Understanding Conservative Vector Fields and Green's Theorem, Study notes of Physics

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Conservative Vector

Fields

Review:

Work C

 F^ ^ d r =^ '( ) C

 F r ^ t dt C

 (^)  Mdx  Ndy  Pdz if F  M i  N j  P k

Outward flux across a simple closed

curve C in the plane is C

 F n ^ ds if C

 (^)  Mdy  Ndx F  M i  N j

is called conservative (or a gradient vector field) if

The function is called the of.

f

f potential

F F  

F

a) if and only if is path independent: C

  f (^)   dr

Fundamental theorem for line integrals :

F F 1 2

= C C

 F^ ^^ dr^  F  dr

= if C is a path from to.

B

C A

 F^ ^^ dr^  F  dr^ A^ B

b) If , then ( ) ( ) C

F   f (^)  F  dr  f B  f A

is also called if represents a velocity vector field. C

 F^ ^ d^ r^ circulation F

= C

 F T ^ ds

Recall:

curl (^) x y z

P N M P N M

y z z x x y M N P

     

 ^ ^  ^ ^ ^   ^  

i j k

F i + j k

Theorem : If F = M i  N j  P k and F   f then curl F  0.

necessary conditions: , ,

M N M P N P

y x z x z y

This condition is "almost" sufficient as well.

But it is always a good test.

In 3 dimensions:

f f f M N P f x y z

F = i j k i j k

Another way to express this, which is easier to remember:

Some terminology for curves C :

A region R is called if every closed curve in R can be shrunk to a point by curves staying in R.

Definition : simply connected

The plane minus the origin is not simply connected.

3-space minus the origin is simply connected.

3-space minus the z-axis is not simply connected.

Important examples :

2 4

so is a gradient field.

y y

y y y x

M x e N y xe

M e N e

 

 

   

     F

  (^)     f x y ,  Mdx  2 x  e ^ y^ dx  x^2  xe  y  G y  

   

 

   

 

   

2, 2,

2 2 2, 2,

,

2

4 2 4 2 4

C

y

Work dr f x y

x xe y



 

  

  

     

 F

  (^)    

Find the work done by the force

, 2 4 along the indicated curve.

y y x y x e y xe

     

Example :

F i j

 ,^     , 

y f (^) y x y xe G y N x y

 (^)     

  4

y y xe G y y xe

 (^)      

G ^  y (^)   4 y

 

2 G y  2 y  C

 

2 2 , 2

y f x y x xe y C

    

need: and

f f M N x y

     

     

2 2 2 2

F  2 xy  3 xz i  2 x y  2 y j  3 x z  2 z j

2 2

2

2

M xy xz

N x y y

P x z z

 

2 2 f  (^)  Mdx  (^)  2 xy  3 xz dx

2 2 3 2 2  x y  2 x z   

G y z ,

2 2

2 x y  2 y  N  f y  2 x y  G y ( , y z ) or^ G^ y ( , y z^^ )^ ^2 y

 

2 2 3 2 2 2 2

the potential is f x y z , ,  x y  2 x z  y  z  C

So F is a gradient field, F   f. Now let's find f.

Determine wether the given vector field is a gradient field.

If so, find a potential function.

Example :

      2 2 2 2

curl 0 6 6 4 4

2 3 2 2 3 2

x y z xz^ xz^ xy^ xy

xy xz x y y x z z

    (^)       

  

i j k

F i + j k = 0

2 2

3 x z  2 z  P  f z  3 x z  H '( ) z

2 2 2 3 2 2 2

or G y z ( , )  y  H z ( ) hence f  x y  2 x z  y  H z ( )

2

or H '( ) z   2 z hence H z ( )   z

need: , and

f f f M N P x y z

        

Compute the work performed by the force 2 2 2 2

as a particle travels around a circle of radius counter clockwise.

y x

x y x y

r

    

Example : F i j

parametrize the circle: x  r cos( ), t y  r sin( ) t dx   r sin( ), t dy  r cos( ) t

