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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.[emoji4]
Thanks for discussion and help.[emoji4]
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.Orodruin said:You don't. Flux integrals require oriented surfaces.
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.collinsmark said: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.
I consider what you said is right, thanks a lot~Orodruin said: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.
Orodruin said: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.
I think your idea is more intuitive.The explanation of mobius strip is fascinating.collinsmark said: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.)
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.collinsmark said: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.
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.collinsmark said: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.
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.collinsmark said:In contrast, in the mobius strip, there are two surface normal vectors -- equal and opposite -- at each part of the strip.
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).collinsmark said:I interpret the OPs question asking about the flux through the total surface of the Mobius strip or Klein bottle itself.
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.collinsmark said:which is guaranteed to be zero
Yes, you always get zero. That's exactly my point. (See more below.)Orodruin said: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.
Yes, but the definition is arbitrary.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.
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.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.
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.collinsmark said:Yes, but the definition is arbitrary.
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.collinsmark said: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.
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.collinsmark said:One needs to break up the surface into many parts as a matter of integration. That's what integration is all about. As a matter of fact, that's the whole point of integration.
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.collinsmark said: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 every point on the surface, there will be a different, separate point that shares the same 3D space.
(Boldface mine.)Orodruin said:collinsmark said:Yes, but the definition is arbitrary.
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.
collinsmark said: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.
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.collinsmark said:I do agree that each point in 3D space corresponds to two, separate points on the Mobius strip's surface space
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.collinsmark said:How is that so different than the center section of the mashed beach-ball where two parts of the beach-ball share the sameplane[planar location in 3D space]?
Again, this is your premise and it is a double covering - indicating that you are not talking about a Möbius strip any more.collinsmark said: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.
collinsmark said: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.
Orodruin said:For example, you are not going to integrate over it twice if you want to know its area.
Electric flux on non-orientable surfaces is a measure of the amount of electric field passing through a surface that cannot be consistently assigned an orientation. This means that the surface cannot be flipped or rotated without changing its orientation.
The electric flux on a non-orientable surface is calculated by taking the dot product of the electric field vector and the surface's normal vector, and then integrating this over the surface. The result is a scalar value that represents the amount of electric field passing through the surface.
Electric flux on non-orientable surfaces is an important concept in electromagnetism as it helps us understand the behavior of electric fields in complex systems. It allows us to analyze electric fields in situations where traditional methods of calculating flux may not apply, such as in surfaces with holes or twists.
The main difference between electric flux on non-orientable surfaces and orientable surfaces is that the latter can be consistently assigned an orientation, while the former cannot. This means that the calculation of flux on non-orientable surfaces requires the use of more advanced mathematical techniques.
Yes, electric flux on non-orientable surfaces can be negative. This occurs when the electric field vector and the surface's normal vector are in opposite directions, resulting in a negative dot product and a negative value for the electric flux. This is similar to the concept of negative flux on orientable surfaces.