Monday, January 02, 2012 ... Français/Deutsch/Español/Česky/Japanese/Related posts from blogosphere

Satyendra Bose: 118th birthday

Because of modest festivities, I didn't have enough time and energy to celebrate the anniversary of Satyendra Bose, an eminent theoretical physicist who was a kind of Indian counterpart of Albert Einstein – because of the Bose-Einstein statistics he is famous for. He also played esraj, the Indian counterpart of violin, the instrument that Einstein played. ;-)

Satyendra Nath Bose was born on January 1st, 1894. Yes, physicists may be born on January 1st, too. (I find it painful to celebrate the January 1st anniversary of the Velvet Divorce of Czechoslovakia. Do you celebrate the anniversary of your divorce with your spouse? Or the January 1st anniversary of the tax hikes in 1960, 1970, 1980, 1990, and 2000?) This piece of data tells you a lot about the personality of his Bengali mother. She may have been an audacious woman who patiently kept her reproductive muscles in check (is that possible?) to avoid a birth in 1893 because she wanted to appear on local TV and in the local newspapers, under the headlines:

The first kid of 1894 is born in Calcutta, British India. Satyendra Nath Bose will become a peer of Albert Einstein.
Unfortunately, his mother forgot that there was no TV in India yet. Moreover, no one knew Albert Einstein who was a less than 15-year-old South German teenager. The young physicist was virtually unknown at that time even though he had already found a big part of relativity by that time. ;-)

Despite her imperfect planning, she didn't give up and she brought Satyendra 6 more younger siblings later. Together, the family could have starred as the Bollywood version of "Snow White And Seven Dwarfs".

Of course, that would be the case assuming that we subtract Satyendra's father who was an engineer in the East Indian Railway Company, later known as the IPCC.

At any rate, Satyendra attended a Hindu school and Calcutta's Presidency College. He learned several languages (and did research in literature, both English and Bengali) and was also good in biotechnology, geology, zoology, anthropology, X-rays, and other sciences.

He is most famous as a theoretical physicist for one important discovery, the Bose-Einstein statistics which will the the main topic of the text below. Let me mention that he could also visit Europe in the mid 1920s, his most productive years in physics, and meet lots of famous physicists such as de Broglie, Curie, and Einstein.

Bosons and their statistics

Bose only made his key discovery because of a mistake he committed during a lecture on the ultraviolet catastrophe at the University of Dhaka. What was the mistake?

Flip two fair coins: what is the probability that you get two tails? Many readers say that the probability is 1/4 but Bose made an ingenious mistake when he said that the probability was 1/3 because there were three options (two heads, two tails, mixed).

Walter Wagner has used a similar reasoning to calculate that there was a 50% probability that the LHC would destroy the Earth. Many other sloppy people think that they can make the same mistake so why aren't they as famous as Bose? Well, Bose not only did the mistake but he also noticed that the result of the mistake agreed with the observations much more accurately than the result of the calculation that is lacking the mistake. ;-)

(The average number of tails that you get by tossing Czech coins is twice as large. That's because instead of heads-and-tails, we have virgins-and-lions, and the lion has two tails.)

In particular, he derived Planck's factor of
\[ \frac{1}{e^{E/kT} - 1} \] using arguments similar to those by Boltzmann which were hybridized with the clever mistake. He was also able to reformulate the mistaken derivation so that the mistake was promoted to a virtue.

And why do we call it Bose-Einstein statistics?

It's because Bose did the same trick I did in Summer 1997 when I finished a small paper on the DLCQ quantization applied to heterotic strings: I asked Lenny Susskind to be a co-author and he agreed, realizing that the paper was correct.

Bose did the very same thing with Albert Einstein. Needless to say, Einstein immediately recognized that Bose's result was correct and important, too. And Bose-Einstein statistics was born.

Later, Einstein, who was nearly 15 years older (recall the teenager story), proved to the the more mature physicist who is also ready to extract useful applications out of an important discovery. Einstein assumed that atoms may be bosons as well. If that's so, they should be able to form Bose-Einstein condensates. So just like the Bose-Einstein statistics was found purely by Bose, Bose-Einstein condensation was discovered purely by Einstein. ;-)

One could say lots of other things about the life and diverse interests of Satyendra Nath Bose. For example, we could say that he died at the age of 80 which is a blessed age. But I am no expert in Bose's life so let me switch to the technical appendix of this blog entry instead. It is pretty much the calculation for which Bose became famous. Note that with the hindsight, we may say that it is totally equivalent to Planck's derivation of the black body curve.

However, without the hindsight, Bose's derivation treating photons and other bosons as a "gas with an unusual mistake" was novel. It was one of the events that gave us the hindsight. Planck's derivation was one bringing the "quantum aspects" to the wave-like description of the electromagnetic field; Bose's derivation was one bringing the "commuting field or wave" aspects (which imply indistinguishability) to the gas-like description of the photons, the quanta of the electromagnetic field. Both approaches had to be found before we were allowed to realize that they're really two descriptions of the same thing.

How to derive Fermi-Dirac and Bose-Einstein distribution using canonical ensemble?

Two weeks ago, Karsus Ren asked:
My textbook says that microcanonical ensemble, canonical ensemble and grand canonical ensemble are essentially equivalent under thermodynamic limit. It also derives Fermi-Dirac and Bose-Einstein distribution from grand canonical ensemble.

