Paradox: Thermodynamic equilibrium does not exist in gravitational fields

[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 Boltzmann distribution. We may not be able to underestimate their contribution if this bouncing-balls-system has somewhat same energy spectrum with Maxwell Boltzmann.

 

[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 Boltzmann distribution. We may not be able to underestimate their contribution if this bouncing-balls-system has somewhat same energy spectrum with Maxwell Boltzmann.

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.

 

[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).