Mathematical Techniques: Vector Algebra and Vector Fields

Home , Vector operator

Lecture notes byW.B.von Schlipp e

Part 1. Vector Algebra.

addition, subtraction, multiplication by scalar factor; asso ciative 1. De nition of vectors;

law of addition; asso ciative, commutative and distributivelaws of multiplication by

scalar factors.

2. Geometrical representation of vectors.

3. Scalar pro duct; distributive and commutativelaw of scalar pro ducts; kinetic energy.

4. Mo dulus of a vector; unit vectors; base vectors; linear indep endence.

5. Vector pro duct of twovectors; distributivelawofvector pro duct; non-commutative

lawofvector pro duct;

Examples: moment of force (torque); angular momentum.

6. Applications: Vector equations of lines and planes.

7. Triple pro ducts; triple scalar pro duct; triple vector pro duct.

transformation prop erties of vectors under rotations; in- 8. Co ordinate transformations;

variance of the dot-pro duct under rotations; re ections; p olar and axial vectors.

9. Exercises and problems.

Part 2. Vector Fields.

1. De nition of scalar and vector elds; eld lines.

2. Gradient of a scalar eld; p otential

3. Fluxofavector eld; divergence.

4. The nabla op erator.

5. Theorems of Gauss and Green.

6. Curl of a vector eld; Stokes' theorem.

7. Application: Maxwell equations of Electromagnetism. 1

Part 1. Vector Algebra.

1. De nition of vectors; addition, multiplication by scalar factor, subtraction; asso ciative

law of addition, commutative and distributivelaws of multiplication by scalar factors.

Avector is a triplet of numb ers. It is represented in the following notation:

~r =(x; y ; z ) or ~a =(a ;a ;a ) (1)

x y z

Here x, y and z are the components of ~r , and similarly a , a and a are the comp onents of

x y z

~a.

Vectors are de ned by the following prop erties:

~ ~

1. Given twovectors, ~a =(a ;a ;a )andb =(b ;b ;b ), the sum of ~a and b is the vector

x y z x y z

~c de ned by

~c = ~a + b =(a + b ;a + b ;a + b ) (2)

x x y y z z

2. The pro duct of vector ~a with a scalar factor is the vector ~c de ned by

~c = ~a =(a ;a ;a ) (3)

x y z

3. Based on de nitions 1 and 2 we can de ne the subtraction of vectors ~a and b as the

~ ~

addition of ~a and (b): thus, the vector di erence of vectors ~a and b is the vector ~c

de ned by

~ ~

~c = ~a b = ~a +(b )=(a b ;a b ;a b ) (4)

x x y y z z

From the ab ove de nitions one gets the following rules:

(i) The addition of vectors is asso ciative:

~ ~ ~

d =(~a + b)+~c = ~a +(b + ~c) (5)

We can therefore drop the brackets in the sum of three vectors and write

~ ~

d = ~a + b + ~c

(ii) The addition of vectors is commutative:

~ ~

~a + b = b + ~a

(iii) The multiplication by a scalar factor is distributive:

~ ~

(~a + b)=~a + b 2

Figure 1: Geometrical representation of vectors.

(iv) The multiplication by a scalar factor is commutative:

~a = ~a

Rules (i) - (iv) follow directly from the corresp onding rules for the vector comp onents, which

are ordinary numbers.

A sp ecial vector is the nul l vector: this is the vector whose three comp onents are all equal

to nought. There is no universally agreed notation for the null vector; since there is usually

no ambiguity,we shall denote it byzero,i.e. write (0; 0; 0) = 0.

2. Geometrical representation of vectors.

The geometrical representation of a vector is that of a directed quantity: thus the vector

~a =(a ;a a ) is represented in a cartesian co ordinate system by the arrow from the origin

x y z

to the p oint P whose co ordinates are a , a and a , resp ectively (Fig. 1). Fig. 1

x y z

The law of addition of vectors leads to the representation of the sum of vectors ~a and b

as the diagonal of the parallelogram constructed from ~a and b as shown in Fig. 2. Fig. 2

To construct the di erence vector ~c = ~a b we can either use the de nition and construct

~ ~

rst the vector b and then use the parallelogram rule to add ~a and (b) (Fig. 3a). Fig. 3a

~ ~

Alternatively we can note that if ~c = ~a b,then~a = b + ~c. Therefore we get ~c bydrawing

the vector from b to ~a (Fig. 3b). Fig. 3b

The vector ~a p oints in the direction of ~a and has a length times the length of ~a.Two

vectors p ointing in the same direction are called col linear.Vectors which p oint in opp osite

directions are called anticol linear. Thus ~a and (~a) are anticollinear, but more generally

also ~a and (~a)if>0.

3. Scalar pro duct.

Distributive and commutative law of scalar products; work and kinetic energy. 3

Figure 2: Addition of vectors ~a and b.

Figure 3: Subtraction of vectors ~a and b. 4

The multiplication of vectors ~a and b is not uniquely de ned. One can multiply the com-

p onents of ~a with the comp onents of b in nine di erentways: a b , a b , ...a b . These

x x x y z z

pro ducts are called dyads. One can also add and subtract dyadsinvarious ways. Twoof

suchcombinations are particularly useful. They are the scalar product and the vector product

of ~a and b.

The scalar pro duct is de ned by

~a b = a b + a b + a b (6)

x x y y z z

Because of the way the scalar pro duct is written it is also called the dot product. In section

8we will see that the scalar pro duct do es not change under a rotation of the co ordinate

axes. This is the prop erty of scalars, and the invariance under rotations is the reason why

the scalar pro duct has its name.

The scalar pro duct is distributive:

~ ~

~a (b + ~c)=~a b + ~a ~c (7)

This follows directly from the the distributivelaw for the comp onents. Thus for the x

comp onents wehave

a (b + c )=a b + a c

x x x x x x x

and similarly for the y and z comp onents.

The scalar pro duct is commutative:

~ ~

~a b = b ~a (8)

This also follows directly from the commutativelaw for the comp onents.

The asso ciativelaw has no meaning in relation to the scalar pro duct. For instance, if

we take the scalar pro duct ~a b, then this is a scalar, and it is meaningless to form its dot

pro duct with a third vector.

The scalar pro duct of ~a with itself is called the modulus squared of ~a, and one writes

2 2 2 2

(9) + a + a a = ~a ~a = a

z y x

The square ro ot of a is called the modulus of ~a and is denoted by j~aj.Thus

2 2 2

j~aj = + a + a (10) a

x y z

Consider the right-angled triangle OAB with sides OA = a , AB = a and hyp otenuse

x y

2 2 2

OB = c (Fig. 4): by Pythagoras' theorem wehave c = a + a . Then consider the Fig. 4

x y

right-angled triangle OB P : OP is the length of ~a, and from the triangle wehave OP =

2 2 2 2 2

c + a = a + a + a . Comparing this with the expression for the mo dulus of ~a we conclude

z x y z

that the modulus of a vector is equal to its length. 5

Figure 4: Mo dulus of vector ~a.

There are three vectors whichhave a sp ecial signi cance, namely the vectors

^{ =(1; 0; 0); |^ =(0; 1; 0) and k (0; 0; 1) (11)

which are of unit length and p oint in the x, y and z direction, resp ectively. Generally any

vector of unit length is called a unit vector.To distinguish unit vectors from general vectors

we shall always put a carat on unit vectors rather than an arrow.

