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Prerequisite of Stoke's Theorem

  1. Jul 14, 2012 #1
    How come the Stoke's Theorem does not require that the manifold on which it takes action, to be simply connected?

    And if this is optional how can we use this theorem to show that irrotational vector field may not be also conservative if they are defined over a multiply connected topological space?
     
    Last edited: Jul 14, 2012
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  3. Jul 15, 2012 #2

    tiny-tim

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    Hi Trifis! :smile:

    Tell us what you think, and then we'll comment! :smile:
     
  4. Jul 15, 2012 #3
    @tiny-tim My knowledge on topology is very limited and I have only seen this theorem proved in close-connected smooth surfaces with boundary.
    By intuition I'd say that the theorem is not applicable in the region of the surface where the hole exists. And as a result when we use the theorem to prove if a field is conservative or not, some information is lost and despite the apparent result we are prohibited from inferring the above property.

    I'd like to here some of those now formally expressed in the mathematical language.
     
  5. Jul 15, 2012 #4

    tiny-tim

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    What theorem are you talking about? :confused:

    If you mean Stoke's theorem, I thought you said the opposite?
     
  6. Jul 15, 2012 #5
    Yeah! That is exactly my problem. If the latter is true, I cannot make the connection with the theorem about the existance of potential (which requires a simply-connected topological space) in my mind.
     
  7. Jul 16, 2012 #6

    tiny-tim

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    let's go back to basics :wink:

    i] what is Stoke's theorem?

    ii] what is an irrotational field?

    iii] what are the definitions of a conservative field?​
     
  8. Jul 16, 2012 #7
    I honestly do not know, if there is a point in doing this or we are just increasing are posts or something...

    Anyway I found this on the net : http://www.math.psu.edu/liu/580f05/580L4.pdf [Broken]

    "Stokes’ Theorem applies to any contour L within a simply connected domain" ... and not any arbitrary manifold. Providing that (the topological condition), we can use the Stoke's theorem: [itex]\iint[/itex](XF)dA=[itex]\oint[/itex]F[itex]\cdot[/itex]dr, which for rotF=0 (irrotational field) yields: [itex]\oint[/itex]F[itex]\cdot[/itex]dr=0 (path independency).

    If there is a version of Stoke's theorem for mutliple-connected spaces I'd like to have a look at its proof...
     
    Last edited by a moderator: May 6, 2017
  9. Jul 16, 2012 #8

    tiny-tim

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    (hmm … it would have been simpler if you'd quoted that at the start)

    that does not say that it only applies as stated

    slightly higher up, it shows a line L "round the hole" of a torus, and it says
    In this case, we do not have a surface S to apply Stokes's Theorem on.​

    Stoke's theorem applies so long as there is a line L and a surface S whose boundary is L …

    in that case, there is clearly no such S, so nothing to apply Stoke's theorem to.

    Stoke's theorem does apply to any circuit L on a torus (or other multiply-connected space) which is the boundary of a surface. :wink:
     
  10. Jul 16, 2012 #9
    No we are clear (and I'm personally content :smile:)

    So the Stoke's Theorem can be applied to multiply-connected spaces but only on regions of their surfaces which are simply-connected and enclosed by a boundary and not on the object as a whole.

    Imagine for example a ring in 2D with inner radius r1 and outer radius r2. We can apply the theorem on any part of the surface between r1 and r2, but we cannot just calculate the line integral over the outer circle and expect to be able to make a statement about our ring (due to the hole in the middle).
     
  11. Jul 16, 2012 #10

    tiny-tim

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    yes, we can only apply Stoke's theorem to a line that doesn't enclose the hole :smile:
     
  12. Jul 17, 2012 #11
    @tiny-tim maybe you could reply to a semi-relevant question of mine in this thread too! I don't wanna open a new one only for this question:

    I am curious about how the topologists prove whether a given space possesses a specific property or not. Is there a formal, mathematical way to prove for example that M1 = {(x,y)[itex]\in[/itex]ℝ2 | |y2+ x2|≤ 10} is not bounded or convex but it's closed and starlike or that M2 = {(x,y)[itex]\in[/itex]ℝ2 | |x2-y2|>1} is not connected? Or do we just draw the sketches and decide with the eye what is what?

    PS: Generally it is easier to prove that an argument does NOT hold, by making use of a counterexample. As far as the M2 region above is concerned, it might be easy to prove that it is not a domain.
    But I have no clue how are we able to formally check all the points of some other arbitrary space (in 2D or higher dimensions) and draw a conclusion about its nature.
     
  13. Jul 17, 2012 #12

    tiny-tim

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    formally…

    for the first one, you take any two points in M1 and prove that the line joining them lies entirely in M1

    for the second, you prove that x = y divides the plane in two, and that there are points on M2 on either side of it​

    but in practice, you just draw the sketches
     
  14. Jul 17, 2012 #13
    So there is always a general formal way of proving the topological properties even if the object is some kind of complex three-dimensional knot-like surface?

    How do you prove that a region IS connected?

    Do we generally start with the definition of the property and try to examine the analytical form of the given region with respect to this property?

