Lectures on Vector Algebra

PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta 1

Vector Algebra & Vector Analysis (Combined Lectures, 1st Ed.)

Lecture Notes prepared by-

Dr. Abhijit Kar Gupta Physics Department, Panskura Banamali College

Panskura R.S., East Midnapore, WB, India, Pin-code: 721152 e-mail: [email protected], [email protected],

Lecture-1

Books to be consulted: 1. Vector Analysis

- Murray R. Spiegel (Schaum Series, McGraw Hill) 2. Mathematical Methods - Merle C. Potter, Jack Goldberg (Prentice Hall of India) 3. Introduction to Mathematical Physics - Charlie Harper (PHI) 4. Mathematical Methods for Physicists - G. Arfken (Academic Pub., Prism Books Pvt. Ltd.) 5. Mathematical Physics - H. K. Dass (S. Chand & Company Ltd.) We begin from the definition of ‘Vector’. Vector: A vector is a quantity having both magnitude and direction. Examples: v (velocity), F (force), m (magnetization) etc. To express a vector A pictorially: We have to know the coordinates of the starting point, the coordinates of the end point with respect to a fixed coordinate system. The arrow (in the picture) indicates the direction and the length of it (measured in the coordinate system) is the magnitude of the vector. Magnitude of a vector A is written as | A | (sometimes only A ). Any vector along the direction of A but having unit magnitude is called unit vector.


Unit Vector: ||

ˆAAu =

Null Vector: Ο

), having magnitude zero.

Parallelogram law for Vector addition: (Sometimes called Triangle law of addition)

The above law can be extended to add any number of vectors:

The resultant vector (sum of all vectors) always starts from the starting point of the 1st vector and ends at the end of the last vector. This is the end-to-end distance of a series of ‘walks’ (as directed by the vectors). Applications: Any polygon of vectors, a model Polymer, Random Walk To construct a Vector Algebra:

• Arithmetic is mathematical operations (summation, subtraction, multiplication and division) with Numbers.

• Algebra is mathematical operations with symbols. • Therefore, Vector Algebra must be mathematical operations with Vectors.

Mathematical operations are to be defined. The ‘unit’ and ‘zero’ entities are already defined. Some basic rules are to be defined to construct the Algebra.

Resultant Vector

A B

BAC +=


Rules of Vector Algebra: 1. ABBA +=+ (Cumulative law for addition)

2. mAAm = (Cumulative law for multiplication) 3. CBACBA ++=++ )()( (Associative law for addition) 4. AmnAnm )()( = (Associative law for distribution)

5. AnAmAnm +=+ )( (Distributive law for addition)

6. BmAmBAm +=+ )( (Distributive law for multiplication). Lecture-2 Let us think of a rectangular Cartesian (right handed) coordinate system:

A vector A has components xA , yA and zA along x-, y- and z-axes respectively.

Therefore, one can write kAjAiAA zyx

)))++= .

Magnitude of A is || A = 222zyx AAA ++ .

Y

Z

i)

j)

k)

X

A

A B

B

AB

C

The three unit vectors along the x-, y- and z-axes are i)

, j)

and k)

respectively.


• Home Work: Prove the above by Pythagoras theorem. Definition of Vector Field: If in a region R(x,y,z) everywhere there is a vector ),,( zyxA defined then we can say that a vector field in R(x,y,z) is defined. Suppose jyixA ˆˆ += and we have ayxA =+= 22|| =constant. Then we have a region of concentric circles for all values of a . The region under each circular ring is a vector filed where a vector of magnitude of the radius of a vector is defined. Lecture-2 Suppose a vector A makes angles α , β and γ with respect to x-, y- and z-axes respectively. We can write:

αcos = || A

Ax , βcos = || A

Ay , γcos = || A

Az .

Therefore, we find

.1||||

||coscoscos

2

2

2

222222 ==

++=++

AA

A

AAA zyxγβα

Problem # 1 Equation of a circle:

A

α

β γ

X

Y

Z


Note: a ‘position vector’ is such that its tip denotes the position of a point and the starting point is at the centre of the coordinate system. Let the position vectors are kzjyixrOP ˆˆˆ

0000 ++=≡

kzjyixrOQ ˆˆˆ ++=≡ ∴ The vector PQ = 0rr −

= )ˆˆˆ()ˆˆˆ( 000 kzjyixkzjyix ++−++

= kzzjyyixx ˆ)(ˆ)(ˆ)( 000 −+−+−

The magnitude of the vector PQ is 2

02

02

0 )()()(|| zzyyxxPQ −+−+−= .

