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Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium

by zi-lao-lan
Tags: boltzmann, entropy, gibbs, thermodynamics
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Dec30-10, 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?
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Dec30-10, 09:26 PM
P: 1,424
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]
Dec31-10, 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?

Dec31-10, 06:09 AM
P: 29
Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium

ln(1/Ω)= ln1-lnΩ lol
Dec31-10, 10:01 AM
P: 1,424
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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