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Let $$\mathbf{B} = \mathbf{\nabla\times A},$$ where $$\mathbf{B}(x,y,z) = B_x(x,y,z)\mathbf{\hat{x}} + B_y(x,y,z)\mathbf{\hat{y}} + B_z(x,y,z)\mathbf{\hat{z}},$$ $$\mathbf{A}(x,y,z) = A_x(x,y,z)\mathbf{\hat{x}} + A_y(x,y,z)\mathbf{\hat{y}} + A_z(x,y,z)\mathbf{\hat{z}},$$ Assume that $$\mathbf{\nabla \cdot A} = 0.$$ This ensures that $\mathbf{A}$ is uniquely defined by the top equation.

Is it possible to calculate $A_x(x,y,z_0)$, $A_y(x,y,z_0)$ if we only know $\mathbf{B}(x,y,z)$ at $z=z_0$?

Here is my naive attempt. Assume that $$A_x(x,y,z_0) = \frac{\partial \phi}{\partial y},$$ $$A_y(x,y,z_0) = -\frac{\partial \phi}{\partial x},$$ where $$\phi=\phi(x,y).$$ Substituing this into the $z$-component of the top equation gives the following 2D Poisson equation $$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\phi(x,y) = -B_z(x,y,z_0).$$ We can solve this for $\phi(x,y)$ which we can use to calculate $A_x(x,y,z_0)$ and $A_y(x,y,z_0)$. Is this a valid solution? There seems to be problems. For example, since $$\mathbf{\nabla\cdot A} = 0,$$ this implies that $$\left.\frac{\partial A_z}{\partial z}\right|_{z=z_0}=0,$$ but $z_0$ is arbitrary so $\partial A_z / \partial z = 0$ $\forall z$?

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    $\begingroup$ The trick here is that if you're assuming a gauge "fixes" a solution, you have no more degrees of freedom to justify that arbitrary choice of $\phi$ but maybe there are more degrees of freedom. I am not so sure anymore. $\endgroup$ Mar 20, 2021 at 0:06

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Given $\mathbf{B}$ on $z=z_0$ we can extend the field to all on $\mathbb{R}^3$ by $$\begin{cases} B_x(x,y,z) := B_x(x,y,z_0) \\ B_y(x,y,z) := B_y(x,y,z_0) \\ B_z(x,y,z) := B_z(x,y,z_0) - z \left( \partial_x B_x(x,y,z_0) + \partial_y B_y(x,y,z_0)\right) \end{cases}$$ This extensin satisfies $\nabla\cdot \mathbf{B} = 0$ but is not the only possible extension.

Then, given $\mathbf{B}$ in all of $\mathbb{R}$ we can construct $\mathbf{A}$ such that $\nabla\times \mathbf{A} = \mathbf{B}$ and $\nabla\cdot \mathbf{A}=0$ by the convolution $$\mathbf{A} = -G * (\mathbf{\nabla\times B}),$$ where $\nabla^2 G = \delta,$ i.e. $G(x,y,z) = \frac{1}{4\pi\sqrt{x^2+y^2+z^2}}.$

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  • $\begingroup$ Thank you for your answer. Just to clarify, you're saying it's impossible to calculate $A_x$, $A_y$ at $z=z_0$ without extending $\mathbf{B}$ in $z$. As a follow-up question. In the OP I calculated $A_x$, $A_y$ by assuming they were of the form $$A_x(x,y,z_0) = \frac{\partial \phi}{\partial y},$$ $$A_y(x,y,z_0) = -\frac{\partial \phi}{\partial x}.$$ Do you know how this restricts the $z$-dependence of $\mathbf{B}$? $\endgroup$
    – Peanutlex
    Mar 24, 2021 at 15:27
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    $\begingroup$ @Peanutlex. At least we need to extend $\mathbf{A}$ and $\mathbf{B}$ to a small neighborhood of $z=z_0$ to be able to define $\nabla\times\mathbf{A}$ and $\nabla\cdot\mathbf{B}$. $\endgroup$
    – md2perpe
    Mar 24, 2021 at 17:59
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It doesn't appear so since we can, for example, add a constant field $\mathbf{A}_0$ to $\mathbf{A}$ so that $$\mathbf{A}_2=\mathbf{A}+\mathbf{A}_0$$ We now have that if $\mathbf{A}$ satisfies $$\nabla \cdot \mathbf{A}=0$$ then also $\mathbf{A}_2$ satisfies $$\nabla \cdot \mathbf{A}_2=0$$ since $$\nabla \cdot \mathbf{A}_2=\nabla \cdot (\mathbf{A}+\mathbf{A}_0)=\nabla \cdot \mathbf{A}+\nabla \cdot \mathbf{A}_0=0+0=0$$ So, you cannot "see" the value of the constant field $\mathbf{A}_0$ from $$\mathbf{B}=\nabla \times \mathbf{A}$$ since also $$\mathbf{B}=\nabla \times \mathbf{A}_2$$ due to $$\mathbf{B}=\nabla \times \mathbf{A}_2=\nabla \times \mathbf{A}+\nabla \times \mathbf{A}_0=\nabla \times \mathbf{A}+0=\nabla \times \mathbf{A}$$

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  • $\begingroup$ That's a good point. I suppose I was wrong when I said that assuming $$\mathbf{\nabla \cdot A}=0$$ ensures $\mathbf{A}$ is uniquely given by $$\mathbf{B}=\mathbf{\nabla \times A},$$ since we can just add a constant to $\mathbf{A}$ and both equations will still be satisfied. $\endgroup$
    – Peanutlex
    Mar 24, 2021 at 15:33

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