Another mathamatician neede to check this CURL-GRAD-DIV

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SUMMARY

The discussion centers on verifying the mathematical identity involving the curl and Laplacian of a vector field F, specifically F = x²z³i. The user initially misapplies the scalar Laplacian instead of the vector Laplacian, leading to incorrect conclusions. The correct application of the vector Laplacian reveals additional terms necessary for proving the identity: ∇×(∇×F) = -∇²F + ∇(∇·F). The final resolution confirms that the user has sorted out the confusion regarding the application of the Laplacian.

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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 that's at fault. Please can someone check this for me.

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

for F=x^{2}z^{3}i


The Attempt at a Solution



Im getting for the LHS:

(\nabla \times F)= 3x^{2}z^{3} j

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

and for the right:

-\nabla^{2}F=-2z^{3} i

(\nabla.F)=2xz^{3}

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

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

\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)
for F=x^{2}z^{3}i
 
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andyc10;2201561 and for the right: [tex said:
-\nabla^{2}F=-2z^{3} i[/tex]

(\nabla.F)=2xz^{3}

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

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

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

isn't \nabla.\nabla = \nabla^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:
andyc10 said:
and for the right:

-\nabla^{2}F=-2z^{3} i
This is not correct.
 
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.
 
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.
 
Vid said:
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 \nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}
 
slider142 said:
This is not correct.

http://en.wikipedia.org/wiki/Vector_Laplacian

f9bc37c3b3575ca8bbc223a9438d437f.png


A_{x}=x^{2}z^{3}

A_{y}=0

A_{z}=0\nabla^{2}A_{x}=2z^{3}

\Rightarrow\nabla^{2}A=2z^{3}i

\Rightarrow-\nabla^{2}F=-2z^{3}i

is this not right?
 
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Last edited by a moderator:
The normal Laplacian is given by \nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}. You already stated that A_x=x^2 z^2. Now all you have to do is apply the normal Laplacian to A_x.
 
  • #10
Thanks guys, sorted it.
 

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