Paradox: Thermodynamic equilibrium does not exist in gravitational field

[Dale]

I think that in the equilibrium, the velocity distribution must be the same whether collisions among the particles occur or not.

With the collisions the distribution of the velocity is uniform in direction and Maxwell-Boltzmann in speed. Can you show that is the case without collisions? I am highly skeptical. I think this is the key. Without this proof you do not have thermal equilibrium and therefore don’t have a well defined temperature in the first place.

 

[Petr Matas]

With the collisions the distribution of the velocity is uniform in direction and Maxwell-Boltzmann in speed. Can you show that is the case without collisions? I am highly skeptical. I think this is the key. Without this proof you do not have thermal equilibrium and therefore don’t have a well defined temperature in the first place.

How about the energy transfer mediated by the small solid object that I mentioned? Isn't that enough to allow reaching the equilibrium (after very long time, of course)?

 

[Dale]

How about energy transfer mediated by the small solid object that I mentioned? Isn't that enough to allow reaching the equilibrium (after very long time, of course)?

Not that I can see. Can you derive the distribution from that? I don't think that I could.

Also, if you can derive the usual distribution, doesn't that invalidate your whole argument? Isn't your argument predicated on the distribution being different from the standard thermal equilibrium? I don't see how you can have it both ways.

 

[Chestermiller]

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

Have you had a course yet in fluid mechanics or Transport Processes?

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.

unless there is no object present.

 

[Petr Matas]

Have you had a course yet in fluid mechanics or Transport Processes?

Unfortunately not. However, I am going to assume the gas is still.

unless there is no object present.

I am going to assume a usual thermal distribution at certain altitude and from there I expect to conclude that it is such everywhere.

Isn't your argument predicated on the distribution being different from the standard thermal equilibrium?

No, the standard thermal distribution is the result that I am expecting to obtain.

 

[Chestermiller]

Unfortunately not. However, I am going to assume the gas is still.


I am going to assume a usual thermal distribution at certain altitude and from there I expect to conclude that it is such everywhere.

Like I said, using continuum fluid mechanics and heat transfer would be much simpler.

 

[Vanadium 50]

ChatGPT says 7

Tell us again how you're not using an AI?

 

[Dale]

No, the standard thermal distribution is the result that I am expecting to obtain

I don’t see how. Can you show it?

I also don’t see the point of adding the little block. What does it do that the walls don’t already do besides make the geometry more complicated?

 

[Petr Matas]

Tell us again how you're not using an AI?

This was the second time . I prefer leaving tedious but simple work to AI (while checking its answers of course) to save time for tasks which are currently beyond the AI capabilities. I mark the AI answers clearly whenever I use them. Or would you prefer me to conceal the use of AI? Or not use AI at all? What is the point?

 

[Petr Matas]

I don’t see how. Can you show it?

I must shut up and calculate.

I also don’t see the point of adding the little block. What does it do that the walls don’t already do besides make the geometry more complicated?

Walls suffice.

 

[Petr Matas]

Like I said, using continuum fluid mechanics and heat transfer would be much simpler.

You mean using the law that heat flows from higher to lower temperature? I want to avoid using this law. Actually, I want to prove (using laws of motion) that it is valid even in gas in gravitation without assuming its validity in the first place.

 

[Chestermiller]

You mean using the law that heat flows from higher to lower temperature? I want to avoid using this law. Actually, I want to prove (using laws of motion) that it is valid even in gas in gravitation without assuming its validity in the first place.

Good luck.

 

[anuttarasammyak]

With the collisions the distribution of the velocity is uniform in direction and Maxwell-Boltzmann in speed. Can you show that is the case without collisions? I am highly skeptical. I think this is the key. Without this proof you do not have thermal equilibrium and therefore don’t have a well defined temperature in the first place.

BTW
1 . When the particles are photons, they do not interact each other but only with walls.
2 . I revisited Laudau Statistical Mechanics chapter II section 27 for relativistic region (27.5)
$$T=constant \ (1-\phi/c^2)$$
The temperature is higher at points in the body where $|\phi|$ is greater. So the things seem to change in relativity.