Now if the magnitude of || PQ is = a (constant) then we can write 22

02

02

0 )()()( azzyyxx =−+−+− . This is an equation of a circle with the centre at ( ),, 000 zyx and radius ‘ a ’. Problem # 2 Equation of a straight line passing through the points ),,( 111 zyxP and

).,,( 222 zyxQ

X

Y

Z

O r

0r

),,( zyxQ

),,( 000 zyxP


Let R ),,( zyx be any point on the straight line which joins ),,( 111 zyxP and

).,,( 222 zyxQ

Now, kzjyixr ˆˆˆ1111 ++=

kzjyixr ˆˆˆ2222 ++=

kzjyixr ˆˆˆ3333 ++=

Let ),,( zyxR be such a point on the line PQ that PQmPR = , where m is a scalar quantity and a fraction here. ∴ )()( 121 rrmrr −=− ⇒ [ ])ˆˆˆ()ˆˆˆ()ˆˆˆ()ˆˆˆ( 111222111 kzjyixkzjyixmkzjyixkzjyix ++−++=++−++ ⇒ [ ]kzzjyyixxmkzzjyyixx ˆ)(ˆ)(ˆ)(ˆ)(ˆ)(ˆ)( 121212111 −+−+−=−+−+− Equating the respective components on both sides,

)()( 121 xxmxx −=− )()( 121 yymyy −=−

)()( 121 zzmzz −=− . The above three equations yield

12

1

12

1

12

1

zzzz

yyyy

xxxx

−−

=−−

=−− .

X

Y

Z

O

),,( 111 zyxP ),,( zyxR

),,( 222 zyxQ 1r

2r

r


Vector in a Rotated Coordinate system:

Let us consider the vector r with respect to the orthogonal Cartesian coordinate system; the components are x and y .

Now, keeping r fixed we can rotate the coordinate system so that the new system rotates by an angle φ with respect to the old one.

Resolving the vector r into the components in the new system we get,

φφ sincos yxx +=′ φφ cossin yxy +−=′

We can write the above in the following fashion:

−

=

′′

yx

yx

φφφφ

cossinsincos

Or in an abstract form: rMr =′ , where M is a Matrix having 2 rows and 2 columns. M is a 22× Matrix. Similarly, the components of the vector form a 12× Matrix (column matrix). In general, any 22× matrix is defined by

2221

1211

aaaa

The determinant of the above matrix is = 12212211 aaaa ⋅−⋅ . Therefore, the determinant of the matrix M is | |M = 1sincos 22 =+ φφ .

X

X ′

Y Y ′

φ

r


Here in this example, the matrix M is called a Rotational matrix. This is also called an orthogonal matrix whose determinant is unity. The magnitude of the vector r and that of r ′ (in the new coordinate system) are same. This means that the magnitude of a vector is an invariant quantity under rotation.

PRODUCT OF TWO VECTORS Dot Product (Scalar Product):

The Scalar or Dot product is defined by θcos|||| BABA =⋅ …………………….(1) Example: Work, W = rF ∆⋅ (Product of Force and displacement). Also, BA ⋅ = )ˆˆˆ()ˆˆˆ( kBjBiBkAjAiA zyxzyx ++⋅++ = zzyyxx BABABA ++ ,…….(2)

where the unit vectors i , j , k satisfy the following relations

1ˆˆˆˆˆˆ =⋅=⋅=⋅ kkjjii and 0ˆˆˆˆˆˆ =⋅=⋅=⋅ kjkiji . Where the three unit vectors are mutually perpendicular. The above relations are called orthogonality conditions. The above orthogonality relations can also be written in a compact form: nm uu ˆˆ ⋅ = mnδ …………………………………………..(3) where the indices m and n take three values, =m 1, 2, 3 and =n 1, 2, 3. The unit vectors

mu or nu means that

1u = i, 2u = j , 3u = k . The symbol in (3) mnδ is called the Kronecker delta defined by

θ A

B


mnδ = 1 for nm = = 0 for nm ≠ From the definition (2), we could write xA = | A |, yA = 0, zA = 0 and xB = | B | θcos to arrive at (1). Also from (2) we can write

AA ⋅ = zzyyxx AAAAAA ++ = 222zyx AAA ++

∴ AA ⋅ = 2|| A ≡ 2A , where we write || A = A , the magnitude of the vector A . Example:

The kinetic energy, E = 2

21 mv = )(

21 vvm ⋅ .

Application: Proof of the Law of Cosines

C = A + B

CC ⋅ = ( ) ( )BABA +⋅+

= BABBAA ⋅+⋅+⋅ 2 ∴ 2C = θcos222 ABBA ++ = )cos(222 φπ −++ ABBA ∴ 2C = φcos222 ABBA −+ . Cross Product (Vector Product):

BAC ×= The vector C will be perpendicular to the plane of A and B . The magnitude

|| C = θsin|||| BA .

Also the vector C is such that A , B and C form a right-handed system. With the above choice,

ABBA ×−=× . Relations among the three unit vectors in the right-handed Cartesian coordinate system:

A

C B

φ θ


0ˆˆˆˆˆˆ =×=×=× kkjjii kji ˆˆˆ =× , ikj ˆˆˆ =× , jik ˆˆˆ =×

kij ˆˆˆ −=× , ijk ˆˆˆ −=× , jki ˆˆˆ −=× . Cross product of two vectors is also defined in the following and more useful way:

C = kCjCiC zyxˆˆˆ ++ = BA× =

zyx

zyx

BBBAAAkji ˆˆˆ

, a 33× determinant.

= )(ˆ)(ˆ)(ˆ yxyxzxzxZyZy ABBAkABBAjABBAi −+−−− .

