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.