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Based on the classical interpretation of probability, the probability for a single particle to be in the $i$th energy state, in an $N$ particle system, should be given by the number of particles in that state, divided by the total number of particles :

$$p(i)= \frac{n_i}{N}$$

Here $n_i$ represents the actual number of particles in the system. However, due to random fluctuations and collisons, the actual number of particles in a particular energy level is never constant and keeps on changing if the total number of particles is finite. By that logic, the 'true' probability of finding a particle at a certain energy level, should be impossible to determine, since we can't sensibly talk about the number of particles in a state, for a finite case anyway.

Hence we use the gibbs/boltzmann distribution, and claim, that the probability of a particle to be in a certain state is given by the following :

$$p(i)=\frac{e^{-\beta E_i}}{Z}$$

However, this is not the exact true probability, is it ? Isn't it technically more like, our best guess of what the true probability of a particle being in state $E_i$ should be ? Since the number of particles in each state keeps on changing, it becomes non-sensical to talk about this 'true' probability of finding the particle in a certain energy state at a time.

So, would it be correct in assuming that the gibbs probability is the average probability or the 'expected' probability of finding a particle in a particular state. Since this is the average probability and not the true one, it becomes impossible to find out the number of particles in that state. Because of this, the number of particles in a state becomes a random variable with a distribution, whose mean is given by $Np(i)$.

So, can we say that in the truest sense of classical probability, the boltzmann probability is average probability of finding a particle in a certain state in the system, only because the true probability keeps on changing as the system undergoes collisions and what not, and the occupancy of a state is never constant ?

In the infinite particle limit, the fluctuations die out, and the actual number of particles become close to the expected number of particles, and so one can claim that the Gibbs probability is approximately equal to the classical probability.

If we could theoretically know the 'true' probability of finding a particle at a certain state, which is impossible ofcourse, we could find the exact number of particles in that state. In that case, the number of particles in the state wouldn't be a distribution, more like a yes or no question, just like picking up coloured balls from a bag - if you know the probability of picking a blue ball from a bag of $100$ balls, you could easily find the number of blue balls. It would be exactly probability multiplied by total number of balls and not some distribution of various possibilities.

But in this case, since you don't know the exact probability, and as the number of particles in the state keep on changing, you can only talk about the expectated number of particles in a state i.e you get a distribution of the total number of particles in a state.

I'm sorry if I'm spending too much time forcing an interpretation of a rather simple problem, but can anyone tell me if my interpretation of the situation correct or not ?

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    $\begingroup$ The classical interpretation of $p(i)$ is the number of times your randomly chosen particle turned out to be in state $i$ divided by the total number of trials. This is different from what you described. Though in non-interacting systems, $p(i)$ can also be computed as the expectation of $n_i$ over $N$. See physics.stackexchange.com/questions/678299/… $\endgroup$ Nov 24 at 17:12
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    $\begingroup$ Probability is an overloaded word, even in technical contexts. I think you know that in your gut, because you seem to be trying to disenangle two or more of the inequivalent definitions by using adjectives like "true." For confirmation, I highly recommend Jaynes' unfinished book Probability Theory: The Logic of Science (2003). I think some of its statements and attitudes are a little off-base, but that's forgivable because Jaynes died before the book was finished. Despite its rough edges, the book helps clarify the relationships between the various concepts we call probability. $\endgroup$ Nov 24 at 17:27
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Your idea makes sense, but I believe it would be better suited if we discussed it in terms of a Bayesian notion of probability instead of a frequentist view.

The definition of probability you presented is "frequentist", in the sense you understand probabilities in terms of the frequency in which a certain outcome will happen. Another possible view is to think of probability in terms of "bets": if I tell you I have a box filled with 120 balls and 80 of them are blue, how much would you be willing to bet on me pulling out a blue ball? If it is blue, I win, otherwise, you win. You would not want to bet more than 1:2 (i.e., if you win I pay you 2 dollars, if I win you pay me 1 dollar), because if I charge you more you are likely to lose money. Notice, however, that if you know I usually lie about the number of balls in a box, your willingness to bet 1:2 will not be the same. Instead, you'll want to bet somewhat less, because you are also taking into consideration some extra information you had beforehand. This shows that the probability is not an absolute concept, but instead something that depends on the information you have.

Let us consider as an example a gas. You want to attribute probabilities to what is the microstate the gas is currently in. How can you do that? Well, you use the Fundamental Postulate of Statistical Mechanics and state that all the available microstates are equally likely.

But suppose now you know your gas is in thermal equilibrium with a reservoir at some temperature $T$. Temperature is just a measure of the mean energy of the particles in your gas, so this is equivalent to saying you have a well-defined mean value for your energy. This time, you have more information than you used to, and you would like to use it to determine your probabilities in the very same way you used it when betting against me. Last time, you used a probability which was uniform, which we can justify by saying that we had no information on why any state should be preferred over some other. This time, we also want to pick a distribution that assumes no extra information than we have, and so we'll pick a probability that maximizes the "desinformation" (AKA entropy) of our system subject to the constraints provided by the information we do have (mean value of energy is fixed).