Avector, divided by its mo dulus, is a unit vector. We can therefore write

a^ =

j~aj

Consider the dot pro ducts of ~a with ^{,^| and k :

~a ^{ = a ; ~a |^ = a ; and ~a k = a

x y z

Now, if ~a makes the angle with the x axis, wehavealsoa = j~aj cos , and similarly

a = j~aj cos and a = j~aj cos ,if and are the angles ~a makes with the y and z axis,

y z

resp ectively.We get therefore the following representation of ~a:

~a = j~aj(cos ; cos ; cos ) (12)

Thus the cosines of the angles , and are seen to de ne the direction of ~a. They are

therefore called the directional cosines of ~a. An imp ortant prop erty of the directional cosines

is found if werecallthat^a is a unit vector. Therefore wehave

2 2 2 2

=^a =1 =cos +cos + cos (13)

j~aj

So there is one relation b etween the three directional cosines and one could conclude that

two directional cosines are sucient to de ne the direction of a vector. But that would b e 6

Figure 5: To derive ~a b = ab cos .

wrong b ecause the relation b etween the directional cosines is quadratic rather than linear.

Therefore two directional cosines de ne the third one only up to its sign.

Consider the dot pro ducts of ^{,^| and k with each other: it follows from their de nitions

that

^ ^

^{ |^ =0; |^ k =0; and k ^{ =0 (14)

Now, ^{ is p erp endicular to| ^ and they are b oth p erp endicular to k . Another word for

p erp endicular is othogonal. We can therefore say that the three vectors ^{, |^ and k are

mutual ly orthogonal. Wehave also seen that ~a ^{ = j~aj cos . Therefore if ~a and ^{ are

orthogonal, then their dot pro duct is equal to zero, b ecause then cos = cos(=2)=0.We

shall see that generally the dot pro duct of anytwovectors ~a and b can b e represented as

~a b = ab cos (15)

~ ~

where wehaveputa = j~aj, b = jbj and we denote the angle b etween ~a and b by .

For twovectors which lie in the xy plane the statement is easily shown. If ~a and b make

angles and with the x axis (Fig. 5), then Fig. 5

~a = a(cos ; sin ; 0); b = b(cos ; sin ; 0)

and hence

~a b = ab(cos cos + sin sin )=ab cos( )=ab cos

For general vectors, which do not lie in a co ordinate plane, we can intuitively see in the

following way that this statement remains true: it is p ossible to rotate the co ordinate frame in

suchaway that after the rotation the vectors lie in a co ordinate plane. This rotation a ects

neither the lengths of the vectors nor the angle b etween them. Therefore the expression

Frequentlythisrelationistaken as the de nition of the dot pro duct. 7

ab cos for the dot pro duct must b e true in any co ordinate frame. Later in this course, in

connection with matrix algebra, we shall present this pro of rigorously.

An imp ortant example of a scalar pro duct in mechanics is the work done by a force F in

moving an ob ject through a distance ~r ,which is de ned by

W = F ~r

Another example is the kinetic energy T of a mass m moving with momentum p~:

T =

where ~p is the dot pro duct ~p ~p .

4. Base vectors; linear indep endence.

In Section 3 wehave mentioned the unit vectors ^{,^| and k ,andwehave shown that the

comp onents of an arbitrary vector ~a could b e written as

a = ~a ^{; a = ~a |^; and a = ~a k

x y z

We can therefore conclude that ~a can b e represented as the linear superposition of the unit

vectors ^{,^| and k :

k (16) ~a = a ^{ + a |^ + a

x y z

where the comp onents of ~a play the role of the superposition coecients. Indeed, if we take

the dot pro duct of the latter equation with ^{ and takeinto account the orthogonalityof^{,^|

and k , then werecover the statement

~a ^{ = a

and similarly for the other two comp onents.

An imp ortantproperty of the unit vectors ^{,^| and k is their linaer independence. This is

the statement that none of the three vectors can b e represented as a linear sup erp osition of

the other two. For ^{,^| and k their linear indep endence can b e proved like this: assume that

they are not linearly indep endent, i.e. assume that there is a linear relationship b etween

them, e.g. ^{ = a|^ + bk or, more symmetrically

k =0 c ^{ + c |^ + c

1 2 3

then, taking the dot pro duct with ^{,we get

c ^{ ^{ + c |^ ^{ + c k ^{ =0

1 2 3 8

but ^{ ^{ =1 and^| ^{ = k ^{ = 0, hence c = 0. Similarly, taking the dot pro ducts with| ^ and

then with k ,we can show that c =0 and c = 0, i.e. there is indeed no linear relationship

2 3

between ^{,^| and k .

The other imp ortant prop erty of the unit vectors ^{,^| and k is their completeness. This

is the prop erty expressed by Eq. (16) whichistrueforany vector ~a.

A set of orthogonal unit vectors like^{,^| and k , which are linearly indep endent and

complete, are called base vectors.

An interesting example of a set of base vectors are the vectors appropriate for p olar

co ordinates:

r^ = (sin cos ; sin sin ; cos );

= (cos cos ; cos sin ; sin );

= (sin ; cos ; 0)

It is left as an exercise to check that these three vectors have all the prop erties required of

a set of base vectors.

The question of linear indep endence has a striking geometrical interpretation. For in-

stance, if three vectors lie in one plane, then they are linearly dep endent. Vectors whichlie

in one plane are said to b e coplanar. Three non-coplanar vectors are linearly indep endent.

More than three vectors in a three-dimensional space are always linearly dep endent. It is

left as an exercise to check all statements made in this paragraph are consistent with the

algebraic de nition of linear (in)dep endence.

5. Vector pro duct of twovectors;

De nition of vector pro duct; distributivelawofvector pro duct; non-commutativelawof

vector pro duct. Examples: moment of force (torque); angular momentum.

The vector pro duct of ~a and b is de ned as the vector ~c, expressed in terms of the

comp onents of ~a and b by

~c =(a b a b ;a b a b ;a b a b ) (17)

y z z y z x x z x y y z

The vector pro duct has its name b ecause it is a vector, i.e. b ecause its comp onents

transform under rotations of the co ordinate axes like the comp onents of a vector. We shall

give the pro of of this statment later on in this course in the chapter on matrices. Another

name for the vector pro duct is cross product b ecause it is usually written in the form of

~c = ~a b

The cross pro duct is distributive:

~ ~ ~

~a (b + ~c)=~a b + ~a b (18) 9

The pro of is straightforward. It can b e done for eachcomponent separately. Thus, for

instance, for the x comp onentwehave

[~c (~a + b)] = c (a + b ) c (a + b )=c a + c b c a c b

x y z z z y y y z y z z y z y

= (c a c a )+(c b c b )

y z z y y z z y

= (~c ~a) +(~c b)

x x

and similarly for the y and z comp onents.

The cross pro duct is not asso ciative:

~ ~

~a (b ~c) 6=(~a b) ~c (19)

The pro of is straight forward. Consider, for instance, the x comp onents of the twovectors:

~ ~ ~

(~a (b ~c)) = a (b ~c) a (b ~c )

x y z z y

and

~ ~ ~

((~a b) ~c) =(~a b) c (~a b) c =(a b a b )c :::

x y z z y z x x z z

where we did not need to write any more terms b ecause we can see that the former expression

do es not contain the x comp onentof~a whereas the latter expression do es contain a .

