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Duda Jarek

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  1. Before transforming into a black hole, it was a neutron star - I am asking about the starting moment of this transformation: when event horizon has just appeared in the center of the neutron star. Then it evolved to finally get out of its surface - from this moment we can call it a black hole. As radius of event horizon is proportional to mass inside, mass is proportional to density times third power of radius, density of matter in the moment of starting event horizon in the center had to reach infinity first. But if baryons are destructible, they should not survive this infinite compression - should be destroyed earlier, creating pressure inside and temporarily preventing the collapse ...
  2. Sure, Hawking radiation does not directly violate baryon number conservation, but only implies that destruction of baryons is possible. If so, it should start happening before getting to infinite density in the center of neutron star, what is required to start forming the event horizon ...
  3. Swansont, Huge amount of baryons form a star, which collapses ... and then "evaporates" into massless radiation. Lots of baryons in the beginning ... pooof ... none at the end - how it is not baryon destruction? Maybe they have just moved to an alternative dimension or something? MigL, so is baryon number ultimately conserved? Could there be created more baryons than anti-baryons in baryogenesis? Can baryons "evaporate" through Hawking radiation?
  4. I am not asking about some specific theory, but the reality. Black hole evaporation requires that baryons are destructible, while formation of event horizon requires reaching infinite density in the center of neutron star - requires that matter can be infinitely compressed, without destruction of its baryons - contradiction.
  5. Ok, it is not exactly a paradox, but just a self-contradiction: if baryons are not indestructible, they should be destroyed before reaching infinite density required to start forming the event horizon.
  6. Swansont, I don't know where exactly these baryons are destroyed. The fact is that initially there were baryons, and finally they are no longer - like in proton decay or baryogenesis, the baryon number is not conserved. About the "massless radiation", they usually expect some EM radiation (anyway, I don't know any massless baryons?)
  7. The hypothetical Hawking radiation means that a set of baryons can be finally transformed, "evaporate" into a massless radiation - that baryons can be destroyed. It requires that this matter was initially compressed into a black hole. If baryons can be destroyed in such extreme conditions, the natural question is: what is the minimal density/heat/pressure required for such baryon number violation? (or while hypothetical baryogensis - creating more baryons than anti-baryons). While neutron star collapses into a black hole, event horizon grows continuously from a point in the center, like it this picture from: http://mathpages.com/rr/s7-02/7-02.htm As radius of event horizon is proportional to mass inside, the initial density of matter had to be infinity. So if baryons can be destroyed, it should happen before starting the formation of event horizon - releasing huge amounts of energy (complete mc^2) - pushing the core of collapsing star outward - preventing the collapse. And finally these enormous amounts of energy would leave the star, what could result in currently not understood gamma-ray bursts. So isn't it true that if Hawking radiation is possible, then baryons can be destroyed and so black holes shouldn't form? We usually consider black holes just through abstract stress-energy tensor, not asking what microscopically happens there - behind these enormous densities ... so in neutron star nuclei join into one huge nucleus, in hypothetical quark star nucleons join into one huge nucleon ... so what happens there when it collapses further? quarks join into one huge quark? and what then while going further toward infinite density in the central singularity of black hole, where light cones are directed toward the center? The mainly considered baryon number violation is the proton decay, which is required by many particle models. They cannot find it experimentally - in huge room temperature pools of water, but hypothetical baryogenesis and Hawking radiation suggest that maybe we should rather search for it in more extreme conditions? While charge/spin conservation can be seen that surrounding EM field (in any distance) guards these numbers through e.g. Gauss theorem, what mechanism guards baryon number conservation? If just a potential barrier, they should be destroyed in high enough temperature ... Is matter infinitely compressible? What happens with matter while compression into a black hole? Is baryon number ultimately conserved? If yes, why the Universe has more baryons than anti-baryons? If not, where to search for it, expect such violation? If proton decay is possible, maybe we could induce it by some resonance, like lighting the proper gammas into the proper nuclei? (getting ultimate energy source: complete mass->energy conversion) Is/should be proton decay considered in neutron star models? Would it allow them to collapse to a black hole? Could it explain the not understood gamma-ray bursts?
