Is Stokes' Theorem Intuitively Satisfying and Accurate?

In summary: Like, what does it tell you about the magnitude of the curl? Or what does it say about the relationship between the curl and the normal?
  • #1
Nikitin
735
27
Stokes' Theorem says the curve integral of any surface S simply equals the counter-clockwise circulation around its boundary-curve C.How can this be right? Let's say you have a hemisphere surface S with centre in origo, and its shadow on the xy plane. Both surfaces will have C as their boundary curve, and so according to the theorem they will have equal circulation. However, what if the curl increases with z? Then more curl will go through the higher surfaces than the bottom, and thus the curl-integral of surface S will not equal the circulation around C.

In addition, can somebody explain to me why the formula for curl, ∇xF, is intuitively pleasing? Ie, why does it make sense?

Thanks :)
 
Physics news on Phys.org
  • #2
Nikitin said:
Stokes' Theorem says the curve integral of any surface S simply equals the counter-clockwise circulation around its boundary-curve C.
Careful there, it does not say that. It says that the surface integral of the curl of a vector field equals the line integral of that field around the boundary. It is a vector version of the fundamental theorem of calculus. Integration and differentiation can cancel leaving only the evaluation on the boundary.


How can this be right? Let's say you have a hemisphere surface S with centre in origo, and its shadow on the xy plane. Both surfaces will have C as their boundary curve, and so according to the theorem they will have equal circulation. However, what if the curl increases with z? Then more curl will go through the higher surfaces than the bottom, and thus the curl-integral of surface S will not equal the circulation around C.

Remember that the two surfaces must have the same boundary and that both the surfaces and the curls have relevant orientation information. Finally remember that not just any vector field is the curl of another vector field. See e.g. Helmholtz decomposition. Every vector field has a solenoidal and conservative component. Your arbitrarily increasing the field w.r.t. z doesn't mean that component is from the curl of another field. Try to construct an example and see what prevents your generating a counter example to Stokes' theorem.


In addition, can somebody explain to me why the formula for curl, ∇xF, is intuitively pleasing? Ie, why does it make sense?

Thanks :)

The "Grad" operator is the (formal) vector of partial derivatives. The cross product tells you how to apply the vector part to the vector valued function F. Compare that with the Divergence. It isn't necessarily intuitively pleasing to everyone. Your intuition could tell you 2+2=5. Your intuition is trained by your experience. You have to do the math to get a "good" intuition about what should happen. Intuition is not knowledge from the aether, it is a "quick calculation" from your past experience. When your intuition leads you counter to facts you need to train it by generating experience. If your intuition tells you 2+2=5, you need to do some arithmetic counting on your fingers until you see that's wrong and your intuition will adjust and tell you 2+2=4. If your intuition tells you Stokes theorem isn't right, you need to work some problems and see how the devil within the details conspires to keep Stokes theorem valid.

It doesn't hurt to also plod through the details of the proof of Stokes' thm. Have you looked at Green's theorem which is the planar form of Stokes'? Do you understand that you can chop a piecewise smooth continuous surface up into tiny pieces which look almost planar? Have you seen how the boundaries of all the tiny pieces cancel each other leaving only integration around the whole boundary?

What is left is to consider how the vector curl dots with the vector surface normal to see how varying the surface doesn't change this relationship.
 
  • #3
Careful there, it does not say that. It says that the surface integral of the curl of a vector field equals the line integral of that field around the boundary. It is a vector version of the fundamental theorem of calculus. Integration and differentiation can cancel leaving only the evaluation on the boundary.
Apologies, I meant curl integral, not curve integral. The latter makes no sense anyway :p

And yeh, I understand Stokes' theorem now. I just needed to look at the surface as a bunch of infinitesimal rectangles, where each has a curl, and where all but the edging rectangles get their curl negated.

But the formula for curl I still struggle with.. I've looked over the proof, and I understand it, but it isn't intuitive at all, unlike the formula for divergence. I dunno, I guess if you say it's not intuitive, I'll have to accept that and just memorize the formula.
 
Last edited:
  • #4
I think it also helps to look at the physical application of fluid flow understanding that the curl of the velocity field gives you vorticity. In that context the "flux of vorticity across the surface equals the flow around the perimeter". But that's really providing one with a better intuition about the meaning of the Curl rather than about the truth of Stokes' thm.
 

1. What is Stokes' theorem?

Stokes' theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field around the boundary of the surface.

2. How is Stokes' theorem used in physics?

In physics, Stokes' theorem is used to calculate the circulation of a fluid flow around a closed loop, which is crucial in understanding the behavior of fluids and their impact on objects moving through them.

3. Can Stokes' theorem be extended to higher dimensions?

Yes, Stokes' theorem can be extended to higher dimensions through the use of differential forms. In three-dimensional space, it is known as the generalized Stokes' theorem or the divergence theorem.

4. What is the relationship between Stokes' theorem and Green's theorem?

Green's theorem is a special case of Stokes' theorem, where the surface is a flat plane in two dimensions. Both theorems involve the connection between a line integral and a surface integral, but Stokes' theorem is more general and can be applied to curved surfaces in three dimensions.

5. Are there any limitations to Stokes' theorem?

Stokes' theorem is only applicable to smooth surfaces and vector fields that are continuously differentiable. Additionally, the surface must have a well-defined boundary for the theorem to hold. In some cases, the boundary may not exist, making Stokes' theorem inapplicable.

Similar threads

Replies
6
Views
1K
Replies
7
Views
1K
Replies
3
Views
2K
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
764
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Replies
9
Views
10K
  • Calculus
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
987
Back
Top