- #1

- 611

- 24

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Mandelbroth
- Start date

- #1

- 611

- 24

- #2

Ben Niehoff

Science Advisor

Gold Member

- 1,879

- 162

You can even relax smoothness of the manifold (and the forms!) if you are careful about using Dirac delta functions. Generally, since a form is something you integrate, they should be thought of as distributions.

The boundary operator can be used in a distributional sense as well. For example, the boundary of a sphere is zero. But it might be helpful to think of a sphere with a point removed, whose boundary is therefore a point; then you can use Stokes' theorem to integrate forms over the sphere.

- #3

- 611

- 24

Could you please expand on this? Thinking of forms as distributions feels foreign, and I don't see where that line of thought would go.You can even relax smoothness of the manifold (and the forms!) if you are careful about using Dirac delta functions. Generally, since a form is something you integrate, they should be thought of as distributions.

- #4

Ben Niehoff

Science Advisor

Gold Member

- 1,879

- 162

[tex]\delta(x,y,z) \, dx \wedge dy \wedge dz[/tex]

Do you have a specific question?

- #5

Ben Niehoff

Science Advisor

Gold Member

- 1,879

- 162

[tex]\omega = \sin \theta \, d \theta \wedge d \phi[/tex]

Now, the sphere ##\Omega## is a closed surface, so ##\partial \Omega = 0##. However, the

Now, it so happens that

[tex]\omega = d \big( - \cos \theta \, d \phi \big) = d \alpha[/tex]

so we can use Stokes' theorem. So

[tex]\int_{\tilde \Omega} \omega = \int_{\partial \tilde \Omega} \alpha = - \int_{\partial \tilde \Omega} \cos \theta \, d \phi[/tex]

To integrate around the "cut" between the north and south poles, we draw a loop around it. On either side there is a vertical part where ##d \phi = 0##, and so these parts do not contribute. Then around the north and south poles, there are tiny circles, at which ##\cos \theta = \pm 1## and ##\phi## runs from 0 to ##2 \pi##. The tiny circles go opposite directions, so each part contributes positively:

[tex]- \int_{\partial \tilde \Omega} \cos \theta \, d \phi = 2 \pi + 2 \pi = 4 \pi[/tex]

So you see, if you are careful about how you cut up a manifold, you can apply Stokes' theorem in all sorts of situations.

In this case, we took a closed surface and removed a set of measure zero to turn it into a surface with boundary. The reason this worked is because the form ##\omega## is smooth on the set of measure zero that we removed. If that were not the case (say ##\omega## had a delta-function-like contribution on the "cut"), then you would have to include an extra piece to account for that.

Share: