On an alternative Stokes theorem

In summary, the conversation is about an alternative form of Stokes' theorem that was discovered while working on a homework problem. The person is curious if the alternative form is true and asks for a proof. Another person responds by explaining that the alternative form is equivalent to the original and provides a way to show this using vector calculus. The original person clarifies that the question is not for their homework but a general question and expresses gratitude for the helpful responses on the forum.
  • #1
curupira
5
0
I find in a homework an alternative Stokes theorem tha i wasn't knew before. I wold like to know that it is really true. Any can give me a proof please?

It is:

Int (line) dℓ′× A = Int (surface)dS′×∇′× A
 
Physics news on Phys.org
  • #2
First off, you cannot modify someone else's theorem and still leave his name attached to it (unless you collaborate with him).
Secondly, I think that this belongs in the Homework sub-forum.
 
  • #3
I can't really read your equation very well but it looks like Stoke's theorem written using alternative notation.

I bet it is equivalent and you can show it with a few lines of manipulation.
 
  • #4
Use the vector field:

[tex]
\vec{F} = \vec{C} \times \vec{A}
[/tex]

where [itex]\vec{C}[/itex] is an arbitrary, but constant vector in the original Stokes' Theorem. Perform some simplifications for:

[tex]
\nabla \times \vec{F}
[/tex]

using nabla calculus and you will arrive at your alternative form of Stokes' Theorem.
 
  • #5
Danger said:
First off, you cannot modify someone else's theorem and still leave his name attached to it (unless you collaborate with him).
Secondly, I think that this belongs in the Homework sub-forum.

Well, First, I think that you don't understood what I say, because i don't speak english, so its hard but I try to post, others understood me and have answered my question. In fact you don't understood nothing.

Second, It is a general question, and don't a homework question, my homework I solved , but there was a little part incomprehensible to me.

I don't want damage the forum, but think that is possible stay posting here without to know english, cause there's people here that help me, and it is usefull to me.
Thanks!
 

Related to On an alternative Stokes theorem

1. What is the alternative Stokes theorem?

The alternative Stokes theorem is a mathematical theorem that relates the integral of a differential form over a manifold to the boundary of the manifold. It is an extension of the classical Stokes theorem and is used in vector calculus and differential geometry.

2. What are the applications of the alternative Stokes theorem?

The alternative Stokes theorem has various applications in physics and engineering, particularly in the fields of fluid mechanics, electromagnetism, and general relativity. It is used to calculate the flux of a vector field through a surface, as well as to derive important equations in these disciplines.

3. How is the alternative Stokes theorem different from the classical Stokes theorem?

The alternative Stokes theorem is a more general form of the classical Stokes theorem. While the classical theorem only applies to oriented surfaces in three-dimensional space, the alternative Stokes theorem can be applied to any n-dimensional manifold without the need for orientation. It also considers more general differential forms, allowing for a broader range of applications.

4. What is the proof of the alternative Stokes theorem?

The alternative Stokes theorem can be proved using techniques from differential geometry and multivariable calculus. It involves breaking down the manifold into smaller pieces and using the fundamental theorem of calculus and the chain rule to evaluate the integral over each piece. These pieces are then combined to form the final result.

5. Are there any limitations to the alternative Stokes theorem?

Like any mathematical theorem, the alternative Stokes theorem has its limitations. It can only be applied to smooth manifolds, and the differential forms involved must satisfy certain conditions. Additionally, the theorem only applies to closed manifolds, meaning that it does not hold for manifolds with boundaries.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
181
  • Calculus and Beyond Homework Help
Replies
26
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Differential Geometry
Replies
11
Views
473
Replies
30
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
623
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
2
Views
2K
  • Quantum Physics
Replies
2
Views
859
Back
Top