In the above, the components ( xC , yC , zC ) of the product vector C are easily identified. Home Work: If ABBA ×=× , what is the relation between the two vectors (assume that they are nonzero vectors)? Lecture-3 Physical Examples of Vector Product: Angular Momentum : prL ×= ,

r is position vector, p is linear momentum. Relation between Linear and Angular velocity: rV ×= ω Numerical EXAMPLES: product of two vectors Given two vectors, kiA ˆˆ −= kjiB ˆˆˆ2 +−=

BA× = 112101

ˆˆˆ

−−kji

= 12

01ˆ1211ˆ

1110ˆ

−+

−−

−−

kji

= )01(ˆ)21(ˆ)10(ˆ −−++−− kji = kji ˆˆ3ˆ −−− 222 )1()3()1( −+−+−=×∴ BA = 11191 =++ ………………………(1)

1121)1()1(021 =−=⋅−+−⋅+⋅=⋅ BA

Therefore, =⋅ BA 1cos||.|| =θBA


We have, || A = 2)1(01 222 =−++ and

|| B = 61)1(2 222 =+−+

∴ θcos = 121

6.21

= ⇒ θsin = θ2cos1− = 1211 .

Now we can also write the cross product of the two vectors,

=×∴ BA θsin||.|| BA = 2 . 6 .1211 = 11 ………………….(2)

The above exercise shows that the definitions of cross products of two vectors in two different ways in (1) and in (2) yield the same result.

1. Using the following vectors

θθ sinˆcosˆ jiP += ,

φφ sinˆcosˆ jiQ −= ,

φφ sinˆcosˆ jiR += prove the following trigonometric identities:

φθφθφθ sincoscossin)sin( +=+ φθφθφθ sinsincoscos)cos( −=+ .

2. Verify that if you have two set of vectors A , B , C and A′ , B′ , C ′ such that 1=⋅′=⋅′=⋅′ CCBBAA , 0=⋅′=⋅′=⋅′=⋅′=⋅′=⋅′ BCACCBABCABA then

CBA

CBA×⋅

×=′ ,

CBAACB×⋅

×=′ ,

CBABAC×⋅

×=′ .

TRIPLE PRODUCT: product of three vectors

Home-Work Problems

Note: A′ , B′ and C ′ are called Reciprocal set of vectors.


Product of three vectors can also be either a vector or a scalar. This can be a combination of dot and cross products or only cross products. Triple Scalar Product:

)()()()( xyyxzzxxzyyzzyx CBCBACBCBACBCBACBA −+−+−=×⋅ )()()( xyyxzzxxzyyzzyx ACACBACACBACACB −+−+−=

)( ACB ×⋅= Thus we can show )( CBA ×⋅ )( ACB ×⋅= )( BAC ×⋅= .

Also, we can show

)( CBA ×⋅ = )()()( CABABCBCA ×⋅−=×⋅−=×⋅− If C = BA× then

)( BAACA ×⋅=⋅ = )()()( xyyxzzxxzyyzzyx BABAABABAABABAA −+−+− = 0

Similarly, 0)( =×⋅=⋅ BABCB . Physical Example:

CBA ×⋅ = Volume of the Parallelepiped by the three vectors A , B and C (taken in

proper order)

A

B C

A

B

C

Cyclically…


Now if CBA ×⋅ = 0 then we can say that the three vectors are coplanar (lie in the same plane). Triple Vector Product:

)()( BACCABCBA ⋅−⋅=×× The above can be proved in a straight forward way, taking kAjAiAA zyx

ˆˆˆ ++= etc. and

using the rule

zyx

zyx

BBBAAAkji

BA

ˆˆˆ

=× and so on.

• Product of Four or more vectors can be obtained by using the rules of

Vector triple products. 1. Prove the following: ( ) ( ) ( )( ) ( )( )CBDADBCADCBA ⋅⋅−⋅⋅=×⋅× and then show that ( ) ( ) ( ) ( )22 BAABBABA ⋅−=×⋅× . 2. Show that ( ) ( ) ( ) ( )DCBACDBADCBA ×⋅−×⋅=××× Lecture-4

Vector Operator

GRADIENT:

Do this yourself and check the formula.

Home-Work Problems


Suppose we have a scalar quantity )(xφ , then the differentiation of this with respect to

the only variable x is dxdφ .

Now if ),,( zyxφ is scalar function whose value depends on the values of the coordinates ),,( zyx then the dependence of φ on each coordinate separately can be expressed

through partial differentiation:

zyxzyx

,

),,(

∂∂φ

≡ x∂

∂φ , zxy

zyx

,

),,(

∂

∂φ≡

y∂∂φ ,

yxzzyx

,

),,(

∂∂φ

≡ z∂

∂φ .

We can now construct a new vector A in the following way:

zyj

xiA

∂∂

+∂∂

+∂∂

=∇≡φφφφ ˆˆ .

Here ∇ is an ‘operator’ (a vector differential operator) which is zy

jx

i∂∂

+∂∂

+∂∂

=∇ ˆˆ .

