Proving Vector Identity Using Standard Identities of Vector Analysis

[madachi]

Homework Statement

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

Homework 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

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!

 

[elect_eng]

Homework Statement

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

Homework 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

The Attempt at a Solution

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

Personally, I would start with identity number 10.

 

[madachi]

Personally, I would start with identity number 10.

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!

 

[elect_eng]

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!

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.

 

[madachi]

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.

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.

 

[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})?

 

[madachi]

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})?

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

 

[elect_eng]

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})?

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.

 

[elect_eng]

\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.

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.