# Vector Field Question

• Apr 23rd 2008, 07:31 PM
Toomin22
Vector Field Question
Okay, here's the question:
Why is the pictured (bitmap file attached) vector field not continuous?

I think I understand the general idea, because I know that there's some point in the middle of the vector field where the vectors rotating around each point are going in the same direction. By this logic, there must be a vector normal to the outer curve.

What I don't understand is how I'm supposed to prove this using technical terms and/or formulas.

Thanks in advance to anyone who can help me out!
• Oct 27th 2008, 12:49 PM
Rebesques
Let the vector field be continuous.

Consider a small disk around one of the singularities. On the boundary of that disk, we have the (fixed) values of the vector field, and by the hypothesis we have a continuous retraction of those values to the inside of the disk. This is a contradiction.
• Oct 27th 2008, 01:40 PM
Laurent
Quote:

Originally Posted by Rebesques
Let the vector field be continuous.

Consider a small disk around one of the singularities. On the boundary of that disk, we have the (fixed) values of the vector field, and by the hypothesis we have a continuous retraction of those values to the inside of the disk. This is a contradiction.

What this proves it that the vector field must be zero somewhere in the disk. And such is indeed the case at the "singularities". This does not answer the question. But perhaps some similar argument could apply.

The fact to prove is (as far as I can guess): if a continuous vector field $x\mapsto \vec{A}(x)$ is zero at exactly two points in a smooth compact domain bounded by a curve $t\mapsto\gamma(t)$, then there is a point $x=\gamma(t)$ on the boundary such that $\vec{\gamma'}(t)\cdot \vec{A}(x)=0$: at this point, the vector field is orthogonal to the boundary. (or: then the scalar product $\vec{\gamma'}(t)\cdot \vec{A}(\gamma(t))$ can't be positive for all $t$).