[ Note: ,x∂∂ ,

y∂∂

z∂∂ are also operators; they are ordinary differential operators.]

φ∇ is called the GRADIENT of the scalar φ . Applications: Let us take the magnitude of the position vector φ = | r | = 222 zyx ++ .

∴ zy

jx

i∂∂

+∂∂

+∂∂

=∇φφφφ ˆˆ ,

where x∂

∂φ = 2/1222 )( zyxx

++∂∂ = 2/1222 )( zyx

x++

= φx .

Similarly, y∂∂φ =

φy and

z∂∂φ =

φz .

∴ φ∇ = φ1)ˆˆˆ( zkyjxi ++ =

φr =

|| rr = n .

Here n is a unit vector in the positive direction of the position vector. We have here rr == ||φ .

So we can write r = n || r = φ∇r = φφ∇ in this case.

Also note that r⋅∇φ = rnn ˆˆ ⋅ = r . Product of the vector on its own unit vector = magnitude of the vector itself.


In general, if φ = |)(| rf = )(rf then x∂

∂φ = xr

rrf

∂∂⋅

∂∂ )( =

rx

drdf

⋅

∴ φ∇ = drdf

rr =

drdfn =

drdn φˆ .

Interpretation of φ∇ : The infinitesimal increment of the position vector dzkdyjdxird ˆˆˆ ++=

∴ ( ) dzz

dyy

dxx

rd∂∂

+∂∂

+∂∂

=⋅∇φφφφ = φd .

So, the above gives an estimate of the change in the scalar function φ due to the change

in position r . Now we can think of a surface where ),,( zyxφ = constant. Example: ,),,( 2222 azyxzyx =++=φ Equation of a Sphere.

If r be the position vector of a point on a surface (as shown in the figure) then the infinitesimal change, rd is on the surface. We can write φd = 0 = rd⋅∇φ .

∴ φ∇ is perpendicular to the vector rd . That means φ∇ is a vector perpendicular to the surface at the point ( ),, zyx . If φ is some kind of potential (electrical or something), the surface is called equipotential surface.

rd φ∇

X

Y

Z φ = constant

rrdr +


Also, along any direction n , the component of φ∇ is ( φ∇ n⋅ ). So this is the rate of change of φ at ( ),, zyx in that direction. The DIVERGENCE: If we have a vector V = zyx VkVjVi ˆˆˆ ++ differentiable at each point ),,( zyx in some region of space, we can define a scalar quantity called Divergence:

V⋅∇ = ( )zyx VkVjViz

ky

jx

i ˆˆˆˆˆˆ ++⋅

∂∂

+∂∂

+∂∂ =

zV

yV

xV zyx

∂∂

+∂

∂+

∂∂

Examples:

• r⋅∇ = ( )zkyjxiz

ky

jx

i ˆˆˆˆˆˆ ++⋅

∂∂

+∂∂

+∂∂ =

zz

yy

xx

∂∂

+∂∂

+∂∂ = 3

• )(rfr⋅∇ = [ ] [ ] [ ])()()( rzfz

ryfy

rxfx ∂

∂+

∂∂

+∂∂

= drdf

rz

drdf

ry

drdf

rxrf

222

)(3 +++ = drdfrrf +)(3

[ ]2222 zyxr ++=Q

• If we have a combination of a vector (V ) and a scalar (φ ) such that VA φ= , then

A⋅∇ = )()()( zyx Vz

Vy

Vx

φφφ∂∂

+∂∂

+∂∂

∴ )( Vφ⋅∇ = z

VV

zyV

Vyx

VV

xz

zy

yx

x ∂∂

+∂∂

+∂

∂+

∂∂

+∂∂

+∂∂ φφφφφφ

=

∂∂

+∂

∂+

∂∂

+

∂∂

+∂∂

+∂∂

zV

yV

xV

Vz

Vy

Vx

zyxzyx φφφφ

= ( ) VV ⋅∇+⋅∇ φφ ( Combination of ‘Divergence’ and ‘Gradient’)

If we would have a vector A = φ∇ then we can write

A⋅∇ = φ∇⋅∇ =

∂∂

+∂∂

+∂∂

⋅

∂∂

+∂∂

+∂∂

zk

yj

xi

zk

yj

xi φφφ ˆˆˆˆˆˆ

= 2

2

2

2

2

2

zyx ∂∂

+∂∂

+∂∂ φφφ = φ2∇ (as it is denoted)


Interpretation: The Divergence signifies the flow of ‘something’ out of a volume. If A⋅∇ = 0 (nothing comes out!), the vector A is called ‘Solenoidal’. The CURL: For the vector V we can define Curl as

V×∇ =

zyx VVVzyx

kji

∂∂

∂∂

∂∂

ˆˆˆ

=

∂∂

−∂∂

+

∂∂

−∂∂

+

∂∂

−∂∂

xyzxyz Vy

Vx

kVx

Vz

jVz

Vy

i ˆˆˆ

Example: If V = )(rfr then we have

V×∇ = [ ] rrfrrf ×∇+×∇ )()( For rV = we can check, r×∇ = 0.

Also we derived earlier, dr

rdfnrf )(ˆ)( =∇ .

