Stokes' Theorem and Curvature on a Torus

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SUMMARY

The discussion centers on the application of Stokes' Theorem in the context of a torus and its curvature. Participants clarify that the object in question is not a torus but rather a cylinder, as the top and bottom edges are not identified, which affects the application of Stokes' Theorem. The integral of the 1-form ##A## along the closed loops ##C_1## and ##C_2## would yield a non-zero result if the top and bottom were not identified. Thus, the boundary of the cylinder is defined by the difference between the two loops, leading to a conclusion that the integral of ##dA## would be zero if it were a torus.

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lichen1983312
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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.
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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?
 
jedishrfu said:
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?
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.
 
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.
 
So this is not a torus? I though the upper side and bottom side are identified as the same edge.
 
Stokes theorem for torus: ##\int_{\mathbb{T}^m} d\omega=0## :) The torus does not have boundary
 
lichen1983312 said:
So this is not a torus? I though the upper side and bottom side are identified as the same edge.

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

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