Curl and Divergence - NDSU - North Dakota State …micohen/oldlecturenotes...Curl and Divergence,...

of 36/36
Curl and Divergence Definition Let ~ F =(F 1 , F 2 , F 3 ) be a vector field. The curl of ~ F is the vector field defined by curl( ~ F )= δF 3 δy - δF 2 δz , δF 1 δz - δF 3 δx , δF 2 δx - δF 1 δy .
  • date post

    09-Jun-2018
  • Category

    Documents

  • view

    217
  • download

    0

Embed Size (px)

Transcript of Curl and Divergence - NDSU - North Dakota State …micohen/oldlecturenotes...Curl and Divergence,...

  • Curl and Divergence

    DefinitionLet ~F = (F1,F2,F3) be a vector field. The curl of ~F is the vectorfield defined by

    curl(~F ) =

    (F3y F2

    z,F1z F3

    x,F2x F1

    y

    ).

    As a mnemonic device, one can think of the curl of ~F as thesymbolic cross product:

    curl(~F ) = ~F = ( x ,y ,

    z ) (F1,F2,F3).

  • Curl and Divergence

    DefinitionLet ~F = (F1,F2,F3) be a vector field. The curl of ~F is the vectorfield defined by

    curl(~F ) =

    (F3y F2

    z,F1z F3

    x,F2x F1

    y

    ).

    As a mnemonic device, one can think of the curl of ~F as thesymbolic cross product:

    curl(~F ) = ~F = ( x ,y ,

    z ) (F1,F2,F3).

  • Curl and Divergence, contd.

    DefinitionAgain let ~F = (F1,F2,F3) be a vector field. The divergence of ~Fis the real-valued function in three variables defined by

    div(~F ) =F1x

    +F2y

    +F3z

    .

    For a mnemonic device, we can think of the divergence as thesymbolic dot product:

    div(~F ) = ~F = ( x ,y ,

    z ) (F1,F2,F3).

  • Curl and Divergence, contd.

    DefinitionAgain let ~F = (F1,F2,F3) be a vector field. The divergence of ~Fis the real-valued function in three variables defined by

    div(~F ) =F1x

    +F2y

    +F3z

    .

    For a mnemonic device, we can think of the divergence as thesymbolic dot product:

    div(~F ) = ~F = ( x ,y ,

    z ) (F1,F2,F3).

  • Physical Significance

    The physical applications of the notions of curl and divergence of a vector field areimpossible to fully capture within the scope of this class (and this slide!). However, wecan give some terse indications in the context of fluid dynamics.

    Think of ~F as representing the velocity field of a three-dimensional body of liquid inmotion.

    Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point

    (x , y , z). Then the curl vector of ~F at (x , y , z) may be imagined as the axis on whichthe fluid makes the wheel spin according to the right-hand rule: that is, if you stickyour right thumb up in the direction of the curl, the wheel will spin in the directionthat your fingers curl. The magnitude of the curl vector is how fast the wheel rotates.

    The divergence of ~F represents the expansion/compression of the fluid at a given

    point (x , y , z). A positive divergence corresponds to fluid expansion, i.e. the fluid is

    generally moving away from the point, while a negative divergence corresponds to fluid

    compression, i.e. the fluid is generally moving toward the point.

  • Physical Significance

    The physical applications of the notions of curl and divergence of a vector field areimpossible to fully capture within the scope of this class (and this slide!). However, wecan give some terse indications in the context of fluid dynamics.

    Think of ~F as representing the velocity field of a three-dimensional body of liquid inmotion.

    Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point

    (x , y , z). Then the curl vector of ~F at (x , y , z) may be imagined as the axis on whichthe fluid makes the wheel spin according to the right-hand rule: that is, if you stickyour right thumb up in the direction of the curl, the wheel will spin in the directionthat your fingers curl. The magnitude of the curl vector is how fast the wheel rotates.

    The divergence of ~F represents the expansion/compression of the fluid at a given

    point (x , y , z). A positive divergence corresponds to fluid expansion, i.e. the fluid is

    generally moving away from the point, while a negative divergence corresponds to fluid

    compression, i.e. the fluid is generally moving toward the point.

