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.


1 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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.