1Unit 1-Mathematical methods in physics

2Circulation of vector field around a loop

Consider a vector field V=V x exV y e y

x

Y

1

2

3

4

c V.dr circulation

x0 x2

, y0 y2 x0

x2

, y0 y2

x0 x2

, y0 y2 x0

x2

, y0 y2

x

y

3Circulation of vector field around a loop

x0

x2

x0 x2

V x x , y0 y02

dx

x

Y

1

2

3

4

x0 x2

, y0 y2 x0

x2

, y0 y2

x0 x2

, y0 y2 x0

x2

, y0 y2

Segment 1 of the path contributes to the integral

Segment 1

4x0

x2

x0 x2

V x x , y0 y02 dxV x x0, y0 y02

x

y0

y2

y0 y2

V y x0 x2 , y dyV y x0 x2 , y0 y

Segment 2

Segment 1

Here we use the approximation. Replace the integrand with thevalue of the vector function at the mid-point and, then, integrate.

5Segment 3

x0

x2

x0 x2

V x x , y0 y2 dxV xx0 , y0 y2 x

Segment 4

y0

y2

y0 y2

V y x0 x2 , y dyV y x0 x2 , y0 y

6c V.dr = Line integral of [segment 1+segment 2+segment 3 +segment 4]

V x x0, y0 y02 x V y x0 x2

, y0 y

V xx0 , y0 y2 x V y x0 x2

, y0 y

c V.dr= V x x0, y0 y02 x V y x0 x2 , y0 y

V xx0 , y0 y2 x V y x0 x2

, y0 y

7c V.dr = Line integral of [segment 1+segment 2+segment 3 +segment 4]

V x x0, y0 y02 x V y x0 x2

, y0 y

V xx0 , y0 y2 x V y x0 x2

, y0 y

c V.dr= V x x0, y0 y02 x V y x0 x2 , y0 y

V xx0 , y0 y2 x V y x0 x2

, y0 y

V x x0, y0 y02 ] x[V y x0x2 , y0

[V x x0 , y0 y2

V y x0 x2 , y0 ] y

8V x x0, y0 y02 ] x y [V x x0 , y0 y2

[V y x0x2 , y0 V y x0 x2 , y0 ] y x

x

y

c V.dr=

c V.d r s

=V x x0, y0 y02 ] [V x x0 , y0 y2

[V y x0x2 , y0 V y x0 x2 , y0 ] x

y x 0 y 0

Now take the limits:

9c V.d r s

=

In the limit x0 y0 s0

Limit s0 [ V y x V x y ]c V.d r s =

c V.d r s

=Limit s0 V . ezc V.d r

s=

For the X-Y plane circulation per area = Z-component Of the CURL of vector V

10

c V.d r s

= V . nLimit s0

Generalizing this idea to an arbitrary rectangle on an arbitrary surface

Stokes theorem

c V.d r=s V . nds

The circulation of a vector field over a closed path C is equal to the Integral of the normal component of the curl of that field over a surface S for which C is a boundary

11

Image from A students guide to Maxwells Equation-Daniel Fleisch

For the adjacent squares in the interior regions, circulations are equal in magnitude and opposite in direction and cancel each other.

12

=

Each arrow represents the integration along that segment. In the interior of rectangle where they share the sides the arrows are inopposite direction. So they cancel each other.

As a result the line integral along the boundary of the surface only survive.

13

Verify Stokes theorem for the square surface shown below for the field

v=2xz3y2 e y4yz2 ez

14

Verify Stokes theorem for the field

where S is the upper half surface of the sphere

and C is its boundary. Let R be the projection of S on the xy-plane

v=2x y ex yz2 e y y

2 z ez

x2 y2z2=1

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