  8. If someone is interested in this subject, I have recently created presentation with new pictures, explanations: https://dl.dropboxusercontent.com/u/12405967/qrsem.pdf Picture-like QR codes is only one of many new possibilities of these extensions of Kuznetsov-Tsybakov problem - when the receiver doesn't need to know the constrains (picture/music/noise characteristics/...), for example there are plenty of new steganograpic applications: - if we would like to encode information in exchanging the least significant bits (LSB) of a picture, but (to be more discrete) would like to change only the least possible amount of bits and the receiver doesn't know the original picture, - if we cannot just manipulate LSB because there is only e.g. 1bit/pixel, we can recreate local grayness by statistics of distribution like in the Lena-like codes above. We could also use more advances dithering methods: getting better picture quality, but at cost of reduced data capacity and algorithm will become more complex (probability for a pixel should depend on already chosen its neighbors). For example e.g. inkjet printers use just 3 colors of nearly identical dots - there are already hidden some basic informations like serial number there, but in a casually looking print we could hide really huge amount of information in precise dot positioning (would require a microscopic precision scanner to decode), - if there is a device producing some noise, we could send information in faking this noise - Kuznetsov-Tsybakov would be required e.g. if the noise characteristics are varying in time, - there are also new possibilities of hiding information in sound, for example to reduce EM noise and target only those e.g. in given public place, some information for your smartphone could be hidden in music you hear ... ... What other applications could you think of?
  9. I've recently realized that this method is extremely close to the base of lossy compression: rate distortion problem - they are kind of dual: while searching for the code, we just need to switch essential bits with the discarded ones. So the rate for rate distortion application is "1-rate" of the original one (last version of http://arxiv.org/abs/1211.1572 ). For example to store distorted/halftone above plane pictures, we need correspondingly 1/8, 1/4 and 1/2 bits/pixel. Intuition is that while the bitmap requires e.g. 1 bit/pixel, such small part of bit stores the "visual aspect", while the rest can be used to store a message (but don't have to be). Correction Trees are perfect for such purposes - can cheaply work about 1% from the theoretical channel limit. Additionally, there appears probably new(?) applications: - extremely cheap storing of halftone pictures, like about 0.18 bits/pixel for Lena above (about 6kB), - dual version of Kuznetsov and Tsybakov problem (gets out of standard nomenclature of rate distortion) - we want to send n bits, but such that only k<n of them are fixed by us - the receiver don't know which, the rest of bits can be random - it occurs it is enough to send a bit more than k bits. Where could we use it - e.g. extremely cheap storing of halftone pictures? ps. Some more devaloped discussion: http://forums.devshed.com/dev-shed-lounge-26/embedding-grayscale-halftone-pictures-in-qr-codes-933694.html
  10. What is this 512x512 picture? Bad Quality Lena indeed, but the trick is that it's black&white - can be directly seen as length 512x512=32kB bit sequence. It occurs that making bit sequence "looking like Lena" reduces the capacity only to about 0.822, what is about 26kB in this case - the visual aspect costs only about 6kB. It has rather too high resolution for practical applications, but here are examples of lower resolution codes looking like a chosen black and white picture: for example the central noisy code contains 800*3/4=600 bytes - making it look like the picture costs only 200 bytes. Here is fresh paper about obtaining it (generalization of Kuznetsov and Tsybakov problem - for constrains known to the sender only): http://arxiv.org/abs/1211.1572 What do you think of replacing today QR codes with nicer looking and more visually descriptive ones? What other applications could you think of for this new steganography for which two colors is finally enough?
  11. Thanks, yes indeed - I had this value in my mind and forgot that it included electron production ... The required positive charge inside neutron to make it stable occurs to be known: http://www.terra.es/personal/gsardin/news13.htm
  12. More interesting nuclear physics questions (waiting for more ): - how short-range strong interaction explains halo nuclei ( http://en.wikipedia.org/wiki/Halo_nucleus )? Like Lithium 11 (8.75ms halftime!), which accordingly to CERN article ( http://cerncourier.com/cws/article/cern/29077 ) comparing to Pb208 should look like that: - mass difference between tritium (3.0160492u) and helium 3 nucleus (3.0160293u) is only 18.6keV - how this decay can be beta decay if the difference between neutron and proton mass is 762keV?
  13. alpha2cen, again the stability of deuteron is obvious in above picture (neutron requires charge for stability, so proton shares its own with neutron), but since they barely calculate single nucleon using lattice QCD, understanding deuteron stability there might be quite distant.
  14. alpha2cen, while it is obvious from the picture above, it seems far nontrivial using QCD: I've just found 4 year old news about calculating nucleon mass on Blue Gene using lattice QCD: 936MeV for both +-22 or 25MeV: http://physicsworld.com/cws/article/news/2008/nov/21/proton-and-neutron-masses-calculated-from-first-principles ps. please remove unnecessary quotes.