The cross pro duct is not commutative. More precisely, the cross pro duct anticommutes:

~ ~

~a b = b ~a (20)

This can b e shown separately for each comp onent; for instance for the x comp onentwehave:

(~a b) = a b a b

x y z z y

whereas

~ ~

(b ~a) = b a b a = (~a b)

x y z z y x

and similarly for the other two comp onents.

Consider the cross pro ducts of the base vectors ^{ =(1; 0; 0),| ^ =(0; 1; 0), and k =(0; 0; 1):

^{ |^ =(i ;i ;i ) (j ;j ;j )=(i j i j ;i j i j ;i j i j )=(0; 0; 1) = k

x y z x y z y z z y z x x z x y y x

^ ^

and similarly| ^ k =^{ and k ^{ =^|.We see in these three examples that the cross pro duct

of ^{ and| ^ is a vector, whose direction is orthogonal to the plane spanned by the vectors ^{

and| ^, and similarly in the other two cases.

Next consider the cross pro duct of ^{ with itself. By direct calculation we nd that

^{ ^{ =0

^ ^

and similarly| ^ |^ = k k =0.

We can show more generally that 10

the cross product of any two vectors ~a and b is orthogonal to the plane spanned

by these two vectors

and

the cross product of any two col linear vectors is the nul l vector.

The former one of these statements can b e shown fairly simply for twovectors that lie

in a co ordinate plane, for instance for the vectors ~a =(a ;a ; 0) and b =(b ;b ; 0). We

x y x y

immediately see that the x and y comp onents of their cross pro duct are equal to nought

b ecause they have factors of a and b in them, which are b oth equal to nought. Thus the

z z

only nonzero comp onentisthe z comp onent, whichis(~a b) = a b a b .

z x y y x

To see that the statement is generally true we app eal again to an intuitive argument: if

the co ordinate system is rotated suchthatthevectors ~a and b do not lie in a co ordinate

plane after the rotation, then the orientation of their cross pro duct in relation to ~a and b

~ ~

remains unchanged, i.e. the cross pro duct of ~a and b is always orthogonal to ~a and b.This

will b e proved rigorously in the chapter on matrix algebra.

The discussion of the direction of the cross pro duct is incomplete without sp elling out

the rule concerning the sense in whichitpoints along the normal to the plane spanned by ~a

and b. This rule is as follows:

~ ~

The cross pro duct of ~a b is orthogonal to the plane spanned by ~a and b such

that, lo oking down from the pro duct vector into their plane, ~a can b e rotated

into the direction of b in anticlockwise direction through an angle of less than

180 .

Toprove the statement that the cross pro duct of collinear vectors is the null vector we

note that collinear vectors are related by b = ~a, where is a scalar factor. Thus wehave

~a b = (~a ~a)

and it is easy to see by direct calculation that all comp onents of the cross pro duct of ~a with

itself are equal to nought.

An interesting formula can b e found for the mo dulus of the cross pro duct:

j~a bj = ab sin (21)

~ ~

where is the angle b etween ~a and b. This statementiseasytoshow for vectors ~a and b that

lie in a co ordinate plane, for instance for ~a =(a ;a ; 0) and b =(b ;b ; 0). Assuming that ~a

x y x y

makes an angle with the x axis we can also write ~a = a(cos ; sin ; 0), and similarly,if

~ ~

is the angle that b makes with the x axis, wehave b = b(cos ; sin ; 0), and hence

~a b = ab(0; 0; cos sin sin cos )= ab(0; 0; sin( )) 11

Figure 6: To derive j~a bj = ab sin .

and since ( )= is the angle b etween ~a and b wehave proved the statement.

If ~a and b do not lie in a co ordinate plane weinvoke the bynow familar argument that

they can b e broughtinto a co ordinate plane by a rotation without changing their magnitudes

and the angle b etween them, so the statement is true generally.

The result that the mo dulus of the cross pro duct of twovectors is equal to the pro duct of

their mo duli times the sine of the angle b etween them gives rise to an interesting geometrical

interpretation. Indeed, consider the vectors ~a and b and draw the parallelogram as shown in

Fig. 6. Fig. 6

The height h of the parallelogram is equal to b sin , and the base is a, and therefore the

area of the parallelogram is ah = ab sin .We can therefore conclude that

The mo dulus of the cross pro duct of ~a and b is equal to the area of the parallel-

ogram spanned by ~a and b.

Let us return to the breakdown of the asso ciativelaw of the vector pro duct. We can see

it nowby noting that the twovectors in Eq. (19) lie in di erent planes: vector ~a (b ~c )

~ ~

lies in the plane spanned by b and ~c, whereas (~a b) ~c lies in the plane spanned by ~a and

6. Applications:

In this section we consider some applications of the concept of vectors in geometry.

6.1. As a rst example consider the vector equation of a line:

~r = n^ + ~a (22) 12

Figure 7: To derive Eq. (22)

where ~a is a constantvector,n ^ is a xed unit vector and is a real parameter that takes

on anyvalue from 1 to 1.To show that this is indeed the equation of a line we pro ceed

like this. Let A and B be two p oints with vectors ~a and b, resp ectively (see Fig. 7). The Fig. 7

straight line through these p oints can b e constructed by rst going to A and then in the

~ ~ ~

direction of (~a b)byavariable distance. Settingn ^ =(~a b)=j~a bj and measuring the

variable distance by we get the required equation.

Let us denote the line by g . The distance of g from the origin is equal to the length of

the p erp endicular, dropp ed from the origin onto g .From the triangle OAP we see that this

length is d = a sin ,where is the angle b etween ~a andn ^ .Buta sin = j~a n^ j,hence

d = j~a n^ j

Exercise: Denote the vector that is p erp endicular to g by ~r . This vector is also p er-

p endicular ton ^ . To lie on g it must satisfy the equation of g with a parameter to b e

determined. We get by imp osing the orthogonality condition:

~r n^ =(~a + n^ ) n^ =0

0 0

hence = ~a n^ . The distance of g from the origin is the length of ~r , i.e.

0 0

j~r j = j~a + n^ j = j~a (~a n^ )^nj

0 0

Show that this expression for the distance is consistent with the one derived previously.

Let us also nd the distance of a p oint M with vector b from g (see Fig. 8). Denote Fig. 8

the fo ot of the p erp endicular dropp ed from M onto g by Q.From the triangle AM Q we see

that MQ = AM sin or, in vector notation,

~ ~ ~

d = j~r bj = j~a bj sin = j(~a b) n^ j

1 13

Figure 8: Distance of p oint M from line g .

6.2. Vector equation of a plane:

The equation of a plane through the p oints given bythevectors ~a, b and ~c is

~r = ~a + b + ~c; where + + =1 (23)

~ ~

This can b e seen in the following way. First note that the vectors ~a b , b ~c and ~c ~a are

collinear. This is obvious from their geometrical representation, but can b e easily checked

by forming their triple scalar pro duct.

~ ~

Then consider the vector ~r = ~a + (~a b )+ (b ~c) where and are arbitrary constants.

~ ~

This vector p oints to a p oint in the plane which is spanned by ~a b and b ~c and passes

through the p ointgiven by ~a. Rearranging weget~r =(1+ )~a +( )b ~c, hence, if we

put =1 , = and = ,we get Eq. (23).

