# Stange proof!

1. Feb 5, 2010

### typhoonss821

One of Maxwell's equations says that $$\nabla\cdot\vec{B}{=0}$$ where B is any magnetic field.
Then using the divergence theore, we find
$$\int\int_S \vec{B}\cdot\hat{n}dS=\int\int\int_V \nabla\cdot\vec{B}dV=0$$.

Because B has zero divergence, there must exist a vector function, say A, such that
$$\vec{B}=\nabla\times\vec{A}$$ .
Combining these two equations, we get
$$\int\int_S \hat{n}\cdot\nabla\times\vec{A}dS=0$$ .
Next we apply Stoke's theorem and the preceding result to find
$$\oint_C\vec{A}\cdot\hat{t}ds=\int\int_S\hat{n}\cdot\nabla\times\vec{A}dS=0$$ .
Thus A is path independent. It follows that we can write
$$\vec{A}=\nabla\psi$$ , where ψ is some scalar fution.

Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that
$$\vec{B}=\nabla\times\nabla\psi=0$$
that is, all magnetic field is zero!

Wow, there must be someting wrong...

My thought is that we can not apply Stoke's theorem in this case because the region we discuss is not simply connected.

But I'm not sure if I am right, please help me check the proof^^

Last edited: Feb 5, 2010
2. Feb 5, 2010

### willem2

The surface you use for the divergence theorem is a closed surface that encloses a volume. The surface is from the Stokes theorem is a surface that has the contour C as edge.