When you toss two pennies there are three possibilities,
- Two Heads
- Two Tails
- One of each
So does that mean there is a 1/3rd chance of getting a double head?
Everyone learning statistics knows that is wrong. In fact, if we label the coins A and B, there are four possibilities,
- Two Heads
- Two Tails
- A=head, B=tail
- A=tail, B=head
so the chance of getting a double head is 1/4 not 1/3.
Anyone of any training knows that. The first list will be sniggered at as a beginner's blunder.
So one day Bose was demonstrating the 'ultra-violet catastrophe' to his students, that is, he was showing that theory predicts a curve far different from experiments into the photoelectric effect. He made the equivalent blunder to predicting that pennies come down double heads one time in three - and the curve came out in accordance with experiment. An embarrassing mistake if you set out to prove that it didn't. An even more embarrassing error when he realized where he'd gone wrong.
Thinking later he realized that if it gave the right answer it might not be a mistake after all. But he could not get it published, as everyone's learned reaction was 'what a silly mistake'. In desperation, he wrote to Einstein, who saw at once what it meant:
- Two photons are fundamentally indistinguishable. Therefore the maths should make no distinction between photon A and photon B, just treat them as 'two photons'.
Einstein wrote a companion paper to Bose's, and sent the two off together, and as it was now endorsed by Einstein, people stopped laughing at Bose's mistake.
These statistics are nowadays called Bose Einstein statistics, and are relevant for other particles besides photons - such particles are called bosons.
If your pennies were bosons, the odds of getting two heads would be 1/3 after all.
In fact, since any two elementary particles of the same type are
indistinguishable (be the pair photons, electrons, protons, or whatever), the odds of getting two heads is never 1/4! In contrast to bosons, fermions obeys the FermiDirac?
statistics and the odds of getting two heads is exactly 0 (e.g., two electrons cannot be in the same state), you will always get one head and one tail.
Recently, a NobelPrize
was awarded for the successful creation of a BoseEinstein
condensate: a big clump of bosons (in this case rubidium atoms). See http://www.colorado.edu/physics/2000/bec/index.html
for a reasonable layman's explanation.
Wow, how fast the NobelCommittee? is now! It took 40 years for Pyotr Kapitsa to get a NobelPrize after his successful creation of a BoseEinstein condensate (in that case helium-4 atoms).
In Feller's book on probability, he gives B-E as an example were you can't use "common sense" to determine probability distributions in advance. Somebody forgot to tell the protons about common sense. -- RobertField
On the topic, anyone know what the fundamental difference between a B-E condensate and a superfluid is?
Superfluid has viscosity of zero. BEC has all atoms synchronized together. It's only a coincidence that both phenomena happen at very cold temperatures. Aren't they pretty much mutually exclusive states? I mean, how can you have flow in a synchronized aggregation of atoms?
There is no fundamental difference between BEC in a gas (rubidium vapors) and in a fluid (liquid helium). But there is a difference in how the BEC phase interacts with non-condensed atoms. For an ideal gas, there is, by definition, only one type of interaction: exchange interaction. For a liquid, the picture is much more complex.
Ok, here is the question. In a superfluid, we get a stream of synchronized atoms, all in the same quantum state, but in a BEC proper, we get a pile of atoms all in the same place. Maybe the difference is simply because of the potential well the condensate is formed in, but then we might expect superfluids to collapse to arbitrary density, and I don't think they do. Why is there a supporting pressure in the one but not in the other?
First, there is no
stream of synchronized atoms, all in the same quantum state in helium. The number of atoms in the condensed state is believed to be significantly smaller than the number of atoms in the superfluid phase of the two-phase liquid model. The superfluid phase is a result of the interaction between the atoms in the condensed state and [some of] the atoms in the excited states.
Second, rubidium vapor becoming BEC doesn't
collapse to arbitrary density. The gaseous BEC in these experiments is actually a pretty big and sparse object. Its size, ~10 um, makes it visible to optical devices, and its density, ~5e14 atoms/cm^3, is five thousand times less than the density of a gas of the room temperature and atmospheric pressure.
Third, the potential well does indeed matter. It is very steep at the border of the fluid and quite homogeneous at the macro level inside the fluid. And remember that in the sole field of a homogeneous potential even a single particle is distributed over the whole space.
Sorry, I was thinking of density in terms of how close the atoms are together, not in terms of particles per volume. My bad. I understand how the potential well works in BEC, I just didn't understand how it's absence works in superfluids. In fact, judging by the above, the problem is I don't have the slightest clue how superfluids work. I don't suppose you know a good reference?
Well, as far as I know, nobody understands exactly how they work at the quantum level. We can write the relevant equation, but we cannot solve it for a fluid. Through experiments with elastic scattering we can observe the BEC fraction itself, but so far this doesn't help us to understand its interaction with the rest of the fluid. That is why a much simpler object for study is so important.