How to get a vector field from its rotation field?

Question

Suppose there is a vector field $\mathbf{V}$ and its rotation field is given as $\mathbf{R}$. The relationship between the two vector fields is as follow: $$\mathbf{R}=\nabla \times\mathbf{V}$$

My question is how to get the original vector field $\mathbf{V}$ from $\mathbf{R}$?

It is not possible: If $K$ is a vector field such that $\nabla\times K=0$, then $\nabla \times (V+K) = R$. You cannot recover $V$ from $R$. — , Mar 03 '14 at 09:54
@John I only want to find a arbitary vector field $\mathbf{V}$ which rotation field is $\mathbf{R}$. Because I want to the calculus ,please see this problem:http://math.stackexchange.com/questions/692558/prove-the-surface-moments-formula — maple, Mar 03 '14 at 10:24
The procedure is called Helmholtz decomposition. See http://en.wikipedia.org/wiki/Helmholtz_decomposition — David H, Mar 03 '14 at 10:45
$V$ is called the vector potential of $R$. See Wikipedia: Vector Potential. — user127096, Mar 03 '14 at 17:00

score 2 · Answer 1 · answered Mar 03 '14 at 11:54

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Formula valid when the domain is star-shaped with respect to the origin (${\rm div}\ {\bf R}=0$ is required): $${\bf V}(x)=\int_0^1 {\bf R}(tx)\times tx\,dt.$$

answered Mar 03 '14 at 11:54

Martín-Blas Pérez Pinilla

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Whoa. Where does this come from? Do you have a proof or a reference? – Mar 03 '14 at 12:09
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The essential trick is using the formula for the curl of a vector product. See http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.pdf. – Martín-Blas Pérez Pinilla Mar 03 '14 at 13:11

score 1 · Answer 2 · answered Mar 03 '14 at 11:40

Fourier transforms make things easier. We then have ${\bf{R(k)}}=i{\bf{k×V(k)}}$. The general form of $\bf{V(k)}$ is ${\bf{V(k)=k×A(k)+k}}Φ(k)$. Now $${\bf{R(k)}}=i{\bf{k×V(k)}}=i{\bf{k×(k×A(k))}}=i{k^2}{\bf{A(k)}}-i{\bf{kk⋅A(k)}}.$$ Without loss of generality we can assume that ${\bf{k.A(k)}}=0$ so $${\bf{A(k)}}=-i{k^{-2}}{\bf{R(k)}},$$ and $${\bf{A(x)∼∫dy(1/|x-y|)R(y)}}.$$ We see that an arbitrary $Φ({\bf{k}})$ is allowed. This is a familiar situation in electrodynamics where $Φ$ and $\bf{A}$ are known as the scalar and vector potential.

How to get a vector field from its rotation field?

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