Will this process gives us the correct probability? I quote DOI: 10.3390/e18070247's adaptation of a quote by Jaynes to answer the question:

If, in a given context, you need to formulate a probability distribution on which to base your bets, choose, among all possible distributions that agree with what you know about the problem, the one having maximum entropy. Why? Is this guaranteed to be the "real" (whatever that may mean) probability distribution? Of course not! In fact you will most likely replace it with a new one as soon as you see the outcome of the next trial—because by then you will have one more piece of information. Why, then? Because any other choice—being tantamount to throwing away some of the information you have or assuming information you don't have—would be indefensible.

Hence, you are correct in the idea that the Boltzmann probability is the "best guess" for how many particles occupy each state given the information available. If you obtain more information, you will likely abandon the Boltzmann distribution as soon as that happens and update your probabilities accordingly, just like someone counting cards on Blackjack updates the probability of getting money with each new card they see. Notice that knowing the "true probability" would mean having complete information about the system, and as you mentioned this ends up leading to a single possibility in which you know exactly how many particles are in each state.

If these ideas interest you, you may want to take a look at this post: Reference for statistical mechanics from information theoretic view.

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  • $\begingroup$ Thank you so much for this. However, I've asked a follow up question regarding this here, regaqrding what distribution I should use for the number of particles in an energy level. $\endgroup$
    – RayPalmer
    yesterday
  • $\begingroup$ My first intuition tells me, binomial distribution is good enough, but once I dig a little deeper, there seems to be another possible distribution, that models the system better. $\endgroup$
    – RayPalmer
    yesterday
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Bayesian statistics, as explained in the answer by @NíckolasAlves, is the principled way of solving the philosophical problem that measuring the probabilities via an infinite number of measurements is impossible. Note however, that Bayesian statistics can be also criticized as being based on belief, encoded in the prior, as our reasonable assumption of what the probability distribution might be. Eventually, both frameworks produce identical results, if applied consistently.

Statistical physics in frequentist framework
However, even within the frequentist framework, the things are somewhat different than the way they are presented in the OP. In particular, the probability is not defined as the ratio $n_i/N$, but rather as the limit $$p_i=\lim_{N\rightarrow +\infty}\frac{n_i}{N}.$$ We need not know the value of this limit - it is sufficient to believe that such a limit exists (and satisfies the basic constraints on the probability, such as it being positive and summing to one). Then, for any finite size sample we have an estimator of what the probability could be: $$\hat{p}_i=\frac{n_i}{N}.$$ Note that the estimator is itself a random variable, whose mean, standard deviation, and other properties (e.g., its consistency, bias, etc.) can be studied by sampling $N$ particles from a theoretical infinitely large system and then averaging over the samples (ensemble averaging).

Statistical mechanics gives us the mean values of such estimators calculated for simplified systems, where the probabilities can be explicitly calculated in the thermodynamic limit, $N\rightarrow+\infty$. Thus, the actual approximations in statistical mechanics have to do not with the assumptions about the size of the system, but with making the approximations that allow evaluating the exact limits.

For example, the well-studied example of an ideal gas neglects the collisions between the atoms/molecules, which allows exact evaluation of its properties. Yet, these collisions are essential for the establishment of the thermodynamic equilibrium. Another example is the ergodicity assumption, which suggests that after a long time the system will explore all the possible configurations, and therefore its state can be described by the ensemble averages (rather than time averages) - this assumption is broken in critical phenomena.

Recommended reading: Statistics chapter from the Review of Particle Physics.

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  • $\begingroup$ I've added a follow up question regarding this. You commented on some other post that the number of particles in a state is a random variable, and can be modelled using the binomial distribution where $p_i$ is the probability of a single particle being in that state $\endgroup$
    – RayPalmer
    yesterday
  • $\begingroup$ However, do you not think, that there can be a better distribution that we can use over the binomial one. For example, can we not consider $N+1$ different systems with different values of $n_i$ and therefore, different 'estimators', and check the likehood of these estimators producing the same outcome as the boltzmann probability ? $\endgroup$
    – RayPalmer
    yesterday
  • $\begingroup$ Based on the boltzmann distribution, the estimated number of particles in a state is $\mu=p_i N$. So from a system of $N$ particles, we can expect to see $\mu$ particles in the particular state. So now, instead of checking the probability of getting exactly $n_i$ particles using the simple binomial distribution, how about we check the probability of getting $\mu$ particles in the state out of $N$, for each of these systems with different values of $n_i$ and henceforth different estimators ? $\endgroup$
    – RayPalmer
    yesterday
  • $\begingroup$ In both cases though, I'd get the same $mean$, so in the end I guess it wouldn't matter. However, it seems to me that this new distribution is more accurate than the simple binomial distribution. If you would kindly take a look at this follow-up question $\endgroup$
    – RayPalmer
    yesterday

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