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Intuitively understanding div(curl F) = 0

  1. Apr 4, 2012 #1
    Hello all!

    I have been reviewing my vector calculus coursework as of late, and this time around, I've been really trying to understand the concepts intuitively/visually instead of just the math. Unfortunately, the identity div(curl F)=0 is giving me trouble.

    I understand divergence is a measure of a vector field's compressibility. I understand curl is a vector field, representing F's rate of rotation. What I'm having a hard time visualizing is why curl F always produces a vector field that is incompressible?

    Thanks for any help you can provide!
  2. jcsd
  3. Apr 4, 2012 #2
    If you know about exterior derivatives, then this identity is equivalent to [itex]d^2 = 0[/itex].

    To give a "physical picture" for this identity, first use Gauss's Theorem for [itex]\mathrm{div}(\mathrm{curl} \, \vec{F})[/itex] to get:
    \int_{\Omega}{d^3x \, \mathrm{div}(\mathrm{curl} \, \mathrm{F})} = \oint_{\Sigma}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})}
    where [itex]\Omega[/itex] is an arbitrary volume element bounded by the closed surface [itex]\Sigma[/itex], with an exterior unit normal [itex]\hat{n}[/itex] everywhere.

    Next, cut the boundary surface [itex]\Sigma[/itex] in two by a plane. Let the intersection of the plane with the boundary surface be a curve C, and let the two halves of the boundary surface be [itex]\Sigma_1[/itex], and [itex]\Sigma_2[/itex]. We would have:
    \oint_{\Sigma}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})} = \iint_{\Sigma_1}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})} + \iint_{\Sigma_2}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})}.
    Now, we can apply Stokes's Theorem for each of these surface integrals. However, notice that if we go along the curve in the clockwise direction, then, according to the right-hand rule, the normal to the surface would always be exterior to one of the surfaces, whereas it would be interior to the other of the surface [itex]\Sigma_1[/itex], and [itex]\Sigma_2[/itex]. Since [itex]\hat{n}[/itex] is the exterior normal on both of them, we need to circumvent the loop C in opposite directions, and the integrals become:
    \iint_{\Sigma_1}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})} + \iint_{\Sigma_2}{da \, \hat{n} \cdot (\mathrm{curl} \, \vec{F})} = \oint_{C^{+}}{d\vec{l} \cdot \vec{F}} + \oint_{C^{-}}{d\vec{l} \cdot \vec{F}} = 0
    since changing the direction of circumvention in a line integral of the second kind changes the sign in front of it.
  4. Apr 4, 2012 #3
    Take a simple but nontrivial vector field with a nonconstant curl.
    [itex] F = \hat{x} y^2[/itex].

    The curl of this is [itex] 2y\hat{z} [/itex].

    This tells you that the curl in this case points in a direction which is perpendicular to the direction of the original vector and perpendicular to the direction containing the spatial variation. (this makes sense if you think of the curl as "nabla cross F")

    The divergence operator on the other hand is measuring variation along the unit vector; how the z component varies along z, and so on.

    The divergence of [itex] 2y\hat{z} [/itex] is zero because the variation of the z component of the vector is zero along the z direction.

    This is not rigorous of course but should give you a sense of the "perpendicularity" of "nabla dot curl".
    Last edited: Apr 4, 2012
  5. Apr 5, 2012 #4
    Dickfore and Antiphon, thank you both so much for your thoughtful replies.

    Antiphon, this was precisely what I was looking for! My gut told me something like this was what was going on when I took the curl of a vector field, but I've always had trouble visualizing things in 3D.

    Thank you both once again!
  6. Apr 5, 2012 #5


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    Dearly Missed

    Or, even more imprecisely:
    The curl tells you how a vector "moves" WITHIN a tiny disk (i.e local rotation in some specified plane the disk lies in), whereas the divergence tells how the vector "moves" ACROSS the disk.
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