Ultracold, superdense atoms become invisible

https://news.mit.edu/2021/atoms-ultracold-scatter-light-1118

by gmays • 5 years ago

141 57 5 years ago

news.mit.edu

There is something missing from this press release.

In their new study, he and his colleagues used techniques they developed previously to first freeze a cloud of fermions — in this case, a special isotope of lithium atom, which has three electrons, three protons, and three neutrons. They froze a cloud of lithium atoms down to 20 microkelvins, which is about 1/100,000 the temperature of interstellar space.

“We then used a tightly focused laser to squeeze the ultracold atoms to record densities, which reached about a quadrillion atoms per cubic centimeter,” Lu explains.

They're freezing lithium 6 vapor and increasing its density. But lithium 6 is normally a solid at room temperature. Solid lithium 6 would contain about

(6.02 * 10^23 / 6.0151 [0]) * 0.460 [1] = 4.6 * 10^22

atoms per cubic centimeter -- a density 7 orders of magnitude greater than what has been attained in this work. Is there something special that prevents this frozen atom cloud from condensing to an ordinary solid? Is it just a very short lived state observed before it condenses to an ordinary solid?

[0] Atomic weight of lithium 6

[1] Specific gravity of lithium 6: https://aip.scitation.org/doi/abs/10.1063/1.1743927

Misdicorl2• 5 years ago

The missing piece is that the coldness of the atoms isn't a random stat. It's an incredibly important feature of this research field. The density in the context of a quantum gas is what's interesting

liquidise• 5 years ago

In general their article's use of "Fermion" clashes with my understanding. I've understood fermions[1] to be a type of sub-atomic particle, like an electron. The article routines refers to them as if they are atoms themselves, including your quoted test.

I presume the MIT literature knows more than i do here, but i'm unsure what i am missing. Anyone have an idea?

1. https://en.wikipedia.org/wiki/Fermion

dragonwriter3• 5 years ago

Read the “Composite Fermions” part of the Wikipedia article you posted. Anything, including many atoms, composed of an odd number of fermions is, itself, a fermion (and anything composed of an even number of fermions is a boson.)

contravariant• 5 years ago

The term "fermion" is often used for everything with a half-integer spin. This is because their wave-functions are still anti-symmetric and therefore the Pauli exclusion principle applies leading to Fermi-Dirac statistics.

A lot of terms are reused for what are technically compound particles, especially since it isn't always too obvious which properties of a particle are intrinsic and which are caused by interactions.

sharikous• 5 years ago

In physics you often find rules for modeling composite objects in terms of simple ones. For example a rigid body can be modeled for many purposes as a point mass, whose value is the rigid body's total mass, located in its barycenter.

In quantum mechanics you can also form composite objects (in the most common formalism it will be a "wavefunction") and you still can determine a total fermion/boson-ness as a function of the elementary objects that form it. So yes, an atom, a proton, a molecule, a table all can be bosons or fermions once you decide to ignore some internal degrees of freedom.

In fact even those subatomic particles you mention could be composite objects for all we know.

mrjangles• 5 years ago

Is there something special that prevents this frozen atom cloud from condensing to an ordinary solid?

Yes. The fact they are Fermions and the Pauli exclusion principle is what stops them from condensing into a solid. Compared to the effect of the Pauli exclusion principle in solids it is very much a case of same same but very different. In solids there are a limited number of 'states' the electrons can occupy. With a gas of free atoms, there are infinite states for the atoms to be in, however the Pauli exclusion principle still affects the probability field of the atoms with respect to each other. Specifically, the Pauli exclusion principle ensures that no atoms are ever stationary with respect to each other. This stops them from colliding with each other.

The best way to think of this is as one of those nbody planet simulations. If you place two stationary planets near each other they start moving together and eventually smash into each other. However, if the planets are moving they just keep zipping around each other, never actually hitting (unless you are very unlucky). The same is true with a cold gas of Fermions. This is often referred to as a 'centrifugal barrier' that keeps them apart.

liuhenry2• 5 years ago

The ELI15 explanation is that at these temperatures, quantum mechanical effects take over and the atoms themselves become indistinguishable from each other. They no longer have a definite position in space and experimentally act as matter waves [1].

This diagram is for bosons, but the top two boxes still apply for fermions (Bose-Einstein condensates are different but closely related): https://condensed-matters-cmcdt.wp.st-andrews.ac.uk/2019/02/...

