# Vecotr identity proof 2

1. Apr 16, 2010

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

Let $F(x,y,z)$ be an irrotational vector field and $f(x,y,z)$ a $C^1$ scalar functions. Using the standard identities of vector analysis (provided in section 2 below), simplify

$(\nabla f \times F) \cdot \nabla f$

2. Relevant equations

Note: The identities below require $f,g,F,G$ to be suitable differentiable, either order $C^1$ or $C^2$.

$1. \nabla (f+g) = \nabla f + \nabla g$
$2. \nabla (\lambda f) = \lambda \nabla f$, where $\lambda$ is a constant
$3. \nabla (fg) = f \nabla g + g \nabla f$
$4. \nabla (\frac{f}{g}) = \frac{g \nabla f - f \nabla g}{g^2}$
$5. \nabla \cdot (F+G) = \nabla \cdot F + \nabla \cdot G$
$6. \nabla \times (F+G) = \nabla \times F + \nabla \times G$
$7. \nabla \cdot (fF) = f \nabla \cdot F + F \cdot \nabla f$
$8. \nabla \cdot (F \times G) = G \cdot (\nabla \times F ) - F \cdot (\nabla \times G)$
$9. \nabla \cdot (\nabla \times F) = 0$
$10. \nabla \times (fF) = f \nabla \times F + \nabla f \times F$
$11. \nabla \times (\nabla f) = 0$
$12. {\nabla}^2 (fg) = f{\nabla}^2 g + g{\nabla}^2 f + 2 \nabla f \cdot \nabla g$
$13. \nabla \cdot (\nabla f \times \nabla g) = 0$
$14. \nabla (f \nabla g - g \nabla f) = f {\nabla}^2 g - g {\nabla}^2 f$

3. The attempt at a solution

I first write $(\nabla f \times F) \cdot \nabla f$
= $\nabla f \cdot ( \nabla f \times F )$

However this doesn't look like any of the vector identity above. Can we write
$\nabla f = f \nabla$ though? As this would enable me to use the identity. Also, is the final answer 0 ?

Thanks!

Last edited: Apr 16, 2010
2. Apr 16, 2010

### elect_eng

3. Apr 16, 2010

How do we start with identity 10? I mean the original expression is not of the form of identity 10, hence how do we use identity 10 to simplify it? Could you show me the first step so I can understand your approach.

Thanks!

4. Apr 16, 2010

### elect_eng

I have only spent about 20 seconds looking at this problem, so keep that in mind. My advice may not lead you far. However, look at the right had side of identity 10. You will see that the last term is helpful to get started. Rearrange this identity.

I'm going out now, but I'll check back later tonight when I get home. I'll see how you made out and put more time in if you are still having trouble.

5. Apr 16, 2010

From identity 10,
$\nabla f \times F = \nabla \times (fF) - f \nabla \times F$

The second term on RHS is zero, since $\nabla \times F =$ as F is irrotational.

So the equation reduces to

$\nabla f \times F = \nabla \times (fF)$

So I'm stuck here, or I did not do as what you have told me to?

Thanks.

6. Apr 16, 2010

### gabbagabbahey

Have you not learned the triple product rule $\textbf{A}\cdot(\textbf{B}\times\textbf{C})=\textbf{B}\cdot(\textbf{C}\times\textbf{A})=\textbf{C}\cdot(\textbf{A}\times\textbf{B})$?

7. Apr 16, 2010

Yeah, I didn't think about that. Thanks!

8. Apr 16, 2010

### elect_eng

That is an elegant method, but doesn't the problem specify the use of the given identities?

One could also use a geometrical argument.

$(\nabla f \times F)$ must be orthogonal to $\nabla f$, since a cross product is orthogonal to both of it's input vectors (assuming it's not zero directly). Hence, the dot product between $(\nabla f \times F)$ and $\nabla f$ must be zero. But, again this would not be using the identities given.

Last edited: Apr 17, 2010
9. Apr 17, 2010

### elect_eng

Yes, you did what I suggested as a starting point. From here you can substitute this result into the original equation. Then, identity 8 can be used, and combined with identity 11 for further simplification. Then, identities 13 and 8 (again) can show that the relation is zero. Perhaps there is a more direct way using only those identities, but this method at least uses only the given identities and proves the result is zero, as expected.

Last edited: Apr 17, 2010