[math-fun] Kerr metric
WDS: As far as I know, an important still-open problem is to find a way to glue the Kerr vacuum metric as the exterior, onto the metric of some realistic hunk of rotating matter as the interior, to get combined metric representing a rotating mass and its gravity field. Seems quite embarrassing that nobody has been able to do that. How hard can it be?
(But there are many successful ways to do this in the non-rotating case.)
Brent Meeker: It's probably pretty hard because you can't just have a rigid rotating hunk of matter (no rigid bodies in relativity). So you'd probably choose a perfect fluid (no viscosity) to model the matter with an appropriate equation of state to model the compressibility. No doubt it can be done numerically. Brent Meeker
--Just because rigid bodies do not exist, does not stop there from being a rigidly rotating hunk of matter (all interpoint-distances preserved as measured by the metric). Further, if you produced a solution with inviscid fluid and not preserving interpoint distances, then I would dispute your solution because viscosity presumably would cause some kind of energy loss. (Although, I suppose you could argue a solution of your sort genuinely was valid for describing a mass made of superfluid liquid helium... but helium is only superfluid in a rather small pressure & temperature set.)
On 10/24/2014 11:43 AM, Warren D Smith wrote:
WDS: As far as I know, an important still-open problem is to find a way to glue the Kerr vacuum metric as the exterior, onto the metric of some realistic hunk of rotating matter as the interior, to get combined metric representing a rotating mass and its gravity field. Seems quite embarrassing that nobody has been able to do that. How hard can it be?
(But there are many successful ways to do this in the non-rotating case.) Brent Meeker: It's probably pretty hard because you can't just have a rigid rotating hunk of matter (no rigid bodies in relativity). So you'd probably choose a perfect fluid (no viscosity) to model the matter with an appropriate equation of state to model the compressibility. No doubt it can be done numerically. Brent Meeker --Just because rigid bodies do not exist, does not stop there from being a rigidly rotating hunk of matter (all interpoint-distances preserved as measured by the metric).
But then the problem is what stress-energy tensor is implied by this hunk - nothing realistic.
Further, if you produced a solution with inviscid fluid and not preserving interpoint distances, then I would dispute your solution because viscosity presumably would cause some kind of energy loss.
Loss to where? The energy can't get out. The effect of viscosity would just be to make the stress-energy tensor more complicated and possibly even turbulent. Brent
(Although, I suppose you could argue a solution of your sort genuinely was valid for describing a mass made of superfluid liquid helium... but helium is only superfluid in a rather small pressure & temperature set.)
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I haven't looked at this, but it's freely downloadable here: < http://www.scirp.org/journal/APM/ >. --Dan
Hum, not that certain, that same person published a paper claiming a new proof of Fermat's Last Theorem, a new general relativity theory, unified field theory, and such and such. That's a lot for 1 person. There are dozen of people claiming the same thing about the Beal conjecture as well, this is not surprising since the AMS is offering 1 million dollar for the proof or a counterexample. just type 'proof of beal conjecture' in google. Bonne Journée, have a nice day. Simon Plouffe
* Dan Asimov <dasimov@earthlink.net> [Oct 25. 2014 09:03]:
I haven't looked at this, but it's freely downloadable here:
Everything under scirp.org/ is what I call fake journals. The operators spam-vertize these "journals" quite a bit. Cf. http://jjj.de/fake-conf.html Best, jj
--Dan
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PUZZLE: Let C denote the Cantor group, i.e., the countable direct product of Z/2Z with itself. Find a field (F,+,*) such that C is isomorphic to the additive group (F,+). --Dan
Adam has pointed out to me that there are various solutions to this puzzle. Aargh. The one I had in mind involves exactly one indeterminate. So ex post facto I'll add that condition: Can you solve the puzzle below using only one indeterminate? --Dan On Oct 26, 2014, at 4:03 PM, Dan Asimov <dasimov@earthlink.net> wrote:
PUZZLE: Let C denote the Cantor group, i.e., the countable direct product of Z/2Z with itself.
Find a field (F,+,*) such that C is isomorphic to the additive group (F,+).
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On the other hand, see http://www.coolissues.com/mathematics/Beal/beal.htm << Beal's Conjecture is disproved for the same reasons Fermat's Last Theorem is proved. >> Simple when you know how, ennit ... WFL On 10/25/14, Dan Asimov <dasimov@earthlink.net> wrote:
I haven't looked at this, but it's freely downloadable here:
< http://www.scirp.org/journal/APM/ >.
--Dan
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We know that there are at most finitely many counter-examples to Beal's conjecture, assuming Mochizuki's proof of the abc conjecture is valid. Specifically, a coprime solution to the Beal equation: x^p + y^q = z^r with p, q, r >= 3 must satisfy the following inequality (since p, q and r cannot all be equal to 3 by the special case of FLT established by Euler): 1/p + 1/q + 1/r <= 1/3 + 1/3 + 1/4 = 11/12 Now define a = x^p, b = y^q, c = z^r, so we obtain: a + b = c with a,b,c coprime. Then rad(abc) <= xyz = a^(1/p) b^(1/q) c^(1/r) < c^(1/p + 1/q + 1/r) <= c^(11/12). Hence, infinitely many counter-examples to Beal's conjecture implies that we can find infinitely many abc triples with c >= rad(abc)^(12/11), which contradicts the abc conjecture. Sincerely, Adam P. Goucher
Sent: Saturday, October 25, 2014 at 1:25 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Announced proof of Beal's conjecture (I)
On the other hand, see http://www.coolissues.com/mathematics/Beal/beal.htm
<< Beal's Conjecture is disproved for the same reasons Fermat's Last Theorem is proved. >>
Simple when you know how, ennit ... WFL
On 10/25/14, Dan Asimov <dasimov@earthlink.net> wrote:
I haven't looked at this, but it's freely downloadable here:
< http://www.scirp.org/journal/APM/ >.
--Dan
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participants (7)
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Adam P. Goucher -
Dan Asimov -
Fred Lunnon -
Joerg Arndt -
meekerdb -
Simon Plouffe -
Warren D Smith