V = S + V V S t r V = S + NV V B t V = S + NB V V ; t as desired.

Ex. 262: We de…ne T dT = + V T ! V ! T t dt ! !:

Ex. 263: Look at T 0 dT 0 0 = 0 + V 0 T ! V ! T 0 t dt ! 0 0 !

d 0 0 ! ! 0 = T J (S (t)) J (S0 (t)) + V !T J V T! J dt 0 0 0 dT 0 @J 0 0 @J dS 0 0 dS 0 ! ! 0 = J J + T J + T J + V !T J V T! J dt 0 @S dt 0 @S 0 dt 0 0 dT 0 0 0 0 0 ! ! 0 = J J + T J V J + T J V J + V !T J V T! J dt 0 0 0 0 0 0 = ::: ??? T 0 = J J t 0 86 CHAPTER 12. CHAPTER 12

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 of time? In that case, would this be a moving surface, and then we would require the derivative in Part III?

Ex. 267: Compute

S dS (S (t)) ! = V S! t dt

@S dS ! = V S! @S dt ! ! = S + S! V V S! r ! ! = S V + V S! V S! r

= S V r = 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

1 0 B = R ; 0 0 so

K = B j j 1 = 0 R = 0: 87

Ex. 271: For a cone, we have

cot 0 B = r ; 0 0 which has determinant zero. Thus,

K = 0:

Ex. 272: For a sphere, we have

1 0 B = R ; 0 1 R so 1 1 K = B = j j R R 1 = R2

Ex. 273: For a torus, we have

cos 0 B = R+r cos ; 0 1 r so cos K = r (R + r cos )

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

1 2 0 r(z)p1+r0(z) B = r00(z) ; 2 0 2 3=2 3 (1+r0(z) ) 4 5 so 1 r (z) K 00 = 3=2 2 2 r (z) 1 + r0 (z) 1 + r0 (z) q r00 (z) = 2 2 r (z) 1 + r0 (z) 88 CHAPTER 12. CHAPTER 12

Ex. 275: We integrate 1 KdS = dS R2 ZS ZS 1 = dS R2 ZS 1 = 4R2 R2 = 4

Ex. 276: We integrate

cos 2 2 cos dS = pSdd r (R + r cos ) r (R + r cos ) ZS Z0 Z0 2 2 cos = r (R + r cos ) dd 0 0 r (R + r cos ) Z 2 Z 2 = cos dd Z0 Z0 = 2 (sin (2) sin (0)) = 0: Part III

Part III

Ex. 291: We have

( ) = arccot (At cot ) ;

= arccot (At cot ) cot = At cot 1 cot = cot At 1 () = arccot cot At

@S (t; S ) J = 0 t @t @ 1 = arccot cot @t At 1 cot = 1 2 2 1 + A2t2 cot At cot = 2 1 2 At + A cot A cot = A2t2 + cot2

@ J 0 = arccot (At cot ) t @t A cot = 1 + A2t2 cot2

Ex. 292:

@Zi A cos V = = @t 0 92

Ex. 293:

@ A cos i @Z @t pcos2 +A2t2 sin2 V i = = 2 3 @t @ A sin 6 @t pcos2 +A2t2 sin2 7 6 7 2A42t sin2 A cos 5 3 2pcos2 +A2t2 sin2 = 2A2t sin2 A sin 2 3 3 2pcos2 +A2t2 sin2 4 A3t sin2 cos 5 3 pcos2 +A2t2 sin2 = A3t sin3 : 2 3 3 pcos2 +A2t2 sin2 4 5

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

Ex. 295: First, note our parametrization:

At cos Zi ( ) = sin Then, compute the shift tensors: @Zi Zi = @S At sin = ; cos so Z = 1 sin cos ; i At since we need i Zi Z = :

Thus,

i V = V Zi A cos = 1 sin cos At 0 = 1 sin : t 93

Ex. 296: Recall A cos 2 2 2 2 Zi () = pcos +A t sin ; 2 A sin 3 pcos2 +A2t2 sin2 4 5 and note that @ 2 cos sin + 2A2t2 sin cos cos2 + A2t2 sin2 = @ 2 cos2 + A2t2 sin2 p A2t2 sin cos cos sin = p cos2 + A2t2 sin2 so p @Zi Zi = @S 2 2 2 2 A2t2 sin cos cos sin cos + A t sin ( A sin ) A cos 1 2 2 2 2 = pcos +A t sin ; 2 2 2 2 2 2 2 2 A2t2 sin cos cos sin cos + A t sin 2p cos + A t sin (A cos ) A sin 3 pcos2 +A2t2 sin2 4 p 5

The result is V = 0, since the motion of a particle on the surface with i constant is normal to tangent space; hence the projection V = V Zi = 0. Ex. 297: We see that V is not a tensor with respect to changes in ambient coordinates.

