How can “absolute zero” be proven?

I thought that “all” atoms above absolute zero are (thought to be) moving.
Yes.
So, in turn, I thought that when atoms get as close to complete stillness as possible, that’s absolute zero.
The math says so, if you were to project the line of decreasing temperature with decreasing movement you would hit the graph's origin of (0,0).

But that is not what physically happens.

Atoms cannot stop moving and still remain atoms. If they did, we could precisely determine both their exact location and their exact momentum simultaneously, and that violates Heisenberg's Uncertainty Principle, which explicitly forbids that exact scenario.

So what the atoms do is sort of stop being atomic. They smear out into a collective mush, meaning what were individual atoms now no longer have a distinct location (or momentum), making it impossible to determine exactly where they are (or how fast they're moving), thus preserving HUP.
 
Not really. There is residual energy in the ground state of many systems but you can't detect any motion directly. Take the vibration of a diatomic molecule like oxygen O=O. The double bond between the atoms is stretchy, so the atoms move apart and together as if connected by a spring. In a classical ball and spring model one would say the energy of vibration changes from kinetic (when the balls are in motion) to potential energy (of the spring) and then back again, over and over again. According to quantum theory there is some energy like this even in the ground state. But all this amounts to is an uncertainty about their position, relative to one another, at any given moment. They are kind of smeared out a bit. This is the wave aspect of wave-particle duality making itself felt. The atoms are not entirely particle-like, but fuzzy due to their wavelike behaviour. So in a way it is not really "motion", or not in a classical sense, just residual kinetic + potential energy that they have.
I'm understanding this better, now. I've been reading about how absolute zero doesn't ''violate'' the uncertainty principle, and it's making more sense now, coupled with your explanation. So, is this why liquid helium doesn't freeze (and become solid) at absolute zero?
 
Yes.

The math says so, if you were to project the line of decreasing temperature with decreasing movement you would hit the graph's origin of (0,0).

But that is not what physically happens.

Atoms cannot stop moving and still remain atoms. If they did, we could precisely determine both their exact location and their exact momentum simultaneously, and that violates Heisenberg's Uncertainty Principle, which explicitly forbids that exact scenario.

So what the atoms do is sort of stop being atomic. They smear out into a collective mush, meaning what were individual atoms now no longer have a distinct location (or momentum), making it impossible to determine exactly where they are (or how fast they're moving), thus preserving HUP.
So, in other words, all particles also act as waves? We can measure the energy of a particle, but the more we attempt to measure it precisely, the more we'll need to let go of knowing its exact position. I get it! I think it's pretty fascinating that Heisenberg figured this out.
 
It's not really a question of whether it exists - we know it does because we know what the limit is, which is zero kinetic energy in the particles of a gas (for example.) As a comparable example we know that the 'speed limit' in the universe is the speed of light, and barring some very radical new discovery, nothing can go faster than that. We don't know this because we have accelerated things to 99.999999% of the speed of light and been thwarted from going any faster - we know this because even when we accelerate something to a few percent of the speed of light we see the exact same effects that will keep us from reaching that speed.

We may never be able to reach either limit with anything material, but we know with great accuracy what the limit _is_.
This is an excellent explanation - thank you! I appreciate everyone's explanations, I was getting tangled up with the 'observer effect,' rather than looking at quantum mechanics for what it is, and that the universe doesn't revolve around human observers. This thread has been humbling. ;)
 
I thought that “all” atoms above absolute zero are (thought to be) moving. So, in turn, I thought that when atoms get as close to complete stillness as possible, that’s absolute zero.
Talking of BEC.....

 
I'm understanding this better, now. I've been reading about how absolute zero doesn't ''violate'' the uncertainty principle, and it's making more sense now, coupled with your explanation. So, is this why liquid helium doesn't freeze (and become solid) at absolute zero?
As I understand it, the issue with helium is that the forces between atoms (called London forces after a German called Fritz London), that cause them to condense into a liquid and then into a solid, are especially weak in the case of helium. So weak that the potential well they create between 2 helium atoms is so shallow that the ground state lies above its top.

