| New Reply |
Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium |
Share Thread |
| Dec30-10, 06:24 PM | #1 |
|
|
Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium
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? |
| Dec30-10, 09:26 PM | #2 |
|
|
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 | #3 |
|
|
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 | #4 |
|
|
Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium
ln(1/Ω)= ln1-lnΩ lol
|
| Dec31-10, 10:01 AM | #5 |
|
|
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.
|
| New Reply |
| Tags |
| boltzmann, entropy, gibbs, thermodynamics |
Similar Threads for: Show how the Boltzmann entropy is derived from the Gibbs entropy for equilibrium
|
||||
| Thread | Forum | Replies | ||
| Gibbs or Entropy | General Physics | 2 | ||
| Equivalence of Boltzmann and Clausius entropy | Classical Physics | 2 | ||
| Change in entropy, Gibbs and Helmholtz in an isothermal compression | Advanced Physics Homework | 2 | ||
| Thermo - Gibbs Free Energy & Entropy | Advanced Physics Homework | 2 | ||
| Entropy of relativistic Boltzmann gas? | Advanced Physics Homework | 1 | ||