Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral Definition of Exterior Derivative?

  1. Dec 5, 2013 #1
    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!
     
  2. jcsd
  3. Dec 6, 2013 #2

    lurflurf

    User Avatar
    Homework Helper

    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.

    $$\left[ \int_\Omega \! 1 \right]\mathrm{d}\omega=\left[ \int_{\partial\Omega} \! \omega \right]$$
     
  4. Dec 6, 2013 #3
    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.
     
  5. Dec 7, 2013 #4
    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.

    I second this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integral Definition of Exterior Derivative?
  1. Exterior Derivative (Replies: 2)

  2. Exterior derivative (Replies: 18)

  3. Exterior derivative (Replies: 6)

  4. Exterior derivative (Replies: 5)

Loading...