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Math Help - Net flow into a tetrahedron?

  1. #1
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    Question Net flow into a tetrahedron?

    I'm working through Advanced Calculus: A Differential Forms Approach at my leisure. In going over two-forms the notions of "flow across an area" and "oriented area" are introduced I hit a brick wall with grasping orientations, though, when asked to find the total flow into a tetrahedron.

    So this is what's confusing me. For a unit flow in the z-direction of xyz-space, the flow across a surface is just the oriented area dxdy of the surface. So all I have to do to answer the question is find the oriented areas of the projections of all four component triangles of the tetrahedron on the xy-plane and add them together. The result should clearly be zero since, intuitively, the flow is uniform so that nothing is accumulating inside the tetrahedron and, therefore, flow in = flow out.

    If I have a regular tetrahedron with vertices PQRS, PQR in the xy-plane and S at point (0,0,1), then the projection looks like a triangle with S at the centroid (attached image).

    Since each triangle is oriented in the direction then we have the four component oriented triangles PQR, PQS, QRS, and PRS.

    Since (where A denotes area), each of the smaller triangles must have an orientation opposite the big triangle so the total sum of oriented areas is zero. But given the orientation of the vertices, QRS, PQR and PQS have the same orientation (all are counter-clockwise) so the total flow into won't sum to zero.

    So clearly I've oriented two of the triangles wrong. I'm confused as to how I should establish orientation if I can't go by the handedness of the direction given by the vertices.
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  2. #2
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    The orientations are actually QRS, -PRS, PQS, and -PQR. The reason is that the edges need to "cancel" so to speak. Imagine an orietation as a path to be walked along. So QRS means "starting at Q, walk to R, then walk to S". Each edge is contained in exactly two faces, so it is transversed twice, and it must be transversed in opposite directions. So the edge PQ is contained in the PQR face and the PQS face, but it is transversed in the same direction both times (P to Q).

    By the way, the oriented faces of a simplex with vertices x^1x^2\cdots x^n are (-1)^k x^1\cdots x^{k-1}x^{k+1}\cdots x^n where  k=1,2,\cdots,n. (do you know what the boundary homomorphism is btw?)
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  3. #3
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    Thanks for the response, maddas.

    Forgive my lack of insight, but I'm still not sure why the edges need to cancel. The next problem in the section is to prove that fact is true for any closed, oriented, polygonal surface in space, but I still don't see why they need to. The only precedent given up to this point in the text is that we follow the vertices to determine the orientation. The handedness of PQR would be to traverse the triangle starting at P, then going to Q, then to R. Given that, I assume that for quadrilateral PQRS I follow P->Q->R->S and can find two oriented sub-triangles PQR and PQS to compose it.

    Why is it not similarly the case where PQRS is now the tetrahedron and since the orientation is P->Q->R->S that I form the triangles PQR, PQS, QRS, and PRS?

    Thanks for any further help you can give me.
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  4. #4
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    Well, what do you think a good definition of orientation is?

    The motivation I guess is that if PQR is an oriented triangle, then you can assign a normal vector to the triangle by the "right hand rule" (ie: a vector n so that if you stand at the top of n and look down at the triangle and read out the vertices in anticlockwise order you get PQR (or QRP or RPQ)). If you look at the normal vectors, the condition that the edges "cancel" ammounts to the fact that the normal vectors all point in the same direction (they all point either to the outside or the inside of the tetrahedron). You know what a Mobius strip looks like? If you cut one up into squares (say) and orient each sqaure, you'll find you can never chuse the orientations so that all the edges cancel..

    Another reason I guess is: consider the fundamental theorem of calculus. The integral of f' over an (oriented!) line is the difference of f on the end points of the line. Break the line up into a bunch of little line segments. Each one is oriented so that the "faces" of the lines cancel. The integrand is approximately the difference of f on the endpoints of the little line segments, and adding them all, all the interior points cancel out and thus leave just the difference of the domain of integration, which is the FTC. You're going to want to generalize this cancelling. I'd explain this better but the emoticons are literally on top of my reply box and I can't see anything. Forgive my spelling errors. I'll edit this later.......

    edit: An orientation on a shape is induced by a consistent (meaning the faces cancel) orientation on its faces. So a quadrilateral PQRS is oriented by an orientation on its faces, the edges PQ, QR, RS, SP. Note that each point occurs once in the first position and once in the second, so they "cancel".
    Last edited by maddas; June 21st 2010 at 07:30 PM.
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  5. #5
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    I couldn't upload anything, but I drew a crappy picture that shows that when the normals are pointing the "same direction" then the edges are transversed in opposite direcitons. The left picture is consistently oriented while the right picture isn't.

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  6. #6
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    Thank you very much, maddas. I see my error now was that I assumed that the order of the vertices described the orientation of the faces of the bounding triangles. Your analogy to the 1-dimensional case helped make sense of the "why" the boundaries cancel. I also now realize that there is a convention that dictates that the normals of the bounding faces of a closed, oriented polygonal solid must always point outwards.

    I appreciate your help and patience in dispelling my confusion!
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