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
57
Part II
10 Chapter 10
59
11 Chapter 11
67
12 Chapter 12
77
III
89
Part III
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
1
Chapter 1
Chapter 1
Ex. 1: We have x = 2x0 , y = 2y 0 . Thus
F 0 (x0 ; y 0 )
= F (2x0 ; 2y 0 )
2
=
(2x0 ) e2y
=
4 (x0 ) e
Ex. 2: Note that the above implies x =
@F 0 0 0
(x ; y )
@x0
@F
(x; y)
@x
Thus,
@F 0
0 0
@x0 (x ; y )
2y 0
2
0
0
1
2 x,
:
y = 12 y. We check
0
2y 0
=
8 (x0 ) e
=
8
=
2( 1 )y
1
2
x e
2
4xey
=
2xey :
= 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
=
a
0
j0
=
0
;
b
where each of the above vectors is taken to be with respect to the standard
basis for R2 . Thus, given point (x; y) in standard coordinates, we have x = ax0 ,
y = by 0 , where (x0 ; y 0 ) is the same point in our new coordinate system. Now,
let T (x; y) be a di¤erentiable function. Then,
rT =
@F
@F
(x; y);
(x; y)
@x
@y
3
4
CHAPTER 1. CHAPTER 1
in standard coordinates
=
@F 0 0 @x0
@F 0 0 @y 0
(x
;
y
)
(x ; y )
(x;
y);
(x; y)
@x0
@x
@y 0
@y
(think of x0 as a function of x).
@F 0 0 1 @F 0 0 1
(x ; y ) ; 0 (x ; y )
@x0
a @y
b
1
@F
1 @F
p
;
a2 + 0 @x0 b2 + 0 @y 0
1 @F
1 @F
p
; p 0 0 0;
j j @y
i0 i0 @x0
=
=
=
as desired.
Ex. 4: Assume
x0
y0
=
a
cos
+
b
sin
sin
cos
x
:
y
Then,
x0
y0
= a + (cos ) x (sin ) y
= b + (sin ) x + (cos ) y
Also,
x0
y0
a
b
=
cos
sin
sin
cos
x
y
=
cos
sin
sin
cos
cos
+sin2
sin
cos2 +sin2
=
cos2
=
cos
sin
sin
cos
x
y
1
x0
y0
a
b
sin
+sin2
cos
cos2 +sin2
0
cos2
x
y0
x0
y0
a
b
Thus,
x = (cos ) (x0 a) + (sin ) (y 0 b)
y =
(sin ) (x0 a) + (cos ) (y 0 b) :
a
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
sin
cos
= cos i+ sin j
cos
sin
=
sin
cos
=
sin i+ cos j
i0
=
j0
1
0
0
1
Now, we have, given a function F , we compute
@F
@F
(x; y) i +
(x; y) j
@x
@y
=
=
=
=
=
=
@F 0 0 @x0
@F
@y 0
@F 0 0 @x0
@F
(x ; y )
(x; y) + 0 (x0 ; y 0 )
(x; y) i +
(x ; y )
(x; y) + 0 (x0 ; y 0
0
@x
@x
@y
@x
@x0
@y
@y
@F 0 0
@F
@F 0 0
@F
(x ; y ) cos + 0 (x0 ; y 0 ) sin
i+
(x ; y ) sin + 0 (x0 ; y 0 ) cos
j
@x0
@y
@x0
@y
@F 0 0
@F 0 0
@F
@F
(x ; y ) cos i
(x ; y ) sin j + 0 (x0 ; y 0 ) sin i + 0 (x0 ; y 0 ) cos j
@x0
@x0
@y
@y
@F 0 0
@F
(x ; y ) (cos i sin j) + 0 (x0 ; y 0 ) (sin i + cos j)
0
@x
@y
@F 0 0
@F 0 0
cos
sin
(x ; y ) i j
+ 0 (x ; y ) i j
sin
cos
@x0
@y
@F 0 0 0 @F 0 0 0
(x ; y ) i + 0 (x ; y ) j
@x0
@y
[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:
p
f (h) =
F (P )2 + h2
h
f 0 (h) = p
F (P )2 + h2
dF (p)
= f 0 (0)
dl
Ex. 8: We have
f (h)
=
f 0 (h)
=
=
1
AP
h
1
2
(AP
1
(AP
h)
2
h)
so
dF (p)
dl
= f 0 (0)
=
7
1
(AP )2
8
CHAPTER 2. CHAPTER 2
Ex. 9: Let
have
denote the measure of angle OP P . By the Law of Sines, we
sin (F (P ))
AP h
=
=
sin
OP
=
sin (
)
OA
sin ( )
OA
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 ))
h
cos (F (P )) sin (F (P ))]
The, from the …rst equation, we have
((OA) h + (OP ) (AP
OP [sin (F (P )) cos (F (P )) cos (F (P )) sin (F
sin (F (P ))
=
AP h
(OA) h
(OA) h sin (F (P )) = (OP ) (AP h) [sin (F (P )) cos (F (P )) cos (F
(OA) h tan (F (P )) = (OP ) (AP h) sin (F (P )) (OP ) (AP h) co
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,
f (h)
=
f 0 (h)
=
=
(OP ) (AP h) sin (F (P ))
(OA) h + (OP ) (AP h) cos (F (P ))
(OP ) sin (F (P )) [(OA) h + (OP ) (AP h) cos (F (P ))]
arctan
(OP ) sin (F (P )) (AP
[(OA) h + (OP ) (AP
(OP ) sin (F (P )) [(OA) h + (OP ) (AP
h) cos (F (P ))]
[(OA) h + (OP ) (AP
h) [(OA)
2
h) cos (F (P ))]
(OP ) sin (F (P )) (AP
2
h) cos (F (P ))] + [(OP ) (AP
h) [(OA)
h) sin (F (P
9
dF (p)
dl
= f 0 (0)
=
=
=
=
(OP ) sin (F (P )) [(OP ) (AP ) cos (F (P ))]
(OP ) sin (F (P )) (AP ) [(OA)
2
(OP ) (AP ) cos (F (P ))]
2
[(OP ) (AP ) cos (F (P ))] + [(OP ) (AP ) sin (F (P ))]
(OP ) sin (F (P )) (OP ) (AP ) cos (F (P )) (OP ) sin (F (P )) (AP ) (OA) + (OP ) sin (F (P )) (AP ) (OP )
2
2
(OP ) (AP )
cos2 (F (P )) + sin2 (F (P ))
(OP ) sin (F (P )) (AP ) (OA)
2
2
(OP ) (AP )
(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,
F (P )
=
F (P )
=
and we have F (P ) = F (P ), so
dF (p)
dl
1
(AB) d
2
1
(AB) d;
2
= 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)
=
f 0 (h)
=
1
(AB) (F (P ) + h)
2
1
AB
2
dF (p)
dl
= f 0 (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]
kR ( + h)
2
R ( )k = 1 + 1
2 cos (h)
by the Law of Cosines. So,
kR ( + h)
kR ( + h)
2
R ( )k
=
2 2 cos (h)
p
R ( )k =
2 2 cos (h)
s
=
=
=
=
2
s
2
r
2 cos 2
2 sin2
4 sin2
h
2
h
2 sin :
2
Ex. 16:
2
2 sin h2
= lim
h!0
h!0
h
lim
1
2
cos h2
;
1
h
2
h
2
11
by L’Hospital’s rule,
=
lim cos
h!0
=
h
2
1:
Ex. 17:
2 sin h2
h!0
h
lim
=
=
lim
h!0
lim
sin h2
h
2
sin 0 +
h!0
h
2
h
2
sin (0)
= 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 ( )
R ( ) R00 ( )
Now, let
=
=
0
1:
be the angle between R ( ), R00 ( ). We thus have
kR ( )k kR00 ( )k cos
kR00 ( )k cos
=
=
1
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 triangles
that kR0 ( + h) R0 ( )k = kR ( + h) R ( )k : Thus,
kR0 ( + h) R0 ( )k
h!0
h
kR ( + h) R ( )k
= lim
h!0
h
= kR0 ( )k
= 1:
kR00 ( )k =
lim
12
CHAPTER 2. CHAPTER 2
Thus, R00 ( ) is of unit length. We then have
cos =
which implies that
R ( ).
=
1;
, or that R00 ( ) points in the opposite direction as
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
0 1 0 1 0 1
0
0
1
@0A ; @1A ; @0A
0
1
0
respectively, and a vector V connecting the origin to a point with coordinates
0 1
x0
@ y0 A
z0
can be expreessed by the linear combination
V =x0 i+y0 j+z0 k:
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 P P 0 = y; hence by the properties of
right triangles, we have
x
r
y
r
=
cos
=
sin ;
or
x = r cos
y = r sin :
Ex. 22: We see from the above that
x2 + y 2
= r2 cos2 + r2 sin2
= r2 ;
hence we may solve for r (taken to be non-negative):
p
r = x2 + y 2 :
Also,
y
x
=
=
so
r sin
r cos
tan ;
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
q
p
x2 + y 2 + z 2 =
r2 sin2 cos2 + r2 sin2 sin2 + r2 cos2
q
= r sin2 cos2 + sin2 sin2 + cos2
q
= r sin2 cos2 + sin2 + cos2
p
= r sin2 + cos2
= r;
so
r (x; y; z) =
Also, z=r = cos ; so
(x; y; z)
=
=
p
x2 + y 2 + z 2 :
z
r
z
arccos p
:
2
x + y2 + z2
arccos
16
CHAPTER 3. CHAPTER 3
Finally,
y
x
=
=
so
r sin sin
r sin cos
tan ;
y
(x; y; z) = arctan :
x
Chapter 4
Chapter 4
Ex. 27:
det J
=
det
"
x
x2 +y 2
y
x2 +y 2
2
2
p
#
y
x2 +y 2
x
x2 +y 2
p
x +y
p
(x2 + y 2 ) x2 + y 2
1
p
:
x2 + y 2
=
=
Ex. 28:
J (1; 1)
=
=
p 1
p 1
12 +12
12 +12
1
1
2
2
1 +1
12 +12
p1
p1
2
2 :
1
1
2
2
Ex. 29:
det J 0
=
det
cos
sin
r sin
r cos
= r cos2 + sin2
= r;
Using the relationship r =
p
x2 + y 2 ; we have det J det J 0 = 1.
17
18
CHAPTER 4. CHAPTER 4
Ex. 30:
J
p
0
2;
cos 4
=
sin
"p 4
4
2
p2
2
2
=
Ex. 31: We evaluate the product
p1
2
1
2
=
p1
2
1
2
=
1
0
J 0 (x; y)
=
cos
sin
=
x
r
y
r
JJ
0
p
p 2 sin 4
2 cos 4
#
1
:
1
"p
2
p2
2
2
1
1
#
0
;
1
as desired.
Ex. 32:
=
2
JJ 0
=
=
"
x
x2 +y 2
y
x2 +y 2
p
x2
x2 +y 2
+
0
y
x
x
2
2
4 x y+y
p 2 2
x +y
p
so
"
r sin
r cos
y
x
3
5;
2
#
y
p 2x 2
2
2
x +y
4 x y+y
x
p 2 2
x2 +y 2
x +y
p
y2
x2 +y 2
0
x2
x2 +y 2
+
y2
x2 +y 2
= I;
similarly, J 0 J = I. Thus, J; J 0 are inverses of each other.
