Identity by using gradient

  • Thread starter Flotensia
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  • #1
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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?
 
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Answers and Replies

  • #2
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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?
 
  • #3
Ray Vickson
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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}##?
 
  • #4
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Scalar triple product means det(a,b,c) or volume of a parallelepiped. Is it a key to solve this problem?
 
  • #5
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When you wrote the components out, what did you get?
 
  • #6
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I wrote
%C0%B9%BF%A2.JPG

by using levi-civita symbol
 
  • #7
Dick
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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?
 
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  • #8
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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?
 
  • #9
Dick
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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.
 

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