Spivak (Calculus on Manifolds) proof of stolkes theorem

In summary, the proof of Stolkes Theorem, given in Spivak, involves a piecewise function that is defined differently for j=1 and j>1. This is due to the fact that the function has compact support on a specific case of M=(0,1]x(0,1)n-1. While f(0,y) and f(x,1) are always equal to 0, f(1,y) may not be 0, leading to the different definitions for j=1 and j>1. The inclusion map and pull back are also used in the integration process.
  • #1
~Death~
45
0
http://planetmath.org/?op=getobj&from=objects&id=4370
that's pretty much the proof of Stolkes Theorem given in Spivak
but I'm having a lot of difficulty understanding the details

specifically...when the piecewise function is defined for j>1 the integral is 0
and for j=1 the integral is nontrivial...why is it defined like that?

Also, I am having difficulty understanding what the inclusion map does (spivak defines it

as I(j,alpha) which is a continuous function or a chain of some sort) but the pull back

I*(j,alpha)fdx^1...dx^n is taken and integrated over in that piecewise function

could someone shed some light on that?

Thanks
 
Physics news on Phys.org
  • #2
~Death~ said:
http://planetmath.org/?op=getobj&from=objects&id=4370
that's pretty much the proof of Stolkes Theorem given in Spivak
but I'm having a lot of difficulty understanding the details

specifically...when the piecewise function is defined for j>1 the integral is 0
and for j=1 the integral is nontrivial...why is it defined like that?

There is no definition involved here.. It is just a fact. Recall that at this stage, they are working on the special case M=(0,1]x(0,1)n-1. For simplicity, let me assume that that n=2 here, so that M=(0,1]x(0,1). The hypothesis is that f has compact support on M. So for any given y in (0,1), f(0,y)=0. And for any x in (0,1], f(x,1)=0. But f(1,y) need not be 0.

Hopefully, you can demonstrate these little assertions if they are not clear to you, and you can generalize my reasoning to higher n, and conclude to the formula for when j=1, and when j>1.
 

1. What is Stokes' Theorem and why is it important?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field to the line integral of its curl. It is important because it provides a powerful tool for computing integrals over surfaces, which has many applications in physics and engineering.

2. What is the main idea behind Spivak's proof of Stokes' Theorem?

The main idea behind Spivak's proof of Stokes' Theorem is to break up the surface into smaller pieces and use the fundamental theorem of calculus to evaluate the line integral over each piece. By summing up these contributions, we can then approximate the surface integral and take the limit as the size of the pieces approaches zero to obtain the exact value.

3. How does Spivak's proof differ from other proofs of Stokes' Theorem?

Spivak's proof is unique in that it uses a partition of unity to break up the surface into smaller pieces, while other proofs may use more geometric or algebraic arguments. Additionally, Spivak's proof is more rigorous and general, as it can be applied to manifolds of any dimension, whereas other proofs may only apply to specific cases.

4. Are there any prerequisites for understanding Spivak's proof of Stokes' Theorem?

Yes, a solid understanding of multivariable calculus and linear algebra is necessary to understand Spivak's proof. Familiarity with topics such as vector fields, line and surface integrals, and the fundamental theorem of calculus will be helpful in understanding the proof.

5. What are some applications of Stokes' Theorem in real-world problems?

Stokes' Theorem has many applications in physics and engineering, such as calculating the work done by a force on a moving object, computing the flow of a fluid through a surface, and finding the electric or magnetic flux through a closed surface. It also has applications in differential geometry, where it is used to study the curvature of surfaces and higher-dimensional manifolds.

Similar threads

Replies
1
Views
1K
  • Calculus
Replies
4
Views
1K
Replies
9
Views
3K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
7
Views
2K
Replies
1
Views
895
  • Calculus
Replies
2
Views
1K
Replies
4
Views
715
Back
Top