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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: 6868243, member: 499340"] The story is about paint, the treatment is as if it's about a solid. Distinguishable balls, half of them blue, the other half red. A coin flip is supposed to have the same probability every time -- you could end up with all blue. So ##M## coin flips yield ##P = \displaystyle \left (1\over 2\right )^M## probability of all heads. If ##M = {1\over 2}N\ ## then ##\ \log P = -{1\over 2}N\log 2## . Placing ##N## balls in ##N## locations can be done in ##A' = N\,!## ways. Placing ##{1\over 2}N## blue balls ##{1\over 2}N## locations (say the top half of the cube) can be done in ##C = \left ( {1\over 2}N\right ) \,!## ways. All the red ones in the other half of the cube idem. So ratio ##Q## (top half blue and bottom half red)/(totally random) is ##C^2/A'##. Stirling: ##\ \log A'= N\log N - N \ ## and ##\ \log C= {1\over 2}N\log {1\over 2}N - {1\over 2}N \ ## so that ##\log Q = N\left(\log {1\over 2}N - \log N\right )= -N\log 2 \quad ##( with thanks to [USER=405866]@mfb[/USER] ) Meaning ##Q=P^2\ ##: with coin flips it's a lot easier than with balls :smile: All this is slightly (well...) repetitive and doesn't get us anywhere in understanding how Penrose messes up, I'm afraid :frown: ##\ ## [/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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