# Proof of $\nabla\times\vec{E}=0\Rightarrow\exists\Phi:\vec{E}=\nabla\Phi$

Would anyone be able to help me prove these two statements involving vector fields? For a suitable region $$\nabla\times\vec{E}=0\Rightarrow\exists\Phi:\vec{E}=\nabla\Phi$$ and $$\nabla\cdot\vec{B}=0\Rightarrow\exists\vec{A}:\vec{B}=\nabla\times\vec{A}$$ Many thanks

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"In a simply-connected region an irrotational vector field has the path independence property. This can be seen by noting that in such a region an irrotational vector field is conservative, and conservative vector fields have the path independence property. The result can also be proved directly by using Stokes' theorem. In a connected region any vector field which has the path independence property must also be irrotational." Start from this sentence from en.wikipedia.org/wiki/Conservative_vector_field – Siminore Dec 3 '12 at 18:13
Well, the first one is quite straightforward. Both of them are simple consequences of the Poincare lemma (if the fields are defined on an appropriate domain). Maybe someone knows a straightforward way to prove the second equality without using such strong tools, but I don't, at least not right away. – Dan Shved Dec 3 '12 at 18:31

The existence of the potentials depend on the shape of your region. More explicitly, they depend on the first and second de Rham cohomology groups of the region vanishing. This is not always the case. It is true for convex sets, and more generally $p$-convex sets, defined below. However, proving it for more general spaces requires more advanced machinery, like homotopy invariance et cetera.

If your region $U\subseteq \mathbb{R}^3$ is $p$-convex for a certain point $p\in U$, meaning that it contains the line segment connecting $p$ to any other point in $U$, then the scalar and vector potentials can be explicitly constructed as follows:

Let $f=(f_1,f_2,f_3)$ be a $C^2$ vector field on a $p$-convex region $U$, where we choose $p$ to be the origin (We can always translate the coordinates to make it so).

1) If $\text{curl}(f)=0$, then there exists a function $F:U\rightarrow \mathbb{R}$ such that $\frac{\partial F}{\partial x_i} = f_i$ for $i=1,2,3$, given by $$F(x_1,x_2,x_3) = \int_0^1 x_1 f(t x)+x_2f_2(tx) + x_3 f_3(tx) \text{d}t$$ where $x=(x_1,x_2,x_3)$.

2) If $\text{div}(f)=0$, then there exists a $F:U\rightarrow \mathbb{R}^3$ with $\text{curl}(F)=f$, given by $$F(x) = \int_0^1 f(tx)\times (tx) \text{d}t$$

I'll let you check that these $F$'s really do give the desired results. The calculations are not very difficult.

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Without loss of generality we may assume that $p=\mathbf{0}$ as you did. – vesszabo Dec 3 '12 at 18:53
Indeed, the formulas I gave for $F$ are only valid when $p=0$. – Espen Nielsen Dec 3 '12 at 19:01
Thank you this is fantastic!! Just one question: is $F_{x_1}=\int_0^1 f_1(tx) + \sum_{i=1}^3 x_it \frac{\partial f_i}{\partial tx_1} dt$ and if so how does this equal $f_1$? for 1. – Freeman Dec 3 '12 at 20:11
You expression is not quite right. We get $F_{x_1}=\int_0^1 f_1(tx)+\sum_{i=1}^3 x_i t \frac{\partial f_i}{\partial x_i}(tx)dt$. If you now calculate $\frac{d}{dt} tf_1(tx)$, plug this into your integral and use the assumption $\text{curl}(f)=0$, you should get what you want. This problem is not a straight path. You have to take some detours and "see" what quantities you want to have in your integral. – Espen Nielsen Dec 3 '12 at 20:20
Thank you very much :) – Freeman Dec 3 '12 at 20:23