I Paradox: Thermodynamic equilibrium does not exist in gravitational fields

[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}&lt;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.

 

[Petr Matas]

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.

That means that in 1D, particle interactions with each other cannot redistribute the kinetic energy and thus equalize the energy spectrum and thus make the system evolve towards the thermodynamic equilibrium. Only collisions with the floor or ceiling will be able to do that.

The situation is quite different in higher dimensions.

 

[A.T.]

That means that in 1D, particle interactions with each other cannot redistribute the kinetic energy and thus equalize the energy spectrum and thus make the system evolve towards the thermodynamic equilibrium. Only collisions with the floor or ceiling will be able to do that.

The one-particle-wide-cylinder I was considering has a floor and ceiling, just like your 3D box. An the 1D math shows that particle-particle-interaction makes no difference to energy redistribution, just like you found for your 3D box (if I understand you correctly).

The situation is quite different in higher dimensions.

Maybe. But 1D seems to get the same (qualitative) result as you got for the 3D case, no?

 

[Petr Matas]

An the 1D math shows that particle-particle-interaction makes no difference to energy redistribution, just like you found for your 3D box (if I understand you correctly).

I meant something else: Interactions push the system towards thermodynamic equilibrium (they only don't for identical particles in 1D, as you have shown) by redistributing energy. If the system has not reached equilibrium yet, then the interactions will be important. In my analysis, I assume that the system is already in equilibrium. Once the equilibrium has been reached, further interactions will make no difference.

 

[snorkack]

I meant something else: Interactions push the system towards thermodynamic equilibrium (they only don't for identical particles in 1D, as you have shown) by redistributing energy. If the system has not reached equilibrium yet, then the interactions will be important. In my analysis, I assume that the system is already in equilibrium. Once the equilibrium has been reached, further interactions will make no difference.

So, for example, the equilibrium profile of density in a vessel should be independent on whether the vessel contains gas in a low density (and they rarely collide with each other, compared to walls) or high density (many collisions between wall and wall). And the collision cross-section should matter for the process of setting up equilibrium - but not for the equilibrium once established. Right?

 

[Petr Matas]

the equilibrium profile of density in a vessel should be independent on whether the vessel contains gas in a low density (and they rarely collide with each other, compared to walls) or high density

You're right. The density $\rho(z)$ as a function of altitude $z$ will be the same up to a multiplicative constant. We should just remember that we are talking about an ideal gas.

high density (many collisions between wall and wall)

Did you mean many collisions between particle and particle?

And the collision cross-section should matter for the process of setting up equilibrium - but not for the equilibrium once established. Right?

Exactly.

 

[snorkack]

Did you mean many collisions between particle and particle?

Ah, I see that it was badly written.
I meant that a particle has many collisions with other particles between collisions of the particle with a wall and a collision of the same particle with a different wall.

 

[Petr Matas]

Ah, I see that it was badly written.
I meant that a particle has many collisions with other particles between collisions of the particle with a wall and a collision of the same particle with a different wall.

I see, no problem. Thinking about this property, there is also a related quantity called mean free path.

 

[anuttarasammyak]

Comparing it with the bouncing-ball-system, I observe the difference, damping is slower in thermal equillibrium.

Please find https://www.wolframalpha.com/input?i2d=true&i=Divide[2,3Sqrt[pi]]Divide[1,x]\(40)Divide[Sqrt[pi],2]-Gamma\(40)Divide[3,2]\(44)x\(41)\(41)+Divide[1,Sqrt[pi]]\(40)Divide[1,2]+Gamma\(40)Divide[1,2]\(44)x\(41)+Divide[1,3]Sum[Power[\(40)-1\(41),n],{n,0,100}]Divide[\(40)Divide[1,2]\(41)!,\(40)Divide[1,2]-n-2\(41)!\(40)n+2\(41)!]Power[x,Divide[\(40)n+1\(41),2]]Gamma\(40)Divide[3,2]-n+2\(44)x\(41)\(41)-Divide[x+1-Power[e,x],x-x*Power[e,x]] wolfram calculation of height averages,

<z>/h of independent bouncing balls - <z>/h of thermal equillibrium $\leq$ 0

where parameter x is $mgh\beta$. They coincide with 1/2 at x=0 but do not coincide for x>0. I do not expect that higher order n terms which I cannot calculate due to compuring resource limit, would contribute so that they coincide.

