Integral Definition of Exterior Derivative?

The exterior derivative is not a scalar, and it cannot be factored out of an integral in this way. Additionally, the limiting process in this definition is not rigorous and needs further justification.
  • #1
bolbteppa
309
41
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!
 
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  • #2
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.

$$\left[ \int_\Omega \! 1 \right]\mathrm{d}\omega=\left[ \int_{\partial\Omega} \! \omega \right]$$
 
  • #3
lurflurf said:
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.
 
  • #4
lurflurf said:
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.

R136a1 said:
Also, moving ##d\omega## out of the integral is very bad.
I second this.
 
  • #5


I am familiar with the concept of the exterior derivative and its use in vector analysis. In short, the exterior derivative is a mathematical operation that generalizes the gradient, divergence, and curl operations in vector analysis. It is a fundamental tool in differential geometry and is used to study the geometry of smooth manifolds. However, while the gradient, divergence, and curl can be defined in integral form, the exterior derivative does not have a rigorous integral definition.

One way to understand the exterior derivative is through the concept of differential forms. Differential forms are mathematical objects that generalize the notion of a vector field. The exterior derivative of a differential form is a new differential form that encodes information about the change of the original form under infinitesimal changes in the manifold. In this sense, the exterior derivative is a measure of the "twisting" or "curling" of the original form.

To answer the question, there is currently no rigorous integral definition of the exterior derivative analogous to the gradient, divergence, and curl in vector analysis. This is because the exterior derivative is a more abstract concept and does not have a direct physical interpretation like the gradient, divergence, and curl. However, there are alternative approaches to defining the exterior derivative, such as using the language of sheaves and cohomology, which provide a rigorous framework for studying the exterior derivative.

Regarding the formulation of the exterior derivative before stating and proving Stokes theorem, it is possible to define the exterior derivative without relying on Stokes theorem. However, Stokes theorem provides a powerful tool for calculating the integral of the exterior derivative over a manifold, which is why it is often introduced in conjunction with the exterior derivative.

As for the beautiful classical argument for the integral definition of the divergence, it is important to note that this argument is only intuitive and does not provide a rigorous proof. To make the limiting process rigorous, one would need to use the language of limits and infinitesimals in a rigorous way. This would involve defining the exterior derivative using the concept of a limit and proving that the one-line proof of the Divergence theorem is justified. However, the details of this process can be quite technical and may require a deeper understanding of differential geometry and the theory of differential forms.

In conclusion, while there is currently no rigorous integral definition of the exterior derivative, it is a fundamental concept in differential geometry and plays a crucial role in many areas of mathematics and physics. Further research and study in this area may lead to a better
 

1. What is the integral definition of exterior derivative?

The integral definition of exterior derivative is a mathematical concept that relates to the study of differential forms and their integration over a specific region. It states that the integral of the exterior derivative of a differential form over a region is equal to the integral of the differential form itself over the boundary of that region.

2. How is the integral definition of exterior derivative useful in mathematics?

The integral definition of exterior derivative is useful in mathematics because it allows for the calculation of integrals over complicated regions by reducing them to integrals over simpler boundaries. This helps in solving many problems in vector calculus, differential geometry, and physics.

3. What are some applications of the integral definition of exterior derivative?

The integral definition of exterior derivative has many applications in various areas of mathematics and physics. Some examples include calculating flux integrals in vector calculus, studying curvature in differential geometry, and solving problems in electromagnetism and fluid dynamics.

4. How does the integral definition of exterior derivative relate to Stokes' theorem?

The integral definition of exterior derivative is directly related to Stokes' theorem, which states that the integral of a differential form over a region is equal to the integral of its exterior derivative over the boundary of that region. In fact, the integral definition of exterior derivative is a direct consequence of Stokes' theorem.

5. Are there any limitations to the integral definition of exterior derivative?

One limitation of the integral definition of exterior derivative is that it only applies to smooth differential forms and regions. It cannot be used to calculate integrals over regions with discontinuous boundaries or forms that are not differentiable. Additionally, it only applies to regions in Euclidean space and cannot be extended to more general spaces without some modifications.

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