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Physics
Classical Physics
Thermodynamics
What is the calculation for value B in Penrose's entropy model?
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[QUOTE="BvU, post: 6867807, member: 499340"] Don't understand how this A' comes about. Penrose introduces intermediate-sized boxes (your 'smaller cubes', I assume) [ATTACH type="full" width="490px" alt="1679352802274.png"]323887[/ATTACH] and has ##N = 10^8,\ k = 10^3,\ n = 10^5##. 'purple' means red/blue in ALL intermediate boxes is between 0.999 and 1.001. He finds (without explaining) there are ##10^{10^{ 2.357\times 10^{25}}} \ ## different arrangements are 'purple' . Our number A . As it happens, ##\left ( 10^{24} \right)\,! ## -- the way to arrange 10[SUP]24[/SUP] balls on 10[SUP]24[/SUP] positions -- is also this number. Well... the whole treatment is supposed to explain (make plausible) that A is extremely close to ##\left ( 10^{24} \right)\,! ## . We all know A ##<< \left ( 10^{24} \right)\,! ## of course, but the whole story is aimed at how much smaller. Penrose then does NOT work out a number for not-uniformly-all-'purple' configurations, but instead states (see #6) a number for 'all blue at the top' that is much too small (we found ##10^{10^{2.326\times 10^{25}}}\ ## in #6 -- meaning the number for not-uniformly-all-'purple' configurations is frighteningly close to ##\left ( 10^{24} \right)\,! ## as well) [INDENT](and once more: sorry for the many mishaps due to boggling mind :nb) )[/INDENT] So, for me the inevitable conclusion is that Penrose does not do a good job here. a while ago I proposed but this too doesn't help us to end up at the ##10^{4.65\times 10^{19}}## in the book -- sorry [USER=733949]@matsu[/USER] ! ##\ ## [/QUOTE]
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Physics
Classical Physics
Thermodynamics
What is the calculation for value B in Penrose's entropy model?
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