Solutions to the wave equation on non orientable surfaces like a mobius strip - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:41:15Z http://mathoverflow.net/feeds/question/29175 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29175/solutions-to-the-wave-equation-on-non-orientable-surfaces-like-a-mobius-strip Solutions to the wave equation on non orientable surfaces like a mobius strip sigoldberg1 2010-06-23T00:05:56Z 2010-06-25T14:43:08Z <p>Given a mobius strip, what do the solutions of the wave equation look like qualitatively? How do they differ from solutions on the equivalent strip glued together as a cylinder? Any refs, particularly to symmetry?</p> http://mathoverflow.net/questions/29175/solutions-to-the-wave-equation-on-non-orientable-surfaces-like-a-mobius-strip/29195#29195 Answer by David Bar Moshe for Solutions to the wave equation on non orientable surfaces like a mobius strip David Bar Moshe 2010-06-23T07:24:30Z 2010-06-23T07:24:30Z <p>The following <a href="http://physics.bgu.ac.il/~dcohen/ARCHIVE/mbs_PRB.pdf" rel="nofollow">paper</a> by: Kousuke Yakubo,Yshai Avishai,and Doron Cohen describes in section II the solution of the Helmholtz equation on a flat rectangular surface having the topology of a Mobius strip. The solution is given in terms of the admissible wave numbers.</p> http://mathoverflow.net/questions/29175/solutions-to-the-wave-equation-on-non-orientable-surfaces-like-a-mobius-strip/29257#29257 Answer by S. Carnahan for Solutions to the wave equation on non orientable surfaces like a mobius strip S. Carnahan 2010-06-23T16:40:32Z 2010-06-23T16:40:32Z <p>Any solution to the wave equation on a Möbius strip lifts to a solution on its orientation double cover, which is a cylinder of equal width but twice the circumference. In order for a solution on the cylinder to descend to the Möbius strip, it is necessary and sufficient that it be invariant under a certain order two symmetry. If the cylinder has coordinates given by $[0,\pi a]$ in the free direction and $[0, 2\pi L]$ in the loop direction, then the solutions are linear combinations of products $AB$, where $A$ has the form $\cos (kn_a t)\cos(\frac{n_a}a x)$ or $\sin (kn_a t)\cos(\frac{n_a}a x)$, and $B$ has the form $\cos(kn_Lt)\cos(\frac{n_L}{L}y)$, $\sin(kn_Lt)\cos(\frac{n_L}{L}y)$, $\cos(kn_Lt)\sin(\frac{n_L}{L}y)$, or $\sin(kn_Lt)\sin(\frac{n_L}{L}y)$. Here, $n_a$ and $n_L$ are nonnegative integers, and $k$ is a constant. Invariance under the symmetry is equivalent to $n_a + n_L$ being an even number. In contrast, solutions on the cylinder of the same dimensions correspond to solutions on the double cover such that only $n_L$ is even.</p>