# Definition of "equally likely" in statistical mechanics

The text I am currently reading (Concepts in Thermal Physics by Blundell) states that,

A system will appear to choose a macroscopic configuration which maximizes the number of microstates.

And that this idea is based on the following assumption:

Each one of the possible microstates of a system is equally likely to occur.

I have trouble understanding what it precisely means for two microstates to be "equally likely." I have a guess as to what it means but I am not sure. My guess was that, given that the system changes its microstate continuously in time, the system spends equal time in each microstate in the long run. So in this case, "equally likely" implies equal long-term frequency for each of the microstates. This would imply that the macrostate that the system spends the most time in is the one that contains the greatest number of microstates. Is this correct meaning of "equally likely" in this context? If not, what does it precisely mean for each one of the possible microstates to be equally likely?

• I'd say that your guess is correct. You can say that equally likely means that the system spends the same time on average on each state. Or (like Níckolas Alves said), you could say that each state has the same probability of being the the true microstate at a given time. It can be shown (under some lax hypotheses) that these two definitions coincide. Aug 22, 2021 at 10:01

Suppose you have a box of volume $$V$$ filled with a mol of an ideal gas with internal energy $$E$$. This defines the macrostate of your system, or intuitively, how your system looks in a macroscopic scale. However, we still don't know how it looks in a microscopic scale, i.e., we don't know how the $$\sim 10^{23}$$ particles over there are behaving individually. There are many different possibilities, which are the microstates of the system. For example, at time $$t = t_0$$ they could have positions $$x_i$$ and velocities $$v_i$$, where the index runs over all the particles. This is one particular microstate. However, the macrostate would be the same if particle $$i=1$$ had position $$x_2$$ and velocity $$v_2$$ while particle $$i=2$$ had $$x_2$$ and $$v_2$$ (I'm assuming things are classical and indistinguishable for simplicity). So which is the correct microstate?

From a macroscopic point of view, we don't know. All we can do is attribute what is the probability of the system being in each possible microstate. The principle you stated implies that both microstates I exemplified are equally likely to be the actual microstate. We don't know which is the right microstate, and all the possible ones are equally likely.

The system moving towards the largest number of microstates is then not only a change of microstates, but also a change of macrostate. If I mix my gas with another box of gas at different temperature or something, the system will reach equilibrium at the macrostate with the most possible microstates. We still won't know what is the right microstate, being able only to attribute probabilities.

Essentially, as OP pointed out in the comments, the idea is that since we do not know which microstate is the correct one, we assign equal probabilities to all of them.

Now, this does have a bit of nuance. Is it always valid to do this? In fact, it depends on the information you have about your system. Instead of an ideal gas, let us pick a generic gas. If the energy is fixed, then all available microstates should have the very same internal energy and there is no reason to prefer one of them over the other ones. We call this the microcanonical ensemble. On the other hand, suppose temperature (which is related to the expectation value of energy) is fixed. In this situation, there could be states with more internal energy than others, as long as the temperature stays the same (for the ideal gas, this won't happen because the energy is proportional to temperature, but let us consider a more general scenario). In this situation, it can be more likely for microstates with lower internal energy to occur, so we won't pick all probabilities to be the same. Instead, they are given by a Boltzmann distribution. This is known as the canonical ensemble.

The key point is that since we do not know what is the true microstate, we can only assign probabilities. We do this according to the information we have (or according to the experimental conditions, if you prefer). For fixed energy, all microstates should have the very same probability of being the true microstate, so they are, in this sense, equally likely.

• Thanks for the answer. But in the sentence "both microstates I exemplified are equally likely to be the actual microstate," what does "equally likely" precisely mean? Aug 22, 2021 at 8:48
• @roflmfao Let me call them $A$ and $B$. "Equally likely" means that the probability $P(A)$ of the true microstate of the system being $A$ and the probability $P(B)$ of the true microstate of the system being $B$ are the same: $P(A) = P(B)$. All microstates have the exact same probability of being the true microstate. We don't know which one is the true one and we treat them all equally Aug 22, 2021 at 9:05
• @Nícholas Alves So is it the fact that we don't know which microstate is the true microstate that justifies assigning equal probabilities to all the microstates? Aug 22, 2021 at 9:28
• @roflmfao Exactly. There are other was of thinking about this and assigning probabilities, but at this point in the book I believe it is precisely what the author had in mind. Eventually he might come to discuss other ways of assigning probabilities (known as different ensembles), but the essential idea is precisely that Aug 22, 2021 at 9:42
• @roflmfao I've updated my answer. Is it clearer now? Aug 22, 2021 at 9:54