The forgotten magnetic scalar potential

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Delta2
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I wonder if there is any book that discusses the possibility of existence of a magnetic scalar potential. That is a scalar potential ##\chi## such that $$\vec{B}=\nabla\chi+\nabla\times\vec{A}$$. From Gauss's law for the magnetic field B we can conclude that it will always satisfy laplace's equation, therefore it will seemingly have no time dependence (even in the case of a time dependent vector potential A):
$$\nabla\cdot\vec{B}=0\Rightarrow \nabla\cdot(\nabla\chi+\nabla\times\vec{A})=0\Rightarrow\nabla^2\chi=0$$

Maxwell's Ampere's law seems to give us no additional information about this potential since it will be :
$$\nabla\times\vec{B}=\nabla\times (\nabla\chi+\nabla\times\vec{A})=\nabla\times(\nabla\chi)+\nabla\times(\nabla\times\vec{A})=0+\nabla\times(\nabla\times\vec{A})$$

So can we prove somehow that in most cases we have ##\chi=0## . Are there any specific systems known that ##\chi\neq 0##?
 

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vanhees71
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The point is that you overdetermine ##\vec{B}## by introducing an additional scalar potential. Of course you can always shuffle between potential and solenoidal parts of a field, but using Helmholtz's theorem together with the usual boundary conditions at infinity makes ##\vec{B}## a pure solenoidal field due to Gauss's Law for the magnetic field, ##\vec{\nabla} \cdot \vec{B}=0##.

Nevertheless, in magnetostatics, it's often convenient to introduce a scalar potential for the magnetic field in the region, where no currents are present, because there
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{j}=0.$$
You usually get a potential that's singular along a plane. Take, e.g., the infinitely long cylindrical wire. Outside of the wire you have ##\vec{j}=0## and thus you can make the ansatz ##\vec{B}=-\vec{\nabla} \chi##. Then you must have ##\vec{\nabla} \cdot \vec{B}=-\Delta \chi=0##. Due to symmetry in cylinder coordinates ##(r,\varphi,z)## you must have ##\chi=\chi(\varphi)##. This leads to
$$\Delta \chi=\frac{1}{r^2} \partial_{\varphi}^2 \chi=0 \; \Rightarrow \; \chi(\varphi)=C \varphi.$$
Then
$$\vec{B}=-C \vec{\nabla} \varphi=-\frac{C}{r} \vec{e}_{\varphi},$$
which is of course the correct solution. The potential is well-defined everywhere except along an arbitrary half-plane with the ##z##-axis as boundary, depending on which interval of length ##2 \pi## you choose for the domain of ##\varphi##.

The integration constant ##C## is defined by the total current through the cylinder, using the integral form of Ampere's Law along a circle in an arbitrary plane perpendicular to the wire
$$\int_C \mathrm{d} \vec{r} \cdot \vec{B}=-2 \pi C=i \; \Rightarrow\; C=-\frac{i}{2 \pi},$$
leading finally to
$$\vec{B}=\frac{i}{2 \pi r} \vec{e}_{\varphi}.$$
 
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