Paradox: Thermodynamic equilibrium does not exist in gravitational field

[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.