1Unit 1-Mathematical methods in physics

2Reference for module vectors and coordinate systems

1) Introduction to electrodynamics: Third edition by D J Griffiths

Chapter 1 : Good source for problems and basic introduction

2) Mathematical methods for Physicists by Arfken and Weber 6th edition

Chapter 1 and 2 Good reference for theory

3) Mathematical Methods in the physical sciences, Mary L Boas

4) Advanced engineering mathematics by Erwin Kreyszig 8th edition

Chapter 8 and 9

3T x , y , zTemperature in the room Height of mountain Z(x,y)

Real physical quantities: depends on (x,y,z) their variation is different in different directions (other examples: Electric potential,

gravitational potential etc.)

How to calculate this rate of change in different directions?

Directional derivatives

4The laws of Nature are but the mathematical thoughts of God. Euclid

The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve. E.P. Wigner (Nobel prize in 1963)

One factor that has remained constant through all the twists and turns of the history of physics is the decisive importance of the mathematical imagination.Freeman Dyson

5Review of derivatives with single variable

f ' x = d fd x

= lim x0

f x x f x x

Geometrical interpretation: Slope of the curve at a given point

d f x =d f x d x

d x

Differential:

x0

Approximate change in when changes from xf x x x

6 f xx0, y0= lim x0

f x0 x , y0f x0 , y0 x

Partial

vary only xx0 x0 x

value of remains constant y= y0

Partial derivative of w.r.t. x

Partial derivatives:

f x , y

7 f yx0, y 0= lim y0

f x0 , y0 y f x0 , y0 y

y0 y0 y

x=x0

Partial derivative of w.r.t. xf x , y

vary only y

value of remains constant Partial

Partial derivative of w.r.t. yf x , y

8How to compute the partial derivatives?

f x

=f x=?

f y

=f y=?

1) Keep x as a variable and y as a constant

2) Treat the function as a single variable of x

3) Do the differentiation w.r.t. x

1) Keep y as a variable and x as a constant

2) Treat the function as a single variable of y

3) Do the differentiation w.r.t. x

9f x , y =Z=9x2 y2

Find partial derivatives at (1,2)

10

f x , y =Z=9x2 y2

Geometrical meaning of derivative of function at (a, b) ?

Let us keep y=2 (x-z plane)

http://moodle.capilanou.ca/mod/book/view.php?id=328667&chapterid=1398

f x ,2=Z=9x222

Geometrical interpretation:

11

f x =Z=5x2

inverted parabola in the X-Z plane

Now find the derivative (partial) of this function at a given x

Let x=1

f x1,2=2xx=1=2

12

Now keep x constant

Let x=1

f 1, y=Z=8 y2

Now find the derivative (partial) of this function at a given y

Let that y=2

g y1,2=2yy=2=4

inverted parabola in the Y-Z plane

13

d f x =d f x d x

d x

Differential for a single variable:

Total differential for two variables:

d f= f x

dx f y

dy

This way you can extend to any dimensions

14Rate of change depends upon the DIRECTION

15

How to calculate the directional derivatives?

d fd s

= lim s0

f rr f r s

= lim s0

f s

We need to get an expression for d fd s

s= x2 y2 z2 s=r

16

d f= f x

dx f y

dy f z

dz

d r=[ exdx e y dy ezdz ]

df= fdr

= [ ex f x e y f y e z f z ][ ex dx e y dy e z dz ]

ex , e y , ez : Unit vectors

17

d r=[ exdx e y dy ezdz ]

f

f= [ e x f x e y f y ez f z ]

df= fdr

: Gradient of a scalar function VECTOR quantity

18

f= [ i f xj f yk f z ]

f= [ i xj yk z ]

= [ i xj yk z ]

19

= [ i xj yk z ]Del operator or nabla operator

Operator: carries an instruction

acts on a function

acts on a scalar function vector function

Example: E= V

20

Find the gradient of f x , y , z=r=x2 y2z2

f= [ e x f x e y f y ez f z ] f x

= xx2 y2z2

Similarly calculate other derivatives and put that in the expression for gradient

f=x e x y e yz ezx2 y2z2

=rr=r

21

Gradient of a function

x

y

f x , y , z=r=x2 y2z2

Circles correspond to the contour. Notice that gradient direction is normal to the contour line (circle)

22

d r=[ exdx e y dy ezdz ]df= fdr

df= fd rcos

df= fds cos

ds=dr

d fd s

= fcos

f= [ e x f x e y f y ez f z ]

23

df= fd rcos

d fd s

= fcos

d fd s

= fu u : Unit vector along the direction of displacement

Component of in the direction

give the directional derivative of f in that direction

f u

24

d fd s

= fcos

d fd s

maximum when =0

d fd s

= f

Rate of change is maximum

if the points along the gradient direction

Direction of GRADIENT points along the maximum rate of change

d r

=0

Magnitude of gradient: rate of change of the function along the maximaldirection

25

If you fix and look around all angles,

i) Find the direction of steepest descent

ii) That will be the direction of the gradient

Imagine water-drops at this point, drops will flow along the direction of gradient

d rGeometrical interpretation of GRADIENT:

direction you will fall if you lose your feet !!

