Consider the position vector field $\vec{r}=(x,y,z)^T$. What would be a vector potential $\vec{A}$ for this field? I was thinking of something like $\vec{A}=(yz,zx,-xy)^T$, which gives $$\nabla\times A=(-2x,2y,0)^T.$$ But that is not exactly correct.

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    $\begingroup$ The vector $\mathbf r$ that you are writing is not solenoidal so it does not have a vector potential associated to it. $\endgroup$
    – Mauricio
    Mar 4 at 10:56
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    $\begingroup$ However it is irrotational, so you can write a potential $V$ such that $\mathbf r =-\nabla V$ $\endgroup$
    – Mauricio
    Mar 4 at 11:00

2 Answers 2


It is not possible to find a vector potential $\mathbf{A}$ such that $$\nabla\times\mathbf{A}=\mathbf{r}. \tag{1}$$

You can prove this by contradiction.
Assume (1) is possible, and apply the divergence operator ($\nabla\cdot$) to it. Then you get $$\underbrace {\nabla\cdot\nabla\times\mathbf{A}}_{=0} =\underbrace {\nabla\cdot\mathbf{r}}_{=3}.$$

This is obviously a contradicction, and hence (1) is not possible.

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    $\begingroup$ I don't see the point in giving three almost identical answers, so I deleted mine because you posted yours some seconds earlier :). To complement this, the OP should note that it is easy to see why the divergence of a curl must be zero, using index notation and the Levi-Civita symbol: $$\nabla\cdot(\nabla\times\vec{A}) = \partial_i(\epsilon^{ijk}\partial_j A_k) = \epsilon^{ijk}\partial_i\partial_j A_k = 0$$ where repeated indices are summed over, and because $\epsilon^{ijk} = -\epsilon^{jik}$ while $\partial_i\partial_j = \partial_j\partial_i$. $\endgroup$
    – Albert
    Mar 4 at 11:18

The linear radial field has $$\nabla \cdot (x,y,z)=3$$ It follows, it can't be the curl of another field $a$ because of $$\nabla\cdot (\nabla \times a) = (\nabla \times \nabla) \cdot a= \sum_{ikl}\ \partial_i \epsilon_{ikl}\partial_k a_l=0 $$ because the sum is symmetric in the derivatives but antisymmetric by $\epsilon$.

Instead, the radial field is the gradient of the scalar potential $\frac{ r^2}{2}$ with the same kind of symmetry/antisymmtry yielding $\nabla \times \vec r =0$ , as for any gradient field from $\nabla \times \nabla =0$.


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