  15. Why neutron requires charge to become stable? Why against Coulomb attraction, nucleus require charge? How this charge is distributed inside nucleon (quarks)? Nucleus? Large nuclei are believed to behave like a liquid – are nucleons freely swimming there, or maybe they are somehow clustered, like in neutron-proton(-neutron) parts? I’m thinking of a topological soliton model of particles (concrete field structures for particles, with quantum numbers as topological charges), which doesn’t only restrict to mesons and baryons like the Skyrme model, but is an attempt to a more complete theory: which family of topological soltions would correspond to our whole particle menagerie. Extremely simple model seems to qualitatively fulfill these requirements: just real symmetric tensor field (like stress-energy tensor), but with Higgs-like potential: preferring some set of eigenvalues (different) – it can be imagined as ellipsoid field: eigenvectors are axes, eigenvalues are radii. Now e.g. simplest charges are hedgehog configurations of one of three axes and topology says there is some additional spin-like singularity required (hairy ball theorem) – we get three families of leptons etc. This is all of nothing approach: a single discrepancy and it goes to trash. Instead it seems to bring succeeding simple answers, like for above questions: Basic structures there are vacuum analogues of Abrikosov vortex: curve-like structure around which (“quantum phase”) two axes makes e.g. pi rotation for ½ spin. The axis along the curve can be chosen in three ways – call them electron, muon or taon spin curve correspondingly. Now baryons would be the simplest knotted structures – like in the figure, two spin curves need to be of different type. The loop around enforces some rotation of the main axis of the internal curve – if it would be 180degrees, this axis would create hedgehog-like configuration, what corresponds to +1 charge. Locally however, such fractional rotation/charge may appear, but finally the sum has to be integer. This picture explains why charge is required for baryon stability, that for this purpose proton can share its charge with neutron. Lengthening the charge requires energy, what makes two nucleons attracting in deuteron - by this strong short-ranged force. Here is fresh paper about this model: http://dl.dropbox.com/u/12405967/elfld.pdf To summarize, this simple model suggests that: - neutron has quadrupole moment (!), - nucleons cluster into “-n-p-“ or “-n-p-n-“ parts sharing charge, - pairs of such clusters can couple, especially of the same type (stability of even-even nuclei), - these clusters are parallel to the spin axis of nucleus(?). How would you answer to above questions? Do these suggestions sound reasonably? Can neutron have quadrupole moment? (Shouldn’t it have dipole or quadrupole moment in quark model?)
  16. Generally, yes. It's important that these bit sequences are uncorrelated and with P(0)=P(1)=1/2 to make it contain maximum amount of information (it would be interesting to consider different cases...) - for example you ask if value is smaller or larger then median of given group. The 30bit length can be sometimes not enough - for simplicity assume you start with infinite one, then cut to required lengths - the average length is expected to be about lg(200)+1.33. So just encoding sequences would take about 200*(lg(200)+1.33) ... but you can subtract lg(200!) bits of information about their order.
  17. Phillip, If for each box there is practically random sequence representing its individual features, to make it distinguishing it indeed has to be about lg(n) bit length in the perfect binary tree case. The randomness makes it a bit larger - this average length becomes D_n ~ lg(n) + 1.3327 So the total amount of information of this population would be nD_n ~ nlg(n) ... but we we don't care about the order of these sequences - all their orderings are in the same equivalence class, so the number of possibilities decreases n! times - the total amount of information is decreased by lg(n!) ~ nlg(n): H_n = nD_n - lg(n!) ~ 2.77544n The amount of information can be further reduced for degree 1 nodes: if all specimens from the population reaching given node, make the same next step (like all tall blonds are slim in given population) - for these nodes we don't need to store this direction, reducing informational content by 1 bit per each degree 1 node. While there are n leaves and n-1 degree 2 nodes, it turns out that the expected number of degree 1 nodes is about lg(e/2)~0.44 So after such reduction, the population requires 2.3327 bits/specimen (calculated in last version of the paper).
  18. 3 features of 0/1 type can distinguish maximum 8 entities. The expected number of features required to distinguish individual (D_n bits) has to grow with approximately lg(n) More precisely: D_n ~ lg(n) - lg(e)/2n + 1.332746 But while calculating entropy of the whole population, we need subtract information about their order: H_n = nD_n - lg(n!) ~ n(lg(n) - lg(e)/2n + 1.332746) - nlg(n) + nlg(e) - lg(2pin)/2 ~ 2.77544 n where we've used Stirling formula. ps. I've just spotted that the amount of bits per element for the reduced minimal prefix tree seem to have the same decimal digits after coma as the 1.332746 above from the D_n formula. It suggests that it is 3/2 + gamma*lg(e) ~ 2.3327461772768672 bits/element Is it just a coincidence or maybe there is some nice and simple interpretation??
