Tensor Calculus with Applications 9789812385055, 9812385053, 9789812385062, 9812385061

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Maks A. Akivis ferusalem Institute of Technology Israel

Vladislav

" l

§ii'i'»8°;,>,

§~§2E:a,

n

.

"S,

* § 3. §?.a~... §-:5*|§l£**

rCacu US

r

.
2

7

15311

=

2731

+

0,31

AS

7

then equation (4) becomes 2

A1231II

+ A2272II2 + A3323II + R

(I)

The new coordinate axes O'ei» are the axes of symmetry of the seconddegree surface. They are called the principal axes of the surface. The point O' is the center of symmetry (the vertex) of the surface. Let I\17é0)

A 2 ; £ 0 , A 3 = 0 , ¢13'7é0~

5.

180

THE GENERAL THEORY oF SECOND-DEGREE

SURFACES

The last condition means that the vector l = a i f e i f is not perpendicular to the vector e 3 » . Then equation (3) can be written in the form A1 lscy

+ 0111 .~\1

2

+ A2 (az2f +

where

R=

)

02 _ f . i2

2

+ 20,31

$83!

R

+ 2631

0,

2

2

- "1' - $2, . a AS AS

),

R If we move the origin O to the point -525, _ 2a3I i.e., if we make the parallel translation of the coordinate system defined by the formulas

o'(-f-g,

$111

=

2311

then equation

+

011 1

$2/I

=

2321

/\1

+

0,21 $3/I

AS '

£1331

+

R 2031

7

becomes 11111

+ I\2£l7g~ + 2a3rQ}3n

=

(II)

Now only the axis 0I63f is the axis of symmetry of the second-degree surface, and the point O' is located at the vertex of the surface. iii) Suppose now that in equation (3) we have

/\1 ;é 0, I\2710, A 3 = 0 ,

0,3/

=0.

The last condition means that the vector l = 05/651 is perpendicular to the vector 831. Then equation (3) can be written in the form A1 (ivy

+

(111 \1 _

)

2

+ A2 Imp' + ( 2\21 ) 2 + R - = 0 ,

where R

a

Ag,

Ag,

/\1

AS

O'(-

all _02I If we move the origin O to the point )1 9 AS 9 0 7 i.e., if we make the parallel translation of the coordinate system defined by the formulas

811!

5811

+

011 7

/\1

$211

=

3721

+ 021 A 2

7

$311

1173:

7

5.2

Reduction

of the General

181

Equation to the Simplest Form

then equation (3) becomes A1¢¢i.,+A2w§,, +R

= 0.

(III)

The surface under consideration is a cylinder whose generators are parallel to the axis O'e3f. This axis is the axis of symmetry of the cylinder. iv) Suppose further that in equation (3) we have

A1 # 0 ,

AS = A 3 = 0 , Ag, +a§, > 0 .

=

The last condition means that the vector l a i f e i f is not collinear with the vector 611. If we rotate the basis {e1f,e2»,e3»} about the vector 61/ through the angle (p determined by the equations cosgo -

0,31

sin up

we + 0,21

use +

'

a

then equation (3) takes the form 2 A1:z:1~ + 2a1»a:1~

+ 2 \lag, + Ag, :v2~ + a

0,

where $11!

=

021Q}21 .|_ 0,31 2231 2311,

$21!

~/a§» +

as»

+ 6211631 \/at + Ag,

-Cl311E21 1

$211

If we move the origin O to the point O'

( ,

- 2A1

above equation takes the form I\1£82/n +2h2.721H

where

2

l 0-al1

va§I

+a2 31

,0)

= 0.

1

the

(IV)

\lag, + Ag, 96 0

h

and $1111

Xl!!

.l.

G11 7

/\1

$2111

z

$211

+

Ala - a

2/\1 ¢a§, + Ag,

7

$3111

=

$311

.

The surface of the fourth type is also a cylinder whose generators are parallel to the axis O'e3», but this cylinder does not have an axis of symmetry which is parallel to the generators, and it does have the plane of symmetry, namely, the plane determined by the point O' and the vectors 62/ and 63/ .

182

5.

THE GENERAL THEORY OF SECOND-DEGREE SURFACES

v) Finally, suppose that in equation (3) we have I\1¢0, I\2=A3-T'-0, Q 2 1 2 6 3 / = 0 .

=

The last conditions mean that the vector l a i f e i f is collinear with the vector 611. In this case equation (3) can be written in the form

A1 (2I1/ + a where

R

1'

2

+R=m

\1

_a

Ag

A1 .

