Timeline for Symmetric black-hole curves
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2013 at 11:21 | vote | accept | Joseph O'Rourke | ||
| Aug 14, 2013 at 1:47 | answer | added | Anton Petrunin | timeline score: 11 | |
| Aug 14, 2013 at 1:24 | comment | added | fedja | At Anton: That's pretty much what I thought too but the devil is in the details and I have no time now. At Yoav: Not quite. You know, the standard "explosion at infinity" solution of the heat equation would be quite a mighty power source too and here the infinity is just brought to a finite point. Needless to say, the curve should have a complicated structure at every scale near $(a,0)$, so no physical considerations can really apply. | |
| Aug 14, 2013 at 0:53 | comment | added | Anton Petrunin | Are you happy with piecewise smooth solution? If YES: An example of black hole for a strip of horizontal light rays can be constructed from two arcs of parabolas with common focus and vertical directrixes. Then you can add horizontal mirrors and take a pair of them with reflections in $x$-axis and get the curve you want. | |
| Aug 13, 2013 at 23:37 | comment | added | Joseph O'Rourke | I was looking for trapping the lightrays (and thus the title "black hole"), regardless of how it is achieved. Perhaps pieces of parabolas and ellipses stitched together ... | |
| Aug 13, 2013 at 23:11 | comment | added | Yoav Kallus | Geometric optics is a particular limit of a wave equation. Certainly time reversibility of that equation and thermodynamics should tell you that the steady state solution where a constant stream of parallel rays approaches from the left must result in an equal amount of power leaving (or else we can time-reverse the solution and have a free power source). Same goes for any finite fraction of rays being captured. | |
| Aug 13, 2013 at 22:25 | comment | added | fedja | OK, add "with finitely many reflections on any finite length piece of trajectory" then :). I guess I may have an idea how to capture the rays coming in across $(0.5,1)$ but the whole interval $(0,1)$ is harder. Or, perhaps, I'm talking nonsense. I need some more time to figure out if I have anything interesting to tell :). | |
| Aug 13, 2013 at 22:09 | comment | added | Ricardo Andrade | @fedja: I guess all that I meant is that it might be hard to define reflection at an instant in time which is a limit point of reflection instants. | |
| Aug 13, 2013 at 21:31 | comment | added | fedja | The latter condition is just infinite trajectory length... | |
| Aug 13, 2013 at 21:24 | comment | added | Ricardo Andrade | @Joseph: Are you looking for a curve which causes an incoming light ray to reflect infinitely many times on it without crossing the line $x=0$? Or are you instead looking for a curve which causes the light to stay trapped inside indefinitely in time? The latter condition may be harder to specify rigorously when the set of instants at which reflections happen has limit points. | |
| Aug 13, 2013 at 20:58 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'geometry' with more specific tags
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| Aug 13, 2013 at 20:52 | comment | added | Yoav Kallus | I think that would violated Liouville's theorem (this one: en.wikipedia.org/wiki/Liouville%27s_theorem_%28Hamiltonian%29). | |
| Aug 13, 2013 at 20:45 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |