Paradox: Thermodynamic equilibrium does not exist in gravitational field

[Philip Koeck]

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
$$ v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}. \tag 3 $$

I'm still trying to make sense of your proof.
Maybe we could go through it step by step.

When you take the first derivative of equation (1), is that with respect to time? If so, why don't you write that?
And then I would also wonder (just like @anuttarasammyak) why you don't include the term 2 g dz/dt, which would give 2 g v_z, right?

Edit: Only z and v_z depend on t so differentiating equation (1) with respect to t gives:

0 = 2 v_z dv_z / dt + 2 g v_z

This means that dv_z / dt = - g, which is correct.

 

[Petr Matas]

When you take the first derivative of equation (1), is that with respect to time? If so, why don't you write that?

It is not a derivative, but rather the total differential. Nevertheless, we can proceed using the total derivative as well, with respect to any variable, for example $u$:
$$
\frac{d}{du}\left(v_{\text z 0}^2\right) = \frac{d}{du}\left(v_{\text z}^2 + 2gz\right)
$$ $$
2v_{\text z 0} \frac{dv_{\text z 0}}{du} = 2v_{\text z} \frac{dv_{\text z}}{du} + 2z \frac{dg}{du} + 2g \frac{dz}{du}
$$ Multiplying this with $\frac{du}{2}$ gives the equation for total differentials
$$ v_{\text z 0} \, dv_{\text z 0} = v_{\text z} \, dv_{\text z} + z \, dg + g \, dz .
$$

And then I would also wonder (just like @anuttarasammyak) why you don't include the term 2 g dz/dt, which would give 2 g v_z, right?

Variable $g$ is constant, as well as $z$ (see post 51), so both $dg$ and $dz$ equal to 0, which results in equation $(3)$.

 

[Philip Koeck]

It is not a derivative, but rather the total differential. Nevertheless, we can proceed using the total derivative as well, with respect to any variable, for example $u$:
$$
\frac{d}{du}\left(v_{\text z 0}^2\right) = \frac{d}{du}\left(v_{\text z}^2 + 2gz\right)
$$ $$
2v_{\text z 0} \frac{dv_{\text z 0}}{du} = 2v_{\text z} \frac{dv_{\text z}}{du} + 2z \frac{dg}{du} + 2g \frac{dz}{du}
$$ Multiplying this with $\frac{du}{2}$ gives the equation for total differentials
$$ v_{\text z 0} \, dv_{\text z 0} = v_{\text z} \, dv_{\text z} + z \, dg + g \, dz .
$$

Variable $g$ is constant, as well as $z$ (see post 51), so both $dg$ and $dz$ equal to 0, which results in equation $(3)$.

Then I'll replace u by t and get my result, which is obviously correct, even if it's not new.

I would say both z and v_z depend on t, but v_z0 doesn't depend on t since it's just a starting value at t=0.

z is the altitude of a particle that's moving upwards in a gravitational field. How can it be constant?

 

[Petr Matas]

Then I'll replace u by t and get my result, which is obviously correct, even if it's not new.

I would say both z and v_z depend on t, but v_z0 doesn't depend on t since it's just a starting value at t=0.

z is the altitude of a particle that's moving upwards in a gravitational field. How can it be constant?

Forget what I said about the total derivative, it's a nonsense.

Let us have a bunch of particles at altitude 0 with vertical velocities in a narrow range from $v_{\rm z 0}$ to $(v_{\rm z 0} + dv_{\rm z 0})$. When it reaches altitude $z$, its vertical velocities will be in the range from $v_{\rm z}$ to $(v_{\rm z} + dv_{\rm z})$. The altitude $z$ is not to be seen as a function of time, but rather as a certain altitude that we are interested in and that is crossed by the bunch some time after crossing the altitude 0. That is why $z$ is held constant. We are interested in the relationship between the two vertical velocity ranges. To get the relationship between $dv_{\rm z 0}$ and $dv_{\rm z}$, we will start from the equation $(1)$ in the proof and apply either the total differential or the derivative with respect to $v_{\rm z 0}$ or $v_{\rm z}$.

