Timeline for Is there a singularity theorem in higher-dimensional Newtonian gravity?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2016 at 18:36 | vote | accept | Adam B | ||
| Sep 13, 2016 at 15:44 | comment | added | Willie Wong | Two comments above it should read, instead of $\sum |x_i|^2$ always remaining bounded, that $\sum|x_i|^2$ always remaining "bounded by a linear function of $t$". | |
| Sep 13, 2016 at 15:31 | comment | added | Willie Wong | ... remain valid, which then implies that all particles must collide at the end). You should think that the proof shows that "collision must happen" by contradiction, but is a bit sharper than a pure contradiction argument because we also get an upper bound on the maximum possible time before collision (we get the bound from the energy). This is a standard technique for showing blow-up of evolution equations (both ordinary and partial differential eqs/systems); an example of its use in PDEs is my recent paper. | |
| Sep 13, 2016 at 15:27 | comment | added | Willie Wong | @AdamB: thanks. Typo fixed. // In regards to your comment, I tried to address it in one of above comments, but I guess it didn't come through clearly. The point is that the breakdown will happen whenever the total gravitational potential energy becomes infinite. This can certainly happen before $\sum |x_i|^2 \searrow 0$. What the argument shows however, is that $\sum |x_i|^2$ always remain bounded (so that the "escape" scenario that can happen for $n$-body 3D gravity cannot happen), and that some collision must happen (because as long as the collision doesn't happen, all the computations... | |
| Sep 13, 2016 at 15:23 | history | edited | Willie Wong | CC BY-SA 3.0 |
fixed typo; thanks @Adam B
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| Sep 13, 2016 at 4:41 | comment | added | Adam B | @WillieWong typo: the $d-2$s in the exponents of Eqs. 1 and 2 should be $2-d$s. | |
| Sep 13, 2016 at 0:56 | comment | added | Adam B | Nice! One thing that confuses me is that on the face of it you have proved too much. I was expecting that two of the particles must collide, but you have apparently shown that they must all collide. Furthermore, according to your formula the collision must happen at the origin, even though that's just an arbitrary place. Is this because what really happens is that you get some two-point collision somewhere, and then the assumptions of proof break down? | |
| Sep 12, 2016 at 16:24 | comment | added | Nawaf Bou-Rabee | That's interesting. | |
| Sep 12, 2016 at 16:04 | comment | added | Willie Wong | @NawafBou-Rabee: to be more precise: We know by definition $R(t) := \sum |x_i|^2 \geq 0$. If $R(0) = 0$, we are done. If $R(0) > 0$, then look at $R'(0)$. If $R'(0) > 0$, by convexity for some $t < 0$ $R(t) = 0$. Vice versa for $R'(0) < 0$. If $R'(0) = 0$ then strict convexity implies that there exists both $t_+ > 0$ and $t_0 < 0$ where $R(t_+) = R(t_-) = 0$. Of course, $x_i = x_j$ for some $i\neq j$ may happen before the predicted singular times from looking at $R(t)$, but we know that if the dynamics is regular until $t_{\pm}$, at $t_{\pm}$ we must have all $x_i = 0$ and collision. | |
| Sep 12, 2016 at 16:00 | comment | added | Willie Wong | @NawafBou-Rabee: a strictly concave function that is positive at one point must equal zero somewhere. Something like $d/dt \sum |x_i|^2$ can NEVER be true in any time-reversible dynamical system. | |
| Sep 12, 2016 at 15:32 | comment | added | Nawaf Bou-Rabee | Why does the inequality after (2) imply that a collision will occur at the origin in finite time? Don't you need something like $d/dt \sum_i |x_i|^2 < 0$? | |
| Sep 12, 2016 at 15:14 | comment | added | Willie Wong | Incidentally, basically the same argument also works for the Hartree equation for a quantum particle interacting with itself through a spherically symmetric potential. See for example arxiv.org/abs/1508.02701 and references therein. | |
| Sep 12, 2016 at 15:00 | history | answered | Willie Wong | CC BY-SA 3.0 |