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curl of a 2D field

Every divergence-free vector field generated from skew-symmetric matrixNon-conservative field with zero curlWhen a vector field can be scaled to form a conservative vector fieldCalculate flux of vector field“Inverse” Helmholtz DecompositionProblem with Deriving Curl in Spherical Co-ordinates.Why is does this vector field have zero-curl everywhere? Plus, broad questions about curl.Calculating push forwards of a vector fieldFinding curl in spherical coordinatesFinding a vector field such that its Curl equals a given vector field

0

$begingroup$

How can I calculate the curl of a 2D field like $textbfF= F_x(x,y)textbfi + F_y(x,y)textbfj$ if the curl is defined is 3D? My book says to apply the definition of curl to the associated 3D field $textbfF = F_x(x,y)textbfi + F_2(x,y)textbfj +0textbfk$, but I don't get the expected result which is $textrot textbfF= (partial_x F_y - partial_y F_x)textbfk$

$left(frac partial F_zpartial y-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial z-frac partial F_zpartial xright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k =beginbmatrixfrac partial F_zpartial y-frac partial F_ypartial z\frac partial F_xpartial z-frac partial F_zpartial x\frac partial F_ypartial x-frac partial F_xpartial yendbmatrix$

since $F_z=0$ the result should be

$textrot mathbfF =left(-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial zright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k $

vector-fields

share|cite|improve this question

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What do you get and how did you get it?
  $endgroup$
  – John Douma
  Mar 12 at 8:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is just a formula. Where is your work?
  $endgroup$
  – John Douma
  Mar 12 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As I said in my answer, your last equation is correct, you just have to consider that, since $vecF$ does not depend on $z$, you have $fracpartial F_xpartial z=0$ and $fracpartial F_ypartial z=0$. Considering this, we get $textrot(vecF)=left(fracpartial F_ypartial x-fracpartial F_xpartial yright)textbfk$
  $endgroup$
  – Gabriele Cassese
  Mar 12 at 10:07
  
  

  
  
  
  

add a comment | 

0

$begingroup$

How can I calculate the curl of a 2D field like $textbfF= F_x(x,y)textbfi + F_y(x,y)textbfj$ if the curl is defined is 3D? My book says to apply the definition of curl to the associated 3D field $textbfF = F_x(x,y)textbfi + F_2(x,y)textbfj +0textbfk$, but I don't get the expected result which is $textrot textbfF= (partial_x F_y - partial_y F_x)textbfk$

$left(frac partial F_zpartial y-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial z-frac partial F_zpartial xright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k =beginbmatrixfrac partial F_zpartial y-frac partial F_ypartial z\frac partial F_xpartial z-frac partial F_zpartial x\frac partial F_ypartial x-frac partial F_xpartial yendbmatrix$

since $F_z=0$ the result should be

$textrot mathbfF =left(-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial zright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k $

vector-fields

share|cite|improve this question

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What do you get and how did you get it?
  $endgroup$
  – John Douma
  Mar 12 at 8:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is just a formula. Where is your work?
  $endgroup$
  – John Douma
  Mar 12 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As I said in my answer, your last equation is correct, you just have to consider that, since $vecF$ does not depend on $z$, you have $fracpartial F_xpartial z=0$ and $fracpartial F_ypartial z=0$. Considering this, we get $textrot(vecF)=left(fracpartial F_ypartial x-fracpartial F_xpartial yright)textbfk$
  $endgroup$
  – Gabriele Cassese
  Mar 12 at 10:07
  
  

  
  
  
  

add a comment | 

$begingroup$

How can I calculate the curl of a 2D field like $textbfF= F_x(x,y)textbfi + F_y(x,y)textbfj$ if the curl is defined is 3D? My book says to apply the definition of curl to the associated 3D field $textbfF = F_x(x,y)textbfi + F_2(x,y)textbfj +0textbfk$, but I don't get the expected result which is $textrot textbfF= (partial_x F_y - partial_y F_x)textbfk$

$left(frac partial F_zpartial y-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial z-frac partial F_zpartial xright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k =beginbmatrixfrac partial F_zpartial y-frac partial F_ypartial z\frac partial F_xpartial z-frac partial F_zpartial x\frac partial F_ypartial x-frac partial F_xpartial yendbmatrix$

since $F_z=0$ the result should be

$textrot mathbfF =left(-frac partial F_ypartial zright)mathbf i +left(frac partial F_xpartial zright)mathbf j +left(frac partial F_ypartial x-frac partial F_xpartial yright)mathbf k $

vector-fields

share|cite|improve this question

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

$endgroup$

share|cite|improve this question

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

share|cite|improve this question

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

edited Mar 12 at 11:45

Gabriele Cassese

1,056315

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

asked Mar 12 at 8:47

Giuliano MalatestaGiuliano Malatesta

295

1 Answer
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1

$begingroup$

The formula you state gives you the correct derivation of $textrottextbfF$, considering that $fracpartial textbfFpartial z=vec0$ and $nabla F_z=nabla 0=vec0$.

In your last equation, you have to consider that $fracpartial F_xpartial z=fracpartial F_ypartial z=0$, since $textbfF$ does not depend on $z$

share|cite|improve this answer

edited Mar 12 at 11:24

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

$endgroup$

add a comment | 

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1

$begingroup$

The formula you state gives you the correct derivation of $textrottextbfF$, considering that $fracpartial textbfFpartial z=vec0$ and $nabla F_z=nabla 0=vec0$.

In your last equation, you have to consider that $fracpartial F_xpartial z=fracpartial F_ypartial z=0$, since $textbfF$ does not depend on $z$

share|cite|improve this answer

edited Mar 12 at 11:24

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

$endgroup$

add a comment | 

1

$begingroup$

The formula you state gives you the correct derivation of $textrottextbfF$, considering that $fracpartial textbfFpartial z=vec0$ and $nabla F_z=nabla 0=vec0$.

In your last equation, you have to consider that $fracpartial F_xpartial z=fracpartial F_ypartial z=0$, since $textbfF$ does not depend on $z$

share|cite|improve this answer

edited Mar 12 at 11:24

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

$endgroup$

add a comment | 

$begingroup$

The formula you state gives you the correct derivation of $textrottextbfF$, considering that $fracpartial textbfFpartial z=vec0$ and $nabla F_z=nabla 0=vec0$.

In your last equation, you have to consider that $fracpartial F_xpartial z=fracpartial F_ypartial z=0$, since $textbfF$ does not depend on $z$

share|cite|improve this answer

edited Mar 12 at 11:24

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

$endgroup$

share|cite|improve this answer

edited Mar 12 at 11:24

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315

answered Mar 12 at 9:07

Gabriele CasseseGabriele Cassese

1,056315