Therefore, )(rfr×∇ = rndrdf

×ˆ = 0. [ n is the unit vector along r ]

Interpretation: The curl signifies the circulation or rotation of ‘something’ around a loop. If V×∇ = 0, then the vector V is called Irrotational. Use the formula )()( BACCABCBA ⋅−⋅=×× and find out

)( A×∇×∇ = ).(2 AA ⋅∇∇+∇− This may also be checked from basic definitions. Try that.

Lecture-5 Applications in Electromagnetic theory: Maxwell’s equations

Note: AAA ×∇+×∇=×∇ φφφ )(

Home-Work Problem


0=⋅∇ E ………………………………(1) 0=⋅∇ H ………………………………(2)

tHE∂∂

−=×∇ …………………………(3)

tEH∂∂

=×∇ ……………………..…….(4)

From (3) we can write

( )

∂∂

−×∇=×∇×∇tHE .

Now, ( ) EEEE 22 )( −∇=⋅∇∇+−∇=×∇×∇ [using equation (1)] Also we can write

( ) 2

2

tE

tE

tH

ttH

∂∂

−=

∂∂

∂∂

−=×∇∂∂

−=

∂∂

−×∇ [using equation (4)]

∴ ……………………….(I) Similarly, from (4) we get

………………………(II) The relations (I) and (II) are called wave equations. Each component of E ( zyx EEE ,, )

and that of H ( ),, zyx HHH satisfy above equations. Therefore, we can write for example,

2

2

2

2

2

2

2

2

tE

zE

yE

xE xzyx

∂∂

=∂∂

+∂

∂+

∂∂

and so on.

That means, in general if φ is any scalar which is any if the components ( zyx EEE ,, ) or

( ),, zyx HHH then 2

2

2

2

2

2

2

2

tzyx ∂∂

=∂∂

+∂∂

+∂∂ φφφφ .

This is a wave equation.

2

22

tEE

∂∂

=∇

2

22

tHH

∂∂

=∇


Additional problems: (i) Express i , j , k in terms of a gradient operator. (ii) Express i , j , k in terms of two gradient operators. Hints: (i) Put φ = x in the expression of the gradient operator and see. (ii) Use the relation i kj ˆˆ =× and then use the result of (i).

Vector Integration We can sum up several vectors and get a new resultant vector. Therefore, we may also integrate a vector which is defined at every point ( zyx ,, ) on a line or on a surface or in a volume. Line Integrals: In a three dimensional Cartesian Coordinate system, the increment of length is

dzkdyjdxird ˆˆˆ ++= .

Let us have a vector zyx AkAjAizyxA ˆˆˆ),,( ++= defined at every point on a continuous curve C .

We have the following type of line integral:

rdAP

P

⋅∫2

1

= rdAC⋅∫ = ( )∫ ++

C zyx dzAdyAdxA .

This represents the integral of the tangential component of A along C from 1P and 2P . If C is a closed curve (simple curve, no intersection by itself), the integral is denoted by

rdA ⋅∫ = ( )∫ ++C zyx dzAdyAdxA .

C

1P

2P


The above is called circulation of A about C . Example: The work done by a force along a path

W = rdFP

P

⋅∫2

1

= dzFdyFdxF zy

P

Px ++∫

2

1

.

Suppose, xjyiF ˆˆ +−= , )0,0(1 =P and )1,1(1 =P . The work done along a path going from point 1P to the point 2P is then

W = ∫ +−1,1

0,0

)( xdyydx = ∫∫ +−1

0

1

0

xdyydx .

Consider the following path as shown:

If we choose another path as shown below:

Therefore, we see that this choice of force, the work done depends on the choice of path.

(0, 0)

(0, 1) (1, 1)

X

Y

W = 0 ∫ +−1

0

.2/1 dx ∫1

2/1

.1 dy = 0.

(0, 0) (1, 0)

(1, 1)

X

Y

∴ W = ∫ ∫+−1

0

1

0

.1.0 dydx = 1.


On the other hand, if rdFWC

⋅= ∫ is independent of the choice of path C joining any

two points 1P and 2P , the force F is called the Conservative force (For any arbitrary

vector A , this is called Conservative vector). If we have A = φ∇ , where φ is a scalar quantity (called, scalar potential),

rdA ⋅ = dzAdyAdxA zyx ++ = dzz

dyy

dxx ∂

∂+

∂∂

+∂∂ φφφ = φd .

∴ rdAP

P

⋅∫2

1

= ∫2

1

P

P

dφ = ).()( 21 PP φφ −

Thus the value of the above integral depends only on the end points 1P and 2P and not on the choice of path connecting them.

Now, as A = φ∇ , we have So, this is also a condition for the vector A to be conservative. • There can be other two types of line integrals, ∫C rdφ and rdA

C×∫ .

Lecture-6

Conservative Force, Potential: If F be a conservative force field, we may write φ∇−=F , φ is a scalar quantity, we

call it potential (The negative sign we choose deliberately for attractive forces).

We can write F = am = 2

2

dtrdm .

