B Warm air goes up....reason on a microscopic scale?

[jbriggs444]

For a single atom in a container of a given temperature, if we repeat this experiment many times, we would find a certain distribution of the energy of the atoms. I.e., our information about the atom will be described by a canonical density matrix with a certain temperature.

So you are now saying that a single atom at a single time does not have a temperature.

That is: You can have a temperature for a bunch of atoms at one time. Or a temperature for a single atom over a long time. But you can't have a temperature for a single atom at a single time.

I would further remark that if you define the temperature of an atom in terms of its long term average behavior in the context of a container then you do not need the atom. Just measure the temperature of the container.

 

[DrDu]

I am not saying that a single atom at a single time has no temperature. If my knowledge is so that it is in a canonical mixed state, the state is basically defined by its temperature and vice versa.

I would further remark that if you define the temperature of an atom in terms of its long term average behavior in the context of a container then you do not need the atom. Just measure the temperature of the container.
[\QUOTE]
That's basically the zeroth law of thermodynamics.

 

[jbriggs444]

I am not saying that a single atom at a single time has no temperature. If my knowledge is so that it is in a canonical mixed state, the state is basically defined by its temperature and vice versa.

What operational test can you propose to determine whether an atom at a particular time is or is not in a canonical mixed state?

That hypothetical atom with 200K worth of kinetic energy in a 100K container -- how do you tell what its temperature is? If you can propose no test at a point in time then I maintain that it has no temperature at a point in time.

 

[DrDu]

It is an ensemble property. It is the same as when I say that a dice has a probability of 1/6 to show a one before throwing it one time.

 

[jbriggs444]

It is an ensemble property. It is the same as when I say that a dice has a probability of 1/6 to show a one before throwing it one time.

So again, you are agreeing that a single molecule at a single time does not have a measurable temperature.

You have posed the definition of temperature so that the molecule only has a temperature if it is in thermodynamic equilibrium with its container. And you have clarified the definition so that it is a measure of a statistical distribution of states the molecule might find itself in. Under those conditions, the "temperature of a molecule" is not really an attribute of the molecule at all. It is, if it exists at all, an attribute of the container.

 

[DrDu]

It is good practice to speak of e.g. hot and cold neutrons. I really don't see a fundamental difference to the temperature concept of atoms and molecules.

 

[jbriggs444]

It is good practice to speak of e.g. hot and cold neutrons. I really don't see a fundamental difference to the temperature concept of atoms and molecules.

Let me try to argue from a different direction and possibly find some common ground.

Suppose that we stick a temperature probe into a small volume of gaseous nitrogen. We obtain a measured result of 100K. Strictly speaking, this is a statistical measurement. We only sampled from a small volume of nitrogen in a large container. So there is some statistical uncertainty in our result. Making up numbers, a result of 100K might be consistent with thermodynamic temperatures between 99.999K and 100.001K at a 90% confidence level.

Of course, there are mundane experimental uncertainties as well. Our instruments may only be good to the plus or minus 0.01K. In this situation we would be comfortable ignoring the statistical uncertainty associated with sampling error. It is down in the noise. One can talk about the sampled temperature as if it were identical to the actual temperature to within the mundane experimental uncertainty.

Now we attempt to probe the temperature by measuring the kinetic energy of a single molecule. We obtain a measured result that corresponds to a temperature of 200K. At the 90% confidence level, what is the range of temperatures that the container might have?

Eyeballing the Stephan Boltzman distribution, it looks like the 90% confidence interval will cover a range from approximately half the mean to approximately double the mean. Roughly speaking then, at the 90% confidence level, a sampled molecule at a kinetic energy of 200K is consistent with temperatures between 100K and 400K.

So I can agree that a single molecule [or neutron] does have a temperature. But that temperature is only measurable to within plus or minus a factor of two at the 90% confidence level.

 

[Rap]

I think it is helpful to look at how much a particle "comes down" between collisions (on average). The amount a particle "comes down" is (1/2)g t^2 where g is the acceleration due to gravity and t is the time between collisions. If t is greater for lower temperature, that would explain it.

t=L/v where L is the distance between molecules and v is the velocity of the particle. L is proportional to n^(1/3) where n is density. v is velocity of the particle, proportional to T^(1/2) where T is the temperature (because T is proportional to average energy which is (1/2)mv^2). Using the ideal gas law, P=nRT where P is pressure, R is the gas constant. Pressure is practically constant (pressure differences are equalized at the speed of sound and I'm assuming the column is not high enough to produce noticeable pressure differences) so we can say that n is proportional to 1/T. So t is proportional to n^(1/3)/T^(1/2) or 1/T^(5/6) and the amount the particles "come down" is proportional to 1/T^(5/3). The higher the temperature, the less they "come down" between collisions, and the low-temperature low-velocity particles wind up near the bottom.