Ex. 33: We use
r (x; y; z)
(x; y; z)
(x; y; z)
p
x2 + y 2 + z 2
z
= arccos p
2
x + y2 + z2
y
= arctan
x
=
y
x
#
3
5
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
@r
@y
@r
@z
@
@x
@
@y
@
@z
=
sin cos
=
sin sin
=
cos
cos cos
r
cos
r sin
sin
r
=
=
=
@
@x
@
@y
@
@z
=
sin
r sin
=
cos sin
=
0:
Thus,
J (r; ; )
=
2 @r
@r
@y
@
@y
@
@y
@x
6@
4 @x
@
@x
2
sin cos
= 4 cos rcos
3
@r
@y
@ 7
@z 5
@
@z
sin sin
sin
r sin
cos
cos
r sin
sin
r
cos sin
0
3
5:
We then compute
J 0 (r; ; )
=
2 @x
@r
6 @y
4 @r
@z
@r
@x
@
@y
@
@z
@
2
sin cos
= 4 sin sin
cos
3
@x
@
@y 7
@ 5
@z
@
r cos cos
r cos sin
r sin
3
r sin sin
r sin cos 5 ;
0
20
CHAPTER 4. CHAPTER 4
So
JJ 0
2
sin cos
= 4 cos rcos
2
= 4
sin sin
cos
r sin
sin
r sin
1
r
2
1
= 40
0
cos
sin
r
cos sin
0
r cos cos
r cos sin
r sin
cos + cos sin + sin2 sin2
cos sin + 1r cos sin + 1r cos cos2 sin
1
cos sin + cos sin sin2
3r
0 0
1 05
0 1
2
2
32
sin cos
5 4 sin sin
cos
2
3
r sin sin
r sin cos 5
0
r cos sin + r cos cos2 sin + r cos s
2
sin2 + (cos ) cos
cos
sin sin + cos
cos
2
r cos cos sin
(cos ) sin sin
[Note: Something may be o¤ with the computation of J]
Ex. 34: We compute
@2f ( ; )
@ 2
=
=
@
@
@
@
@
@
@
+
@
+
@F @A @F @B
+
+
@a @
@b @
@F @A @F @
+
@a @
@a @
@F @B
@F @
+
@b @
@b @
@F @
@F @C
+
@c @
@c @
@F @C
@c @
@A
@
@B
@
@C
@
by the product rule. We continue:
@f ( ; )
@ 2
=
@ 2 F @A
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
+
@a2 @
@a@b @
@a@c @
@
@a @ 2
2
2
2
@ F @C @B
@F @ 2 B
@ F @A @ F @B
+
+
+
+
@a@b @
@b2 @
@b@c @
@
@b @ 2
@ 2 F @A
@ 2 F @B
@ 2 F @C @C
@F @ 2 C
+
+
+
+
:
2
@a@c @
@b@c @
@c @
@
@c @ 2
21
Similarly,
@2f ( ; )
@ @
@f ( ; )
@ 2
@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 @
@
2
2
2
@ F @B
@ F @C @A @F @ 2 A
@ F @A
+
+
+
=
@a2 @
@a@b @
@a@c @
@
@a @ @
2
2
2
@ F @A @ F @B
@ F @C @B
@F @ 2 B
+
+
+
+
@a@b @
@b2 @
@b@c @
@
@b @ @
2
2
2
@ F @A
@ F @B
@ F @C @C
@F @ 2 C
+
+
+
+
@a@c @
@b@c @
@c2 @
@
@c @ @
=
=
@
@
@
@
@ 2 F @A
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
+
2
@a @
@a@b @
@a@c @
@
@a @ 2
@ 2 F @C @B
@F @ 2 B
@ 2 F @A @ 2 F @B
+
+
+
+
2
@a@b @
@b @
@b@c @
@
@b @ 2
@ 2 F @A
@ 2 F @B
@ 2 F @C @C
@F @ 2 C
+
+
+
+
:
2
@a@c @
@b@c @
@c @
@
@c @ 2
Ex. 35: We compute
@3f ( ; )
@2 @
@ 2 F @A
+
@a2 @
@
@ 2 F @A
+
@
@a@b @
@
@ 2 F @A
+
@
@a@c @
= I + J + K;
=
@
@
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
@a@b @
@a@c @
@
@a @ @
2
2
@ F @B
@ F @C @B
@F @ 2 B
+
+
+
2
@b @
@b@c @
@
@b @ @
2
2
@ F @B
@ F @C @C
@F @ 2 C
+
+
+
2
@b@c @
@c @
@
@c @ @
22
CHAPTER 4. CHAPTER 4
for sake of clarity,
I
=
=
=
=
=
=
=
@ 2 F @A
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
+
@a2 @
@a@b @
@a@c @
@
@a @ @
@ 2 F @A
@ 2 F @B
@ 2 F @C @A
+
+
@a2 @
@a@b @
@a@c @
@
@ 2 F @A
@ 2 F @B
@ 2 F @C @ @A
@ @F @ 2 A
@F
@3A
+
+
+
+
+
@a2 @
@a@b @
@a@c @
@
@
@
@a @ @
@a @ 2 @
2
2
2
@ @ F @A
@ F @B
@ F @C @A
+
+
2
@
@a @
@a@b @
@a@c @
@
@ 2 F @A
@ 2 F @B
@ 2 F @C @ 2 A
+
+
+
@a2 @
@a@b @
@a@c @
@ @
@ 2 F @A
@ 2 F @B
@ 2 F @C @ 2 A
@F @ 3 A
+
+
+
+
2
@a @
@a@b @
@a@c @
@ @
@a @ 2 @
@ @ 2 F @A
@ 2 F @B
@ 2 F @C @A
+
+
2
@
@a @
@a@b @
@a@c @
@
2
2
2
@ F @A
@ F @B
@ F @C @ 2 A
@F @ 3 A
+2
+
+
+
2
@a @
@a@b @
@a@c @
@ @
@a @ 2 @
@ @ 2 F @A @ 2 F @ @A
@ @ 2 F @B
@ 2 F @ @B
@ @ 2 F @C
@2F
+
+
+
+
+
2
2
@
@a
@
@a @
@
@
@a@b @
@a@b @
@
@
@a@c @
@a@c
2
2
2
3
2
@ F @A
@ F @B
@ F @C @ A
@F @ A
+2
+
+
+
@a2 @
@a@b @
@a@c @
@ @
@a @ 2 @
@ @ 2 F @A @A @ 2 F @ @A @A
+
@
@a2 @ @
@a2 @
@
@
2
2
@ @ F @B @A
@ F @ @B @A
+
+
@
@a@b @ @
@a@b @
@
@
2
2
@ @ F @C @A
@ F @ @C @A
+
+
@
@a@c @ @
@a@c @
@
@
2
2
2
@ F @A
@ F @B
@ F @C @ 2 A
@F @ 3 A
+2
+
+
+
@a2 @
@a@b @
@a@c @
@ @
@a @ 2 @
3
3
3
@ F @A
@ F @B
@ F @C @A @A @ 2 F @ 2 A @A
+
+
+
@a3 @
@a2 @b @
@a2 @c @
@ @
@a2 @ 2 @
3
3
3
@ F @A
@ F @B
@ F @C @B @A
@ 2 F @ 2 B @A
+
+
+
+
@a2 @b @
@a@b2 @
@a@b@c @
@ @
@a@b @ 2 @
3
3
3
@ F @A
@ F @B
@ F @C @C @A
@ 2 F @ 2 C @A
+
+
+
+
@a2 @c @
@a@b@c @
@a@c2 @
@ @
@a@c @ 2 @
2
2
2
2
@ F @A
@ F @B
@ F @C @ A
@F @ 3 A
+2
+
+
+
@a2 @
@a@b @
@a@c @
@ @
@a @ 2 @
@
@
@
@
23
J
=
=
=
=
=
@ 2 F @A @ 2 F @B
@ 2 F @C @B
@F @ 2 B
+
+
+
@a@b @
@b2 @
@b@c @
@
@b @ @
2
2
2
@ F @A @ F @B
@ F @C @B
+
+
2
@a@b @
@b @
@b@c @
@
@ 2 F @A @ 2 F @B
@ 2 F @C @ @B
@ @F @ 2 B
@F @
@2B
+
+
+
+
+
@a@b @
@b2 @
@b@c @
@
@
@
@b @ @
@b @
@ @
2
@ 2 F @C @B
@ @ F @A @ 2 F @B
+
+
@
@a@b @
@b2 @
@b@c @
@
@ 2 F @A @ 2 F @B
@ 2 F @C @ 2 B
@ 2 F @A @ 2 F @B
@ 2 F @C @ 2 B
@F @ 3 B
+
+
+
+
+
+
+
@a@b @
@b2 @
@b@c @
@ @
@a@b @
@b2 @
@b@c @
@ @
@b @ 2 @
@ @ 2 F @A
@ 2 F @ @A
@ 2 F @ @B
@2F @
@ @ 2 F @B
@ @ 2 F @C
+
+
+
+
+
2
2
@
@a@b @
@a@b @
@
@
@b
@
@b @
@
@
@b@c @
@b@c @
2
2
2
2
2
2
2
2
@ F @A @ F @B
@ F @C @ B
@ F @A @ F @B
@ F @C @ B
@F @ 3 B
+
+
+
+
+
+
+
2
2
@a@b @
@b @
@b@c @
@ @
@a@b @
@b @
@b@c @
@ @
@b @ 2 @
@ @ 2 F @A @B
@ 2 F @ @A @B
+
@
@a@b @ @
@a@b @
@
@
@
@
@
@
2
@2F
@ 2 F @ @B @B
@B
+
@b2
@
@b2 @
@
@
2
@ F @C @B
@ @C @B
+
@b@c @ @
@
@
@
@ 2 F @A @ 2 F @B
@ 2 F @A @ 2 F @B
@ 2 F @C @ 2 B
@ 2 F @C
+
+
+
+
+
+
@a@b @
@b2 @
@b@c @
@ @
@a@b @
@b2 @
@b@c @
3
3
2
3
2
@ F @B
@ F @C @A @B
@ F @ A @B
@ F @A
=
+
+
+
@a2 @b @
@a@b2 @
@a@b@c @
@ @
@a@b @ @ @
@
@
@
+
@
+
@2B
@F @ 3 B
+
@ @
@b @ 2 @
2
@ 3 F @A @ 3 F @B
@ 3 F @C
@B
@ 2 F @B @B
+
+ 2
+
2
3
@a@b @
@b @
@b @c @
@
@b2 @ @ @
@ 3 F @B
@ 3 F @C @C @B
@ 2 C @B
@ 3 F @A
+ 2
+
+
+
@a@b@c @
@b @c @
@b@c2 @
@ @
@ @ @
2
2
2
2
2
@ F @A @ F @B
@ F @C @ B
@ F @A @ 2 F @B
@ 2 F @C
+
+
+
+
+
+
@a@b @
@b2 @
@b@c @
@ @
@a@b @
@b2 @
@b@c @
+
@2B
@F @ 3 B
+
@ @
@b @ 2 @
@C
@
24
K
CHAPTER 4. CHAPTER 4
=
=
=
=
=
@ 2 F @A
@ 2 F @B
@ 2 F @C @C
@F @ 2 C
+
+
+
2
@a@c @
@b@c @
@c @
@
@c @ @
2
2
2
@ F @A
@ F @B
@ F @C @C
+
+
@a@c @
@b@c @
@c2 @
@
2
2
2
@ F @A
@ F @B
@ F @C @ @C
@ @F @ 2 C
@F @
@2C
+
+
+
+
+
2
@a@c @
@b@c @
@c @
@
@
@
@c @ @
@c @
@ @
2
2
2
@ @ F @A
@ F @B
@ F @C @C
+
+
@
@a@c @
@b@c @
@c2 @
@
2
2
2
@ F @A
@ 2 F @A
@ F @B
@ F @C @ 2 C
@ 2 F @B
@ 2 F @C @ 2 C
@F @
+
+
+
+
+
+
+
@a@c @
@b@c @
@c2 @
@ @
@a@c @
@b@c @
@c2 @
@ @
@c @
2
2
2
2
2
@ @ F @A
@ F @ @A
@ F @ @B
@2F
@ @ F @B
@ @ F @C
+
+
+
+
+
@
@a@c @
@a@c @
@
@
@b@c @
@b@c @
@
@
@c2 @
@c2
2
2
2
2
2
2
2
2
@ F @A
@ F @B
@ F @C @ C
@ F @A
@ F @B
@ F @C @ C
@F @
+
+
+
+
+
+
+
@a@c @
@b@c @
@c2 @
@ @
@a@c @
@b@c @
@c2 @
@ @
@c @
2
2
@ @ F @A @C
@ F @ @A @C
+
@
@a@c @ @
@a@c @
@
@
@ @ 2 F @B @C
@ 2 F @ @B @C
+
+
@
@b@c @ @
@b@c @
@
@
@
@
@
@
2
@2F
@C
@ 2 F @ @C @C
+
2
@c
@
@c2 @
@
@
2
2
2
2
@ F @A
@ F @B
@ F @C @ C
@ 2 F @A
@ 2 F @B
@ 2 F @C
+
+
+
+
+
+
2
@a@c @
@b@c @
@c @
@ @
@a@c @
@b@c @
@c2 @
@ 3 F @A
@ 3 F @B
@ 3 F @C @A @C
@ 2 F @ 2 A @C
=
+
+
+
2
2
@a @c @
@a@b@c @
@a@c @
@ @
@a@c @ @ @
3
3
3
@ F @A
@ F @B
@ F @C @B @C
@ 2 F @ 2 B @C
+
+ 2
+
+
@a@b@c @
@b @c @
@b@c2 @
@ @
@b@c @ @ @
+
@
@
@2C
@F @
+
@ @
@c @
2
@ 3 F @A
@ 3 F @B
@ 3 F @C
@C
@ 2 F @ 2 C @C
+
+
+
@a@c2 @
@b@c2 @
@c3 @
@
@c2 @ @ @
@ 2 F @B
@ 2 F @C @ 2 C
@ 2 F @A
@ 2 F @B
@ 2 F @C
@ 2 F @A
+
+
+
+
+
+
2
@a@c @
@b@c @
@c @
@ @
@a@c @
@b@c @
@c2 @
+
@2C
@F @
+
@ @
@c @
25
Ex. 36
@f ( ; )
@ 2
@2f ( ; )
@ @
2
@ 2 F @A
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
+
@a2 @
@a@b @
@a@c @
@
@a @ 2
2
2
2
@ F @A @ F @B
@ F @C @B
@F @ 2 B
+
+
+
+
@a@b @
@b2 @
@b@c @
@
@b @ 2
2
2
2
@ F @A
@ F @B
@ F @C @C
@F @ 2 C
+
+
+
+
@a@c @
@b@c @
@c2 @
@
@c @ 2
2
j
2 i
i
@ F @A @A
@F @ A
=
+ i
i
j
@a @a @ @
@a @ 2
=
@ 2 F @A
@ 2 F @B
@ 2 F @C @A @F @ 2 A
+
+
+
2
@a @
@a@b @
@a@c @
@
@a @ @
2
2
2
@ F @A @ F @B
@ F @C @B
@F @ 2 B
+
+
+
+
2
@a@b @
@b @
@b@c @
@
@b @ @
2
2
2
@ F @A
@ F @B
@ F @C @C
@F @ 2 C
+
+
+
+
2
@a@c @
@b@c @
@c @
@
@c @ @
j
i
2 i
2
@F @ A
@ F @A @A
+ i
=
@ai @aj @ @
@a @ @
=
@f ( ; )
@ 2 F @Aj @Ai
@F @ 2 Ai
=
+ i
2
i
j
@
@a @a @ @
@a @ 2
Ex. 37 We may generalize the above three equations, setting
= , to yield
1
=
,
@2f ( ; )
@ 2 F @Aj @Ai
@F @ 2 Ai
=
+ i
:
i
j
@ @
@a @a @ @
@a @ @
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]
dx
d
(sin arccos x)
[arccos x]
dx
d
[arccos x]
dx
= 1
= 1
=
1
(sin arccos x)
26
CHAPTER 4. CHAPTER 4
By examining triangles with unit hypotenuse, we obtain
p
1 x2 ;
sin arccos x =
so
d
[arccos x] =
dx
p
1
x2
1
Ex. 40: We know that f; g satisfy
g 0 (f (x)) f 0 (x) = 1:
Di¤erentiating both sides, we obtain
d 0
d 0
[g (f (x))] f 0 (x) + g 0 (f (x))
[f (x)] = 0
dx
dx
g 00 (f (x)) f 0 (x) f 0 (x) + g 0 (f (x)) f 00 (x) = 0
2
g 00 (f (x)) [f 0 (x)] + g 0 (f (x)) f 00 (x) = 0
as desired.
