Integral Definition of Exterior Derivative?

  • Thread starter bolbteppa
  • Start date
310
38
Is there a rigorous integral definition of the exterior derivative analogous to the way the gradient, divergence & curl in vector analysis can be defined in integral form?

Furthermore can it be formulated before stating & proving Stokes theorem?

Finally, a beautiful classical argument for the integral definition of the divergence intuitively existing is given in Purcell (page 78), & it leads to the quickest (one line, coordinate independent!) proof of the Divergence theorem I've ever seen. How does one make the limiting process rigorous, & will making the limiting process rigorous justify that one-line proof?

Really appreciate any help with this - thanks!
 

lurflurf

Homework Helper
2,417
122
Yes it is the same. The proofs are not really shorter though. It just moves the hard work into the definition. Be careful learning math from intro physics books, they are not written for that purpose.

$$\mathrm{d}\omega=\lim_{\Omega\rightarrow\Omega_0}\left[ \int_{\partial\Omega}\! \omega \right]/ \left[\int_{\Omega}\! 1\right]$$

In the limit we shrink the region to a single point. Informally we can imagine the Stokes formula

$$\left[ \int_\Omega \! \mathrm{d}\omega \right]=\left[ \int_{\partial\Omega} \! \omega \right]$$

If the region is small (in the limit) the exterior derivative is constant and can factor out.

$$\left[ \int_\Omega \! 1 \right]\mathrm{d}\omega=\left[ \int_{\partial\Omega} \! \omega \right]$$
 
341
51
Yes it is the same. The proofs are not really shorter though. It just moves the hard work into the definition. Be careful learning math from intro physics books, they are not written for that purpose.

$$\mathrm{d}\omega=\lim_{\Omega\rightarrow\Omega_0}\left[ \int_{\partial\Omega}\! \omega \right]/ \left[\int_{\Omega}\! 1\right]$$

In the limit we shrink the region to a single point. Informally we can imagine the Stokes formula

$$\left[ \int_\Omega \! \mathrm{d}\omega \right]=\left[ \int_{\partial\Omega} \! \omega \right]$$

If the region is small (in the limit) the exterior derivative is constant and can factor out.

$$\left[ \int_\Omega \! 1 \right]\mathrm{d}\omega=\left[ \int_{\partial\Omega} \! \omega \right]$$
This reply makes no sense. The exterior derivative ##d\omega## is a differential form, and not (in general) a scalar. Your definition of
$$\mathrm{d}\omega=\lim_{\Omega\rightarrow\Omega_0}\left[ \int_{\partial\Omega}\! \omega \right]/ \left[\int_{\Omega}\! 1\right]$$
implies that it's a scalar.

Also, moving ##d\omega## out of the integral is very bad.
 
612
23
Yes it is the same. The proofs are not really shorter though. It just moves the hard work into the definition. Be careful learning math from intro physics books, they are not written for that purpose.

$$\mathrm{d}\omega=\lim_{\Omega\rightarrow\Omega_0}\left[ \int_{\partial\Omega}\! \omega \right]/ \left[\int_{\Omega}\! 1\right]$$
lurflurf, I'm not sure if your RHS actually means anything. In your denominator, it looks like you're attempting to integrate a 0-form on a manifold of degree greater than 0. Plus, your RHS looks like it isn't a differential form of degree greater than 0.

Also, moving ##d\omega## out of the integral is very bad.
I second this.
 

Related Threads for: Integral Definition of Exterior Derivative?

Replies
19
Views
6K
  • Posted
Replies
6
Views
3K
  • Posted
Replies
2
Views
5K
  • Posted
Replies
18
Views
8K
Replies
10
Views
1K
  • Posted
Replies
1
Views
2K
  • Posted
Replies
3
Views
4K
  • Posted
Replies
6
Views
4K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top