 

[Petr Matas]

Theorem:
A gas column in thermodynamic equilibrium in a classical homogeneous gravitational field has the same temperature everywhere.

Proof using the laws of motion:

Let at time 0 a point particle with mass $m$ is located at height 0 and its velocity is $\mathbf{v}_0 = (v_{\text x 0}, v_{\text y 0}, v_{\text z 0})$. Its total energy is
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z 0}^2). $$
Let us assume the particle moves without collisions. At time $t$ it is at height $z$, its velocity is $\mathbf{v} = (v_{\text x 0}, v_{\text y 0}, v_{\text z})$ and it has the same total energy
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z}^2) + mgz,
$$ where $g$ is the gravitational acceleration.

Comparison of the two equations yields
$$
v_{\text z 0}^2 = v_{\text z}^2 + 2gz, \tag 1
$$ $$
v_{\text z 0} = \pm \sqrt{v_{\text z}^2 + 2gz}. \tag 2
$$
Differentiation of equation $(1)$ yields
$$ v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}. \tag 3 $$
Let $\rho(z, v_\text z)$ is the density of particles at height $z$ with vertical velocity $v_\text z$.

Consider at time 0 a bunch of particles at height 0 with velocities in range from $v_{\text z 0}$ to $(v_{\text z 0} + dv_{\text z 0})$; they contribute to $\rho(0, v_{\text z 0})$. The number of such particles, which cross a unit-area horizontal boundary at this height in unit time, is
$$ dN = \rho(0, v_{\text z 0}) v_{\text z 0} dv_{\text z 0}. \tag 4$$
At time $t$, the same particles are at height $z$, their velocities range from $v_{\text z}$ to $(v_{\text z} + dv_{\text z})$ and they contribute to $\rho(z, v_{\text z})$. The same number of particles will cross a unit-area horizontal boundary at this height.
$$ dN = \rho(z, v_{\text z}) v_{\text z} dv_{\text z}. \tag 5$$
Equations $(2)$, $(3)$, $(4)$, $(5)$ together yield
$$ \rho(z, v_{\text z}) = \rho\left(0, \pm \sqrt{v_{\text z}^2 + 2gz}\right). \tag 6$$
Let the velocity distribution at height 0 is Maxwell–Boltzmann (i.e. thermal) with temperature $T$. That means the density at that height is
$$ \rho(0, v_{\text z}) = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right), \tag 7
$$ where $C$ is a constant.

From equations $(6)$ and $(7)$ we get
$$ \begin{align}
\rho(z, v_{\text z}) \nonumber
& = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2 + 2gz}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot \rho(0, v_{\text z}). \nonumber
\end{align} $$
We can see that the velocity distribution is the same at all heights $z$ (i.e. the temperature is the same everywhere) and the density decreases exponentially with height $z$. Interaction between the particles cannot change this distribution, because they are already in equilibrium.

End of proof.

This result may be generalized to non-homogeneous classical gravitational fields by replacing $V(z) = gz$ with a generic gravitational potential $V(x, y, z)$.

 

[anuttarasammyak]

Differentiation of equation (1) yields
(3)vz0dvz0=vzdvz.
Let ρ(z,vz) is the density of particles at height z with vertical velocity vz.

By differentiation of (1) I got (3)
$$v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}+gdz. \tag 3$$
Am I wrong ?

 

[Petr Matas]

By differentiation of (1) I got (3)
$$v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}+gdz. \tag 3$$
Am I wrong ?

At height 0, let us have a narrow velocity interval from $v_{\text z 0}$ to $(v_{\text z 0} + dv_{\text z 0})$. The goal is to find what velocity interval at height $z$ it is mapped to. Therefore $z$ is a constant. Any ideas how to express it more clearly?

 

[anuttarasammyak]

Thanks. So have you conquered your losing-speed intuition by your calculation which tells same velocity distribution but different number density by height ?

 

[Petr Matas]

Thanks. So have you conquered your losing-speed intuition by your calculation which tells same velocity distribution but different number density by height ?

Exactly.