Ex. 41: We compute
f 0 (x) = ex
f 00 (x) = ex
1
g 0 (x) =
x
g 00 (x)
=
1
;
x2
So
2
g 00 (f (x)) [f 0 (x)] + g 0 (f (x)) f 00 (x)
=
=
=
as desired.
1 2x
1
e + x ex
e2x
e
1+1
0;
(4.1)
27
Ex. 42: We compute
f 0 (x)
=
f 00 (x)
=
=
0
g (x) =
g 00 (x) =
p
1
x2
1
2x
1
1
2
3=2
x2
3=2
x 1 x2
sin (x)
cos (x) ;
So
2
g 00 (f (x)) [f 0 (x)] + g 0 (f (x)) f 00 (x)
1
cos (arccos x)
1 x2
p
x 1 x2
x
+ p
3
1 x2
1 x2
x
x
+
1 x2
1 x2
=
=
=
=
sin (arccos (x))
x 1
3=2
x2
0;
as desired.
Ex.
obtain
43:
We di¤erentiate both sides of the second-order relationship to
d
2
g 00 (f (x)) [f 0 (x)] + g 0 (f (x)) f 00 (x)
= 0
dx
h
i
d
d 0
d
d 00
2
2
[g (f (x))] [f 0 (x)] + g 00 (f (x))
[f 0 (x)] +
[g (f (x))] f 00 (x) + g 0 (f (x)) f 00 (x) = 0
dx
dx
dx
dx
3
g (3) (f (x)) [f 0 (x)] + g 00 (f (x)) 2f 0 (x) f 00 (x) + g 00 (f (x)) f 0 (x) f 00 (x) + g 0 (f (x)) f (3) (x) = 0
3
g (3) (f (x)) [f 0 (x)] + 3g 00 (f (x)) f 0 (x) f 00 (x) + g 0 (f (x)) f (3) (x)
Ex.
Ex.
Ex.
Ex.
44:
45:
46:
47:
Ex. 48: We begin with the identity (note the top indices should be considered "…rst")
0
Jii0 Jji = ij ;
and write out the dependences on unprimed coordinates:
0
Jii0 (Z 0 (Z)) Jji (Z) =
i
j
(Z)
(4.2)
=
0
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
@
@ h i
i
0
i0
J
(Z
(Z))
J
(Z)
=
(Z)
0
i
j
k
@Z
@Z k j
i
h
0
@
Jii0 (Z 0 (Z)) Jji (Z) = 0
k
@Z
i
@
@ h i0
i
0
i0
i
0
J
(Z)
= 0;
J
(Z
(Z
(Z))
J
(Z)
+
J
(Z))
0
0
j
i
@Z k i
@Z k j
since di¤erentiation passes through the implied summation over i0 . Then, using
the de…nition of the Jacobian,
@
@Z k
@Z i0
@Z i
@
@Z i
0
(Z) +
(Z 0 (Z))
0 (Z (Z))
i
j
@Z
@Z
@Z i0
@Z k
@Z i0
(Z)
@Z j
0
= (4.3)
0
0
@Z k
@Z i0
@Z i
@2Z i
@2Z i
0
0
(Z
(Z))
(Z)
(Z)
+
(Z
(Z))
(Z)
0
0
0
@Z k @Z i
@Z k
@Z j
@Z i
@Z k @Z j
= (4.4)
0;
applying the chain rule to the …rst term, and implying summation over new
index k 0 . Then, if we de…ne the "Hessian" object
Jki 0 ;i0 :=
@2Z i
(Z 0 )
@Z k0 @Z i0
0
i
, we write (4.3) concisely:
with an analogous de…nition for Jk;i
0
0
0
i
Jki 0 ;i0 Jkk Jji + Jii0 Jk;j
= 0;
or, using a renaming of dummy indicex k 0 to j 0 and a reversing of the order of
partial derivatives,
0
0
i0
Jii0 j 0 Jji Jkj + Jii0 Jjk
= 0:
This above tensor relationship represents n3 identities.
Ex. 49: Since each Jii0 is constant for a transformation from one a¢ ne
coordinate system to another, each second derivative vanishes, and hence each
Jii0 j 0 = 0.
Ex. 50: Begin with the identity derived in Ex. 48:
0
0
0
i
Jii0 j 0 Jji Jkj + Jii0 Jjk
= 0;
29
Then, letting k 0 be arbitrary, we multiply both sides by Jkk0 , implying summation
over k:
h
i
0
0
i0
Jii0 j 0 Jji Jkj + Jii0 Jjk
Jkk0 = 0
0
0
0
i
Jkk0
Jii0 j 0 Jji Jkj Jkk0 + Jii0 Jjk
0
but, Jkj Jkk0 =
j0
k0 ;
=
0
so
Jii0 j 0 Jji
0
j0
k0
0
i
Jkk0 = 0:
+ Jii0 Jjk
0
Note that we have jk0 = 1 if and only if j 0 = k 0 , so the …rst term is equal to
0
Jii0 k0 Jji . After re-naming k 0 = j 0 , we obtain
0
0
i
Jii0 j 0 Jji + Jii0 Jjk
Jjk0 = 0:
(4.5)
Ex. 51: Let k 0 be arbitrary, and multiply both sides of (4.5) by Jkj0 ,
implying summation over j:
h
i
0
i0 k
Jii0 j 0 Jji + Jii0 Jjk
Jj 0 Jkj0 = 0
0
0
i
Jii0 j 0 Jji Jkj0 + Jii0 Jjk
Jjk0 Jkj0
0
Jii0 j 0 ik0
Jii0 j 0
i0
k0
+
+
i0 k j
Jii0 Jjk
Jj 0 Jk 0
0
i
Jjk
Jii0 Jjk0 Jkj0
=
0
=
0
=
0:
Rename the dummy index in the second term i0 = h0 . Then,
Jii0 j 0
i0
k0
0
h i
+ Jjk
Jh0 Jjk0 Jkj0 = 0:
Noting that the …rst term is zero for all i0 6= k 0 , and setting k 0 = i0 :
0
h i
Jii0 j 0 + Jjk
Jh0 Jjk0 Jij0 = 0:
We then may re-introduce k 0 as a dummy index:
0
k i k j
Jii0 j 0 + Jjk
Jk0 Jj 0 Ji0 = 0:
Then, switch the roles of j; k as dummy indices:
0
k i j k
Jii0 j 0 + Jkj
J k 0 J j 0 J i0 = 0
30
CHAPTER 4. CHAPTER 4
Ex. 52: Return to
0
0
0
i
Jii0 j 0 Jji Jkj + Jii0 Jjk
= 0;
and write out the dependences
0
0
0
i
(Z)
Jii0 j 0 (Z 0 (Z)) Jji (Z) Jkj (Z) + Jii0 (Z 0 (Z)) Jjk
0
=
0
=
0
0
@Z i
@2Z i
@Z j0
@Z i
@2Z i
0
0
(Z)
(Z) +
(Z)
0 (Z (Z))
0
0 (Z (Z))
j
i
j
k
i
@Z @Z
@Z
@Z
@Z
@Z j @Z k
Then, di¤erentiate both sides with respect to Z m :
"
#
0
0
@
@2Z i
@Z i
@Z j0
@Z i
@2Z i
0
0
0 =
(Z (Z))
(Z)
(Z) +
(Z (Z))
(Z)
@Z m @Z i0 @Z j 0
@Z j
@Z k
@Z i0
@Z j @Z k
"
#
0
0
@
@Z i
@Z j0
@2Z i
@
@Z i
@Z j0
@2Z i
0
0
=
(Z
(Z))
(Z)
(Z)
+
(Z
(Z))
(Z)
(Z) +
0
0
0
0
@Z m @Z i @Z j
@Z j
@Z k
@Z i @Z j
@Z m @Z j
@Z k
@
"
#
0
0
@2Z i
@
@Z i
@Z i
@
@2Z i
0
0
+ m
(Z) +
(Z)
0 (Z (Z))
0 (Z (Z))
i
j
k
i
m
@Z
@Z
@Z @Z
@Z
@Z
@Z j @Z k
"
#
0
0
0
@Z m
@2Z i
@Z j0
@2Z i
@Z j0
@3Z i
@Z i
0
0
(Z
(Z))
(Z
(Z))
=
(Z)
(Z)
(Z)
+
(Z)
(
0
0
0
0
0
@Z i @Z j @Z m
@Z m
@Z j
@Z k
@Z i @Z j
@Z m @Z j
@Z k
"
#
0
0
0
@2Z i
@Z m
@2Z i
@Z i
@3Z i
0
0
+
(Z
(Z))
(Z)
(Z)
+
(Z
(Z))
(Z)
@Z i0 @Z m0
@Z m
@Z j @Z k
@Z i0
@Z j @Z k @Z m
0
0
0
0
0
0
0
0
0
0
j
m i j
i
m i
i
= Jii0 j 0 m0 Jm
Jj Jk + Jii0 j 0 Jjm
Jkj + Jii0 j 0 Jji Jkm
+ Jii0 m0 Jm
Jjk + Jii0 Jjkm
;
so, setting k 0 = m0 as a dummy index:
0
0
0
0
0
0
0
0
0
0
j
k
i
i
k
i
Jii0 j 0 k0 Jji Jkj Jm
+ Jii0 j 0 Jkj Jjm
+ Jii0 j 0 Jji Jkm
+ Jii0 k0 Jjk
Jm
+ Jii0 Jjkm
= 0: (4.6)
m
Then, multiply both sides by Jm
0 , implying summation over m:
0
0
0
0
0
0
0
0
0
0
k m
i
m
i
i j
m
i
i
k m
i i
m
Jii0 j 0 k0 Jji Jkj Jm
Jm0 + Jii0 j 0 Jkj Jjm
Jm
0 + J i0 j 0 J j J
km Jm0 + Ji0 k0 Jjk Jm Jm0 + Ji0 Jjkm Jm0
0
Jii0 j 0 k0 Jji Jkj
0
0
k
m0
0
0
0
0
0
i
m
i
i j
m
i
i
+ Jii0 j 0 Jkj Jjm
Jm
0 + J i0 j 0 J j J
km Jm0 + Ji0 k0 Jjk
0
k
m0
0
i
m
+ Jii0 Jjkm
Jm
0
This holds for all m0 , so speci…cally for m0 = k 0 , the above identity reads
0
0
0
0
0
0
0
0
j
i
i
i
Jii0 j 0 k0 Jji Jkj + Jii0 j 0 Jkj Jjm
Jkm0 + Jii0 j 0 Jji Jkm
Jkm0 + Jii0 k0 Jjk
+ Jii0 Jjkm
Jkm0 = 0 (4.7)
=
0
=
0;
31
k
0
Next, in an analogous manner, multiply both sides by Jm
0 for arbitrary m :
0
0
0
0
0
0
0
0
0
0
0
0
0
0
j i
i
k
i
i j
m k
i
m k
i
k
i i
m k
Jii0 j 0 k0 Jji Jkj Jm
0 + J i0 j 0 J
k Jjm Jk0 Jm0 + Ji0 j 0 Jj Jkm Jk0 Jm0 + Ji0 k0 Jjk Jm0 + Ji0 Jjkm Jk0 Jm0
Jii0 j 0 k0 Jji
0
0
j
m0
m k
i i
i
k
i
i j
m k
i
k
i
Jkm0 Jm
+ Jii0 j 0 Jkj Jjm
0 + J i0 j 0 J j J
km Jk0 Jm0 + Ji0 k0 Jjk Jm0 + Ji0 Jjkm Jk0 Jm0
=
0
=
0;
rename the dummy index h0 = j 0 in all but the …rst term:
Jii0 j 0 k0 Jji
0
j0
h0 i0
i
m k
i
i0 h0
m k
i
i0 k
i i0
m k
m0 +Ji0 h0 Jk Jjm Jk0 Jm0 +Ji0 h0 Jj Jkm Jk0 Jm0 +Ji0 k0 Jjk Jm0 +Ji0 Jjkm Jk0 Jm0
= 0;
then, as in the previous exercises, set m0 = j 0 :
0
0
0
0
0
0
0
i
h
i
i
Jii0 j 0 k0 Jji +Jii0 h0 Jkh Jjm
Jkm0 Jjk0 +Jii0 h0 Jji Jkm
Jkm0 Jjk0 +Jii0 k0 Jjk
Jjk0 +Jii0 Jjkm
Jkm0 Jjk0 = 0;
(4.8)
j
Finally, multiply both sides by Jm
0 :
0
0
0
0
0
0
j
h0
Jii0 h0 Jji Jkm
Jkm0 Jjk0 Jm
0
i0 k j
Jii0 k0 Jjk
Jj 0 Jm0
0
0
j
i
h i
m k j
i i
m k j
i
i
k j
i
i h
m k j
Jii0 j 0 k0 Jji Jm
0 + Ji0 h0 Jk Jjm Jk 0 Jj 0 Jm0 + Ji0 h0 Jj Jkm Jk 0 Jj 0 Jm0 + Ji0 k 0 Jjk Jj 0 Jm0 + Ji0 Jjkm Jk 0 Jj 0 Jm0
0
Jii0 j 0 k0 im0
+
0
j
i0
Jii0 h0 Jkh Jjm
Jkm0 Jjk0 Jm
0
+
+
+
j
i0
Jii0 Jjkm
Jkm0 Jjk0 Jm
0
rename the dummy index i0 to g 0 in all but the …rst term:
Jii0 j 0 k0
0
0
g 0 h0
g0 k j
i0
i
h0 g
m k j
i g
m k j
i
i
m k j
m0 +Jg 0 h0 Jk Jjm Jk0 Jj 0 Jm0 +Jg 0 h0 Jj Jkm Jk0 Jj 0 Jm0 +Jg 0 k0 Jjk Jj 0 Jm0 +Jg 0 Jjkm Jk0 Jj 0 Jm0
Then, set m0 = i0 :
0
0
0
0
0
0
g
g
g k j
h
Jii0 j 0 k0 +Jgi 0 h0 Jkh Jjm
Jkm0 Jjk0 Jij0 = 0:
Jkm0 Jjk0 Jij0 +Jgi 0 h0 Jjg Jkm
Jj 0 Ji0 +Jgi 0 Jjkm
Jkm0 Jjk0 Jij0 +Jgi 0 k0 Jjk
(4.9)
Ex. 53: We have the following relationship
00
Z Z 0 Z (Z)
= Z;
or for each i,
00
Z i Z 0 Z (Z)
= Z i.
