A Stokes' theorem on a torus?

1. Apr 27, 2017

lichen1983312

I am now looking at a physics problem that should be a use of stokes' theorem on a torus. The picture (b) here is a torus that the upper and bottom sides are identified as the same, so are the left and right sides. $A$ is a 1-form and $F = dA$ is the corresponding curvature. As is shown in the equation, the author says the integration of $F$ over the whole torus is the same thing as the difference between the two line integral along C1 and C2. Is this a case of stokes' theorem? I don't understand how C1 and C2 is the boundary of S. Please help.

2. Apr 27, 2017

Staff: Mentor

Notice they using radial measure on the horizontal axis and they mention that the difference is $2\pi$

That means C1 is a closed loop and C2 is a closed loop.

Does that make sense?

3. Apr 27, 2017

lichen1983312

Sorry for the confusion, the horizontal axis is parameterized with angle, both C1 and C2 are closed loops because the left and right sides are identified.

4. Apr 27, 2017

Staff: Mentor

So what don't you understand about the boundary of S? It's like a ring with C2 as the outer boundary and C1 as the inner boundary.

When they plot it using radial measure S looks like a rectangle.

5. Apr 27, 2017

lichen1983312

So this is not a torus? I though the upper side and bottom side are identified as the same edge.

6. Apr 28, 2017

zwierz

Stokes theorem for torus: $\int_{\mathbb{T}^m} d\omega=0$ :) The torus does not have boundary

7. Apr 28, 2017

lavinia

I do no think this is a torus. Rather, it seems to be a cylinder. The two vertical edges seem identified but the top and bottom do not. If so, then it is a case of Stokes Theorem because the boundary of this cylinder is $C_1-C_2$ or $C_2-C_1$ depending on the orientation.

If the top and bottom were also identified then you would have a torus but then the integral of $dA$ would be zero since as zwierz pointed out, the boundary of a torus is empty.

Another way to think of it is that if $C_1$ and $C_2$ are identified to make a torus then the integrals of $A$ along them would be the same except for a sign and would cancel out to give zero.

Last edited: Apr 28, 2017
8. Apr 28, 2017

lichen1983312

Hi guys, Thanks for the help, I think it really is a ring not torus.