I should go back a bit to explain......

We had a thread that covered some of this a couple of years ago: https://www.sciforums.com/threads/infinite-potential.165930/

The bit that may be helpful in this context is this:-

QUOTE
These are features of any bound state, i.e. any state in which a system is confined by a potential of any shape. In chemical bonds, for instance, the shape of the potential confining the atoms so that they can't break apart is asymmetric, with a right hand side that does not go to infinity, because the bond can be broken by adding sufficient energy. It looks more like this:


bond-20energy-gif.192688



But when you solve the equations, you still get a series of allowed levels and a ground state that is not at the bottom of the well, i.e. with a zero point energy. So the features of the simpler particle in a box scenario are retained, at least qualitatively.

P.S. A good model for the potential illustrated above is the Morse potential. Here is a link to the Wiki article on it. In the diagram you can see the energy levels calculated for this potential: https://en.wikipedia.org/wiki/Morse_potential

UNQUOTE

This is a graph of energy versus distance between 2 atoms. You can see that it is at a minimum at a certain distance, and rises as you push them together or move them further apart. The degree to which this happens depends on how strong the attraction is between the atoms, i.e. a stronger attraction produces a deeper "potential well" at the optimum separation distance, requiring more energy to be put in to separate them or push them further together. But the only stable states, according to QM, are at certain fixed separations. So you can draw a series of horizontal lines on this diagram, at different energy levels, and the system can only occupy one of those, and nothing in between. Importantly, the ground state, i.e. the lowest allowed level, is not at the very bottom of the well, but a bit above it. The energy remaining at that bottom level is the "zero point energy".

In the caseof helium atoms the attraction is very weak and so the shape of the well is very very shallow, barely a depression at all. It turns out the ground state is above the top. So that means helium will never freeze.
 
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I guess one simpler way to put it is that heat energy, by definition, is transferable. Since zero point energy is not transferable to any other system, it contributes nothing to a system's thermal energy. So absolute zero means "the state of any system with no transferable kinetic energy."

Maybe this is just me, but I've always had a little problem with calling 2.7k the "temperature of empty space" or "the void" because it seems to muddle concepts. Say we find a truly empty cubic centimeter of space, perhaps in one of those big voids between supergalaxies. Well, yes, if a photon passes through, on average it could convey 2.7k to a cold atom or hadron bit that's lying around. But that bit of space is empty, so it doesn't have any baryonic matter that could be jiggled up to 2.7k. So the photon passes on through and doesn't part with any of its energy. I guess what I'm saying is that it's a bit like standing near a stove - if you are there, you will be warmed up by it. If the stove is just radiating into a void and no one's there warming themself, then nothing is warmed and all the radiation moves on. So 2.7k is kind of like "potential temperature in space." Hard vacuum doesn't really have a temperature at all. All the CMB radiation just passes on through.
 
Talking of BEC.....

Omg, that is incredibly amazing! Thanks for sharing.

From the article:

Their BEC, cooled to just five nanoKelvin, or about -459.66°F, and stable for a strikingly long two seconds, is made from sodium-cesium molecules. Like water molecules, these molecules are polar, meaning they carry both a positive and a negative charge. The imbalanced distribution of electric charge facilitates the long-range interactions that make for the most interesting physics, noted Will.

Research the Will lab is excited to pursue with their molecular BECs includes exploring a number of different quantum phenomena, including new types of superfluidity, a state of matter that flows without experiencing any friction. They also hope to turn their BECs into simulators that can recreate the enigmatic quantum properties of more complex materials, like solid crystals.

"Molecular Bose-Einstein condensates open up whole new areas of research, from understanding truly fundamental physics to advancing powerful quantum simulations," he said. "This is an exciting achievement, but it's really just the beginning."


