Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibriumby zilaolan Tags: boltzmann, entropy, gibbs, thermodynamics 

#1
Dec3010, 06:24 PM

P: 4

1. The problem statement, all variables and given/known data
Show how the Boltzmann entropy is derived from the Gibbs entropy for systems in equilibrium. 2. Relevant equations Gibbs entropy S=  [tex]\int[/tex] [tex]\rho[/tex](p,q) (ln [tex]\rho[/tex](p,q)) dpdq where [tex]\rho[/tex](p,q) is the probability distribution Boltzmann entropy S= ln[tex]\Omega[/tex] where [tex]\Omega[/tex] is the number of microstates in a given macrostate. 3. The attempt at a solution 1. Well, when the system is in equilibrium (ie when the Boltzmann entropy can be used) all microstates have equal probability. So this means that each microstate has a probability of 1/[tex]\Omega[/tex] and the probability distribution [tex]\rho[/tex] will have a constant value regardless of what p and q are. 2. I tried putting [tex]\rho[/tex]=1/[tex]\Omega[/tex] and subbing it into the Gibb's equation S=  [tex]\int[/tex] 1/[tex]\Omega[/tex] (ln [tex]\1/[tex]\Omega[/tex]) d[tex]\Omega[/tex] using d[tex]\Omega[/tex] since we want to add up over all the microstates and there are [tex]\Omega[/tex] of them. But I can see that this won't give me the Boltzmann entropy. Any ideas? 



#2
Dec3010, 09:26 PM

P: 1,406

Your problem is in setting [tex]dp dq = d\Omega[/tex], because with omega, you mean a fixed number, not a variable! It is clearer in this manner:
[tex] \int \rho \ln \rho dp dq =  \int \frac{1}{\Omega} \ln \frac{1}{\Omega} dp dq = \left(  \frac{1}{\Omega} \ln \frac{1}{\Omega} \right) \int dp dq = \left( \frac{1}{\Omega} \ln \Omega \right) \Omega = \ln \Omega[/tex] 



#3
Dec3110, 05:55 AM

P: 4

Thanks for the reply :) But I'm still not sure how you get to the last step.
That means the the integral of dpdp =  omega, but I can't see why that is. Is it something to do with normalising it? 



#4
Dec3110, 06:09 AM

P: 29

Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium
ln(1/Ω)= ln1lnΩ lol




#5
Dec3110, 10:01 AM

P: 1,406

Well, in a discrete system, omega is the number of microstates, but we're working with a continuous system here: then omega is the phase space volume, by which I mean the "volume" in (p,q)space formed by all the available (p,q)points. I think it's just defined that way actually.



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