The process is 1) cool until the de Broglie wavelength is larger than the typical interparticle distance and 2) compress.

If you did 2) before getting 1) then you would risk forming a classical liquid/solid. More info on the process: [2].

[1] https://en.wikipedia.org/wiki/Matter_wave [2] https://www.uni-muenster.de/Physik.AP/Demokritov/en/Forschen...

fooqux2• 5 years ago

If I understand this correctly, couldn't this explain dark matter?

The article didn't say how dense this cloud was, but if it was still a cloud it couldn't be that dense right? Surely something that could exist in the cold vastness of space.

gpm• 5 years ago

Near absolute zero states of matter don't really work the same way. I wouldn't read too much into them saying "cloud".

ben_w2• 5 years ago

The cosmic microwave background is 136000 times hotter, so I don’t think so, no.

fooqux2• 5 years ago

Fair, but the article implied that it's a somewhat linear effect- you don't need absolute zero for it to make the cloud dim. The article didn't do a great job of saying what temps they were testing at.

Still, I'm not surprised. I doubt some armchair scientist is going to solve the riddle of dark matter. I was curious and got some good answers. Thanks!

dtgriscom• 5 years ago

"136000 times hotter" sounds a lot more impressive than "2.72546 degrees hotter"...

toxicFork• 5 years ago

Is that temperature uniformly spread?

jwuphysics• 5 years ago

No. The experimentalists used lasers to condense the atoms to high densities.

klyrs• 5 years ago

Sounds like a fun conspiracy. Aliens using space lasers to hide, what, 25% of the matter in the universe from us?

fcsonline• 5 years ago

Dark matter?

zelienople• 5 years ago

It's matter and it's dark, but that means it's dark matter, not Dark Matter. That matters.

gmuslera• 5 years ago

Didn't understand if it stops but do not bounce back light or if it lets light to pass as there was nothing there. Those are different ways to be invisible.

Shadonototra• 5 years ago

.. to our current technology

That's the missing point, invisible is subjective

ergocoder• 5 years ago

Physics is a highly speculative field due the limitation of our technology.

Physicists know this. But laypeople don't.

With the anti and pro science folks, this part is getting worse where you are called multiple insults for even casting doubt on any physics theory/law like "nothing can be faster than speed of light" or "energy is always conserved".

gus_massa2• 5 years ago

It's invisible only for the light of colors with photons with low energy, because the trick only blocks some interactions of the material and light.

If you use light with photons with more energy [perhaps ultraviolet] then other interactions are possible and it is visible.

So "invisible" is a simplification/exaggeration in the press article.

danwills• 5 years ago

Agree it's an exaggeration, but still seems fascinating - that photons with the right energies go straight through, as if the matter wasn't there?! Does sound quite dark-matter-ish, but as someone else pointed out, the temperatures seem off. I prefer to believe dark matter is mostly something we don't yet quite understand about gravity/matter/space.

peter_d_sherman• 5 years ago

If this is true, then consider the following:

If we know that:

A) Superconductors can conduct a theoertical infinite amount of electricity when supercooled (super-cooling being the key to this phenomena)...

B) That there is a corollary; a relationship between electricity and information; that is, that electricity can carry information (unless you're reading this in the far future, the computer you are reading this on is proof of that! <g>), and a theoretical infinite amount of electricity can carry a theoretical infinite amount of information...

C) If A and B are true, then that would imply that super-cooling -- is the key to being able to transfer a theoretically infinite amount of information, that is, there's a link, a relationship between super-cold things (super low temperatures) -- and the ability to pass information through it...

So, that brings us to the title of this article: "Ultracold, superdense atoms become invisible".

If this article is true, then not only is there a link between super-cold temperatures, the ability to pass theoretically infinite information through a superconducting material (which really just consists of atoms of the superconducting substance, which really just consists of probable repeating structure of that substance's atoms in space...) but also atom invisibility...

So, if true, we have ultra-cold, invisibility, and theoretical infinite information... all at the same place, at the same time...

OK, so with that background, I'm going to go for "full crackpot" here... <g>

If atoms (or heck, any subparticle really, this would include any and every subparticle) -- become invisible when super-cooled -- then:

Question #1: Are they really there, at all? (When super-cooled?)