∴ dt

rdF ⋅ = dt

rddt

rdm ⋅2

2

=

⋅

dtrd

dtrd

dtdm

2 = ( )2

2v

dtdm

∴ rdF ⋅ = )(2

2vdm

Therefore, Work done along any arbitrary path from a point A to the point B is

W = rdFB

A

⋅∫ = )(2

2∫B

A

vdm = 2m B

Av2 = 22

21

21

AB mvmv − .

On the other hand,

0=×∇ A


rdFB

A

⋅∫ = rdB

A

⋅∇− ∫ φ = ∫−B

A

dφ = )()( BA φφ − .

∴ )()( BA φφ − = 22

21

21

AB mvmv − .

Therefore, we can identify )(Aφ as the potential energy at A and )(Bφ as that at B . So, the negative sign in F = φ∇− is also explained in the sense of conservation of total energy:

)(Aφ + 2

21

Amv = )(Bφ + 2

21

Bmv = constant.

A Standard Problem: If zyx AkAjAiA ˆˆˆ ++= is a conservative vector, (i) Find the scalar potential, (ii) Find the integration of A along a curve from some point 1P ( ,1x ,1y 1z ) to

2P ( ,2x ,2y 2z ). Soln.

(i) We can write A = zy

jx

i∂∂

+∂∂

+∂∂

=∇φφφφ ˆˆ = zyx AkAjAi ˆˆˆ ++ .

∴ x∂

∂φ = ),,( zyxAx ……………………(1)

y∂∂φ = ),,( zyxAy ……………………(2)

z∂

∂φ = ),,( zyxAz ……………………(3)

Integrating (1), (2) and (3) respectively, φ = ),,(1 zyxf + ),(1 zyc φ = ),,(2 zyxf + ),(2 zxc φ = ),,(3 zyxf + ),(3 yxc Choose the integration constants 1c , 2c and 3c such that φ turns out to be the same in all the expressions. Example:

kxzjxizxyA ˆ3ˆˆ)2( 223 +++= Here we have,

32 zxyx

+=∂∂φ ………………………………………(1)

2xy=

∂∂φ ……………………………………………… (2)


23xzz=

∂∂φ ……………………………………………..(3)

Integrating (1), (2) and (3) respectively, ),(1

32 zycxzyx ++=φ

),(

),(

33

22

yxcxz

zxcyx

+=

+=

φ

φ

So, we choose 0),(1 =zyc , 32 ),( xzzxc = and yxyxc 2

3 ),( = such that 32),,( xzyxzyx +=φ .

One may also add some arbitrary constant (independent of x, y, z) with it. Surface Integrals: Consider the area element .dS This area element can be treated as a vector dSnSd ˆ≡ , where n is a unit normal vector to indicate the positive direction.

There are two conventions: (i) If the surface is CLOSED, we take the outward normal as positive (ii) If the surface is OPEN surface, the positive normal depends on the direction in which the perimeter of the open surface is traversed (which is dependent on the handedness of the coordinate system).

X

Y

Z

Sd


If a vector A is defined on the surface, the surface integral SdA ⋅∫ can be interpreted as a flow or flux through the given surface. How to solve a surface integral: Suppose, we have an arbitrary surface S . If we can project the surface S over any plane, and the area of the projected surface is R , then the integration of a vector A over S is equal to the integration of A over R .

SdA ⋅∫ is the flow of “something” through the surface S . So, if we consider ABCD be an imaginary pipe, then the flow through S must be equal to the flow through R or any other cross-section in that. Surface area elements in the XY -plane is dxdy , YZ -plane is dydz , ZX -plane is dzdx .

X

Y

Z A

B C

D

R

Sd

X

Z

Y


In this example, to calculate the flow of A through R , we need to estimate the projected area of dS on R , which is dxdy . We can write,

Therefore,

dSnAS

ˆ⋅∫∫ = )ˆˆ(

ˆkn

dxdynAR ⋅

⋅∫∫

[ Note: Double integrals have been used as we have to integrate over two variables ] Lecture-7 Example: Surface Integral

Evaluate dSnAS

ˆ⋅∫∫ , where A = ( )jyix ˆˆ41

+ and S is the surface of the cylinder

1622 =+ yx included in the first octant bounded between 0=z and 5=z .

n

k

dS

dxdy

)ˆˆ( kndSdxdy ⋅=

n

j

X

Y

Z

0=x

R xz-plane

4=x

5=z

0=z


Area element on the xz -plane is dxdz . The entire area of the 1st quadrant of the cylinder can be projected onto the xz -plane, where x ranges from 0=x to 4=x and z ranges from 0=z to 5=z . So, the area element dS of the given surface is related to corresponding area element dxdz on xz -plane by

dS = ( )jndxdz

ˆˆ ⋅.

Therefore,

( )jndxdznAdSnA

RSˆˆ

ˆˆ⋅

⋅=⋅ ∫∫∫∫

The normal vector to the surface 1622 =+ yx at any point ),( yx is )( 22 yx +∇ = jyix ˆ2ˆ2 + . [Consider this is a equipotential surface.] The unit normal is

( )jyixyxjyixn ˆˆ

41

)2()2(

ˆ2ˆ2ˆ22

+=+

+= .