2 2 2 2 ^  C C

y x dr dx dy x y x y

 ^  

 F^  i^ j^ i^ j

2 2 ^ 

sin( ) cos( ) = ( sin( )) ( cos( ) C

r t r t r t r t dt r r

 ^ ^ ^ 

 i^ j^ i^ j

 

2 2 2 2 2 2 0

= r sin ( ) t r cos ( ) t dt r



^ 

2

0

= 1 dt 2



 ^  

is conservative since with arctan (check it!)

y f f x

     (^)    

Puzzle 2 : F F

Solution : F is not defined at (0, 0), and the plane minus the origin is not simply connected!

(for any radius r !)

 

   

(^2 2 2 )

2 2 2 2 2 2 2 2

is conservative since

x^1 x^ y^ x^^2 x x y x x y (^) x y x y

 ^ ^ ^ ^ ^     ^   (^)    (^)  

Puzzle 1 : F

 

   

(^2 2 2 )

2 2 2 2 2 2 2 2

1 2 and

y x^ y^ y^ y x y y x y (^) x y x y

 ^  ^  ^ ^ ^     ^   (^)    (^)  

Example : Gravitational vector field

2

GmM  

r F | r | | r | 11 2 2

is the vector from the center of the sun to the planet is the mass of the sun is the mass of the planet is the gravitational constant G = 6.674 10 (from 1798)

M m G  ^ N m kg 

r

 

3 2 2 2 3/

x y z

x y z

 

r i j k F | r |

 

is conservative with potential f i e.. f ( , x y z , ) 2 2 2 1/ x y z

      

Claim : F | r |

 

2 2 2 1/ Indeed: f ( , x y z , )  x y z

    

and similarly for and

f f

y z

 

 

Example : The work necessary to "escape" the force field from a point p :

= ( ) ( ) C p |^ |

dr dr f f p p





 F^^ ^  F ^ ^  ^ 

   

2 2 2 3/ hence (^22 2 2) 3/ 2

f x x x y z x (^) x y z

         (^)  

   GmM

Greens Theorem

Closed Curve Line Integral C

^ Pdx^  Qdy

Closed Curve Orientation:

Counter-clockwise

Clockwise

C

^ Pdx^  Qdy

C

Pdx  Qdy

 C

  Pdx  Qdy 

Green’s Theorem (in the plane = 2 dim.)

Suppose that C is a simple piecewise smooth closed curve.

C is the boundary of a region R. If^ P Q P ,^ ,^ y^ , and^ Qx^ are continuous on^ R , then

 (^) x y 

R C

Q  P dA  Pdx  Qdy  

 (^) x y  C R

Pdx  Qdy  Q  P dA  

If Qx  Py 1, then

R

 dA  R

 The area of the

interior region

If you have the parametrization of a closed curve and want to find the enclosed

area then you can use this consequence of Green's Theorem to set up the line integral.

choose: , 2 2 2

P   y Q  x  Qx  Py 

We can use Green's theorem to compute areas:

2 R

area R xdy ydx 

 (^)  

stands for the boundary of the region

R R



2

R

 dA  (^) x y   C R

Pdx  Qdy  Q  P dA  

Pdx  Qdy  xdy  ydx

 a a

b

 b

2 2 The parametrization of the ellipse 2 2 1:

x y

a b

 

x  a cos t

y  b sin t

Area (^2) C

 (^)  xdy  ydx

0  t  2 

   

2

0

c

cos sin 2

a t b os t dt b t a sin tdt

  (^)     

dx   a sin tdt

dy  b cos tdt

2 2 2

0

cos sin 2

ab t ab t dt

  (^)    

2 2 2

0

cos sin 2

ab t t dt

  (^)  

2

ab dt

  (^)  2

ab

  ab

2 2 Find the area enclosed by the ellipse 2 2 1.

x y

a b

Example :  

positively oriented