# A digression on statistics and a party with Ms. Fermi-Dirac and Mr. Bose – Post #5

To explain the next standard candle, I need to digress a little into the math of statistics of lots of particles. The most basic kind is the statistics of distinguishable particles.

Consider the following scenario. You’ve organized a birthday party for a lot of different looking kids (no twins, triplets, quadruplets, quintuplets …). Each kid has equal access to a large pot of M&Ms in the center of the room. Each kid can grab M&Ms from the pot and additionally when they bounce off each other while playing, can exchange a few M&Ms with each other. After a long while, you notice that all the M&Ms in the central pot are gone.  Let’s suppose there are truly a ${\underline {large}}$ number of kids $(K)$ and a truly ${\underline {humongous}}$ number of M&Ms $(N)$.

Interesting question – how many M&Ms is each kid likely to have? A simpler question might be – how many kids likely have 1 M&M? How many likely have 2?..How many likely have 55?… How many likely have 5,656,005?…

How do we answer this question?

If you use the notation $n_i$ for the number of kids that have $i$ M&Ms?, then, we can easily write down

$\sum\limits_{i} n_i = K$

is the total number of kids.

$\sum\limits_i i n_i = N$

is the total number of M&Ms.

But that isn’t enough to tell! We need some additional method to find the most likely distribution of M&Ms (clearly this wouldn’t work if I were there; I would have all of them and the kids would be looking at the mean dad that took the pot home, but that’s for a different post). The result, that Ludwig Boltzmann discovered, at the end of the 19th century, was ${\bf not}$ simply the one where everybody has an equal number of M&Ms. The most likely distribution is the one with the most number of possible ways to exchange the roles of the kids and still have the same distribution. In other words, maximize the combinatoric number of ways

${\it \Omega} = \frac {K!} {n_1! n_2! n_3! ...n_{5,005,677}! ...}$

which is the way of distributing these kids so that $n_1$ have $1$ M&M, $n_2$ have $2$ M&Ms, $n_3$ have $3$ M&Ms…, $n_{5,005,677}$ have $5,005,677$ M&Ms and so on.

Boltzmann had a nervous breakdown a little after he invented the statistical mechanics, which is this method and its consequences, so don’t worry if you feel a little ringing in your ears. It will shortly grow in loudness!

How do we maximize this ${\it \Omega}$?

The simplest thing to do is to maximize the logarithm of ${\it \Omega}$, which means we maximize

$\log \Omega = \log K! - \sum\limits_{i} \log n_i!$

but we have to satisfy the constraints

$\sum\limits_{i} n_i = K, \hspace{5 mm} \sum\limits_i i n_i = N$

The solution (a little algebra is required here) is that $n_i \propto e^{-\beta i}$ where $\beta$ is some constant for this ‘ere party. For historical reasons and since these techniques were initially used to describe the behavior of gases, it is called the inverse temperature. I much prefer “inverse gluttony” – the lower $\beta$ is, the larger the number of kids with a lot of M&Ms.

Instead of the quantity $i$, which is the number of M&Ms the children have, if we considered $\epsilon_i$, which is (say) the dollar value of the $i$ M&Ms, then the corresponding number of kids with $"value" \: \epsilon_i$ is $n_i \propto e^{-\beta \epsilon_i}$

Few kids have a lot of M&Ms, many have very few – so there you go, Socialists, doesn’t look like Nature prefers the equal distribution of M&Ms either.

If you thought of these kids as particles in a gas and $\epsilon_i$ as one of the possible energy levels (“number of M&Ms”) the particles could have, then the fraction of particles that have energy $\epsilon_i$ would be

$n(\epsilon_i) \propto e^{- \beta \epsilon_i}$

This distribution of particles into energy levels is called the Boltzmann distribution (or the Boltzmann rule). The essential insight is that for several ${\bf distinguishable}$ particles the probability that a particular particle is in a state of energy $\epsilon$ is proportional to $e^{-\beta \epsilon}$.

After Boltzmann discovered this, the situation was static till the early $1920s$ when people started discovering particles in nature that were ${\bf indistinguishable}$. It is a fascinating fact of nature that every photon or electron or muon or tau particle is $exactly$ identical to every other photon or electron or muon or tau particle (respectively and for all other sub-atomic particles too). While this fact isn’t “explained” by quantum field theory, it is used in the construction of our theories of nature.

Back to our party analogy.

Suppose, instead of a wide variety of kids, you invited the largest $K$-tuplet the world has ever seen. $K$ kids that ${\bf ALL}$ look identical. They all have the same parents (pity them ${\bf please})$, but hopefully were born in some physically possible way, like test-tubes. You cannot tell the kids apart, so if one of them has 10 M&Ms, its indistinguishable from $any$ of the kids having 10 M&Ms.

Now what’s the distribution of the number of kids $n_i$ with $\epsilon_i$ value in M&Ms? The argument I am going to present is one I personally have heard from Lubos Motl’s blog (I wouldn’t be surprised if its more widely available, though given the age of the field) and it is a really cute one.

There are a couple of possibilities.