Ex. 298: We have i V = V Zi; so V Zj = V j; con…rming that V i is a tensor (V is an invariant and Zj is a tensor).

i Ex. 299: Both V and Zi are tensors with regard to ambient coordinate i changes; hence the contraction V = V Zi is a tensor with regard to ambient coordinate changes.

Ex. 300: This was done before.

Ex. 301: Both parametrizations use Cartesian ambient coordinates. Thus, i0 the Jacobians Ji are the identity. We compute

i i0 i i0 i i V Ji + Z Ji Jt = V + Z Jt A cos A cot 1 sin = + At : 0 A2t2 + cot2 cos 94

[To be continued]

Ex. 302: Write

V 0 = V i0 Z 0 i0 = V iJ i0 + Zi J i0 J Z J j J 0 i i t j i0 = V iJ i0 Z J j J 0 + Zi J i0 J Z J j J 0 i j i0 i t j i0 i 0 j i 0 j = V Zj J i + Z Jt Zj J i

j 0 j 0 = V Zj J + Z Jt Zj J

0 0 = V J + Jt J

0 0 = V J + J Jt ; as desired.

Ex. 303: Let the unprimed coordinates denote the …rst parametrization. Then, note d J 0 = (arccot (At cot )) d 1 = At csc2 1 + A2t2 cot2 At csc2 = 1 + A2t2 cot2

1 At csc2 A cot At csc2 V J 0 + J 0 J = sin + t t 1 + A2t2 cot2 A2t2 + cot2 1 + A2t2 cot2 =

[perhaps I am using the wrong Jacobians]

Ex. 304: We have t cos Zi = ; sin so d d S = (t cos ) i + (sin ) j; d d 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 cos N i = ; t sin and Ni = cos t sin ; since our ambient space is in Cartesian coordinates. Thus,

i C = V Ni cos = cos t sin 0 = cos2 :

Ex. 305: As before, compute

d cos d sin S = i + j d cos2 + t2 sin2 ! d cos2 + t2 sin2 ! 2 2 p1 2 2 p2 t sin cos cos sin 2 2 2 t sin cos cos sin = 2 cos + t sin ( sin ) cos i+ cos + t sin (cos ) sin j cos2 + t2 sin cos2 + t2 sin2 ! cos2 + t2 sin2 ! p p sin t2 sin cos2 cos2 sin p cos t2 sin2 cos cos sin2 p = i+ j; 2 3 2 3 cos2 + t2 sin cos2 + t2 sin2 ! cos2 + t2 sin cos2 + t2 sin2 ! p p p p so

cos cos sin2 +t2 cos sin2 sin cos2 sin +t2 cos2 sin Ni = 3 + 3 ; pcos2 +t2 sin2 (cos2 +t2 sin2 ) 2 pcos2 +t2 sin2 (cos2 +t2 sin2 ) 2 h i and

i C = V Ni t sin2 cos 3 cos cos sin2 +t2 cos sin2 sin cos2 sin +t2 cos2 sin pcos2 +A2t2 sin2 = 3 + 3 3 pcos2 +t2 sin2 (cos2 +t2 sin2 ) 2 pcos2 +t2 sin2 (cos2 +t2 sin2 ) 2 t sin 2 3 3 h i pcos2 +A2t2 sin2 = 4 5 96

[Note: Shouldn’tboth be 1 by geometric considerations?]

Ex. 306: We have

0 0 0 V = V J + J Jt

@T (t; S0) @T (t; S) = + J t; @t @t t r so compute

@T (t; S0) 0 _ T (t; S0) = V (t; S0) T r @t r 0

@T (t; S) 0 0 = + J t V TJ ; @t t r r since T has the tensor property. But, r 0 0 0 0 V = V J + J Jt ; so @T (t; S) _ 0 0 T (t; S0) = + Jt t V J + J Jt TJ r @t r r a0 @T (t; S) = + J t V + J T @t t r t r @T (t; S) = + J t V T J T @t t r r t r @T (t; S) = V T @t r @T (t; S) = V T; @t r so _ T does not depend on changes in suface coordinates. r Ex. 307: This follows from the sum rule for partial derivatives and the covariant derivative

Ex. 308: This follows from the product rule for partial derivatives and the covariant 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 have i S = Z Zi:

Ex. 313: Begin with

i i _ R = V V Z Zi r and j V = V Zj ; so

i j i _ R = V V Z Z Zi r j i j i i = V V j N Nj Zi i i j i = V V + V N Nj Zi j i = V N NjZi; as desired.

