Deriving the vector field from a curl

Question

If $B= (x/r^3,y/r^3,z/r^3)$ find a vector $A$ such that $curl A = B$. My attempt:

$∂A_3∂Y−∂A_2∂Z=x/r^3$

$∂A_1∂Z−∂A_2∂y=y/r^3$

$∂A_2∂x−∂A_1∂y=z/r^3$

I assumed $A_3=0$, ending up having to solve:

$A_2=\int xr^{-3}dz$

Here, I got stuck. Is there an easier way of doing this?

Answer 1

I assume your book asks for any solution even though physically meaningless. Take for instance $A_3 = 0$. Then

\begin{equation} B_1 = \frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z} = \frac{x}{(\sqrt{x^2+y^2+z^2})^3} \end{equation} Thus, \begin{equation} A_2 = -\int \frac{x}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{-xz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}} \end{equation} Similarly, \begin{equation} B_2 = \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x} = \frac{y}{(\sqrt{x^2+y^2+z^2})^3} \end{equation} Thus, \begin{equation} A_1 = \int \frac{y}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{yz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}} \end{equation}

---

Solving $\nabla \times \mathbf{A} = \mathbf{B}$ is a coupled partial differential equation, which is not complete unless boundary conditions are given to make solution $\mathbf{A}$ unique. Otherwise, the following answer assumes $\mathbf{A}$ is zero on boundary (decays at infinity).

By Helmholtz decomposition, every vector field $\mathbf{B}$ is decomposed to a curl-free and divergence-free component, by \begin{equation} \mathbf{B} = \nabla \times \mathbf{A} - \nabla \phi, \end{equation} where $\phi$ is the potential (the curl-free part, since $\nabla \times (\nabla \phi) = 0$), and $\mathbf{A}$ is vector potential (the divergence-free part, since $\nabla \cdot (\nabla \times \mathbf{A}) = 0$). Here

\begin{equation} \mathbf{B} = \mathbf{0} - \nabla \left(\frac{1}{\sqrt{x^2 + y^2 + z^2}} \right) = -\nabla \left(\frac{1}{r}\right). \end{equation}

It means $\mathbf{B}$ can be decomposed into a potential field $\phi=r^{-1}$ and no vector potential $\mathbf{A}$.

The potential is unique upto a constant, $\phi = \frac{1}{r}+c$ for any constant $c$. The vector potential is unique up to a gradient field, $\mathbf{A} = \mathbf{0} + \nabla \psi$ for any scalar field $\psi$.

Indeed, $\mathbf{B} = \frac{1}{r^2}\mathbf{e}_r$ is the Coulomb force, which is curl free and represented by potential above.

---

Answer 2

It's not possible to construct a solution of $\nabla \times \mathbf A = \mathbf r/r^3$ that would be valid in $\mathbb R^3 \setminus \{(0, 0, 0)\}$. Otherwise, taking $\mathcal S$ to be a sphere centered at the origin, we would get $$\int_{\mathcal S} \mathbf B \cdot d\mathbf S = 4 \pi = \int_{\partial \mathcal S} \mathbf A \cdot d\mathbf s = 0.$$ If we want a solution in a region not containing a closed surface around the origin, we can take $\mathbf A = A_\theta \mathbf e_\theta$: $$\nabla \times (A_\theta \mathbf e_\theta) = -\frac 1 {r \sin \theta} \frac {\partial A_\theta } {\partial \phi} \mathbf e_r + \frac 1 r \frac {\partial (r A_\theta)} {\partial r} \mathbf e_\phi.$$ The second term disappears if $A_\theta = f(\theta, \phi)/r$. Then we can set $f(\theta, \phi) = -\phi \sin \theta$.