7. Triple pro ducts;

Triple scalar pro duct; triple vector pro duct.

7.1. The scalar pro duct of a vector ~a with the cross pro duct b ~c is a scalar. It is called

~ ~

the triple scalar product of the vectors ~a, b and ~c. If the angle b etween b and ~c is wehave

b ~c =^nbc sin

wheren ^ is the unit vector normal to the plane spanned by b and ~c, and weknow (see section

~ ~

6) that jb ~cj is the area of the parallelogram based on b and ~c.Taking the dot pro duct of

~a with b ~c weget

~a (b ~c)=(~a n^ )bc sin

but h = ~a n^ = a cos is the height of the parallelepip ed based on the vectors ~a, b and ~c (see

Fig. 9), and therefore wehave the result that the triple scalar pro duct ~a (b ~c)isthe Fig. 9 14

Figure 9: Triple scalar pro duct.

volume of this parallelepip ed.

Our discussion is de cient in so far as we can get a negativevalue for the volume. This

is the case when is greater than =2, i.e. when the vectors ~a, b and ~c, in that order, form

a left-handed system. Thus, to get the volume as a p ositive quantity irresp ectiveofthe

handedness of the three vectors, wesay that the volume of the parallelepip ed is equal to the

modulus of the triple pro duct:

V = j~a (b ~c)j (24)

parallelepip ed

An interesting symmetry prop erty of the triple scalar pro duct is expressed by the equa-

tions

~ ~ ~

~a (b ~c)=b (~c ~a)=~c (~a b) (25)

which can b e veri ed by direct evaluation of the three triple pro ducts. We can express

this prop ertyinwords bysaying that the triple product does not change under a cyclic

permutation of the vectors. Because of this symmetry one frequently uses the simpli ed, and

more symmetric, notation (~a; b; ~c), i.e.

~ ~ ~ ~

(~a; b; ~c )=~a (b ~c)= b (~c ~a)=~c (~a b) (26)

Note however that the sign of the triple pro duct changes under an o dd p ermutation of the

vectors, e.g.

~ ~

(~a; b; ~c )=(~a; ~c; b) (27)

From the latter equation we get the result that the triple scalar pro duct vanishes if it contains

one vector twice, for instance

~ ~

(~a; b; b )=0 (28)

~ ~ ~ ~ ~

Indeed, if we set ~c = b in Eq. (27), we get (~a; b; b)= (~a; b; b), and since the only number

that is equal to minus itself is zero, we get the result of Eq. (28). 15

More generally we can show that the triple scalar pro duct vanishes if the three vectors

are coplanar. Indeed, if they are coplanar, then we can write, for instance,

~c = ~a + b

and hence

~ ~ ~ ~ ~ ~

(~a; b; ~c )=(~a; b; ~a + b)=(~a; b; ~a)+(~a; b; b)=0 (29)

b ecause b oth scalar triple pro ducts on the right-hand side are equal to noughtby Eq. (28).

The converse statement, that the vectors ~a, b and ~c are coplanar if their triple scalar

pro duct is equal to nought, can also b e proved. Thus the vanishing of the triple scalar

pro duct is the necessary and sucient condition for the vectors to b e coplanar.

Exercise: Prove that the vectors ~a, b and ~c are coplanar if their triple scalar pro duct is

equal to nought.

An interesting extension of the scalar triple pro duct arises if we take the dot pro duct of

two cross pro ducts. Then one can prove the following identity

~ ~ ~ ~ ~ ~

(~a b) (~c d )=(~a ~c)(b d) (~a d )(b ~c ) (30)

This identity, known as Lagrange's identity, can b e proved by explicit evaluation. This is

straight forward but tedious. More interesting is the following pro of.

We rst note that the left-hand side is a scalar, made of the four vectors ~a to d, and that it

dep ends linearly on eachofthefourvectors. Thus, if wewant to rewrite it in terms of dot

pro ducts, wemust consider the following combinations:

~ ~ ~ ~ ~ ~

(~a b)(~c d); (~a ~c)(b d) and (~a d )(b ~c) (31)

~ ~ ~

Next wenotethat(~a b) (~c d)changes sign under the replacement ~a $ b and under

~c $ d . Therefore wemust discard the rst one of the three expressions in Eq. (31), which

changes into itself under either one of these transformations. The second and third ones of

these expressions change into each other. We therefore get a change of sign under either

one of the replacements if we take the di erence of these two expressions. There remains an

overall sign to x. This can b e done bycho osing a particular case, for instance by setting

~ ~

~a = ~c =^{ and b = d =^|.

7.2. The triple vector pro duct (~a b) ~c has the interesting prop erty that

~ ~ ~

(~a b) ~c =(~a ~c )b (b ~c)~a (32)

~ ~

Pro of: wenotethat(~a b) is orthogonal to the plane spanned by ~a and b, and therefore

its cross pro duct with ~c lies in that plane. Therefore the triple cross pro duct is a linear

combination of ~a and b, i.e.

~ ~

(~a b) ~c = ~a + b 16

where the scalar factors and must dep end linearly on b and ~c and on ~a and ~c, resp ectively.

They must therefore b e of the form of (b ~c)and (~a ~c ), where and are numerical

factors. Now, the triple cross pro duct changes sign under the replacement ~a $ b. Therefore

wemust have = , i.e.

~ ~ ~

(~a b) ~c = [(~a ~c)b (b ~c)~a]

The remaining numerical factor can b e found by considering a particular case, e.g. ~a = ~c =^{

and b =^|.

Exercise: Use the identity (32) to prove that

~ ~ ~

~a (b ~c)+b (~c ~a)+~c (~a b)= 0

~ ~

An interesting extension of the triple vector pro duct is the expression (~a b) (~c d),

for whichwe can deduce the following identities:

~ ~ ~ ~ ~ ~ ~

f (~a b) (~c d) = b(~a; ~c; d ) ~a(b; ~c; d )

~ ~ ~ ~

= ~c(~a; b; d ) d (~a; b; ~c ) (33)

~ ~

whichshows that the vector f lies b oth in the plane spanned by ~a and b and in the plane

spanned by ~c and d, i.e. it lies along the intersection of these two planes.

8. Co ordinate transformations.

Transformation prop erties of vectors under rotations; invariance of the dot pro duct under

rotations; re ections; p olar and axial vectors.

In this section we discuss co ordinate transformations in two dimensions. The generaliza-

tion to the three dimensional case will b e done in the chapter on matrix algebra.

0 0

Thus consider two Cartesian co ordinate frames (xy )and(x ;y ) with common origin O

0 0

(see Fig. 10). The x axis (y axis) makes the angle with the x (y ) axis. The base Fig. 10

0 0

vectors in the xy frame are denoted ^{ and| ^, those in the primed frame are ^{ and| ^ .Thus

wehave the following set of dot pro ducts:

2 2 0 2 0 2 0 0

^{ = |^ =^{ =^| =1; ^{ |^ =^{ |^ =0;

0 0

^{ ^{ = |^ |^ = cos ;

^{ |^ = cos + = sin (34)

=sin ^{ |^ = cos

2 17

Figure 10: Rotation of co ordinate frame.