Using the total differential:
$$
\begin{align}
d \! \left( v_{\rm z 0}^2 \right) &= d \! \left( v_{\rm z}^2 + 2gz \right) \nonumber \\
2v_{\rm z 0} \, dv_{\rm z 0} &= 2v_{\rm z} \, dv_{\rm z} + 2z \, dg + 2g \, dz \nonumber \\
\end{align}
$$ Plugging in $dg = 0$ and $dz = 0$ and dividing by 2 yields the equation $(3)$ in the proof.

Using the derivative with respect to $v_{\rm z}$:
$$
\begin{align}
\frac{d}{dv_{\rm z}} \left( v_{\rm z 0}^2 \right) &= \frac{d}{dv_{\rm z}} \left( v_{\rm z}^2 + 2gz \right) \nonumber \\
2v_{\rm z 0} \, \frac{dv_{\rm z 0}}{dv_{\rm z}} &= 2v_{\rm z} + 2z \frac{dg}{dv_{\rm z}} + 2g \frac{dz}{dv_{\rm z}} \nonumber \\
\end{align}
$$ Plugging in $\frac{dg}{dv_{\rm z}} = 0$ and $\frac{dz}{dv_{\rm z}} = 0$ and multiplying with $\frac{dv_{\rm z}}{2}$ yields again the equation $(3)$ in the proof.

 

[Philip Koeck]

Using the derivative with respect to $v_{\rm z}$:
$$
\begin{align}
\frac{d}{dv_{\rm z}} \left( v_{\rm z 0}^2 \right) &= \frac{d}{dv_{\rm z}} \left( v_{\rm z}^2 + 2gz \right) \nonumber \\
2v_{\rm z 0} \, \frac{dv_{\rm z 0}}{dv_{\rm z}} &= 2v_{\rm z} + 2z \frac{dg}{dv_{\rm z}} + 2g \frac{dz}{dv_{\rm z}} \nonumber \\
\end{align}
$$ Plugging in $\frac{dg}{dv_{\rm z}} = 0$ and $\frac{dz}{dv_{\rm z}} = 0$ and multiplying with $\frac{dv_{\rm z}}{2}$ yields again the equation $(3)$ in the proof.

I would say v_z and z are not independent of each other, no matter whether you are considering 1 particle or a range of particles.

dv_z0 / dv_z should be 1, I would say.

 

[anuttarasammyak]

Please find attached a preliminary phase space diagram of bouncing balls between floor and ceiling via
$$v_0^2=v^2+2gz$$
$$z=-\frac{v^2}{2g}+\frac{v_0^2}{2g}$$
Three colored cycles represent motion of three near-by-speed balls. The variances of speed seem
$$\sigma_{floor} < \sigma_{ceiling}$$
I hope this will help confirm/improve the idea in the discussion.

 

[Petr Matas]

I would say v_z and z are not independent of each other,

Unless you interpret equation $(1)$ in the following way: A particle crosses altitude $z$ with vertical velocity $v_{\rm z}$. What was its vertical velocity at altitude 0?

no matter whether you are considering 1 particle or a range of particles.

You are right that your issue has nothing to do with the distinction between one particle and a group of particles.

dv_z0 / dv_z should be 1, I would say.

Can you see why $dv_{\rm z 0}$ and $dv_{\rm z}$ may not be equal?

Also note that the symbols $z$ and $v_{\rm z}$ are used in two different meanings:

1. as a coordinate
2. as a specific value independent of time – this is the meaning used in my equations. I think I should have labelled them $z_1$ and $v_{\rm z 1}$ instead.

 

[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 ?

I have taken some snap shot copies of familiar Windows start/shut down moving icons from Web with no intention of selection. Many of them seem that upper half contains more dots than the lower. It seems to be in accord with the above said.

 

[Petr Matas]

I have taken some snap shot copies of familiar Windows start/shut down movng icons. It seems that upper half contains mor dots than the lower. It seems to be in accord with the above.

The dots are not in equilibrium – all of them have the same total energy.

Your observation agrees with my observation 2:

As a bunch of particles rises and loses speed, they come closer together in space, but farther apart in velocity.

 

[anuttarasammyak]

There are many balls.

The dots are not in equilibrium – all of them have the same total energy.

So may I expect that we will observe more balls under side when we gather spectrum of energy balls each of which conserve its energy ?

 

[Petr Matas]

So may I expect that we will observe more balls under side when we gather spectrum of energy balls each of which conserve its energy ?

Exactly (for thermal spectrum).

 

[anuttarasammyak]

Thanks. Then as far as ball reach the ceiling, my argument seems to apply, i.e. they spend more time in upper side. The low and dense comes from low spectrum energy balls which do not reach ceiling, I suspect. How do you think of it ?

 

[Philip Koeck]

Thanks. Then as far as ball reach the ceiling, my argument seems to apply, i.e. they spend more time in upper side. The low and dense comes from low spectrum energy balls which do not reach ceiling, I suspect. How do you think of it ?

It might be worth estimating how many particles won't reach the ceiling given a certain height of the container (for example 1000 m) and a certain T (maybe 300 K).

 

[Demystifier]

In equlibrium, the probability density that a particle is in the state $({\bf x},{\bf p})$ in the one-particle phase space is
$$\rho({\bf x},{\bf p})\propto e^{-\beta H({\bf x},{\bf p})}$$
where $\beta=1/(kT)$ is constant. Neglecting interparticle interactions (ideal gas), in a uniform gravitational field we have
$$H({\bf x},{\bf p})=\frac{{\bf p}^2}{2m}+gz$$
so
$$\rho({\bf x},{\bf p})\propto e^{-\beta \frac{{\bf p}^2}{2m} } e^{-\beta gz}$$
The first factor is the usual Maxwell distribution of kinetic energies. The second factor shows that the gas density decreases with altitude.

 

[anuttarasammyak]

In equlibrium

My interest in this thread is whether the equilibrium state you explained is realized or interpreted by ensemble of independent particles whose energy is conserved independently, just by adjusting initial energy spectrum.

 

[Demystifier]

My interest in this thread is whether the equilibrium state you explained is realized or interpreted by ensemble of independent particles whose energy is conserved independently, just by adjusting initial energy spectrum.

It isn't. Equilibration requires some interaction between particles. However, the interaction may be weak, so that it can be neglected in the formula for distribution in the phase space.

 

[pines-demon]

In equlibrium, the probability density that a particle is in the state $({\bf x},{\bf p})$ in the one-particle phase space is
$$\rho({\bf x},{\bf p})\propto e^{-\beta H({\bf x},{\bf p})}$$
where $\beta=1/(kT)$ is constant. Neglecting interparticle interactions (ideal gas), in a uniform gravitational field we have
$$H({\bf x},{\bf p})=\frac{{\bf p}^2}{2m}+gz$$
so
$$\rho({\bf x},{\bf p})\propto e^{-\beta \frac{{\bf p}^2}{2m} } e^{-\beta gz}$$
The first factor is the usual Maxwell distribution of kinetic energies. The second factor shows that the gas density decreases with altitude.

Is this thread related to the problem of the partition function diverging when changing $mgz\to-GmM/r$?

 

[Demystifier]

Is this thread related to the problem of the partition function approach not working when changing $mgz\to-GmM/r$?

This thread is not related to it. I am not aware that there is a problem with the partition function in your case, but I guess the problem only occurs when you let $r\to 0$, and it goes away if you consider a realistic body of a finite size, inside which the gravitational potential takes a different form without a divergence at $r\to 0$.

 

[pines-demon]

This thread is not related to it. I am not aware that there is a problem with the partition function in your case, but I guess the problem only occurs when you let $r\to 0$, and it goes away if you consider a realistic body of a finite size, inside which the gravitational potential takes a different form without a divergence at $r\to 0$.

Oh my bad. It also diverges at $r\to \infty$ actually.

 

[Demystifier]

It also diverges at $r\to \infty$ actually.

That's trivial, even the partition function for the free particle diverges in this limit. That's because the partition function is proportional to the volume $V$, which diverges if you let $r\to \infty$. This benign IR divergence is removed by putting the system into a large but finite volume $V$.

 

[anuttarasammyak]

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.

It isn't.

@Demistifier Thanks for encouraging post.

 

[anuttarasammyak]

It might be worth estimating how many particles won't reach the ceiling given a certain height of the container (for example 1000 m) and a certain T (maybe 300 K).

Even in high temperature most of the particles have almost zero energy in Maxwell Boltzman distribution. We may not be able to underestimate their contribution if this bouncing-balls-system has somewhat same energy spectrum with Maxwell Boltzman.

 

[Petr Matas]

Thanks. Then as far as ball reach the ceiling, my argument seems to apply, i.e. they spend more time in upper side. The low and dense comes from low spectrum energy balls which do not reach ceiling, I suspect. How do you think of it ?

You're right.

It might be worth estimating how many particles won't reach the ceiling given a certain height of the container (for example 1000 m) and a certain T (maybe 300 K).

As anuttarasammyak said, consider that the distribution has its maximum at zero velocity, so there are always plenty of particles that don't reach the ceiling.

Now let's tackle your question: Which particles starting from altitude 0 won't reach altitude $z_1$? Those with insufficient energy of their vertical movement:
$$ \frac{1}{2} m v_{\rm z 0}^2 < m g z_1 $$
What fraction of particles at altitude 0 do they constitute? It is
$$
d(z_1)
= \frac {\text{slow particles}} {\text{all particles}}
= \frac {\int_{v_{\rm z 0}^2 < 2g z_1} \rho(0, v_{\rm z 0}) \, dv_{\rm z 0}} {\int_{-\infty}^{+\infty} \rho(0, v_{\rm z 0}) \, dv_{\rm z 0}}
$$
Can you do the maths?

My interest in this thread is whether the equilibrium state you explained is realized or interpreted by ensemble of independent particles whose energy is conserved independently, just by adjusting initial energy spectrum.

It is.

It isn't. Equilibration requires some interaction between particles. However, the interaction may be weak, so that it can be neglected in the formula for distribution in the phase space.

Interaction with the walls will mediate the energy transfer between particles. After equilibration you may disable even this process.

 

[Philip Koeck]

Even in high temperature most of the particles have almost zero energy in Maxwell Boltzman distribution. We may not be able to underestimate their contribution if this bouncing-balls-system has somewhat same energy spectrum with Maxwell Boltzman.

Are you talking about the Boltzmann distribution or the Maxwell-Boltzmann distribution?

Google images for Maxwell-Boltzmann energy distribution.

 

[Petr Matas]

Are you talking about the Boltzmann distribution or the Maxwell-Boltzmann distribution?

In equilibrium:
The velocity $\mathbf{v} = (v_{\rm x}, v_{\rm y}, v_{\rm z})$ and its components $v_{\rm x}$, $v_{\rm y}$, $v_{\rm z}$ obey the Boltzmann distribution.
The speed $|\mathbf{v}| = \sqrt{v_{\rm x}^2 + v_{\rm y}^2 + v_{\rm z}^2}$ obeys the Maxwell–Boltzmann distribution.

Thank you for your question. I see I got it wrong before:

Let the velocity distribution at height 0 is Maxwell–Boltzmann (i.e. thermal) with temperature $T$. That means the density at that height is
$$ \rho(0, v_{\text z}) = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right), \tag 7
$$

 

[anuttarasammyak]

You're right.

Based on this idea, I would like to estimate average height <z> of particles.
The trajectory
$$z=v_0t-\frac{1}{2}gt^2$$

1. lower energy particle which does not touch the ceiling at height h.
maximum height
$$\frac{v_0^2}{2g}<h$$
time to reach maximum
$$T=\frac{v_0}{g}$$
$$<z>=\frac{1}{T}\int_0^T z(t) dt$$
$$=\frac{v_0^2}{3g}$$

2. higher energy particle which touches the ceiling
$$\frac{v_0^2}{2g}>h$$
time to reach the ceiling
$$T=\frac{v_0-\sqrt{v_0^2-2gh}}{g}$$
$$<z>=\frac{1}{T}\int_0^T z(t) dt$$
$$=\frac{(v_0-\sqrt{v_0^2-2gh})(2v_0+\sqrt{v_0^2-2gh})}{6g}$$
$$=\frac{v_0^2+2gh-v_0\sqrt{v_0^2-2gh}}{6g}$$

I do not keep enough energy to carry out ensemble integral with weight of Maxwell way Gaussian spectrum for $v_0$ to know whether it coincides with the result of statistical mechanics for thermal equillibrium state of
$$<z>=-h\frac{\partial}{\partial (\beta mgh)}\log{\frac{1-e^{-\beta mgh}}{\beta mgh}}$$
or not. If you are tough enough, I should appreciate your effort.

 

[Petr Matas]

Based on this idea, I would like to estimate average height <z> of particles.

Just a few thoughts for now:

Consider a particle that does not reach the ceiling. Height $z(t)$ of the particle is a quadratic function of time and its graph is a parabola. For its segment from height $z(0) = 0$ to the top $z_{\rm p}$ of the parabola, the average height is $\left< z \right> = \frac{2}{3} z_{\rm p}$.

A particle, that just touches the ceiling, has average height at $\frac{2}{3}$ of the box height.

A particle with infinite vertical velocity has average height at $\frac{1}{2}$ of the box height.

A particle, that has finite speed and bounces off the ceiling, is somewhere in between, i.e. its average height is between $\frac{1}{2}$ and $\frac{2}{3}$ of the box height.

Next, we have already derived the density
$$ \rho(z) \propto \int_{-\infty}^{+\infty} \rho(z, v_{\rm z}) \, dv_{\rm z} \propto \exp(-\beta m g z)
$$ using the laws of motion, so we can start from there.

 

[Demystifier]

A few words about the intuition that particles at smaller $z$ should have larger kinetic energy and hence larger temperature. It this was so, kinetic energy would be correlated with $z$, they would not be independent, by knowing one you would also know something about the other. This would reduce the available phase space of the particle, which would reduce entropy. But recall that the Boltzmann distribution is obtained precisely by requiring that the entropy is maximal. If the entropy is not maximal, the system is not in the statistical equilibrium.

Another source of wrong intuition is that one thinks of a particle (in a gas) as being free, so that its kinetic energy increases when its gravitational potential energy decreases. But it's not free. It interacts with all the other particles in the gas. When the particle attains kinetic energy by falling down, it almost immediately transfers this extra kinetic energy to the environment. You can think of it as having a strong friction with the environment, so that its kinetic energy does not increase during the fall. A particle in the gas is more like a falling leaf than like a falling apple. Or even better, it is like a balloon with the same density as the air, so that, in average, it neither falls down nor moves up.

 

[dextercioby]

We could have used some input from @vanhees71, but he's retired from this community.

 

[Petr Matas]

Another source of wrong intuition is that one thinks of a particle (in a gas) as being free, so that its kinetic energy increases when its gravitational potential energy decreases. But it's not free. It interacts with all the other particles in the gas. When the particle attains kinetic energy by falling down, it almost immediately transfers this extra kinetic energy to the environment. You can think of it as having a strong friction with the environment, so that its kinetic energy does not increase during the fall. A particle in the gas is more like a falling leaf than like a falling apple. Or even better, it is like a balloon with the same density as the air, so that, in average, it neither falls down nor moves up.

The question of the gas column's equilibrium temperature profile was resolved in the first nine posts of this thread – the temperature is the same everywhere, as you say. It has been claimed multiple times in this thread, that the interactions between the particles are a game changer, but it is not true. Even if the particles do not interact with each other, temporary interaction with e.g. the floor will be sufficient to bring them into thermodynamic equilibrium.

The observation, that a particle loses kinetic energy when moving up, is correct. However, only particles with high total energy will reach high altitude. This selection effect causes the average kinetic energy to be the same at all altitudes. That is the true reason why the kinetic energy does not change with altitude in average, even though it does for individual particles. Therefore, even if the particle interactions are ignored, the analysis using the laws of motion will lead to the correct result if done right.

 

[anuttarasammyak]

1. lower energy particle which does not touch the ceiling at height h.
maximum height
v022g<h
time to reach maximum
T=v0g
<z>=1T∫0Tz(t)dt
=v023g

Integration with weight
$$f(v_0)=\sqrt{\frac{m\beta}{2\pi}}e^{-\frac{m\beta}{2}v_0^2}$$
is rather easier. Contribution of the balls which do not reach the ceiling to <z> is
$$2 \int_0^\sqrt{2gh} dv_0 \frac{v_0^2}{3g} \sqrt{\frac{m\beta}{2\pi}}e^{-\frac{m\beta}{2}v_0^2}$$
$$=\frac{2}{3\sqrt{\pi}}\frac{1}{m\beta g}\int_0^{m\beta gh}\sqrt{t}e^{-t} dt$$
$$=h\frac{2}{3\sqrt{\pi}}\frac{1}{m\beta gh} (\frac{\sqrt{\pi}}{2}-\Gamma(\frac{3}{2},m\beta gh))$$
ref. https://www.wolframalpha.com/input?i=2/(3sqrt(pi))(integration+of+√ｘe^(-x)+from+0+to++x)divided+by+x

I observe in both high and low tenperature limit, the contribution tends to zero. It has a peak of about 0.15 h around $mg\beta h=1$.

 

[anuttarasammyak]

2. higher energy particle which touches the ceiling
v022g>h
time to reach the ceiling
T=v0−v02−2ghg
<z>=1T∫0Tz(t)dt
=(v0−v02−2gh)(2v0+v02−2gh)6g
=v02+2gh−v0v02−2gh6g

$$=\frac{h}{3a}(1+a-\sqrt{1-a})$$
$$=h(\frac{1}{2}+\frac{1}{3}\sum_{n=2}^\infty \binom{\frac{1}{2}}{n}(-)^n a^{n-1})$$
$$=h(\frac{1}{2}+\frac{a}{24}+\frac{a^2}{48}+\frac{5a^3}{384}+...)$$
where
$$a=\frac{2gh}{v_0^2}<1$$
To investing the contribution according to a^n terms, integration of a^n with weight
$$f(v_0)=\sqrt{\frac{m\beta}{2\pi}}e^{-\frac{m\beta}{2}v_0^2}$$
for region a<1 is
$$\sqrt{\frac{m\beta gh}{\pi}}\int_0^1 a^{n-\frac{3}{2}} e^{\frac{-m\beta gh}{a}}da$$
$$=\sqrt{\frac{(m\beta gh)^n}{\pi}}\int_0^\frac{1}{m\beta gh} t^{n-\frac{3}{2}} e^{-\frac{1}{t}}dt$$
$$=\sqrt{\frac{(m\beta gh)^n}{\pi}}\Gamma(\frac{1}{2}-n, m\beta gh)$$
https://www.wolframalpha.com/input?i=integrate+x^(n-3/2)+e^(-1/x)+from+0+to+x

Thus contribution to <z> from balls touching the ceiling is
$$\frac{h}{\sqrt{\pi}}[\frac{1}{2}\Gamma(\frac{1}{2},b)+\frac{1}{3}\sum_{n=2}^\infty \binom{\frac{1}{2}}{n} (-)^n b^{\frac{n-1}{2}}\Gamma(\frac{3}{2}-n,b)]$$
where
$$b=mgh\beta$$
The contribution to <z> by the lowest four power term is shown in https://www.wolframalpha.com/input?i=plot+y=1/(sqrt(pi))(1/2+Gamma(1/2,x)+1/3+sqrt{x}+1/8+\Gamma(-1/2,x)+1/48+x+Gamma(-3/2,x)+5/384+x+sqrt(x)+Gamma(-3/5,x)))

In combination of 1 and 2 we get <z>/h plot wrt $mgh\beta$ which is in the aproximation above said as shown in https://www.wolframalpha.com/input?...,2]]Gamma\(40)Divide[3,2]-n+2\(44)x\(41)\(41)

Due to restriction of computing resource I could get n=0,1,2,3 but not value for higher orders.

I do not keep enough energy to carry out ensemble integral with weight of Maxwell way Gaussian spectrum for v0 to know whether it coincides with the result of statistical mechanics for thermal equillibrium state of
<z>=−h∂∂(βmgh)log⁡1−e−βmghβmgh
or not.

Plot of <z>/h in thermal equillibrium is https://www.wolframalpha.com/input?i=differentiate+-+log+(1-e^(-x))+++log+x

Comparing it with the bouncing-ball-system, I observe the difference, damping is slower in thermal equillibrium.
I should apreciate it if you would check and confirm/claim my calculation.

 

[Petr Matas]

We have already derived the density
$$ \rho(z) \propto \int_{-\infty}^{+\infty} \rho(z, v_{\rm z}) \, dv_{\rm z} \propto \exp(-\beta m g z)
$$ using the laws of motion, so we can start from there.

Let us have a vertical column of ideal gas.

$\rho(z)$ is the density of particles at altitude $z$,
$\beta = \frac {1} {k_{\rm B} T}$ is the thermodynamic beta,
$k_{\rm B}$ is the Boltzmann constant,
$T$ is the thermodynamic temperature,
$m$ is the particle mass,
$h$ is the column height,
$g$ is the gravitational acceleration.

The average altitude of particles above the column floor (i.e. the altitude of the column's center of mass) can be written as
$$
\begin{align}
\left< z \right> &= \frac {\int_0^h z \, \rho(z) \, dz} {\int_0^h \rho(z) \, dz} \nonumber \\
\nonumber \\
&= \frac {\int_0^h z \, e^{-\beta m g z} \, dz} {\int_0^h e^{-\beta m g z} \, dz} \nonumber \\
\nonumber \\
&= \frac {1} {β m g} - \frac {h} {e^{β m g h} - 1} \nonumber
\end{align}
$$ (calculation).

Some interesting values:
$$
\begin{align}
& \lim_{h \to \infty} \left< z \right> = \frac {1} {β m g} \nonumber \\
\nonumber \\
& \lim_{T \to 0} \left< z \right> = 0 \nonumber \\
\nonumber \\
& \lim_{T \to \infty} \left< z \right> = \frac{h}{2} \nonumber
\end{align}
$$

 

[Petr Matas]

Based on this idea, I would like to estimate average height <z> of particles.

$$
\left< z \right> = \frac {1} {β m g} - \frac {h} {e^{β m g h} - 1}
$$ $$
\lim_{h \to \infty} \left< z \right> = \frac {1} {β m g}
$$

Let us continue. The particle mass
$$ m = M / N_{\rm A},
$$ where

$M$ is the molar mass,​

$N_{\rm A} \approx 6.022 \times 10^{-23} \, \text{mol}^{-1}$ is the Avogadro constant.​

$$ \beta m = \frac {M / N_{\rm A}} {k_{\rm B} T} = \frac {M} {R T}
$$ where $R = N_{\rm A} k_{\rm B} \approx 8.314 \, \rm{J \, K^{-1} \, mol^{-1}}$ is the molar gas constant.

For

$T = 300 \, \rm{K}$,​

$M = M_{\rm N_2} \approx 28 \, \rm{g/mol}$,​

$g = 9.80665 \, \rm{m/s^2}$,​

$h \to \infty$:​

$$ \left< z \right> = \frac {RT} {Mg} \approx 9.08 \, \rm{km} $$

 

[A.T.]

It has been claimed multiple times in this thread, that the interactions between the particles are a game changer, but it is not true. Even if the particles do not interact with each other, temporary interaction with e.g. the floor will be sufficient to bring them into thermodynamic equilibrium.

My naive way to think about it, is to consider a perfectly elastic collision of two equal masses in 1D:

https://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

Two identical particles simply swap their velocities in elastic 1D collision. So if you don't label them, and don't look, then afterwards you cannot tell if there was a collision, or if they just passed through each other. Therefore in a very thin, one particle wide cylinder of gas, the velocity distribution is not affected by whether the particles interact or just pass through each other.

I have not tried to generalize this to 3D, but the above mentioned velocity swap still applies along the contact normal, while the velocities perpendicular to it remain unchanged.