Di¤erentiating both sides with respect to Z j , we obtain
@
Z i (Z 0 (Z 00 (Z))) =
@Z j
i
j:
= 0:
=
0
=
0;
32
CHAPTER 4. CHAPTER 4
Then, we apply the chain rule twice:
@Z i @
[Z 0 (Z 00 (Z))] =
@Z i0 @Z j
0
i
@Z i @Z i @ h i00
Z
(Z)
=
0
00
@Z i @Z i @Z j
0
00
@Z i @Z i @Z i
=
@Z i0 @Z i00 @Z j
or
0
00
Jii0 Jii00 Jji =
i
j:
i
j
i
j
i
j;
Chapter 5
Chapter 5
Ex. 61: ij .
Ex. 62: Assume U is an arbitrary nontrivial linear combination U i Zi of
coordinate bases Zi . Since
U U >0
and
U U = Zij U i U j ;
or in matrix notation
U U = U T ZU ,
this condition implies Z = Zij is positive de…nite.
Ex. 63:
kV k =
p
V V
q
=
Zij V i V j
Ex. 64: Put Z = Zij . Thus, Z 1 = Z ij by de…nition.
arbitrary nontrivial vector. Then, de…ne y = Z 1 x. We have
yT
=
Z
1
1 T
= xT Z
= xT Z
33
T
x
1
;
Let x be an
34
CHAPTER 5. CHAPTER 5
since Z
1
is symmetric. Then, since Z is positive de…nite, note
0 < y T Zy
= xT Z 1 ZZ
= xT Z 1 x.
1
x
Since x was arbitrary, this implies that Z = Z ij > 0.
Ex. 65:
Zi Zj
= Z ik Zk Zj
= Z ik Zkj
by de…nition. But,
Z ik Zkj =
i
j;
Zi Zj =
i
j
so we have
Ex. 66, 67: [Not sure - which coordinate system are we in (if any?)]
Ex. 68: Use the de…nition
Zi
Zik Zi
= Z ij Zj
= Zik Z ij Zj
= jk Zj
= Zk :
Thus,
Zk
= Zik Zi
= Zki Zi ;
since Zik is symmetric.
Ex. 69: Examine
Zki Zi Zj
= Zki Z in Zn 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 Z ij 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:
Z ij Zjk
= Zi Zj Zjk
= Zi Zk ;
from 5.17
=
i
k
from 5.16.
Ex. 72: We compute
Z1 Z2
=
=
=
=
=
1 1
i
j j
3 3
1
1
i j
j j
3
3
1
1
2
kik kjk cos
kjk
3
3 3
1
1
1 2
(2) (1)
1
3
2
3
0;
thus, Z1 ; Z2 are orthogonal. We further compute
Z2 Z1
1 4
i+ j
3 3
=
1
4
2
kik + kik kjk cos
3
3
3
1
4
1
(4) + (2) (1)
3
3
2
=
=
=
so Z2 ; Z1 are orthogonal.
i
0;
36
CHAPTER 5. CHAPTER 5
Ex. 73:
Z1 Z1
1 1
i
j
3 3
=
i
1
1
2
kik
kik kjk cos
3
3
3
4 2 1
=
3 3 2
= 1:
1 4
i+ j j
=
3 3
1
4
2
kik kjk cos + kjk
=
3
3
3
2 1
4
=
+
3 2
3
= 1:
=
Z2 Z2
Ex. 74:
Z1 Z2
=
=
=
=
=
1 1
i
j
3 3
1
2
kik +
9
4 4
+ +
9 9
3
9
1
:
3
1 4
i+ j
3 3
4
1
kik kjk cos + kik kjk cos
9
3
9
3
1 4
9 9
Ex. 75: Let R denote the position vector. We compute
Z3
=
=
@R (Z)
Z3
@R (r; ; z)
@z
for cylindrical coordinates
= lim
h!0
R (r; ; z + h)
h
R (r; ; z)
:
4
2
kjk
9
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)
h
R (r; ; z)
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
2
3
1 0 0
Zij = 40 r2 05 .
0 0 1
Now, since Z ij is de…ned to be the inverse of
2
1
0
Z ij = 40 r 2
0
0
Zij , we may easily compute
3
0
05 ;
1
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
= Z 3j Zj
= 0Z1 + 0Z2 + 1Z3
= Z3
38
CHAPTER 5. CHAPTER 5
Chapter 6
Chapter 6
Ex. 87: Look at
0
0
0
0
Z ij Jii Jjj Zj 0 k0 = Z ij Jii Jjj Zjk Jjj0 Jkk0
by the tensor property of Zjk
0
= Z ij Zjk Jii Jkk0
=
=
=
i i0 k
k Ji Jk 0
0
Jki Jkk0
i0
k0 ;
0
0
so, in linear algebra terms, we have that Z ij Jii Jjj is the matrix inverse of Zj 0 k0 .
0
0
0 0
By uniqueness of matrix inverses, this forces Z ij Jii Jjj = Z i j , as desired.
Ex. 88: Let Z,Z 0 be two coordinate systems. Write the unprimed coordinates in terms of the primed coordinates
Z = Z (Z 0 ) :
Then,
@F (Z)
@Z i0
@F (Z (Z 0 ))
@Z i0
@F @Z i
@Z i @Z i0
@F i
J 0;
@Z i i
=
=
=
so
@F
@Z i
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
Sij
=
Si0 j 0
=
@Ti
Zj
@Ti0
Z j0
so
Si0 j 0
=
=
@Ti0
@Z j 0
@
Ti Jii0 ;
@Z j 0
since T is a covariant tensor,
@
Ti (Z 0 (Z)) Jii0 (Z 0 )
Z j0
@T i @Z j i
=
J 0 + Ti Jii0 j 0
@Z j @Z j 0 i
= Sij Jjj0 Jii0 + Ti Jii0 j 0
=
6= Sij Jjj0 Jii0
(except in the trivial case where Ti = 0). Thus, in general, the collection
@Ti
Zj
is not a covariant tensor.
Ex. 90: Compute
Si0 j 0
=
=
@Ti0
@Tj 0
0
j
@Z
@Z i0
@Ti j i
J 0 J 0 + Ti Jii0 j 0
@Z j j i
@Tj i j
J 0 J 0 + Ti Jji0 i0
@Z i i j
by the above, interchanging the rolls of i0 ; j 0 for the second term:
=
@Ti j i
J 0 J 0 + Ti Jii0 j 0
@Z j j i
@Tj j i
J 0 J 0 + Ti Jii0 j 0 ;
@Z i j i
41
since Jji0 i0 = Jii0 j 0 ,
@Tj j i
@Ti j i
J 0 J 0 + Ti Jii0 j 0
J 0J 0
@Z j j i
@Z i j i
@Ti
@Tj
Jjj0 Jii0
@Z j
@Z i
=
=
Ti Jii0 j 0
= Sij Jii0 Jjj0 ;
so this skew-symmetric part Sij is indeed a covariant tensor.
Ex. 91: Put
S ij =
@T i
;
@Z j
so
S
0
i0 j 0
=
=
@T i
@Z j 0
@ h i i0 i
T Ji ;
@Z j 0
since T is a contravariant tensor,
h 0
i
@
i
0
i0
i @
i
0
T
(Z
(Z))
J
+
T
J
(Z
(Z
))
i
@Z j 0
@Z j 0 i
0
i
j
i
@T j 0 i0
@J @Z
=
Jj Ji + T i ij
j
@Z
@Z @Z j 0
0
0
i0 j
= S ij Jii Jjj + T i Jij
Jj 0
=
0
6= S ij Jii Jjj
0
except in the trivial case.
Ex. 92: We have
k
ij
= Zk
@Zi
;
@Z j
so in primed coordinates,
k0
i0 j 0
= Zk
0
= Zk
0
@Zi0
@Z j 0
@Zi i j
J 0 J 0 + Zi Jii0 j 0
@Z j i j
42
CHAPTER 6. CHAPTER 6
by our work done earlier (note that Zi is a covariant tensor)
Zk Jkk
0
@Zi i j
J 0 J 0 + Zi Jii0 j 0
@Z j i j
since Z k is a contravariant tensor,
=
Zk
=
Zk
=
Zk
6=
Zk
@Zi
@Z j
@Zi
@Z j
@Zi
@Z j
@Zi
@Z j
0
0
Jkk Jii0 Jjj0 + Zk Zi Jkk Jii0 j 0
0
Jkk Jii0 Jjj0 +
0
k k0 i
i J k J i0 j 0
0
Jkk Jii0 Jjj0 + Jik Jii0 j 0
0
Jkk Jii0 Jjj0 =
k k0 i j
ij Jk Ji0 Jj 0
except in the trivial case.
Ex. 93: Compute
@Ti0 j 0
@Z k0
=
=
=
=
=
=
i
@ h
i j
T
J
0 Jj 0
ij
0
i
@Z k
@
@
@ h ji
j
i j
i
i
[T
]
J
J
J0
0 Jj 0 + Tij
0 Jj 0 + Tij Ji0
ij
0
0
i
i
@Z k
@Z k
@Z k0 j
@
[Tij ] Jii0 Jjj0 + Tij Jii0 k0 Jjj0 + Tij Jii0 Jjj0 k0
@Z k0
@
[Tij (Z (Z 0 ))] Jii0 Jjj0 + Tij Jii0 k0 Jjj0 + Tij Jii0 Jjj0 k0
@Z k0
@Tij @Z k i j
J 0 J 0 + Tij Jii0 k0 Jjj0 + Tij Jii0 Jjj0 k0
@Z k @Z k0 i j
@Tij k i j
J 0 J 0 J 0 + Tij Jii0 k0 Jjj0 + Tij Jii0 Jjj0 k0 :
@Z k k i j
43
Thus, from 5.66,
k0
i0 j 0
=
=
=
=
=
=
=
=
=
1 k0 m0 @Zm0 i0
@Zm0 j 0
@Zi0 j 0
+
Z
2
@Z j 0
@Z i0
@Z m0
1 km k0 m0 @Zm0 i0
@Zm0 j 0
@Zi0 j 0
+
Z Jk Jm
2
@Z j 0
@Z i0
@Z m0
1 km k0 m0 @Zmi j m i
m
i
m i
J 0 J 0 J 0 + Zmi Jm
Z Jk Jm (
0 j 0 Ji0 + Zmi Jm0 Ji0 j 0
2
@Z j j m i
@Zmj i m j
j
m
m j
J 0 J 0 J 0 + Zmj Jm
+
0 i0 Jj 0 + Zmj Jm0 Ji0 j 0
@Z i i m j
@Zij m i j
J 0 J 0 J 0 Zij Jii0 m0 Jjj0 Zij Jii0 Jjj0 m0 )
@Z m m i j
1 km k0 m0 @Zmi j m i
@Zmj i m j
@Zij m i j
Z Jk Jm
Jj 0 Jm0 Ji0 +
Ji0 Jm0 Jj 0
J 0J 0J 0
j
i
2
@Z
@Z
@Z m m i j
0
1
m0
m
i
m i
+ Z km Jkk Jm
Zmi Jm
0 j 0 Ji0 + Zmi Jm0 Ji0 j 0
2
0
1
j
m0
m j
m
+ Z km Jkk Jm
Zmj Jm
0 i0 Jj 0 + Zmj Jm0 Ji0 j 0
2
1 km k0 m0
Z Jk Jm Zij Jii0 m0 Jjj0 + Zij Jii0 Jjj0 m0
2
@Zmj
@Zij
1 km k0 j i @Zmi
Z J k J j 0 J i0
+
j
i
2
@Z
@Z
@Z m
0
1
m0
m
i
m i
+ Z km Jkk Jm
Zmi Jm
0 j 0 Ji0 + Zmi Jm0 Ji0 j 0
2
0
1
j
m0
m j
m
+ Z km Jkk Jm
Zmj Jm
0 i0 Jj 0 + Zmj Jm0 Ji0 j 0
2
1 km k0 m0
Z Jk Jm Zij Jii0 m0 Jjj0 + Zij Jii0 Jjj0 m0
2
k k0 j i
ij Jk Jj 0 Ji0
0
1
m0
m
i
m i
+ Z km Zmi Jkk Jm
Jm
0 j 0 Ji0 + Jm0 Ji0 j 0
2
0
1
j
m0
m j
m
+ Z km Zmj Jkk Jm
Jm
0 i0 Jj 0 + Jm0 Ji0 j 0
2
0
1 km
m0
Z Zij Jkk Jm
Jii0 m0 Jjj0 + Jii0 Jjj0 m0
2
k k0 j i
ij Jk Jj 0 Ji0
0
1
m0
m
i
m i
+ ki Jkk Jm
Jm
0 j 0 Ji0 + Jm0 Ji0 j 0
2
0
1
j
m0
m
m j
+ kj Jkk Jm
Jm
0 i0 Jj 0 + Jm0 Ji0 j 0
2
0
1 km
m0
Z Zij Jkk Jm
Jii0 m0 Jjj0 + Jii0 Jjj0 m0
2
k k0 j i
ij Jk Jj 0 Ji0
1 0 m0 m
m i
+ Jik Jm
Jm0 j 0 Jii0 + Jm
0 J i0 j 0
2
1 0 m0
j
m
m j
+ Jjk Jm
Jm
0 i0 Jj 0 + Jm0 Ji0 j 0
2
0
1 km
m0
Jii0 m0 Jjj0 + Jii0 Jjj0 m0
Z Zij Jkk Jm
2
k k0 j i
ij Jk Jj 0 Ji0
1 0 m0 m
1 0 m0 m i
Jm0 j 0 Jii0 + Jik Jm
Jm0 Ji0 j 0
+ Jik Jm
2
2
1 0 0
1 0 0
j
j
44
CHAPTER 6. CHAPTER 6
=
k k0 j i
ij Jk Jj 0 Ji0
1 0
1 0
m0 m
Jm0 j 0 + Jik Jii0 j 0
+ Jik Jii0 Jm
2
2
1 0
1 0
m0 m
+ Jjk Jjj0 Jm
Jm0 i0 + Jjk Jij0 j 0
2
2
0
0
0
1 km
1 km
m i
m0 i j
Z Zij Jkk Jm
Z Zij Jkk Jm
Ji0 m0 Jjj0
Ji0 Jj 0 m0
2
2
1 k0 m0 m
1 0 m0 m
k k0 j i
k0 i
=
J 0 0 + kj 0 Jm
Jm0 i0
0 J
ij Jk Jj 0 Ji0 + Ji Ji0 j 0 +
2 i m mj
2
0
0
k k j i
k i
=
ij Jk Jj 0 Ji0 + Ji Ji0 j 0 + 0:
Thus, we have
k0
i0 j 0
=
k k0 j i
ij Jk Jj 0 Ji0
0
1 km
m0 i
Z Zij Jjj0 Jkk Jm
Ji0 m0
2
0
+ Jik Jii0 j 0 :
Ex. 94: We show the result for degree-one covariant tensors. The generalization to other tensors is then evident.