**Deleted my earlier questions/comment, because I found the answers.

So, BEC is considered another state of matter. But, beyond the fundamental states of matter such as liquids, solids, and gas.

So I’ve been wondering what would BEC look like if we could see it for ourselves? Here’s what I found:

A Bose-Einstein condensate (BEC) can look like a dense lump at the bottom of a magnetic trap or bowl, similar to a drop of water condensing on a cold surface. When it first forms, it can also look like a pit in a cherry because it's still surrounded by normal gas atoms.
 
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I guess one simpler way to put it is that heat energy, by definition, is transferable. Since zero point energy is not transferable to any other system, it contributes nothing to a system's thermal energy. So absolute zero means "the state of any system with no transferable kinetic energy."

Maybe this is just me, but I've always had a little problem with calling 2.7k the "temperature of empty space" or "the void" because it seems to muddle concepts. Say we find a truly empty cubic centimeter of space, perhaps in one of those big voids between supergalaxies. Well, yes, if a photon passes through, on average it could convey 2.7k to a cold atom or hadron bit that's lying around. But that bit of space is empty, so it doesn't have any baryonic matter that could be jiggled up to 2.7k. So the photon passes on through and doesn't part with any of its energy. I guess what I'm saying is that it's a bit like standing near a stove - if you are there, you will be warmed up by it. If the stove is just radiating into a void and no one's there warming themself, then nothing is warmed and all the radiation moves on. So 2.7k is kind of like "potential temperature in space." Hard vacuum doesn't really have a temperature at all. All the CMB radiation just passes on through.
I see your point, though I'm not sure it isn't a semantic one.

Another way of what you're saying is that, in a vacuum, there is no heat transfer by convection or conduction - but there certainly is by radiation.
 
Yes.

The math says so, if you were to project the line of decreasing temperature with decreasing movement you would hit the graph's origin of (0,0).

But that is not what physically happens.

Atoms cannot stop moving and still remain atoms. If they did, we could precisely determine both their exact location and their exact momentum simultaneously, and that violates Heisenberg's Uncertainty Principle, which explicitly forbids that exact scenario.

So what the atoms do is sort of stop being atomic. They smear out into a collective mush, meaning what were individual atoms now no longer have a distinct location (or momentum), making it impossible to determine exactly where they are (or how fast they're moving), thus preserving HUP.
Thank you for this! It's making sense, now.

In addition to Pinball1970 's post #25, I've been reading that AI may rely on ''manipulating'' quantum states. That's both exciting and a little scary. Maybe manipulate is the wrong word choice.
 
Thank you for this! It's making sense, now.

In addition to Pinball1970 's post #25, I've been reading that AI may rely on ''manipulating'' quantum states. That's both exciting and a little scary. Maybe manipulate is the wrong word choice.
I'm not sure what AI and qubits have to do with each other. I mean, sure, a powerful computer configured as AI might be run on a quantum computer, but it's kind of the equivalent to saying "Windows may rely on manipulating RAM memory addresses".
 
I see your point, though I'm not sure it isn't a semantic one.

Another way of what you're saying is that, in a vacuum, there is no heat transfer by convection or conduction - but there certainly is by radiation.
Surely the point is that temperature is, properly speaking, a bulk property of matter, proportional to the average thermal kinetic energy of the atoms or molecules that comprise it. It seems to me a bit false to describe radiation, even a black body radiation distribution, as having a temperature, when there is no thermal kinetic energy present.

Although you could stick a thermometer into a portion of empty space and it would eventually reach 2.7K, once its own thermal radiation and the CMBR came to equilibrium.
 
As I understand it, the issue with helium is that the forces between atoms (called London forces after a German called Fritz London), that cause them to condense into a liquid and then into a solid, are especially weak in the case of helium. So weak that the potential well they create between 2 helium atoms is so shallow that the ground state lies above its top.

I should go back a bit to explain......

