Proving Nabla-Cross(A x B) Equation

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Homework Help Overview

The discussion revolves around proving the equation \nabla\times(A\times B) and involves vector calculus concepts, particularly the properties of the curl operation and vector identities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the validity of the "bac-cab" equation and its conditions. There are attempts to manipulate the equation and questions about specific terms, such as whether (B.\nabla)A-(A.\nabla)B equals zero. Some suggest using specific vector representations to check equality through algebraic manipulation.

Discussion Status

Multiple approaches are being discussed, including algebraic verification and the use of index notation. Some participants express uncertainty about the validity of certain equations, while others share resources and suggest methods for understanding the topic better.

Contextual Notes

There are references to external resources for learning index notation and vector identities, indicating that some participants may be seeking foundational knowledge to aid their understanding of the problem.

rado5
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Homework Statement



Prove that [tex]\nabla\times(A\times B)= (B.\nabla)A-(A.\nabla)B-B(\nabla.A)+A(\nabla.B)[/tex]

Homework Equations



bac-cab [tex]\nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B[/tex]

The Attempt at a Solution



I know that [tex]B(\nabla.A)=(\nabla.A)B[/tex] and [tex]A(\nabla.B)= (\nabla.B)A[/tex]

So what about [tex](B.\nabla)A-(A.\nabla)B=?[/tex] Does it equal to zero? Or maybe bac cab is not related to this problem!
 
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rado5 said:

Homework Equations



bac-cab [tex]\nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B[/tex]

When and how is this ["bac - cab"] equation valid? The above equation is valid only as long as "a" is...?
 
So how can I prove it? Please help me!
 
rado5 said:
So how can I prove it? Please help me!

Prove it by letting [itex]\vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle[/itex] and just work out both sides to check they are equal. It's easy, just a little algebra.
 
Another interesting way would be to do to use a gem of a trick that I learned off of the Feynman Lectures. Refer to Feynman Lectures, Volume II, Lecture 27, Field Energy and Field Momentum.
 
LCKurtz said:
Prove it by letting [itex]\vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle[/itex] and just work out both sides to check they are equal. It's easy, just a little algebra.

Thank you very much for your help. I actually proved it in the way you suggested me, but only for the x-component, and it was a lot of algebra!
 
anirudh215 said:
Another interesting way would be to do to use a gem of a trick that I learned off of the Feynman Lectures. Refer to Feynman Lectures, Volume II, Lecture 27, Field Energy and Field Momentum.

Thank you very much. I will try to study it.
 
Do you know index notation? Vector identities are quite easy with it.
 
Pengwuino said:
Do you know index notation? Vector identities are quite easy with it.

Please tell me about "index notation". I went to http://en.wikipedia.org/wiki/Index_notation but I didn't completely understand your point of view!
 
  • #10
http://www.physics.ucsb.edu/~physCS33/spring2009/index-notation.pdf

Give this a try. Unfortunately, when I learned it, it was during lectures and not in our textbook so I can't tell you what book you can learn it out of. This should be enough though.
 
Last edited by a moderator:
  • #11
Pengwuino said:
http://www.physics.ucsb.edu/~physCS33/spring2009/index-notation.pdf

Give this a try. Unfortunately, when I learned it, it was during lectures and not in our textbook so I can't tell you what book you can learn it out of. This should be enough though.

Thank you very much. I downloaded it and I will read it.
 
Last edited by a moderator:

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