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