3
$\begingroup$

Gauss's law for magnetism is $\vec{\nabla} \cdot \vec{B} = 0$, and I read that this implies we can write $\vec{B} = \vec{\nabla} \times \vec{A}$, for some vector potential $\vec{A}$.

I understand that having $\vec{B} = \vec{\nabla} \times \vec{A}$ implies the statement of Gauss's law. But I am having trouble proving the converse; that $\vec{\nabla} \cdot \vec{B} = 0$ implies the magnetic field must take the form $\vec{B} = \vec{\nabla} \times \vec{A}$.

My attempt: By the Helmholtz decomposition theorem (https://en.wikipedia.org/wiki/Helmholtz_decomposition), we can write any vector field $\vec{B} = -\nabla \phi + \vec{\nabla} \times \vec{A}$. Then $\vec{\nabla} \cdot \vec{B} = 0 \Rightarrow -\nabla^2 \phi = 0$. However, this does not imply that $\phi \equiv 0$. So why can't magnetic fields be described in terms of some vector potential $\vec{A}$ but also some scalar potential $\phi$ that has $0$ Laplacian?

I don't think this has anything to do with the existence of a magnetic monopole or not, since I'm taking $\vec{\nabla} \cdot \vec{B} = 0$, but I may also be wrong.

$\endgroup$
2
  • $\begingroup$ Actually pseudoscalar potential $\phi$ is widely used to solve magnetostatics problems. $\endgroup$ – Alex Trounev Nov 17 '20 at 22:24
  • $\begingroup$ look at the webpage carefully again, you will see that when $\nabla \cdot \mathbf{F}=0$ then $\Phi=0$, for EM $\mathbf{B} = \mathbf{F}$ and there follows $\phi =0$ $\endgroup$ – hyportnex Nov 17 '20 at 22:52
2
$\begingroup$
  1. Use Hodge duality to rewrite the magnetic vector field $\vec{B}$ as a 2-form $B=\frac{1}{2}\epsilon_{ijk}B^i\mathrm{d}x^j\wedge\mathrm{d}x^k$.

  2. Gauss's law for magnetism then becomes the fact that the 2-form is closed $\mathrm{d}B=0$.

  3. It then follows from Poincare Lemma that the 2-form $B=\mathrm{d}A$ is exact, where $A=A_i\mathrm{d}x^i$ is a 1-form.

  4. This translates back into the sought-for formula $\vec{B}=\vec{\nabla}\times \vec{A}$.

  5. (Poincare Lemma fails for a magnetic monopole because the punctured 3-space $\mathbb{R}^3\backslash\{0\}$ is not contractible.)

$\endgroup$
1
  • $\begingroup$ Thanks, I see that $dB = 0$ implies the existence of some $A$ s.t. $B = dA$, but I guess my question is, how do we know we can't add some extra term (not of the form $dA$) to $B$ such that $dB = 0$ still holds? I am not too familiar with Hodge duality; it might help me to see what the Hodge dual for the expression $\vec{\nabla} \phi$ would be? $\endgroup$ – gwtw14 Nov 20 '20 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.