Point of stokes theorem - conservative field

Question

I am working through electrodynamics at the moment and I have a rather elementary question - which I apologize for. But after some research on google I am still not sure if I have understand "the point" of stokes' theorem correctly:

So let's say we have a given electric field and a given curve for which we should estimate the work done.

• Stokes' theorem can only be used for closed curves

• Stokes' theorem can not be used for open curves

• Stokes's theorem can only be used in non-conservative fields (otherwise curl=0)

• Since closed curves in conservative fields imply W=0 and got no curl, it only makes sense to have a method for closed curves embedded in non-conservative fields.

• The given electric field must have been produced by a magnetic field, because electric fields created by charges do not have curl.

Have I understood that correctly ?

I know that this question is extremely elementary but I am trying to get the whole idea of this and trying to understand the "joke" about stokes' theorem.

Answer 1

Stokes' theorem van be given in various forms, but in electrodynamics you probably encountered the 'classical' form $$ \iint\limits_A \left(\vec\nabla\times \vec F\right) \cdot\text{d}\vec s=\int\limits_{\partial A} \vec F \cdot \text{d}\vec l \,. $$ Hence, to use the theorem, you need

• a surface $A$ and
• a vector field $\vec F$ which is defined and differentiable on $A$.

The the theorem tells you that the integral of the curl of $F$ is equal to the line integral of the field itself along the boundary, $\partial A$.

This should already answer your first two questions: The curve on the right-hand side of the eqaution is not just any curve, but the boundary of a surface, and boundaries are always closed. On the other hand, the field may have zero curl, in which case the left-hand side of the equation gives zero, and consequently the rhs is also zero -- this is just the statement of the theorem.

To your last question: You are essentially correct here.

Two further points:

• If the surface is closed, the boundary is empty $\partial A=\emptyset$, and the rhs is zero. This is in a way a situation dual to a curl-free field.
• There are many different surfaces $A_1\neq A_2$ having the same boundary curve, $\partial A_1=\partial A_2$. Can you see how the preceding statement implies that this is not a problem?
• Finally, being curl-free is neccessary for conservativity, but not sufficient, if the field has singularities is certain locations.