 

[anuttarasammyak]

Congratulations. Let me understand more. Say a bouncing ball goes up and down between floor and ceiling. By gravity

ball speed near floor > ball speed near ceiling

time duration ball staying in 10cm layer from floor < time duration ball staying in 10cm layer from ceiling

ball number desity in 10cm layer from floor < ball number density in 10cm layer from ceiling

This contradicts the result. Where am I wrong ?

 

[Petr Matas]

This contradicts the result. Where am I wrong ?

There are many balls.

 

[anuttarasammyak]

Each ball has high probability to be near ceiling and the sum of them, though there is no interaction between, show high number density near floor. Very interesting to me.

 

[Petr Matas]

Another observation: Trajectories of individual particles are parabolas. A particle never appears above the top of its parabola.

 

[Petr Matas]

Another observation 2: As a bunch of particles rises and loses speed, they come closer together in space, but farther apart in velocity. These opposite effects on density cancel out, which can be seen from equation (6): It is an equality between densities of the same bunch at different heights; all multiplicative factors disappeared.

 

[anuttarasammyak]

Each ball has high probability to be near ceiling and the sum of them, though there is no interaction between, show high number density near floor. Very interesting to me.

Statistical/thermal mechanics explains that maximum entropy state of vertical column gas has throughout same temperature as Maxwell said. Newton Mechanics of bouncing balls seem to have different nature to me where energy conservation is applied individually to the balls not to the whole system and with no concept of entropy.
Anyway it is your thread not mine. If you have no ambiguity I should shut up.

 

[Petr Matas]

Statistical/thermal mechanics explains that maximum entropy state of vertical column gas has throughout same temperature as Maxwell said. Newton Mechanics of bouncing balls seem to have different nature to me where energy conservation is applied individually to the balls not to the whole system and with no concept of entropy.
Anyway it is your thread not mine. If you have no ambiguity I should shut up.

Analysis using Newton mechanics may be too difficult to apply on large ensembles, but it must give the same results as thermal laws. In our case its application was rather easy and it showed me where exactly my intuition went wrong. This allows me to attain deeper understanding.

 

[Philip Koeck]

Let $\rho(z, v_\text z)$ is the density of particles at height $z$ with vertical velocity $v_\text z$.

Consider at time 0 a bunch of particles at height 0 with velocities in range from $v_{\text z 0}$ to $(v_{\text z 0} + dv_{\text z 0})$; they contribute to $\rho(0, v_{\text z 0})$. The number of such particles, which cross a unit-area horizontal boundary at this height in unit time, is
$$ dN = \rho(0, v_{\text z 0}) v_{\text z 0} dv_{\text z 0}. \tag 4$$
At time $t$, the same particles are at height $z$, their velocities range from $v_{\text z}$ to $(v_{\text z} + dv_{\text z})$ and they contribute to $\rho(z, v_{\text z})$. The same number of particles will cross a unit-area horizontal boundary at this height.
$$ dN = \rho(z, v_{\text z}) v_{\text z} dv_{\text z}. \tag 5$$

Could you define dN and ρ? What are the units?

 

[Petr Matas]

Could you define dN and ρ? What are the units?

$\rho(z, v_{\rm z}) = \frac{d}{dV} \frac{d}{dv_{\rm z}} N_1$ is the density of particles at height $z$ with vertical velocity $v_{\rm z}$. Unit: 1/(m³•ms^–1)

$N_1$ is the number of particles. Unit: 1

$V$ is the volume. Unit: m³

$v_{\rm z}$ is the vertical velocity. Unit: ms^–1

$dN = \rho(z, v_{\text z}) v_{\text z} dv_{\text z}$ is the number of particles with vertical velocity in range from $v_{\text z}$ to $(v_{\text z} + dv_{\text z})$, which cross a unit-area horizontal boundary at height $z$ in unit time. Unit: 1/(m²•s)

 

[Philip Koeck]

Let us assume the particle moves without collisions. At time $t$ it is at height $z$, its velocity is $\mathbf{v} = (v_{\text x 0}, v_{\text y 0}, v_{\text z})$ and it has the same total energy
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z}^2) + mgz,
$$ where $g$ is the gravitational acceleration.
......
We can see that the velocity distribution is the same at all heights $z$ (i.e. the temperature is the same everywhere) and the density decreases exponentially with height $z$.

