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

1. May 16, 2009

### andyc10

1. The problem statement, all variables and given/known data

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:
$$\nabla\times(\nabla \times F)=-\nabla^{2}F+\nabla(\nabla.F)$$

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

3. 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 $$F=x^{2}z^{3}i$$

2. May 16, 2009

### Dvsdvs

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: May 16, 2009
3. May 16, 2009

### slider142

This is not correct.

4. May 16, 2009

### Cyosis

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. May 16, 2009

### Vid

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. May 16, 2009

### cristo

Staff Emeritus
Presumably, given that we are in Cartesian coordinates, you use the fact that $$\nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}$$

7. May 17, 2009

### andyc10

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

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

$$A_{y}=0$$

$$A_{z}=0$$

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

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

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

is this not right?

Last edited by a moderator: Apr 24, 2017
8. May 17, 2009

### cristo

Staff Emeritus
This isn't correct. $$\nabla^2 F_x=\nabla^2x^2z^3=2z^3+6zx^2$$

Last edited by a moderator: Apr 24, 2017
9. May 17, 2009

### Cyosis

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. May 17, 2009

### andyc10

Thanks guys, sorted it.