# Flux through a Mobius strip

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A friend of mine asked me what is the flux of the electric field (or any vector field like $$\vec r=(x,y,z)\mapsto \frac{\vec r}{|r|^3}$$ where $|r|=(x^2+y^2+z^2)^{1/2}$) through a Mobius strip. It seems to me there are no way to compute it in the "standard" way because the strip is not orientable, but if I think about the fact that such a strip can indeed be built (for example using a thin metal layer), I also think that an answer must be mathematically expressible.

Searching on wikipedia I found that

http://en.wikipedia.org/wiki/Mobius_resistor

A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. It provides a resistor which has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time.

How can I relate the highlighted phrase to some known differential geometry (physics, analysis?) theorem?

Thanks a lot!

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user tetrapharmakon
asked Jan 5, 2011
retagged Dec 8, 2014
Hello tetrapharmakon and welcome to the site. Imho this is a nice question since one has to think about some fundamental concepts. You might also want to look at prb.aps.org/pdf/PRB/v79/i3/e035321 for another application of the Möbius strip :)

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user Robert Filter
This is a great question!

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
do you know the experimental setting flair.monash.edu.au/publications/pdfs/BBVIV5_Leweke_etal.pdf ?

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user janmarqz

## 6 Answers

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As you say, there is no standard definition of the flux through a nonorientable surface. I will try to convince you that you shouldn't want to define this.

There are two standard facts about flux. The first is that, if $V$ is a three dimensional volume, with boundary $\partial V$, and $F$ is a vector field, then $\int_{\partial V} F \cdot n = \int_V \nabla \cdot F$. But a Mobius band cannot be embedded in a closed surface which contains a volume $V$, so this formula doesn't apply. Physically speaking, what I am saying is that you would never care about Gauss's law for a Mobius band, because you can't make sense of the notion of the band being part of a surface which encloses a charge.

The other key mathematical fact is that, if $S$ is a surface with boundary $\partial S$ and $F$ a vector field, then $\int_{\partial S} F = \int_{S} \nabla \times F$. I will show that there is no hope of a formula like this for a Mobius strip. Specifically, I will show that there is a vector field $F$ which has $\nabla \times F=0$ everywhere on the Mobius strip $M$, yet $\int_{\partial M} F \neq 0$.

Let the $z$ axis pass up through the center of the Mobius strip, with $\partial M$ winding twice around this axis and staying outside a cylinder of radius $\epsilon$ around the $z$-axis. Take the field at point $(x,y,z)$ to be $(y/(x^2+y^2), -x/(x^2+y^2), 0)$, for any $(x,y,z)$ with $x^2+y^2 \geq \epsilon^2$, and interpolate however you like within that vertical cylinder. (Physically, this is the $B$-field from a current along the $z$-axis. Or, if it isn't, then replace my formula by that one.) Then $\nabla \times F=0$ outside the vertical cylinder, and in particular everywhere on the Mobius strip. But $\int_{\partial M} F = 4 \pi$.

I previously had a physical example here, but I think I got some of the details wrong, so it is gone now.

I don't know about the Mobius resistor question. You might get a better answer at physics.stackexchange.com.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user David Speyer
answered Jan 5, 2011 by (110 points)
I think instead of "now hope" in the third paragraph you mean "no hope".

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user Noah Stein
Fixed, thanks. Makes a difference, doesn't it?

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user David Speyer
The vector field you describe is the $B$-field for some amount of current moving in the negative $z$ direction along a wire of radius at most $\epsilon$.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user S. Carnahan
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If you cut a Möbius band the short way, to produce a rectangle, and orient that rectangle, then the flux through that rectangle is well-defined. The same procedure on a cylinder produces a constant function, since the flux does not depend on the cut. Because the orientation on the rectangle disagrees with itself across the cut for a Möbius band, the flux is not constant. It is a continuous function which might be viewed as spin-$1/2$, or antiperiodic with antiperiod $2\pi$: The flux reverses sign when you continuously change the cut and orientation around the Möbius band since that reverses the orientation of the rectangle.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user Douglas Zare
answered Jan 5, 2011 by (60 points)
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Assuming the material of the strip is a good conductor any charges, and currents, on the strip must end up on the boundary. This is true at least in the limit of small applied e.m.f.s. So its not really Gauss', but Ampere's law you're looking for here:

$$\int_{\partial M} I \cdot dl = \Phi_B$$

i.e. the line integral of the current $I$ around the boundary $\partial M$ of the Mobius strip $M$. This quantity is actually a topological invariant of the manifold (in this case $M$) under question. For the Mobius strip it is as can be seen to be zero by starting at any given point on the boundary and evaluating the integral as one moves along the edge for one complete cycle.