(-

0)

all If we move the origin O to the point O' , 0, , i.e., if we make the /\Q parallel translation of the coordinate system defined by the formulas

$111

2611

+

al'

M

a

$21f

=

£82/7

$31!

=

$31

,

then equation (3) reduces to the form A12111

+ R=

(V)

Later on we will show that a surface of this type is a pair of planes (real, imaginary or coinciding) located symmetrically relative to the plane O'e2»e3» . 3. We combine the results obtained in this section in the following theorem, in which we will omit all primes in the indices in equations (I)-

(V)Theorem 1. The general equation (1)

of

a second-degree surface given

relative to some rectangular Cartesian coordinate system can be reduced to one of the following five types by means of a rotation and a parallel translation of the underlying coordinate system:

A,1¢§+A21¢0,A2¢0,

(11) (III)

A107h'¢07

(IV)

A1¢07

(V)

A1m§+A2 0, q > 0), we tities and arrive at the equation

-3-3

~;

2

2

p

q

E + ii

=: 2183 ,

(III)

which is the canonical equation of an elliptic paraboloid. 112. /\1 and A2 have opposite signs. Then, if we assume that A1 is opposite in sign to 03 and denote ._.23. A by p and 5;-5 by q (p > 0, q > 0), we obtain the canonical equation 2 181

2 .732

p

q

2273

of

(III)

a hyperbolic paraboloid.

III. The simplest equation of surfaces of this type is A1a:2 + A2a:§ + R

= 0.

(III)

In these case we have five different possibilities:

1111. A1 and A2 are of the same sign, and R has the opposite sign. Then equation (III) can be written in the form 2

B.2 + 01

2 $2

-5 az

1,

(III1)

-3-

where Ag = and Ag = - . Equation ( I I I ) is the canonical equation of an elliptic cylinder.

1112. A1, A2, and R are of the same sign. Then setting R A1

:a2 17

R ,\2

=a227

5.4

Classification

of Second-Degree

Surfaces

191

we reduce equation (III) to the canonical form

23. - .et = 1. at Ag

(1112)

We obtained the equation of a surface having no points with real coordinates. This surface is called an imaginary elliptic cylinder.

1113. A1 and A2 are of opposite signs, and R 75 0. Then if A1 is R R, Ag = opposite in sign to R, then setting Ag = - -X1 /\2 1 we reduce equation (III) to the canonical form 2

2

Q

$2

2 02

2 01

1.

(1113)

This is the canonical equation of a hyperbolic cylinder.

1114. A1 and A2 are of opposite signs, and R = 0. Denoting Ag Ag = , we obtain the canonical equation 2

2

972

321

_

1

I*1l a

7

'

0

(III )

02

01

two intersecting planes. 1115. A1 and A2 are of the same sign, and R = 0. Denoting again 2 2 1 1 $1 ' IA1l' 02 - I I I ' we reduce equation (III) to the canonical form of

2

2 911 2 01

372 - + -02

=0.

(III )

Equation (III5) is satisfied by the coordinates of points lying on the straight line az; = 2:2 0. We shall say that equation (III5) defines a pair of imaginary planes having a common real straight

=

line. IV. The simplest equation of surfaces of type (IV) is

Ale? +

21132

= 0,

AS ¢ 0, h ¢ 0.

(IV)

In this case, there is only one type of surfaces defined by the equation

g where p = parabolic cylinder.

I

213332,

(IV )

Equation (IV1) is the canonical equation of a

192

5.

THE GENERAL THEORY OF SECOND-DEGREE SURFACES

V. The surfaces of type (V) have the simplest equation

A1Q§+R=0, A1;é0.

(V)

Here three possibilities exdst:

VI. A1 and R are of opposite signs. Then (V) takes the form

ii = az

= -%. Equation

where 02 parallel planes.

(VI)

1

(VI) is the canonical equation of two

V2. A1 and R are of the same sign. Then denoting at the canonical equation 2 "3I1

G

02

R As we arrive

lllllllll

2

(V2)

of two imaginary parallel planes.

V3. R = 0. Equation (V) takes the form 2 X1=0o

(Va)

This equation is the canonical equation of two coinciding planes. Thus, altogether we obtained 17 possible, geometrically distinct types of second-degree surfaces . As to second-degree curves in the plane, they are separated into nine types: 11- Ellipse:

2 $1 '

+

2 382

1.

Ag

01

2

2

331 12. Imaginary ellipse: - 7

982

2

la. Hyperbola:

&2 al

2 322

'E

1.

7

az

01

1.

02

2

14. Two intersecting straight lines:

2.12 01

.

