Timeline for Anti-gravity in an infinite lattice of point masses
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| S Feb 7, 2017 at 5:58 | history | suggested | Helder Velez | CC BY-SA 3.0 |
update video link
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| Feb 7, 2017 at 2:32 | review | Suggested edits | |||
| S Feb 7, 2017 at 5:58 | |||||
| Aug 28, 2014 at 3:07 | comment | added | Bob Jarvis - Слава Україні | The remaining masses are not "repelled" from the hole - they're attracted to where the hole isn't. | |
| Jan 12, 2013 at 21:52 | history | edited | Qmechanic♦ |
retagged;
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| S Jan 12, 2013 at 20:38 | history | suggested | Helder Velez | CC BY-SA 3.0 |
new link, because youtube was removed
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| Jan 12, 2013 at 17:41 | review | Suggested edits | |||
| S Jan 12, 2013 at 20:38 | |||||
| Sep 18, 2011 at 4:13 | answer | added | Helder Velez | timeline score: -1 | |
| Sep 16, 2011 at 19:01 | answer | added | Ron Maimon | timeline score: 13 | |
| Dec 29, 2010 at 12:04 | vote | accept | Sklivvz | ||
| Dec 26, 2010 at 23:16 | comment | added | Sklivvz | @Jerry Schirmer: Regarding the first part, actually: if you take a series of open circular intervals that tends to infinity and take the sum over the enclosed point masses, the force at the center is always zero, so the limit is zero. Regarding the second part: I never said the system would contract or shrink. I meant that since all points are equivalent, if there is any acceleration it would be an accelerating translation of all points at the same time (which is the contradiction) | |
| Dec 26, 2010 at 14:21 | comment | added | Zo the Relativist | @Sklivvz: The first part of what you said is basically right--if you are careful about how you take the limit to an infinite sum, you can sometimes get a finite number when you subtract two infinite quantities. You point out a symmetry of the problem that enables one to do this to this sum. The second contraction is wrong, however--systems can expand and contract without having a center--look at expanding (or contracting) cosmologies. | |
| Dec 26, 2010 at 10:36 | comment | added | Sklivvz | To me it's fairly simple to get over the contradiction. The contribution to the force at (x,y) is exactly the opposite of the one at (-x,-y). I know that mathematically is not very sound... but any other value would also not conserve energy. A non-zero net force means acceleration (and this would apply to every point mass equally). The whole system would be accelerating out of nowhere. | |
| Dec 26, 2010 at 0:00 | comment | added | Zo the Relativist | You can regularize the sum by assembling the lattice in a particular way, and not have to worry about the formal divergences. | |
| Dec 23, 2010 at 18:19 | answer | added | kennytm | timeline score: 1 | |
| Dec 23, 2010 at 17:22 | history | edited | Sklivvz | CC BY-SA 2.5 |
added 105 characters in body
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| Dec 23, 2010 at 17:19 | comment | added | Noldorin | @Sklivvz: Intuitively, yes perhaps! I wouldn't even like to guess, mathematically. | |
| Dec 23, 2010 at 17:09 | comment | added | Sklivvz | @Noldorin: it was not a proof, hence the should | |
| Dec 23, 2010 at 16:46 | comment | added | Noldorin | @Sklivvz: That's not considering it rigourously enough. This is quite a mathematically challenging question, and involves the Euler-Maclaurin formula for sums. | |
| Dec 23, 2010 at 16:41 | comment | added | Sklivvz | @kalle43: the force on a point goes to zero as $o(r^{-2})$ as r->infinity, so the sum should converge | |
| Dec 23, 2010 at 16:35 | comment | added | TROLLHUNTER | The question is if the vector sum of all forces converge to 0 or diverge, if it diverges the question is meaningless. | |
| Dec 23, 2010 at 16:32 | answer | added | Zo the Relativist | timeline score: 3 | |
| Dec 23, 2010 at 16:28 | comment | added | unsym | I think it should be correct because of superposition principle, but the sign of force might be positive. | |
| Dec 23, 2010 at 16:17 | comment | added | Noldorin | Interesting question! :) | |
| Dec 23, 2010 at 15:59 | history | asked | Sklivvz | CC BY-SA 2.5 |