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Just a simple question, from the title of the thread, is a vector field = 0 if the divergence is 0 and the curl is 0? I had trouble finding an answer anywhere online, proof of why or why not would be helpful, thanks!

$$\nabla \cdot \vec{F} = 0$$ $$\nabla \times \vec{F} = 0$$ $$\vec{F} = ?$$

Spaderdabomb
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  • sorry i dont know how to write the $del $symbol. if $\Delta .F=0$ and $\Delta*F=0$ implies $$F_x+F_y+F_z=0$$ it does not imply $F=0$ – Suraj M S Dec 07 '13 at 06:21
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    @ Surah M.S.: latex for $\Delta$, is "\Delta"; for $\nabla$, it's "\nabla", the centralized dot, $\cdot$, is "\cdot". So $\nabla \cdot F$ is "\nabla \cdot F". You might find it helpful to check out http://en.wikibooks.org/wiki/LaTeX/Mathematics – Robert Lewis Dec 07 '13 at 06:37

4 Answers4

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No. To get a feel for what's up, let $\phi$ be a non-constant harmonic function so $\nabla^2 \phi = 0$ but $\nabla \phi \ne 0$. Set $\mathbf F = \nabla \phi \ne 0$; then Then $\nabla \times \mathbf F = \nabla \times \nabla \phi = 0, \,$ since the curl of a gradient always vanishes. Also, $\nabla \cdot \mathbf F = \nabla \cdot \nabla \phi = \nabla^2 \phi = 0$. The divergence and curl of $\mathbf F$ both vanish, but not $\mathbf F$!

This line of reasoning can, like tape or film, be re-wound and run "backwards": if $\mathbf F \ne 0$ and $\nabla \times \mathbf F = 0$, then (locally at least) there is a function $\phi$ with $\mathbf F = \nabla \phi \ne 0$; if now we also have $\nabla \cdot \mathbf F = 0$, then $\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \mathbf F = 0$, and $\phi$ is harmonic.

The classic examples of such a field may be found in the elementary theory of electromagnetism: in the absence of sources, that is, charges and currents, static (non -time varying) electric fields $\mathbf E$ and magnetic fields $\mathbf B$ have vanishing divergence and curl: $\nabla \times \mathbf B = \nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf B = \nabla \cdot \mathbf E = 0$; the electrostatic potential function $\phi$ such that $\mathbf E = -\nabla \phi$ with $\nabla^2 \phi = 0$ exists by virtue of these facts; a similar assertion holds for the $\mathbf B$ field.

Hope this helps. Cheerio,

and as always (as ol' James Clerk M. has taught us),

Fiat Lux!!!

Robert Lewis
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  • Yes, "Fiat lux" is appropriate here. Could you explain why it keeps appearing elsewhere? – dfeuer Dec 07 '13 at 06:20
  • @ dfeuer: tell you what: check out the late W. Thurston's ansswer to this question: http://mathoverflow.net/questions/43690/whats-a-mathematician-to-do and get back to me! Regards, RKL – Robert Lewis Dec 07 '13 at 06:23
  • thanks! great answer, I even noticed I came across one of these types of fields shortly after in some E&M problems I was working on with constant potential boundaries =) – Spaderdabomb Dec 07 '13 at 08:00
  • @Spaderdabomb: and I thank you! Glad to help out. And yeah, these fields are all over the place! P.S. muchas gracias for the "acceptance" as well. – Robert Lewis Dec 07 '13 at 08:03
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No. Just consider a nonzero constant vector field, i.e. $$V=(a,b,c)$$ for some nonzero constants $a, b, c$. Then the curl of $V$ is zero vector, and the divergence of $V$ is $0$.

Paul
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For example $\vec{u} = (x+y, x-y, 0)$ has $\vec{\nabla} \cdot \vec{u}=0$ and $\vec{\nabla} \times \vec{u}=0$. See picture here: http://newtechno.altervista.org/documenti/Corso_di_fisica/campo1.gif

sTertooy
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Vincenzo Fiorentini
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No. For example, if we consider simple 2 dimension field,

$\vec{V} = a (x,-y) $ , for any scalar $a$ then

$ \vec{\nabla} . \vec{V} = 0 $ and $ \vec{\nabla} \times \vec{V} = 0 $

bjcolby15
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user196937
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