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

[zi-lao-lan]

Homework Statement

Show how the Boltzmann entropy is derived from the Gibbs entropy for systems in equilibrium.

Homework 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.

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/\Omega) d\Omega<br /> using d\Omega since we want to add up over all the microstates and there are <br /> \Omega of them. But I can see that this won&#039;t give me the Boltzmann entropy.<br /> <br /> Any ideas?

 

[nonequilibrium]

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

 

[zi-lao-lan]

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?

 

[astrozilla]

ln(1/Ω)= ln1-lnΩ lol

 

[nonequilibrium]

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.