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westernguy339 t1_jd1qpnp wrote

No actually. Phases of matter really are how that matter behaves in relation to itself. A solid liquid or gas can only be defined because of the relationships atoms have with one another. A single uranium atom in water is a liquid, in air its a gas, and in a rock is a solid.

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SatanScotty t1_jd4dere wrote

Could you measure the kinetic energy of the entire atom and estimate “this atom has a level of energy consistent with a solid”?

Could you also note whether or not it’s ionized in a way consistent with plasma?

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Not_Pictured t1_jd4ebj3 wrote

Not really since kinetic energy in the case of a single atom is the same as saying velocity. And velocity is relative.

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Zalack OP t1_jd4el25 wrote

Wait. Does that mean a single atom essentially doesn't have a temperature?

🤯

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florinandrei t1_jd4ubra wrote

Technically, you can associate a temperature to the velocity of the atom measured relative to the container, and therefore obtain a "temperature" for that atom. But a lot of concepts become quite strained when you reduce things to single atoms, and temperature is one of them. A single atom does not have a temperature in the normal sense.

To your initial question: the phases of matter are only defined for molecular or atomic collectives. Single molecules or atoms do not have a clearly defined phase of aggregation. Even for large molecular collectives it is not always clear whether they are solid, liquid, or gas. For example, on geologic time scales, even some "solids" can flow.

The phases of matter are more like convenience concepts. We use them to simplify discussions that would otherwise be complex. There's nothing fundamental about them. Do not get stuck in rigid categorizations there, because there's no point in doing that.

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OPossumHamburger t1_jd5ei1k wrote

Phases of matter are more than convenient classifications of multiple atoms.

Because of their states, they exhibit different physical properties including changes in electrical conductance.

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yakbrine t1_jd5k2pr wrote

His point to my understanding is that is kind of the point. There’s probably tons of variables like this for every solid and liquid. And the sole fact they are solid or liquid does not give them said properties or everything would be identical. The point being everything is extremely nuanced and we’ve created these categorizations so we don’t have to define everything as a mathematical equation instead of ‘solid gas liquid’

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florinandrei t1_jd6br1i wrote

> they exhibit different physical properties including changes in electrical conductance

Of course they do. I'm just saying - the borders between them are far more fuzzy than most people imagine.

E.g. consider the changes that occur in tar or pitch when cooled from the boiling point of water to the boiling point of nitrogen. It's liquid at one end. It's solid at the other. The changes are smooth, without any sharp transitions.

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NeverPlayF6 t1_jd5w6tb wrote

Since states are defined by how "they exhibit different physical properties," how many N2 atoms does it take to behave like a solid/liquid/gas in any given system?

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Angdrambor t1_jd7xac3 wrote

It sounds extremely convenient to me, to be able to model bulk properties like conductance instead of modeling individual atoms.

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Dr-Luemmler t1_jd5fsbi wrote

Maybe I dont get what you are saying about temperature, but what you are saying doesnt make sense to me. If a single atom wouldnt have a temperature, because it cant have a velocity alone, what happens if we drop a second atom in the void? Does now (kinetic) energy spawn from nothing? Besides that, temperature itself isnt relative as we have a true zero. Even if it is just theoretical.

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6strings10holes t1_jd5htjc wrote

You can't establish energy really, only changes or relative amounts to a reference frame.

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Dr-Luemmler t1_jd5io43 wrote

Thats my point. Ofc, in a labratory you need a reference to measure the velocity of a single atom. The reference frame obviously can be broken down to other atoms if you want, but that doesnt mean a single atom cant have kinetic energy by itself.

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florinandrei t1_jd6bwq6 wrote

It definitely does have a kinetic energy.

The only thing is - when you go from kinetic energy to temperature, you run into all sorts of trouble if you do it for single entities.

Temperature is an inherently collective measure. If it's single particles, stick to kinetic energy.

What is the "temperature" of this marble I'm throwing? ;) (not the temperature of the glass, but the "temperature" of the marble as a single particle with some kinetic energy)

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Dr-Luemmler t1_jd75stk wrote

Defining temperature by kinetic energy, you could calculate it for a single marble. If you want to use the advanced definition of temperature via entropy, sure lets do it:

$T = dE/dS $

So temperature is the change of internal energy when changing the entropy. In statistical thermodynamics, one can now define entropy by the number of availible states $\Omega$ with its degrees of freedom.