Now consider the p oint P with radius vector ~r . In the xy frame it is represented by

0 0 0 0 0 0 0

~r = x^{ + y |^, and in the x y frame by ~r = x ^{ + y |^ , and since these are di erent

representations of the same vector wehave the identity

0 0 0 0

~r = x^{ + y |^ = x ^{ + y |^

Taking the dot pro ducts rst with ^{ and then with| ^ weget

0 0 0 0

x = x ^{ ^{ + y |^ ^{

0 0 0 0

y = x ^{ |^ + y |^ |^

and hence, with Eq. (34), weget

0 0

x = x cos y sin

0 0

y = x sin + y cos (35)

and similarly we get the formulas for the inverse transformation:

x = x cos + y sin

y = x sin + y cos (36)

Another imp ortant transformation is the re ection at the origin. This is de ned as the

transformation

0 0 0

x ! x = x; y ! y = y; and z ! z = z (37)

Under this transformation the vector ~r changes its sign:

~r ! ~r = ~r (38)

where ~r and ~r refer to the same physical vector.

Wecannow restate our original de nition of a vector more precisely: 18

avector is a triplet of numbers, the components of the vector, which transform

under rotations of the coordinate axes and under re ections like the components

of ~r

Let us examine various quantities whichwehave considered in this chapter as to their

transformation prop erties.

Consider how the mo dulus of ~a transforms under rotations:

2 2 2 0 2 0 2

j~aj = a + a ! a + a

x y x y

2 2

= (a cos + a sin ) +(a sin + a cos )

x y x y

2 2

= a + a

x y

and one says that the mo dulus of a vector is invariant under rotations. Similarly one

can show that the mo dulus is invariant under re ections. These invariance prop erties are

characteristic of scalars.Onede nes a scalar as a quantity that is invariant under rotations

and re ections.

It is left as an exercise to verify that the dot pro duct of twovectors is invariant under

rotations and re ections. Thus the dot pro duct of twovectors is a scalar.

Next consider the cross pro duct of twovectors which lie in the xy plane. This vector has

only a z comp onentwhichwe exp ect to b e una ected by a rotation in the xy plane. This

can b e checked using the transformation formulas (35). The question as to the full three-

dimensional rotations must b e p ostp oned, but anticipating the result of that calculation we

can say that the cross pro duct of twovectors transforms under rotations likea vector. This

is the justi cation for calling it a vector.

Let us also consider the b ehaviour of the cross pro duct under re ections. Wehave

0 0

~ ~ ~ ~

~a b ! ~a b =(~a) (b )=~a b

i.e. we get the imp ortant result that under re ections the cross pro duct of twovectors

do es not behavelikea vector, i.e. it do es not change its sign! This result gives rise to the

distinction b etween two classes of vectors: those that do and those that do not change sign

under re ections. The former are called polar vectors, whereas the latter are called axial

vectors (or sometimes pseudo vectors). These names have their origin in applications of

vector calculus in mechanics.

Of fundamental imp ortance in mechanics are the p olar vectors ~r , the momentum p~ and

the force F . Examples of axial vectors in mechanics are the moment of force (or torque)

~ ~ ~

M = ~r F and the angular momentum J = ~r ~p.

Finally let us sharp en up the statement that the cross pro duct of twovectors is an axial

vector. In this statementitwas tacitly assumed that the twovectors were p olar vectors.

for the time b eing we consider rotations only in the xy plane; re ections are done in three dimensions. 19

One can easily check that the cross pro duct of two axial vectors is an axial vector and that

the cross pro duct of an axial vector and a p olar vector is a p olar vector. Let us denote for

~ ~

the sake of this discussion a general p olar vector by P and a general axial vector by A. Then

we can formally express this result by the following scheme:

~ ~ ~ ~ ~ ~ ~ ~

P P = A A = A; P A = P

The distinction b etween p olar and axial vectors has also a b earing on the notion of

scalars. The statement that the dot pro duct of twovectors is invariant under re ections also

needed the tacit assumption that the twovectors were p olar vectors. If, on the other hand,

we consider the dot pro duct of a p olar vector with an axial vector, then we get a quantity

whichchanges sign under re ections (but is invariant under rotations). We therefore also

distinguish b etween two classes of scalars, those which do not and those whichdochange

sign under re ections. The former are called scalars, whereas the latter are called pseudo

scalars. 20

9.) Exercises and problems.

~ ~

1) Evaluate the mo duli of the vectors ~a and b, the scalar pro ducts ~a b and nd the angles

between the ~a and b if

1.1 ~a =(2; 3; 1), b =(4; 7; 5);

(3; 4; 0), b = 1.2 ~a = (2; 1; 3)

~ ~

2) Evaluate and compare the vector expressions ~a (b + ~c)and~a b + ~a ~c if

2.1 ~a =(3; 2; 1), b =(2; 3; 7), ~c =(4; 1; 3);

2.2 ~a =(2:3; 1:2; 1:7), b =(0:2; 3:5; 0:7), ~c =(2:4; 3:1; 1:3).

~ ~

3) Evaluate the vector pro ducts ~a b for the vectors of Q1. Using the expression j~a b j =

ab sin nd the angles theta b etween the vectors ~a and b and compare the results with

the angles found in Q1.

4) Using the results of Questions 1 and 3 show that

2 2 2 2

~ ~

(~a b) +(~a b) = a b

5) Show that the kinetic energy T = of a particle of mass m travelling with momentum

~p can b e represented as

2 2

p J

T = +

2m 2mr

~r ~p

where p = is the radial momentum and J = ~r ~p is the angular momentum of

j~r j

the particle.

6) Use the triple scalar pro duct to nd the equation of a circle that lies on the intersection

of a unit sphere with a plane through the centre of the sphere and passes through

twopoints given by the unit vectors ~a =(sin cos ; sin sin ; cos )andb =

a a a a a

(sin cos ; sin sin ; cos ). The equation should b e in the form of a relation

b b b b b

between the p olar angle and azimuth of a general p oint on the circle.

Find the minimum and maximum values of if = =3, =0, =2=3 and

a a b

= =2.

o o

Answer: 2 [50:77 ; 129:23 ]orj cot j 2=3. 21

Part 2. Vector Fields

1. De nition of scalar and vector elds; eld lines.

2. Gradient of a scalar eld; p otential

3. Fluxofavector eld; divergence.

4. The nabla op erator.

5. Theorems of Gauss and Green.

6. Curl of a vector eld; Stokes' theorem.

7. Exercises and problems.

1. De nition of scalar and vector elds.

In physics one calls a function of x; y ; and z a eld. We shall study in this section

scalar and vector elds.

Examples of scalar elds are the gravitational and the electrostatic p otentials. Exam-

ples of vector elds are the gravitational eld, the electrostatic eld and the velo city

eld of a uid.

We shall usually denote scalar elds by(x; y ; z )or(~r ), and vector elds by A(~r )or

E (~r )etc.

A scalar eldischaracterised byasinglevalue at eachpointinspace.

Avector eld is characterised by its magnitude and direction at eachpoint in space.

To get an intuitive picture of scalar and vector elds one draws equipotential lines for

scalar elds and eld lines for vector elds.

In physics one nds empirically relations b etween di erent elds, for instance b etween

magnetic eld and current. Here we study the mathematical form of such relationships.

2. Gradient of a scalar eld; p otential.

Consider a scalar eld, such as the gravitational p otential or the electrostatic p otential.

