Paradox: Thermodynamic equilibrium does not exist in gravitational fields

[Petr Matas]

TL;DR Summary

A gas in a graviational field tends to an adiabatic temperature gradient, which leads to an indefinite heat flow in an isolated system.

I've come across a paradox I can't resolve.

Let's have an isolated system: A gas in a box in a homogeneous gravitational field. When thermodynamic equilibrium is reached, the gas should have an adiabatic temperature gradient (temperature decreases with increasing height). The walls are in thermal contact with the adjacent gas layers, so the temperatures of the upper and lower walls should be different. The upper and lower walls are black on the inside, so they are in thermal contact mediated by radiation (the gas is transparent). Their temperatures should therefore be equal, which conflicts the previous result. So it appears that no state of thermodynamic equilibrium exists. Significant macroscopic energy flows occur spontaneously in the system indefinitely, which shouldn't be possible.

I think that in a state of thermodynamic equilibrium, the parts of the system must be in equilibrium internally and with each other.

I considered the effect of gravitational red shift, but it doesn't seem to resolve the paradox, because it may be 10 orders of magnitude weaker than the adiabatic gradient.

I also tried to resolve the paradox using ChatGPT (in Czech), which concluded that the system only reaches a quasi-stationary state because it takes too long to reach equilibrium. However, I don't think this resolves the paradox either, because the paradox consists in the conclusion that no state of thermodynamic equilibrium exists whatsoever. However, an isolated system should have such a state, shouldn't it?

 

[Frabjous]

Why are you assuming adiabatic?

 

[anuttarasammyak]

TL;DR Summary: A gas in a graviational field tends to an adiabatic temperature gradient, which leads to an indefinite heat flow in an isolated system.

When thermodynamic equilibrium is reached, the gas should have an adiabatic temperature gradient (temperature decreases with increasing height

In equilibrium temperature is constant through the system. If not heat transfer from high to low. It is not equilibrium.

 

[Petr Matas]

Why are you assuming adiabatic?

Gas in a graviational field in thermodynamic equilibrium is cooler at the top and hotter at the bottom. This effect is called adiabatic temperature gradient. It is also present in the Earth's atmosphere, where the temperature decreases with increasing height, unless disturbed by external energy flows.

 

[Vanadium 50]

It's always "It's a paradox!", never "something I don't understand."

Arguing with ChatGPT by proxy is not something we do here.

 

[Petr Matas]

In equilibrium temperature is constant through the system.

This may not apply when a gravitational field is present. Let's assume the density of the gas is low, so particle collisions are rare. When a particle flies from the top to the bottom (with no collisions, for simplicity), its graviational potential energy is converted to its kinetic energy. Therefore the kinetic energy of the particles at the bottom is higher than at the top, which leads to the temperature gradient.

 

[Petr Matas]

It's always "It's a paradox!", never "something I don't understand."

Yes, a paradox is an unexpected or erroneous result, which demonstrates our lack of understanding.

 

[Frabjous]

Gas in a graviational field in thermodynamic equilibrium is cooler at the top and hotter at the bottom. This effect is called adiabatic temperature gradient. It is also present in the Earth's atmosphere, where the temperature decreases with increasing height, unless disturbed by external energy flows.

Please provide a reference for this. Earth’s atmosphere is not in thermodynamic equilibrium.

 

[Petr Matas]

Please provide a reference for this. Earth’s atmosphere is not in thermodynamic equilibrium.

Ok, Wikipedia proves you right:
Lapse rate in an isolated column of gas

It also seems that the effect is better known as adiabatic lapse rate in English.

But why does the lapse rate vanish in equilibrium? Why doesn't it form according to the following thought experiment?

Let's assume the density of the gas is low, so particle collisions are rare. When a particle flies from the top to the bottom (with no collisions, for simplicity), its graviational potential energy is converted to its kinetic energy. Therefore the kinetic energy of the particles at the bottom is higher than at the top, which leads to the temperature gradient.

 

[Frabjous]

Ok, Wikipedia proves you right:
Lapse rate in an isolated column of gas

It also seems that the effect is better known as adiabatic lapse rate in English.

But why does the lapse rate vanish in equilibrium? Why doesn't it form according to the following thought experiment?

In equilibrium, there is no convection.

If there are no collisions, you do not have interacting particles which means you do not have a well-defined pressure or temperature. Hard to call this a gas. In real life, we expect density to be higher at the surface. In your experiment, density will be higher at the top since the particles will be moving slower and spend more time in that region.

 

[Chestermiller]

Suppose I have an insulated vertical cylinder of length L and cross sectional area containing an ideal gas at uniform temperature T and pressure $P_0$ at its base. The gas is at "hydrostatic equilibrium." That means that $$\frac{dP}{dz}=-\rho g =- \frac{PM}{RT}g$$where M I the molecular weight. The solution to this equation, subject to the pressure condition at the base is $$P=P_oe^{-\frac{Mgz}{RT}}$$The weight of the gas is then $$W=P_oA(1-e^{-\frac{MgL}{RT}})$$So, if we know the weight of the gas, we can determine the pressure at the base. In the gravitational field, even though the local density varies with z, we have local thermodynamic equilibrium at each elevation. How is this not thermodynamic equilibrium? Certainly there are no convection currents.