  • Physical Significance

    The physical applications of the notions of curl and divergence of a vector field areimpossible to fully capture within the scope of this class (and this slide!). However, wecan give some terse indications in the context of fluid dynamics.

    Think of ~F as representing the velocity field of a three-dimensional body of liquid inmotion.

    Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point

    (x , y , z). Then the curl vector of ~F at (x , y , z) may be imagined as the axis on whichthe fluid makes the wheel spin according to the right-hand rule: that is, if you stickyour right thumb up in the direction of the curl, the wheel will spin in the directionthat your fingers curl. The magnitude of the curl vector is how fast the wheel rotates.

    The divergence of ~F represents the expansion/compression of the fluid at a given

    point (x , y , z). A positive divergence corresponds to fluid expansion, i.e. the fluid is

    generally moving away from the point, while a negative divergence corresponds to fluid

    compression, i.e. the fluid is generally moving toward the point.

  • Linearity of Curl and Divergence

    If ~F , ~G are any two vectors fields then

    curl(~F + ~G ) = curl(~F ) + curl(~G ) and

    div(~F + ~G ) = div(~F ) + div(~G ),

    and if c is any constant then

    curl(c~F ) = c curl(~F ) and div(c~F ) = c div(~F ).

    In other words curl and div are linear transformations.

  • Boundary Orientations

    Let G (u, v) be a smooth one-to-one parametrization with domainD of a two-dimensional surface S in R3. Define the boundary ofS, denoted S , to be the image of D under G .

    We informally fix a boundary orientation on S as follows: if youare a normal vector standing on the surface S walking along theboundary with the correct orientation, then the surface is on yourleft and the void is on your right. (See picture.)

  • Stokes Theorem

    Theorem (Stokes Theorem)

    Let S be a surface parametrized by a smooth one-to-one functionG (u, v) with domain D, where D is comprised of simple closedcurves. Then

    S~F ds =

    Scurl(~F ) dS.

  • Example

    Verify Stokes theorem for the field ~F = (y , 2x , x + z) and theupper unit hemisphere S, with outward-pointing normal vectors.

  • SolutionKnown: ~F = (y , 2x , x + z)

    First we will just compute the line integral about the boundary S , which is the unitcircle in the xy -plane oriented counterclockwise. We can parametrize S in the usualway:

    ~c(t) = (cos t, sin t, 0) on the domain [0, 2].

    Now compute:

    S~F ds =

    20

    ~F (~c(t)) ~c (t)dt

    =

    20

    ( sin t, 2 cos t, cos t) ( sin t, cos t, 0)dt

    =

    20

    (sin2 t + 2 cos2 t)dt

    =

    20

    (1 + cos2 t)dt

    =

    20

    (3

    2+

    1

    2cos 2t

    )dt

    =

    [3

    2t +

    1

    4sin 2t

    ]20

    = 3.

  • SolutionKnown: ~F = (y , 2x , x + z)

    First we will just compute the line integral about the boundary S , which is the unitcircle in the xy -plane oriented counterclockwise. We can parametrize S in the usualway:

    ~c(t) = (cos t, sin t, 0) on the domain [0, 2].

    Now compute:

    S~F ds =

    20

    ~F (~c(t)) ~c (t)dt

    =

    20

    ( sin t, 2 cos t, cos t) ( sin t, cos t, 0)dt

    =

    20

    (sin2 t + 2 cos2 t)dt

    =

    20

    (1 + cos2 t)dt

    =

    20

    (3

    2+

    1

    2cos 2t

    )dt

    =

    [3

    2t +

    1

    4sin 2t

    ]20

    = 3.

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    Now we will compute the flux integral over S. First we parametrize the upper unithemisphere in the usual way:

    G(, ) = (cos sin, sin sin, cos) on the domain [0, 2] [0, 2

    ].

