# Vector potential of position field

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.

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

It is not possible to find a vector potential $$\mathbf{A}$$ such that $$\nabla\times\mathbf{A}=\mathbf{r}. \tag{1}$$
Assume (1) is possible, and apply the divergence operator ($$\nabla\cdot$$) to it. Then you get $$\underbrace {\nabla\cdot\nabla\times\mathbf{A}}_{=0} =\underbrace {\nabla\cdot\mathbf{r}}_{=3}.$$
• 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 $\partial_i\partial_j = \partial_j\partial_i$. 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 $$\nabla \times \nabla =0$$.