# Why does Gauss's law for magnetism imply existence of a magnetic vector potential?

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.

• Actually pseudoscalar potential $\phi$ is widely used to solve magnetostatics problems. Nov 17, 2020 at 22:24
• 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 $\phi =0$ Nov 17, 2020 at 22:52
• Where does it say that on the page? Mar 21 at 16:52

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.)

• 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? Nov 20, 2020 at 18:49
• Is there any other proof of this without hodge duality, using standard vector calculus, as I'm struggling to understand why $\phi$ with a laplacian of 0 couldn't be added? Is it due to the fact that we aren't taking into account the curl? Or the fact that we assume the field vanishes at infinity? Mar 22 at 15:27
• Hi jensen paull. Thanks for the feedback. The issue is not Hodge duality -- it can in principle be formulated in either language. The point is that it is possible to choose an $A$ (as a function of $B$) so that $\phi$ is not needed. Note however that $A$ does generically not satisfy any particular BCs. Mar 22 at 18:20