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

by zi-lao-lan
Tags: boltzmann, entropy, gibbs, thermodynamics
 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= - $$\int$$ $$\rho$$(p,q) (ln $$\rho$$(p,q)) dpdq where $$\rho$$(p,q) is the probability distribution Boltzmann entropy S= ln$$\Omega$$ where $$\Omega$$ 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/$$\Omega$$ and the probability distribution $$\rho$$ will have a constant value regardless of what p and q are. 2. I tried putting $$\rho$$=1/$$\Omega$$ and subbing it into the Gibb's equation S= - $$\int$$ 1/$$\Omega$$ (ln $$\1/[tex]\Omega$$) d$$\Omega$$ using d$$\Omega$$ since we want to add up over all the microstates and there are $$\Omega$$ of them. But I can see that this won't give me the Boltzmann entropy. Any ideas?
 P: 1,424 Your problem is in setting $$dp dq = d\Omega$$, because with omega, you mean a fixed number, not a variable! It is clearer in this manner: $$- \int \rho \ln \rho dp dq = - \int \frac{1}{\Omega} \ln \frac{1}{\Omega} dp dq = \left( - \frac{1}{\Omega} \ln \frac{1}{\Omega} \right) \int dp dq = \left( \frac{1}{\Omega} \ln \Omega \right) \Omega = \ln \Omega$$