Ex. 314: Compute

d d NB dS = S dS dt dt r ZS ZS _ = ( S ) dS CB S dS S r r S r Z Z @ ( S ) = r V S dS CB S dS S @t r r S r Z Z @ ( S ) i = r V Z S dS CB S dS @t i r r r ZS ZS

[ask about integral problems]

Ex. 315: By the above,

NB dS ZS is constant. Since our surface is of genus zero, we may smoothly append our surface 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 equal to 2 NdS ZS for all t, since for a sphere,

2 B = : R

But, S NdS = 0 (our surface is closed), so we have that S NB dS vanishes. R R Ex. 316: Need to show: d @F F d = d + CF dS; dt @t Z Z ZS i.e. d A2 b(t;x) F (x; y) dydx = CF dS; dt ZA1 Z0 ZS @F since @t = 0. Clearly, C is zero on all of S except for the portion given by the graph of b. Let B denote the surface given by this graph. Then,:

CF dS = CF dB ZS ZB

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

S = i + bxj, so 1 N = ( bxi + j) ; 2 1 + bx and p 0 V i = ; bt so

i C = V Ni b = t : 2 1 + bx p 99

2 Now, at each t, our surface has line element 1 + bx, so A2 p CF dB = btF (x; b (t; x)) dx ZB ZA1 A2 d b(t;x) = F (x; y) dydx; dt ZA1 Z0 by FTC, d A2 b(t;x) = F (x; y) dydx dt ZA1 Z0 Ex. 317: We have @T i0 (t; S ) U i0 = 0 @t @ = T i (t; S) J i0 (Z (t; S)) @t i @ @ = T i (t; S (t; S )) J i0 + T i J i0 (Z (t; S)) @t 0 i @t i @T i @T i @ @J i0 @ = + S J i0 + T i i Z (t; S) @t @S @t i @Zj @t @T i @T i @S @Zj @Zj @S = + J i0 + T iJ i0 + @t @S @t i ji @t @S @t @T i = U i + J J i0 + T iJ i0 V j + Zj J @S t i ji t @T i = U iJ i0 + J J i0 + T iJ i0 V j + T iJ i0 Zj J ; i @S t i ji ji t which is the desired result.

Ex. 318: This follows similarly to the above.

Ex. 319: Write @T 0 (t; S ) U 0 = 0 @t @ = T (t; S) J 0 (t; S) @t @ @ = T (t; S) J 0 + T J 0 (t; S) @t @t @T @T @S @J 0 @J 0 @S = + J 0 + T + @t @S @t @t @S @t ! @T = U J 0 + J J 0 + T J 0 + T J 0 J : @S t t t 100

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

Ex. 321: Note

i0 i i0 V T = V T J r r i @T i = V (t; S) + i T jZj J i0 @S jk i @T i = V j0 Z (t; S) + i T kZj J i0 j0 @S jk i @T i = V jJ j0 + Zj J j0 J (t; S) + i T kZj J i0 j j t @S jk i @T i @T i = V jJ j0 + V jJ j0 i T kZj + Zj J j0 J + Zj J j0 J i T kZj J i0 j @S j jk j t @S j t jk i @T i @T i = V jJ j0 J i0 + V jJ j0 i T kZj J i0 + Zj J j0 J J i0 + Zj J j0 J i T kZj J i0 ; j @S i j jk i j t @S i j t jk i so given our work above,

i0 i i @T i0 @T i0 @T i0 i i0 j i i0 j V T = J + J J + T J V + T J Z J @t r @t i @S t i ji ji t @T i @T i V jJ j0 J i0 + V jJ j0 i T kZj J i0 + Zj J j0 J J i0 + Zj J j0 J i T kZj J i0 j @S i j jk i j t @S i j t jk i

[not …nished] Chapter 13

Chapter 16

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

i _ T = _ T Zi r r i i = _ T Zi + T _ Zi; r r by commutativity with contraction and the product rule,

i = T Zi; since the second term would be zero by the metrinilic property.

Ex. 326: Compute @Z @Z (Z (t)) i = i @t @t @Z @Zj = i @Zj @t k j = ijZkV j k = V ijZk; as desired.