Denote the scalar eld by(~r )=(x; y ; z ). Keep z xed at some value z , i.e. consider

the scalar eld in a plane parallel to the (x; y ) plane, and consider the relation

(x; y ; z )= c

0 22

where c is a constant. This is an implicit equation of y as a function of x.Ifwe represent

this function in the (x; y ) plane, we get a line at xed p otential or equipotential line.

We get a set of equip otential lines if we let c assume di erentvalues (Fig. 1). Such

a set of equip otential lines are also called \contour lines". If now z is allowed to vary

continuously we get a set of equipotential surfaces.

Examples:

1. Let (~r )=r , then the equation of the equip otential surfaces takes on the form

2 2 2 2

(~r )=r = x + y + z = c

i.e. the equation of a family of spherical surfaces.

2. Let

1 1

(~r )=

j~r ~r j j~r + ~r j

0 0

i.e. the p otential of two unlikepointcharges at ~r and ~r . The contour lines for

0 0

z = 0 are as shown in Fig. 2.

If wemovealongacontour line the p otential do es not change. If wemoveatanangle

to an equip otential line we observeachange of the p otential. The fastest variation of

the p otential is observed when wemove at right angles to the contour lines. This is an

intuitivelyobvious statement. Wearenow going to put it on a rigorous mathematical

fo oting. Todothisweintro duce the concept of the gradient of a scalar eld.

Consider the di erential of a scalar eld (~r ). By the rules of calculus of functions of

several variables wehave

@ @ @

d(~r )= dx + dy + dz (39)

@x @y @z

and we see that this expression has the form of a scalar pro duct of twovectors,

@ @ @

(40) ; ;

@x @y @z

and

(dx; dy ; dz ) d~r

of (~r ) and is denoted grad (~r ). Thus The vector (40) is called the gradient

@ @ @

grad (~r )= ; ; (41)

@x @y @z 23

and we can rewrite Eq. (39) as

d(~r ) = grad(~r ) d~r (42)

Examples:

@r @r @r

; ; grad r =

@x @y @z

x @r y x @r

2 2 2

= , and similarly = = but r = x + y + z , hence

2 2 2

@x r @y r

x + y + z

@r z

and = ,hence

@z r

x y z 1 ~r

grad r = ; ; = (x; y ; z )= =^r

r r r r r

1 1 1

@ @ @

r r r

grad = ; ;

r @x @y @z

but

1 1 1

@ @ @

1 @r 1 x 1 @r 1 y 1 @r 1 z

r r r

= = ; = = ; = = ;

2 2 2 2 2 2

@x r @x r r @y r @y r r @z r @z r r

hence

~r r^ 1

= = grad

3 2

r r r

Prop erties of the gradient.

If u(~r ) and v (~r )aretwo scalar functions and a is a constant, then

(i)

grad(u(~r )+v (~r )) = grad u(~r ) + grad v (~r ) (43)

(ii)

grad(au(~r )) = a grad u(~r ) (44)

(iii)

grad(uv )=u grad v + v grad u (45)

(iv)

f (u)

gradf (u(~r )) = grad u (46)

du 24

(v)

grad(~r ) is p erp endicular to the equip otential surface (~r )= c: (47)

The pro ofs of prop erties (i) to (iv) are left to the reader as an exercise: they involve

no more than the rules of partial di erentiation of functions of several variables. The

pro of of prop erty (v) pro ceeds as follows:

consider the equation of the equip otential surface (~r )=c. Taketwo p oints on this

surface, separated by an in nitesimal displacement ~r , hence

d(~r )=grad(~r ) ~r =0

i.e. grad is p erp endicular to ~r . But in the limit when the two p oints coalesce,

~r ! d~r lies in the tangential plane of the equip otential surface, and hence grad

is normal to the surface = c. Note that this is what we said ab out the gradient

intuitively at the b eginning of this section.

The directional derivativeof(~r ).

Related to the gradientisthe directional derivative of a scalar eld.

Consider the scalar eld (~r ), and lets ^ b e an arbitrary unit vector. The expression

@ (~r ) (~r + "s^) (~r )

=lim

"!0

@s "

is called the (directional) derivativeof(~r ) in the direction ofs ^.To nd the relation-

ship b etween the directional derivative and the gradient, represent the unit vectors ^ in

terms of its directional cosines:s ^ = (cos ; cos ; cos ), hence

(~r + "s^)= (x + " cos ; y + " cos ; z + " cos )

and expanding this ab out " =0 weget

@ (~r ) @ (~r ) @ (~r )

(~r + "s^)=(~r )+ " cos + " cos + " cos + O (" )

@x @y @z

and hence

(~r + "s^) (~r ) @ (~r ) @ (~r ) @ (~r )

= cos + cos + cos + O (")

" @x @y @z

and in the limit of " ! 0 the left-hand side b ecomes the directional derivative, and on

the right-hand side we get the scalar pro duct of (~r ) withs ^,thus

@ (~r )

=^s grad (~r ) (48)

@s 25

Now, if the normal unit vector to the equip otential surface is denoted by^n,wecan

write grad =n ^ jgrad j, and hence

@ (~r )

= jgrad (~r )j cos

where cos =^s n^ . In the particular case whens ^ =^n, i.e. when the directional

derivative is taken in the direction normal to the equip otential surface, wehave

@ (~r )

= jgrad (~r )j

and b ecause of cos <1we can also conclude that

@ (~r )

jgrad (~r )j = max

i.e. the p otential changes most rapidly in the direction of the normal to the equip o-

tential surface (\direction of steep est descent").

Finally, let us remark that in physics one calls a vector eld, which is the gradientof

a scalar eld, a conservative vector eld.

3. Flux of a vector eld; divergence.

Let S b e a surface z = f (x; y ) and A(~r )a vector eld. Consider a surface element dS

whose normal isn ^ .Theintegral

A(~r ) ndS^

is called the ux of the vector eld A(~r )throughS . frequently one uses alternative

~ ~

notations. Observing that A (~r )=A(~r ) n^ is the normal comp onentof A(~r ) one can

write for the ux

A (~r )dS

and denotingndS ^ by dS one can also write

~ ~

A(~r ) dS

In the particular case when the surface S is closed, the integral is usually denoted by

the symbol . The surface integral

I I

~ ~ ~

A(~r ) dS = A(~r ) ndS^

S S

is the net ux of A out of the closed surface S ,if^n is the outward normal vector to

dS . If all eld lines of A b egin and end outside of the volume V enclosed by S , then

we exp ect the net ux to b e zero. In this case one says that there are no sources of 26

~ ~

A in V . On the other hand, if there is a source of A in V ,such as for instance an

electric charge that gives rise to a eld E , then the net ux has a p ositivevalue whose

magnitude dep ends on the strength of the source. If the net ux out of V takes on a

negativevalue, one says that the surface S contains a sink of the vector eld A. In the

caseofelectriccharge this corresp onds to a negativecharge.

~ ~

A dS and let the volume V shrink to a p oint P with Consider the expression

co ordinates ~r ,then

~ ~ ~

A dS (49) div A(~r ) = lim

V !0

is called the divergence of A at the p oint P .From our preceding discussion we see that

the divergence is a lo cal measure of the strength of the sources of the eld A.