  19. Once again, it's the lower boundary - we can approach it in computer science, but in biology I totally agree there is usually used much larger amount. From the other side ... DNA contains muuuch more more information, but the question is the perspective we look from - I'm talking about e.g. sociobiological level - of interactions ... entropy corresponds to complexity of the society. Important are not genotypes, but phenotypes - precisely: how other specimens react on them - does the behavior depend on which individual it is? For bacterias not really ... what are the simplest colonies which distinguish individuals in their population?
  20. This is only a lower boundary - in reality there is used much more. But maybe there can be found a transition - for example ant's behavior doesn't rather depend from which specimen from its family it is interacting - the entropy/complexity of their society grows let say with log(n) ... Maybe there are some species where the distingctiguishabily starts and so the complexity start growing linearly with e.g. 2.77544 coefficient. If there would be e.g. essential ordering of individuals, entropy would grow faster: with log(n!)~nlog(n) If there would be essential interactions for different pairs, it would grow with n^2 If interactions within larger groups, even with 2^n... And remember that distinguishness is weaker than ordering: if you would just write these sequences of minimal distinguishing features, it would grow like H_n + lg(n!) = nD_n these 3 bits you are talking about is D_n - average number of distinguishing bits - the shortest unique prefix. You need to subtract the ordering information. Bits carries the most information (are the most distinguishing) if they have probability 0.5 - so e.g. the first bit could say if someone is taller than median height...
  21. Even individuality is devaluating in our times - to about 2.3327464 bits per element/specimen (thanks to James Dow Allen): http://groups.google.com/forum/#!topic/comp.compression/j7i-eTXR14E But this is the final minimal price tag - it cannot be reduced further. Specifically, for the minimal prefix tree, a random sequence (representing individual features of a specimen) has about 0.721 probability of being identified as belonging to the population ... so if we are interested only in distinguishing inside the population, we can afford increasing this probability up to 1. To reduce the amount of information in the minimal prefix tree, let us observe that if there appears degree 1 node inside the tree, all sequences from the population going through that node will certainly go in the corresponding direction - we can save 1 bit about the information which exactly is this direction. In standard prefix tree these degree 1 nodes were the place where it could turn out that an outsider does not belong to the population - removing this information would raise false positive probability from 0.721 to 1. So if for sequences (0101.., 1010.., 1001..), the minimal prefix tree remembers (without ordering!): (0....., 101..., 100...), such reduced one remembers only (0....., 1.1..., 1.0...) What decreases its asymptotic cost from 2.77544 bits/specimen to about 2.332746.
  22. Imagine there is a population/database/dictionary and we would like to distinguish its elements. So for each element, let us somehow encode its individual features (e.g. using a hash function) as a bit sequence - the most dense way is to use sequence of uncorrelated P(0)=P(1)=1/2 bits. We can now create minimal prefix tree required to distinguish these sequences, like in the figure below. For such ensemble of random trees of given number of leaves ([math]n[/math]), we can calculate Shannon entropy ([math]H_n[/math]) - the amount of information it contains. It turns out that it asymptotically grows with at average 2.77544 bits per element [math](\frac{1}{2}+(1+\gamma)\lg(e))[/math]. The calculations can be found here: http://arxiv.org/abs/1206.4555 Is it the minimal cost of distinguishability/individuality? How to understand it? ps. related question: can anyone find [math]D(n)=\sum_{i=0}^{\infty} 1-(1-2^{-i})^n [/math] ? Clarification: The first thought about this distinctiveness is probably n! combinatorial term while increasing the size of the system, but n! is about the order and its logarithm grows faster than linearly. Distinctiveness is something much more subtle. It is well seen in equation (10) from the paper (which should like): [math]H_n+\lg(n!)=nD_n[/math] where [math]D_n[/math] is the average depth of leaves of such n leaf tree - average amount of bits distinguishing given element. So this equation says: distinctiveness/individualism + order = total amount of information distinctiveness grows linearly with n (2.77544 asymptotic linear coefficient) information about their ordering grows faster: [math]\lg(n!)\approx n\lg(n)-n\lg(e)[/math].