Suppose there was a funny rule (made up by Ms. Fermi-Dirac, a well known and strict party host) that said that there could be at most $1$ kid that had, say $\epsilon_i$ value in M&Ms (for every $i$). Suppose $P_0(\epsilon_i)$ were the probability that ${\underline {no}}$ kid had $\epsilon_i$ of value in M&Ms. Then the probability that that 1 kid has $\epsilon_i$ of value in M&Ms is $P_0 e^{-\beta \epsilon_i}$ – remember the Boltzmann rule! Now if no other possibility is allowed (and if one kid has $i$ M&Ms, it is indistinguishable from any of the other kids, so you can’t ask which one has that many M&Ms)

$P_0(\epsilon_i) + P_0(\epsilon_i) e^{-\beta \epsilon_i} = 1$

since there are only two possibilities, the sum of the probabilities has to be 1.

This implies

$P_0(\epsilon_i) = \frac {1}{1 + e^{-\beta \epsilon_i}}$

And we can find the probability of there being $1$ kid with value $\epsilon_i$ in M&Ms. It would be

$P_1({\epsilon_i}) = 1 - P_0({\epsilon_i}) = \frac {e^{-\beta \epsilon_i}}{1 + e^{-\beta \epsilon_i}}$

The $expected$ number of kids with value $\epsilon_i$ in M&Ms would be

${\bar{\bf n}}(\epsilon_i) = 0 P_0(\epsilon_i) + 1 P_1({\epsilon_i}) = {\bf \frac {1}{e^{\beta \epsilon_i}+1} }$

But we could also invite the fun-loving Mr. Bose to run the party. He has no rules! Take as much as you want!

Now, with the same notation as before, again keeping in mind that we cannot distinguish between the particles,

$P_0(\epsilon_i) + P_0(\epsilon_i) e^{-\beta \epsilon_i} + P_0(\epsilon_i) e^{-2 \beta \epsilon_i} + .... = 1$

which is an infinite (geometric) series. The sum is

$\frac {P_0(\epsilon_i) }{1 - e^{-\beta \epsilon_i} } = 1$

which is solved by

$P_0(\epsilon_i) = 1 - e^{-\beta \epsilon_i}$

The expected number of kids with value $\epsilon_i$ in M&Ms is

${\bar{\bf n}}(\epsilon_i) = 0 P_0(\epsilon_i) + 1 P_0(\epsilon_i) e^{-\beta \epsilon_i} + 2 P_0(\epsilon_i) e^{-2 \beta \epsilon_i} + ...$

which is

${\bar{n}}(\epsilon_i) = P_0(\epsilon_i) \frac {e^{-\beta \epsilon_i} } {(1 - e^{-\beta \epsilon_i})^2} = {\bf \frac {1}{e^{\beta \epsilon_i} -1}}$

Now, here’s a logical question. If you followed the argument above, you could ask this -could we perhaps have a slightly less strict host, say Ms. Fermi-Dirac-Bose-2 that allows up to 2 kids to possess a number of M&Ms whose value is $\epsilon_i$? How about a general number $L$ kids that are allowed to possess M&Ms of value $\epsilon_i$ (the host being the even more generous Ms. Fermi-Dirac-Bose-L). More about this on a different thread. But the above kinds of statistics are the only ones Nature seems to allow in our 4 – dimensional world (three space and one time). Far more are allowed in 3 – dimensional worlds (two space and one time) and that will also be in a different post (the sheer number of connections one can come up with is fantastic!).

The thing to understand is that particles that obey Fermi-Dirac statistics (a maximum of one particle in every energy state) have a “repulsion” for each other – they don’t want to be in the same state as another Fermi-Dirac particle, because Nature forces them to obey Fermi-Dirac statistics.  If the states were characterized by position in a box, they would want to stay apart. This leads to a kind of outwards pressure. This pressure (described in the next post) is called Fermi-degeneracy pressure – its what keeps a peculiar kind of dense star called a white dwarf from collapsing onto itself. However, beyond a certain limit of mass (called the Chandrasekhar limit after the scientist that discovered it), the pressure isn’t enough and the star collapses on itself – leading to a colossal explosion.

These explosions are the next kind of “standard candle”.

${\bf {Addendum}}$:

I feel the need to address the question I asked above, since I have been asked informally. Can one get statistics with choosing different values of $L$ in the above party? The answer is “No”. The reason is this – suppose you have $K L$ kids at the party, with a maximum of $L$ kids that can carry M&Ms of value $\epsilon_i$. Then we should be able to divide all our numbers by $L$ (making a scale model of our party that is $L$ times smaller) that has $K$ kids, with a maximum of $1$ kid that is allowed to hold M&Ms of value $\epsilon_i$. You’d expect the expected number of kids with M&Ms of value $\epsilon_i$ to be, correspondingly, $L$ times smaller! Then, the expected number of particles in a state (with a limit of $L$ particles in each state) is just $L$ times the expected number with a limit of $1$ particle in each state.

So all we have are the basic Fermi-Dirac and Bose statistics (1 or many), in our three-space-dimensional party!