Curl and Divergence etc algebra manipulating quick question

  • Thread starter binbagsss
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  • #1
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##\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
 

Answers and Replies

  • #2
954
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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)
 
  • #3
Charles Link
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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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