May 2012 - mathfun@mailman.xmission.com

[math-fun] my biggest lie yet
by Bill Gosper 28 May '12

28 May '12

http://www.tweedledum.com/rwg/stanfordn3.pdf (p 14) (Also: Gosper, Strip mining in the abandoned orefields of Nineteenth Century mathematics, in Computers in Mathematics (D. Chudnovsky & R. Jenks, eds.) pp 261-283, m. dekker, N. Y.) "8. Continued fractions This is three N - K planes, j = −1, 0, 1, joined by two J matrices. In the top (j = 1) plane, the k direction computes a continued fraction, and on the bottom (j = −1), it’s the n direction. [...] Thus, to get a relation between two continued fractions, we must j-hop between planes before switching directions. If k runs from c to ∞ and n runs from d to ∞, and the corresponding continued fractions converge, then they are insensitive to whatever subsequent matrices accrue on the right by way of path closure. I.e., the two paths can be (j,k,n) = (0,c,d) → (1,c,d) → (1,∞,d) and (0,c,d) → (−1,c,d) → (−1,c,∞). This yields the identity [an undersimplified version of that symmetric CF identity I sent the other day]" All of a sudden, I'm deriving false identities with this! E.g., with K and N matrices (c85) [km(k,n),nm(k,n)] [ 4 n 4 k 11 1 ] [ n k 13 1 ] [ --- + --- + -- - ] [ - + - + -- - - ] [ 5 5 30 5 ] [ 5 5 60 5 ] (d85) [[ ], [ ]] [ k 1 1 ] [ k 1 1 ] [ - - - - k + - ] [ - + - n + - ] [ 2 6 3 ] [ 2 6 4 ] Check path invariance: (c81) CHECKM(); [ 0 0 ] (d81) [ ] [ 0 0 ] Confirm convergence of a dozen N terms: (c82) (PRUD(NM(0,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d82) [- 1.12262566430759d0, - 1.12264219198252d0] This says that as a function of the tail t, the PRUD (matrix product) computes (A t + B)/(a t +b), with A/a ~ B/b ~ -1.12262. This is the "insensitivity" claimed above. Continuing to wander in the +n and +k directions shouldn't change this value much. But tacking on 69 k steps: (c88) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,69),DFLOAT(%%[1]/%%[2])); (d88) [0.30872240339868d0, 1.27956902429756d0] This is batbleep! Not only was the value bashed, the convergence was destroyed. Note that by path invariance, it doesn't matter how we got here: (c89) (PRUD(KM(K,0),K,0,69) . PRUD(NM(70,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d89) [0.30872240339868d0, 1.27956902429756d0] There's nothing obviously wrong with the K matrix except for being a bit slow to converge: (c86) (PRUD(KM(K,0),K,0,11),DFLOAT(%%[1]/%%[2])); (d86) [1.10980568563031d0, 1.48744247508643d0] (c87) (PRUD(KM(K,0),K,0,69),DFLOAT(%%[1]/%%[2])); (d87) [1.62173336398064d0, 1.62175009332196d0] Except it's not even the same sign as the N product! There's nothing magic about 69 terms of K. Using 11 instead still fails bigtime; (c83) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,11),DFLOAT(%%[1]/%%[2])); (d83) [- 0.77240969070225d0, - 1.11033801208252d0] (c84) (PRUD(KM(K,0),K,0,11) . PRUD(NM(12,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d84) [- 0.77240969070225d0, - 1.11033801208252d0] Maybe the kids can help get to the bottom of this. --rwg (j-hopping to force continued fraction form is just cosmetic, and only adds to the confusion above.)

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[math-fun] Black hole paradox...
by Warren Smith 28 May '12

