Absolute Entropy (Statistical)

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The discussion focuses on the derivation of entropy expressed as S = -k. SUM P.lnP and the confusion surrounding the equations U = SUM E.P and dU = SUM E.dP - SUM P.dE, particularly the presence of the minus sign. Participants clarify that the minus sign arises from the application of the product rule in calculus. The transition from dS = -k. SUM lnP.dP to dS = -k.d(SUM P.lnP) is explained by recognizing that the whole expression is treated as a differential, where the last term becomes zero due to the constraint that the sum of probabilities equals one. The derivation ultimately hinges on understanding these calculus principles and the behavior of probability distributions.
Master J
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I was looking at a derivation of entropy expressed as an absolute probability:

S = -k. SUM P.lnP
(What is the name of this by the way?)

In the derivation, it makes the following statements which I really just don't get!

U = SUM E.P

so therefore dU = SUM E.dP - SUM P.dE

Where does the minus sign come from? Should it not be a plus??

Then, it goes from dS = -k. SUM lnP.dP to dS = -k.d(SUM P.lnP)

How is that true?? First it was just P that was a differential element, now its the whole expression in the bracket??
 
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How does the whole derivation go?

Master J said:
U = SUM E.P
so therefore dU = SUM E.dP - SUM P.dE
Where does the minus sign come from? Should it not be a plus??
I agree.

Master J said:
Then, it goes from dS = -k. SUM lnP.dP to dS = -k.d(SUM P.lnP)

How is that true?? First it was just P that was a differential element, now its the whole expression in the bracket??
Check what the whole expression is equal to. By the product rule:
\mathrm{d}(\sum P_i\ln P_i)=\sum(\ln P_i\mathrm{d}P_i+P_i\mathrm{d}\ln P_i)=\sum\ln P_i\mathrm{d}P_i+\sum P_i\frac{\mathrm{d}P_i}{P_i}=\sum\ln P_i\mathrm{d}P_i+\sum\mathrm{d}P_i
The last term is zero since
\sum P_i=1
and hence
\sum \mathrm{d}P_i=\mathrm{d}(\sum P_i)=\mathrm{d}1=0
 
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