Boltzmann equation/ Statistical Mechanics

AI Thread Summary
The discussion focuses on demonstrating that the function f(Ω) in the entropy equation S = k*f(Ω) is ln(Ω). By applying Stirling's approximation to the multiplicity Ω, it simplifies to a form that leads to the conclusion that ln(Ω) can be expressed in terms of probabilities. The relationship between the entropies and multiplicities of two combined systems is established, showing that total entropy is additive while multiplicity is multiplicative. The logarithmic function is confirmed as the only function that satisfies the requirement for combining multiplicities. This understanding clarifies the role of the logarithm in statistical mechanics and entropy calculations.
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Homework Statement



If we assume entropy is a function of the multiplicity, \Omega, (S=k*f(\Omega)) show that that function f(\Omega) is ln(\Omega).

Homework Equations


The Attempt at a Solution



\Omega can be written as N!/ni!. By using stirling's approximation, this becomes \Omega= ((N/e)^N)/((n1/e)^n1*(n2/e)^n2*...(ni/e)^ni). We know that the probability pi=N/ni so this reduces to W=1/(p1^n1*p2^n2*...*pi^ni). To make this user friendly take the log so ln(\Omega)=-\Sigmapi*ln(pi).

I just started down the road of trying to use definition of multiplicity and probabilities and I did get to ln(\Omega), but it doesn't seem like I'm really doing a solid proof and I'm not sure what's missing/ how to tie it together.
 
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Consider two systems, with entropies S1 and S2, multiplicities W1 and W2. What can you say about the entropy of the combined system? What can you say about the multiplicity of the combined system?
 
the total entropy s=s1+s2 and the multiplicity w=w1*w2. Is the log just out of convenience then?
 
No -- it's the only function that would fit the requirement that f(w1)+f(w2)=f(w1*w2).
 
Makes sense...thank you. It's been a long week. Nice to finally know where that log came from.
 
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