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Calculus: I can't understand why curl of gradient of a scalar is zero

  1. Sep 5, 2008 #1
    (Sorry, the title should read "...why curl of gradient of a scalar "function" is zero)

    Of course I know how to compute curl, graident, divergence. Algebrically I know curl of gradient of a scalar function is zero.

    But I want to know the reason behind this...and also the reason why gradient of divergence of a vector function is always zero.

    This really makes me feeling bad for a long time. Thanks in advance.
  2. jcsd
  3. Sep 5, 2008 #2


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    The gradient of a scalar function would always give a conservative vector field. Now think carefully about what curl is. If you've done an E&M course with vector calculus, think back to the time when the textbook (or your course notes) derived [tex]\nabla \times \mathbf{H} = \mathbf{J}[/tex] using Ampere's circuital law. What is the closed path integral of a conservative field?

    I'm wondering about your second question too...
  4. Sep 5, 2008 #3


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    Is this true? In Gauss's law, the divergence of the electric field is to equal an arbitrary charge density. I would be surprised if the gradient of an arbitrary charge density is zero.
  5. Sep 5, 2008 #4

    Ben Niehoff

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    [tex]\nabla(\nabla \cdot \vec F) = 0[/tex]

    is certainly false. I think you mean

    [tex]\nabla \cdot (\nabla \times \vec F) = 0[/tex]

    which is true.
  6. Sep 5, 2008 #5


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    Hi chingcx! :smile:

    (curl grad f)x = ∂/∂y(∂f/∂z) - ∂/∂z(∂A/∂y) = 0

    Similarly, div curl A = 0
    But (grad div A)x = ∂/∂x(∂Ax/∂x) + ∂/∂x(∂Ay/∂y) + ∂/∂x(∂Ax/∂z) ≠ 0 :smile:
  7. Sep 5, 2008 #6


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    Hey you're right. Wow I can't believe I didn't even bother thinking about whether it might be correct, as opposed to why it might be correct.
  8. Sep 5, 2008 #7
    ya, sorry, I mean divergence of curl of vector function is always zero. Why is that true?
  9. Sep 5, 2008 #8
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