Ex. 327: Write i T = TiZ ; so i i @Ti i @Z i _ TiZ = _ T = Z + Ti V TiZ ; r r @t @t r 101 102 CHAPTER 13. CHAPTER 16

but

@Zi @Zi @Zj = @t @Zj @t = i ZkV j jk = i V jZk; jk so we have @Ti j k _ Ti = V Ti V Tk; r @t r ij after index renaming in the second term of the above expression.

i Ex. 328: We know _ T is invariant, and since _ T = _ T Zi and Zi is a tensor, by the quotient law,r _ T i must be a tensor. Anr argumentr using changes in coordinates would followr similarly to the covariant derivative computations.

i Ex. 329: This also follows from the fact that _ T = _ TiZ and by the quotient law. An argument using changes in coordinatesr r would also follow similarly.

Ex. 330: Put i Sj = Tj Zi: then, clearly, i _ Sj = _ T Zi: r r j

Dot both sides with Zk:

k i k _ Sj Z = _ T Zi Z r r j k k _ Sj Z = _ T : r r j

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 this result to arbitrary indices.

i j Ex. 332: Assume S ;Ti are arbitrary tensors. Put

j i j U = S Ti : 103

Now, compute

@ SiT j i j i i j n j i j _ S T = V S T + V S T r i @t r i nk i j @U j n j i j = V U + V S T ; @t r nk i since both the partial derivatives and the covariant surface derivatives commute with contraction.

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

Ex. 334: Note i k i T = Z kT ; r j r j and compute

i i @Tj (t; Z (t; S)) i k i m k m i _ T (t; S) = V T (t; Z (t; S)) + V T (t; Z (t; S)) V T (t; Z (t; S)) r j @t r j km j kj m i i k @Tj @Tj @Z k i k i m k m i = + V Z kT + V T V T @t @Zk @t r j km j kj m i i @Tj @Tj k k i k i m k m i = + V V Z kT + V T V T @t @Zk r j km j kj m i i @Tj @Tj i m m i k k i = + k + kmTj kjTm V V Z kTj @t @Z ! r i @Tj i k k i = + kT V V Z kT @t r j r j i @Tj k k i = + V V Z kT @t r j i @Tj k i = + CN kT ; @t r j

k k where the last step follows from the observation that V V Z is the normal component.

@Zi Ex. 335: Note that @t = 0; and the second term of the above is also zero by the metrinilic property for covariant derivatives. Hence, _ Zi = 0: the other results follow similarly. r 104 CHAPTER 13. CHAPTER 16

Ex. 336: Compute @S (Z (t; S)) @S @Zi = @t @Zi @t @S = V i @Zi @S @S = V i @S @Zi @2R = Z V i @S @S i @S = Z V i @S i = [not sure]

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

Ex. 338: Compute

(V S + CN) = V S + V S + CN+C N r r r r r = V S + V NB + N C + C B S r r = V S + V NB + N C CB S : r r

Ex. 339: Use

@T @S _ T = S + T V T S V T NB r @t @t r

@T = S + T V S + V NB + N C CB S V T S V T NB @t r r r

@T = S + V T S + V T NB + T N C CB T S V T S V T NB @t r r r

@T = S + T V S + T CN T CB S V T S @t r r r

@T = S + T V S + T CN T CB S V T S : @t r r r

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 = _ S r @S ! ! ! ! ! = V !S ( V CB ) S! V CB S ! @t r r r @S ! ! ! ! = V S! + CB S! V S ! + CB S ! @t r r @S = V V + CB + CB @t r r @S = V V + 2CB ; @t r r so @S = V + V 2CB : @t r r

The contravariant case follows similarly.

Ex. 342: First, examine

@S = SS ; @S using the properties of the cofactor matrix. Then, compute

@S @S @S = @t @S @t

= SS ( V + V 2CB ) r r = S V + V 2CB r r = 2S ( V CB ) : r

Then,

@pS 1 @S = @t 2pS @t

= pS ( V CB ) ; r 106 CHAPTER 13. CHAPTER 16 and

1 @ pS 1 @pS = @t 2 @t pS

1 = pS ( V CB ) S r 1 = ( V CB ) : pS r

Ex. 343: Note " = pSe ; so

@" @pS @e = e + pS @t @t @t @pS = e ; @t since e does not depend on t.

= pS V CB e r

= " V CB : r

Ex. 344: Note 1 " = pS e ; so

1 @" @ pS @e = e + pS @t @t @t 1 @ pS = e @t 1 = V CB e pS r = " V CB : r 107

Ex. 345: Simply write

_ " = _ " S S r r = _ " S S ; r by the metrinilic property, = 0; by the above.