26

With ordinary vectors we define three operations:

1) Multiplication with a scalar:

2) Multiplication of two vectors

A a

AB

AB

Dot product giving a scalar

cross product giving a vector

Correspondingly we have three ways the operator

1) On a scalar function :

2) on a vector function via the dot product DIVERGENCE

via the cross product CURL

gradient

A :A

A :A

27

How to calculate the directional derivatives?

d fd s

= lim s0

f rr f r s

d fd s

= fu

Gradient of a scalar function: A vector whose component along will give rate of change along that particular direction

u

28

f= [ e x f x e y f y ez f z ]

Gradient in Cartesian form:

= [ i xj yk z ]Del operator or nabla operator

29

If you fix and look around all angles,

i) Find the direction of steepest descent

ii) That will be the direction of the gradient

Imagine water-drops at this point, drops will flow along the direction of gradient

d rGeometrical interpretation of GRADIENT:

direction you will fall if you lose your feet !!

30

Exercise 1:Given function

Find the rate of change of this function at (1,3,1) in

the direction

Exercise 1:

Temperature distribution is given by

I) In which direction the temperature increasing most rapidly At (-1, 2, 3)?

ii) What is the max. rate of increase at this point?

T=x2 y2xyz273

f=x2 y z4xyz2

2 ex e y2 e z .

31

f x , y =z=x2 y2 Paraboloid

How does the gradient this of this function look like?

32

The minimum of the function is locatedat the origin

f x , y =z= x2 y2

2 f=r

Gradient graph: Minimum of f(x,y) corresponds to point where arrows are pointing away

If we move along radial direction of f(x,y), you will see the maximum rate of change (i.e. along the direction of gradient)

33

f

f x , y , z=const.

is perpendicular to the level surface

Heat flow: perpendicular to surfaces with constant temperature (ISOTHERMAL)

Electric field of a point charge: perpendicular to the equipotential surface (radial) (spherical)

df=0

f x

dx f y

dy f z

dz=0

fdr=0

Level surface:

34

V=V x x

V y y

V z y

r=3

Divergence of radial vector

V x=V x x , y , zV y=V y x , y , z

V z=V z x , y , z

F=x x y y

F=2

Divergence

35

Positive divergence Zero divergence

Positive divergence Negative divergence

36

Positive divergence (diverging)

Negative divergence (converging)

Presence of Source Presence of Sink

37

Physical meaning of divergence : Velocity of a compressible fluid : density of the fluid

x

y

z

x

y

zA

C

G

D

E

F

B

H

x , y , z , t

dV=dx dy dz

v x , y , z

J=v [J ]=ML2T1

38

Rate of inward flow of the fluid through ABCD (along Y-axis)

= [ v y ] y0 x z

(other components have no contribution in this direction) v x ,v z

Rate of outward flow of the fluid through EFGH (along Y-axis)

= [ v y ] y0 y x z

Net flow along Y-direction = [v y ] y0 y [ v y ] y0 x z

= [v y ] y0 y [ v y ] y0

y x z y

39

Net flow along Y-direction

= [v y ] y0 y [ v y ] y0

yV

Net flow along x-direction

Net flow along Z-direction

=[ J y ] y0 y [J y ] y 0

yV

=[J x ]x0x [J x ]x0

xV

=[J z ] z0 x[J z ]z0

zV

= [v y ] y0 y [ v y ] y0

y x z y

40

Total outward flow

Total outward flow / V

= [ [ J x ]x0 x [J x ]x0 x [J y ] y0 y[ J y ] y0 y

[ J z ]z0 y [J z ] z0 z ]V

= [ [ J x ]x0 x [J x ]x0 x [J y ] y0 y[ J y ] y0 y

[J z ]z0 y[ J z ]z0 z ]

41

x0 y0 z0

Total outward flow / V

= [ [ J x ]x0 x [J x ]x0 x [J y ] y0 y[ J y ] y0 y

[ J z ]z0 y [J z ] z0 z ]V

x

y

z

x

y

zA

C

G

D

E

F

B

H

reduces to a POINT

42

lim x 0[J x ]x0 x[ J x ]x0

x= J x x

= J x x

J y y

J z z

= J

DIVERGENCE: Net outward flow per unit volume evaluated at a point

Net outward flow per unit volume evaluated at a point

43

THE CURL

V= x y z x y zV x V y V zV x=V x x , y , z

V y=V y x , y , zV z=V z x , y , z

V x , y , z =V x exV y e yV z e z

VV

f V =f V f V

44

v=x ex y e y v= y exx e y

v=0 v=2 e z

45

46

Curl of central force field

r f r =f r r[ f r ]r

r= x y z x y zx y z

=0 f rr= f

rrr=0

Using the relation

We get

r f r =0

f r =r f r

f V =f V f VShow that

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