    Now compute the normal:

    ~T = ( sin sin, cos sin, 0) and ~T = (cos cos, sin cos, sin);

    ~n = ~T ~T = (cos sin2 , sin sin2 , sin cos).

    At this moment we verify mentally or graphically that we have chosen the correct

    orientation for our normal vector ~n. (If we had accidentally parametrized S in such away that ~n faced the wrong direction, then we could fix the problem by just flipping

    the sign on ~n.)

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    Now we will compute the flux integral over S. First we parametrize the upper unithemisphere in the usual way:

    G(, ) = (cos sin, sin sin, cos) on the domain [0, 2] [0, 2

    ].

    Now compute the normal:

    ~T = ( sin sin, cos sin, 0) and ~T = (cos cos, sin cos, sin);

    ~n = ~T ~T = (cos sin2 , sin sin2 , sin cos).

    At this moment we verify mentally or graphically that we have chosen the correct

    orientation for our normal vector ~n. (If we had accidentally parametrized S in such away that ~n faced the wrong direction, then we could fix the problem by just flipping

    the sign on ~n.)

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    Now we will compute the flux integral over S. First we parametrize the upper unithemisphere in the usual way:

    G(, ) = (cos sin, sin sin, cos) on the domain [0, 2] [0, 2

    ].

    Now compute the normal:

    ~T = ( sin sin, cos sin, 0) and ~T = (cos cos, sin cos, sin);

    ~n = ~T ~T = (cos sin2 , sin sin2 , sin cos).

    At this moment we verify mentally or graphically that we have chosen the correct

    orientation for our normal vector ~n. (If we had accidentally parametrized S in such away that ~n faced the wrong direction, then we could fix the problem by just flipping

    the sign on ~n.)

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    Now we will compute the flux integral over S. First we parametrize the upper unithemisphere in the usual way:

    G(, ) = (cos sin, sin sin, cos) on the domain [0, 2] [0, 2

    ].

    Now compute the normal:

    ~T = ( sin sin, cos sin, 0) and ~T = (cos cos, sin cos, sin);

    ~n = ~T ~T = (cos sin2 , sin sin2 , sin cos).

    At this moment we verify mentally or graphically that we have chosen the correct

    orientation for our normal vector ~n. (If we had accidentally parametrized S in such away that ~n faced the wrong direction, then we could fix the problem by just flipping

    the sign on ~n.)

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    ~n = (cos sin2 , sin sin2 , sin cos)

    Next compute curl(~F ):

    curl(~F ) = det

    ~i ~j ~kx

    y

    z

    y 2x x + z

    =

    (

    y(x + z)

    z(2x),

    x(x + z) +

    z(y),

    x(2x)

    y(y)

    )= (0,1, 3).

  • Solution, contd.

    Known: ~F = (y , 2x , x + z)S~F ds = 3

    ~n = (cos sin2 , sin sin2 , sin cos)

    Next compute curl(~F ):

    curl(~F ) = det

    ~i ~j ~kx

    y

    z

    y 2x x + z

    =

    (

    y(x + z)

    z(2x),

    x(x + z) +

    z(y),

    x(2x)

    y(y)

    )= (0,1, 3).

  • Solution, contd.Known: ~F = (y , 2x , x + z)S~F ds = 3

    ~n = (cos sin2 , sin sin2 , sin cos)

    curl(~F ) = (0,1, 3)

    Finally we are ready to compute the flux of curl(~F ) through S and show that it equals3 as per Stokes theorem.

    Scurl(~F ) dS =

    /20

    20

    curl(~F )(G(, )) ~n(, )dd

    =

    /20

    20

    (0,1, 3) (cos sin2 , sin sin2 , sin cos)dd

    =

    /20

    20

    ( sin sin2 + 3 sin cos)dd

    = 3 2 /2

    0sin cosd

    = 6 [

    1

    2sin2

    ]/20

    = 3[1 0] = 3.

  • Solution, contd.Known: ~F = (y , 2x , x + z)S~F ds = 3

    ~n = (cos sin2 , sin sin2 , sin cos)

    curl(~F ) = (0,1, 3)

    Finally we are ready to compute the flux of curl(~F ) through S and show that it equals3 as per Stokes theorem.