28 May '12

> Salamin: > No, baryon number and lepton number are not preserved under processing through a black hole. --ah, but in the paradox I outlined, Joe "never falls in" to the hole in Mary's view (even though he falls in his own view) hence after black hole evaporation, Joe was never "processed" hence his Lepton & Baryon numbers ARE conserved. In Mary's view. So anyhow, Salamin's statement here (which last I heard was the agreed physics view too and had been stated by Hawking) encapsulates the contradiction I was speaking of. > Hawking radiation peaks at a wavelength on the order of the size of the black hole, and exponentially dies off at shorter wavelength, like the Planck black body spectrum.? If baryon number is to be preserved, then the black hole must emit baryons as part of the Hawking radiation.? But the lowest mass baryon has a mass of 1 GeV, so baryons cannot begin to be emitted until the black hole has shrunk to about 1 fm in size.? But the mass of a black hole? is proportional to its radius, and 1 fm / 10 km = 1e-19.? So by the time a black hole is ready to emit baryons, its remaining mass is about 1e-18 solar masses, and it's too late to conserve baryon number. --Agreed! And I like this argument even better than other arguments I'd seen before. > Applying the same reasoning to each of the three lepton numbers (e, ?, ?), the lowest mass lepton is the neutrino.? Even with a neutrino mass of 1 ?eV, the black hole size must shrink to 1 m. > It is possible that lepton number fails to be conserved in the first place.? In the old days when neutrinos were massless, the spin of a neutrino was always antiparallel to its momentum, and always parallel for an antineutrino, or maybe the other way around, I can never remember which.? Now that neutrinos have mass, you can go fast enough to overtake a neutrino, reverse the direction of its momentum, and thus reverse its helicity.? You now have one of two possibilities. (1) The neutrino is still a neutrino, but has the "wrong" helicity.? Its interaction via the weak force is vastly much weaker.? Or, (2) the neutrino has become an antineutrino.? In this case, lepton number is not conserved.? In the first case, the neutrino is said to be a Dirac particle, in the second case a Majorana particle.? It is not known which case is true. --Agreed. > There exist even-even nuclei that are more stable than the two adjacent even-odd isobars (equal mass nuclei).? These cannot beta decay the ordinary way.? But if there is a more stable isobar two steps away, a double beta decay is possible, with the emission of two electrons and two neutrinos.? Measured double beta decay half lives range from 7.0e18 years for Mo-100 to 3.5e24 years for Te-128 ( http://www.nndc.bnl.gov/bbdecay/list.html ).? Experiments have been underway for a long time to search for neutrinoless double beta decay, the confirmation of which would demonstrate that neutrinos are Majorana particles. > > ? --? Gene --excellent and helpful answer, but my original Joe+Mary statement of the paradox, remains un-addressed. In Mary's view, Joe never quite falls in to the hole, hence his baryon number is preserved after hole evaporates. But in Joe's view, he fell in, hence his baryon number is not preserved. Which? Incidentally, re previous complaints Mary would effectively stop being able to see Joe since he'd become too redshifted (and also since the photons Joe emitted would arrive at Mary less and less often and become spaced too far apart in time -- a different symptom of time-distortion)... these complaints become moot after evaporation complete, since now Joe's photons can reach Mary without any redshift. --- Using Salamin's "it's too late for Baryon conservation" argument, it seems to me we can argue that even if Joe never fell in, then IF at the end of the process Joe's baryons are still there, THEN we will contradict conservation of mass-energy, since all but 10^(-18) part of the energy already was radiated away before the hole started to radiate baryons and whatever was left of Joe managed to become visible to Mary again. I.e, it is too late for Joe's baryons to be preserved because there is not enough mass-energy left.

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[math-fun] Square lattice percolation
by Allan Wechsler 28 May '12

28 May '12

If the cells of a square grid are randomly colored black and white, with p being the probability of a cell being black, then for p small enough we can ask the expected size F(p) of the black polyomino including a given cell. (If that cell happens to be white, the size is zero.) The reference cell is white with probability (1-p); this contributes 0*(1-p)=0 to the expected polyomino size. It is black, and all its neighbors are white, with probability p(1-p)^4; this contributes p - 4p^2 + 6p^3 - 4p^4 + p^5 to the expected size. All bigger polyominoes only contribute quadratic or higher terms, so we know the power series for F(p) begins 0 + p + ... If you perform a census of the ways the starting cell can be part of a domino, you learn that the next term must be 4p^2. Repeating this exercise with the trominoes, if I have done it right, gives us the cubic term, 12p^3. I conjecture that all the coefficients of this power series are positive. If I did my counting right, the quartic coefficient is 24. The sequence incipit 0, 1, 4, 12, 24, only yields one hit on OEIS, and it isn't this. I tried to calculate the quintic coefficient and made a blunder somewhere. This must be known stuff. What are some more terms of this series? Each oriented polyomino with one distinguished cell contributes n p^n (1-p)^m to the power series, where n is the number of cells in the polyomino, and m is the number of cells immediately adjacent to it.