I don't mean to imply a particle has a temperature, so please append "on average" to any statement that is deemed offensive.

 

[SirCurmudgeon]

Interesting (at least to me) wet air (air saturated with water vapor) is lighter that dry air.
Wet air goes up...

 

[NFuller]

Instead of a kinetic approach, I would try with a statistical approach. In general, a particle has a kinetic energy (due to motion) and a potential energy (due to gravitational force acting on the particle). In statistical physics there is an equipartition theorem, which, loosely speaking, says that energy likes to be distributed equally in all forms. So if particle has a lot of kinetic energy (corresponding to high temperature), then it also likes to have a lot of potential energy (and hence likes to go up). The only question is - why do particles like to distribute energy equally? That's because it maximizes entropy. When there is a lot of particles, then there is more phase space available for equidistributed energy than for all energy distributed in one form only.

I think this is a pretty good answer except that the equipartition theorem only applies at equilibrium. The difficulty in explaining why hot air rises at the microscopic level, is that this system is not at equilibrium. The temperature is not uniform through the gas and not all of the particles have the same energy. Eventually the system will reach equilibrium, at which point we can apply equilibrium statistical mechanics, but this after the warm gas has risen and come to the same temperature of the cool gas. It may be possible to develop a kinetic model for the motions of the warm and cool gas but I doubt it would be easy.

 

[sophiecentaur]

Interesting (at least to me) wet air (air saturated with water vapor) is lighter that dry air.
Wet air goes up...

Yes. The standard description and explanation really doesn't help. Air is described as a sort of SPONGE which, 'of course', would get heavier when full of water. But air is a mixture of gases. H₂O molecules are much lighter than N₂ and O₂ molecules so a volume of 'wet air' will contain more light molecules than a volume of 'dry air' (Same pressure and temperature) so it will float in air.

 

[Brendan Graham]

It's satisfactory.

What happens to the gas which is cool? It needs less room. That's why it comes down?

Assuming the base of the enclosing vessel is fixed, both hot and cool gas molecules under an external gravitational field have a fixed floor beyond which they cannot travel. Rather than coming down, more appropriately, molecules of a cooler gas just do not reach up (against the force of gravity) as far as those that are hotter therefore the probability distribution of finding cooler gas molecules near the floor is greater.

 

[Demystifier]

I think this is a pretty good answer except that the equipartition theorem only applies at equilibrium. The difficulty in explaining why hot air rises at the microscopic level, is that this system is not at equilibrium. The temperature is not uniform through the gas and not all of the particles have the same energy. Eventually the system will reach equilibrium, at which point we can apply equilibrium statistical mechanics, but this after the warm gas has risen and come to the same temperature of the cool gas. It may be possible to develop a kinetic model for the motions of the warm and cool gas but I doubt it would be easy.

How about my post #40, where I further elaborated my explanation? You are of course right that equipartition theorem is only valid in the equilibrium, but my explanation is based on the idea that the system evolves towards the equilibrium. In other words, the hot gas goes up because that makes the system closer to the equilibrium.

 

[Ilythiiri]

In passing, I would just love to have some gadget capable of displaying such a rising air current!

http://www.instructables.com/id/Schlieren-Imaging-How-to-see-air-flow/

Here you go (:

 

[Dr. Courtney]

I'm not sure buoyancy has a microscopic explanation.

 

[Zaya Bell]

You need to first understand that gravitation acts on molecules, and they have pressure. For hotter object, the spaces between them increases, thus reducing pressure, while the cold molecules' spaces decreases and pressure increases. Thus if the cold molecules were up they would apply more pressure downward and displace the hotter molecules upward. It's analogous to putting your hand into a water container. The higher pressure (F/A) of your hand displaces the lower pressure of the water molecules and your hand now occupies the space the water did.
Got it?

 

[jbriggs444]

gravitation acts on molecules, and they have pressure.

Individual molecules do not have temperature or pressure. Groups of molecules have temperature and pressure. If one invokes the emergent properties of temperature and pressure to describe convection, one has abandoned the microscopic explanation and gone for a macroscopic explanation.