    Scurl(~F ) dS =

    /20

    20

    curl(~F )(G(, )) ~n(, )dd

    =

    /20

    20

    (0,1, 3) (cos sin2 , sin sin2 , sin cos)dd

    =

    /20

    20

    ( sin sin2 + 3 sin cos)dd

    = 3 2 /2

    0sin cosd

    = 6 [

    1

    2sin2

    ]/20

    = 3[1 0] = 3.

  • Higher-Dimensional Boundary Orientations

    Now let W denote a closed bounded region in R3 enclosed by asmooth parametrized surface S = W, the boundary of W.

    We orient W by requiring that the normal vectors ~n pointoutward, away from W.

  • Divergence Theorem

    Theorem (Divergence Theorem)

    Let S = W be a smooth parametrized surface enclosing a regionW in R3. Let ~F be a vector field defined on W. Then

    W~F dS =

    W

    div(~F )d(x , y , z).

  • ExampleLet ~F = (x2, z4, ez) and let S be the boundary of the box [0, 2] [0, 3] [0, 1]in R3. Use the divergence theorem to calculate the flux of ~F through S.

    Solution.First compute the divergence of ~F :

    div(~F ) = x

    (x2) + y

    (z4) + z

    (ez) = 2x + ez .

    Now apply the divergence theorem to convert the flux integral to a volumeintegral:

    S

    ~F dS = 1

    0

    30

    20

    (2x + ez)dxdydz

    =

    10

    30

    [x2 + xez ]2x=0dydx

    =

    10

    30

    (4 + 2ez)dydz

    =

    10

    (12 + 6ez)dz

    = 12 + 6e 6 = 6(1 + e).

  • ExampleLet ~F = (x2, z4, ez) and let S be the boundary of the box [0, 2] [0, 3] [0, 1]in R3. Use the divergence theorem to calculate the flux of ~F through S.

    Solution.First compute the divergence of ~F :

    div(~F ) = x

    (x2) + y

    (z4) + z

    (ez) = 2x + ez .

    Now apply the divergence theorem to convert the flux integral to a volumeintegral:

    S

    ~F dS = 1

    0

    30

    20

    (2x + ez)dxdydz

    =

    10

    30

    [x2 + xez ]2x=0dydx

    =

    10

    30

    (4 + 2ez)dydz

    =

    10

    (12 + 6ez)dz

    = 12 + 6e 6 = 6(1 + e).

  • ExampleLet ~F = (x2, z4, ez) and let S be the boundary of the box [0, 2] [0, 3] [0, 1]in R3. Use the divergence theorem to calculate the flux of ~F through S.

    Solution.First compute the divergence of ~F :

    div(~F ) = x

    (x2) + y

    (z4) + z

    (ez) = 2x + ez .

    Now apply the divergence theorem to convert the flux integral to a volumeintegral:

    S

    ~F dS = 1

    0

    30

    20

    (2x + ez)dxdydz

    =

    10

    30

    [x2 + xez ]2x=0dydx

    =

    10

    30

    (4 + 2ez)dydz

    =

    10

    (12 + 6ez)dz

    = 12 + 6e 6 = 6(1 + e).

  • Consequences of Stokes and Divergence Theorems

    FactConservative vector fields have zero curl. That is:

    If ~F = V , then curl(~F ) = ~0.

    Proof.Cross-partials property!

    CorollaryIf ~F is a conservative vector field, then the circulation of ~F about any simpleclosed curve C is 0.

    Proof.Stokes theorem implies that the circulation

    C~F ds is equal to some surface

    integral of the field curl(~F ) = ~0, which is always 0.

    (We also already know this from the fundamental theorem for conservativevector fields.)

  • Consequences of Stokes and Divergence Theorems

    FactConservative vector fields have zero curl. That is:

    If ~F = V , then curl(~F ) = ~0.

    Proof.Cross-partials property!

    CorollaryIf ~F is a conservative vector field, then the circulation of ~F about any simpleclosed curve C is 0.