.

.

.

.

2 $2

-5

0.

62

.

.

x?

: v2 2 _ 0

01

"2

15. Two 1rnag1nary stralght lxnes intersecting at a real polnt: - + "5

5.5

Classification

II.

Parabola:

of Second-Degree Surfaces

193

kg = 219122, p 75 0.

mg = 02, a 75 0. 1112. Two imaginary parallel straight lines: _mg = a2, a 96 0. 1113. Two coinciding straight lines: mg = 0. 1111. Two parallel straight lines:

The reader can easily obtain the above classification and the corresponding canonical equations.

5.5

Application of the Theory of Invariants Classification of Second-Degree Surfaces

1. In this section we characterize each of 17 types of second~degree surfaces obtained in Section 5.4 by means of the invariants 11,12,13,14 and the semiinvariants $3, K2. Note that we have proved in Section 5.3 that the semiinvariant $3 is an invariant for surfaces of types (III), (IV), and (V) , and the semiinvariant $2 is an invariant for surfaces of type (V). The table below gives necessary and sufficient conditions for a second-degree surface to belong to one of the 17 types obtained in Section 5.4. No.

Name of surface

Characterization

Canonical equation

Surfaces of type (I), la ¢ 0

1

Ellipsoid

I40,/1/3>0

2

Imaginary ellipsoid

I4>0,/2>0,I1/3>0

3

Hyperboloid of one sheet

I4>0,/2 < 00T/1/3 0, I1K3 > 0

$1

01

11

12

Hyperbolic cylinder

Two intersecting planes

12 < 0, K3 # 0

12 < 0, K3

=0

2 031

2 01

2

212 01

13

Two imaginary intersecting planes

/2 > 0, Kg

3

0

2

--2

2 071 2 01

Q

1

2 02

IN

i'

1

02

2

2_ = 0 2 02

2

$2 +0 2 02

5.5

Classification

of Second-Degree Surfaces

195

Table 1 (continuation)

No.

Name of surface

Surfaces of type (IV), fs

14

Parabolic cylinder

Surfaces of type (V), /3

Canonical equation

Characterization

= 0, 14 = 0, 12 = 0, K3 75 0 K3¢0

22i

= 219272,

p# 0

= 0, 14 = 0, 12 = 0, K3 :=. 0 -

15

Two parallel planes

K2 0

_

17

Two coinciding planes

K2

=0

= a2 2

971

-0

Table 1 (continuation) First, we note that all "characterizations" indicated in Table 1 are invariant with respect to the general transformations of the coordinate system, since 11,12,13,14 are invariant relative to such transformations, and K2 and K3 are also invariant relative to such transformations for the surfaces of types in whose description they occur. Moreover, these "characterizations" are invariant relative to multiplication of the left-hand side of the surface equation by a number S ¢ 0, since after such a multiplication the quantities fi , /27 13, 14, K2 and K3 are

multiplied by s, s2, 83, 8'*, s2, and 83, respectively. It follows that, when we prove the necessity of the conditions indicated in Table 1, we can start from the canonical equations of surfaces of each type. Let us check the necessity of the conditions indicated in Table 1, for example, for the ellipsoid; the hyperboloid of one sheet; the hyperbolic paraboloid; the imaginary elliptic cylinder; the parabolic cylinder; and two parallel planes.

5.

196

THE GENERAL THEORY oF SECOND-DEGREE

23

2

$2 x 1. Ellipsoid : EQ+ -3* +

11

_1+ Ag

SURFACES

1. We have

1

Ag

1

+

Ag'

1

1

:=

1

+ a§a3 + a a g '

080g

= -I

la

l

12

,

aagag

4

We can see from these formulas that 12 > 0, 1113 > 0, 14 < 0. x2

3 . Hyperboloid of one sheet: -% + 01

1

11

1

I1111111D

al2

+ -~ 022

5;

1 l l l l l _

Ag

_I4

= 1. We have 1

I2

-1

la

-3

2

1_(L+.;) 2 2 2

e

2 2 a1a2

=

_

a3

al

a2

1 2 2 2' 010203

If /2 > 0, then 1 1


0, then 1

Ag

>

1

1

01

Ag

-2 +

and 12

1 2 2 (*10*2

1 1 1 -+2 2 a32 012 622 < 0102 1

( )

1 +1 "

2

al

Ag

2

1 + 1 +1

0%

2 2 01162

0%

Thus, in this case the relations 12 > 0 and 11la > 0 are incompatible, and hence we have either 12 _< 0 or 1113 _< 0.

< 0.

