# Finding a vector from the curl of a vector

• user1139
In summary, the equation ##\nabla\phi=\nabla\times \vec{A}## can be rewritten in component form, but solving for ##\vec{A}## is not straightforward due to the coupling of equations. However, an integral solution similar to the Biot-Savart equation may be used, along with potential boundary conditions, to solve for ##\vec{A}##.f

#### user1139

Consider the following

\begin{equation}
\nabla\phi=\nabla\times \vec{A}.
\end{equation}

Is it possible to find ##\vec{A}## from the above equation and if so, how does one go about doing so?

[Moderator's note: moved from a homework forum.]

Last edited by a moderator:
Consider the following

\begin{equation}
\nabla\phi=\nabla\times \vec{A}.
\end{equation}

Is it possible to find ##\vec{A}## from the above equation and if so, how does one go about doing so?

[Moderator's note: moved from a homework forum.]
What do you get if you write this in components?

Working with Cartesian coordinates, I will be able to equate the respective components on the LHS and RHS. The problem comes when I want to find ##A_x##, ##A_y## and ##A_z## since the equations are now coupled.

Working with Cartesian coordinates, I will be able to equate the respective components on the LHS and RHS. The problem comes when I want to find ##A_x##, ##A_y## and ##A_z## since the equations are now coupled.
Yes, but it is still a linear equation system.

Do you know how to solve it? I ran out of ideas.

You have ##c_1=z-y\, , \,c_2=x-z\, , \,c_3=y-x## which is ##\begin{bmatrix}
0&-1&1\\1&0&-1\\-1&1&0
\end{bmatrix}\cdot \begin{bmatrix}
x\\y\\z\end{bmatrix}=\begin{bmatrix}
c_1\\c_2\\c_3
\end{bmatrix}##
so invert the matrix and solve it. Of course you only get ##x\triangleq\dfrac{\partial A_1}{\partial x_1} ,\ldots##

The matrix is singular and hence, cannot be inverted.

The matrix is singular and hence, cannot be inverted.
Then you cannot get ##A.##

Consider the following

\begin{equation}
\nabla\phi=\nabla\times \vec{A}.
\end{equation}

Is it possible to find ##\vec{A}## from the above equation and if so, how does one go about doing so?

[Moderator's note: moved from a homework forum.]

Let $A = \nabla \times F + \nabla \psi$. Then $$\nabla \times A = \nabla(\nabla \cdot F) - \nabla^2F = \nabla \phi.$$ Adding a gradient to $F$ does not change its curl, which is all we care about, but does change its divergence, so we can assume $\nabla \cdot F = 0$. That results in $F$ satisfying Poisson's equation $$\nabla^2 F = - \nabla \phi.$$ This is three decoupled equations in the cartesian components of $F$. There is no way to determine $\psi$ since $\nabla^2 \psi = \nabla \cdot A$ is not specified.

• PeroK
If there is any hope, I think you would also need some boundary conditions.

There is an integral solution to ## \nabla \times A =\nabla \phi ##. It is basically a Biot-Savart integral type solution to ## \nabla \times B=\mu_o J ##, to solve for ##B ##. There is also the possibility of a homogeneous solution, (## \nabla \times A=0 ##), so that the solution for ## A ## is not unique by this method.

• PeroK
To write out the above Biot-Savart type solution, ## A(x)=\int \frac{\nabla \phi \times (x-x')}{4 \pi |x-x'|^3} \, d^3x' ##, where ##x ## and ## x' ## are 3 dimensional coordinate vectors.