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[math-fun] Black hole paradox (Hawking radiation)
by Warren Smith 28 May '12

28 May '12

I looked through quite a few papers that tried to solve the Hawking information-loss puzzle and the one I liked the best was this one: http://arxiv.org/abs/hep-th/0507287

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[math-fun] Send "Bignum Arithmetic" ringtone to your cellphone
by Henry Baker 28 May '12

28 May '12

FYI -- That is, if you want a 46-minute ringtone that is more likely to put you to sleep... http://beemp3.com/download.php?file=9448349&song=1658-How+to+implement+bign…

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[math-fun] "email trolling"
by Bill Gosper 28 May '12

28 May '12

> Two technical notes ("email trolling"): > > Your mail claims > Content-Type: text/plain; charset="iso-8859-7" > but then you use the UTF-8 code for "Pi": > Char: π (960, #o1700, #x3c0, file ...) point=1485 of 1704 (87%) > column=34 > This renders the character either corrupted or > "invisible" (happens for me (mutt)) in MUAs that behave correctly. > > > Further you use > Content-Transfer-Encoding: base64 > This means the content is just one big lump > of a hex string in the mail-box, making grep useless. > I suggest to use > Content-Transfer-Encoding: 8bit > > cheers, jj You must be confusing the right way to do things with the GMail way to do things. Despite these irregularities, normally touchy SquirrelMail finds nothing wrong with this GMailing other than receiving two (identical?) simultaneous copies, even though it completely begibbered the πs in my *Subject: * [math-fun] Sum csc($B&P(B k coth 1)/(k log k) converges? [Was: Teacher trolling] (and the Japanese in my Engrish impertinence.) Seriously, can you tell me what to tell GMail? --rwg

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[math-fun] Black hole paradox
by Warren Smith 28 May '12

28 May '12

OK, I read S.Hawking's http://arxiv.org/pdf/hep-th/0507171v2.pdf Phys.Rev.D72 (2005) 084013 but do not understand it. It supposedly shows "information is not lost during black hole formation+evaporation." This also was supposedly shown by J.Maldecena's "AdS-CFT correspondence" which I do understand (for a very poor value of "understand" -- I understand his top level argument plan). Neither tells me the answer to this: "Is Joe's lepton number preserved when he falls into a black hole and it later evaporates?" Neither convinces me they're right, either. In fact, if that exact verbatim paper had been written by me, not Hawking, I feel 100% confident Physical Review D would not have published it -- i.e. this is yet another example of the failure of the refereeing process in Science. Now here's a little thought experiment that suggests Hawking is wrong. Consider the big bang that created our universe. Did it generate lots of information from {little or none}? If so, we would seem to have a counterexample to Hawking's "physics is information-preserving (unitary)" claim. Another: the correct theory of quantum gravity is not known. If it is possible to create two such theories, both self-consistent, but one features black hole info loss, other doesn't, then I would conclude that the power of pure thought simply cannot tell us whether "info loss" happens. And I think that is the situation we are in, or anyhow neither Maldecena nor Hawking has ruled that out. Instead they each have worked within one partial quantum theory of gravity, and since not the same theory, evidently they disagree.

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Re: [math-fun] Black hole paradox
by Henry Baker 28 May '12

28 May '12

I asked a physicist precisely the same question back in the early 1990's; so far as I could tell, he didn't understand the question. Now that computer scientists & physicists are at least speaking the same language (at least when talking about quantum computers), the question needs to be asked again. My best understanding of physics & information is the following: Since the evolution of physical reality is "unitary", the volume of "volume elements" of phase space are conserved. This means that time-reversal can be contemplated, at least in theory. If one believes in "collapse of the wave function" due to "measurement", then physics still has to preserve some of these properties. In particular, in order to conserve phase space volume, if an amount of information of N bits is created, then an equal # of bits of information has to be destroyed. Otherwise, thermodynamics/entropy/heat engines/Maxwell's Daemons don't work right. So even if information can be created/destroyed in our universe, the _amount_ of such information must remain constant. In particular, Hawking/Bekenstein tell us that the _amount_ of information within a physical volume is precisely determined by the surface area surrounding that volume. So for the universe to "inflate", a gigantic amount of information has to be quickly "pumped into" the universe. Try convincing your local creationist about the absence of a deity after you tell him/her about this little problem... At 05:00 PM 5/27/2012, Warren Smith wrote: >Now here's a little thought experiment that suggests Hawking is wrong. Consider >the big bang that created our universe. Did it generate lots of information from >{little or none}? If so, we would seem to have a counterexample to Hawking's >"physics is information-preserving (unitary)" claim.

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[math-fun] black hole paradox (Hawking radiation)
by Warren Smith 27 May '12

