Another mathamatician neede to check this CURL-GRAD-DIV

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Homework Help Overview

The discussion revolves around verifying a vector calculus identity involving the curl, gradient, and divergence of a vector field F, specifically for the case where F is defined as x²z³i. Participants are examining the equality: ∇×(∇×F) = -∇²F + ∇(∇·F).

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute both sides of the equation but expresses confusion over discrepancies in their results. Some participants question the correctness of applying the scalar Laplacian to a vector field and suggest the need for the vector Laplacian instead. Others raise concerns about the interpretation of the terms involved and the calculations performed.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided guidance on the use of the vector Laplacian, while others have pointed out potential errors in the original poster's calculations. There is no explicit consensus yet on the resolution of the problem.

Contextual Notes

Participants are working under the assumption that the problem is set in Cartesian coordinates, and there is a focus on ensuring the correct application of vector calculus identities. The original poster's repeated attempts suggest a struggle with the underlying concepts.

andyc10
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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?
 
Last edited by a moderator:
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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