I kT in systems other than ideal gas

  • Thread starter DaTario
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Is kT still meaningful in other general thermodynamic samples (those with a huge number of elements)?
Hi All,

When dealing with the kinetic theory of gases in thermodynamics, we obtain the result that the mean kinetic energy per atom is (3/2) kT. In considering different samples like 200g of liquid water or a solid cube of lead with one cubic meter, does kT still play an important role in revealing some statistical property of these samples in thermal equilibrium?

Best wishes,

DaTario
 

hilbert2

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It doesn't have as easy an interpretation for systems with interacting components, but the idea is that when considering energy values several times higher than ##kT##, it becomes unlikely to find that amount of energy in a microscopic degree of freedom of the system.

The equation ##E = \frac{3kT}{2}## for the energy per constituent particle only applies for point-particle ideal gases. If the gas molecules have a nonzero length, there is energy also in the rotational motion.
 
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Mister T

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When dealing with the kinetic theory of gases in thermodynamics, we obtain the result that the mean kinetic energy per atom is (3/2) kT.
That's for a monatomic ideal gas. The energy associated with each degree of freedom is ##\frac{1}{2}kT##.
 

vanhees71

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More precisely: The mean energy associated with each phase-space degree of freedom entering the Hamiltonian quadratically is ## k T/2##. This holds true in the strict sense only for a gas of non-interacting particles (or rather where the mean free path of each gas molecule is very large compared to the typical duration of a collision, i.e., in the dilute-gas limit) or for particles being bound by forces ##\propto## their distance, i.e., when you can describe the system by harmonic oscillator modes. Each mode provides 3 such degrees of freedom to the kinetic energy ##\propto p^2## and 3 to the potential energy ##\propto x^2##. That can be used to derive the Dulong-Petit law that the speficic heat of a solid is ##6 N k## or the molar heat ##6 N_{\text{A}} k=6R##.

Note that this holds only in the classical limit. At low temperatures, when quantum statistics becomes relevant, it's not valid anymore.
 
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Thanks, vanhees71.
Now, just for the sake of clarity, in the parenthesis quoted bellow, how are we to handle this comparison between mean free path (a length) and the typical duration of collisions (a time interval)? Is it correct to say that the mean velocity is implicitly evoked here?

Best wishes.

(or rather where the mean free path of each gas molecule is very large compared to the typical duration of a collision, i.e., in the dilute-gas limit)
 

DrClaude

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Now, just for the sake of clarity, in the parenthesis quoted bellow, how are we to handle this comparison between mean free path (a length) and the typical duration of collisions (a time interval)? Is it correct to say that the mean velocity is implicitly evoked here?
Use the mean time between collisions for the former.
 

vanhees71

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Basically it's saying that the Boltzmann equation is valid, if the collision time is large in comparison to the time it takes for the "particles to become on-shell", i.e., the collision can be described well by assuming that the particles are in asymptotic free in-states before the collision and also reach the asymptotic free out-states after the collision. Then and only then a particle interpretation and the use of scattering cross sections rather than off-equilibrium transition rates make sense.

An example, where quantum-coherence effects are important is the Landau-Pomeranchuk Migdal effect for Bremsstrahlung. Then you have to coherently sum up multi-particle scattering events to get a finite infrared limit for the Bremsstrahlung. That's most intuitively seen in a approach using the Langevin equation, as detailed in

https://arxiv.org/abs/hep-ph/9510417v1
 
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Thank you, vanhees71 and DrClaude.

A last question concerning vanhees71's observation in #4:

Is the case where particles are bound by forces proportional to their distance, i.e., "when you can describe the system by harmonic oscillator modes" an exception? Is it an exceptional case where Boltzmann constant still can be associated with measurable thermodynamic parameters of the system?
 

vanhees71

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Yes. Then you can map the corresponding lattice vibrations to socalled quasiparticles, which are called phonons, which is because these lattice vibrations are nothing else than sound waves. This is anyway an important finding by L. D. Landau: Often you can describe a condensed matter system effectively by a (dilute) gas of quasiparticles, which are quantized descriptions of collective modes of the many-body system.
 

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