5.5

Classification

of Second-Degree Surfaces £§_.=z§

8. Hyperbolic paraboloid

p

q

Then

233.

1

I3=0,/4= 10. Imaginary elliptic cylinder: -5

pq

+ 5;

0, 12

Kg

>0.

-1. We have

2

I 3 = 0 , /4

197

=

1

'

11

1

=

+

1 02

12 > 0, /1 K3 > 0. 14. Parabolic cylinder: 1:21

12 15. Two parallel planes:

= 2p kg, I3

p 96 0. In this case, we have

= 14 = 0 ,

K3=-P2?é0~

mg = a2, a 56 0. Here

I2=I3=I4=K3=0,

K2=-G2 - u 1 .

It follows that U1

--u=.f(:n1 -re),

U2

U1

.f ( 0 2 2 - 0 3 1

If we add these two equations, we obtain U2

f (331 - xv) + .1°(a82 - sol).

u

But U2 - u = f ( $ 2 - ac) can be written as follows: f(A231

f (Am1 + Aaz2). Hence the previous equation

+

A332)

= f(A2J1) + f(A=U2)»

and this coincides with the first condition in the definition of a linear vectorvalued function. To prove the second condition, we note that it follows from the first condition that f(nA:1':) = ~f(A=») for an integer n. Next if m is an integer, then m

(Aw)

Are

m

A

mf

mM

which implies

f

1 not( A w)

Ax

no

a

Comparing previous equations, we easily find that

fl

n ¢1111111

m

Ax

)

-fm),

7

238

6.

APPLICATIONS

TO MECHANICS AND PHYSICS

and this is the second condition in the definition of a linear vector-valued function for rational factors a 1. But since the function f is continuous, the second condition is satisfied for any real a:

=

II

f@=A=~n

0,

u(x01 + Ax1, kg + Ax2, x + Ax3) > u(x?, xg, xg), and the function 'U.(2I1 , xg, xg) has a local minimum at the point Mo (mg , x8, xg) . On the other hand, if the quadratic form is negative definite, then Au > 0,

u(x01 + Ax1,

xg + Ax2, x

+ Ax3) < u(xi,

mg, xg),

and the function u(x1 , xg, a:3) has a local maximum at the point M0(xQ, xg, xg) . As to necessary and sufficient conditions of the form to be positive or negative definite, such conditions are known-these are the conditions listed in Sylvester's criteria (see Section 4.6, pp. 164-165). Thus for a function u(x1,x2,x3) to have a local minimum or maximum at a stationary point M0($011 xg, =v8), it is sufficient that the following conditions are satisfied:

Q) for local minimum,

011

> 0,

011

0,12

021

022

> 0,

G11

012

6/13

021

022

C123

031

032

033

011

012

013

021

022

023

G31

032

033

> 0;

B) for local maximum,

011

< 0,

G11

G12

021

G22

> 0,

< 0.

If neither the condition a) nor the condition B) is satisfied, then two cases are possible: the form (Aa:, Arc) can be indefinite, i.e., in a neighborhood of a stationary point M0 In, kg) the form can take values of different signs; or the form (A:v, Aw) can be semidefinite, i.e., in a neighborhood of a stationary point M0(xi, kg, kg) the form takes values of the

g,

268

7. FOUNDATIONS

oF TENSOR ANALYSIS

=

II

same sign but it can vanish not only for A1121 A272 = Azv3 0 (in particu l a , the form can be identically zero) . In the First case, the function u(:v1,x2,a:33 does not reach an extreme value at the stationary point M0($°172=8»$§)» and in the second case an extreme value could be reached or it is not reached at the stationary point M0 lag, avg): in this case an investigation of third-order terms of Taylor's formula similar to that which was done for second-order terms is required for those values Ax; for which the second-order terms vanish (in particular, for arbitrary Ami when all second-order partial derivatives vanish at the stationary point M0($I9 kg, :1:g)).

g,

Remark 4. For a function u(:z:) of one variable, the quadratic form d>(Aa:, Acc) reduces to one term 'U"(!lI0)A2I2 7

where a:0 is a stationary point. This form is positive definite if u"(:v0) > 0 and negative definite if u"(:v0) < 0. In the first case, the function 'u.(:z:) has a local minimum at the point :no, and in the second case it has a local maximum at this point. Remark 5. For a function u(a:1, 232) of two variables, the investigation of its behavior in a neighborhood of a stationary point M0 (kg, are) is reduced to the investigation of the quadratic form of two variables Axel and A232: (Aa:, Aan)

where again air

= G11(A$1)2 + 2G12A271 A122 + 022(A$E2)2,

= 6 @2';~ t

J

M0

. Sylvester's

criterion (see Section 4.6, pp. 164-

165) shows that the function in question has a local minimum at the point M0 if a 11 a 12 > 0, (111

