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I Electric flux on non-orientable surfaces

  1. Mar 13, 2016 #1
    How to define and calculate electric flux on mobius strip or klein bottle?These surfaces are non-orientable, so I feel confused about that.
    Thanks for discussion and help.
     
  2. jcsd
  3. Mar 13, 2016 #2

    collinsmark

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    Hello @Chaos,

    Welcome to PF! :welcome:

    Fascinating question. I never thought about it until now.

    Consider the differential flux on one particular, small part of the mobius strip or Klein bottle. Let's call that small, little area vector [itex] \vec{dA_1} [/itex].

    [tex] d \Phi_{E1} = \vec E \cdot \vec{dA_1} [/tex]

    But elsewhere on the surface somewhere there also exists another differential area vector in exactly the same place and same orientation except in the exact opposite direction,

    [tex] d \Phi_{E2} = - \vec E \cdot \vec{dA_1} [/tex]

    So what do you get when you integrate over the entire surface? :wink:

    [tex] \Phi_E = \int_S \vec E \cdot \vec{dA} [/tex]
     
    Last edited: Mar 13, 2016
  4. Mar 13, 2016 #3

    Orodruin

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    You don't. Flux integrals require oriented surfaces.
     
  5. Mar 13, 2016 #4

    collinsmark

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    I don't think there's anything keeping you from doing it. The math and physics still work. The total flux is just guaranteed to be zero is all. :smile:
     
  6. Mar 13, 2016 #5

    Orodruin

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    No it is not. It depends on which part of the surface you consider to be oriented in which direction and this is an arbitrary choice. Flux integrals deal specifically with oriented surfaces.
     
  7. Mar 13, 2016 #6
    I consider what you said is right, thanks a lot~
     
  8. Mar 13, 2016 #7

    collinsmark

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    Sure, it's an arbitrary choice, but there's an arbitrary choice of surface normal direction for any surface that does not enclose a finite volume. Once the choice is made all that's necessary is consistency in application and interpretation. The mobius strip is special because when described mathematically, any given point on the mobius strip the surface will pass through that point twice, once with the surface normal oriented in the original, chosen direction and again with the surface normal oriented in the opposite direction.

    (For orientable open surfaces, each point on the surface corresponds to a single surface normal direction. However, the side that the surface normal vector points is an arbitrary choice. In contrast, in the mobius strip, there are two surface normal vectors -- equal and opposite -- at each part of the strip.)
     
  9. Mar 13, 2016 #8
    I think your idea is more intuitive.The explanation of mobius strip is fascinating.
    I come up with the explanation of klein bottle. since the klein bottle doesn't have the distinction between insides and outsides, if we place a charge the klein bottle will not enclose the charge, so it is reasonable to say the flux is zero.
    @Orodruin 's explanation is simple, but I think it is too simple and non-intuitive, so it is hard to accept that.
     
  10. Mar 13, 2016 #9

    Orodruin

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    No it won't. Not unless you integrate over the surface twice - once with each normal direction. If you do this you always get zero regardless of whether the surface is oriented or not.

    This is not true. In the set of oriented surfaces (or, more appropriately, ##N-1##-chains), each surface comes with a well defined normal direction. The normal direction is part of the definition of the surface. The set of oriented surfaces is what flux integrals are defined for. The fact that you need to select an orientation to make a more loosely defined unoriented (but orientable) surface into an oriented one is a different matter.

    The strip does not have several parts. It is an unoriented (an non-orientable) surface. For example, you are not going to integrate over it twice if you want to know its area.
     
  11. Mar 13, 2016 #10

    collinsmark

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    I agree with you on this, if the question were rephrased to ask, "what is the flux through a particular closed, finite path?" Then I agree that a Mobius strip or a Klein bottle doesn't make any sense to use. That's not because we can't find the flux through a Mobius strip or Klein bottle (which is guaranteed to be zero), but rather its because a Mobius strip or Klein bottle cannot enclose the path. It wouldn't make any sense for that application.

    Let me elaborate for those unfamiliar with flux through a closed path.

    Suppose the closed path is a simple circle and the electric field is uniform. You could calculate the flux through it by integrating over the dot product of electric field and the differential area of the flat circle, over the entire area of the circle. Another way to do it is to take a larger sphere and cut out a little hole with the same shape as original circle, and integrating the dot product of the electric field and the differential area of the sphere, over the area of sphere (well, the part of the sphere where the hole is not cut out). You'll get the same answer either way.

    You can't use this method with a Mobius strip because it is topologically impossible for the Mobius strip to fully enclose the finite path.

    But the way I interpreted the OPs question, it was a different question. I interpret the OPs question asking about the flux through the total surface of the Mobius strip or Klein bottle itself. Not through a closed path enclosed by a surface (The Mobius Strip or Klein bottle won't enclose a closed path).
     
