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F = ∇·F

  1. Jun 13, 2014 #1
    A scalar field can be the exact form of a vector field (potential form)? It's make sense?
     
  2. jcsd
  3. Jun 13, 2014 #2

    Matterwave

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    Nope, doesn't make sense. A divergence is not an exterior derivative. A scalar field is a 0-form. It can't be an exact form because an exact n-form must be the exterior derivative of of a n-1 form. There are no -1 forms, so a 0-form cannot be considered exact.
     
  4. Jun 13, 2014 #3
    You have a lot of knowledge in several areas of science, impressive!
     
  5. Jun 14, 2014 #4
    But the operation ##\vec{\nabla} \cdot \vec{F} = F## exist! But still so ##\vec{F}## isn't a potential form and ##F## isn't an exact form?
     
  6. Jun 14, 2014 #5

    Matterwave

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    You can't call it that. Those terms you used have very specific meanings. You call ##\vec{F}## the vector field and ##F## its divergence.
     
  7. Jun 14, 2014 #6
  8. Jun 14, 2014 #7

    Matterwave

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    The analogy this article is talking about is if ##\vec{\nabla}\cdot\vec{F}=0## we call ##\vec{F}## incompressible analogous to closed. If ##\vec{F}=\vec{\nabla}\times\vec{A}## then we call ##\vec{F}## solenoidal analogous to exact. The final analogy is that exact implies closed ##\vec{\nabla}\cdot(\vec{\nabla}\times\vec{A})=0##.

    But these are analogies. You can NOT call ##\vec{F}## a (closed or exact) form. A form has a specific mathematical definition.

    Give you an example. Say I have a bicycle and a car. If I put a motor on my bicycle, it turns into a motorized vehicle which is somewhat analogous to a car. But I DON'T call my motorized bicycle a car. I call it a motorized bicycle, or a motorcycle. In the same way F above is analogous to a closed or exact form but I CANNOT call F a closed or exact form. I call it an incompressible or solenoidal vector field.
     
  9. Jun 26, 2014 #8
    Maybe what OP was asking for is the following identities for the Cartan derivatives ##d## of ##k##-forms in ##\mathbb R ^3## with canonical metric:
    $$0) \, d \phi = (\nabla \phi)^T \cdot d \bar x$$
    $$1) \, d A = (\nabla \times \bar A)^T \cdot \star d \bar x
    \quad, A:= \bar A^T \cdot d \bar x$$
    $$2) \, d B = (\nabla \cdot \bar B) \star 1 \quad, B:= \bar B^T \cdot \star d \bar x $$
    Here ##{}^T## is the transpose and ##\star## the Hodge-operator.
     
    Last edited: Jun 26, 2014
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