We had a thread that covered some of this a couple of years ago: https://www.sciforums.com/threads/infinite-potential.165930/

The bit that may be helpful in this context is this:-

QUOTE
These are features of any bound state, i.e. any state in which a system is confined by a potential of any shape. In chemical bonds, for instance, the shape of the potential confining the atoms so that they can't break apart is asymmetric, with a right hand side that does not go to infinity, because the bond can be broken by adding sufficient energy. It looks more like this:


bond-20energy-gif.192688



But when you solve the equations, you still get a series of allowed levels and a ground state that is not at the bottom of the well, i.e. with a zero point energy. So the features of the simpler particle in a box scenario are retained, at least qualitatively.

P.S. A good model for the potential illustrated above is the Morse potential. Here is a link to the Wiki article on it. In the diagram you can see the energy levels calculated for this potential: https://en.wikipedia.org/wiki/Morse_potential

UNQUOTE

This is a graph of energy versus distance between 2 atoms. You can see that it is at a minimum at a certain distance, and rises as you push them together or move them further apart. The degree to which this happens depends on how strong the attraction is between the atoms, i.e. a stronger attraction produces a deeper "potential well" at the optimum separation distance, requiring more energy to be put in to separate them or push them further together. But the only stable states, according to QM, are at certain fixed separations. So you can draw a series of horizontal lines on this diagram, at different energy levels, and the system can only occupy one of those, and nothing in between. Importantly, the ground state, i.e. the lowest allowed level, is not at the very bottom of the well, but a bit above it. The energy remaining at that bottom level is the "zero point energy".

In the caseof helium atoms the attraction is very weak and so the shape of the well is very very shallow, barely a depression at all. It turns out the ground state is above the top. So that means helium will never freeze.
I skimmed the first couple of pages of that link, it was helpful - thanks! I won't relive that thread here, in terms of examining 'infinite potential,' because we'll be going off track, but...interesting to note, that the ''proper'' way to explore the topic of energy in general, is that ''energy is never lost, it just becomes unavailable for extraction.'' (James R post #6) So energy will remain ''constant'' if it's not influenced in some way.
 
I skimmed the first couple of pages of that link, it was helpful - thanks! I won't relive that thread here, in terms of examining 'infinite potential,' because we'll be going off track, but...interesting to note, that the ''proper'' way to explore the topic of energy in general, is that ''energy is never lost, it just becomes unavailable for extraction.'' (James R post #6) So energy will remain ''constant'' if it's not influenced in some way.
Yes, the first 2 pages were full of Write4U's misconceptions so best not to go down that rabbit hole. What I wanted to get from it was really the diagram showing the Morse potential, which in that thread I had gone to some trouble to find and copy. It's a handy diagram for showing how the energy of a molecule changes when a chemical bond is stretched or compresses, as happens when the molecule vibrates, or when the atoms are pulled completely apart to break the bond.
 
I'm not sure what AI and qubits have to do with each other. I mean, sure, a powerful computer configured as AI might be run on a quantum computer, but it's kind of the equivalent to saying "Windows may rely on manipulating RAM memory addresses".
Not really. Maybe I didn't articulate my point well, but I was trying to convey that quantum computing relies on the principles of quantum mechanics, which offers us the potential for super-fast computer systems.
 
Yes, the first 2 pages were full of Write4U's misconceptions so best not to go down that rabbit hole. What I wanted to get from it was really the diagram showing the Morse potential, which in that thread I had gone to some trouble to find and copy. It's a handy diagram for showing how the energy of a molecule changes when a chemical bond is stretched or compresses, as happens when the molecule vibrates, or when the atoms are pulled completely apart to break the bond.
No, I appreciate the diagram - It sounds like helium is an anomaly because of its unique properties, and it doesn't form ''normal'' chemical bonds like those described by Morse Potential. Not that it matters entirely, but I'm not sure I'll ever understand 'the math' behind all of this.
 
