Timeline for Dynamics of electrons on a sphere
Current License: CC BY-SA 3.0
8 events
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| Sep 17, 2015 at 19:53 | comment | added | user21349 | @NoamD.Elkies: non-unique solutions can happen but only when the system starts at a singularity or reaches a singularity I don't think this is true, since Norton's dome would seem to be a counterexample. It certainly doesn't have any singularity of the particular type you describe. Or did you have in mind some definition of singularity more general than the example you gave? If so, what is it -- a singularity in what function? The indeterminacy of Norton's dome is usually described not as an issue involving a singularity but as one involving a lack of Lipschitz continuity. | |
| Sep 17, 2015 at 18:01 | comment | added | Noam D. Elkies | @BenCrowell non-unique solutions can happen but only when the system starts at a singularity or reaches a singularity. Here the electrons start in different (albeit close) positions, and they can never crash into each other because the exert repulsive forces on each other that blow up as $r \to 0$. | |
| Sep 17, 2015 at 16:27 | comment | added | user21349 | This seems closely analogous to packing of hard spheres, where it is known that random packing is always worse than the optimal packing. So I think it's very unlikely that such a system achieves the global minimum in all cases. | |
| Sep 17, 2015 at 16:25 | comment | added | user21349 | @NoamD.Elkies: The answer to Q1 must be Yes, because the differential equation has a unique solution so it must retain the initial symmetry. I don't think this really constitutes a proof, since there are counterexamples such as the famous Norton's dome: en.wikipedia.org/wiki/Norton%27s_dome . The problem is basically that solutions to the equations of motion in Newtonian mechanics may be nonunique, i.e., Newtonian mechanics is not actually deterministic in all cases. | |
| Mar 18, 2015 at 11:30 | comment | added | user35593 | I did a Matlab computation which showed that for $n>4$ the regular $n$-gon on the equator is locally unstable. m=20; f3=zeros(1,m); for n=2:m ind=(0:n-1)*2*pi/n; x=[cos(ind);sin(ind);zeros(1,n)]; eps=10^(-4); y=[cos(eps);0;sin(eps)]; %force F=zeros(3,1); for i=2:n xi=x(:,i); F=F+(y-xi)/norm(y-xi)^3; end %projection F=F-(F'*y)*y; f3(n)=F(3)/eps; end f3 | |
| Mar 18, 2015 at 10:39 | comment | added | Joseph O'Rourke | @NoamD.Elkies: Nice question on stability of equatorial $n$-gons. Thanks! | |
| Mar 18, 2015 at 1:25 | comment | added | Noam D. Elkies | The answer to Q1 must be Yes, because the differential equation has a unique solution so it must retain the initial symmetry. For Q2 the particles must approach a stable local minimum, but in general they don't have to find a global minimum, and the probability of success might depend also on the strength of the damping. (Is a regular $n$-gon on the equator locally stable once $n>3$?) | |
| Mar 18, 2015 at 0:58 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |