# Equal a priori probabilties in statistical physics

In Statistical physics by F.Mandl he introduces the postulate of equal a priori probabilities, which states that for an isolated system all microstates compatible with the given constraints of the system are equally likely to occur i.e. have equal a priori prbabilities. He then goes on saying that as a result of this postulate, "the probability that the system is in the macrostate specified by $$(E,V,N,a)$$ is proportional to $$\Omega(Ε,V,N,a)$$. It's the result of this postulate that I'm finding hard to understand. It makes sense to me but what im struggling is to understand is the correlation between the two, meaning how we determined the result from the postulate he introduced.

Suppose we have a volume with two identical compartments. Initially, the compartments are separated, and each contains a different gas: Compartment $$A$$ contains the gas $$a$$ and compartment $$B$$ contains the gas $$b$$.

At time $$t=0s$$ we open a valve, so that the two gases are no longer separated. Shortly after we opened the valve, the gases will not been mixed completely. However, with time passing by, this imbalance will decrease. The thermal equilibrium state is reached, when the two gases are completely mixed.

The described scenario is intuitive. However, the physical explanation for the mixing is given by the law a equal priori probability. To emphasise this, let's consider some questions:

• What is the probability that a specific molecule of gas $$a$$, which we call $$a_1$$, is in the compartment $$A$$? The law of equal prob. tells us that we have $$P(a_1 \in A)=1/2 = P(a_1 \in B)$$. We can ask this question for each specific molecule, and the answer will always be the same.
• Assuming independence between the location of each molecule, we can calculate the probability that $$a_1$$ is in compartment $$A$$, $$a_2$$ is in compartment $$A$$, and $$a_3$$ is in compartment $$B$$. The law of equal probability tells us, that this is $$P(a_1 \in A) \cdot P(a_2 \in A) \cdot P(a_3 \in B) = 1/2^3$$.
• If we specify $$N$$ particular molecules, we would obtain the probability $$1/2^N$$. So, as long as we specify the molecules, every configuration has the same probability.
• The probability changes, if we consider the molecules as indistinguishable. If we are no longer interested in which compartment which molecule is, but only how many molecules of gas $$a$$ are in compartment $$A$$, we obtain a probability distribution, which is sharply peaked around the $$50\%:50\%$$ case.

Maybe a simplified model is also useful, so let's consider a system which consists of three spins and a magnetic field $$B$$. Each spin can be either

• up: $$|\uparrow\rangle$$ = parallel to B-field, or
• down: $$|\downarrow\rangle$$ = anti-parallel to B-field.

Suppose we also know that the total energy of the system is $$E = - \vec \mu \cdot \vec B = - \mu B$$. Hence, we know that two spins are up and one is down. The system occupies one of the following allowed configurations: $$|\uparrow\rangle, |\uparrow\rangle, |\downarrow\rangle \\ |\uparrow\rangle, |\downarrow\rangle, |\uparrow\rangle \\ |\downarrow\rangle,|\uparrow\rangle, |\uparrow\rangle$$ So, in this system it is not true that each spin points with the same probability up as it points down. However, each of the three allowed configurations are have the same probability.

• I understand your explanation but what i'm not asking that exactly. You just explained to me the postulate of equal a priori probabilities, but what im asking is lets say for your 3 spin model, why the probabilty of the macrostate with energy $E=-\mu B$ is proportional to its statistical weight, and how you can derive this with the postulate of equal a priori probabilities. Apr 20 '20 at 22:28
• That does not make sense: If we use two down spins and one up spin, we obtain again three different micro states. However, the probability that the system is in one of these micro states is zero, given that the energy is $E=-\mu B$. In your question, you stated the following variables (E, V, N, a). These variables define restrictions, under which we maximise the entropy $S = k_B \ln(\Omega)$. Apr 20 '20 at 22:42
• ok fine the example i made was bad but i still don't understand this phrase then "the probability that the system is in the macrostate specified by (E,V,N,a) is proportional to Ω(E,V,N,a)" Apr 20 '20 at 22:45
• The three spin example is not a macro state, so let's consider the gas example. Here the macro state is "approx 50% of the $a$ atoms in compartment $A$ (and analog for $b$ atoms)". Maybe, plot the probability that you have 20 atoms and how they are distributed among the two compartments. Apr 20 '20 at 22:49
• Let's consider a system with 3 dipoles like your example with a fixed magnetic field and volume and also lets say we don't know the energy value. According to the postulate and its consequence each macrostate has a probability of occuring that is proportional to the statistical weight of the macrostate. Im trying to understand how this postulate leads us to this conclusion Apr 20 '20 at 23:00

This is a problem in probability theory. A probability space is rhoughly speaking a set of outcomes $$\mathcal{S}$$ together with a probability measure $$\mathbb{P}$$ which, for every subset $$\Sigma\subset \mathcal{S}$$ of the outcomes gives you the probability that the outcome you are interested in be there. There is a technical aspect that such $$\mathbb{P}$$ must be defined in a $$\sigma$$-algebra of subsets of $$\mathcal{S}$$ but you don't need to bother with this right now.

The probability measure $$\mathbb{P}$$ has to obey:

1. For every $$\Sigma\subset \mathcal{S}$$ we have $$\mathbb{P}(\Sigma)\in [0,1]$$ so that probabilities lie between $$0$$ and $$1$$;

2. Some outcome has to occur, so that $$\mathbb{P}(\mathcal{S})=1$$;

3. If $$\{\Sigma_i\}$$ is a discrete collection of pairwise disjoint subsets $$\Sigma_i\cap \Sigma_j =\emptyset$$ then the probability of the union is the sum of probabilities $$\mathbb{P}\left(\bigcup_{i}\Sigma_i\right)=\sum_i \mathbb{P}(\Sigma_i).$$

In your case $$\mathcal{S}$$ is the set of all possible microscopic states and $$\mathbb{P}$$ gives the probability that the actual microscopic state realized be in some subset of $$\mathcal{S}$$. We further assume $$\mathcal{S}$$ to be finite.

Now let be given a tuple $$\mathbf{X}$$ of variables describing the macroscopic state. In your question you take $$\mathbf{X} = (E,V,N,a)$$ but it could be anything really.

We can group microscopic states according to values of $$\mathbf{X}$$. This means that we can consider $$\Sigma(\mathbf{X})$$ to be the set of all states in $$\mathcal{S}$$ compatible with the macroscopic state $$\mathbf{X}$$. Notice that $$\Omega(\mathbf{X})$$, the number of microscopic states compatible with $$\mathbf{X}$$, is the number of elements in the set $$\Sigma(\mathbf{X})$$. In mathematical notation we write this as $$\Omega(\mathbf{X})=|\Sigma(\mathbf{X})|$$.

Now we introduce your postulate and derive its consequence:

Postulate: All microscopic states are equally likely.

Mathematically the postulate tells us that if we take any two distinct singleton sets, $$\{s\}\subset {\cal S}$$ or $$\{s'\}\subset {\cal S}$$ with $$s\neq s'$$, containing two distinct states $$s$$ and $$s'$$, then $$\mathbb{P}(\{s\})=\mathbb{P}(\{s'\}).$$

Now let $$\Sigma(\mathbf{X})$$ be given. Decompose it as a union of singletons for each of its elements:

$$\Sigma(\mathbf{X})=\bigcup_{i=1}^{\Omega(\mathbf{X})}\{s_i\},\quad s_i \in \Sigma(\mathbf{X}).$$

Now take the probability $$\mathbb{P}$$ and use property (3). Moreover also use that $$\mathbb{P}(\{s_i\})=p$$ because of the postulate.

$$\mathbb{P}(\Sigma(\mathbf{X}))=\sum_{i=1}^{\Omega(\mathbf{X})}\mathbb{P}(\{s_i\})=p \sum_{i=1}^{\Omega(\mathbf{X})}1=p \Omega(\mathbf{X}).$$

Therefore the probability that the system is in a microscopic state compatible with the macroscopic state $$\mathbf{X}$$ is proportional to $$\Omega(\mathbf{X})$$ with proportionality constant being the common probability to all microscopic states.

Finally we can determine $$p$$. Decompose $$\cal S$$ in the union of its constituents, take the probability and apply conditions (2) and (3). The exact same derivation as above shows that $$p = 1/N$$ where $$N$$ is the total number of states.

• So if i understand this correctly every microstate has the same probability to occur but it just so happens that some microstates give the same macrostate? Apr 20 '20 at 22:40
• Also the postulate states that the microstates who give the same macrostate have equal probability, not that all the microstates are equaly likely, so shouldn't it be $\{s\} \subset \Sigma$ ? Does that change anything? Apr 20 '20 at 22:53
• As for the first question, yes, the postulate states that every microstate has the same probability to occur, but given a macrostate there is a whole set of microstates which are compatible with it. The number $\Omega(\mathbf{X})$ is the number of such microstates. Regarding the second question saying that "all microstates are equally likely" is exactly the same thing as saying that "all microstates occur with same probability".
– Gold
Apr 21 '20 at 0:15
• Regarding my second question. Isn't it something different to say that all microstates occur with the same probability and something else to say that given a specific macrostate, the microstate that give the macrostate have the same probability to occur? Apr 21 '20 at 10:21