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Vector Calculus HW question

  1. Dec 18, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    The basic identities of vector calculus and how scalar fields and vector fields interact

    3. 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!
     
  2. jcsd
  3. Dec 18, 2015 #2

    blue_leaf77

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    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.
     
  4. Dec 18, 2015 #3
    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?
     
  5. Dec 18, 2015 #4

    blue_leaf77

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    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).
     
  6. Dec 18, 2015 #5
    yeah i got the answer thanks alot!
     
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