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With the Coulomb gauge $\nabla\cdot{\vec A}=0$, and $\nabla\times\vec A=\vec B$, the vector potential satisfies the Poisson's equation, $$\nabla^2{\vec A} = -\mu_0{\vec J},$$ which for $\vec J\to0$ at infinity, leads to the solution (Griffiths' electrodynamics, Eq. 5.65, (see [1]): $${\vec A}({\vec r})= \frac{\mu_0}{4\pi}\int\frac{{\vec J}({\vec r}\,')\,d^3r'}{|{\vec r}-{\vec r}\,'|}.$$ Since the Coulomb gauge does not fix ${\vec A}$ uniquely (see here), is the above solution unique? I think that the solution is actually $${\vec A}({\vec r})= \frac{\mu_0}{4\pi}\int\frac{{\vec J}({\vec r}')d^3r'}{|{\vec r}-{\vec r}'|}+\nabla\phi(\vec r)$$ where $\phi(\vec r)$ is a solution of $\nabla^2\phi=0$. I think that to set $\phi=0$, we need additional conditions on $\vec A$. Please explain what condition do we need to invoke, such that we can get Griffiths' solution.

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    $\begingroup$ $\phi$ needs to satisfy $\nabla^2 (\nabla\phi) = 0$? $\endgroup$
    – jim
    Jun 13, 2022 at 17:44
  • $\begingroup$ There are cases where griffiths solution yields incorrect results for example the magnetic vector potential of an infinitely straight wire. I'm pretty sure the boundary condition that you need to evoke to get griffiths solution is that $\vec{A}(\infty)=0$ I will try to find a question that may give you some insight. But you are correct, that there are infinitely many potentials that satisfy the coulomb gauge. $\endgroup$ Jun 13, 2022 at 18:18
  • $\begingroup$ @jensenpaull Does $\vec A\to 0$ at $|\vec r|\to \infty$ also make $\phi=0$ (or at least a constant) everywhere in space? $\endgroup$ Jun 13, 2022 at 18:23
  • $\begingroup$ physics.stackexchange.com/questions/684001/… look at J Murray answer, where he explains where you can get some issues with a divergent A, that is not valid! This is because the standard formula assumes A vanishes at infinity which is not the case for a situations. He then explains another way to solve for A given new boundary conditions. $\endgroup$ Jun 13, 2022 at 18:23
  • $\begingroup$ Well your question was "is the above solution unique", in general no. There are infinitely many solutions to A in the coulomb gauge. As for griffiths formula for $\vec{J}(\infty) = 0$, I don't think he claims it to be unique, since any solution will do provided your situation matches boundary conditions). However you are correct that any $\nabla \phi $function can be added provided $\nabla^2 \phi = 0$ and is still a solution whilst also satisfying the boundary conditions that $\vec{J}(\infty) = 0$. To make another condition that A vanishes at infinity aswell would require $\phi = 0$... $\endgroup$ Jun 13, 2022 at 18:31

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The Helmholtz Theorem says that a vector field $\mathbf{A}$ is uniquely determined by ${\rm div}\,\mathbf{A}$, ${\rm curl}\,\mathbf{A}$, and $\mathbf{A}\rightarrow0$ as $r\rightarrow\infty$. So, when dealing with a localized source, as described in the question, there is only a single transverse $\mathbf{A}$ that will vanish at spatial infinity.

This can be seen directly from the harmonic property of the $\phi$ in the question. If $\nabla\phi$ is to be added to $\mathbf{A}$ without changing $\nabla\cdot\mathbf{A}$, then, as pointed out, $\phi$ must satisfy $\nabla^{2}\phi=0$. However, if we are to have $\nabla\phi\rightarrow0$ as $r\rightarrow\infty$, then $\phi$ goes to a constant at spatial infinity. Since $\phi$ is a solution of Laplace's Equation, it has no local extrema. If $\phi$ goes to a constant value in every direction for $r\rightarrow\infty$, the only way it can have no local maxima or minima is if it is constant; thus, the $\nabla\phi$ added to $\mathbf{A}$ vanishes.

However, there are many reasonable source configurations for which the source $\mathbf{J}$ does not vanish at infinity, and so neither does $\mathbf{A}$. The simplest example is an infinite wire. There is no vector potential $\mathbf{A}$ for an infinite wire than vanishes as $r\rightarrow\infty$, since the source extends to spatial infinity. Thus, a point can be near to the source for arbitrarily large $r$. There are also situations where we may be given not $\mathbf{J}$ but $\mathbf{B}$. For example, a constant background magnetic field $\mathbf{B}_{0}$ everywhere cannot be written as the curl of a $\mathbf{A}$ that vanishes at infinity. It is because we want to account for the kinds of configurations discussed in this paragraph that we consider the Coulomb gauge not to be a complete gauge.

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  • $\begingroup$ Would like to point out, it is entirely possible for you to use the coulomb gauge to calculate the correct magnetic vector potential of an infinite wire. $\endgroup$ Jun 13, 2022 at 18:51

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