  12. Mar 13, 2016 #11

    Orodruin

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    Regardless, the answer to the question is the same. The flux integral is defined for oriented surfaces only and it does not make sense to talk about flux integrals for non-oriented surfaces. As soon as you start considering only a part of the Möbius strip, you can define a flux integral because the small part becomes orientable (of course, you then also have to select an orientation, otherwise the integral is not well defined).

    Your argument still requires integrating twice over the Möbius strip. The result of this is equivalent to considering an oriented surface with two parts, one which points in each direction and therefore the integrals will cancel.
     
  13. Mar 13, 2016 #12

    collinsmark

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    Yes, you always get zero. That's exactly my point. :smile: (See more below.)

    Yes, but the definition is arbitrary.

    Suppose you have a flat piece of paper oriented such that one side of the paper points East and the other side points West. Which direction does the surface normal point? It's an arbitrary choice.
    Agreed.

    One needs to break up the surface into many parts as a matter of integration. That's what integration is all about. That's all I meant.

    You only need to integrate over the surface of the Mobius strip once.

    It's just that for every point in 3D space on the Mobius strip there are two correspond points on the strip's surface space. You only need to integrate over the surface once, even though it is guaranteed that for each point on the surface space, there will be a different, separate point that shares the same 3D space (even though they are distinct points on the surface space). And the two points are guaranteed to have surface normals in opposite directions.
     
    Last edited: Mar 13, 2016
  14. Mar 13, 2016 #13

    Orodruin

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    No it is not. You are not paying attention here. In the set of oriented surfaces, the orientation is part of the definition of the surface. Once it is defined, the orientation is obviously not arbitrary - it is what it is. The flux integral is only defined for oriented surfaces.

    A piece of paper is not an oriented surface. The flux integral is not defined until you select a normal direction and thereby orient it! In this particular case you have to select to compute the flux from east to west or vice versa - this defines the surface normal and makes the piece of paper an oriented surface through which you can define the flux. Obviously the different choices of orientation will result in the exact opposite result.

    Yes, but your chosen break-up is essentially equivalent to integrating over the surface twice! This is not how you integrate. On an oriented surface, the orientation of the different parts you integrate over are correlated.

    This is simply not true. You have here implicitly made a double covering of the Möbius strip which you are integrating over. This double covering is orientable.
     
  15. Mar 13, 2016 #14

    collinsmark

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    (Boldface mine.)

    Yes, we are both in agreement on that. (I don't see anything in my previous posts that contradict this. I don't understand the objection.)

    You can choose to define surface normal of the finite plane represented by the flat piece of paper to correspond to the East facing side of the paper, or you can choose to define it to correspond to the West facing side of the piece of paper. Once the choice is made, consistency is important.

    But in the process of creating the definition, the choice is arbitrary. Either possible definition will do, so far as consistency is maintained after the definition is chosen.

    I really don't think we are in disagreement on this. Maybe we should move on to other aspects.

    I do agree that each point in 3D space corresponds to two, separate points on the Mobius strip's surface space. But I do not agree that that is the same thing as integrating twice over the surface space.

    Consider a second example of an orientable surface where at various places the surface intesects itself. Take a beach ball for example, and mash it flat in the middle so that it sort of looks like a torus, but with the center section containing a plane [plane-like section] with two sides [one side facing to the left and the other to the right]. Each point in that center section in 3D space will contain two points in the surface space. You can find the total flux through this surface if want. Or, if you cut a hole in the shape (through one side and one side only [if the hole happens to be in the co-planar section -- don't cut two holes, is what I mean]) and use it to find the flux through the path defined by that hole (in this case, the object is able to adequately surround and enclose the path). My point with this example is that there is nothing in the rules which say that an object's points in surface space cannot share the same 3D space.

    The difference with the Mobius strip (compared to my example above) is that 1) all points on the Mobius strip's 3D space each correspond to two, individual points in the surface space, not just some of the the points. And 2), the Mobius strip is not suitable to find the flux through a closed path since the Mobius strip is not capable of surrounding and enclosing a path (i.e, that application doesn't make any sense).

    Yes, they are correlated in 2D surface space, but not necessarily for points in 3D space.

    Recall that mashed beach-ball example above. There are many points in 3D space that each correspond to two separate points in 2D surface space (i.e., points in surface space share the same points in 3D space -- in the mashed ball example, this is everything in the middle, mashed section). And in this example, in the case where a 3D point is in the middle, mashed section, it will correspond to two surface normals, and the resulting differential flux elements will cancel because the surface normals are in opposite directions. There are two surface normals because each surface normal corresponds to a different element of the surface regardless of the fact that the two surface elements share the same location in 3D space. There's nothing in the rules that say there is anything wrong with this.