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Surely the point is that temperature is, properly speaking, a bulk property of matter, proportional to the average thermal kinetic energy of the atoms or molecules that comprise it. It seems to me a bit false to describe radiation, even a black body radiation distribution, as having a temperature, when there is no thermal kinetic energy present.
The point about black body radiation temperatures is that black body radiation is typically produced in a situation where energy is constantly being transferred back and forth between matter and photons, so that the two are in a thermal equilibrium. If that's the case, then it makes some sort of sense to describe the "temperature of the radiation" as being the same as the temperature of the matter that emitted the radiation in the first place.

Planck's original model for blackbody radiators was very simple. It just assumed that there were some oscillators (i.e. matter of some unspecified type) that could store energy and exchange it back and forth with (classical) electromagnetic waves. Planck showed that the observed features of the spectrum of light emitted by an ideal black body only depends on the temperature of the emitter. That determines the intensity of the emitted light, its peak wavelength and the distribution and relative intensities of all the frequencies emitted.

It makes no sense to talk about the "temperature of space" or the "temperature of the vacuum", because - as you said - temperature is a statistical property of an ensemble of atoms or molecules (or something else that isn't a vacuum).

Although you could stick a thermometer into a portion of empty space and it would eventually reach 2.7K, once its own thermal radiation and the CMBR came to equilibrium.
Yes. When we say that the cosmic microwave background radiation has a temperature of 2.7K, it really means that its spectrum looks like it was emitted by matter at a temperature of 2.7K. That was not true back at the time when the CMBR was first free to travel through space without being absorbed by matter, of course. The expansion of space since then has caused a 'cosmic red shift' of the initial black body spectrum.
 
Do you mean "something still" or "still would be detected"?

Anyway,do you mean you could bounce a signal or two off whatever you thought was moving and then you could tell if it was moving either with respect to you or with respect the the region that you suspected might be at absolute zero?

I don't think there are any regions anywhere at absolute zero although I cannot say whether it is possible such a region could exist inside a black hole.

Apparently black holes are very, very, very cold ,unlike my own preconception of them as very dynamic and energetic regions
This is why commas are so important. lol I meant “still would be detected.” But, it’s a moot point now since what I was initially asking, doesn’t have much to do with the concept of absolute zero.
 
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No, I appreciate the diagram - It sounds like helium is an anomaly because of its unique properties, and it doesn't form ''normal'' chemical bonds like those described by Morse Potential. Not that it matters entirely, but I'm not sure I'll ever understand 'the math' behind all of this.
None of the inert gases (Helium, Neon, Argon, Krypton,Xenon, Radon) easily forms normal chemical bonds. All of them are monatomic gases. That is because all of them have complete outer shells of electrons. There are no "vacancies" in the outermost shell, ready to be filled by sharing electrons with another atom, which is what happens in chemical bonding.

This rule gets a bit weaker, however, with the larger members of the series, e.g. Xenon, which has some empty electron orbitals, not too much higher in energy, into which electrons can be promoted, leaving behind a gap which can be used to form bonds.

This Morse potential can however also be applied to the much weaker attraction due to London "dispersion" forces. These are not chemical bonds, but a feeble attraction due to a sort of flickering of tiny partial electric charges on neighbouring atoms due to the motion of their electrons (or to uncertainty in their position). There is a Wiki article on this here: https://en.wikipedia.org/wiki/London_dispersion_force. in which you will see the good old Morse potential, or something very similar, pops up yet again.

These dispersion forces produce an attraction between molecules that causes molecular substances to condense into liquid and finally the solid state.
 
It makes no sense to talk about the "temperature of space" or the "temperature of the vacuum", because - as you said - temperature is a statistical property of an ensemble of atoms or molecules (or something else that isn't a vacuum).
I would say the photon density in an area not space itself.
If you were looking for Absolute Zero, would you shield the experiment from the affects of the CMB photons? If yes, then why?


I was reading a book on helium and I couldn't put it down.
 
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