# Finding a vector potential for a solenoidal vector field

I have to find a vector potential for $$F = -y \hat{i} + x \hat{j}$$

This is what I have done:

We know that, if $$\nabla \cdot F = 0$$, we can construct the following:

$$F= \nabla\times G$$

Where $$G$$ is the vector potential we want to find out.

We know what F is, so it is just about doing the following:

$$\frac{\partial G_3}{\partial y} - \frac{\partial G_2}{\partial z} = -y$$

$$\frac{\partial G_1}{\partial z} - \frac{\partial G_3}{\partial x} = x$$

Noting that the partial derivatives with respect to $$z$$ are zero in this case, we get:

$$G = \frac{-x^2-y^2}{2}+C$$

Where $$C$$ is just the gradient of any scalar.

I am given a whole list of possible vector potentials:

Now I could use the most brute method: Trial and error with each possible vector potential given, using the equation:

$$G_n = \frac{-x^2-y^2}{2}+C$$

Solving for $$C$$ and seeing whether it holds.

This is pretty tedious; is there any brightest method?

Thanks.

EDIT

$$\frac{\partial G_2}{\partial z} = y$$

$$\frac{\partial G_1}{\partial z} = x$$

$$\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} = 0$$

I get:

$$G = $$

Which indeed satisfies:

$$F= \nabla\times G$$

But this option is not in the list...

Now let's set $$G_2 = 0$$:

$$\frac{\partial G_3}{\partial y} = -y$$

$$\frac{\partial G_1}{\partial z} - \frac{\partial G_3}{\partial x} = x$$

$$\frac{\partial G_1}{\partial y} = 0$$

I get:

$$G = <0,0,\frac{-x^2 - y^2}{2}>$$

Which indeed satisfies:

$$F= \nabla\times G$$

• Your assumption that the partial derivatives with respect to $z$ are $0$ is incorrect, and you can clearly see that by seeing that the answers have nonzero partial derivatives with respect to $z$. Take a look at this to see whats going on in the background: galileo.math.siu.edu/Courses/251/S12/vpot.pdf . In fact you will find that you can make $G_3 = 0$ by a choice of constant – Hushus46 Mar 24 at 13:45
• I mean $G_2 = 0$, and the if you want to know which answer is correct it is choice G where $G = <xz,0,-\frac{1}{2}y^2>$ – Hushus46 Mar 24 at 13:53
• @Hushus46 Thank you, that pdf is really helpful! I will delve into it and if I don't get what is going on I will ping you again. – JD_PM Mar 24 at 14:08
• Yea sure, let me know. You will see on the last page of the PDF the two exercises to repeat when $G_2 = 0$ and $G_1 = 0$. The point is that there are multiple $G$'s with their $\nabla \times G$ equal to $F$. Also I think there is a typo in the middle of the first page where it is explaining the choice of $H$, it should be $f(x,y,z)$ instead of $f(x,y,x)$ – Hushus46 Mar 24 at 14:10
• If the difference between any two potentials $G$ has gradient $0$, then those two $G$'s are basically the same and differ by a "constant" in the same sense from one-dimensional calculus that there are multiple choices for $\int f = F +C$ – Hushus46 Mar 24 at 14:14

As has been shown to OP already, this link gives a certain method to determine $$G$$.

If $$\mathbf{G}=(G_1,G_2,G_3)$$ can be decomposed into another potential $$\mathbf{H}=(H_1,H_2,H_3)$$ and the gradient of a scalar function $$f(x,y,z)$$, i.e

$$\mathbf{G} = \mathbf{H} + \nabla f$$

This implies that

$$\nabla \times \mathbf{G} = \nabla \times (\mathbf{H} + \nabla f) = \nabla \times \mathbf{H} + \nabla \times(\nabla f) = \nabla \times \mathbf{H} + \mathbf{0} = \nabla \times \mathbf{H}$$

Hence $$\mathbf{G}$$ is not unique and one can make specific choices to determine $$\mathbf{G}$$.

If we make the choice such that $$\frac{\partial f}{\partial z} = -H_3$$

Then $$\mathbf{G}=(H_1,H_2,H_3) +(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},-H_3) = (H_1+\frac{\partial f}{\partial x},H_2+\frac{\partial f}{\partial y},0) = (G_1,G_2,0)$$.

