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Another mathamatician neede to check this! CURL-GRAD-DIV

  • Thread starter andyc10
  • Start date
  • #1
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Homework Statement



I have repeated this problem over and over again so either the question is wrong or my likely its my method thats at fault. Please can someone check this for me.

Prove this is true:
[tex]\nabla\times(\nabla \times F)=-\nabla^{2}F+\nabla(\nabla.F)[/tex]

for [tex]F=x^{2}z^{3}i[/tex]


The Attempt at a Solution



Im getting for the LHS:

[tex](\nabla \times F)= 3x^{2}z^{3} j[/tex]

[tex]\nabla\times(\nabla \times F)= -6x^{2}z i + 6xz^{2}k[/tex]

and for the right:

[tex]-\nabla^{2}F=-2z^{3} i[/tex]

[tex](\nabla.F)=2xz^{3}[/tex]

[tex]\nabla(\nabla.F)=2z^{3}i + 6xz^2k[/tex]

[tex]-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k[/tex]

[tex]\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)[/tex]
for [tex]F=x^{2}z^{3}i[/tex]
 

Answers and Replies

  • #2
24
0
andyc10;2201561 and for the right: [tex said:
-\nabla^{2}F=-2z^{3} i[/tex]

[tex](\nabla.F)=2xz^{3}[/tex]

[tex]\nabla(\nabla.F)=2z^{3}i + 6xz^2k[/tex]

[tex]-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k[/tex]

[tex]\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)[/tex]
for [tex]F=x^{2}z^{3}i[/tex]
isn't [tex]\nabla[/tex].[tex]\nabla[/tex] = [tex]\nabla[/tex]^2= d^2/dx^2+d^2/dy^2+etc.

ugh the text never comes out right why is line 1 not equal to negative line 3?
 
Last edited:
  • #3
1,013
65
and for the right:

[tex]-\nabla^{2}F=-2z^{3} i[/tex]
This is not correct.
 
  • #4
Cyosis
Homework Helper
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It seems you've taken the scalar Laplacian and then made a vector out of it. Make sure you use the vector Laplacian and you will get the extra term you need to proof the equality.
 
  • #5
Vid
401
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This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.
 
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
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This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.
Presumably, given that we are in Cartesian coordinates, you use the fact that [tex]\nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}[/tex]
 
  • #7
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Last edited by a moderator:
  • #8
cristo
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Last edited by a moderator:
  • #9
Cyosis
Homework Helper
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The normal Laplacian is given by [tex]\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}[/tex]. You already stated that [itex]A_x=x^2 z^2[/itex]. Now all you have to do is apply the normal Laplacian to A_x.
 
  • #10
5
0
Thanks guys, sorted it.
 

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