 

[Petr Matas]

@anuttarasammyak: Why are you taking such a complicated approach? Doesn't post #104 answer your question?

 

[Petr Matas]

$$H({\bf x},{\bf p})=\frac{{\bf p}^2}{2m}+gz$$

I think that $m$ is missing at the end of the formula. Hamiltonian of a point mass in homogeneous gravitational field is
$$H({\bf x},{\bf p})=\frac{p^2}{2m}+mgz.$$

 

[Demystifier]

Of course, thanks for the correction!

 

[anuttarasammyak]

1. Let us see System of independent bouncing balls energy spectrum of which is same as Maxwell distribution.
hf(r) : time average height of each ball trajectory with floor speed v_0 is
f(r)=\frac{2}{3r} for r>1
f(r)=\frac{1}{3r}(1+r+\sqrt{1-r})
for 0<r<1 where h is height of ceiling and
r=\frac{2gh}{v_0^2}
is ratio of potential energy at ceiling to energy of ball. Using theta function
f(r)=\frac{2}{3r}\theta(r-1)+\frac{1}{3r}(1+r+\sqrt{1-r})\theta(r)\theta(1-r)

2. average height of the group is
h\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty}dv_0 f(r) e^{-\frac{1}{2}mv_0^2\beta}
Recalling that for each trajectory
v_0^2=v^2+2gz
this integral is exressed as
\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty}dv\int_0^h dz \ f(\frac{2gh}{v^2+2gz})e^{-(\frac{1}{2}mv^2+mgz)\beta}
where integration wrt z and integration wrt v are not separatively done.

3. Corresponding value for theramal equillibrium is
\frac{1}{h}\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty} dv \int_0^h dz\ z e^{-(\frac{1}{2}mv^2+mgz)\beta}
where integratio wrt z and integration wrt v can be carried out independently.

4. This shows that system of independent bouncing balls energy spectrum of which is same as Maxwell distribution, and system of thermal equillibrium are not same. My previous posts where I calculated values for the both and observe the difference confirm it.

 

[Petr Matas]

2. average height of the group is
h\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty}dv_0 f(r) e^{-\frac{1}{2}mv_0^2\beta}

Maybe I have found a mistake. If I read this expression correctly, you are integrating over the particles, which are currently at the floor. I agree, that every particle bounces off the floor regularly, but they do so with different time periods $t_{\rm P}(v_0)$ and with different vertical velocities $v_0$ (i.e. the time spent at 0–1 μm above the floor depends on $v_0$). Both of these affect the particle's contribution to the velocity distribution at the floor and I think that they have to be compensated for in your integral by giving weight
$$
w(v_0) = \left| v_0 \, t_{\rm P}(v_0) \right|
$$ to the particle:
$$
\left< z \right> = \frac {h \int_{-\infty}^{+\infty}dv_0 \, f(r) \, e^{-\frac{1}{2}mv_0^2\beta} \, w(v_0)} {\int_{-\infty}^{+\infty}dv_0 \, e^{-\frac{1}{2}mv_0^2\beta} \, w(v_0)}
$$
Obviously, your normalization constant $\sqrt{\frac{m\beta}{2\pi}}$ has to change accordingly – it is replaced with the denominator above.

 

[anuttarasammyak]

I have some argument with your point but before that, is your formula in accordance with thermal equilibrium ?

 

[Petr Matas]

I have some argument with your point but before that, is your formula in accordance with thermal equilibrium ?

I didn't evaluate the modified formula, because it is too complicated. The approach $\rho(0, v_0) \mapsto \rho(z, v)$ stands on the same assumptions (i.e. non-interacting bouncing balls) as your approach, it is easier to apply and it leads to a result in accordance with equilibrium. Although your approach is complicated, it should lead to the same result if applied correctly, shouldn't it?

