How to correctly think about number of microstates of a system?

I am currently learning statistical mechanics and am finding it a bit difficult to intuitively grasp the concept of the number of microstates for a given macrostate. As a simple example, consider the free expansion of one mole of gas from a volume $$V_1$$ into a larger volume $$V_1 + V_2$$. The change in entropy for this process is $$\Delta S = R \ln \left(\frac{V_1 + V_2}{V_1}\right) = k_B N_A \ln \left(\frac{V_1 + V_2}{V_1}\right) = k_B \ln \left[\left(\frac{V_1}{V_1 + V_2}\right)^{-N_A}\right].$$ The standard argument then given in my book is that the probability of a single molecule being in the original volume $$V_1$$ is $$\frac{V_1}{V_1+V_2}$$ so that the probability of the initial state in which all $$N_A$$ molecules are in volume $$V_1$$ is $$p_1=\left(\frac{V_1}{V_1+V_2}\right)^{N_A}.$$ The probability of the final state in which each of the the molecules is in any place within the volume $$V_1 + V_2$$ is $$p_2 = 1$$. Hence, the ratio of probabilities is $$\frac{p_2}{p_1} = \left(\frac{V_1}{V_1 + V_2}\right)^{-N_A}.$$ The claim made at this point is that the ratio of the number of microstates for the the final state ($$W_2$$) to the number of microstates for the initial state initial state ($$W_1$$) is given by $$\frac{W_2}{W_1} = \frac{p_2}{p_1} = \left(\frac{V_1}{V_1 + V_2}\right)^{-N_A}.$$ Therefore, we have $$\Delta S = k_B \ln\left(\frac{W_2}{W_1}\right)$$ and this is the justification given for the Boltzmann formula. From a purely mathematical point of view, this derivation makes a lot of sense. What I am confused about is the exact nature of $$W_1$$ and $$W_2$$. It makes sense that if $$W_1$$ and $$W_2$$ are both finite quantities, then the ratio $$\frac{W_2}{W_1} = \frac{p_2}{p_1}$$ must hold. However, suppose I begin with a system in a given microstate for state $$1$$ in which all the particles are contained within the volume $$V_1$$. Then, since the volume $$V_1$$ contains an uncountably infinite number of points and I can place each of the $$N_A$$ molecules at any of these points, it feels like $$W_1$$ should be uncountably infinite. Similarly, $$W_2$$ should be infinite and the ratio $$\frac{W_2}{W_1}$$ would not make sense. Also, not every state for which all the molecules are in the volume $$V_1$$ would correspond to the same macrostate (for example, all of the molecules occupy one fourth of that volume) so I'm not sure why all of these possible states are counted in $$W_1$$. I'm hoping someone can explain the correct way to think about the quantities $$W_1$$ and $$W_2$$ and why both of these quantities are (large) finite numbers.

• As you correctly observe, $W_i$ as "number of states" only really makes sense if the phase space is quantized in some way, having countable discrete states. A more general notion is that of phase space volume, and this is the correct concept in the case of a classical gas, as here.
– ummg
Nov 11, 2023 at 18:00

For a quantized system, i.e. one with discrete configurations, the quantity $$W$$ is simply a count of the number of microscopic states compatible with a particular macroscopic state. For a system with continuous degrees of freedom this must be generalized, because there will generally be an uncountable set of microstates compatible with a particular macroscopic state.

If your system consists of $$N$$ particles in 3 dimensions, then it will be described by $$3N$$ position variables $$q_1, \dotsc, q_{3N}$$ and $$3N$$ momentum variables $$p_1, \dotsc p_{3N}$$. The phase space volume $$W$$ of a particular macrostate is then defined as

$$W = C \int dq_1 \dotsc dq_{3N} dp_1 \dotsc dp_{3N},$$

where the integral is taken over all configurations (microstates) compatible with the macrostate. The proportionality constant $$C$$ depends both on convention and whether or not the particles are distinguishable, but you are safe to ignore that subtlety for now.