You assume that the particles don't collide. This means that each individual particle loses kinetic energy when it gains potential energy. So your result that the velocity distribution is the same at all heights contradicts your assumption, doesn't it?

Then I'm also wondering about the connection you make between temperature and velocity distribution.
There was a discussion about this on PF and I believe the general agreement was that T is proportional to the average total energy and not to the average kinetic energy.
So the same velocity distribution wouldn't imply the same temperature anyway.
I have to check that thread, though.

Edit: Here's the result of checking: Look at post 77 and the discussion leading to it in this thread:
https://www.physicsforums.com/threads/interpretation-of-temperature-in-liquids-solids.1050908/page-3

 

[Petr Matas]

You assume that the particles don't collide. This means that each individual particle loses kinetic energy when it gains potential energy.

Exactly. Therefore intuition says that average kinetic energy is lower at higher altitudes. However, it is only intuition, so I don't use it as an assumption in the formal analysis using laws of motion.

So your result that the velocity distribution is the same at all heights contradicts your assumption, doesn't it?

I didn't assume that average kinetic energy changes with height. It just shows that the intuition was wrong. And it's why I call it a paradox, i.e. an observation that the correct result is counterintuitive.

Interaction between the particles cannot change this distribution, because they are already in equilibrium.

I assumed there are no collisions and I obtained a thermal distribution. Collisions just thermalize the distribution. Therefore if we turn on the collisions in this state, nothing will happen, because the distribution is already thermal.

Then I'm also wondering about the connection you make between temperature and velocity distribution.

Under certain conditions (someone will surely name them) temperature is also proportional to the average energy per degree of freedom. Then the definitions using total energy and kinetic energy are equivalent. In my analysis I assumed equilibrium and chose vertical velocity as the degree of freedom to examine.

Additionally, under certain conditions, temperature can also be defined locally. For example, you can measure temperature of an object like the atmosphere at different positions.

 

[Petr Matas]

Another observation 3: Average total energy of particles at height $z$ depends on $z$, but average kinetic energy doesn't.
[Edited]

 

[Philip Koeck]

Another observation 3: Average total energy of particles at height $z$ depends on $z$, but average total energy of all particles doesn't.

In your model you assumed no collisions, didn't you?
That means every single particle has constant total energy and therefore the average total energy is also constant.

Actually I'm not sure what you are saying here. Obviously the average total energy of all particles doesn't depend on z. How could it? The particles are at a range of different altitudes.

 

[Petr Matas]

In your model you assumed no collisions, didn't you?
That means every single particle has constant total energy and therefore the average total energy is also constant.

Yes, let's have a look into it. If the particles don't exchange energy at all, then they can have any total energy distribution indefinitely. However, if the walls meditate the energy transfer between particles (as is the case for photons), then the energy of individual particles will change on collisions with the walls, but it will stay constant during the flight. Interaction with the walls is sufficient to reach equilibrium distribution.

 

[Petr Matas]

Obviously the average total energy energy of all particles doesn't depend on z.

It's a tautology, isn't it? I was a bit uneasy about it upon writing as well. Actually I meant to say:

Average total energy of particles at height $z$ depends on $z$, but average kinetic energy doesn't.

 

[Philip Koeck]

It's a tautology, isn't it? I was a bit uneasy about it upon writing as well. Actually I meant to say: Average total energy of particles at height z depends on z, but average kinetic energy doesn't.

But from basic mechanics it should be the other way round. The average total energy should be the same at every height. Collisions with the wall shouldn't change that on average if the wall has the same temperature as the gas.
The average kinetic energy should decrease with height.

T should be the same at all heights if you make sure thermal equilibrium is maintained.
So you actually see that T is given by average total energy and not kinetic energy. I think!

 

[Petr Matas]

The average total energy should be the same at every height.

There is a selection effect at play: Only particles with high total energies reach high altitudes. See observation 1 and consider that the distribution has a maximum at zero kinetic energy.

The average kinetic energy should decrease with height.

The kinetic energy decreases with height for individual particles, but the average kinetic energy of particles at height $z$ is independent of height.

It is so counterintuitive!