The degrees of freedom a single atom has are $3N-3$ = 0. That basically means, this atom only has the translation dofs and the electronic ones. Lets neglect the electronic ones (even though they might be important, as with then we might be able to measure the temperature) then the temperature of a single atom is solely defined by its kinetic energy.

Can we access it in labratory without using the interaction with other atoms? No! But in simulations we can. Or what kinds of problems do we have?

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Quantum_Quandry t1_jde90h1 wrote

A single atom most definitely cannot have kinetic energy all by itself. SR/GR makes it abundantly clear that you must have something to reference against to make a measurement, and the answer changes depending on which reference point you're using. This should be obvious to anyone who has driven a car. Let's say you have three cars, yourself going 50mph north, a second car ahead of you and to your left going 45mph north, and a third car going 50mph ahead of you headed south. You have to swerve left or right due to an obstacle ahead, which do you choose? Obviously you're going to swerve left, ignoring the velocity of your swerve itself, you're going to overtake the car on your left a a relative 5mph and if you go right you'd be moving 100mph relative. Or you could split the difference and drive directly into the obstacle which is going 50mph relative to you.

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TheArmitage t1_jd5hp7t wrote

>what happens if we drop a second atom in the void?

In doing so, you've introduced energy into the system. That atom had to get there somehow, and that takes energy.

>Besides that, temperature itself isnt relative as we have a true zero.

Yes, it is. It's just self-referential. Thermal motion is the motion of atoms in a substance relative to each other. So if all atoms in a substance have zero motion relative to each other, it has a temperature of 0K.

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Acewasalwaysanoption t1_jd5ll20 wrote

Nobody said that we can't have a reference point, just that we have a single atom of an element, as opposed to a macroworld-sized amount to easily determine its phase.

Like if I'm the last person on the world, I can't tell if I'm handsome or if I'm rich, without other people to compare myself. But I know how fast I am, because I don't need other people for a reference system.

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TheArmitage t1_jd5mdnn wrote

It is inherent in the definition of temperature that the substance is compared to itself. You cannot have an external reference point for temperature, because then it's not temperature.

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Acewasalwaysanoption t1_jd5oj3o wrote

Sorry, I may have misread something.

New question: what you exactly mean by "compared to itself"? It can't be literally itself in the same state, as it would be the same, all the time. Can't be a chunk of the material, or any material that has the same temperature in its core and surface would be at 0 difference and...incomperable?

Also, using thermometers isn't using an external point if reference in general? Originally nercury's change in volume to tell a completely different material's temperature. Works because energy transfer.

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Purplestripes8 t1_jd6moch wrote

By compared to itself it means the motions of the atoms within an object relative to each other. The object itself can have any velocity depending on the observer but no matter which direction it's moving as a whole or how fast, the atoms within still have the same motions relative to each other, which is signified by temperature.

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8426578456985 t1_jd534z5 wrote

How fast would that be? Being a single atom in a glass jar?

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Noggin01 t1_jd5acka wrote

Depends. Is the glass moving?

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8426578456985 t1_jd5aonm wrote

I want to say no, but relative to what? Say the jar is floating in space with no radiation hitting it and the hydrogen atom is just teleported into the jar so their relative speeds are zero at the start.

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KarlSethMoran t1_jd4fn4k wrote

It doesn't. You need internal degrees of freedom to define temperature. An isolated atom has zero internal degrees of freedom due to Galilean invariance.

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lizardweenie t1_jd4l91d wrote

An atom has internal degrees of freedom due to electronic and spin transitions, so it can certainly be excited. In general though as previous users mentioned, temperature is an ensemble property, so the atom wouldn't have a well defined temperature.

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Dr-Luemmler t1_jd7e31c wrote

An ensemble can also be one atom or one molecule. So yes, it certainly has a well defined temperature.

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lizardweenie t1_jd92qsl wrote

As other posters have mentioned, temperature is a property of a distribution. It tells us the probability of populating an excitation of a given energy. This isn't up for debate, it's just a matter of definitions. If you want to come up with some new concept that is well defined for a single particle, that's cool, but temperature doesn't work for single particles.

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Dr-Luemmler t1_jd9m4t4 wrote

Oh, the distribution works if there are multiple available states at a given energy level. And there are. You can even calculate it for a single particle in a box. We actually did in class...

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lizardweenie t1_jd9xip3 wrote

I don't mean to be rude, but it really seems like you haven't learned about temperature in a rigorous way (Like you would in a statistical mechanics class). It sounds like you've at least had some sort of exposure to undergrad level quantum mechanics, which is great. But recognize that your knowledge may not apply to this, and consider taking a statistical physics class.