Assume Ti = 0. Then, since Ti0 = Ti Jii0 , we have Ti0 = 0.
Ex. 95: Note that in such a coordinate change,
0
0
Jii = Aii ;
so the "Hessian" object
0
i
Jij
= 0;
0
since Aii is assumed to be constant with respect to Z. This means that all but
i0
@T
0
i0
, @T , @Zi kj0 , and even ik0 j 0 are
the …rst terms in the above computations of @T
Z j 0 @Z j 0
zero, and hence we have that each of these objects have this "tensor property"
with respect to coordinate changes that are linear transformations.
0 0
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,
cAijk
1 km
Z Zij Jii0 Jk
2
45
is a tensor. Compute
0
0
cAij 0 k0 = cAijk Jii Jjj0 Jkk0 ;
i
since Aijk is a tensor. Thus, by the above, we have if each A (n)jk is a tensor,
then the sum
N
X
i
cn A (n)jk
n=1
is a tensor. Thus, linear combinations of tensors are tensors.
Ex. 97: We have that
Si T ij
= Si ik T kj
= ik Si T kj ;
which is a tensor, since both ik and Si T kj are tensors by the previous section
and by the fact that the product of two tensors is a tensor.
i
Ex. 98:
i = n by the summation convention.
dimension of the ambient space.
Thus,
Ex. 99: We have
Vij = Vijk Zk ;
so
Vij Zm
= Vijk Zk Zm
= Vijk m
k
= Vijm
so substituting m = k, we have an expression for the components
Vijk = Vij Zk
So,
0
Vik0 j 0
= Vi0 j 0 Zk
0
0
= Vij Jii0 Jjj0 Zk Jkk ;
since both Vij ; Zk are tensors
= Vij Zk Jii0 Jjj0 Jkk
0
i
i
returns the
46
CHAPTER 6. CHAPTER 6
by linearity
= Vijk Jii0 Jjj0 Jkk
0
as desired.
Ex. 100: Fix a coordinate system Z i
Tkij = Tkij Jii Jjj Jkk ;
so
0
0
Tkij Jii Jjj Jkk0
0
= Tkij
0 0
i0 j 0 k
i j k0
= Tki0j ;
as desired.
0
= Tkij Jii Jjj Jkk Jii Jjj Jkk0
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
A = eijk ai1 aj2 ak3 :
Note that switching the roles of 1; 2 in the above equation yields
eijk ai2 aj1 ak3
= eijk aj1 ai2 ak3
= ejik ai1 aj2 ak3
=
eijk ai1 aj2 ak3
=
A
Generalizing, we let (r; s; t) be a permutation of (1; 2; 3). We may then see
that
A = eijk erst air ajs akt
(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
X
3!A =
eijk erst air ajs akt
p ermutations
(r;s;t)
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:
3!A = eijk erst air ajs akt ;
or
A=
1 ijk
e erst air ajs akt
3!
51
52
CHAPTER 9. CHAPTER 9
We may similarly show that for aij , we have
A=
1
eijk ai1 aj2 ak3 :
3!
Ex. 184: We have
123 r s t
rst a2 a2 a3
=
123 s r t
srt a2 a2 a3
after index renaming
123 r s t
srt a2 a2 a3
123 r s t
rst a2 a2 a3 :
=
=
Since
123 r s t
rst a2 a2 a3
=
123 r s t
rst a2 a2 a3 ;
we need
123 r s t
rst a2 a2 a3
The result for
132 s r t
srt a2 a3 a2
= 0:
follows similarly.
Ex. 185: Note
123
rst
= e123 erst = 1 erst = erst ;
so
123 r s t
rst a1 a2 a3
= erst ar1 as2 at3 = A:
132 r s t
rst a2 a3 a1
t r s
= 132
rst a1 a2 a3
r s t
= 132
str a1 a2 a3
123 r s t
=
str a1 a2 a3
=
estr ar1 as2 at3
= A:
Also,
53
[Note: Is there an error somewhere - should this be
Ex. 186: De…ne
Air =
We check that
A?]
1 ijk rst
e e ajs atk :
2!
@A
= Air :
@air
Check
@A
@alu
=
=
=
=
=
=
=
1 ijk rst @ (air ajs akt )
e e
3!
@alu
1 ijk rst @air
@ajs
@akt
e e
ajs akt + air
akt + air ajs
3!
@alu
@alu
@alu
i
1 ijk rst h l u
l u
l u
e e
i r ajs akt + air j s akt + air ajs k t
3!
1 h ijk l rst u
ijk
ijk l rst u
e
s akt + air ajs e
je
r ajs akt + air e
ie
3!
1 ljk ust
e e ajs akt + air eilk erut akt + air ajs eijl ersu
3!
1 ljk ust
e e ajs akt + eilk erut air akt + eijl ersu air ajs
3!
1
3eljk eust ajs akt ;
3!
after an index renaming,
1 ljk ust
e e ajs akt
2!
= Alu
=
as desired. Similarly, if we de…ne
Air =
we have
1
eijk erst ajs atk ;
2!
@A
= Air
@air
by a similar argument.
Ex. 187: In cartesian coordinates,
Zij =
i
j;
l rst u
t
ke
i
54
CHAPTER 9. CHAPTER 9
so
Z
Thus,
= jZ j
= jIj
= 1:
p
Z = 1:
In polar coordinates,
[Zij ] =
1 0
;
0 r2
Z
1 0
0 r2
so
=
= r2 ;
hence
p
Z = r:
In spherical coordinates,
2
1 0
[Zij ] = 40 r2
0 0
3
0
0 5;
r2 sin2
so
Z
=
1 0
0 r2
0 0
0
0
r2 sin2
= r4 sin2 ;
thus,
p
Z = r2 sin :
55
Ex. 188: We compute, using the Voss-Weyl formula,
ri ri F
=
=
=
=
=
1 @
p
Z @Z i
1
r2 sin
1
r2 sin
1
r2 sin
1
r2 sin
p
@
@r
@
@r
@F
@Z j
r2 sin (1)
@F
@r
+
@
@
r2 sin r2
@F
@
+
@
@
r2 sin
@F
@2F
+ sin
@
@ 2
@F
@2F
@2F
+ r2 cot
+ r2 2 + r2 sin2
@
@
@ 2
@F
@2F
@2F
+ r2 2 + r2 cot
+ r2 sin2
@
@
@ 2
@F
@2F
+ r2 2
@r
@r
1
r2
=
2 @F
@2F
+
r @r
@r2
=
@2F
2 @F
+
@r2
r @r
+
r2
sin
cos
=
=
=
=
=
p
1 @
@F
p
ZZ ij
i
j
@Z
@Z
Z
1 @
@F
@
@F
r (1)
+
r r2
r @r
@r
@
@
2
2
@ F
1 @
@F
@ F
r
+ r3 2 + r 2
r @r
@r
@z
@
2
2
@2F
1 @F
@ F
@ F
+ r 2 + r3 2 + r 2
r @r
@r
@z
@
2
2
2
@ F
@ F
1 @F
@ F
+ r2 2 +
+
:
@r2
r @r
@z 2
@
+
@
@z
r (1)
@F
@z
@F
@
@F
@
@F
@
+ r4 sin3
+ r2 sin2
Ex. 189: We compute, for cylindrical coordinates
ri ri F
r2 sin r2 sin2
@
@
@F
@F
+
r4 sin
+
r4 sin3
@r
@
@
@
@
@F
@F
@
@
sin
r2
+ r4
sin
+ r4 sin3
@r
@r
@
@
@
@F
@2F
@F
@2F
sin
2r
+ r2 2 + r4 cos
+ sin
@r
@r
@
@ 2
=
2r
ZZ ij
@2F
@ 2
@2F
@ 2
56
CHAPTER 9. CHAPTER 9
Part II
Part II
57
Chapter 10
Chapter 10
Ex. 213: Note that
Z i Zj
Zi (S
=
S
=
S (S
=
6
=
(S
i
j;
Zj )
Zj ) Zi
Zi
Zj ) S
since
(S
Zj ) S
is merely the projection of Zj onto the tangent space. [ASK]
EARLIER ATTEMPT:
Z i Zj
S
Zi Zj S
S S
=
i
Z Zj
S Zj
S Zj
=
=
i
j
i
j
i
j
i
j
i
=
i
= Zj Zi ;
Z
Z
which forces
S Zj = Zj
??? [Not sure - maybe a dimensional argument?]
59
60
CHAPTER 10. CHAPTER 10
Ex. 214: We have
T i = T Zi :
Now,
T i Zi
= T Z i Zi
= T
= T ;
as desired.
Ex. 215: We show that (V
(V
P) N = 0. Compute
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
Pji Pkj
= N i Nj N j Nk
= N i (1) Nk
= Pki ;
as desired.
Ex. 217:
compute
We show that V
(V
T) S
T is orthogonal to the tangent plane.
=
=
=
=
=
(V
V
V
V
0:
(V
S
S
S
S )S ) S
(V S ) S S
(V S )
V S
We
61
Ex. 218: Similarly to 216, we have, given de…nition Tji = N i Nj
Tji Tkj
= N i Nj N j Nk
= N i Nk
= Tki :
[Note: This seems like the exact same problem - do we mean to de…ne
Tji = T i Tj ?]
Ex. 219: We have [Note that this implies 213 additionally]
i
j:
N i Nj + Z i Zj =
Contract both sides with Ni :
i
j Ni
N i Nj Ni + Z i Zj Ni
=
Ni N i Nj + Ni Z i Zj
= Nj
i
Ni N Nj + 0 = Nj
Ni N i Nj = Nj ;
where the third line follows from Ni Z i = 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
Ni N i = 1;
as desired.
Ex. 220: Using similar manipulations of indices to the earlier discussion of
the Levy-Civita symbols, we derive
1
4
ijk t s
rst Tj Tk
1
4
=
=
1
4
ijk s t
rts Tj Tk
ijk s t
rst Tj Tk ;
so
N i Nr
=
1
4
=
2
=
1
2
ijk s t
rst Tj Tk
1
4
1
4
ijk s t
rst Tj Tk
ijk s t
rst Tj Tk :
ijk t s
rst Tj Tk
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
J
0
0
@S
:
=
@S
Ex. 222: From before, we have
@Zij
= Zli
@Z k
From the analogous de…nitions of S
@S
@S
l
jk
+ Zlj
l
ik :
; we have
=S
+S
compute
=
=
=
=
=
1 ! @S!
@S!
S
+
2
@S
@S
1 !
S
S!
+S
2
1 !
S
S!
+S
2
1
+ S !S
2
1
2
2
;
@S
@S !
!
+S
!
+S
!
!
+S
!
+S
!
+
+S
!
S
S
!
S
!
!
S
!
S
+S
S
!
!
!
S
!
S
as desired.
Ex. 223: Assume the ambient space is re¤ered to a¢ ne coordiantes. We
have
@Z i
+ ijk Z Z j Z k
@S
@Z i
= Zi
+ 0;
@S
= Zi
since
i
jk
= 0 in a¢ ne coordinates.
!
