Sound in a gas of non-interacting particles?

  • #1
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Hi.

In some statistical approaches (e.g. canonical ensemble), the particles of an ideal gas are non-interacting. Still, it's possible to derive the ideal gas law and other thermodynamic relations.

Wikipedia gives an equation for the speed of sound in an ideal gas. How can there be waves in a medium of non-interacting components?

Or do statistical approaches not apply here because we have no thermal equilibrium and we need a kinetic approach?
 

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  • #2
nasu
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The particles in an ideal gas interact when they collide and exchange momentum and energy.
Without collisions there will be no thermal equilibrium.
 
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  • #3
boneh3ad
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Generally, when assuming the particles are non-interacting, this doesn't discount collisions. It just means they are being treated as essentially rigid spherical particles that collide but don't repel or attract each other.
 
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  • #4
256bits
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the particles of an ideal gas are non-interacting
Non-interacting means that behave independent from one another, and exchange momentum only upon collision.

At the top of the wiki,
Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure,[1] as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them
.
 
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  • #5
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The particles in an ideal gas interact when they collide and exchange momentum and energy.
Without collisions there will be no thermal equilibrium.
Generally, when assuming the particles are non-interacting, this doesn't discount collisions. It just means they are being treated as essentially rigid spherical particles that collide but don't repel or attract each other.
Non-interacting means that behave independent from one another, and exchange momentum only upon collision.
But the Hamiltonian used to derive all thermodynamics in the canonical ensemble is purely kinetic:
$$H\left(\left\{\mathbf{q}_i\right\},\left\{\mathbf{p}_i\right\}\right)=\sum_i \frac{\mathbf{p}^2_i}{2m}$$
It's separable, there is no interaction at all.
 
  • #6
vanhees71
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Check vol. X of Landau and Lifshitz, where the Boltzmann equation is derived. Then you'll see, how this comes about: The particles are assumed to move almost always freely and that only from time to time collisions occur, i.e., the mean free path of the particles is assumed to be much smaller than the interaction range. The relaxation time, i.e., the time scale upon which small perturbations of the gas from equilibrium relaxes to equilibrium is parametrically given by
$$\tau_{\text{eq}} \sim \frac{1}{n v_{\text{rel}} \sigma},$$
where ##n## is the particle-number density, ##v_{\text{rel}}## the average relative velocity between the particles, and ##\sigma## the (elastic) scattering cross section for ##2 \rightarrow 2## collisions.

The gas is the better described by hydrodynamics (i.e., local thermal equibrium) the smaller ##\tau_{\text{eq}}## is. So a gas is the better a perfect fluid the larger the cross section is!
 
  • #7
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So my assumptions from the start is correct that one needs to take a kinetic approach and the ideal gas as modeled in the canonical ensemble is not capable of transmitting sound?
 
  • #8
vanhees71
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Sure it is. You have to consider a small perturbation (in density in this case) to get the speed of sound (linear-response theory).
 
  • #9
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Does it mean that, on the limit as the collision cross-section shrinks to zero, sound becomes impossible as the damping of sound diverges?
 

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