∴ nA ˆ⋅ = 1)(161 22 =+ yx

jn ˆˆ ⋅ = jjyix ˆ4

ˆˆ⋅

+ = 4y

Hence the surface integral is

dxdzyR

∫∫4 = ∫∫

== −

5

0

4

0216

14zx

dzdxx

= dxx∫ −

4

0216

120

= 20π = 10π ∴ The Integral of the vector A over the entire surface of the cylinder is

dSnAS

ˆ⋅∫∫ = π104× = π40 .

However, the area of the surface of the cylinder is = rlπ2 = 542 ××π = π40 . Therefore, we can say that the flow or flux of a unit normal vector (note here, nA ˆ≡ ) through the entire surface is just the actual area of the surface.

There are other different kinds of surface integrals which we encounter very commonly: ∫ Sdφ , SdA×∫ .

Note: Substitute θcos4=x , the integral reduces to

2

0

2

πθπ

=− ∫ d


Homework Problems: 1. Evaluate dSnA

S

ˆ⋅∫∫ over the entire surface S of the region bounded by the cylinder

922 =+ zx , 0=x , 0=y , 0=z and 8=z , if A = kxjyxiz ˆˆ)2(ˆ6 −++ . 2. Evaluate dxdyyx

R∫∫ + 22 over the region R in the xy -plane bounded by

3622 =+ yx . Volume Integrals: We encounter the following Volume integrals: τdA⋅∇∫ , τφd∫∇ , τdA×∇∫ , where τd is the differential volume. Let us consider an Integral of first kind.

A⋅∇ is the divergence of vector A (or flow out of some volume). ∴ τdA⋅∇∫ is the divergence or flow out of some volume ∫ τd .

If Sd is the vector area element of this volume then the flow through the entire area of that volume is also SdA ⋅∫ , If we think it physically, the flow of something out of some volume has to come out through the entire surface enclosing that volume. Therefore, τdA⋅∇∫ = SdA ⋅∫ . Thus we correspond between volume and surface integrals. This is Gauss’ Divergence formula. This is not proved here (see any text book).

Calculation of ordinary Surface and Volume Integrals

Area of a Circle: 222 ayx =+

A = ∫∫dxdy = dxdya

ax

xa

xay∫ ∫−=

−

−−=

22

22

= 2 dxxaa

a∫−

− 22 = 2.2

.2 πa = 2aπ .

Volume of a Sphere: 2222 azyx =++

V = dxdydza

ax

xa

xay

yxa

yxaz∫ ∫ ∫−=

−

−−=

−−

−−−=

22

22

222

222

= dxdyyxaa

a

xa

xa∫ ∫−

−

−−

−−

22

22

2222

[ Substitute θsin22 xay −= , the integral in the third bracket becomes 2/ ππ


Home Work Problem:

# Let us take a vector )ˆˆˆ(1 kzjyixa

A ++= to evaluate dSnAS

ˆ⋅∫∫ over a sphere

2222 azyx =++ above the xy-plane. Hints: • Solve the problem either directly or by applying Gauss’ Divergence theorem. • To solve the integral directly one has to calculate the unit normal vector to the surface

S (the hemisphere above the XY-plane).

Unit normal, n = || φ

φ∇

∇ .

Here, φ∇ = kzjyixzyx ˆ2ˆ2ˆ2)( 222 ++=++∇ .

∴ n = 222 )2()2()2(

ˆ2ˆ2ˆ2zyx

kzjyix++

++ = ( )kzjyixa

ˆˆˆ1++

∴ 1ˆ =⋅ nA , azkn =⋅ ˆˆ

The projection of the infinitesimal area dS onto the xy-plane is dxdy . k is the unit normal vector to the xy-plane.


Therefore, dSnAS

ˆ⋅∫∫ = ( )( )kndxdynA

Rˆˆ

ˆ⋅

⋅∫∫ [ as shown earlier in Lecture-6]

= dxdyz

a∫∫1

= dydxyxa

aa

ax

xa

xay∫ ∫−=

−

−−= −−

22

22222

1 .

Lecture-8 GREEN’S FORMULA: Let C be a simple closed curve in the XY-plane. R is the region bounded by the closed curve. Suppose, ),( yxM and ),( yxN are two differentiable functions in the region R. We can then evaluate the following integral,

1I = dxdyy

yxM

R∫∫ ∂∂ ),( = dxdy

yMb

ax

xy

xyy∫ ∫= =

∂∂)(

)(

2

1

Here )(1 xyy = represents the lower portion APB of the closed curve C and )(2 xyy = represents the upper portion AQB of the curve.

∴ 1I = [ ] dxyxMxy

xyy

b

ax

)(

)(

2

1

),(==

∫ = [ ]dxyxMyxMb

a∫ − ),(),( 12

Y

X

P

Q

AB

q

p

b a

C

)(1 xy

)(2 xy

R


= ∫ ∫−−b

a

a

b

dxyxMdxyxM ),(),( 21 = ∫−C

dxyxM ),(

∴ ∫C

dxyxM ),( = dxdyy

yxM

R∫∫ ∂∂

−),( …………………………….(1)

Similarly,

2I = dxdyx

yxN

R∫∫ ∂∂ ),( = dydx

xNq

py

yx

yxx∫ ∫= =

∂∂)(

)(

2

1

Here )(1 yxx = represents the left portion PAQ of the closed curve C and )(2 yxx = represents the right portion PBQ of the curve.

