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Vector Calculus - Use of Identities

  1. Jan 11, 2015 #1
    1. The problem statement, all variables and given/known data

    By using a suitable vector identity for ∇ × (φA), where φ(r) is a scalar field and A(r) is a vector field, show that
    1. ∇ × (φ∇φ) = 0,
      where φ(r) is any scalar field.

    2. Relevant equations

    ∇×(φA) = (∇φ)×A+φ(∇×A)?

    3. The attempt at a solution

    I honestly have no idea how to even start this one. Any help would be hugely appreciated.
     
  2. jcsd
  3. Jan 11, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    You can apply the relevant equation to the expression you are asked to evaluate.
     
  4. Jan 11, 2015 #3
    Hi mfb, thank you for your reply. That is what I tried to do before but thought I must have been doing it wrong as i ended up with something that didn't cancel.

    Okay so can I assume A = ∇φ
    and the identity is ∇×(φA) = (∇φ)×A+φ(∇×A)

    and follow that through like this:?

    ∇ x (φ∇φ) = ∇φ x ∇φ + φ∇ x ∇φ

    I feel as if there is another identity that I must need to complete the question but have searched through my notes and online and can't find anything. Does ∇φ x ∇φ = -(φ∇ x ∇φ) in some way?
     
  5. Jan 11, 2015 #4
    I've just had a thought. Are you able to say

    A = ∇φ and therefore -A = φ∇

    Allowing us to complete the question like this.

    ∇ x (φ∇φ) = (A x A) + (-A x A)
    ∇ x (φ∇φ) = (A x A) - (A x A)
    = 0
     
  6. Jan 11, 2015 #5

    mfb

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    Staff: Mentor

    So you had an idea where to start, and even a start. Why didn't you write that down? Then it would have been possible to see what went wrong.

    I would probably add brackets to make it more clear: (∇φ) x (∇φ) + φ(∇ x ∇φ)
    What is the cross-product of something with itself?

    No. The right equation is not even well-defined.
     
  7. Jan 11, 2015 #6
    Sorry mfb, I thought my working was complete rubbish and irrelevant so didn't want to embarrass myself on here.

    I think I understand now, I didn't realise that you can put a set of bracket in like that. And the cross product of something with itself is zero. Thanks!
     
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