Why does $\vec \nabla \times \vec A = 0$ imply $\vec A = - \nabla B$

Question

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.

Possible duplicate of Does zero curl imply a conservative field? — , May 27 '18 at 20:09
Possible duplicate of Why curl free field implies existence of potential function?. — A.Γ., May 27 '18 at 20:12

score 1 · Accepted Answer · answered May 27 '18 at 20:18

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.

answered May 27 '18 at 20:18

J.G.

115,835

N.B this only applies if your domain is simply connected! – May 28 '18 at 03:27
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. – J.G. May 28 '18 at 06:31

Why does $\vec \nabla \times \vec A = 0$ imply $\vec A = - \nabla B$

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