∴ 2I = [ ] dyyxNyx

yxx

q

py

)(

)(

2

1

),(==

∫ = [ ]dyyxNyxNq

p∫ − ),(),( 12

= ∫ ∫+q

p

p

q

dyyxNdyyxN ),(),( 12 = ∫C

dyyxN ),(

∴ ∫C

dxyxN ),( = dxdyx

yxN

R∫∫ ∂∂ ),( …………………………….(2)

Adding (1) and (2),

∫ +C

NdyMdx = dxdyy

MxN

R∫∫

∂∂

−∂∂ …………………………..(3)

This is Green’s Formula in the xy-plane. Green’s Formula (3) can be written in the vector notation: We can write

( ) rdAjdyidxjyxNiyxMdyyxNdxyxM ⋅=+⋅+=+ )ˆˆ(ˆ),(ˆ),(),(),( Here jyxNiyxMA ˆ),(ˆ),( +=

dyjdxird ˆˆ +=

Then A×∇ =

0

ˆˆˆ

NMzyx

kji

∂∂

∂∂

∂∂ on a plane


=

∂∂

−∂∂

+∂∂

+∂∂

−y

MxNk

zMj

zNi ˆˆˆ

∴ ( )

∂∂

−∂∂

=⋅×∇y

MxNkA ˆ .

Therefore, formula (3) can be written as

( ) dRkArdARC

ˆ⋅×∇=⋅ ∫∫∫ ,……………………………………(4)

where dxdydR = . The above formula (4) when generalized for an arbitrary surface S bounded by a simple curve C (both sides of the surface are open) can be written as

( ) dSnArdASC

ˆ⋅×∇=⋅ ∫∫∫ ……….This is Stoke’s Formula.

-------------------------------------------------------------------------------------------------------- Problems: (To solve by using Green’s Formula and Stokes Formula) #1. Use Green’s Theorem to evaluate dyyxdxxyx

C

)()( 222∫ +++ where C is the Square

formed by the lines 1±=y , 1±=x . Solution

∫ +C

NdyMdx )( = dxdyy

MxN

∫∫

∂∂

−∂∂

= ( ) ( ) dxdyxyxy

yxx∫ ∫

− −

+

∂∂

−+∂∂1

1

1

1

222 = ∫ ∫− −

1

1

1

1

xdxdy = ∫ ∫− −

1

1

1

1

dyxdx = 0.

#2. Using Stoke’s Theorem to evaluate [ ]∫ −−−

C

zdzydyyzdxyx 22)2( , where C is a

circle 122 =+ yx corresponding to a sphere of unit radius. Solution Here we identify the vector, A = kzyjyziyx ˆˆˆ)2( 22 −−− and kA ˆ=×∇ .

∴ rdAC

⋅∫ = ∫∫ ⋅ dSnk ˆˆ = ∫∫dS = ∫∫ ⋅ )ˆˆ( kndxdy = ∫∫dxdy = π .


#3. Use Stoke’s Theorem to evaluate [ ]∫ −+−++C

dzzydyzxdxyx )()()2( where C is the

boundary of the triangle with vertices (2,0,0), (0,3,0) and (0,0,6) oriented in the anti-clockwise direction. Solution

rdAC

⋅∫ = [ ]∫ −+−++C

dzzydyzxdxyx )()()2(

= [ ] [ ]dzkdyjdxikzyjzxiyxC

ˆˆˆˆ)(ˆ)(ˆ)2( ++⋅−+−++∫

Here, A = kzyjzxiyx ˆ)(ˆ)9ˆ)2( −+−++

∴ A×∇ =

zyzxyxzyx

kji

−−+∂∂

∂∂

∂∂

2

ˆˆˆ

= ki ˆˆ2 − .

The equation of the plane ABC (a triangle) is 1632=++

zyx .

The normal vector φ

φ

∇

∇=n , where =φ

632zyx

++ .

φ∇ = )ˆˆ2ˆ3(61 kji ++ , ∴ n = )ˆˆ2ˆ3(

141 kji ++

∴ nA ˆ)( ⋅×∇ = 145

O

(2,0,0)

(0,3,0)

(0,0,6)

X

Y

Z

A

B

C


Now applying Stoke’s theorem:

rdAC

⋅∫ = ( ) dSnAS

ˆ⋅×∇∫∫ = ∫∫S

dS145 = ∫∫ ⋅R kn

dxdy)ˆˆ(14

5

= ∫∫++⋅R kjik

dxdy

)ˆˆ2ˆ3(141ˆ14

5

= ∫∫R

dxdy5 = 5 × projection of the area ABC on the XY-plane

= 5 × Area of triangle OAB = )32(215 ××× =15.

Lectures on Vector Algebra

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