Question #2: If the answer to Question #1 is that they are not (again, when super-cooled), that they somehow collapse (for the time that they are super-cooled) and then re-emerge when heated (boy, wouldn't this solve all problems of all kinds of atoms and sub-particles "disappearing" and "reappearing" if true?) -- then here's "full crackpot":

If all of that were true -- then couldn't any single atom or sub-particle (any of them, no matter how small) -- be modeled and/or viewed as INERTIA -- but INERTIA RELATIVE TO SCALE, INERTIA RELATIVE TO TEMPERATURE... or more specifically INERTIAL FIELDS RELATIVE TO TEMPERATURE, or even more specifically INERTIAL FIELDS RELATIVE TO TEMPERATURE, RELATIVE TO SCALE...

Well, if one or more things, one or more aspects of the above logic is false, then the answer is 'No'...

But who knows?

Maybe it's possible...

And then again, maybe it isn't... <g>

In other words, if true -- all atoms, particles and subparticles, all of them -- are INERTIA (force) -- at their relative scale...

gus_massa2• 5 years ago

They are only invisible for a laser with a color with low energy that can only "move" the atoms. When light colides with an atom and changes direction, the atom has some speed after the colission.

If the atoms can't move because there are other atoms blocking them, then the can't redirect the light. [I oversimplified a few details here.]

If you use a laser with other color, then the light can be absorbed by the electrons in the atoms, and the electrons can jump and later emit the light. So it will not be visible with another laser. [Probably an ultraviolet laser is enough.]

Question #1: Are they really there, at all? (When super-cooled?)

Yes.

byteware• 5 years ago

A) there is a critical current density above which superconductivity collapses, so no infinite current, not even theoretically

trhway• 5 years ago

“An atom can only scatter a photon if it can absorb the force of its kick, by moving to another chair,” explains Ketterle, invoking the arena seating analogy. “If all other chairs are occupied, it no longer has the ability to absorb the kick and scatter the photon. So, the atoms become transparent.”

by that logic neutron stars should be highly transparent/invisible too.

NickNameNick• 5 years ago

Maybe? But wouldn't you expect at least some non-degenerate matter on the surface of a neutron star?

I doubt we'll find any 'naked' neutron stars.

datameta• 5 years ago

The difference being that neutron stars can revolve tens or hundreds of times a second and have a surface temperature of ~600K degrees.

mr_toad• 5 years ago

by that logic neutron stars should be highly transparent

Neutron stars aren’t uniformly crushed to the limit where they couldn’t interact with photons. That would be the point at which they’d be on the verge of collapsing into a black hole, and photon scattering would be irrelevant.

tsimionescu• 5 years ago

Perhaps the core of the neutron star, but light would have to reach there first, through the many layers of less dense matter above, right?

denton-scratch3• 5 years ago

I think a neutron star is fully degenerate; effectively it's all core. It's not made of atoms; it's more like a huge atomic nucleus. It's much more dense than the "super-dense" gas these researchers are playing with.

A newly-formed neutron star should contain about as much heat as the stellar core that collapsed to form it, but in a much smaller space; so I'd expect it to be super-hot. WP says it's temperature is around 600,000K.

So rather than becoming transparent, I'd guess that a neutron star would be super-reflective. It's surface is ultra-smooth, and I can't see a photon getting far into a neutron star without encountering a neutron. Newly-formed neutron stars lose heat very quickly, but they do that by emitting huge numbers of neutrinos, not photons.

excerionsforte• 5 years ago

Crazy thought I has was what if you could make a supercooled wall to the point you could see through it then you could realistically only see it if you perturbed it and what you'd be able to see is the "wave" as the energy travels from the perturbation. Kinda like those anime shields. Maybe a little too crazy of a thought.

867-5309• 5 years ago

supercooled and ultrasqueezed

sound like there may be a link to black holes?

ketralnis• 5 years ago

No, a sibling notes that this is still way less dense than regular old solid lithium

tchbos• 5 years ago

Makes sense now with all the UFOs showing up on radar as very cold objects!

idiocrat• 5 years ago

Trying to seed a black hole.

jhoechtl• 5 years ago

I would assume this is because of Heisenberg. If the Atom does no longer move you should be able to nail it down to a specific position yet at this tinyness we are entering the land of probable positions instead of definite ones so the atom has to disappear so laws of physics still hold.

simonh• 5 years ago

I'm afraid not, it's a consequence of the Pauli exclusion principle and the way photons interact (or not) with the atom's electrons.

DeathArrow• 5 years ago

Physics of cold can get quite fascinating. It's a whole new world of discoveries.

Ultracold, superdense atoms become invisible

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