My question is then: How to derive Fermi-Dirac and Bose-Einstein distribution using canonical ensemble. The expression of canonical ensemble
\[\rho\propto e^{-\beta E}\] seems to imply Boltzmann distribution only.
Your humble correspondent answered:

Your general Boltzmann Ansatz says that the probability of a state \(n\) depends on its energy as
\[ p_n = C \exp(-\beta E_n) \] where \(\beta = 1/kT\). Fermions are identical particles that, for each "box" or one-particle state they can occupy (given e.g. by \(nlms\) in the case of the Hydrogen atom-like states), admit either \(N=0\) or \(N=1\) particles in it. Higher numbers are forbidden by the Pauli exclusion principle. The energies of the multi-particle state with \(N=1\) and \(N=0\) in a particular one-particle state \(nlms\) differ by \(\epsilon\). Consequently,
\[ \frac{p_1}{p_0} = \frac{C\exp(-\beta (E+\epsilon))}{\exp(-\beta E)} = \exp(-\beta \epsilon) \] where I used the Boltzmann distribution. However, the probabilities that the number of particles in the given one-particle state is equal to \(N=0\) or \(N=1\) must add to one,
\[ p_0 + p_1 = 1.\] These conditions are obviously solved by
\[ p_0 = \frac{1}{1+\exp(-\beta\epsilon)}, \qquad p_1 = \frac{\exp(-\beta\epsilon)}{1+\exp(-\beta\epsilon)},\] which implies that the expectation value of \(n\) is equal to the right formula for the Fermi-Dirac distribution:
\[\langle N \rangle = p_0\times 0 + p_1 \times 1 = p_1= \frac{1}{\exp(\beta\epsilon)+1}\] The calculation for bosons is analogous except that the Pauli exclusion principle doesn't restrict \(N\). So the number of particles (indistinguishable bosons) in the given one-particle state may be \(N=0,1,2,\dots\). For each such number \(N\), we have exactly one distinct state (because we can't distinguish the particles). The probability of each such state is called \(p_n\) where \(n=0,1,2,\dots\).

We still have
\[\frac{p_{n+1}}{p_n} = \exp(-\beta\epsilon)\] and
\[ p_0 + p_1 + p_2 + \dots = 1 \]
These conditions are solved by
\[ p_n = \frac{\exp(-n\beta\epsilon)}{1+\exp(-\beta\epsilon)+\exp(-2\beta\epsilon)+\dots } \] Note that the ratio of the adjacent \(p_n\) is what it should be and the denominator was chosen so that all the \(p_n\) from \(n=0,1,2\dots\) sum up to one.

The expectation value of the number of particles is
\[ \langle N \rangle = p_0 \times 0 + p_1 \times 1 + p_2\times 2 + \dots\] because the number of particles, an integer, must be weighted by the probability of each such possibility. The denominator is still inherited from the denominator of \(p_n\) above; it is equal to a geometric series that sums up to
\[ \frac{1}{1-q} = \frac{1}{1-\exp(-\beta\epsilon)} \] Don't forget that \(1-\exp(-\beta\epsilon)\) is in the denominator of the denominator, so it is effectively in the numerator.

However, the numerator of \(\langle N \rangle\) is tougher and contains the extra factor of \(n\) in each term. Nevertheless, the sum is analytically calculable:
\[ \begin{align} \sum_{n=0}^\infty n \exp(-n \beta\epsilon) = - \frac{\partial}{\partial (\beta\epsilon)} \sum_{n=0}^\infty \exp(-n \beta\epsilon) =\dots \\
\dots = - \frac{\partial}{\partial (\beta\epsilon)} \frac{1}{1-\exp(-\beta\epsilon)} = \frac{\exp(-\beta\epsilon)}{(1-\exp(-\beta\epsilon))^2} \end{align} \] This result's denominator has a second power. One of the copies gets cancelled with the denominator before and the result is therefore
\[ \langle N \rangle = \frac{\exp(-\beta\epsilon)}{1-\exp(-\beta\epsilon)} = \frac{1}{\exp(\beta\epsilon)-1}\] which is the Bose-Einstein distribution.

You could also obtain another version of the Boltzmann distribution for distinguishable particles by a similar calculation. For such particles, \(N\) could take the same values as it did for bosons. However, the multiparticle state with \(N\) particles in the one-particle state would be degenerate because the particles are distinguishable. There would actually be \(N!\) multiparticle states with \(N\) particles in them. The sum would yield a Taylor expansion for the same exponential.

The omission of combinatorial factors such as this \(N!\) was Bose's original mistake that was soon turned into a virtue. For identical particles such as bosons, there is no \(N!\) because by permuting the individual bosons in the same state, you get literally the same state. Bosons are twins who don't have any social security cards to be told apart.

A technicality added later

The derivation above was for \(\mu=0\). When the chemical potential is nonzero, all appearances of \(\epsilon\) have to be replaced by \((\epsilon-\mu)\). Of course that one may only talk about a well-defined value of \(\mu\) when we deal with a grand canonical potential; it is impossible to derive a formula depending on \(\mu\) from one that contains no \(\mu\) and assumes it's ill-defined.

The derivation above was meant to show that the difficult \(1/(\exp\pm 1)\) structures do appear from a simpler \(\exp(-\beta E)\) Ansatz because I think it's the only nontrivial thing to be shown while discussing the relations between the Boltzmann and BE/FD distributions. If that derivation proves the same link as the textbook does, then I apologize but I think there is "nothing else" of a similar kind to be proven.


You may have heard that the ATLAS detector of the LHC was built out of LEGO parts for €2,000 which could have saved about $10 billion.

However, the ATLAS detector is useless without the accelerator part of the experiment. This omission has been fixed, too. Here is the LEGO gadget that picks and accelerates the protons:

Video via Jorge Pullin

Add to Digg this Add to reddit

snail feedback (0) :