If you did take such a class, you would learn that beta (which is propositional to 1/T) can be defined in terms of the partition function of the system of interest, but the entire concept relies on having multiple particles, (not simply one particle that transitions from state to state).

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Dr-Luemmler t1_jdaglj7 wrote

I dont want to be rude, but you just need different states the particle could be, which you get with the quatification of impulse for each direction and the electronic states. Having multiple undistinguishable particle, and measuring their states is just one way to calculate the partition function. Another one is to track the trajectory of a single particle. In other words, we just need different states with different probabilities. I see no reason why that would not hold for a single atom. Temperature itself is also not a relative measurement as you can also see temperature dependent radiation from only a single atom.

The temperature in thermodynamics is defined as $T = dE/dS$. As $S \approx log(\Omega)$, the amount of "accessible" states need to increase with increasing temperature to hold the first formula. As a single atom has three dofs, we fullfill it.

Sorry, I really see no reason you could be right. I have also studied a bit advanced statistical thermodynamics and wrote my BA in that field. But I can be wrong, I cant say I was excellent in that field and some years have past since then. Maybe you can give me some hints for proper literature.

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lizardweenie t1_jdamc26 wrote

No worries, you're not being rude. As for references, this a matter of basic definitions so I'd recommend some good textbooks, depending on your background.

I'd say that Chandler's book is pretty good: (I used it at the beginning of my PhD) http://pcossgroup.xmu.edu.cn/old/users/xlu/group/courses/apc/imsm_chandler.pdf

If you're looking for a different perspective, I've heard good things about Reichl: "A Modern Course in Statistical Physics"

Fun fact about this statement: > the amount of "accessible" states need to increase with increasing temperature to hold the first formula

This need not be the case. In certain scenarios, you can actually obtain negative temperatures which are perfectly valid. https://en.wikipedia.org/wiki/Negative_temperature

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lizardweenie t1_jdapyy8 wrote

I just thought of a reasonable thought experiment that might clarify your confusion:

Say you have a bath of non interacting hydrogen atoms (consider for a moment, only electronic excitation), and we are able to measure the state of each atom.

Say we measure this bath and find that f_0 fraction are in the ground electronic state E0, and f_1 are in the first excited state E1. We could then infer a temperature by comparing these populations to a Boltzmann distribution, which tells us the relative probability of finding an atom in a state at a given energy (for a given temperature). In this case temperature is a well defined and meaningful concept.

Now say instead that we have a single hydrogen atom, we measure its state, and we find that it's in the first excited state. What then is the temperature? If we try to infer a temperature from this, (using a Boltzmann distribution), we get -inf. Say instead we measure it, and it's in E0. In this case, our inferred temperature will be 0. So for this single atom system, any temperature that we try measure can only give two values, (0, or negative infinity). In this system, clearly temperature isn't behaving how we would like it to.

This troublesome result points to a larger problem with the question: asking "what is the probability distribution for state occupation" doesn't really work well for the example: the atom was measured and determined to be in state E1, its probability distribution is a delta function, which is an inherently non-thermal distribution.

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avoid3d t1_jd4urxm wrote

Hmm, in physics we learned that a more nuanced way of reasoning about temperature is relating it to the change in entropy as heat is added.

If I understand you correctly you are arguing that heat cannot be added to a single atom since there are no inter molecular forces to create oscillations to store the heat.

I’d argue that heat can be added since there are other kinds of energy states that are possible in a single atom such as electric phenomena.

Is there something I’m misunderstanding?

edit This lovely commenter explains this topic very well:

https://www.reddit.com/r/askscience/comments/11x4f9t/comment/jd4r58z/

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RevengencerAlf t1_jd4xdut wrote

I think this is both true and kind of not and it gets weirdly philosophical. It doesn't have temperature as we're taught about it in HS physics class, sure, since that is generally the internal kinetic energy of molecules vibrating and bumping into each other, but atoms themselves have internal degrees of freedom at the quantum level that can reasonably be used to describe temperature. The excitement state of an atom's electrons is the most obvious one.

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KarlSethMoran t1_jd4ys58 wrote

Sure. That is the electronic temperature. I was coming from the perspective of a classical point-particle picture and the kinetic, not thermodynamic, definition of temperature.

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MarzipanMission t1_jd520vi wrote

How is thermodynamic temperature different from viewing it from a kinetic perspective?

