Vector Calculus HW question

  • #1

Homework Statement


http://faculty.fiu.edu/~maxwello/phz3113/probs/set1.pdf

I'm working on problem 2. Trying to prove that the dot product between a vector field and its curl is zero.

Homework Equations


The basic identities of vector calculus and how scalar fields and vector fields interact

The Attempt at a Solution


My only real attempt is expanding what was given using an identity. What i have now is that

f(del X A) = A X del(f)

In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument. I do have a question though. In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!
 

Answers and Replies

  • #2
blue_leaf77
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In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument.
I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with ##\mathbf{A}## being always perpendicular to ##\nabla \times \mathbf{A}## because the cross product in between the last expression makes it perpendicular to ##\mathbf{A}##. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with ##\mathbf{A}##.
In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!
No, they are just vector multiplied with a scalar.
 
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  • #3
I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with ##\mathbf{A}## being always perpendicular to ##\nabla \times \mathbf{A}## because the cross product in between the last expression makes it perpendicular to ##\mathbf{A}##. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with ##\mathbf{A}##.

No, they are just vector multiplied with a scalar.
Yeah what i mean is if A X B = C then C must be perpendicular to A and B so if your doing Del X A = C then C must be perpendicular to A but maybe that is not the case when your talking about del?
 
  • #4
blue_leaf77
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but maybe that is not the case when your talking about del?
One of the identities involving curl is ##\nabla \times (f\mathbf{A}) + \mathbf{A}\times \nabla f= f(\nabla \times \mathbf{A}) ##. Try multiplying by dot product both sides with ##\mathbf{A}##. An example in electromagnetism is one of the Maxwell equations, ##\nabla \times \mathbf{E} = -\partial \mathbf{B} /\partial t##, in most cases electric and magnetic fields are perpendicular, except for certain cases like that in the wave propagation in waveguides (for example see http://physics.stackexchange.com/qu...ric-and-magnetic-fields-are-not-perpendicular).
 
  • #5
One of the identities involving curl is ##\nabla \times (f\mathbf{A}) + \mathbf{A}\times \nabla f= f(\nabla \times \mathbf{A}) ##. Try multiplying by dot product both sides with ##\mathbf{A}##. An example in electromagnetism is one of the Maxwell equations, ##\nabla \times \mathbf{E} = -\partial \mathbf{B} /\partial t##, in most cases electric and magnetic fields are perpendicular, except for certain cases like that in the wave propagation in waveguides (for example see http://physics.stackexchange.com/qu...ric-and-magnetic-fields-are-not-perpendicular).
yeah i got the answer thanks alot!
 

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