63
Ex. 224 [Still Working]
Ex. 225 We compute, given
Z1 ( ; )
Z2 ( ; )
Z3 ( ; )
Zi
=
=
@Z i
@S
2
0
0
we have
Zi =
Ni
@Z 1
@S 2
@Z 2 7
@S 2 5
@Z 3
@S 2
3
0
05 ;
1
3
0
1
05 =
0
1
2
0
0 4
1
1
0
1
0
3
@Z 1
@S 12
6 @Z
4 @S 1
@Z 3
@S 1
2
0
= 41
0
then, note that since
= R
=
=
0
0
1
0
2 3
0
= 415
0
=
i
0
0
j
1
0
0
;
1
0
1
2 3
0
405
1
k
0
1
= i2 3
1
= 405
0
S1
S2
=
=
=
=
=
=
Z1i Zi
Z12 Z2
R cos cos i + R cos sin j R sin k
Z2i Zi
Z23 Z3
R sin sin i + R sin cos j
64
CHAPTER 10. CHAPTER 10
S
=
S
p
R2 cos2 cos2 + R2 cos2 sin2 + R2 sin2
R2 cos cos sin sin + R2 cos sin sin cos
=
R2
0
0
R2 sin2
R 2
0
0
R 2 sin
s
R2
0
=
0 R2 sin2
=
S
R2 cos cos sin sin + R2 cos sin
R2 sin2 sin2 + R2 sin2 cos2
[ASK - should this be the same as when the ambient coordinates are Cart.?
2
= R2 sin :
Now, recall the Christo¤el symbols for the ambient space (in spherical coords):
Now, setting
1
22
1
33
=
=
2
12
=
2
33
=
3
13
=
2
23
=
as coord. 1 and
= Zi
r
r sin2
1
2
21 =
r
sin cos
1
3
31 =
r
2
=
cot
:
32
as coord. 2, and using
@Z i
+
@S
i
jk Z
Zj Zk;
65
we compute
1
11
@Z1i
+
@S 1
@Z 2
Z21 11 +
@S
2
@Z
Z21 11 +
@S
@Z12
@S 1
0
1
12 = 0 +
= Zi1
i
1 j k
jk Z Z1 Z1
=
2
1 j k
jk Z2 Z1 Z1
=
=
=
=
1
21
2
1 2 2
22 Z2 Z1 Z1
i
1 j k
jk Z Z2 Z1
2
1 j k
jk Z2 Z2 Z1
2
1 3 2
32 Z2 Z2 Z1
=
=
= cot (1) (1) (1)
= cot
i
1 j k
=
jk Z Z2 Z2
1
22
2
1 j k
jk Z2 Z2 Z2
2
1 3 3
33 Z2 Z2 Z2
=
=
=
2
11
sin cos
=
=
2
21
=
= 0
2
=
12 =
=
2
22
i
2 j k
jk Z Z1 Z1
3
2 j k
jk Z3 Z1 Z1
3
2 2 2
22 Z3 Z1 Z1
i
2 j k
jk Z Z2 Z1
3
2 j k
jk Z3 Z2 Z1
3
2 3 2
32 Z3 Z2 Z1
=
= 0
i
2 j k
=
jk Z Z2 Z2
3
2 3 3
=
33 Z3 Z2 Z2
= 0;
(note
@Z i
@S
vanishes in each computation).
66
CHAPTER 10. CHAPTER 10
Ex. 226: We have
q
2
2
(x0 (s)) + (y 0 (s)) = 1;
since this is an arc-length parametrization. Thus,
Ni
y 0 (s)
x0 (s)
2
S
S
p
=
S
2
= (x0 (s)) + (y 0 (s))
= 1
= 1
= 1
[ASK why x0 x00 + y 0 y 00 = 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)
q
2
2
2
r0 (s) + r (s) 0 (s) = 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, 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
=
r S
@S
@S
@S
=
@S
@S
=
@S
@S
=
@S
@S
=
@S
@S
=
@S
= 0:
=
1
S
2
!
@S!
@S!
+
@S
@S
S
S
S
S
@S
@S !
1 ! @S!
@S!
@S
S
+
S
2
@S
@S
@S !
1 ! @S!
@S!
@S
1 !
+
2
@S
@S
@S !
2
1 @S
@S
@S
1 @S
+
2 @S
@S
@S
2 @S
1 @S
1 @S
2 @S
2 @S
Similarly, we may show that in the contravariant case,
r S
= 0:
67
1 ! @S!
@S!
@S
S
+
2
@S
@S
@S !
@S!
@S!
@S
+
@S
@S
@S !
@S
@S
+
@S
@S
S
68
CHAPTER 11. CHAPTER 11
For the Levy-Civita symbols, note
"
=
"
=
p
Se
1
p e
S
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 ambient
case.
Ex. 233: We compute, using
S
=
R 2
0
R
2
0
sin
2
;
the following:
r r F
=
=
=
=
=
=
1 @
p
S @S
1
2
R sin
1
2
R sin
1
R2 sin
1
R2 sin
1
2
R sin
p
@
@
@
@
@
@
@
@
@
@
@F
@S
SS
@F
@
@F
1
R2 sin S 2
+ 2
@S
R sin @
@S
@
1
@F
@F
R2 sin S 11
+ 2
R2 sin S 22
@
R sin @
@
1
@
@F
R2 sin R 2
+ 2
R2 sin R 2 sin
@
R sin @
@F
1
@
1 @F
sin
+ 2
@
R sin @
sin @
@F
1
@2F
sin
+ 2 2
:
@
R sin @ 2
R2 sin S 1
Ex. 234: For the surface of a cylinder, note
S
p
=
S
R 2
0
= R;
0
1
2
@F
@
69
so
r r F
p
@F
1 @
p
SS
@S
@S
S
1 @
@F
1 @
RS 11
+
R@
@
R @z
@2F
1 @2F
+
:
R2 @ 2
@z 2
=
=
=
RS 22
@F
@z
Ex. 235: Note
S
p
=
S
(R + r cos )
0
2
0
r
2
= r (R + r cos ) ;
and compute
r r F
p
1 @
@F
p
SS
@S
@S
S
1
@
@F
=
r (R + r cos ) S 1
r (R + r cos ) @
@S
@
@F
1
r (R + r cos ) S 2
+
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 ) @
@
2
1
@ F
1
@
@F
=
(R + r cos )
2
2 + r 2 (R + r cos ) @
@
@
(R + r cos )
=
Ex. 236: We have
"
S
p
S
2
r (z)
0
=
1
0
1+r 0 (z)2
q
2
= r (z) 1 + r0 (z) ;
#
:
70
CHAPTER 11. CHAPTER 11
Thus,
r r F
1 @
p
S @S
p
@F
@S
1
@ p 11 @F
1
@
q
q
=
SS
+
@
@
@z
2
2
r (z) 1 + r0 (z)
r (z) 1 + r0 (z)
q
1
@
2
2 @F
q
=
r (z) 1 + r0 (z) r (z)
@
@
2
r (z) 1 + r0 (z)
!
q
@
1
@F
1
2
q
+
r (z) 1 + r0 (z)
2
2 @z
1 + r0 (z) @z
r (z) 1 + r0 (z)
0
1
1
r (z)
1 @2F
@ @
@F A
q
q
=
+
2
2 :
2 @z
2 @z
r (z) @
r (z) 1 + r0 (z)
1 + r0 (z)
=
SS
p
SS 22
Ex. 237: These were computed earlier.
Ex. 238: We compute:
r Zi
=
=
=
=
@Zi
Z j kij Zk
@S
@Zi @Z m
Z j kij Zk
@Z m @S
@Zi m
Z
Z j kij Zk
@Z m
k
m
Z j kij Zk
im Zk Z
= Zm
= 0;
after index renaming.
Zij = Zi Zj ; we have
k
im
Zj
k
ij
Zk
The contravariant case follows similarly.
r Zij = 0
by the product rule. Similarly,
r Z ij = 0:
[Levy-Civita Symbols to come]
Also, since
@F
@z
71
Ex. 239: Begin with equation 10.41:
Ni N i = 1;
and compute take the surface covariant derivative of both sides:
0
= r
Ni N i
= r Ni N i + Ni r N i
= r N j Zij Nk Z ik + Ni r N i
r N j Zij Nk Z ik + Ni r N i ;
=
by the metrinilic property,
= r N j Nk
k
j
+ Ni r N i
= r N k Nk + Ni r N i
= Nk r N k + Ni r N i
= 2Ni r N i ;
after index renaming. Thus,
Ni r N i = 0:
72
CHAPTER 11. CHAPTER 11
Ex. 240: We compute
"ijk "
Zj Nk
= "ijk "
=
=
=
=
=
=
=
=
1
"kmn " Z m Z n
2
Zj
1 ijk
" "kmn " " Zj Z m Z n
2
1 ijk
Zj Z m Z n
2 kmn
1 ijk
Zj Z m Z n
2 mkn
1 ijk
Zj Z m Z n
2 mnk
1 ij
2
Zj Z m Z n
2 mn
ij
Zj Z m Z n
mn
j i
m n
i j
m n
i j
m n
j i
m n
Zj Z m Z n
=
Zj Z i Z j
=
Zi
= Zi
Zj Z m Z n
Zi
Zj Z j Z i
Zj Z i Z j +
Zi +
Zi
i j
m n
Zj Z m Z n
Zj Z m Z n +
Zj Z j Z i
Zi
Zi + Zi
[Some factors of 2 needed?]
Ex. 241: Note that for a general covariant-contravariant tensor, we have
r Tji = Z k rk Tji :
Thus,
r u = Z k rk u
and
r u=Z
k
rk u:
Thus,
r r u
= r
Z
k
k
rk u
= r Z r k u + Z k r rk u
= B N k rk u + Z k Z m rm rk u:
= B N k rk u + Z
= B N k rk u + Z
k
k
Zn Z mn rm rk u
Zn S Z mn rm rk u
j i
m n
Zj Z m Z
73
Now, set
=
and contract:
r r u
= B N k rk u + Z
= B N k rk u +
k
Zn S Z mn rm rk u
Z k Zn Z mn rm rk u
= B N k rk u + Z k Zn Z mn rm rk u:
Now,
N k Nn + Z k Zn =
k
n;
so
Z k Zn =
k
n
N k Nn :
k
n
N k Nn Z mn rm rk u
We substitute in the above:
= B N k rk u +
r r u
= B N k rk u + Z km rm rk u N k Nn Z mn rm rk u
= B N k rk u + rm rm u N m N k rm rk u;
or after renaming dummy indices,
r r u = B N i ri u + ri ri u
N i N j ri rj u;
or
N i N j r i r j u = ri ri u
Ex. 242: Let
r r u + B N i ri u
Z i (s)
be the parametrization of the line normal to the surface, emanating from point
Z0i . Note that we have
Z i (h)
dZ i
(0) = lim
h!0
ds
h
Z0i
= N i:
also, compute
d dZ i
Zi
ds ds
=
=
=
d2 Z i
dZ i dZi
(Z (s))
Zi +
2
ds
ds ds
d2 Z i
Zi +
ds2
d2 Z i
Zi +
ds2
dZ i @Zi dZ k
ds @Z k ds
dZ i n
dZ k
Z
;
n
ds ik
ds
74
CHAPTER 11. CHAPTER 11
so at s = 0;
d dZ i
Zi js=0
ds ds
=
=
=
=
d2 Z i
Zi + N i nik Zn N k
ds2
d2 Z i
Zi + N i N k nik Zn
ds2
d2 Z i
Zi + N j N k njk Zi
ds2
d2 Z i
+ N j N k ijk Zi
ds2
Now, examine the LHS of the above. Since
d dZ i
Zi
ds ds
represents the second derivative of a line, the LHS vanishes. Thus,
d2 Z i
=
ds2
NjNk
i
jk ;
Then, de…ne
F (s) = u (Z (s)) ;
so
F 0 (s)
=
F 00 (s)
=
=
dZ i
@u
(Z (s))
(s)
i
@Z
ds
d @u
dZ i
@u
d dZ i
(Z
(s))
(s)
+
(Z
(s))
(s)
ds @Z i
ds
@Z i
ds ds
@2u
dZ i
dZ j
@u
d2 Z i
(Z
(s))
(s)
(s)
+
(Z
(s))
(s) :
@Z i @Z j
ds
ds
@Z i
ds2
at s = 0 :
F 00 (0)
=
=
now, note
@2u
@u
d2 Z i
i j
(Z
(0))
N
N
+
(Z
(0))
(0)
@Z i @Z j
@Z i
ds2
@
d2 Z i
[ri u] N i N j + ri u 2 (0)
j
@Z
ds
75
rj ri u
@ri u
@Z j
@ri u
@Z j
=
= rj ri u +
k
ij rk u
k
ij rk u;
so
F 00 (0)
d2 Z i
(0)
ds2
d2 Z i
= rj ri uN i N j + kij rk uN i N j + ri u 2 (0)
ds
i j
i j k
j k i
= N N rj ri u + N N ij rk u N N jk ri u
=
rj ri u +
k
ij rk u
N i N j + ri u
= N i N j rj ri u;
thus
after renaming indices.
@2u
= F 00 (0) = N i N j ri rj u
@n2
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
0:
@
@S
@
@S
@
@S
!
+
+
!
!
!
!
Ex. 245: This was done for the …nal exam.
Ex. 246: Compute
R
= R
=
R
=
R
77
!
!
(12.5)
(12.3)
(12.5).
!
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
G
= R
S
= R
S
1
RS S
2
1
R ;
2
so
G
= R
= R
= R
= 0;
S
R
R
R
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 partial
derivatives. Now, we consider the covariant case:
(r r
r r )Ti
= r r Ti
=
=
=
=
=
=
=
@ r Ti
@S
r r Ti
r Ti
@ r Ti
@S
r Ti
@ r Ti
@ r Ti
@S
@S
i
@T
@
@T i
@
+ Z k ikm T m
+ Zk
@S
@S
@S
@S
@2T i
@
@
@2T i
+
Z k ikm T m
@S @S
@S
@S @S
@S
@
@
Z k ikm T m
Z k ikm T m
@S
@S
@Z k i
@ i
@T m
m
+ Z k km + Z k ikm
km T
@S
@S
@S
i
@Z k i
@
@T m
m
Z k km T m Z k ikm
km T
@S
@S
@S
0 [not sure yet]
!
i
m
km T
Zk
i
m
km T
80
CHAPTER 12. CHAPTER 12
Ex. 252: Look at
R
+R
as desired.
