Why does $\vec \nabla \times \vec A = 0$ imply $\vec A = - \nabla B$ where $\vec A$ is a vector field and $B$ is a scalar field?

I see this in my Electricity and Magnetism textbooks all over the place and I just took it for granted. Is this a theorem or does it come from one?

I can verify in my head that this is valid, but I would love some context.


1 Answer 1


If $\vec{A}$ has a simply connected domain, any closed loop $C$ encloses a surface $S$ on which Stokes's theorem gives $\int_C \vec{A}\cdot d\vec{r}=\int_S\vec{\nabla}\times \vec{A}\cdot d\vec{S}=0$. But these vanishing loop integrals imply any infinitesimal $\vec{A}\cdot d\vec{r}$ is following a scalar field's value through space, i.e. $\vec{A}=-\vec{\nabla}B$ as required.

  • $\begingroup$ N.B this only applies if your domain is simply connected! $\endgroup$
    – user223391
    May 28, 2018 at 3:27
  • $\begingroup$ As explained here, $(x\vec{j}-y\vec{i})/r^2$ on $\mathbb{R}^3\backslash O$ is a famous example of when the result fails for a domain that isn't simply connected. $\endgroup$
    – J.G.
    May 28, 2018 at 6:31

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