Finding a vector from the curl of a vector

[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.]

 

[fresh_42]

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?

 

[user1139]

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.

 

[fresh_42]

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.

 

[user1139]

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

 

[fresh_42]

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 $x\triangleq\dfrac{\partial A_1}{\partial x_1} ,\ldots$

 

[user1139]

The matrix is singular and hence, cannot be inverted.

 

[fresh_42]

The matrix is singular and hence, cannot be inverted.

Then you cannot get $A.$

 

[pasmith]

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.

 

[robphy]

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

 

[Charles Link]

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.

 

[Charles Link]

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.