So we can choose $$\mathbf{G}$$ such that it can be either

\begin{align} &(0,G_2,G_3) \text{ or}\\ &(G_1,0,G_3) \text{ or}\\ &(G_1,G_2,0) \end{align}

So let us see what these choices can produce. We have the equations of $$\nabla \times \mathbf{G} = \mathbf{F}$$,

\begin{align} &\frac{\partial G_3}{\partial y} - \frac{\partial G_2}{\partial z} = -y \\ &\frac{\partial G_1}{\partial z} - \frac{\partial G_3}{\partial x} = x \\ &\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} = 0 \\ \end{align}

If $$G_1 = 0$$, then we have

\begin{align} &\frac{\partial G_3}{\partial y} - \frac{\partial G_2}{\partial z} = -y \\ -&\frac{\partial G_3}{\partial x} = x \Rightarrow G_3 = -\frac{x^2}{2}+C_3(y,z)\\ &\frac{\partial G_2}{\partial x} = 0 \Rightarrow G_2 = C_2(y,z) \\ \end{align}

substituting the last two equations into the first, we get

$$\frac{\partial G_3}{\partial y} - \frac{\partial G_2}{\partial z} = [C_3(y,z)]_y - [C_2(y,z)]_z = -y$$

Here, for simplicity, we can choose $$C_3(y,z)=0$$, because if $$C_2(y,z)=0$$ then two components of $$\mathbf{G}$$ are $$0$$, which never happens in the given possiblities. So, $$-[C_2(y,z)]_z = -y \Rightarrow C_2(y,z)=yz$$

Then $$\boxed{\mathbf{G} = (0,-yz,-\frac{x^2}{2})}$$ which can be verified to satisfy $$\nabla \times \mathbf{G} = \mathbf{F}$$

If $$G_2 = 0$$, then we have

\begin{align} &\frac{\partial G_3}{\partial y} = -y \Rightarrow G_3 = -\frac{y^2}{2} + C_3(x,z)\\ &\frac{\partial G_1}{\partial z} - \frac{\partial G_3}{\partial x} = x \\ -&\frac{\partial G_1}{\partial y} = 0 \Rightarrow G_1 = C_1(x,z) \\ \end{align}

then we get

$$\frac{\partial G_1}{\partial z} - \frac{\partial G_3}{\partial x} = [C_1(x,z)]_z - [C_3(x,z)]_x = x$$

Here, for simplicity, we can choose $$C_3(x,z)=0$$, for the same reason being we dont want two components of $$\mathbf{G}$$ to be $$0$$ $$[C_1(x,z)]_z = x \Rightarrow C_1(x,z)=xz$$

Then $$\boxed{\mathbf{G} = (xz,0,-\frac{y^2}{2})}$$ which can be verified to satisfy $$\nabla \times \mathbf{G} = \mathbf{F}$$

If $$G_3 = 0$$, then we have

\begin{align} -& \frac{\partial G_2}{\partial z} = -y \Rightarrow G_2 = yz + C_2(x,y)\\ &\frac{\partial G_1}{\partial z} = x \Rightarrow G_1 = xz +C_1(x,y) \\ &\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} = 0 \\ \end{align}

then we get

$$\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} = [C_2(x,y)]_x - [C_1(x,y)]_y = 0 \Rightarrow [C_2(x,y)]_x = [C_1(x,y)]_y$$

Here, for simplicity, we can choose $$C_2(x,y)=C_1(x,y)=0$$, and so $$\boxed{\mathbf{G} = (xz,yz,0)}$$ which can be verified to satisfy $$\nabla \times \mathbf{G} = \mathbf{F}$$

Of the three boxed solutions, only $$\mathbf{G} = (xz,0,-\frac{y^2}{2})$$ is on our list, and hence it is our answer.

However, given that we have a list of options, one can arrive to this answer earlier by realizing that certain solutions of the form $$(xz, f(y),0)$$ will never satisfy the curl equation, so the answer will be in the form$$(G_1,0,G_3)$$

• I have to write that this is an enlightening answer. – JD_PM Mar 24 at 20:02