Let me support my claim, that interactions don't matter in equilibrium, with another argument: The equilibrium state, as described by well-known formulas, is independent of collision cross-section $\sigma$. The limit $\sigma \to 0$ corresponds to non-interacting particles (bouncing balls).

 

[anuttarasammyak]

I didn't evaluate the modified formula, because it is too complicated.

Obviously, your normalization constant mβ2π has to change accordingly – it is replaced with the denominator above.

I am afraid that these properties suggest your formula as well as mine does not coincide with the thermal equilibrium.

My argument to your point:

Maybe I have found a mistake. If I read this expression correctly, you are integrating over the particles, which are currently at the floor. I agree, that every particle bounces off the floor regularly, but they do so with different time periods tP(v0) and with different vertical velocities v0 (i.e. the time spent at 0–1 μm above the floor depends on v0). Both of these affect the particle's contribution to the velocity distribution at the floor and I think that they have to be compensated for in your integral by giving weight

We do not have to think of bouncing time period here because time average in period has been already considered in getting f(r).

time to reach maximum
T=v0g
<z>=1T∫0Tz(t)dt
=v023g

 

[Petr Matas]

We do not have to think of bouncing time period here because time average in period has been already considered in getting f(r).

I see, $t_{\rm P}(v_0) = 2T$, where $T$ is the time to reach maximum. It seems to me that you incorporated it correctly into your formula for average height of individual particle $\langle z(v_0) \rangle = h f(r)$. The problem is averaging this average not only over the particles currently at the floor, but over those in the entire volume:
$$
\langle z \rangle = \frac 1 N \int_N \langle z(v_0) \rangle \, dN,
$$ where $N$ is the number of particles in the volume.

The Boltzmann distribution for $z = 0$ is the distribution of $v_0$ for particles currently at the floor, not in the entire volume.

 

[anuttarasammyak]

The problem is averaging this average not only over the particles currently at the floor, but over those in the entire volume:

Why do you think that I have focused on the particles currently at the floor ? I used v_0^2 is to express energy of a trajectory wherever on the trajectory the ball is.

 

[Petr Matas]

Why do you think that I have focused on the particles currently at the floor ? I used v_0^2 is to express energy of a trajectory wherever on the trajectory the ball is.

I am aware that $v_0^2$ characterizes the entire trajectory. So you have a population of trajectories. Where does its distribution come from?

 

[anuttarasammyak]

Where does its distribution come from?

It is coming from you. You stated that energy of bouncing-balls have same distribution spectrum as Maxwell's.

 

[Petr Matas]

It is coming from you. You stated that energy of bouncing-balls have same distribution spectrum as Maxwell's.

Let me explain. Assuming that particles currently at the floor have thermal spectrum of velocities $v_0$ and that they move without collisions, I derived the spectrum of velocities $v$ at altitude $z$. I found that the spectrum is the same as at the floor and that the density is lower. But the derived spectrum isn't the spectrum of velocities $v_0$ at the bottom of the trajectory, but of velocities $v$ at altitude $z$. Velocity $v_0$ is greater than $v$. At altitude $z$ (unlike at the floor), there are no particles with $|v_0| < \sqrt{2gz}$. In other words, different altitudes have the same spectra of $v$, but different spectra of $v_0$.

 

[anuttarasammyak]

. Assuming that particles currently at the floor have thermal spectrum of velocities v0 and that they move without collisions, I derived the spectrum of velocities v at altitude z.

Recalling that for each trajectory
v0^2=v^2+2gz
this integral is exressed as

Is the spectrum
Ce^{-(\frac{1}{2}mv^2+mgz)\beta}=Ce^{-\frac{1}{2}mv_0^2\beta}
specified by total energy of the trajectory $v_0^2$ ?

 

[Petr Matas]

Is the spectrum
Ce^{-(\frac{1}{2}mv^2+mgz)\beta}=Ce^{-\frac{1}{2}mv_0^2\beta}
specified by total energy v_0^2 ?

