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!



Flux through a Mobius strip
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](http://en.wikipedia.org/wiki/Gauss%2527s_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](https://physics.stackexchange.com/).