Not that there is anything wrong with that.

 

[Zaya Bell]

Individual molecules do not have temperature or pressure. Groups of molecules have temperature and pressure. If one invokes the emergent properties of temperature and pressure to describe convection, one has abandoned the microscopic explanation and gone for a macroscopic explanation.

Not that there is anything wrong with that.

They sure do. They are not point particles which means they have area. Thus P=mg/A. Even a photon has pressure.

 

[jbriggs444]

They sure do. They are not point particles which means they have area. Thus P=mg/A. Even a photon has pressure.

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

 

[Zaya Bell]

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

I don't think you know what you're saying. Check this up anyway.
https://www.photonworld.de/nc/en/magazine/featurearticle/the-power-of-photons/fseite/1.html

 

[jbriggs444]

I don't think you know what you're saying.

I am saying that a single molecule does not have a pressure. A bunch of molecules do. A single photon does not have a pressure. A bunch of photons do.

Can you calculate the pressure which you believe is associated with a single molecule of nitrogen at 500 meters/second?

Edit: Note that your calculation will at least implicitly use a reference frame against which that 500 meters/second is determined. This means that your calculated pressure will be a property of a molecule of nitrogen in the context of a reference frame, not an intrinsic property of the molecule.

 

[NFuller]

my explanation is based on the idea that the system evolves towards the equilibrium. In other words, the hot gas goes up because that makes the system closer to the equilibrium.

I think your explanation is appropriate for a 'B' level thread, like this one. I was just pointing out that its difficult to justify the behavior of a dynamical system using results from equilibrium statistical mechanics. In your post #40, it looks like you are saying that the convection of the warm gas is driven by an increase in entropy. However, entropy is a rather useless variable for a non-equilibrium system and we aren't even guaranteed that the entropy is increasing as the gas flows upward. The only thing we do know is that when the gasses finally come to equilibrium, that will be a maximum entropy state.

 

[Zaya Bell]

I am saying that a single molecule does not have a pressure. A bunch of molecules do. A single photon does not have a pressure. A bunch of photons do.

Can you calculate the pressure which you believe is associated with a single molecule of nitrogen at 500 meters/second?

Edit: Note that your calculation will at least implicitly use a reference frame against which that 500 meters/second is determined. This means that your calculated pressure will be a property of a molecule of nitrogen in the context of a reference frame, not an intrinsic property of the molecule.

One molecule does have pressure. It's pressure is different from the macroscopic pressure of the gas which is the speed at which they strike the container and the area of movement. But rather, classically, as a result of their mass and acceleration. As much as one photon has energy and momentum it sure can exert pressure. If I shoot an electron, it also exert pressure on whatsoever it collides with. Generally, all particle exert pressure.

 

[jbriggs444]

One molecule does have pressure. It's pressure is different from the macroscopic pressure of the gas which is the speed at which they strike the container and the area of movement. But rather, classically, as a result of their mass and acceleration. As much as one photon has energy and momentum it sure can exert pressure. If I shoot an electron, it also exert pressure on whatsoever it collides with. Generally, all particle exert pressure.

That is not pressure. That is impulse. It is not an attribute of the molecule. It is an attribute of the interaction.

 

[hasahinivo]

Think of molecules as balls. Some are heavier, and some are lighter.

When 2 molecules collide, they interchange momentum and energy. Because energy and momentum is conserved, both molecules get similar amounts of momentum. I didn't said "the same momentum", because it depends on the relative speeds, but "similar momentum" is good enough to understand it. Here is a better explanation.

Because the lighter ball has less mass, it needs more speed for the same momentum, so the lighter ball gets more speed out of the collision (it also needs a lot more energy, because energy scales squared with speed).
Here are some illustrative video:





So, the lighter molecules run much faster after thermal equilibrium is achieved. Faster in all directions. If they were in zero gravity, the lighter molecules would escape faster in all directions, at the speed of sound.

In non zero gravity, molecules are bounded in all directions, except up, so they move up faster than the heavier molecules.
Going up requires more energy, because is necessary to fight against gravity, so not all molecules with the same weight go up. Only the ones with enough energy go up.

Because the lighter gas has more energy per molecule, it also has more molecules up.

