Entropy as log of omega (phase space volume)

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SUMMARY

The discussion centers on deriving the relationship between entropy (S) and phase space volume (Ω) in statistical mechanics, specifically showing that S is proportional to the logarithm of Ω. The solution involves recognizing the additivity of S and the multiplicative nature of Ω, leading to the conclusion that S can be expressed as S = c * log(Ω). The approach discussed includes using derivatives and differential equations to arrive at this result, confirming the validity of the method presented in Pathria's textbook.

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diegzumillo
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Homework Statement


I've seen this problem appear in more than one textbook almost without any changes. It goes like this:
Assume the entropy ##S## depends on the volume ##\bar{\Omega}## inside the energy shell: ##S(\bar{\Omega})=f(\bar{\Omega})##. Show that from the additivity of ##S## and the multiplicative character of ##\bar{\Omega}##, it follows that ##S=const \times log \bar{\Omega}##

Homework Equations

The Attempt at a Solution


I've found a couple of solutions already (one in Pathria), that consists of considering two subsystems and calculating the derivatives of S in respect to the omegas, plus a bunch of assumptions. But from the problem statement I can't help but think there's got to be a simpler way. I'm trying to expand S as a sum of ##f(\Omega_i)##, expand each f as a power series, and then use the fact that omega is the product of the omegas of the subsystems. But that leads nowhere.

Slightly off-topic: I'm taking a grad course on statistical mechanics but my previous knowledge on stat mech is very weak, so I'll probably be on these forums frequently throughout the semester. Is this the appropriate forum for stat mech homework questions?
 
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I don't see how power series would be easier than the described approach.

Looking at two subsystems together leads to ##f(xy)=f(x)+f(y)## (where x,y are the Ω of two different systems). Calculating the derivative with respect to x gives ##yf'(xy)=f'(x)##, the derivative of this with respect to y leads to ##f'(xy)+xyf''(xy)=0## or (using z=xy) ##f'(z) = -zf''(z)##. This differential equation is solved by ##f'(z)=\frac{c}{z}## which leads to ##f(z)=c \log(z)##.

diegzumillo said:
Is this the appropriate forum for stat mech homework questions?
It is.
 
Thanks :) That makes perfect sense. I solved it in a similar way but made some unnecessary turns here and there and it made things look more complicated.
 

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