    Proof.Stokes theorem implies that the circulation

    C~F ds is equal to some surface

    integral of the field curl(~F ) = ~0, which is always 0.

    (We also already know this from the fundamental theorem for conservativevector fields.)

  • Consequences of Stokes and Divergence Theorems

    FactConservative vector fields have zero curl. That is:

    If ~F = V , then curl(~F ) = ~0.

    Proof.Cross-partials property!

    CorollaryIf ~F is a conservative vector field, then the circulation of ~F about any simpleclosed curve C is 0.

    Proof.Stokes theorem implies that the circulation

    C~F ds is equal to some surface

    integral of the field curl(~F ) = ~0, which is always 0.

    (We also already know this from the fundamental theorem for conservativevector fields.)

  • Consequences of Stokes and Divergence Theorems

    FactConservative vector fields have zero curl. That is:

    If ~F = V , then curl(~F ) = ~0.

    Proof.Cross-partials property!

    CorollaryIf ~F is a conservative vector field, then the circulation of ~F about any simpleclosed curve C is 0.

    Proof.Stokes theorem implies that the circulation

    C~F ds is equal to some surface

    integral of the field curl(~F ) = ~0, which is always 0.

    (We also already know this from the fundamental theorem for conservativevector fields.)

  • Consequences of Stokes and Divergence Theorems, contd.

    FactCurl fields have zero divergence. That is:

    If ~F = curl(~A), then div(~F ) = 0.

    Proof.A consequence of Clairauts theorem! Remember this theorem says mixedsecond-order partial derivatives are equal for continuously differentiablefunctions. Try working through this on your own.

    CorollaryIf ~F is the curl field of some vector field ~A, then the flux of ~F through anyclosed surface is 0.

    Proof.The divergence theorem says that the flux of ~F is equal to a volume integral of

    the function div(~F ), again a 0 function which gives a 0 integral.

  • Consequences of Stokes and Divergence Theorems, contd.

    FactCurl fields have zero divergence. That is:

    If ~F = curl(~A), then div(~F ) = 0.

    Proof.A consequence of Clairauts theorem! Remember this theorem says mixedsecond-order partial derivatives are equal for continuously differentiablefunctions. Try working through this on your own.

    CorollaryIf ~F is the curl field of some vector field ~A, then the flux of ~F through anyclosed surface is 0.

    Proof.The divergence theorem says that the flux of ~F is equal to a volume integral of

    the function div(~F ), again a 0 function which gives a 0 integral.

  • Consequences of Stokes and Divergence Theorems, contd.

    FactCurl fields have zero divergence. That is:

    If ~F = curl(~A), then div(~F ) = 0.

    Proof.A consequence of Clairauts theorem! Remember this theorem says mixedsecond-order partial derivatives are equal for continuously differentiablefunctions. Try working through this on your own.

    CorollaryIf ~F is the curl field of some vector field ~A, then the flux of ~F through anyclosed surface is 0.

    Proof.The divergence theorem says that the flux of ~F is equal to a volume integral of

    the function div(~F ), again a 0 function which gives a 0 integral.

  • Consequences of Stokes and Divergence Theorems, contd.

    FactCurl fields have zero divergence. That is:

    If ~F = curl(~A), then div(~F ) = 0.

    Proof.A consequence of Clairauts theorem! Remember this theorem says mixedsecond-order partial derivatives are equal for continuously differentiablefunctions. Try working through this on your own.

    CorollaryIf ~F is the curl field of some vector field ~A, then the flux of ~F through anyclosed surface is 0.

    Proof.The divergence theorem says that the flux of ~F is equal to a volume integral of

    the function div(~F ), again a 0 function which gives a 0 integral.

  • Summary of Vector Field Operations

    f ~F curl ~G div g

    real-valued function vector field vector field real-valued function

    Doing any two consecutive operations in a row results in zero, i.e. curl(f ) = ~0 anddiv(curl(~F )) = 0.

  • Thanks

    Thanks for a great semester!