On 28.01.2012 23:26 meekerdb said the following:

On 1/27/2012 11:47 PM, Evgenii Rudnyi wrote:

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You disagree that engineers do not use thermodynamic entropyYes. I disagreed that information "has nothing to do with thermodynamic entropy", as you wrote above. You keep switching formulations. You write X and ask if I agree. I disagree. Then you claim I've disagreed with Y. Please pay attention to your own writing. There's a difference between "X is used in place of Y" and "X has nothing to do with Y".

`A good suggestion. It well might be that I express my thoughts unclear,`

`sorry for that. Yet, I think that my examples show that`

1) There is information that engineers employ. 2) There is the thermodynamic entropy. 3) Numerical values in 1) and 2) are not related to each other.

`Otherwise I would appreciate if you express the relationship between`

`information that engineers use and the thermodynamic entropy in your own`

`words, as this is the question that I would like to understand.`

`I understand you when you say about the number of microstates. I do not`

`understand though how they are related to the information employed by`

`engineers. I would be glad to hear your comment on that.`

Evgenii

but you have not shown yet how information in engineering is related with the thermodynamic entropy. Form the Millipede examplehttp://en.wikipedia.org/wiki/Millipede_memory"The earliest generation millipede devices used probes 10 nanometers in diameter and 70 nanometers in length, producing pits about 40 nm in diameter on fields 92 µm x 92 µm. Arranged in a 32 x 32 grid, the resulting 3 mm x 3 mm chip stores 500 megabits of data or 62.5 MB, resulting in an areal density, the number of bits per square inch, on the order of 200 Gbit/in²." If would be much easier to understand you if you say to what thermodynamic entropy corresponds the value of 62.5 MB in Millipede.The Shannon information capacity is 5e8 bits. The thermodynamic entropy depends on the energy used to switch a memory element. I'd guess it must correspond to at least few tens of thousands of electrons at 9v, so S ~ [5e8 * 9e4 eV]/[8.6e-5 eV/degK * 300degK]~17e15 So the total entropy is about 17e15+5e8, and the information portion is numerically (but not functionally) negligible compared to the thermodynamic. BrentThe only example on Thermodynamic Entropy == Information so far from you was the work on a black hole. However, as far as I know, there is no theory yet to describe a black hole, as from one side you need gravitation, from the other side quantum effects. The theory that unites them seems not to exist. EvgeniiMy example would be Millipede http://en.wikipedia.org/wiki/Millipede_memory I am pretty sure that when IBM engineers develop it, they do not employ the thermodynamic entropy to estimate its information capabilities. Also, the increase of temperature would be destroy saved information there. Well, I might be deliberately obtuse indeed. Yet with the only goal to reach a clear definition of what the information is. Right now I would say that there is information in engineering and in physics and they are different. The first I roughly understand and the second not. EvgeniiBrent

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