2
$\begingroup$

From electrodynamics we know that $\boldsymbol{\nabla}\mathbf{B}=\mathbf 0$ hence we can introduce a vector potential such that $\mathbf{B}=[\boldsymbol \nabla\times \mathbf{A}]$.

What is the general mathematical proof that a vector field with zero divergence can be expressed as a curl of another vector field (that this field would necessarily exist)? What are the restrictions on the fields?

I suspect this question was asked before, I just didn't manage to find a clean answer.

$\endgroup$
5
  • 5
    $\begingroup$ The answer is topological and basically given by de Rham’s theorem: zero divergence on a particular domain gives a vector potential on that domain if and only if the second homology of the domain is trivial. See also math.stackexchange.com/questions/3865242/… $\endgroup$ Oct 9, 2021 at 19:17
  • $\begingroup$ Is ther an accessible proof I can read? I don't know much about topology, my background is in physics with standard calculus, vector analysis etc $\endgroup$
    – Al Guy
    Oct 9, 2021 at 19:19
  • 4
    $\begingroup$ The proof of the “if and only if” is a deep result and quite difficult. Poincaré’s lemma gives a sufficient condition and is easier to prove but is not as strong. You might wish to glance at the American Mathematical Monthly paper referenced in the answer to the question I linked to above. $\endgroup$ Oct 9, 2021 at 19:20
  • $\begingroup$ Another way to put it is to identify sufficient conditions for$$\vec{\nabla}\cdot\vec{B}=0,\,\vec{A}\left(t,\,\vec{r}\right):=\frac{1}{4\pi}\int\frac{\vec{\nabla}\times\vec{B}\left(\vec{r}^\prime\right)}{\left|\vec{r}-\vec{r}^\prime\right|}d^3\vec{r}^\prime\implies\vec{\nabla}\times\vec{A}=\vec{B}.$$ $\endgroup$
    – J.G.
    Oct 9, 2021 at 20:08
  • 2
    $\begingroup$ Let me also be more precise: my comment above elides some quantifier subtleties. On a domain $D$, consider the property $P$ that “every divergence-free field on $D$ admits a vector potential on $D$.” There is a theorem (consequence of de Rham) that says $D$ has property $P$ if and only if it has trivial second homology. In the case that the second homology isn’t trivial for a particular $D$, this only means that some divergence-free fields won’t have vector potentials; others will. So for a particular field on a nontrivial $D$, it could still go either way. $\endgroup$ Oct 10, 2021 at 7:48

1 Answer 1

0
$\begingroup$

The framework of vector-analysis provides certain concepts and identities regarding how 'vectors' can be manipulated.

One of them being: a divergence-less $[\nabla.\vec{X}=0]$ vector field should wind upon itself, or simply be solenoidal $ [\vec{X} \text{ is curl of some other field}\implies \vec{X}=\nabla\times \vec{Y}]$ since $\forall \vec{Y}\,\nabla . (\nabla \times \vec{Y})=0$. (Unlike many other identities in vector analysis, this one's quite easy to visualize.)

And since EM deals with vector fields representing E and B in space, the divergence-lessness of $\vec{B}$ gives its solenoidal character, being $\vec{B}=\nabla \times \vec{A}$

But as @symplectomorphic mentioned in the comment, the restrictions are quite intricate indeed, in general.

$\endgroup$
1
  • 1
    $\begingroup$ Unfortunately this is not an answer to the question. Zero divergence does not imply the existence of a vector potential. Take the electric field of a point charge at the origin in 3-space. Its divergence is zero on its domain (3-space minus the origin), but there is no vector potential for this field. If there were, Stokes’s theorem would tell us that the flux of the field around the unit sphere containing the origin is zero (because the sphere has no boundary). But in fact it’s not zero (just do the integral, which is easy). $\endgroup$ Oct 9, 2021 at 20:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .