I Paradox: Thermodynamic equilibrium does not exist in gravitational fields

[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!

 

[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} &lt; \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}&lt;h
time to reach maximum
T=\frac{v_0}{g}
&lt;z&gt;=\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}&gt;h
time to reach the ceiling
T=\frac{v_0-\sqrt{v_0^2-2gh}}{g}
&lt;z&gt;=\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
&lt;z&gt;=-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.