Superfluidity - Really exciting but really incomprehensible too!

[TouK]

Well the last days i was trying to comprehend what is going on with the spectacular phenomenon of superfluidity.
I saw some experiments with helium cooled to the temperature of 2.17K. I read about zero viscosity and all these exciting abilities it has but i couldnt find how helium acquires zero viscosity.
In addition i read that it has something to do with bosons and the Bose-Einstein Condensation and if yes what exactly is that?

I would appreciate it if anyone could explain me some basics things beacuse i am really interested in this phenomenon and quantum physics generally.

 

 

Thank you for your time and sorry for and mistakes regarding my english as i am 16 years old from greece

 

[Enthalpy]

He-4 gets superfluid around 2K while He-3 needs some 1mK. This is a strong suggestion of boson versus fermion behaviour - the other difference being the magnetic moment of He-3.

 

A BEC is suggested as the cause of superfluidity, yes.

[TouK]

thank you

[Enthalpy]

en.wiki tells "is related to the Bose–Einstein condensation, but it is not identical"

http://en.wikipedia.org/wiki/Superfluidity

http://en.wikipedia.org/wiki/Superfluid_helium-4

 

For instance the narrow peak of heat capacity at a clear transition temperature tells something more happens than just a Bose-Einstein condensation.

[Jake1]

The way Benjamin Schumacher explains it, as uncertainty in energy decreases, uncertainty in position increases. So when helium is at such a low temperature, uncertainty in energy is very low, so uncertainty in position increases and is greater than the space between the helium atoms. Thus the atoms form one uniform liquid with zero friction because there is no definite space between the atoms.

 

Or something like that. Someone should correct my misconceptions.

[Enthalpy]

It's a common explanation, and I'm dissatisfied with it.

 

The delocalization of helium atoms is computed as if they were alone in free space. But in a liquid, they have neighbours, which obviously prevent them from passing through an other. So the delocalization computed i vacuum does not apply.

 

Quite naturally, if one atom was behind an other, it won't get in front of the other by passing through. Neither with fermion atoms nor with bosons. This is a possibility with a gaseous Bose-Einstein Condensate because atoms have room around each other to "pass" by (or rather to exist, since there is no movement) but not with a liquid.

 

More generally, any theory of superfluidity should explain why

(1) Helium keeps a volume, and better, nearly the same volume, when becoming superfluid

(2) What the heat capacity is, at the superfluid transition. This is not a Bose-Einstein condensation.

 

I suppose good theories of superfluidity (which I haven't read) address these points, but their explanations for non-specialists miss the useful aspects.

[md65536]

The delocalization of helium atoms is computed as if they were alone in free space. But in a liquid, they have neighbours, which obviously prevent them from passing through an other. So the delocalization computed i vacuum does not apply.

 

Quite naturally, if one atom was behind an other, it won't get in front of the other by passing through. Neither with fermion atoms nor with bosons. This is a possibility with a gaseous Bose-Einstein Condensate because atoms have room around each other to "pass" by (or rather to exist, since there is no movement) but not with a liquid.

 

This sounds like a classical explanation of quantum mechanical effects, or even a requirement that quantum mechanics obeys "obvious" classical reasoning. If a helium atom's position is that uncertain, can you really claim that it is in a position to block other atoms from passing, or that its behavior is the same as "normal" where atoms cannot pass through each other? Or I guess... is the obviousness of your argument based on quantum mechanics, or classical?

[Enthalpy]

It results from the repulsion between atoms, which don't interpenetrate. QM doesn't change that.

 

Essentially, quantum equations keep the forces known classically but add a wave behaviour. For instance, q²/(eps*4*pi*R) is written just as usual in Schrödinger's equation.

 

That strong repulsion can't be overcome by tunnelling, because atoms (or even protons) are too heavy. Electrons can tunnel to one atomic radius distance (this defines the atomic radius) with a few eV negative energy there; a similar distance and energy deficit, but with an atom 90² times heavier than an electron makes the probability exp(90) time smaller, that is, inexistant.

 

So, reversed: each atom's position is constrained by the neighbours, which it can't pass through. The vacuum computation of one atom's delocalization doesn't apply, because that computation supposes no other interaction.

[md65536]

So, reversed: each atom's position is constrained by the neighbours, which it can't pass through. The vacuum computation of one atom's delocalization doesn't apply, because that computation supposes no other interaction.

Does this mean that if an atom's position is uncertain, it will "block" atoms (from passing) over a larger space, reducing fluidity?

Or does it mean that the position isn't uncertain as hypothesized, ie. the uncertainty when not in vacuum doesn't increase near absolute zero?

Or simply that the "uncertainty of position" explanation doesn't make sense?

 

 

Edit: If the uncertainty principle does indeed apply to position and energy, and the position of atoms in a solid can truly be constrained by neighboring particles, then the energy must be uncertain. If the average energy truly was very close to absolute zero, what would that require? The possibility of negative energy? Or that a very few atoms would have quite a high energy?

Edited January 30, 2014 by md65536

[Enthalpy]

In a dense state like a liquid, every helium atom (or liquid molecule molecule) is closely surrounded by other atoms that can't be passed through and are too closely packed to be passed by. So movements are collective, and are mainly shear.

 

Each helium atom is blocked by its neighbours, but collective wobble remains possible. Just as in a solid, phonons permit a delocalization of each atom, without the need that one atom passes through a neighbour. And the ground state keeps some minimum delocalization, or uncertainty of position, which is given as minimum energy of each mode if expressed in terms of phonons.

 

In crystals, some phonon states (typically atoms bound only locally to a single other atom, where the mode moves a small mass hence at a high frequency) have a significant energy and are not excited at cold, which explains the small heat capacity at cryogenic temperature. Though, the global vibration modes have a very small frequency, hence continue to store heat at cryo temperatures.

Edited January 31, 2014 by Enthalpy