2 Scalar and Vector Field

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brief description of scalar and vector fields required for mathematical analysis

Transcript of 2 Scalar and Vector Field

Scalar and Vector FieldsScalar Field :

A scalar quantity, smoothly assigned to each point of a certain region of space is called a scalar fieldExamples :i) Temperature and pressure

distribution in the atmosphereii) Gravitational potential around the

earth

iii) Assignment to each point, its distance from a fixed point

222 zyxr

O

O

),,( zyx

),,( zyxf

x

z

y

Once a coordinate system is set up, a scalar field is mathematically represented by a function : )(),,( rfzyxf

is the value of the scalar assigned to the point (x,y,z)

),,( zyxf

A smooth scalar field implies that the function , is a smooth or differentiable function of its arguments, x,y,z.

),,( zyxf

Since the scalar field has a definite value at each point, we must have

),,(),,( zyxfzyxf

O),,( zyxf

Consider two coordinate systems.

x

y

z

O’

z

y

x

),,( zyxf

Vector Fields :

A vector quantity, smoothly assigned to each point in a certain region of space is called a vector field

Examples :

i) Electric field around a charged bodyii) Velocity variation within a steady flow of fluid

iii) Position vector assigned to each point

Or

Once a coordinate system is fixed, a vector field is mathematically represented by a vector function of position coordinates : )(),,( rForzyxF

O

)(rF

r

Resolving the vector at each point into its three components, the vector field can be written as :

kzyxFjzyxFizyxFzyxF zyxˆ),,(ˆ),,(ˆ),,(),,(

A smooth vector field implies that the three functions, , are smooth or differentiable functions of the three coordinates x,y,z.

zyx FFF &,

),(sin),(cos),( yxFyxFyxF yxx

),(cos),(sin),( yxFyxFyxF yxy

x

yF

x

y

How Are the Component Functions in Two Frames Related to One another?

In Three Dimensions :

zyxx FRFRFRF 131211

zyxy FRFRFRF 232221

zyxz FRFRFRF 333231

MatrixRotationRRRRRRRRR

R

333231

232221

131211

Gradient of a Scalar Field

dzzfdy

yfdx

xfrfrdrfdf

)()(

O

r rdr

rd

kdzjdyidxkzfj

yfi

xf ˆˆˆˆˆˆ

rdf

Where, we have the shorthand notation :

fkzfj

yfi

xf

ˆˆˆ

Since is a vector assigned to each point

f

it defines a vector field. This vector field is called the gradient of the scalar field

f

We have

df )ˆ(ˆ ndsrddsnfrdf

nfdsdf

n

ˆˆ

That is, the rate of change of a scalar field in any direction at a point, is the component of the gradient of the field in the given direction, at that point.

Thus, gradient of a scalar field at any point may be defined as a vector, whose direction is the direction in which the scalar increases most rapidly, and whose magnitude is the maximum rate of change

fdsdfnf

dsdf

n

maxˆ

ˆ

And the maximum value occurs when is in the same direction as

n̂f

Ex. 1.3Find the gradient of the scalar field :

222)( zyxrrf

and show that it has the properties as stated.

Prob. 1.12The height of a certain hill (in feet) is given by :

)122818432(10),( 22 yxyxxyyxh

where x & y are distances (in miles) measured along two mutually perpendicular directions from a certain point. (c) How steep is the hill at the place (1 mi,

1 mi)(d) In which direction, must one move at this point so that the slope is 220 ft/mi

x

y

),( yxh

x

y

),( yxh

Further Examples :

)()( rFrf b)

Let Then )()() rrFrrfi

)()() rrFrrfii

gdgdfrgfa

))(()

Prob. 1.13

Let r be the separation vector from a fixed point to the point and let r be its length. Find :

),,( zyx ),,( zyx

a) (r2)

b) (1/r)

c) (rn)

The ‘Del’ Operator

The gradient of a scalar field can be thought of as the result of the vector differential operator :

zk

yj

xi

ˆˆˆ

acting on the scalar field :f

zk

yj

xif

ˆˆˆ

The gradient (2-dim) will be a vector field, if, at every point in space, we have :

yf

xf

xf

sincos

yf

xf

yf

cossin

Is the Gradient of a Scalar Field a Vector Field?

Prob. 1.14

Important Properties of the Gradient1. The line integral of the gradient of a scalar field from one point to another, is independent of the path of integration (it depends only on the two points)

)()( 12

21

PfPfldfldfCC

1C2C

1P

2P

2. Line integral of the gradient around a closed path is zero

0ldf

3. If the line integral of a vector field from one point to another is independent of the path joining them, then that vector field must be the gradient of a scalar field.

),,( zyx

:),,( zyxffieldscalartheConstruct

),,(

)0,0,0(

),,(zyx

ldFzyxf

),,(int zyxpothetoorigintheconnectingpatharbitraryanyispaththewhere

Proof :Let be a vector field whose line integral is path independent.