> 0, M2

021

022

and it has a local rnaudmum at the point Mo if G11

< 0, M2

G11

012

021

G22

> 0,

If M2 < 0, then the quadratic form (A:c, Am) is indefinite, and the function u(:z:1,:z:2) does not have an extreme value at the point Mo~ But if M2 = 0, then the quadratic form (Ax, Arc) is semidefinite, and the further investigation is needed to clear up the question on the presence of an extreme value at the point Mo.

7.1

Tensor Fields and Their Differentiation

6. Consider now the differentiation tensor field Go

of

269

a vector field, i.e., a first-order

x 0i(2511$272I3)~

The absolute derivative of this field equals

as;

Gm

627k

This second-order tensor is called the gradient of a vector field. Next we write the formula relating the absolute differential day and the absolute derivative Go,kz day = am dfvk 1 (7)

.

=

>

where daze are the coordinates of the vector dsc M M ' . As we know (see Section 3.4, p. 95), the second-order tensor (li,k produces the linear transformation Arc, y

-

or in the coordinate form, Yi

0i,k27k°

Hence formula (7) can be written in the form do

= A(m) dar.

(8)

>

Since do w a(M') - a ( M ) and do: = M M ' , up to second-order in{inites~ iras, the linear transformation A(M), by acting on the infinitesimal dis> placements M M ' = dar, produces the corresponding increment of the vector field a(M): a(M') - a ( M ) = Aa(M) z A(M) dar. (9) Equation (9) shows that the linear transformation A(M) determines the principal linear part of the increment of the vector field a at the point M. Consider further the trace of the linear transformation A(M):

tr A(M)

G,i3,j.

Since the derivative ii,/c is a second-order tensor, then the above trace is an invariant called the divergence of the vector field a. It is denoted by div a: Gm

div a.

The field of this invariant is a scalar field in the same region V in which the original vector field a was defined. Since the invariant div a is obtained

270

7. FOUNDATIONS OF TENSOR ANALYSIS

from the vector field a ( M ) by differentiation, it is called a differential invariant of the field a ( M ) . Next we consider the vector z whose coordinates are obtained by the contraction of the tensor a with the discriminant tensor Et , namely, we set Zi = -6ijk01j,k. (10) The vector z is called the curl of the vector Held a(M): z

curls.

Writing formulas (10) in detail, we find the following expressions for the coordinates ii of the vector z: 21

(0,2

-

02,3)6»

Z2

(0l1,3

-

013,1)6»

Z3

(012,1

-

G1,2)61

where the quantity 6 equals +1 in a right-handed coordinate system and equals -1 in a left-handed coordinate system. This proves, that up to a factor, the coordinates of the vector curl a coincide with the components of the doubled alternated tensor a s . Hence with any vector field a(M) defined in a region V, the new vector field curl a defined in the same V is invariantly associated. We recall that a vector field a is said to be solenoidal in a region V if div a = 0 in V, and a is called irrotational in V if curl a = 0 in V. Finally, we find the divergence of a vector Held a which is the gradient of some scalar field (p(M) : a

grad up

(P,i€i

6(p i i . 611%

As we already know, to compute the divergence of the vector Held a, we must find the absolute derivative a of the field a and contract it with respect to the indices i and k. But Gm = (p.ik, so that div a

The operator

32up 3x§

7.1

271

Tensor Fields and Their Differentiation

is called the Laplace operator (or Laplacian). By means of this operator the divergence of the vector Held grad (p can be written in the form

= Ago.

div grad up

If the scalar field go(M) satisfies the condition Ago = 0, then it is called harmonic. The equation

629° + 6290 + 0% __ 0,

avg

Org

3:132

"

which the function go(a;1, :r:2,:r3) defining a harmonic vector field satisfies, is called the Laplace equation. The function (p(:r:1,:v2,:v3) satisfying the Laplace equation is called harmonic. 7. Up to now, we studied tensor fields whose tensors depend on the location of a point in a space but do not depend on a point of time at which the field is considered. Such fields are called stationary. If the tensor of a field depends not only on the location of a point in a space but also on a point of time, then the field is called nonstotionary. The components of a nonstationary tensor field are functions of the coordinates Ii of the point M and time t. For instance, for a third-order tensor field, this dependence can be written as Geek

= Clijk(231,fU2,

$37

U-

The rate of change of a tensor field at a fixed point M is described by the partial derivatives 801i.ik It is easy to see that these derivatives form 6t again a tensor field of the same order as the original field clijk- Suppose that a nonstationary tensor field describes a certain property of a material medium whose particles are in motion. We find how the components of the tensor Click connected with a fixed particle transform when the particle is moving. Suppose that a trajectory of the particle is described by the equation Gui

= I1Ii(t) .