Does that mean that the movement of atoms relative to each other, in the kinetic sense of temperature, is not what the temperatures talk about in thermodynamics? So temperature is not a universal concept then? It is context dependent, and has many definitions?

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sticklebat t1_jd58ks6 wrote

Thermodynamic temperature is defined as the rate at which the internal energy of a system changes as its entropy changes.

In contrast, temperature from kinetic theory is essentially a measure of the average translational kinetic energy of the particles in a system.

The two are sometimes, but not typically, equal. The temperature that you know and love is the second one, but thermodynamic temperature is also widely used in science.

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wnoise t1_jd5kly2 wrote

The second is special case of the first. The statistical mechanics temperature really is the fundamental one.

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sticklebat t1_jd5ppqo wrote

Yep. But in most scenarios corresponding to human experience the first is reasonably applicable and much easier to understand.

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Dr-Luemmler t1_jd7c3ex wrote

>So temperature is not a universal concept then? It is context dependent, and has many definitions?

Yes it has, but the definitions are all different sides of the same coin. Or in other words, they add in complexity, but are more or less the same. They are not contradicting.

Temperature IS the average kinetic energy of a systems particles. Thats not just the classical definition that is also the result if you combine quantum theory and statistical thermodynamics. This kinetic energy just is not only translational but also rotation and vibration. A single atom though, does not have the degrees of freedom to rotate or vibrate. Besides its spin, but that is not important here.

There is another dof, and thats the electronic one. Yes, here energy can also be stored. Also probably neglectable here, but OP was very unprecise with his thought experiment here...

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glurth t1_jd4n4io wrote

"You need internal degrees of freedom to define temperature."

This sounds inaccurate. Wouldn't the excitement states of a single atom's electrons be an internal degree of freedom?

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KarlSethMoran t1_jd4x47b wrote

That would define the electronic temperature, not the temperature of the atom in the classical, point particle picture.

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Merakel t1_jd4oogr wrote

It sounds weird to me too, but I also at first thought he was talking about fahrenheit so I'm probably not worth listening to :/

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blacksideblue t1_jd52p3b wrote

so in a zero pressure environment like space, does that mean all matter is out of phase until gravity or some form of surface tension groups atoms together?

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KarlSethMoran t1_jd6vmwt wrote

I don't get what you mean by "out of phase". Gravity is exceedingly weak at the atomic scale, it can be safely ignored.

Atoms feel van der Waals attraction. It's a very weak interaction, but billions of billions times stronger than gravity at this scale still. It will get even noble gases into a crystal when there's sufficiently little motion.

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Mutants_4_nukes t1_jd57n5m wrote

Temperature is a macroscopic value. It is a measure of the kinetic energy of a large collection of atoms or molecules. Individual atoms do not.

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ANNOYING_TOUR_GUIDE t1_jd5egbp wrote

No. Once you get down to that scale, our macroscopic concepts don't apply. Quantum physics is strange and unintuitive.

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leadsynth t1_jd6g4dz wrote

Like how a single atom doesn’t have a color?

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my58vw t1_jd5szy4 wrote

An atom has kinetic energy due to the random movement of electrons within the atom. The movement of electrons are theoretically relative to the nucleus, and single KE is a function of mass and velocity, the components of atoms have a speed and this a temperature. A self contained particle nearing absolute zero could again be stationary.

Temperature in a normal situation relies are other particles, but then as others have said breaks standard conventions of physics and chemistry

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ZeBeowulf t1_jd4xbqw wrote

No, because of zero-point energy atoms even at absolute zero still have some kinetic energy and thus a temperature.

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Dr-Luemmler t1_jd5f45s wrote

Ehm, what? I know what you are saying, but just because you need some kind of interaction to measure ANYTHING. Or in other words with that logic you couldnt even measure the impuls of a flying particle because to measure it, the particle would need to interact with another particle somehow. In a simulation you for example could measure the energy level of a single particle and then determine its state. So here for md. For dft simulations you could also use the electron probability densities to determine the distance to other particles.

This kind of access to the physics are not availible for a single atom ofc

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RedditAtWorkIsBad t1_jd51cph wrote

And to add the comment about how velocity is relative, even if you have a large mass of material moving quickly, this doesn't make it hotter. So, velocity isn't by itself the metric you need but variance in velocity, where velocity is a vector quantity. This way you can get a picture of the range of differences in velocity amongst the particles. Temperature is directly related to this (and would only be related to this for simple point masses that only react like billiard balls.)

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Solesaver t1_jd578qg wrote

One other wrench to throw into this thesis is that phase of matter is actually determined by a combination of temperature and pressure. The pressure of a single atom is also not a meaningful metric.