F

x y z

zyx zdzyxFydyxFxdxFzyxf0 0 0

),,()0,,()0,0,(),,(

),,( zyx

)0,0,(x)0,,( yx

Choosing the path as above

To show : Ff

),,( zyxFzf

z

Similarly, choosing the paths as below,

),,( zyx

)0,0,(x

),0,( zx ),,( zyx

),,0( zy),0,0( z

xy FxfF

yf

&

fF

),,( zyx

),,( 000 zyx

),,(

),,( 000

),,(zyx

zyx

ldFzyxg

However, the scalar field, whose gradient is the vector field , is not unique.

F

We still have : gF

),,(

)0,0,0(

000

),,(),,(zyx

ldFzyxgzyxf

Czyxg ),,(

However,

That is, the two scalar fields, which give us the same vector field as their gradient, differ from each other by a constant.

If the scalar field is constructed as :

),,(

),,( 000

),,(zyx

zyx

ldFzyxf

then :0),,( 000 zyxf

Since are arbitrary, the zero of the scalar field is arbitrary.

000 &, zyx

Level or constant surfaces of a scalar field

Given a scalar field , the locus of all points which

satisfy the condition :

),,( zyxf

Czyxf ),,(

defines a surface, known as the level or constant surface

of the field

Czyxf ),,(

1C2C

3C

),,(),,( 000 zyxfzyxf

Through every point in space, one can draw a level surface of :

),,( 000 zyxf

Level surfaces of the scalar field :222),,( zyxzyxf

4. The gradient of a scalar field, at each point in space, is perpendicular to the level surface (constant surface) of the scalar field, passing through that point

.),,( Constzyxf

Pf

P

Ex.: Find the unit normal to the surface :

22 yxz

Given the three components of a vector field, , one can construct nine first derivatives :

zyx FFF &,

zF

yF

xF zxx

.,..........,,

What scalar and vector fields can be constructed out of these nine first derivatives?

Divergence and Curl of a Vector Field

Correction!

x

y

'x

'y

),( 00 yx

Coordinate Transformation Equations

;)(sin)(cos 00 yyxxx

)(cos)(sin 00 yyxxy

;sincos 0xyxx 0cossin yyxy

It turns out that only one scalar field and one vector field can be constructed out of these nine first derivatives :Divergence :

zF

yF

xFF zyx

Curl :

kyF

xF

jxF

zFi

zF

yFF xyzxyz ˆˆˆ

kFjFiFz

ky

jx

iF zyxˆˆˆˆˆˆ

kFjFiFz

ky

jx

iF zyxˆˆˆˆˆˆ

zyx FFFzyx

kji

ˆˆˆ

Example :

Let the vector field be : rrF

)(

3 F

0

F

Surface Integral of a Vector Field

F

ad

S

S

adF

i) Gauss’ Divergence Theorem :

S V

dFadF

VS

Here, is any vector field and , any closed surface

F

S

Two Important Theorems

ad

A Simple Application of Divergence Theorem

General Proof of Archimedes’ Principle

S S

b adPFdF

xy

zFd dah )()( zhgzP

ii) Stokes’ Curl Theorem

ld

ad

C S

adFldF )(

Here, C is any closed loop (planar or otherwise), and S is any surface bounded by the loop.

Second DerivativesfieldVectorf

2

2

2

2

2

2

)(zf

yf

xff

i) f2

Where is a second derivative operator, known as the Laplacian.

2

2

2

2

2

22

zyx

ii) 0

f

fieldscalarAF

F

a legitimate vector field, which is not much used.

fieldvectorAF

0) Fi

FFFii

2)

Certain Product Rules

)()()(.1 fggffg

)()()(.2 ABBABA

ABBA)()(

)()()(.3 fFFfFf

)()()(.4 BAABBA

Prob. 1.60Show that

V S

adTdTa

)()(

Hence show that :0

S

adaProb. 1.61b :

Prob. 1.61c : a is the same for all surfaces sharing the same boundary.

Prob. 1.61a : Find the vector area of a hemispherical bowlBack to Prob. 1.60

V S

adFdFb )()(

V S

adTUUTdTUUTd

)()()( 22

Prob. 1.61d

Show that : ldra

21

Proof :

C S

adrAldrA )]([)(

Use the product rule :BAABBA)()()(

)()( ABBA

Two Important Results

i) If the Curl of a vector field vanishes, then that vector field must be the gradient of a scalar field :

fFtsfF

..0

We have seen : 0)0) Fiifi

The converse results are also true.

ii) If the Divergence of a vector field vanishes, then the vector field must be the Curl of a scalar field :

GFtsGF

..,0

Theorem I : The following statements are equivalent.(A) Vector field is such that its line integral around any closed path is zero

F

(C) Vector field is the gradient of a scalar field

F

(B) Vector field is such that its line integral from one point to another is independent of the path joining the two.

F

)()()()( DCBA Or

)()()()()( ADCBA

(D) Vector field is such that it has a vanishing curl

F