A vector field $\mathbf{F}$ is called solenoidal if $\mathbf{\nabla} \cdot \mathbf{F}=0$, and this is an equivalent condition to the existance of some vector field $\mathbf{A}$ such that $\mathbf{\nabla} \times \mathbf{A}=\mathbf{F}$. Why is this true? How would we go about finding $\mathbf{A}$?
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1See this http://math.stackexchange.com/questions/129104/how-to-find-a-vector-potential-inverse-curl?rq=1 – rlartiga May 12 '14 at 21:45
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1And this http://math.stackexchange.com/questions/81405/anti-curl-operator – rlartiga May 12 '14 at 21:46
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And also http://en.wikipedia.org/wiki/Vector_potential – pointer May 12 '14 at 21:49
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While it's true that $\nabla \cdot \left(\nabla \times \mathbf{A} \right) = 0$, for any vector field $\mathbf{A}$, only under the correct assumptions does $\nabla \cdot \mathbf{F} = 0$ imply that there exists a vector field $\mathbf{A}$ such that $\nabla \times \mathbf{A} = \mathbf{F}$.
If $\mathbf{F}$ is defined on a contractible subset of $\Bbb{R}^3$ (in particular, all of $\Bbb{R}^3$ works), then one can find a vector potential $\mathbf{A}$. (This is the essence of the Poincaré Lemma for differential forms. In fact, the failure for some solenoidal vector fields to have any vector potential is the starting point of de Rham cohomology and algebraic topology.)
For instance, the field $\mathbf{F}$ defined on $\Bbb{R}^3 \setminus \{ \mathbf{0} \}$ by $$ \mathbf{F}(x, y, z) = \left( \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \right), $$ where $r^2 = x^2 + y^2 + z^2$ is solenoidal, but has no vector potential.
Sammy Black
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