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Usually it is said that the Maxwell equation $\vec \nabla . \vec{B}=0$ is solved by introducing the vector potential according to $\vec B=\vec \nabla \times \vec A$.

In principle, we could write the more general decomposition $\vec B=\vec \nabla \times \vec A+\vec \nabla f$ and require $\nabla^2 f=0$, right?

Why is this never done? Is it a particular case of Gauge invariance?

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    $\begingroup$ This freedom is already present in the gauge freedom of $A$, so there's no point in writing it as a separate term. $\endgroup$
    – knzhou
    Commented Apr 6, 2017 at 20:27
  • $\begingroup$ Gauge transformations $\vec A\to \vec A+\vec\nabla g$ leave the magnetic field invariant and purely rotational. My question involves a magnetic field which has a gradient term, so its not simply the curl of $\vec A$. $\endgroup$
    – thedude
    Commented Apr 6, 2017 at 22:55
  • $\begingroup$ In other words, suppose I have a vector potential $\vec A$ which produces the magnetic field $\vec B$. Is there another potential which is simply a Gauge transform of $\vec A$ and produces the field $\vec B+\vec\nabla f$? $\endgroup$
    – thedude
    Commented Apr 6, 2017 at 22:57
  • $\begingroup$ No, the point is that the gauge transformation keeps $B$ the same. If you change $B$, you've got a different physical situation. $\endgroup$
    – knzhou
    Commented Apr 6, 2017 at 22:58
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    $\begingroup$ Oh, I see the real issue. Generally this freedom is removed by boundary conditions. For example, if you're working in $\mathbb{R}^3$ demanding the field vanish quickly at infinity rules out $f$. $\endgroup$
    – knzhou
    Commented Apr 6, 2017 at 23:17

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It follows from the fact that $\mathrm{div} B = 0 \iff B = \mathrm{rot} A$, unless you wish to work with some nontrivial manifolds where $dA = 0$ is not the same as $A = dB$.

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  • $\begingroup$ Although I am aware that $dB=0$ should imply $B=dA$, I have difficulty understanding the present situation. If I write $\vec B=\vec\nabla\times \vec A+\vec\nabla f$ and impose $\nabla^2 f=0$, does it not still follow that div$B=0$? $\endgroup$
    – thedude
    Commented Apr 7, 2017 at 0:37
  • $\begingroup$ Yes, it does. It also follows that $\vec B=\vec\nabla\times \vec A+\vec\nabla f = \vec\nabla\times \vec A_1$ So, presenting solution in your way is just unnecessarily difficult, as presenting solution of x - 2 = 0 as x = 2 + (3 - 3). That is why people usually choose the easiest way. $\endgroup$
    – user108687
    Commented Apr 7, 2017 at 0:43
  • $\begingroup$ Could you please write $A_1$ explicitly in terms of $A$ and $f$? $\endgroup$
    – thedude
    Commented Apr 7, 2017 at 0:50
  • $\begingroup$ No, I could not. I can only claim, that such $A_1$ exists. Do you have a problem with understanding that in 3D $dA = 0 \iff A = dB $ is the same as $\mathrm{div} B = 0 \iff B = \mathrm{rot} A$? $\endgroup$
    – user108687
    Commented Apr 7, 2017 at 1:03
  • $\begingroup$ No, I don't. I have thought some more about it and I'm fine. The fact that $A_1$ does not have an obvious form in terms of $A$ and $f$ was troubling me, but as you say it exists and that's it. Ok. Thanks. $\endgroup$
    – thedude
    Commented Apr 7, 2017 at 1:15

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