Timeline for What is the most efficient information storage?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 24, 2016 at 16:55 | comment | added | user121330 | $\delta E_n >> \Delta E_n$. The proposed system would require measurement better than theory allows. | |
| Nov 6, 2014 at 6:05 | comment | added | Selene Routley | @user27542 Please see the end of my updated answer. I think your explanation solves the problem and Jerry's encoding wouldn't even come near to violating the Bekenstein bound, even if we could measure energy with infinite precision. | |
| Nov 6, 2014 at 6:03 | comment | added | Selene Routley | @Charles Deary me I'm having a truly bad hair day. See my final calculation at the end of my updated answer. | |
| Nov 6, 2014 at 0:47 | comment | added | Charles | @WetSavannaAnimalakaRodVance: I don't think it scales as $n$ but rather as $n^2$ (making the volume scale as $n^6$), see physics.stackexchange.com/q/144819/2818 | |
| Nov 5, 2014 at 22:36 | comment | added | Selene Routley | @RBarryYoung Actually, in modern explanations of Gödel's theorem and uncomputability, you simply take the Gödel number to be exactly as you say - the integer value of the binary string of (say ASCII) characters encoding a mathematical proposition. Gödel worked his encoding scheme out before computers, and before the concept of an array in memory and its binary value had become internalised by scientists the world over. | |
| Nov 5, 2014 at 21:47 | comment | added | Charles | @WetSavannaAnimalakaRodVance: Doesn't the radius scale as $n^2$? But in any case having $n^3$ distinguishable states isn't a problem, since they only contain $\log_2(n^3)\sim k\log n$ bits of information. | |
| Nov 5, 2014 at 15:14 | comment | added | RBarryYoung | Since you are starting with bits, theres no real need to go to string encoding, then Godel encoding, you can just say "the integer value of the binary string of bits", which is unambiguous. | |
| Nov 4, 2014 at 20:11 | comment | added | Zo the Relativist | @KyleKanos: no, the whole point of Godel numbers is that they're unique -- given a coding scheme, the relation between the number and the string is one-to-one (this is different than hashcodes). The fact that the atom would get arbitrarily large is a complaint about this. | |
| Nov 4, 2014 at 19:10 | comment | added | Kyle Kanos | Aren't Godel number non-unique? In which case, the received information from observing the nth level could mean a multitude of things. | |
| Nov 4, 2014 at 9:31 | comment | added | user65081 | an atom occupies a volume in space, the larger the number of bits the larger the atom (because the more exited the state, the larger becomes the electron cloud). Whatever system you would like to imagine, will fall short of the ultimate computer defined by the most efficient one you can have (which includes all possible degrees of freedom allowed by the laws of physics) just before it collapses into a black hole. | |
| Nov 4, 2014 at 8:45 | comment | added | user27542 | The size of the excited atom is going to be very big. If you measure efficiency as "bits per atom", surely it's the way to go. If you measure efficiency as "bits per volume of space", not at all. | |
| Nov 4, 2014 at 7:16 | comment | added | user65081 | yes, if you are allowed to get rid of noise you can also compute using neural networks over the reals, which are also impractical but demonstratively more powerful than a Turing computer. The Bekestein Bound is related to digital storage, which is the only paradigm we know to work above thermal noise (I mean, you could code using other codes beyond binary, such as using n states of an atom, but there is an upper limit to n, n cannot not be arbitrary as in your example | |
| Nov 4, 2014 at 7:12 | comment | added | Zo the Relativist | @WetSavannaAnimalakaRodVance: well, you're kind of cheating the Bekenstein bound with this example, since you're basically utilizing a zero entropy state of a single atom to store your information. There's no reason to expect the thing to collapse to a black hole, since it is still just a hydrogen atom with a very nearly ionized electron. | |
| Nov 4, 2014 at 6:50 | comment | added | Selene Routley | +1 Most interesting and a neat, clean example. I'm not sure how this fits with the Bekenstein bound (you'd be better qualified than I to comment) if the latter indeed holds; maybe the H energy states would undergo a change in a full quantum description of spacetime such that the infinite number of very high $n$ ones actually become finite in number. I guess this comment would just be one of the many loose ends needing to be resolved with a full resolution of the black hole information paradox. | |
| Nov 4, 2014 at 6:36 | history | answered | Zo the Relativist | CC BY-SA 3.0 |