+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;
=
!
!;
!
!
!
!
!
!;
!
!;
!;
!
!
!;
!;
!
!
!
81
Ex. 253: Compute
r" R
+r R
"
+r R
"
@R
@S "
@R
+
@S
@R
+
@S
@R
=
@S "
@R
+
@S
@R
+
@S
@R
=
@S "
@R
+
@S
@R
+
@S
@R
=
@S "
+:::
!
=
"
"
!
"
"
!
"
"
!
"R !
" R!
!
R!
!
R!
as desired.
R
!
R!
!
R!
"
"
!
R!
!
R!
"
@
!;
"
R
! "
!
R
!"
!
R
!"
!
"
R
!
!
"R
!
R
! "
!
R
!"
+
!
"R
!
!
!
"
R
!
!
R
"!
!
"R
!
!
!
R
!"
+
!
"
R
!
!
"
R
!
+
!
R
!"
!
"R !
" R!
!
!
!
"R !
"
@S
1
R1212
" "
" "
4
S
1
R1212
=
(2) (2)
4
S
R1212
=
S
= K;
=
"
" R!
Ex. 254: Compute
1
" "
4
"
!
"R
!
R
! "
!
R
!"
@
!;
@S
+
;!
;
!
!
"
@
; !
@S
82
CHAPTER 12. CHAPTER 12
Ex. 255: Compute
K (S
S
S S
)
=
=
=
=
=
=
Ex. 256: Note
1
R
2
1
" " R
(S S
4
1
" " S S R
4
1
" " S S R
+
4
1
1
R
+
R
4
4
1
1
(2) R
+ (2) R
4
4
R
:
=
1
R!
2
S! S ;
so
1
R
2
1
R! S ! S
2
1
K"! " S ! S
=
2
1
K
=
2
= K
=
Ex. 257: This follows since
r S =r S ;
hence
NB
= NB
;
or
B
=B
:
S S
1
" "
4
1
" "
4
)
S S
S S
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
jB j =
:
Now,
B B := C :
In linear algebra terms, we have
C =B2
then,
= tr B 2
=
1 + 2;
B B
where 1 ; 2 are the eigenvalues of B 2 . But, since
2
by the properties of eigenvalues, we have
B B
=
=
1
=
2
2
2 jB j
by the above.
Ex. 259: We compute, given
r (z)
= a cosh
z
b
a
(z b)=a
= a
a (z
e
2
=
+ e(b
2
a
b)=a
+ e(b
2
1 (z
e
2
b)=a
1 (z
e
2a
b)=a
r0 (z) =
r00 (z) =
e
1 (b
e
2
+
1 (b
e
2a
z)=a
z)=a
z)=a
z)=a
:
2
and
2
=(
2
) =
84
CHAPTER 12. CHAPTER 12
Compute
r00 (z) r (z)
2
r0 (z)
=
=
=
1 (z
e
2a
1 2(z
e
4
1
b)=a
b)=a
+
1 (b
e
2a
1 2(b
e
4
a (z
e
2
z)=a
1 2(z
e
4
z)=a
Thus,
r00 (z) r0 (z)
2
r0 (z)
1=0
2
B
r00 (z) r0 (z) r0 (z)
q
2
r (z) 1 + r0 (z)
=
=
0;
as desired.
Ex. 260: We have
V
as desired.
dR
(S (t))
dt
@R dS
=
@S dt
= S V
= V S
=
1
b)=a
b)=a
a
+ e(b z)=a
2
1
1 + e2(b z)=a
4
1 (z
e
2
b)=a
1 (b
e
2
z)=a
85
Ex. 261: Compute
A
=
=
=
=
=
=
=
=
=
=
=
dV
dt
d
[V S ]
dt
dS
dV
S +V
(S (t))
dt
dt
dV
@S dS
S +V
dt
@S dt
dV
@S
S +V V
dt
@S
dV
r S +
S +V V
S
dt
dV
S +V V
S +V V r S
dt
dV
S +V V
S +V V r S
dt
V
S +V V r S
t
V
S + NV V B
t
V
S + NB V V ;
t
as desired.
Ex. 262: We de…ne
T
dT
=
+V
t
dt
!T
!
V
!
T!:
Ex. 263: Look at
0
0
T
t
=
=
=
=
=
=
dT
0
0
+V
0
!T
!
0
V
!
0
T!
0
dt
0
0
0
d
!
T J (S (t)) J 0 (S 0 (t)) + V
V ! 0 T! J
!T J 0
dt
0
@J 0 0 dS 0
dT
0
0
@J dS
!
J J 0 +T
J 0 +T
+V
J
!T J 0
dt
@S dt
@S 0
dt
dT
0
0
0
0
0
!
J J 0 + T J V J 0 + T J 0 0V J + V
V
!T J 0
dt
::: ???
T
0
J J 0
t
!
V
!
0
0
T! J
T! J
0
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
t
=
=
=
dS (S (t))
V ! S!
dt
@S dS
V ! S!
@S dt
r S + ! S! V
V
= r S V +V
!
S!
V
!
!
S!
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 =
1
R
0
0
;
0
so
K
= jB j
=
0
=
0:
1
R
87
Ex. 271: For a cone, we have
cot
r
B =
0
;
0
0
which has determinant zero. Thus,
K = 0:
Ex. 272: For a sphere, we have
1
R
B =
0
1
R
0
;
so
K
1
R
= jB j =
=
1
R
1
R2
Ex. 273: For a torus, we have
cos
R+r cos
B =
0
1
r
0
so
K=
;
cos
r (R + r cos )
Ex. 274: For a surface of revolution, we have
3
2
p1
0
2
0
r(z) 1+r (z)
5;
B =4
r 00 (z)
0
0 (z)2 3=2
1+r
(
)
so
K
=
r (z)
=
r00 (z)
1
q
2
2
1 + r0 (z)
1 + r0 (z)
r00 (z)
2
r (z) 1 + r0 (z)
2
3=2
88
CHAPTER 12. CHAPTER 12
Ex. 275: We integrate
Z
KdS
=
S
=
=
=
Z
1
dS
2
R
S
Z
1
dS
R2 S
1
4 R2
R2
4
Ex. 276: We integrate
Z
Z 2 Z 2
p
cos
cos
Sd d
dS =
r (R + r cos )
S r (R + r cos )
0
0
Z 2 Z 2
cos
=
r (R + r cos ) d d
r (R + r cos )
0
0
Z 2 Z 2
=
cos d d
0
=
=
0
2 (sin (2 )
0:
sin (0))
Part III
Part III
89
91
Ex. 291: We have
( ) = arccot (At cot ) ;
so
= arccot (At cot )
= At cot
cot
1
cot
At
( )
Jt
=
=
=
=
=
Jt
0
=
cot
=
arccot
1
cot
At
@S (t; S 0 )
@t
1
@
arccot
cot
@t
At
1
1 + A21t2 cot2
cot
At2 + A1 cot2
A cot
A2 t2 + cot2
=
=
@
arccot (At cot )
@t
A cot
1 + A2 t2 cot2
Ex. 292:
V =
cot
At2
@Z i
A cos
=
0
@t
92
Ex. 293:
Vi
2
@
6 @t
i
=
@Z
=6
4@
@t
@t
2
= 4
2
= 4
2
2
p2A
2
p2A
p
p
p
A cos
cos2 +A2 t2 sin2
p
A sin
cos2 +A2 t2 sin2
t sin2
cos2
2
A cos
+A2 t2
2
t sin
3
sin2
A sin
3
cos2 +A2 t2 sin2
A3 t sin2
cos
cos2 +A2 t2 sin2
A3 t sin3
cos2 +A2 t2 sin2
3
3
3
3
7
7
5
5
3
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:
Zi ( ) =
At cos
sin
Then, compute the shift tensors:
Zi
@Z i
@S
At sin
cos
=
=
;
so
1
At
Zi =
sin
cos
;
since we need
Zi Z i =
:
Thus,
V
= V i Zi
=
1
At
=
1
t
sin
sin
cos
:
A cos
0
93
Ex. 296: Recall
Zi ( ) =
and note that
@ p
@
cos2 + A2 t2 sin2
2
A cos
2 t2 sin2
4 cos2 A+A
sin
p
cos2 +A2 t2 sin2
p
=
=
so
Zi
=
=
@Z i
@S
cos2
1
+ A2 t2 sin2
3
5;
2 cos sin + 2A2 t2 sin cos
p
2 cos2 + A2 t2 sin2
A2 t2 sin cos
cos sin
p
2
2
2
cos + A t sin2
2p
cos2 + A2 t2 sin2 ( A sin ) A cos
4p
cos2 + A2 t2 sin2 (A cos ) A sin
A2p
t2 sin cos
cos sin
cos2 +A2 t2 sin2
A2p
t2 sin cos
cos sin
cos2 +A2 t2 sin2
The result is V = 0, since the motion of a particle on the surface with
constant is normal to tangent space; hence the projection V = V i Zi = 0.
Ex. 297: We see that V
coordinates.
is not a tensor with respect to changes in ambient
Ex. 298: We have
V = V i Zi ;
so
V 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 Zi are tensors with regard to ambient coordinate
changes; hence the contraction V = V i 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,
0
the Jacobians Jii are the identity. We compute
0
0
V i Jii + Z i Jii Jt
= V i + Z i Jt
A cot
A cos
=
+ 2 2
0
A t + cot2
1
At
sin
cos
:
3
5;
94
[To be continued]
Ex. 302: Write
V
0
0
= V i Zi0
0
0
0
0
= V i Jii Zj Jij0 J
= V i Zj J
0
= V j Zj J
0
= V J
= V J
0
0
j
i
0
0
+ Z i Jii Jt Zj Jij0 J
+ Z i Jt Zj J
+ Z j Jt Zj J
+
0
Zj Jij0 J
V i Jii + Z i Jii Jt
=
Jt J
0
0
j
i
0
0
0
+ J Jt ;
as desired.
Ex. 303: Let the unprimed coordinates denote the …rst parametrization.
Then, note
J
0
d
(arccot (At cot ))
d
1
At csc2
1 + A2 t2 cot2
At csc2
1 + A2 t2 cot2
=
=
=
V J
0
0
+ J Jt
=
=
At csc2
1
sin
t
1 + A2 t2 cot2
+
A cot
At csc2
A2 t2 + cot2 1 + A2 t2 cot2
[perhaps I am using the wrong Jacobians]
Ex. 304: We have
Zi =
so
S =
t cos
sin
;
d
d
(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
t sin
Ni =
;
and
Ni = cos
t sin
;
since our ambient space is in Cartesian coordinates. Thus,
C
= V i Ni
cos
=
cos2 :
=
Ex. 305: As before, compute
!
cos
d
d
p
i+
S
=
2
d
d
cos2 + t2 sin
=
cos2
p
=
cos2 + t2 sin2
!
j
!
t2 sin cos
cos sin
p
cos2 + t2 sin ( sin ) cos
2
2
cos + t sin2
!
t2 sin cos2
cos2 sin
cos
i+ p
p
3
cos2 + t2 sin2
cos2 + t2 sin2
sin
cos2 + t2 sin2
so
h
cos
Ni = pcos2 +t2 sin2
sin
p
p
1
+ t2 sin2
cos
0
t sin
2
cos sin2 +t2 cos sin2
(cos2
+t2
sin2
p
3
)2
cos2
sin
+t2 sin2
cos2
+
sin +t2 cos2
(cos2
+t2
sin2
t2 sin2 cos
cos sin2
p
3
cos2 + t2 sin2
sin
3
)2
and
C
p
i+ cos2 + t2 sin2 (cos
i
;
= V i Ni
=
=
h
p
cos
cos2 +t2 sin2
2
cos sin
2
2
+t cos sin
3
(cos2 +t2 sin2 ) 2
p
sin
cos2 +t2 sin2
+
2
cos
2
2
sin +t cos
sin
3
(cos2 +t2 sin2 ) 2
!
i
2
4
p
p
t sin2
cos
cos2 +A2 t2 sin2
t sin3
cos2
+A2 t2
sin2
3
3
3
5
96
[Note: Shouldn’t both be 1 by geometric considerations?]
Ex. 306: We have
0
V
@T (t; S 0 )
@t
0
0
= V J + J Jt
@T (t; S)
=
+ Jt r t;
@t
so compute
_ (t; S 0 )
rT
=
=
0
@T (t; S 0 )
V (t; S 0 ) r 0 T
@t
0
0
@T (t; S)
+ Jt r t V r T J ;
@t
since r 0 T has the tensor property. But,
V
0
=V J
0
0
+ J Jt ;
so
_ (t; S 0 )
rT
=
=
=
=
=
0
0
@T (t; S)
+ Jt r t
V J + J Jt r T Ja0
@t
@T (t; S)
+ Jt r t
V
+ Jt
r T
@t
@T (t; S)
+ Jt r t V r T Jt r T
@t
@T (t; S)
V r T
@t
@T (t; S)
V r T;
@t
_ does not depend on changes in suface coordinates.
so rT
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
S = Z i Zi :
Ex. 313: Begin with
_ = Vi
rR
V Z i Zi
and
= V j Zj ;
V
so
_
rR
=
Vi
V j Zj Z i Zi
=
Vi
Vj
=
Vi
V i + V j N i Nj Zi
i
j
N i Nj
Zi
= V j N i Nj Zi ;
as desired.
