Proving Nabla-Cross(A x B) Equation

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SUMMARY

The forum discussion centers on proving the Nabla-Cross(A x B) equation, specifically \nabla\times(A\times B)= (B.\nabla)A-(A.\nabla)B-B(\nabla.A)+A(\nabla.B). Participants suggest using algebraic manipulation by substituting \vec A = \langle f,g,h\rangle and \vec B = \langle u,v,w\rangle to verify the equality. Additionally, they reference the Feynman Lectures, Volume II, Lecture 27, for alternative methods and insights. The discussion also touches on the utility of index notation for simplifying vector identities.

PREREQUISITES
  • Understanding of vector calculus and vector identities
  • Familiarity with the Nabla operator and its applications
  • Basic knowledge of algebraic manipulation in vector equations
  • Introduction to index notation for vectors
NEXT STEPS
  • Study the Feynman Lectures, Volume II, Lecture 27, on Field Energy and Field Momentum
  • Learn about index notation and its application in vector calculus
  • Practice proving vector identities using algebraic methods
  • Explore advanced topics in vector calculus, such as curl and divergence
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on vector calculus and electromagnetism, will benefit from this discussion.

rado5
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Homework Statement



Prove that \nabla\times(A\times B)= (B.\nabla)A-(A.\nabla)B-B(\nabla.A)+A(\nabla.B)

Homework Equations



bac-cab \nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B

The Attempt at a Solution



I know that B(\nabla.A)=(\nabla.A)B and A(\nabla.B)= (\nabla.B)A

So what about (B.\nabla)A-(A.\nabla)B=? 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 \nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B

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 \vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle 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 \vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle 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.
 
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  • #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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