    And after all, do the mathematicians and the topologists bother to prove all those attributes for each space, or do they just make the graphs and study from then on more interesting things (like homeomorphisms or aspects of differential geometry) ?
     
    Last edited: Jul 17, 2012
  15. Jul 18, 2012 #14
    M1 is actually both bounded and convex. I think you would benefit from seeing a specific example of how such properties can be proved formally, so I'll prove this claim. I should also mention that I know very little topology, so I can't answer you question about what topologists do in practice. Here we go...

    Proving that M1 is bounded is easy. Note that for [itex] (x,y)\in\mathbb{R}^{2} [/itex], [itex] \left|x^{2}+y^{2}\right| = \left\|(x,y)\right\|^{2} [/itex], so

    [itex] (x,y)\in M_{1} [/itex]

    iff [itex]\left\|(x,y)\right\|^{2} \leq 10[/itex]

    iff [itex]\left\|(x,y)\right\| \leq \sqrt{10}[/itex].

    Hence, by definition, M1 is bounded. (If you're using the sup norm, use the fact that the sup norm of a point is always ≤ the euclidean norm, so the sup norms of all the points in M1 are also ≤ √10.

    To prove that M1 is convex, let [itex] a,b\in M_{1} [/itex]. We want to prove that the entire line segment [itex]\left\{a+t(b-a) : t\in[0,1]\right\} [/itex] is contained in M1. But for [itex]t\in[0,1][/itex], we have

    [itex]\left\|a+t(b-a)\right\|[/itex]

    [itex]= \left\|(1-t)a+tb\right\|[/itex]

    [itex]\leq (1-t)\left\|a\right\|+t\left\|b\right\|[/itex] (triangle inequality)

    [itex]\leq (1-t)\sqrt{10}+t\sqrt{10}[/itex] (from the fact that a and b are in M1 and the observation made in the first proof that the points in M1 are exactly those points with norm ≤√10.)

    [itex]= [(1-t)+t]\sqrt{10}[/itex]

    [itex]= \sqrt{10}.[/itex]

    That is, [itex]\left\|a+t(b-a)\right\| \leq \sqrt{10}[/itex]. Therefore, [itex]a+t(b-a)\in M_{1}.[/itex] Done.
     
  16. Jul 18, 2012 #15
    You are absolutely right! I meant to write M1 = {(x,y)[itex]\in[/itex]ℝ2 | |y2-x2|≤ 10} ...
    But nevertheless I got an idea of what's all about...

    I believe that for a 3D surface those proofs would be a lot more different (and complicated). If you don't have the time to clarify all the principles, I'd appreciate it, if someone could suggest a good book or paper to me, which gives answers to questions like those of post #13 and could all in all induct me in this area of mathematics. I've only dealed with such things within the framework of analysis or geometry so far.
     
  17. Jul 24, 2012 #16
    Really I do not know where I should start with :frown:

    Are there no book suggestions at all?

    At least a guideline? Is it topology or differential geometry which deals with such proofs on more complex objects?
     
  18. Jul 24, 2012 #17
    To determine whether a region is simply connected, it is usually easiest to draw and think about it visually (in the same sense that sometimes humans are smarter than computers), and in some cases as above, there are slick proofs like the using the parametrized line above. If you want to know the name of an abstract theory which you might teach a robot so that it was smart enough to determine if an object was simply connected, or more generally shared some topological property with another object, you would study the subject of ALGEBRAIC TOPOLOGY. In other words, this subject looks to determine global properties of a space, such as knots et cetera. It is a very strange subjects in which they try to draw as much as possible, because the algebra gets daunting quickly. But the algebra does indeed appear to be very powerful. I am NOT an expert. In the subject the aim is sometimes to take two spaces, and associate with each an algebraic object, and so instead of the difficult task of comparing the two spaces, we compare the two algebraic structures. You might google Betti numbers and torsion coefficients, which I think relate to some of the early days of the theory.

    (In other words, it is a long way off to learn the relevant machinery, so drawing is a HUGE shortcut. Also, there is a free book online that is easy to find.)


    As for the hole in your region, notice that if you take a second concentric ring a little away from the singularity, then in the resulting annular region with boundary, Stokes theorem will work again. The problem with the hole at the singularity, is that for Stokes theorem to hold, the differential (-form) field must be defined on the boundary (and nearby). But if the hole is your boundary, and the field has a singularity there, then... no dice.
     
  19. Jul 25, 2012 #18
    Thanks for the reply!

    Algebraic topology it is then! As the name suggests its study should require good backround knowledge of both abstract algebra and topology. It is the latter though which I'm still struggling with...
    But as you've already mentioned, graphs become a very useful tool for the study of complex multidimensional structures, where algebra and differential geometry become impractical. I'm just wondering, up to what level of formality do those graphs qualify in this theory!
     
  20. Jul 26, 2012 #19
    Uh oh, red flag! Assuming I know what you meant when you said "algebra and differential geometry become impractical", and when you use the word "graphs", I am not sure what you mean when you say "up to what level of formality do these graphs qualify in this theory".
     
  21. Jul 26, 2012 #20
    I simply meant that maths have always needed rigorous proofs...
    Graphs serve as a powerful tool to gain insight and motivation but I have no idea how these theories would embody them in a formal way.
     
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