The entropy of the system in a canonical ensemble is given by the Gibbs entropy formula. This formula depends on the choice of microstates. For example, if I model $\mathbb R^3$ by a lattice and do the simplifying assumption that each molecule is on a grid point, then the entropy is dependent on the density of the lattice.

On the other hand, change in entropy is also defined by $$dS=\frac{dQ_{rev}}{T}.$$ This seems independent of the choice of microstates.

So for example, suppose that a gas is in some box and then it all flows out of that box. The entropy goes from $S$ to zero. When we use the Gibbs entropy formula, then the difference between initial and final entropy in the box depends on the choice of microstates. But when we use the other equation, it only depends on the number of heat that flowed out.


1 Answer 1


The thermodynamic definition of entropy present here is independent of such things as microstates as it's defined as a macroscopic quantity (like all quantities in thermodynamics), a state variable that quantifies the system as a whole. The definition of entropy from statistical mechanics, however, $S =k_b \log W$, does obviously depend on the number of microstates $W$.

While these two quantities have the same name and can be linked to each other in some situations (most importantly, in the link between classical thermodynamics and statistical mechanics), they are not equal to each other because they exist in different contexts.

This brings up the question of how exactly one can calculate the number of microstates of a system, but that's a different story.

  • $\begingroup$ I'll ask a new question for that. $\endgroup$
    – Riemann
    Oct 19, 2022 at 16:28

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