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I have a problem with the scalar potential. At the moment, I'm working with electromagnetism (Maxwell theory) and there's a vector field $w$ with $\vec{\nabla}\times w=0$, for which one can find a scalar potential. I know that a curl-free field implies the existence of a potential function and I have a proof for the case that $\nabla \times (\cdot) = 0$ is defined on all $\mathbb{R}^{3}$, as can be seen here.

My question is how can you prove you can find a potential function if there is a singularity in the vector field? For example, my vector field $w(x)$ is defined for all $x \ne x_0$ and has a singularity at $x_0$.

Can someone help me? Thanks!

eranreches
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lissy1234567
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1 Answers1

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You can't, because in general this is false.

Lets look for example at the following vector field

$$\vec{F}=\left(-\frac{y}{x^{2}+y^{2}},\frac{x}{x^{2}+y^{2}},0\right) \tag{1}$$

that is defined on $\mathbb{R}^{3}\setminus\{\left(0,0,0\right)\}$. It is easy to see that

$$\vec{\nabla}\times\vec{F}=\begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\\partial_{x}&\partial_{y}&\partial_{z}\\-\frac{y}{x^{2}+y^{2}}&\frac{x}{x^{2}+y^{2}}&0\end{vmatrix}=\hat{z}\left(\partial_{x}\frac{x}{x^{2}+y^{2}}+\partial_{y}\frac{y}{x^{2}+y^{2}}\right)=\hat{z}\left(\frac{x^{2}+y^{2}-2x^{2}}{\left(x^{2}+y^{2}\right)^{2}}+\frac{x^{2}+y^{2}-2y^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=\vec{0} \tag{2}$$

So you might think that there exists a potential function $V$ such that $\vec{F}=-\vec{\nabla}V$, but this would imply that

$$\int_{C}\vec{F}\cdot\vec{{\rm d}\ell}=0 \tag{3}$$

for every closed curve $C$. If we choose $C$ to be the unit circle in the $\rm XY$ plane, namely

$$\left(x,y,z\right)=\left(\cos\theta,\sin\theta,0\right),\:\: \theta\in\left[0,2\pi\right) \tag{4}$$

then

$$\int_{C}\vec{F}\cdot\vec{{\rm d}\ell}=\int_{C}\left(-\frac{y{\rm d}x}{x^{2}+y^{2}}+\frac{x{\rm d}y}{x^{2}+y^{2}}\right)=\int_{0}^{2\pi}\left(\frac{\sin^{2}\theta}{\cos^{2}\theta+\sin^2\theta}+\frac{\cos^{2}\theta}{\cos^{2}\theta+\sin^2\theta}\right){\rm d}\theta=2\pi\tag{5}$$

which clearly contradicts Eq. $3$, and therefore a potential function does not exist. This fact is related to objects known as differential forms. The vector field in Eq. $1$ corresponds to the differential form

$$\omega=-\frac{y}{x^{2}+y^{2}}{\rm d}x+\frac{x}{x^{2}+y^{2}}{\rm d}y \tag{6}$$

and the fact that $\vec{F}$ is curl-free (Eq. $2$) is equivalent to saying that the exterior derivative of the form vanishes

$${\rm d}\omega=0 \tag{7}$$

Such a form that satisfies Eq. $7$ is called closed form. The existence of potential, in the jargon of differential forms, is written as

$$\omega={\rm d}\alpha\tag{8}$$

and one then refers to $\omega$ as an exact form. It is always true that an exact form is closed, but the opposite is false as you've seen above. Nevertheless, if one assumes that the domain in question is a star domain (note that one may require an alternative condition), then the statement is iff. In the example we studied the domain is $\mathbb{R}^{3}\setminus\{\left(0,0,0\right)\}$, and it is obiously not a star domain.

eranreches
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  • Thanks a lot! So, but if there's a star domain given. How can i proof it? – lissy1234567 Nov 30 '17 at 07:58
  • @lissy1234567 You should take the star point of your domain and define your potential function at $x$ as the line integral of the field on the straight path connecting the star and $x$. Then you should show that this is indeed a potential function by calculating the gradient. – eranreches Nov 30 '17 at 10:11
  • thank you for your help! – lissy1234567 Nov 30 '17 at 10:27