Then the rate of change of the components of the tensor with the particle equals ddijk

adijk

II

dt aaijk

615

+

045 k

connected

a a i j k d1171 61111

dt

%'

But 3231 are the components of the velocity vector of a a i j k , l 7 and material medium. Denote the latter by pa. Then the above formula takes the form daijk

6cI4jk

II

dt

it

+ (Iiik,I vl-

(11)

272

7.

FOUNDATIONS OF TENSOR ANALYSIS

The first term on the right-hand side of (11) describes the change of the components of the tensor click in the fixed point M, and the second term is connected with the motion of the particle in space. The second term is called the transient term. Obviously, formulas of type (11) are valid for nonstationary tensor fields of any order. For a scalar field (p = (,o(M,t), the formula similar to (11) is

610 + (AM.

QE

(12)

it

dt

If we denote the velocity vector by v, then

3so + v - g r a d

div

IIIIIIIIIIIII

it

dt

a ( M , t ) , the analogous formula to (11)

For a nonstationary vector field a is

up.

day + -- as; a» 'U . it at z,k '°

The last formula is equivalent to the relation

do dt

_ pa + A (M)'v» - at

where A ( M ) is the linear transformation determined by the tensor flue.

PROBLEMS 1. Find local extreme values of the following functions of two and three variables:

a) u b) u

c)

few

mg + $11/22 + kg - 3aa:1 - 3bx2; mi' + $3 - 2:1: + 4551182 - 2x§; e ' " " § ( o f + brag), a > 0, b > 0;

d) u

cosxl cosxg cos(x1 -l-x2), 0 _< 11 $ vr, 0
0,x2>0,:v3>0;

(am + b$2 + c:v3) e""21""°§'"°23.

2. Prove the following formulas:

=

a) grad (go + ¢') grad go + grad do, div (a + b) = div a + div b, A((p+ go) = A(p-|- A 1/,', curl(a-l-b) c u r l - I - c u r l b , curl (curl (a + b)) curl (curl a) + curl (curl b) ;

=

=

7.1

273

Tensor Fields and Their Differentiation

-

b) grad (cpzb) = (pgrad Tb + 1/:grad up, div (cps) = Equation (2) means that the vector MM' = A:z: goes into the vector >NN' = Ay by means of the linear transformation E + V(M)At, where V(M) is the linear transformation determined by the absolute derivative tensor of the vector field v ( M ) . It follows that, up to second-order infinitesimals with respect to AA: and At, a small neighb or food of every point of the deformable medium undergoes a homogeneous deformation. Next, we write the decompositions of the vectors AA: and Ay relative to the basis { € 1 , € 2 , € 3 } 2 As:

Axiei, AY =

A3/i€i~

Then equation (2) can be written in the following coordinate form: A3/i = (

where all the components of the tensor Gink and all the entries 'YM of the transformation matrix are taken at the same point M .

302

7.

FOUNDATIONS OF TENSOR ANALYSIS

We noted in Section 7.1 that for a tensor field, all algebraic operations on the tensor of this field are performed separately at each point. Therefore all such operations are automatically transferred to the case of a tensor field: they should be performed at each point M relative to that local moving frame which is attached to this point. 5. Next, we find out how the quantities We 3 Ouija and F , transform under the passage from one orthogonal curvilinear coordinate system to another one. Formulas (2) imply that the quantities ii are the coordinates of a vector dM relative to the basis {€17 et, e3}. Thus the set of quantities w1,w2,w3 form a first-order tensor field. Hence, under the passage from one orthogonal curvilinear coordinate system to another, this tensor field undergoes the transformation Wif = 'We wi(15) Next, differentiating (14) and using (5) and similar relations for the vectors et' 7 dei/

Wife/ é j f

I

7

we find that

+ 'YM wig 65,

swf' € j ' = d%'i€i

or applying (14) and changing the summation index i for j in the first term, we obtain weft' 'YM 61 = (d'yw

+ 'Yrs we) et -

Since the vectors 61 are linearly independent, it follows that (Je ri:

'YE-Ij

=

d')'i/j

+ 'Ye/i W,j

_

Multiplying both members of this equation by Vi and summing up with respect to the index j , we get (4)1 /j1