 

[russ_watters]

Ok, Wikipedia proves you right:
Lapse rate in an isolated column of gas

It also seems that the effect is better known as adiabatic lapse rate in English.

But why does the lapse rate vanish in equilibrium? Why doesn't it form according to the following thought experiment?

The lapse rate exists in the atmosphere because it is not in thermodynamic equilibrium. The lapse rate does not exist in your original thought experiment because it is in thermodynamic equilibrium.

 

[Petr Matas]

Now I believe that in thermodynamic equilibrium the temperature profile is isothermic, but I don't think that such conclusion comes from your arguments. I think that any temperature profile may be in hydrostatic equilibrium, although some of them would have Rayleigh–Taylor instabilities.

The fact that in thermodynamic equilibrium the temperature does not change with height is somewhat surprising given that the kinetic energy of a particle changes as it moves vertically. I think this fact has to be proven using statistical mechanics.

 

[Chestermiller]

Now I believe that in thermodynamic equilibrium the temperature profile is isothermic, but I don't think that such conclusion comes from your arguments. I think that any temperature profile may be in hydrostatic equilibrium, although some of them would have Rayleigh–Taylor instabilities.

The fact that in thermodynamic equilibrium the temperature does not change with height is somewhat surprising given that the kinetic energy of a particle changes as it moves vertically. I think this fact has to be proven using statistical mechanics.

I assert that any disturbance to the equilibrium will be damped out without any instabilities.

 

[Petr Matas]

I assert that any disturbance to the equilibrium will be damped out without any instabilities.

It is a bit off-topic, but here is a counterexample: Let the gas temperature change sharply at certain height with cold gas at the top and hot at the bottom. Although it is initially in hydrostatic equilibrium, this configuration is unstable and convection will start to form soon – hot gas will rise and cold gas will descend.

 

[russ_watters]

The fact that in thermodynamic equilibrium the temperature does not change with height is somewhat surprising given that the kinetic energy of a particle changes as it moves vertically.

Particles don't move vertically in the aggregate if in equilibrium. On average if one particle is moving up another must be moving down.

It's really weird that your initial thought was shown to be right and now you're arguing against that. Why can't you just accept the win?

 

[Chestermiller]

It is a bit off-topic, but here is a counterexample: Let the gas temperature change sharply at certain height with cold gas at the top and hot at the bottom. Although it is initially in hydrostatic equilibrium, this configuration is unstable and convection will start to form soon – hot gas will rise and cold gas will descend.

The system will re-equilibrate at the average temperature.

 

[Petr Matas]

Particles don't move vertically in the aggregate if in equilibrium. On average if one particle is moving up another must be moving down.

I agree, one particle is moving up and another is moving down. However, the one moving up is losing kinetic energy (due to conversion to potential energy), the one moving down is gaining kinetic energy. This leads naive intuition to the incorrect conclusion that particles lower in the column have higher kinetic energy. I only recognized this mistake after the following post:

Please provide a reference for this.

It's really weird that your initial thought was shown to be right and now you're arguing against that. Why can't you just accept the win?

I am not sure which thought you are referring to. I just want to know where the naive intuition went wrong. I believe that a proof using statistical mechanics can enlighten it.

 

[Dale]

I just want to know where the naive intuition went wrong.

I think it was probably in assuming a non-interacting gas. But this is not my area of expertise

 

[anuttarasammyak]

I just want to know where the naive intuition went wrong.

You may be able to investigate the idea that only kinetic energy matters. For an example, in ensemble of harmonic oscillators, does each oscillator is heated when it coming back to the center and cooled down when it goes far ?

 

[Petr Matas]

You may be able to investigate the idea that only kinetic energy matters. For an example, in ensemble of harmonic oscillators, does each oscillator is heated when it coming back to the center and cooled down when it goes far ?

Isn't the local temperature of a still gas given by the average kinetic energy of particles nearby?
I would say that temperature is defined for an ensemble, not for individual particle or oscillator. In equilibrium, each oscillator has different total energy. Those with higher total energy spend less time near the center and therefore their higher kinetic energy during their motion near the center contributes less to the average kinetic energy in that region. On the other hand, those with lower total energy never appear far from the center and therefore they do not contribute in that region. This idea arose from the following statement:

In your experiment, [...] at the top [...] the particles will be moving slower and spend more time in that region.

 

[anuttarasammyak]

Isn't the local temperature of a still gas given by the average kinetic energy of particles nearby?

Local temperature is a concept in non equilibrium systems.here we discuss equilibrium state.

 

[Petr Matas]

Local temperature is a concept in non equilibrium systems.here we discuss equilibrium state.

