# Are the Gibbs and Boltzmann forms of Entropy equivalent?

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1. Aug 10, 2016

### bananabandana

1. The problem statement, all variables and given/known data
Are the Gibbs and Boltzmann entropies always equivalent?

2. Relevant equations
$$S=k_{B}ln\Omega$$ [Boltzmann entropy, where $\Omega$ is the number of available microstates

$$S=-k_{B}\sum_{i}p_{i} ln(p_{i})$$ [Gibbs entropy, where $p_{i}$ is the probability of a particle being in the $i^{th}$ microstate.

3. The attempt at a solution
I would say no - since Boltzmann implicitly assumes that all of the microstates have equal probability. This works in a system where we can apply the fundamental postulate - i.e the microcanonical ensemble. But that definitely doesn't apply to the Canonical or Grand Canonical ensembles! (as far as I can see)

However, my textbook seems to be suggesting otherwise - i.e that the fundamental postulate always applys, and therfore the Gibbs and Boltzmann entropies are always equal... Are they mistaken?

2. Aug 10, 2016

### ShayanJ

The definition of equilibrium is that the system wouldn't have net changes in its macrostate but only fluctuates around it. This is only possible if the probabilities associated to the microstates giving that macrostate are the maximum among the probabilities of all of the microstates possible for that system, because at any time the system goes to the direction of more probable microstates. Equilibrium is when this evolution stops and so there should be no direction for a net change, which means all directions should be equally probable. So I think Gibbs and Boltzmann forms of entropy are equivalent for a system in equilibrium.

3. Aug 26, 2016

### bananabandana

Sorry for the slow reply - I understand what you are saying for the definition of equilibrium - what you say seems intuitively sensible. However, is it not a result that for a given microstate $j$ in the Boltzmann distribution, we have $p_{j} = \frac{e^{-\beta j}}{Z}$ - so how can the probabilities all be the same? Have I fundamentally misunderstood something?

<Moderator's note: LaTeX fixed>

Last edited by a moderator: Aug 26, 2016
4. Aug 26, 2016

### ShayanJ

That's correct but irrelevant. The point is that the the microscopic states we're talking about here are equivalent to each other as far as macroscopic quantities(like energy) are concerned. But the Boltzmann factor is giving the probability for the system's macroscopic quantity(energy) to have a particular value, which means its giving the probability for the system to be in any of those equivalent microstates.