 

[NFuller]

One molecule does have pressure. It's pressure is different from the macroscopic pressure of the gas which is the speed at which they strike the container and the area of movement. But rather, classically, as a result of their mass and acceleration. As much as one photon has energy and momentum it sure can exert pressure. If I shoot an electron, it also exert pressure on whatsoever it collides with. Generally, all particle exert pressure.

Pressure is an average macroscopic quantity of an entire system. I think you are envisioning a box with a single particle in it and saying that the walls of the box will experience some average pressure over long time scales as the particle bounces around inside. What @jbriggs444 is telling you is that the particle itself does not have pressure, rather only the walls of the box experience pressure.

Pressure is a property of the system not the particles alone. This is easy to see if you shrink the size of the box while keeping the temperature constant. The pressure will increase on the walls even though the molecules all still have the same kinetic energy.

 

[Demystifier]

However, entropy is a rather useless variable for a non-equilibrium system

I strongly disagree. Entropy is only useful when it changes, and it changes only in non-equilibrium.

 

[Brendan Graham]

Does that not beg the question? What do you mean by thermodynamic equilibrium for a single molecule within a container?

If one has a container at 100K and a molecule with a kinetic energy equivalent to 200K, is that molecule in thermodynamic equilibrium with the container? Is there enough information to tell?

 

[NFuller]

I strongly disagree. Entropy is only useful when it changes, and it changes only in non-equilibrium.

Entropy is a state variable only for systems at equilibrium. It is difficult to even define entropy for a dynamical system. Saying entropy is only useful for non-equilibrium systems strikes me as naive, since most texts on statistical mechanics and thermodynamics use entropy only when describing equilibrium states.

There have been attempts to define and use entropy for non-equilibrium systems, which have produced mixed results. Here's a short discussion on the subject https://arxiv.org/pdf/1305.3912.pdf.

 

[Demystifier]

Entropy is a state variable only for systems at equilibrium. It is difficult to even define entropy for a dynamical system. Saying entropy is only useful for non-equilibrium systems strikes me as naive, since most texts on statistical mechanics and thermodynamics use entropy only when describing equilibrium states.

Nonsense. Every thorough statistical mechanics textbook introduces Boltzmann and Gibbs entropy in some of the initial chapters, before even mentioning the concept of equilibrium.

Even you in your insight article https://www.physicsforums.com/insights/statistical-mechanics-part-equilibrium-systems/ have introduced Gibbs entropy before mentioning equilibrium and canonical ensemble.

My coworkers and me have been calculating entropy far from equilibrium in a dynamical system in this work: https://arxiv.org/abs/1011.4173v5

There have been attempts to define and use entropy for non-equilibrium systems, which have produced mixed results. Here's a short discussion on the subject https://arxiv.org/pdf/1305.3912.pdf.

In this paper authors require entropy to satisfy some additional properties. The standard Boltzmann and Gibbs entropy (which turn out to be equal only in equilibrium) do not need to satisfy all these properties.

 

[NFuller]

Every thorough statistical mechanics textbook introduces Boltzmann and Gibbs entropy in some of the initial chapters, before even mentioning the concept of equilibrium.

Even you in your insight article https://www.physicsforums.com/insights/statistical-mechanics-part-equilibrium-systems/ have introduced Gibbs entropy before mentioning equilibrium and canonical ensemble.

Right, but I can't use it to solve for the state of a system until I impose the equilibrium requirement. Even in Zwanzig's book on non-equilibrium systems, I don't think the word entropy even appears in the book; it's not shown in the glossary/appendix.

In this paper authors require entropy to satisfy some additional properties. The standard Boltzmann and Gibbs entropy (which turn out to be equal only in equilibrium) do not need to satisfy all these properties.

I'm not sure about this. I think if the entropy doesn't satisfy these properties, it isn't useful for determining the state of a system. Anyways, I will read through your paper to get a better understanding of your view. I think we are deviating from the intention of this thread.

 

[Demystifier]

Right, but I can't use it to solve for the state of a system until I impose the equilibrium requirement. Even in Zwanzig's book on non-equilibrium systems, I don't think the word entropy even appears in the book; it's not shown in the glossary/appendix.

I'm not sure about this. I think if the entropy doesn't satisfy these properties, it isn't useful for determining the state of a system. Anyways, I will read through your paper to get a better understanding of your view. I think we are deviating from the intention of this thread.

I agree that out of equilibrium, entropy is less useful to describe the state. Nevertheless it can be defined any may be very useful for studying the second law.