Potential magnetic field lines and Stokes theorem....

[octopode]

Hi,

A potential magnetic field has no curl. According to the "curl theorem" or stokes theorem, a vector field with no curl does not close. Yet, Maxwell's equation tell us we shall not have magnetic monopoles, so the loops have to close... ? What am I missing to remove this apparent paradox of a no-curl magnetic field with unclosed field lines on the one hand (Stokes) and magnetic field lines which have to close somewhere on the other hand (Maxwell)? Do we allow this like in classical optics where two parallel lines cross only at infinity? So the no-curl loops actually close at infinity?

Thanks

 

[andrewkirk]

To what field are you referring?

If you are referring to the magnetic scalar potential then the curl for that is not defined, as it is not a vector field. So Stokes' theorem does not apply.

If you are referring to the magnetic vector potential then from where did you get the statement that the field has no curl? My EM is very rusty but according to wikipedia the curl of the mag vector potential field is the observed mag field, which is not zero.

Perhaps you are thinking of gauge adjustments, whereby one can add a curl-free vector field to the mag vector potential field and still generate the same mag field. But adding curl-free components doesn't change the curl of the mag vector potential field, which remains nonzero. I see no reason why the field lines of a gauge field cannot close. Their closing would not (I think) imply anything about magnetic monopoles.

 

[vanhees71]

The magnetic field is never rotation free everywhere (except it's 0), because according to the Ampere-Maxwell Law you have (in vacuo and in Heaviside Lorentz units)
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_c=\frac{1}{c} \vec{j}$$
and according to Gauss's Law for the matgnetic field
$$\vec{\nabla} \times \vec{B}=0.$$
So it's a solenoidal field, and its natural potential is a vector potential
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
which is defined by $\vec{B}$ up to an arbitrary gradient field (gauge invariance).

However, locally in regions, where $\vec{\nabla} \times \vec{B}=0$, you can as well describe it via a scalar potential, which then is necessarily a harmonic potential, i.e., fulfilling
$$\vec{\nabla} \cdot \vec{B}=-\Delta \Phi_{M}=0.$$
As I said, this can never be globally true, and usually $\Phi_{M}$ is not unique, because usually you have something like current loops involved, i.e., the region where $\vec{B}$ is curl free, is not simply connected.