It is frequently convenient to use a representation of the divergence in terms of partial

derivatives of the comp onents of A:

@A @A @A

x y z

div A(~r )= + + (50)

@x @y @z

To derive this formula cho ose V in the form of a rectangular volume with faces parallel

to the co ordinate planes, see Fig. NN. The surface integral b ecomes a sum of integrals

over the six faces of V . Consider the two faces parallel to the yz -plane:

Z Z

~ ~

A(x; y ; z ) (x^)dy dz + A(x + x;y;z) xdy^ dz =

(1) (2)

(" # )

@A (x; y ; z )

= A (x; y ; z )+ x + O (x ) A (x; y ; z ) dy dz

x x

where wehave expanded A (x + x; y; z ) up to the linear term in x. The terms

A (x; y ; z ) cancel and we are left with

# "

Z Z

z +z y +y

@A (x; y ; z ) @A (x; y + y; z + z)

x x

dz dy x = xyz

@x @x

z y

where 2 [0; 1] and 2 [0; 1]. If we divide this by dV = xyz and let the volume

shrink to a p ointwe are left with @A (x; y ; z )=@ x. Similarlywe get from the other two

pairs of faces @A (x; y ; z )=@ y and @A (x; y ; z )=@ z , and hence for the entire surface S

y z

@A (x; y ; z ) @A (x; y ; z ) @A (x; y ; z )

x y z

div A(~r )= + +

@x @y @z

Prop erties of divergence.

~ ~

(i) The divergence is a linear op eration, i.e. given the vector elds A and B and the

scalar constant c the following identities hold true:

~ ~ ~ ~ ~ ~

div (A + B ) = div A + div B; and div (c A)=c div A (51) 27

(ii)

div ~r =3 (52)

(iii)

df (r )

+3f (r ) (53) div f (r )~r = r

(iv)

div f (~r )~r = ~r div f (~r )+3f (~r ) (54)

(v)

n n

div (r ~r )=(n +3)r (55)

(vi)

~ ~ ~

div [f (~r )A(~r )] = A(~r ) grad f (~r )+f (~r )divA(~r ) (56)

4. The nabla op erator. In the preceding sections wehave encountered expressions in

which partial derivatives w.r.t. x, y and z play an imp ortant role. One can see that it

is convenienttocombine the partial derivative operators @=@x, @=@y and @=@z into a

vector op erator

@ @ @

r ; ; (57)

@x @y @z

called nabla (or sometimes \del").

Using the nabla op erator the gradient and divergence are written as

grad (~r )= r (~r ) (58)

and

~ ~

div A(~r )=rA(~r ) (59)

resp ectively. Thus the nabla op erator can b e handled likea vector, except that one

must rememberthatitdoesnotcommute with eld functions like an ordinary vector

b ecause it is also a di erential op erator that op erates on the function to the right.

Consider Eq. (49) 5. Theorems of Gauss and Green.

~ ~ ~

div A(~r ) = lim A dS

V !0

whichwehave written down as the de nition of the divergence of A(~r ). Realising that

the integral on the r.h.s. and the volume V are in nitesimal in the limit, we can also

write

~ ~ ~

A dS div A(~r )=

~ ~ ~

A dS = div A(~r ) dV d

S 28

Integrating this over the a nite volume V enclosed by the surface S we get

I Z

~ ~ ~

A dS = div A(~r ) dV (60)

S V

whichisknown as Gauss' theorem or divergence theorem.

The derivation of Gauss' theorem given here may seem rather formal. A more rigorous

and more transparent derivation pro ceeds as follows. Consider a volume V enclosed by

the surface S . Divide the volume into in nitesimal volume elements V , i =1; 2; :::; n

enclosed by surfaces S . Then, by de nition of the divergence, wehave

~ ~ ~

(div A ) V = A dS

i i

and summing over all volume elements weget

n n

X X

~ ~ ~

(div A ) V = A dS

i i

i=1 i=1

On the r.h.s. eachvolume element either shares a surface with a neighb ouringvolume

element or has an external surface. The external surface elements together makeupthe

surface S . The contributions of all internal surfaces, on the other hand, cancel since

they enter into the sum with opp osite signs for the two adjacentvolume elements.

Therefore, letting n !1 and at the same time the maximum linear dimensions of V

tend to zero, we get Gauss' theorem.

Of interest is frequently the divergence of vector elds which are the gradients of scalar

elds. Thus, if (~r ) is a scalar eld, wewant to nd the divergence of its gradient.

Using the nabla op erator we write this in the following form:

! ! !

@ @ (~r ) @ @ (~r ) @ @ (~r )

div grad (~r ) = r(r(~r )) = + +

@x @x @y @y @z @z

2 2 2 2 2 2

@ (~r ) @ (~r ) @ @ @ @ (~r )

+ + = + + (~r ) =

2 2 2 2 2 2

@x @y @z @x @y @z

where in the last step wehave written the expression in the form of a di erential

op erator op erating on the scalar function. This di erential op erator,

2 2 2

@ @ @

+ + (61) r =

2 2 2

@x @y @z

is called the Laplacian op erator. Formally it is given by the scalar pro duct rr.

Therefore we can write the equivalent forms

div grad = r(r ) = (r r) = r 29

Consider the scalar elds (~r ) and (~r ). The gradients of and are vectors, and

so is the expression A = [(r ) (r)]. Take the divergence of this vector:

2 2

div A = r[(r ) (r)] = [(r ) (r )]

and if we apply the divergence theorem to this equation we get

I Z

2 2

[(r ) (r)] dS = [(r ) (r )]dV (62)

S V

which is known as Green's theorem. Thus Green's theorem is a generalization of Gauss'

divergence theorem.

6. Curl of a vector eld; Stokes' theorem.

The last of the vector op erations to consider is the curl of a vector eld. This concept

arises in uid dynamics in connection with the mathematical description of vortices

and in electromagnetism for instance in the relationship b etween magnetic elds and

currents.

Considering a vortex one observes that the uid rotates ab out the centre of the vortex

like a solid disk. The linear velo city of small volume lements of the uid increases

linearly from the centre and is directed p erp endicularly to the radius vector if we put

the origin of the co ordinate system at the centre of the vortex. If we put the xy -plane

normal to the vortex axis, we can write for the linear velo city

~v = !r'^

where ! is the angular velo city, which is indep endentof r ,and'^ isaunitvector

p erp endicular to the radius vector and p ointing in the direction of increasing azimuth '.

Thus' ^ =(x^ sin ' +^y cos '), and therefore, recalling that x = r cos ' and y = r sin ',

weget~v = ! (y x^ + xy^).

Now consider the line integral

I = ~v d~r

where C is a closed circular path around the origin (see Fig. NN). Since r = const. on

C ,wehave d~r = r 'd'^ , and hence

2 2

I = !r d' =2r !

and noting that A = r is the area of the circle enclosed by C we nally get

I =2A!

We see that the integral characterizes the strength of the vortex; it is called the circu-

lation of the vector eld ~v (Lord Kelvin). 30

If wetake a di erent closed path in the vortex eld we can repro duce the same result

even if the path do es not enclose the centre of the vortex (see exercise NN).

In the expression for the circulation wehave the arbitrary factor A. A b etter charac-

teristic of the vortex is therefore the following expression

~v d~r

which can b e used to get a lo cal measure of the circulation if we let the path C contract

into a p oint. Assuming that the surface A is already small enough to b e at, wecan

assign a normaln ^ to it and de ne the comp onent of the curl of ~v in the direction ofn ^ :

(curl ~v ) =lim ~v d~r (63)

A!0

Here weusetheconvention that, lo oking down from the unit vectorn ^ , the integration

along C go es in anticlo ckwise direction (\mathematically p ositive" sense).