27 May '12

> From: Mike Stay <metaweta(a)gmail.com> > http://en.wikipedia.org/wiki/Thorne-Hawking-Preskill_bet --I fail to see the relevance. I ask "did Joe fall in, or not?" > From: Henry Baker <hbaker1(a)pipeline.com> > Story 3: > Scenario 1. I'm looking out at the universe from the Earth, and in every direction I see a red shift which is larger, the "further away" it is. I conclude that I am in an expanding universe. > > Scenario 2. I'm looking out at the universe from the Earth, but at some very large distance from Earth, there is a gigantic black hole, which bends _all_ the light that I see, so that no matter in which direction from Earth I look, I'm actually seeing the same black hole, but from different directions. Furthermore, objects & radiation going into this black hole become red-shifted due to the enormous gravity of the black hole. > > How do I tell these two scenarios apart? Does it even matter; perhaps they are equivalent? > > I got this idea by considering what our universe looks like to Joe as he approaches the black hole; the black hole actually fills up _more_ than 50% of the sky, due to the bending of the light rays. Just before Joe falls in, the black hole fills up 99.9% of Joe's sky, and the universe that he's exiting looks like a very distant black hole to him. --Interesting claims, but I doubt true. First, the "universe Joe is exiting" is blue shifted in his view not redshifted. Second, light rays from the black hole horizon surface cannot reach Joe (while outside hole) at all. What Joe sees is not immediately obvious to me, though I think somebody has computed a movie and there is known math on this. I think Joe's view will be very distorted like thru a funhouse mirror but he'll see the outside-the-hole universe in over 50% of his perceptual sky. > From: Robert Munafo <mrob27(a)gmail.com> > Since I am totally unable to understand Hawking, I suppose I am > qualified to give the obvious answer, and not feel too bad about being > wrong (-: --you don't need to understand Hawking, you just need to know the general claimed properties of Hawking radiation (I think). > Mary, while watching the black hole shrink, sees the red-shifted image > of Joe disappear. This happens after a sufficient length of time that > Joe's image has already been redshifted so far that the uncertainty > principle (viewed as the energy-time relation deltaEdeltat > h) prevents Mary > from distinguishing when Joe's red-shifted image became invisible. --true. Joe gets exponentially red shifted in Mary's view, which soon overcomes even Mary's ability to wait 10^70 years. So Mary's "view" of Joe "never falling in" effectively gets blacked out, she can't see him with normal light. If however Joe were transmitting magic nonphoton information-particles which just carry info not energy, at lightspeed, then Mary would continually "see" Joe never falling in. > So the "paradox" created by quantum effects (i.e. Hawking evaporation) > is also resolved by quantum effects. --not really. I mean, Joe still never falls in, even though Mary has trouble seeing him. --A better objection is this: in Mary's view, Joe gets ultra-near the horizon without fall in. But as the black hole Hawking-shrinks, the horizon shrinks, ultimately to far smaller than an atomic nucleus, with Joe still incredibly close to (but outside) it. Hence Joe really is crushed, even in Mary's view where he never falls in. But even that does not really resolve the paradox. Here is why: Joe's component particles (neutrons, etc) stay outside, in Mary's view. Their characteristic counts (e.g. "lepton number") hence are preserved. But in Joe's view, those particles all fell in. The Hawking radiation then knows only about the mass and charge of the hole, and will NOT preserve lepton number, etc. So Joe still is preserved (Mary's view) versus not (Joe's view) for some weaker notion of "preserved." This still seems a contradiction. > On 5/27/12, Warren Smith <warren.wds(a)gmail.com> wrote: >> Joe jumps into a Schwarzschild black hole. >> Mary stays behind and watches him fall. >> >> The standard stories are then as follows: >> [...] >> 4. NOW, new addition to above stories. The black hole gradually is >> shrinking due to "Hawking radiation." It will vanish completely in about 10^70 >> years (let's say). >> >> So Mary, who is very long-lived, will then say "the new state of the >> universe is: >> there is no black hole anymore, Joe never quite fell in, and so Joe is >> still with us. In fact, I'm going to go meet him now, since he's only 1 hour older by his >> personal clock." >> >> But Joe will say "I fell in, I was crushed into a point, and my mass-energy >> was Hawking-radiated away. I'm gone." >> >> How do you resolve this paradox? > > -- > Robert Munafo -- mrob.com > Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - > mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com > > > > ------------------------------ > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > End of math-fun Digest, Vol 111, Issue 25 > ***************************************** -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step) and math.temple.edu/~wds/homepage/works.html

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Re: [math-fun] Black hole movies
by Henry Baker 27 May '12

27 May '12

I believe the motion of the movie (mostly), but I don't believe this is the image that Joe would see. Light orbits the black hole, as well, which should create a sort of laser effect; I don't see that here. --- I'm amazed that with all the $$$ James Cameron has made, he hasn't yet funded a sci-fi movie with true relativistic/GR graphics. We've now got the graphics horsepower to do this in high def, so we should be able to screw the heads of the current crop of teenagers on straight w.r.t. relativistic effects. At 02:10 PM 5/27/2012, Warren Smith wrote: >http://casa.colorado.edu/~ajsh/schw.shtml > >gives computer-made movies showing what you'd see falling into a hole >in various ways. It helps to read the text explanation of what you >are about to see on each movie.

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