Ex. 346: Use Gauss’Theorema Egregium:

K = B j j 1 = "" B B ; 2 so

2 _ K = _ " " B B + " _ " B B + " " _ B B + " " B _ B r r r r r = " " _ B B + " " B _ B ; r r since the …rst two terms vanish,

= " " C + CB B B + " " B C + CB B r r r r = " " B C + C" " B B B + " " B C + C" " B B B r r r r = B C + B C + B B B C + B B B C r r r r = B C + B C + B B B C + B B B C r r r r = B C + B C + B B B C + B B B C r r r r B C + B C + B B B C + B B B C r r r r = B C + B C + B B B C + B B B C r r r r B C + B C + B B B C + B B B C r r r r

= 2B C 2B C + 2B B B C 2B B B C r r r r

= 2 B C B C + B B B B B B C r r r r = 2 B C B C + B KC ; r r r r 108 CHAPTER 13. CHAPTER 16

which gives us the desired result [Note: I believe the last equality follows from Gauss’Theorema Egregium]. Chapter 14

Chapter 17

Ex. 347: Note that u is an invariant with respect to ambient indices, and thus @u iu = i . Write r @Z @ @ @u iu = @tr @t @Zi @ @u = @Zi @t @ = i u; r @t

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

Ex. 348: First, compute, given polar coordinates

@u iu = r @Zi @ J0(r) = @r pJ1() 2 @ J0(r) 3 @ pJ1()

4 J1(r) 5 = pJ1() " 0 #

Now, since for polar coordinates,

ij 1 0 Z = 2 ; 0 r

109 110 CHAPTER 14. CHAPTER 17 so

i ij u = Z ju r r J1(r) 1 0 pJ () = 2 1 0 r 0 " # J1(r) = pJ1() : " 0 # thus, 2 2 i J1 (r) iu u = 2 : r r J1 ()

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

2 i iu u = r r

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

@x i @t V = @y @t a cos = b sin

With respect to the Cartesian basis,

S = (1 + at) sin i + (1 + bt) cos j

1 (1 + bt) cos Ni = ; 2 2 (1 + at) sin (1 + at) cos2 + (1 + bt) sin2 q at t = 0, cos N = i sin

i C = V Ni = a cos2 + b sin2 ; 111

thus, by the Hadamard formula,

2 2 = a cos2 + b sin2 d 1 Z0 2 2 = a cos2 + b sin2 d Z0 2 2 = a cos2 + b 1 cos2 d Z0 2 2 = a cos2 + b cos2 d 2b2 Z0 2 2 = (a + b) cos2 d 2b2 Z0 2 1 2 = (a + b) + sin 2 2b2 2 4 0 2 = (a + b) 2b2 = 2 (a + b) 2b2 = 2 (a + b) :

Since = 2, we have the desired result.

Ex. 349: [Not sure]

Ex. 350: Want to show:

i i 1 = u1Ni u Ni u1 dS: r r ZS Dirichlet: i 1 = C iu udS r r ZS Neumann:

1 =

Ex. 354: We have

_ C + 2V C + B V V = B r r _ V + V V C C CV B = 0: r r r 112 CHAPTER 14. CHAPTER 17

Write

i V = V Zi

and

i C = V Ni;

Begin with the second, and contract with S :

_ V S + V V S C CS CV B S = 0 r r r

_ (V S ) V _ S + V (V S ) V V S C CS CV B S = 0 r r r r r

_ (V S ) V N C + V (V S ) V V B N C CS CV B S = 0: r r r r

Then, manipulate the …rst:

_ C + 2V C + B V V = B ; r r and multiply by N:

_ CN + 2V CN + B V V N = B N r r _ (CN) C _ N + 2V CN + B V V N = B N r r r _ (CN) C ( S C) + 2V CN + B V V N = B N r r r _ (CN) + CS C + 2V CN + B V V N = B N r r r _ (CN) + 2V CN + V V B N+C CS = B N: r r r 113

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

_ (V S ) + _ (CN) + V CN + V (V S ) CV B S = B N r r r r _ V + V CN + V (V S ) CV B S = B N r r r _ V + V (CN) V C N + V (V S ) CV B S = B N r r r r _ V+V V V C N CV B S = B N r r r i _ V+V V V C N Zi CV B S = B N r r r i _ V+V V V C N Zi CV B S = B N [note the metrinilic property] r r r i _ V+V V + V CZ B Zi CV B S = B N r r i _ V+V V + V CB Z Zi CV B S = B N r r _ V+V V + CV B S CV B S = B N r r _ V+V V = B N r r