But

' Y j f j 'YkIj

'7j'j

'Y1u 5

:

'YA='i d'Yi'j + 'YM 'Ywj waf-

55:k1 • Thus we finally obtain www

= 'YW 6i7i3i + W e 'We wig-

(16)

We can see from (16) that the quantities Week are not transformed according to the tensor law transformation: there are additional terms 'Ye' 5 d'yif j that do not vanish necessarily since the quantities 'Yi' j are changing from point to point. Note that using (2), we can represent the differentials d'7i'i in the form d'Yi' 5

3

0% _ii du 6114¢

k

6% j 1 We ha

a=l 5110!

7.4

303

Moving Frame and Tensor Fields

or setting 3'Yi'i 1

nua

ha

7i'jQ 7

in the form

d'Yi'j = "Yifjz wlUsing (7) and (17), we find from (16) that

(17)

1`i'k'/' We = 'YW 'Wil Wz + 'YM W k Fm: wlExpressing the forms we in terms of We by the formulas WE

= 'YW We 1

which are inverse to formulas (15), and taking into account that 'Yu' we finally obtain

'YUM

Firm/ll = To 'YM 'Yifjz + 'We 'Ykfk 'Yul Fikz-

(18)

It follows from (18) that the quantities Fun do not form a tensor, too. The fact that the quantities win and Fm are not tensors can be also explained by the following geometric considerations. We proved earlier that these quantities are identically equal to zero in the rectangular Cartesian coordinate system, and that in the cylindrical and spherical coordinate systems there are nonzero quantities among them. However, for tensors, such a situation is impossible: if all coordinates of a tensor vanish in one coordinate system, then (because of a linear homogeneous law of transformation of coordinates) the same is true in any other admissible coordinate system.

PROBLEMS 1. Find the quantities h i , ° J i , w i 5 , and Fink in the orthogonal curvilinear coordinate systems indicated in the Problems 2 a)-e) and 3 c)-f) of Section 7.3. 2. Prove

a) formula (15) directly using (14) and the formulas We = ii • dM;

Win

=

e /

- dM,

b) formulas (16) and (18) directly using (6), (8), (13), and the formulas Ldiljf = 6.1/ • d é j l , Win = iii ' d€_j. 3. Find the expressions of the components 'y,-»¢(M) of the matrix determining the passage from the moving frame associated with the curvilinear coordinate system (u1,u2, up), to the moving frame associated with the curvilin-

ear coordinate system ('U»1' , U21, U31 ), in terms of the partial derivatives and the Lark coefficients he and ha of both systems.

Zig?

304

7.

FOUNDATIONS OF TENSOR ANALYSIS

4. Find the expressions of the quantities 'Ywjz occurring in equations (17) in terms of the second derivatives of the new coordinates with respect to the old coordinates and the Lark coefficients hi and hi' of both systems (see Problem 3).

7.5

Differentiation of a Tensor Field in Curvilinear Coordinates

1. We now turn to consideration of the operation of differentiation of a tensor field in curvilinear coordinates. First, we consider the differentiation of a scalar jield. Suppose that in a curvilinear coordinate system (up, up, Us) defined in some region V of the space $3, a scalar field SO

= (P(U1,U2/Usl

is given. Its differential is defined by the equation div

3(/)

due

$11i

3

09o 1 w - " ` °

a=13ua

ha

. as

here we applied formulas (2) of Section 7.4 (p. 297). Next, denoting the in wa by ( A A , 3cp 1 (As , (1) 3140 ho,

coe{'Hcient

we obtain

do

(2)

(P,i Wi-

The set of quantities (P,1, up 2 , (P,3 is called the covariant (absolute) derivative of the scalar field (p. Since d o as well as (p are scalar fields, and Hz is a first-order tensor field (see Section 7.4, p. 304), the coefficients (Pg in (2) form a first-order tensor field, too. The vector with the coordinates (P,i is an invariant vector not depending on the choice of the coordinate system in the region V in question of the space Ea- Let us prove that this vector is the gradient of the scalar field ( p , i.e., (3) grad (P = SO,i eiIn fact, if we pass to the rectangular Cartesian coordinate system, we obtain

n¢=1,

(PJ

_ 6(p 623i

7.5 Diferentiotion

of a Tensor Field

in Curvilinear Coordinates

305

and 6(p et

= grad go,

by the definition of the gradient of a scalar field given in Section 7.1. But the last equation does not depend on the choice of the coordinate system in the region V. Hence it is valid in any curvilinear coordinate system. We find now the expression of the gradient of a scalar field in the cylindrical and spherical coordinate systems. Example 1. Cylindrical coordinate system. For this coordinate system, we had (see Example 2 in Section 7.4, p. 301)

hi = 1 ,

h3=1.