I think it is well defined in the equilibrium as well. My question is: How do we prove that its value is the same everywhere?

 

[Frabjous]

You are approaching this wrong. Coming up with ideas that need to be refuted is a never ending process.

Isn't the local temperature of a still gas given by the average kinetic energy of particles nearby?
I would say that temperature is defined for an ensemble, not for individual particle or oscillator. In equilibrium, each oscillator has different total energy. Those with higher total energy spend less time near the center and therefore their higher kinetic energy during their motion near the center contributes less to the average kinetic energy in that region. On the other hand, those with lower total energy never appear far from the center and therefore they do not contribute in that region. This idea arose from the following statement:

This is beginning to approach a word salad. If you are feeding our answers into an AI, you need to stop. It is not helping you.

You mechanistic explanations have so far ignored a force in the problem, namely the pressure.

 

[Petr Matas]

If you are feeding our answers into an AI, you need to stop. It is not helping you.

I am not. I used AI before the initial post in an attempt to resolve the paradox, which was not successful. That's why I came here. You managed to convince me that even in gravitation, the equilibrium temperature profile is isothermic, not adiabatic, which resolved the paradox. However, it is somewhat counterintuitive. That's why I am asking for a proof.
And yes, it is difficult to teach me, because I want to be convinced.

Currently I am trying the following approach using statistical mechanics and laws of motion. Let us describe the state of the gas column using density $f(z, v_z)$ of particles at height $z$ with vertical velocity $v_z$. Assuming thermodynamic equilibrium, find the partial differential equations that $f(z, v_z)$ must satisfy.

You mechanistic explanations have so far ignored a force in the problem, namely the pressure.

The pressure can be obtained from $f$.

 

[anuttarasammyak]

Ok, Wikipedia proves you right:
Lapse rate in an isolated column of gas

It is a fun for me to find in the reference that Maxwell explained clearly in his Heat Theory p.330

 

[anuttarasammyak]

Currently I am trying the following approach using statistical mechanics and laws of motion.

For your reference I got partition function of
$$Z(\beta)=\hbar^{-3N}((\frac{2m\pi}{\beta})^{\frac{3}{2}} \pi r^2 \frac{1-e^{-\beta mgh}}{\beta mg})^N$$
where r is radius and h is height of column. With no gravity as usual
$$Z(\beta)=\hbar^{-3N}((\frac{2m\pi}{\beta})^{\frac{3}{2}} \pi r^2 h)^N$$

 

[Petr Matas]

It is a fun for me to find in the reference that Maxwell explained clearly in his Heat Theory p.330

Yes, I came across this result in the Wikipedia article. It is the same argument as the one from the initial post:

The upper and lower walls are black on the inside, so they are in thermal contact mediated by radiation (the gas is transparent). Their temperatures should therefore be equal,

I would like to reach the same result using statistical mechanics.

 

[Dale]

Isn't the local temperature of a still gas given by the average kinetic energy of particles nearby?

Not exactly. The temperature is given by the average energy per internal degree of freedom in thermal equilibrium. Thermal equilibrium means that the energy is evenly distributed amongst those internal degrees of freedom. In the case of a non-interacting gas it is only the vertical velocity that has the gradient.

Because the energy is not evenly distributed amongst the available degrees of freedom, I don’t think that the gas can be said to be in thermal equilibrium. Therefore it does not have a well defined temperature.

On the other hand, if you allow interactions then there will not be a velocity gradient. And the gas will have a thermal equilibrium and thus a well defined temperature.

It seems to me like your whole line of reasoning requires an assumption that prevents the gas from having a defined temperature and then claiming that it has a temperature gradient. But again, this is not my area of expertise.

 

[Chestermiller]

Even in an ideal gas, the molecules collide with each other to exchange energy. What is the mean free path of an oxygen molecule in a gas that is at room temperature and a pressure of 1 bar?

 

[Petr Matas]

ChatGPT says 71 nm

 

[Chestermiller]

ChatGPT says 71 nm

In other words, each molecule experiences a multitude of collisions and energy transfers per unit time, which translates into significant heat conduction within the gas.

 

[Petr Matas]

In other words, each molecule experiences a multitude of collisions and energy transfers per unit time, which translates into significant heat conduction within the gas.

I agree, but I believe that my approach is independent of the assumption about collisions.

 

[Dale]

I agree, but I believe that my approach is independent of the assumption about collisions.

I certainly don’t think so. How does your objection even exist if the gas molecules collide with each other?

 

[Petr Matas]

I certainly don’t think so. How does your objection even exist if the gas molecules collide with each other?

I think that in the equilibrium, the velocity distribution must be the same whether collisions among the particles occur or not. Let's assume there is a small solid object somewhere in the gas column. Every particle will hit this object from time to time. Even if the particles don't interact with each other, collisions with this object will mediate energy transfer among the particles. I think this allows me to ignore the collisions among the particles in the analysis of the equilibrium state.