Curl in Cylindrical Vector Fields: Exploring Zero Curl

[superg33k]

If a vector field has any component in a circular direction how can its curl be zero? If I imagine a vortex of water, it makes sense that it will be easier to go with the water in a circle than it would be to go against the water in a circle. Or more mathsy:

A vector field in cylindrical coords is defined by:
$$\mathbf{E} = \frac{A}{r} \mathbf{ \hat{e}_{\phi} }$$
where A is a constant. Therefore the curl is zero.
$$\mathbf{curlE} = \frac{1}{r} \frac{\partial}{\partial r} r \frac{A}{r} \mathbf{ \hat{e}_{z} } = \mathbf{0}$$
So taking a line integral in a circle, taking it back to the same point should also be zero.
$$\int^{2 \pi}_{0} \mathbf{E \cdot} r \mathbf{\hat{e_{\phi}}} \partial \phi = \int^{2 \pi}_{0} \frac{A}{r} \mathbf{ \hat{e}_{\phi} } \mathbf{\cdot} r \mathbf{\hat{e_{\phi}}} \partial \phi =\int^{2 \pi}_{0} A \partial \phi = 2 \pi A \neq 0$$

Thanks in advance

 

[arildno]

Quite correctly calculated!

The reason is that your function E has a singularity in the region bounded by your curve (i.e, at the origin).

The theorem you had in mind specifically requires, in order to be valid, that such singularities do not exist, so you have misapplied it.