fl2=U1,

Thus, by (1), we find that 3(p

(P,1

7

61/,1

1 3(p Aug 7

(P,2

90,3

U1

W 6'U,3

By (3), it follows that grad (p

=

6(p 6114

61 +

1 6(,o Aug

U1

62

6cp 6 + 6113 3-

Example 2. Spherical coordinate system. In the spherical coordinate system, the Lark coefficients were determined by the formulas (see Example 3 in Section 7.4, p. 302)

hi

1,

ha

h2=U1,

= U1 sin $2 .

Thus, by (1), we find that 90,1

350 - 3u1 1

(P,2

1 3cp U1 3u27

90,3

U1

1 sin $2

0(/)

3143

.

By (3), it follows that grad up

3c,o

- 011.1

61 +

1 6(p U 1 6'U2

62

+

U1

1 (0, 1707 . . , -1), . 7 (0, 0, . . . , 1, -1). The dimension equals n 1.

.

..

.

..

.

-

13. Any n linearly independent solutions of the equation form a basis, and the space is of dimension n. The coordinates of an arbitrary solution with respect to any basis are just the coefficients of the expansion of the solution with respect to the elements of the basis. Section 1.4

1. a) In the triangle ABC represent f

:RP

c) In the rhombus ABCD we have

R - H, and then find IBCI2; R = XI + 1 , 1 = 1 - R. Now

in the form

b) In the parallelogram ABCD we have find + l@l2;

IAé12 : 1,T1312, a.nd

hence

R-z=(Es'-7175)-(Aé+E5)=0; d) In the rectangle ABCD we have

V + i?l2 = 1,U§ - 112,

R i.e.,

-1

= 0, and hence l,?l2 = l%l2 or l / l = l%l;

e) The proof is similar to a); f) The median AD of the triangle ABC is given by find

1»i512, using

the result of Problem 1 a);

ET :

+ ,T*§).

Now

319

SELECTED HINTS AND ANSWERS g) Let AAA and BB1 be equal medians of the triangle ABC. Then

> IAA112 = 1831> P, and

+ R12 = Ii

hences

+

QP or

(E5+IR5+1l+T3?)-(Az§+R*-B/l-1)=0,

=

)

that Ea' CON 0, where CC; is the other median of the triangle; h) In the trapezoid ABCD we have SO

M + ? + @ = M , I = z § + W , @7>'=8A+,T/3, and

hence

l,R*l2+IBz5l2 : l/T1§+1,*l2+IBtl+E5l2 = lE3'l2+22Tzl~1 +IB75l2+IE'l2 . iN + l»U)'l" : 17al12 + U I " + 2 { l , I 2 + Hz . I i * - Zi`))} =l/U5l2+l1l2+2Azl-(,+§é-A'z8) = 1239 + 15312 + 2, . 53 = 12612 + l?2?12 +2l/nWI

-m

i) In a regular tetrahedron A1 A2A3A4 we have > > > > > > ) > > A3A4 A1A4 - A 1 A 3 » A1A2 -A3A4 A1A2 • A1A4 - A1A2 ~A1Aa > > . or A1A2 A3A4 12 cos - 12 0.

9. In En the inequality is the same as in Problem 2, except that now i , j , k :- 1,2,. . . , n instead of i , j , k 1,2,3. In C'[a, b] we have

=

b

s

f(¢)9(¢)d¢ 10. 90°, 60°, 30°

\lab

f2(¢)d¢ I/abg2(t)dt

.

14. Take the scalar product of the vector iv;

+ :fig + ... + :no with itself.

s lwl2 +2l==IIyI + 11/122 , 2 11v12 - 2l=vllyl + Iyl ,

15. la:-I-yl2 = : c a : + 2 : n y + y - y { by the Cauchy-Schwarz inequality. I

16.

I

fl'f2(i)d¢-

_< \ / f " [ f ( ¢ )

f"92(0)d¢

+ 9(¢)]2 it S \/f" f m )

l

17. Calculate k

Do o

•

et) et

9

i=1

noting that k

Pre

st: ' a n

:c= Z ( m - e i ) e i . i=1

dt+

\/f" gm) dt.

320

SELECTED HINTS AND ANSWERS

= (t2 - 1)*, and

18. a) Let U4¢(t)

prove that uI"(i1)

:

0 if