In the particular case when C lies in the yz -plane (or in a plane parallel to the yz -

plane) and the normal p oints along the x axis we get (curl ~v ) . Let us evaluate this

comp onentofthecurlcho osing the path C in the form of a rectangle whose sides are

parallel to the y and z axes, see Fig. NN. The integral then b ecomes the sum of four

integrals along the four sides of the rectangle:

I Z Z Z Z

~v d~r = v (x; y +y; z )dz v (x; y ; z +z )dy v (x; y ; z )dz + v (x; y ; z )dz

z y z y

C (1) (2) (3) (4)

In the integrals along sides (1) and (2) we expand the argumentabouty and z ,

resp ectively,andweget

Z Z

z +z y +y

@v @v

z y

dz dy y z

@y @z

z y

For suciently small y and z the integrands are constant and can b e taken outside

the integrals, hence

Z Z I

z +z y +y

@v @v @v @v

y y z z

y z yz z y dz dy = ~v d~r =

@y @z @y @z

z y C

but A = yz is the area enclosed by C , hence, dividingby A and letting all sides

of the rectangle tend to zero, weget

@v @v

y z

(curl ~v ) =

@y @z

and similarly weget

@v @v @v @v

x z y x

(curl ~v ) = and (curl ~v ) =

y z

@z @x @x @y 31

and hence, nally,

@v @v @v @v @v @v

y x z y x z

curl ~v = ; ; : (64)

@y @z @z @x @x @y

Stokes' theorem equates the integral of the curl of a vector eld ~v (~r )over a surface S

to the line integral of ~v over the b oundary ` of S :

Z I

~ ~

(curl ~v ) dS = ~v d` (65)

S `

The surface S need not b e a plane surface. To get the correct sign in Stokes' theorem

the unit normal vectorn ^ must satisfy the corkscrew rule with the direction, in which

the b oundary ` is traversed.

The idea of the pro of of Stokes' theorem is similar to that of Gauss' theorem. Divide

S into n cells S =^n S . In the limit of S ! 0we get for the ith cell, by de nition

i i i i

of the curl,

(curl ~v ) = ~v d`

Taking the scalar pro duct withn ^ S and summing over i we get

i i

n n

X X

(curl ~v ) S = ~v d`

i i

i=1 i=1

If wetake the limit of n !1 and at the same time let the maximum diameter of S

tend to nought, then we get on the left-hand side (curl ~v ) dS . On the right-hand

side the contributions from all internal lines cancel, since they are traversed twice in

opp osite directions for each of the neighb ouring cells, and only the contributions from

the external lines remain, which add up to yield the line integral ~v d`.

The curl of a vector eld takes on a compact form when it is expressed in terms of the

nabla op erator. Indeed, consider the cross pro duct r~v :by de nition of nabla we

have

! !

@ @ @v @v @v @v @v @v @

z y x z y x

; ; (v ;v ;v )= ; ; r~v =

x y z

@x @y @z @y @z @z @x @x @y

i.e. the curl of ~v .

Prop erties of the curl.

(i)

r~r =0

(ii)

r (r(~r ))=0

i.e. apotential eld is irrotational. 32

(iii)

r(r~v )=0

i.e. a vortex eld has no sources.

(iv)

r(~u ~v )=~v (r~u) ~u (r~v )

(v)

r(r~v )=r(r~v ) r ~v

The pro of of (i) to (iv) is straightforward. The pro of of (v) is also straightforward but

lengthy. Here it is done for the x comp onent:

@ @

(r~v ) (r~v ) [r(r~v )] =

z y x

@y @z

! !

@ @v @v @v @ @v

x z y x

@y @x @y @z @z @x

2 2 2 2

@ v @ v @ v @ v

x z x y

+ =

2 2

@y@x @z@x @y @z

2 2

this expression do es not change if we add @ v =@ x to the rst pair of terms and subtract

it from the second pair. This gives us

! !

2 2 2

@v @v @v @ @ @ @ @

x y z

+ + + + v = (r~v ) r v

x x

2 2 2

@x @x @y @z @x @y @z @x

and similarly for the y and z comp onents.

7. Application: Maxwell equations of Electromagnetism.

Maxwell's equations in SI units are of the following form:

@ D

curlH = ~| + Ampere-Maxwell law (66)

@ B

curlE = Faraday-Lenz law (67)

rD = Gauss' law (68)

rB = 0 no magnetic charges (69)

~ ~

D = "E (70)

~ ~

B = H (71)

~ ~ ~

Here H is the magnetic eld strength, B is the magnetic ux density, E is the electric

eld strength and D is the electric displacement; these four vectors are functions of

space and time. and ~| are the charge density and current density, resp ectively.

French: Systeme International d'Unites 33

" = " " is the permittivity,with" the relative permittivity (also called dielectric

r 0 r

constant)and" the permittivity of freespace or electric constant; " is dimensionless,

0 r

" =8:854 pF m . = is the permeability with the relative permeability

0 r 0 r

(which is dimensionless) and =4 10 Henry p er meter the permeability of free

space or magnetic constant.

Note that

" = c (72) 1=

0 0

is the sp eed of light in free space.

From the Ampere-Maxwell law together with Gauss' lawwe get immediately the conti-

nuity equation: take the divergence of Eq. (66), use the vector identity r(rH )=0,

and substitute the divergence of D from Eq. (68), hence

+ r~| =0 (73)

In free space " = =1, = 0 and ~| =0. We can therefor rewrite the Maxwell

r r

~ ~

equations in terms of, for instance, H and D , eliminatingalso " and ,thus

0 0

@ D

curlH = (74)

@ H 1

(75) curlD =

c @t

rD = 0 (76)

rH = 0 (77)

whichcanbeshown to have plane wave solutions

i(k ~r !t)

~ ~

D = D e (78)

and

i(k ~r !t)

~ ~

H = H e (79)

~ ~

~ ~ ~ ~ ~

with D ? H , D ? k and H ? k . The direction of D at di erent times t de nes a

plane, transverse to the direction of propagation of the wave. This plane is the plane

of polarization, and the wave is said to b e a transverse wave. The sup erp osition of two

~ ~

waves travelling in the same direction with vectors D and D , which are mutually

1 2

orthogonal and whose phases di er by =2, pro duces an el liptical ly polarised wave; in

~ ~

the particular case of equal amplitudes, jD j = jD j, the waveissaidtobecircularly

01 02

polarised. 34

~ ~

Exercise: Assuming that D is a plane wave, show that H is also a plane wave with

the same phase as D , and that the electric and magnetic elds are orthogonal to each

other and to the direction of propagation.

~ ~

We can deduce wave equations for D and H from the Maxwell equations, for instance

by taking the curl of Eq. (74) and substituting the curl of D from (75), hence, applying

the identity

r(r~a(~r )) = r(r~a(~r )) r ~a(~r ) (80)

and using Eq. (77), we get the wave equation

1 @

r H =0 (81)

2 2

c @t

substitutinginto which the plane wave function (79) wegetthedispersion law of

electromagnetic waves in free space,

! = ck (82)

The wave equation for D is obtained similarly. 35