# Questions about the derivation of the gibbs entropy

1. Oct 22, 2013

### V0ODO0CH1LD

In statistical mechanics the macro-state of a system corresponds to a whole region in the microscopical phase-space of that same system, classically, that means that an infinity of micro-states relate to a single macro-state. Similarly, given a hamiltonian, a whole surface in the microscopical phase space of a system can correspond to a single energy level (the surfaces of constant energy).

I've seen different derivations of the gibbs entropy, one in which a region in the microscopic phase-space corresponding to a macro-state of a system is divided into subregions of constant energy, and another where regions of constant energy get divided into subregions that correspond to different macro-states.

Although the procedure is aways to find the configuration of micro-states in that region that maximize the number of ways in which the micro-states can be organized into the subregions, I don't see how that makes sense in either case.

In the case where the outer region corresponds to a macro-state, does organizing the micro-states into subregions of constant energy means changing the hamiltonian? Because given a hamiltonian, a point in phase-space has a definite energy, right? Also, how does it even make sense to rearrange the micro-states into the subregions? For example, given a hamiltonian if all micro-states but one have the same energy, there is still only one way I can arrange those micro-states in that configuration, because the hamiltonian dictates that configuration.

I don't even know how to think about the other approach..

My other question comes later, and it has to do with the "transformation" of the expression that represents the number of ways you can organize the micro-states into the subregions
$$\frac{N!}{\prod\limits_in_i!}$$
into the expression that later gets simplified into the gibbs entropy
$$k_bln\left(\frac{N!}{\prod\limits_in_i!}\right).$$
Is that an empirical adaptation? Or is there a logical step involved? I'm okay if it's just an empirical thing, but I would like to know what that "thing" is.

EDIT: In the expressions above the N represents the number of micro-states in the outer region, and the n's represent the number of micro-states in each subregion. I know that those number are all infinite classically, but this derivation is an adaptation of a quantum mechanics derivation where those number are finite because the states of the system are discrete. In this case those numbers are just representations.

Last edited: Oct 22, 2013
2. Oct 22, 2013

### jfizzix

Traditionally, one divides the phase space into regions the size of minimum position-momentum uncertainty boxes since this is thought to be the smallest volume of phase space a single system can be in due to the Heisenberg uncertainty principle. Historically, the size of these discrete regions was considered arbitrary since the coarse-graining would only define a zero value for the entropy; maximizing the entropy to find equilibrium conditions are largely independent of where the zero of entropy is. In this case, all formulas for entropy and so forth would be defined with respect to "standard values".

The reasons for taking the logarithm are twofold
1.) we define the entropy as an extensive parameter; only the logarithm of the number of microstates would be extensive (i.e. growing linearly with the "extent" of the system).
2.) the entropy without dimensions is proportional to the number of bits on average it would take to unambiguously describe the particular microstate the system is in; it is an information entropy which is very useful in statistical calculations.

Boltzmann's constant is an historical artifact of having different units for temperature and energy. entropy can easily be thought of as a dimensionless quantity, and a thermodynamic temperature be defined as Boltzmann's constant times the usual temperature.

3. Oct 22, 2013

### V0ODO0CH1LD

So the volume occupied by a state of the system in phase-space is the size of one of this uncertainty boxes? And those volumes are the subregions that I divide the outer region defined by the macro-state of the system into? Still, how does the rearranging of the system's micro-states looks like in that picture? What the $n$ in those equations represent in that case?

4. Oct 23, 2013

### jfizzix

Yes, the volume of phase space occupied by a microstate of the system is the size of one of these uncertainty boxes.

One of these boxes in 1D phase space would be of area $dqdp=\frac{\hbar}{2}$.
If we have a single particle in 3D phase space, the box will be of volume $(\frac{\hbar}{2})^{3}$
If we have $N$ particles in 3D phase space, the box will be of volume $(\frac{\hbar}{2})^{3N}$

Again, if you define entropy relative to a "standard state" you don't need to worry about the size of your boxes.

Forgive me, I think I need to know more about exactly what you are asking. Is the Gibbs entropy you refer to the entropy in the grand canonical ensemble (system connected to temperature and particle number reservoir) or just the canonical ensemble (system connected to temperature reservoir)?