Note: Some of the arrows are incorrect. Think of the strip as a freeway with the vertical axis running through its center, where cars drive only on the edges. You can divide the edge into parts "upper" and "lower", then the current should be circulating one way in the "upper" half and the opposite way in the "lower" half, changing directions at the place where the "interchange" is located. Label accordingly.

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
answered Jan 5, 2011 by (1,985 points)
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The currents in the picture are correct. The "inner" part is in contact with the + , the "outer" with the -. At the twist, there is a switch between "inner" and "outer".

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user Raskolnikov
The arrows are incorrect if you take the current in the strip to be running along the edges, instead of in the interior of the strip itself. So, I guess they are correct in the original context of the figure, but incorrect for my purposes.

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
I mean, along the axis through an infinite line crossed by current

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user tetrapharmakon
@Tetrapharmakon yes that is the orientation of the strip. The strip lies in the xy plane and the z-axis is the vertical direction.

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
I was saying, are you considering an "infinite" line crossed by current and the boundary of the strip intertwining it twice? Or something like the boundary itself is crossed by current?

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user tetrapharmakon
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Well, it only makes sense to talk about what is possible to be made into a resistor. So I talk precisely about the resistor that is presented on the wiki page. That is, two metallic strips separated by a dielectricum and then the ends are joined together after twisting so that the "outer" strip connects to the "inner" strip and visa versa. But it makes sense to consider a loop that goes twice round but taking once the "inner" path and then the "outer". Not sure anymore what space_cadet has in mind.

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user Raskolnikov
it makes sense to consider a loop that goes twice round but taking once the "inner" path and then the "outer" - @raskolnikov that is what I'm talking about, if that helps. And, thanks for the clarification about whether or not there is a current in the "inside of the strip".

This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
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David Speyer already explained it very well. But if you want more details of the explicit calculations, you can look here:

http://www.math.uwaterloo.ca/~karigian/teaching/multivariable-calculus/moebius.pdf

I made this handout many years ago for a multivariable calculus class. Enjoy.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user Spiro Karigiannis
answered Jan 8, 2011 by (60 points)
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You probably know this, but if you want to define numerical value of flux through a surface, you need to specify a "plus" direction and a "minus" direction, so we can say whether a small chunk of flow through the surface contributes positively or negatively to the total. Since a Möbius strip has no consistent choice of normal direction, this can't be done. The physicality of a Möbius strip doesn't help you define flux, but it does drive home the point that you can't paint such an object with two different colors with no color discontinuities.

It is possible to consider a notion of gross flow, where you integrate the absolute value of flux. I don't know what people do with it, but it doesn't seem to be stable against high-frequency perturbations.

The claims about the Möbius resistor seem to be overselling the merits a bit. This probably depends on your chosen application, but I'm pretty sure you can minimize residual self-inductance with counterrotating coils of resistive wire.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user S. Carnahan
answered Jan 5, 2011 by (190 points)
@Scott-Carnahan, the Mobius resistor concept missing the fact that the only resistance being provided between two leads connected to it is the internal resistance of the conductor. In electrical circuit modeling, you take the resistance between two components connected to the same lead to be $0$ (zero). There's no need to worry about "self-inductance" with a mobius resistor because you're also connecting to the same side of a capacitor. It might act a little bit like an antenna perhaps, but not as a resistor in any useful sense of the word. A mobius resistor is a short circuit.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user sleepless in beantown
Perhaps I was too generous in my assessment. I figured the strip of conductor would be made out of something with nontrivial resistance.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user S. Carnahan
Scott Carnahan: There's some evidence that your original assumption was right. A 1964 /Time/ article about the Möbius resistor says, "Davis made a Möbius loop out of a strip of nonconducting plastic that had metal foil bonded to both sides to serve as an electrical resistance." Since the plastic core is non-conducting, it appears that all the resistance is supposed to be provided by the foil coating.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user Vectornaut
Oops! Forgot the link: time.com/time/printout/0,8816,876181,00.html

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user Vectornaut
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The Mobius resistor concept seems like bunk/hokum because there is only one side to the resistor: both leads connecting to such a resistor would be connected to the same side and would theoretically have $0$ (zero) resistance for circuit-modeling purposes.

The three answers prior to my comment here by Douglas Zare, David Speyer, and Scott Carnahan should all be sufficient to convince you that since you can't orient this surface, you cannot define a positive/negative orientation for the direction of flow through this surface and thus cannot define a flux through a mobius strip.

This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user sleepless in beantown
answered Jan 5, 2011 by (30 points)

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