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garrettj100 t1_jd59lci wrote

Another issue is the energy of an atom doesn’t determine its temperature. Not exactly.

The high school definition of temperature as the average kinetic energy of the particles is merely an approximation appropriate only for gasses. Thusly the “ideal gas law”.

It’s better to think of temperature as a thermodynamic arrow. Heat flows from higher temperatures to lower ones. The rigorous definition of temperature is the inverse of the derivative of the entropy with respect to energy:

T = 1 / δS/δE

As you add more energy to a system, it gets more entropy, but because entropy is logarithmic it grows slower. So the derivative gets smaller. Thusly the temperature rises.

The flow of energy from high temperatures to low temperatures means that total entropy rises, because the system with lower temperature gains more entropy from the infinitesimal of energy. It’s how the universe obeys the second law of thermodynamics: Entropy always increases.

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HaikuBotStalksMe t1_jd5kgi1 wrote

Water evaporates even below 100° so it's hard to use any single value for measuring a phase.

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cracksmack85 t1_jd4k75m wrote

Semi-related question - when I took gen chem in college, if a solid was dissolved in a liquid it was always denoted as “aqueous” and we were told to treat it as a liquid, but it was never really clear to me whether that substance is a liquid or a solid. Can you explain the “aqueous” designation?

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[deleted] t1_jd4mf0y wrote

Aqueous here means that the solution which constitutes a solvent and solute, has water as the chosen solvent.

Example being Sodium Chloride. Sodium Chloride (aq) simply means that you have a saline solution. The individual atoms of Sodium Chloride are dissolved in liquid water.

I wouldn't really call NaCl a liquid if you have NaCl (aq) because you're no longer dealing with pure NaCl. The aqueous solution is a liquid though.

This leads to an interesting question about solids that do not dissolve in a liquid. Here you're dealing with a type of colloid (effectively one phase suspended in another phase) And the dispersed compound does not have to be solid btw- it can be liquid (such as milk).

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cdcformatc t1_jd5d0dx wrote

Liquid salt to me makes me think of molten salt heated past it's melting temperature, which is very different than an aqueous solution.

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Angdrambor t1_jd7y889 wrote

It gets even more fun when you look at materials like Ice or Steel, which have like a dozen solid phases each.

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octonus t1_jd54mix wrote

We discuss states of matter in terms of how a molecule interacts with its neighbors. If the solvent is a liquid (water in the case of aqueous solution), all interactions that you would care about are liquid like. In the case of a solid solution ie. bronze they would be solid-like.

The key reason we note it as being a solution rather than a liquid is to point out that the neighbors a molecule interacts with are the solvent, rather than molecules of the same type.

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RestlessARBIT3R t1_jd4n0l2 wrote

Aqueous just means that it’s dissolved in a liquid. Ionic things dissolved in a liquid get covered in water molecules because the water molecules are polar and the ions are charged. Non-ionic things dissolved in liquid are usually just polar and can form hydrogen bonds with the water molecules so that it looks like a single solution as opposed to when things don’t dissolve

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titsngiggles69 t1_jd4kedq wrote

What about Bose einstein condensate?

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MrCompletely t1_jd5ljhj wrote

BEC is also an emergent, a phase of matter that only exists in aggregate. The key characteristic of a BEC is that multiple atoms can be described as a single quantum mechanical entity.

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titsngiggles69 t1_jd5muo4 wrote

Ah... So if I had two atoms, could I be right half the time? 😂

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ewanatoratorator t1_jd4iwl0 wrote

Interesting. Does this extend to metals with their delocalised electrons too?

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UnshelteredInstincts t1_jd4vffx wrote

When we talk about delocalized electrons in metals, that really applies when they're metallically bonded - ergo, when there's multiple. So yes, it does apply to metals!

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RollWave_ t1_jd54l9p wrote

What if an astronaut comes across it in the vacuum of space?

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RonPaul42069 t1_jd5i3k0 wrote

So if a single atom moves so fast it loses its electron, it isn’t plasma?

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pants_pantsylvania t1_jd4y7me wrote

What if it was a single atom frozen in other types of matter? Couldn't you say it was frozen/moving slow in relation to the matter around it regardless of type?

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ashpens t1_jd5ze5j wrote

Sort of similar concept to cells and whether or not they constitute a tissue or an organism. If you're zoomed all the way in on one cell in isolation, it loses the relativity you would need to determine higher classifications.

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