Scalar Fields and the Relationship between Gradient and Triple Product

[Flotensia]

Homework Statement

Let us consider three scalar fields u(x), v(x), and w(x). Show that they have a relationship such that f(u, v, w) = 0 if and only if

(∇u) × (∇v) · (∇w) = 0.

Homework Equations

The Attempt at a Solution

I could think nothing but just writing the components of (∇u) × (∇v) · (∇w) = 0. How could i prove this?

 

[Flotensia]

Homework Statement

Let us consider three scalar fields u(x), v(x), and w(x).
Show that they have a relationship such that f(u, v, w) = 0 if and only if
(∇u) × (∇v) · (∇w) = 0.

Homework Equations

The Attempt at a Solution

I can do nothing but just writing components of (∇u) × (∇v) · (∇w). How can I prove this identity?

 

[Ray Vickson]

Homework Statement

Let us consider three scalar fields u(x), v(x), and w(x). Show that they have a relationship such that f(u, v, w) = 0 if and only if

(∇u) × (∇v) · (∇w) = 0.

Homework Equations

The Attempt at a Solution

I could think nothing but just writing the components of (∇u) × (∇v) · (∇w) = 0. How could i prove this?

If $\vec{A} = \nabla u$, $\vec{B} = \nabla v$ and $\vec{C} = \nabla w$, what does the given condition say about the directions of the vectors $\vec{A}, \vec{B}, \vec{C}$?

 

[Flotensia]

Scalar triple product means det(a,b,c) or volume of a parallelepiped. Is it a key to solve this problem?

 

[Brian T]

When you wrote the components out, what did you get?

 

[Flotensia]

I wrote

by using levi-civita symbol

 

[Dick]

Scalar triple product means det(a,b,c) or volume of a parallelepiped. Is it a key to solve this problem?

One way is easy. If f(u,v,w)=0 then to show the triple product is zero you just have to show that the gradients of u, v and w are linearly dependent. Use the chain rule. The other way is harder, you need to have some sort of integrability theorem like Frobenius. I'm a little fuzzy on that. What have you got?

 

[mfb]

Sorry, am I missing something? There is always such a function f(u,v,w) - just define it to be zero everywhere. Are we looking for a non-trivial (!) linear function f?

 

[Dick]

Sorry, am I missing something? There is always such a function f(u,v,w) - just define it to be zero everywhere. Are we looking for a non-trivial (!) linear function f?

Good point. I'm guessing the problem should state that f has at least one nonzero partial derivative at points where you want to show the triple product is zero. Just a guess.