    How is that so different than the center section of the mashed beach-ball where two parts of the beach-ball share the same plane [planar location in 3D space]?

    (Yes, I know there is a difference in-so-far that the Mobius strip only has a single side, and the mashed beach ball has both an inner and outer side. You need not point that out. [However, if this difference is your fundamental objection, I'd like to learn more about why that makes everything invalid.])

    What I'm saying is that if you were to make a Mobius strip out of a really thin (ideally thin) slice of paper, and paste on many, many, tiny, ideal flux detectors all around its entire one-and-only, single side (and assume each, tiny, ideal flux detector can detect its own orientation, knows its own surface area, and can detect the magnitude and direction of the surrounding E field such that it can calculate its own little flux), and then sum the flux readouts together (of all detectors), I predict the sum will be zero -- not undefined, or "error," but just zero.
     
    Last edited: Mar 13, 2016
  16. Mar 13, 2016 #15

    Orodruin

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    The point is that without the choice, the flux integral is not defined. Flux integrals deals with oriented surfaces and an oriented surface has a well defined surface normal.
    This is the point - it does not and so your premise is wrong. If you map two points on your surface to the Möbius strip in 3D, you do not have a Möbius strip, you have an orientable double cover of the Möbius strip. If you embed the Möbius strip into three dimensions, it has to be a one-to-one mapping just like any embedding. The two-to-one mapping you are thinking of is not an embedding - it is a two-to-one map which covers the Möbius strip twice.

    You do not integrate over coordinates mapping to the same point on the manifold twice if you want to integrate over the manifold. This is true for the sphere as well as for the Möbius strip. It is also not clear what you mean by the "center section". If it is what I think you mean, any time you map something to a three-dimensional space such that some points are mapped by two points on the surface, you do not have an embedding.

    Again, this is your premise and it is a double covering - indicating that you are not talking about a Möbius strip any more.

    And you would be wrong. The reason you are wrong has already been explained and is based on the fact that you are not considering the Möbius strip but an oriented double covering of the Möbius strip.
     
  17. Mar 13, 2016 #16

    collinsmark

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    Okay, rereading your previous posts, and found something that I missed the first time. Perhaps this explains my misunderstanding:

    If I was asked to find the area of a 1 cm by 10 cm rectangle, I would answer 10 cm^2. It's understood that it's not necessary to include the "back side" of the rectangle. Only a single side is required.

    If I was then asked to take one of the short sides, flip it over in 3D, and attach it to the other short side forming a Mobius strip (which only has one side, of course), and asked to find the surface area of that resulting, single side, I would have answered 20 cm^2.

    My logic is I were to take a pen, trace a line around the long side of the surface, and measure the distance from where I started to where I ended, I would have traced out a 20 cm length. The width remains 1 cm. So the area would be (20 cm)(1 cm) = 20 cm^2.

    So I assume now that my approach is incorrect. But where did I go wrong? (I haven't studied Mobius strips, so this is new to me. I'd like to learn something new.)
     
  18. Mar 13, 2016 #17

    collinsmark

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    'Doing some more reading here.
    https://en.wikipedia.org/wiki/Orientability

    In the mean time while I continue learning, how about I retract my previous statement where I claimed that the total electric flux through the surface of a Möbius strip is zero, and revise that to say,

    "The total electric flux through the oriented double covering of the Möbius strip is zero."

    I can live with that. :smile:
     
  19. Mar 13, 2016 #18

    collinsmark

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    Yes, @Orodruin is correct.

    From https://en.wikipedia.org/wiki/Orientability#Orientable_double_cover,

    A closely related notion uses the idea of covering space. For a connected manifold M take M*, the set of pairs (x, o) where x is a point of M and o is an orientation at x; here we assume M is either smooth so we can choose an orientation on the tangent space at a point or we use singular homology to define orientation. Then for every open, oriented subset of M we consider the corresponding set of pairs and define that to be an open set of M*. This gives M* a topology and the projection sending (x, o) to x is then a 2-1 covering map. This covering space is called the orientable double cover, as it is orientable. M* is connected if and only if M is not orientable.
    Another way to construct this cover ...
    [...]
    In the former case, one can simply take two copies of M, each of which corresponds to a different orientation.​

    That seems to be exactly what I inadvertently did.

    So yeah, although the electric flux through the orientable double cover of the Möbius strip is zero, it doesn't make sense to calculate the flux through the Möbius strip itself.

    'Like Orodruin says.
     
  20. Mar 13, 2016 #19

    Orodruin

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    I never doubted me for a moment! :wink:

    Jokes aside, this is how we learn new stuff. :smile:
     
  21. Mar 13, 2016 #20

    collinsmark

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    Yep!

    I learned something new about topology today! Win-win for all. :woot:
     
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