Ex. 314: Compute
Z
Z
d
d
NB dS =
r S
dt S
dt
Z S
_ (r S
r
=
S
Z
@ (r S
=
@t
ZS
@ (r S
=
@t
S
dS
) dS
Z
CB r S dS
Z
V r r S dS
CB r S dS
S
Z
V i Zi r r S dS
CB r S dS
S
)
)
S
[ask about integral problems]
Ex. 315: By the above,
Z
NB dS
S
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
Z
2 NdS
S
for all t, since for a sphere,
B =
But,
R
S
2
:
R
NdS = 0 (our surface is closed), so we have that
Ex. 316: Need to show:
Z
d
Fd
dt
i.e.
d
dt
Z
A2
A1
Z
=
Z
@F
d +
@t
Z
S
NB dS vanishes.
CF dS;
S
b(t;x)
F (x; y) dydx =
0
R
Z
CF dS;
S
since @F
@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,:
Z
Z
CF dS =
CF dB
S
B
Clearly, S is the vector (given relative to the ambient Cartesian basis)
S = i + bx j,
so
and
1
N= p
( bx i + j) ;
1 + b2x
Vi =
0
;
bt
so
C
= V i Ni
bt
= p
:
1 + b2x
99
p
Now, at each t, our surface has line element 1 + b2x , so
Z
Z A2
CF dB =
bt F (x; b (t; x)) dx
B
A1
A2
Z
=
d
dt
A1
by FTC,
=
d
dt
Z
A2
A1
Z
Z
b(t;x)
F (x; y) dydx;
0
b(t;x)
F (x; y) dydx
0
Ex. 317: We have
Ui
0
0
=
=
=
=
=
=
=
@T i (t; S 0 )
@t
0
@
T i (t; S) Jii (Z (t; S))
@t
0
0
@ i
@
T (t; S (t; S 0 )) Jii + T i
Jii (Z (t; S))
@t
@t
0
0
@T i @
@T i
@J i @
+
S
Jii + T i ij Z (t; S)
@t
@S @t
@Z @t
0
@T i
@T i @S
@Z j
@Z j @S
i0
+
Jii + T i Jji
+
@t
@S @t
@t
@S @t
i
0
0
@T
i
Ui +
J
Jii + T i Jji
V j + Z j Jt
@S t
0
0
@T i
i0 j
i0 j
J J i + T i Jji
V + T i Jji
Z Jt ;
U i Jii +
@S t i
which is the desired result.
Ex. 318: This follows similarly to the above.
Ex. 319: Write
U
0
=
=
=
=
@T
0
(t; S 0 )
@t
0
@
T (t; S) J (t; S)
@t
0
0
@
@
T (t; S) J + T
J (t; S)
@t
@t
!
0
0
0
@T
@T @S
@J
@J @S
+
J +T
+
@t
@S @t
@t
@S @t
= U J
0
+
@T
J J
@S t
0
+T J
0
t
+T J
0
Jt :
100
Ex. 320: The covariant case is analogous to the above.
Ex. 321: Note
V r Ti
0
0
= V r T i Jii
0
@T i
(t; S) + ijk T j Z j Jii
= V
@S
0
0
@T i
(t; S) + ijk T k Z j Jii
= V j Zj 0
@S
0
0
0
@T i
=
V j Jjj + Z j Jjj Jt
(t; S) + ijk T k Z j Jii
@S
i
0 @T
0
0
0
0
@T i
+ V j Jjj ijk T k Z j + Z j Jjj Jt
+ Z j Jjj Jt ijk T k Z j Jii
=
V j Jjj
@S
@S
i
0
0
0
0 @T
0
0
0
@T i i0
Jii + V j Jjj ijk T k Z j Jii + Z j Jjj Jt
Ji + Z j Jjj Jt ijk T k Z j Jii ;
= V j Jjj
@S
@S
so given our work above,
@T i
@t
0
V r Ti
0
[not …nished]
=
0
@T i i0
@T i
i0 j
i0 j
Ji +
Jt Jii + T i Jji
V + T i Jji
Z Jt
@t
@S
i
0
0
0
0 @T
0
0
@T i i0
Jii + V j Jjj ijk T k Z j Jii + Z j Jjj Jt
Ji + Z j Jjj Jt
V j Jjj
@S
@S
i
k j i0
jk T Z Ji
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
_
_ T i Zi
rT
= r
_ i Zi + T i rZ
_ i;
= rT
by commutativity with contraction and the product rule,
= T i Zi ;
since the second term would be zero by the metrinilic property.
Ex. 326: Compute
@Zi
@t
=
=
=
@Zi (Z (t))
@t
@Zi @Z j
@Z j @t
k
j
ij Zk V
= Vj
k
ij Zk ;
as desired.
Ex. 327: Write
T = Ti Zi ;
so
i
_ i Zi = rT
_ = @Ti Zi + Ti @Z
rT
@t
@t
101
V r Ti Zi ;
102
CHAPTER 13. CHAPTER 16
but
@Zi
@t
=
=
=
@Zi @Z j
@Z j @t
i
k j
jk Z V
i
j k
jk V Z ;
so we have
_ i = @Ti
rT
@t
V r Ti
Vj
k
ij Tk ;
after index renaming in the second term of the above expression.
_
_ =rT
_ i Zi and Zi is a
Ex. 328: We know rT
is invariant, and since rT
i
_
tensor, by the quotient law, rT must be a tensor. An argument using changes
in coordinates would follow similarly to the covariant derivative computations.
_ =rT
_ i Zi and by the
Ex. 329: This also follows from the fact that rT
quotient law. An argument using changes in coordinates would also follow
similarly.
Ex. 330: Put
Sj = Tji Zi :
then, clearly,
_ j = rT
_ i Zi :
rS
j
Dot both sides with Zk :
_ j Zk
rS
_ j Zk
rS
_ ji Zi Zk
= rT
_ jk :
= rT
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.
Ex. 332: Assume S i ; Tij are arbitrary tensors. Put
U j = S i Tij :
103
Now, compute
_ SiT j
r
i
=
=
@ S i Tij
@t
@U j
@t
V r
S i Tij + V n
V r Uj + V n
j
i j
nk S Ti
j
i j
nk S Ti ;
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
r Tji = Z k rk Tji ;
and compute
_ i (t; S)
rT
j
=
=
=
=
=
=
=
@Tji (t; Z (t; S))
V r Tji (t; Z (t; S)) + V k ikm Tjm (t; Z (t; S))
@t
@Tji
@Tji @Z k
i
+
V Z k rk Tji + V k ikm Tjm V k m
kj Tm
@t
@Z k @t
@Tji
@Tji k
i
+
V
V Z k rk Tji + V k ikm Tjm V k m
kj Tm
@t
@Z k
!
@Tji
@Tji
i
m
m i
k
+
+ km Tj
V Z k rk Tji
kj Tm V
@t
@Z k
@Tji
+ rk Tji V k V Z k rk Tji
@t
@Tji
+ V k V Z k rk Tji
@t
@Tji
+ CN k rk Tji ;
@t
where the last step follows from the observation that V k
component.
V Z k is the normal
i
Ex. 335: Note that @Z
@t = 0; and the second term of the above is also zero
_ i = 0: the other
by the metrinilic property for covariant derivatives. Hence, rZ
results follow similarly.
Vk
m i
kj Tm
(t; Z (t; S))
104
CHAPTER 13. CHAPTER 16
Ex. 336: Compute
@S (Z (t; S))
@t
=
=
=
=
=
=
@S @Z i
@Z i @t
@S i
V
@Z i
@S @S
Vi
@S @Z i
@2R
Z Vi
@S @S i
@S
Z Vi
@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
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
@t
@T
@t
@T
@t
@T
@t
@T
@t
S +T
S +T
@S
@t
V r T S
r V S + V NB
V T NB
+ Nr C
CB S
V r T S
S + r V T S + V T NB
+ T Nr C
CB T S
S + T r V S + T r CN
T CB S
V r T S
S + T r V S + T r CN
T CB S
V r T 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:
V T NB
V T NB
105
Ex. 341: Note
0
_
= rS
@S
=
@t
@S
=
@t
@S
=
@t
@S
=
@t
V ! r! S
(r V !
CB ! ) S!
r V ! S! + CB ! S!
r V !S
r V
r V + CB
+ CB
r V
r V + 2CB
;
r V!
!
+ CB ! S
so
@S
@t
=r V +r V
2CB
:
The contravariant case follows similarly.
Ex. 342: First, examine
@S
= SS
@S
;
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,
p
@ S
@t
1 @S
p
2 S @t
p
=
S (r V
=
CB ) ;
)
CB ! S
!
!
106
CHAPTER 13. CHAPTER 16
and
@
p
1
p
@ S
1
p 2 @t
S
p
1
S (r V
CB )
S
1
p (r V
CB ) :
S
S
=
@t
=
=
Ex. 343: Note
"
=
p
Se
;
so
@"
@t
p
@ S
e
@t
p
@ S
e
@t
=
=
since e
+
;
does not depend on t.
p
S r V
=
= "
"
CB
e
CB
:
r V
Ex. 344: Note
=
p
p @e
S
@t
1
S
e
;
so
@"
@t
@
=
=
=
=
p
1
S
@t
p
@
S
@t
e
+
p @e
S
@t
1
e
1
p r V
S
"
r V
CB
e
CB
:
107
Ex. 345: Simply write
_
r"
_ " S S
= r
_
= 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
_
2rK
_
= r"
"
= " "
_
B B + " r"
_
rB
B +" "
_
rB
B +" "
B B +" "
_
B rB
;
_
B rB
since the …rst two terms vanish,
= " "
r r C + CB B
= " "
B r r C + C" "
=
B
r r C + CB B
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 +" "
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 r r C+
B B B C+
B r r 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
=
2 B r r C
B r r C+ B B B
=
2 B r r C
B r r C + B KC ;
2B B B C
B B B
C
B B B C
B B B C
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
ri u = @Z
Write
i.
@ @u
@t @Z i
@ @u
=
@Z i @t
@
= ri u;
@t
@
ri u
@t
=
under smoothness assumptions (hence the partials commute), and since
an invariant.
Ex. 348: First, compute, given polar coordinates
ri u
=
=
=
@u
i
@Z
2
@
4 @r
"
J ( r)
p0
J1 ( )
J ( r)
p0
J1 ( )
@
@
p
J1 ( r)
J1 ( )
0
#
Now, since for polar coordinates,
Z ij =
1
0
;
0 r 2
109
3
5
@u
@t
is
110
CHAPTER 14. CHAPTER 17
so
ri u
= Z ij rj u
#
"
J ( r)
p 1
1
0
J
(
)
1
=
0 r 2
0
"
#
p
=
J1 ( r)
J1 ( )
:
0
thus,
2
ri uri u =
2
J1 ( r)
2
J1 ( )
:
Now, at t = 0, our surface yields r = 1, so
2
ri uri u =
Next, compute C for our surface evolution. Consider, in Cartesian coordinates,
Vi
=
@x
@t
@y
@t
=
a cos
b sin
With respect to the Cartesian basis,
S =
Ni = q
(1 + at) sin i + (1 + bt) cos j
1
2
(1 + at)
cos2
2
2
+ (1 + bt) sin
at t = 0,
Ni =
C
cos
sin
= V i Ni
= a cos2
+ b sin2 ;
(1 + bt) cos
(1 + at) sin
;
111
thus, by the Hadamard formula,
Z 2
a cos2
=
1
0
Z
2
=
2
+ b sin2
d
2
a cos2
+ b sin2 d
a cos2
+b 1
a cos2
+ b cos2 d
0
Z
2
=
2
cos2
d
0
Z
2
=
2
0
2
=
(a + b)
Z
2b
2
2
cos2 d
2b
2
0
2
=
(a + b)
2
+
1
sin 2
4
2
=
(a + b)
2
=
=
Since
=
2
2b
(a + b) 2b
2
(a + b) :
2
2b
2
0
2
2
, we have the desired result.
Ex. 349: [Not sure]
Ex. 350: Want to show:
Z
1 =
S
u 1 N i ri u
Dirichlet:
1
=
Z
Ni ri u1 dS:
Cri uri udS
S
Neumann:
1
=
Ex. 354: We have
_ + 2V r C + B
rC
_
rV
+V r V
Cr C
V V
=
CV B
=
B
0:
112
CHAPTER 14. CHAPTER 17
Write
V
= V i Zi
and
C = V i Ni ;
Begin with the second, and contract with S :
_ (V S )
r
_ (V S )
r
_
rV
S +V r V S
Cr CS
CV B S
=
0
V V r S
Cr CS
CV B S
=
0
V V B
Cr CS
CV B S
=
0:
_
V rS
+ V r (V S )
V Nr C + V r (V S )
N
Then, manipulate the …rst:
_ + 2V r C + B
rC
V V
=
B ;
and multiply by N:
_
rCN
+ 2V r CN + B
V V N =
B N
_ + 2V r CN + B
C rN
V V N =
B N
C ( S r C) + 2V r CN + B
V V N =
B N
_ (CN) + CS r C + 2V r CN + B
r
V V N =
B N
_ (CN)
r
_ (CN)
r
_ (CN) + 2V r CN + V V B
r
N+Cr CS
=
B N:
113
Now, add the results of both manipulations to each other:
_ (V S ) + r
_ (CN) + V r CN + V r (V S )
r
CV B S
=
B N
_ + V r CN + V r (V S )
rV
CV B S
=
B N
V Cr N + V r (V S )
CV B S
=
B N
_
rV+V
r V
CV B S
=
B N
CV B S
=
B N
V Cr N i Zi
CV B S
=
B N [note the metrinilic property]
_
rV+V
r V + V CZ i B Zi
CV B S
=
B N
_
rV+V
r V + V CB Z i Zi
CV B S
=
B N
_
rV+V
r V + CV B S
CV B S
=
B N
=
B N
_ + V r (CN)
rV
_
rV+V
r V
_
rV+V
r V
V Cr N
V Cr
N i Zi
_
rV+V
r V