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Curl and Divergence etc algebra manipulating quick question

  1. May 29, 2016 #1
    ##\nabla p = \rho \nabla \phi ##

    My textbook says that by taking the curl we get:

    ## 0=\nabla \rho X \nabla \phi ## **

    I don't follow. I understand the LHS is zero, by taking the curl of a divergence.
    But I'm unsure as to how we get it into this form, from which it is clear that the gradients of ##\rho## and
    ##\phi## are parallel, since I get:

    ##\nabla X \rho \nabla \phi ##, I know that the curl acting on a scalar field doesn't make sense, I would get ##\rho \nabla X \nabla \phi ##, taking the scalar field ##\rho## to the left since it can not be operated on by a curl. I don't see how you would get **

    Many thanks in advance
     
  2. jcsd
  3. May 29, 2016 #2
    It is rather straightforward to prove that
    [tex]
    \nabla \times (a \mathbf{B}) = (\nabla a) \times \mathbf{B} + a (\nabla \times \mathbf{B})
    [/tex]
    where ##a## is a scalar field and ##\mathbf{B}## is a vector field. This formula is also readily available in most lists of vector calculus identities (although I would strongly recommend you try to prove it yourself)
     
  4. May 29, 2016 #3

    Charles Link

    User Avatar
    Homework Helper

    One quick comment: The LHS is the curl of a "gradient". (The curl of a gradient is zero and so is the divergence of a curl).
     
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