  23. In last week Nature Physics there is nice experiment about controlling in the future if photons in the past are entangled or not: Victor chooses later (QRNG) if Alice-Bob photons are correlated (left) or not (right): so obtaining |R>|R> means that more probably Victor has chosen entanglement, |R>|L> separation - there is nonzero mutual information, so once again I don't see a reason it couldn't be used to send information? Here is good informative article with link to the paper: http://arstechnica.c...-beforehand.ars ps. If someone is anxious about the "conflict" of fundamental time/CPT symmetry with our 2nd law-based intuition, it should be educative to look at very simple model: Kac ring - on a ring there are black and white balls which mutually shift one position each step. There are also some marked positions and when a ball goes through it, this ball switches color. Using natural statistical assumption ("Stoßzahlansatz"): that if there is p such markings (proportionally), p of both black and white balls will change the color this step, we can easily prove that it should leads to equal number of black and white balls - maximizing the entropy ... ... from the other side, after two complete rotations all balls have to return to the initial color - from 'all balls white' fully ordered state, it would return back to it ... so the entropy would first increase to maximum and then symmetrically decrease back to minimum. Here is a good paper with simulations about it: http://www.maths.usy...ts/kac-ring.pdf The lesson is that when on time/CPT symmetric fundamental physics we "prove" e.g. Boltzmann H theorem that entropy always grows ... we could take time symmetry transformation of this system and use the same "proof" to get that entropy always grows in backward direction - contradiction. The problem with such "profs" is that they always contain some very subtle uniformity assumption - generally called Stoßzahlansatz. If underlying physics is time/CPT symmetric, we just cannot be sure that entropy will always grow - like for Kac ring and maybe our universe also ...
  24. I apology for digging this thread up, but finally there is practical implementation and it beats modern state of arts methods in many application. It can be seen as greatly improved convolutional code-like concept – for example no longer using convolution, but carefully designed extremely fast operation allowing to work on much larger states instead. Other main improvements are using bidirectional decoding and heap (logarithmic complexity) instead of stubbornly used stack (linear complexity). For simplicity it will be called Correction Trees (CT). The most important improvement is that it can handle larger channel damage for the same rate. Adding redundancy to double (rate ½) or triple (rate 1/3) the message size theoretically should allow to completely repair up to correspondingly 11% or 17.4% damaged bits for Binary Symmetric Channel (each bit has independently this probability to be flipped). Unfortunately, this Shannon limit is rather unreachable - in practice we can only reduce Bit Error Rate (BER) if noise is significantly lower than this limit. Turbo Codes (TC) and Low Density Parity Check (LDPC) are nowadays seen as teh best methods – here is comparison of some their implementations with CT approach – output BER to input noise level: We can see that CT still repairs when the others has given up – making it perfect for extreme application like far space or underwater communication. Unfortunately repairing such extreme noises requires extreme resources – software implementation on modern PC decodes a few hundreds bytes per second for extreme noises. Additionally, using more resources the correction capability can be further improved (lowering line in the figure above). From the other side, CT encoding is extremely fast and correction for low noises is practically for free – like up to 5-6% for rate ½. In comparison, TC correction always requires a lot of calculation, while LDPC additionally requires also a lot of work for encoding only. So in opposite to them, CT is just perfect for e.g. hard discs – everyday work uses low noise, so using CT would make it extremely cheap. From the other hand, if it is really badly damaged, there is still correction possible but it becomes costly. Such correction itself could be also made outside, allowing for further improvement of correction capabilities. Paper: http://arxiv.org/abs/1204.5317 Implementation: https://indect-project.eu/correction-trees/ Simulator: http://demonstrations.wolfram.com/CorrectionTrees/
  25. You could also just absorb these photons/electrons with their angular momentum - I think it's doable experiment: Shoot with polarized electrons at object floating on surface of conductive liquid - it would absorb polarized electrons, then release unpolarized ones - so the difference of average angular momentum should make it rotate. I think you couldn't rotate it this way, but maybe I'm wrong ? My sarcasm was only about "universal answer" when physicists don't understand something: "it's quantum.". While quantum mechanism is kind of extension of classical one with the wave nature of particles and in first approximation (h=0) of QM you still get classical mechanics - quantum concepts are not unreachable for our minds, if only we stop basing on such imaginary limit of our understanding. We shouldn't just "shut up and calculate", but still try to understand e.g. dynamics behind wavefunction collapse, field configurations behind particles ... and many other important fields ignored because of "it's quantum" universal answer. Have a good weekend.
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