Now, phase space volume behaves much like a microstate count, which is why it is often safe to think about it as such. For example, suppose that the $$q_i$$ are just the $$x$$-, $$y$$-, and $$z$$-coordinates of the particles in a classical gas, and suppose they are confined to a container of volume $$V$$. Then

$$\int dq_1 \dotsc dq_{3N} = \left( \int dx \, dy \, dz \right)^N = V^N.$$

Hence if you start with a container of volume $$V_1$$ and change its volume to $$V_1 + V_2$$, keeping all else fixed, you will find that

$$\frac{W_2}{W_1} = \frac{C \, (V_1 + V_2)^N \int dp_1 \dotsc dp_{3N}}{C \, V_1^N \int dp_1 \dotsc dp_{3N}} = \left( \frac{V_1 + V_2}{V_1} \right)^N,$$

which, as you can see, is consistent with the discretized argument outlined in the original question.

In the above I tried to make the presentation as simple as possible while retaining the most important points, but in doing so I brushed over some technical details. At the request of the OP I will now go into those details.

The macroscopic state of the system has a certain total energy $$E$$ associated with it, so the integral in the definition of $$W$$ needs to be restricted to those values of $$q = (q_1, \dotsc, q_{3N})$$ and $$p = (p_1, \dotsc, p_{3N})$$ that satisfy $$H(q,p) = E$$, where $$H$$ is the Hamiltonian of the system. There is a technical issue here, because this condition determines a $$(6N-1)$$-dimensional region (a hypersurface), and hence the integral will be zero. The region will have zero width, so to speak; think of the volume of a surface in 3D, the area of a line in 2D, the length of a point in 1D. What we really want to measure is the, let's say, "hyperarea" of this hypersurface. To do so we adopt the more technical definition

$$W = C \int \delta(H(q,p) - E) \, dq_1 \dotsc dq_{3N} dp_1 \dotsc dp_{3N},$$

where $$\delta$$ is the Dirac delta function.

To retain the results above, consider an ideal gas, having Hamiltonian

$$H(q,p) = \sum_{i=1}^{3N} \frac{p_i^2}{2 m}.$$

This Hamiltonian is independent of the position variables, hence

$$W = C \int dq_1 \dotsc dq_{3N} \int \delta(H(q,p) - E) \, dp_1 \dotsc dp_{3N},$$

and you can check for yourself that we once again get

$$\frac{W_2}{W_1} = \left( \frac{V_1 + V_2}{V_1} \right)^N$$

in the OPs scenario. (Note however that this result holds because of the specific form of the Hamiltonian, which lets us separate the integral into a position part and a momentum part. This is not always possible, but it is possible for the ideal gas.)

• Thank you! This helps a lot. One quick question: in the integrals in the ratio W_2/W_1, what region is each integral over (I assume this is a region of phase space with the position bounds corresponding to the volume V_1 and V_1 + V_2, but I'm not sure what the momentun region would be) and why do the integrals involving momenta cancel? Nov 12, 2023 at 12:56
• The momentum region is determined by the energy $E$ in the system. For a noninteracting gas the energy is just the sum of the kinetic energies of the particles, i.e. $E = \sum_{i=1}^{3N} p_i^2 / 2 m_i$. This condition constrains the momenta, giving the region of integration. The momentum integrals cancel because they are unaffected by the container volume.
– ummg
Nov 12, 2023 at 15:01
• There is an extra subtlety here that I brushed over. A condition like $E = \sum_i^{3N} p_i^2 / 2 m_i$ for the momentum integral actually determines a $(3N-1)$-dimensional (hyper-) surface rather than a $3N$-dimensional volume. Hence the integral, as written, becomes zero. But there are ways to "measure" this surface as well, which is what we are interested in; they are just a little more technical than what I wanted to include in the answer.
– ummg
Nov 12, 2023 at 15:07
• What value and units does constant $C$ have? Nov 12, 2023 at 16:37
• Entropy value can be disambiguated only via quantum statistics, where we postulate there is one available state per phase space volume $h^{3N}$. Nov 13, 2023 at 1:59