Yes, but the phase space has two dimensions ($z$ and $v$), so we must integrate over both,
$$
\begin{align}
\langle z \rangle &= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, z \nonumber \\
&= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, \langle z(v_0) \rangle, \nonumber
\end{align}
$$ or compensate for it with the weight.

 

[Petr Matas]

Yup, that's peace for ya. Relativity

I dropped the relativity altogether after finding that it does not resolve the paradox.

 

[anuttarasammyak]

Yes, but the phase space has two dimensions (z and v), so we must integrate over both,
<br /> \begin{align}<br /> <br /> \langle z \rangle &amp;= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, z \nonumber \\<br /> <br /> &amp;= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, \langle z(v_0) \rangle, \nonumber<br /> <br /> \end{align}

=C \int_0^h dz \ e^{-mgz\beta} \int_{-\infty}^{+\infty} dv \ e^{-\frac{1}{2}mv^2\beta} &lt;\mathbb{z}(\sqrt{v^2+2gz})&gt;
which is equivalent to my

this integral is exressed as
\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty}dv\int_0^h dz \ f(\frac{2gh}{v^2+2gz})e^{-(\frac{1}{2}mv^2+mgz)\beta}
where integration wrt z and integration wrt v are not separatively done.

Now we get the same result which is different from thermal equilibrium.

 

[pines-demon]

=C \int_0^h dz \ e^{-mgz\beta} \int_{-\infty}^{+\infty} dv \ e^{-\frac{1}{2}mv^2\beta} &lt;z(\sqrt{v^2+2gz})&gt;
which is equivalent to my

Now we get the same result which is different from thermal equilibrium.

Now that the problem is solved with statistical mechanics, one could check if we can exhaust the methods and see if one gets the same solutions with Langevin equations of with Fokker-Planck equation.

 

[DrDu]

The problem is also how you define temperature. In thermodynamics, systems in thermal equilibrium are defined to have the same temperature. However, this is not the temperature you measure, say, with a mercury thermometer which is the "local temperature". Hence you cannot run a carnot engine between two states at different heights, if they are in thermal equilibrium, although they have different local temperatures.

 

[anuttarasammyak]

The problem is also how you define temperature. In thermodynamics, systems in thermal equilibrium are defined to have the same temperature. However, this is not the temperature you measure, say, with a mercury thermometer which is the "local temperature". Hence you cannot run a carnot engine between two states at different heights, if they are in thermal equilibrium, although they have different local temperatures.

Maxwell refferred in post #26 says that temperature is same at high and low positions. What you mention is the progress after him ?
I find the relation in (27.2) of Landau-Rifshitz Statistial Mechanics
T\sqrt{g_{00}}=const.
So in thermal equillibrium
T_{@high}&lt;T_{@low}

 

[DrDu]

I would call $T\sqrt{g_{00}}$ the thermodynamical temperature and $T$ the local temperature. thermodynamical temperature is a consequence of the zeroth law of thermodynamics (transitivity of thermal equilibrium). There is no reason to think that it does not hold in gravitational systems. The situation is comparable to the concept of mass, given your interest, in special relativity there is rest mass, longitudinal mass and transversal mass. All three concepts coincide in the non-relativistic Galilean limit. So in gravitational fields, it might be useful to introduce several temperature concepts. This is ok, so long as you make clear which one you are using.

 

[anuttarasammyak]

I would call $T\sqrt{g_{00}}$ the thermodynamical temperature and $T$ the local temperature. thermodynamical temperature is a consequence of the zeroth law of thermodynamics (transitivity of thermal equilibrium). There is no reason to think that it does not hold in gravitational systems.

It resembles with
\omega\sqrt{g_{00}}=const.
along world line of a photon which means red shift of going up light. They have the same reason.
Photon gas in vertical cylinder at equilibrium show blueish at bottom end and reddish at top end.

 

[Demystifier]

The temperature involving the factor $\sqrt{g_{00}}$ is known as the Tolman law. I have written a short insight article about it: https://www.physicsforums.com/insights/tolman-law-in-a-nutshell/