Boltzmann collision term irreversibility

AI Thread Summary
The discussion focuses on the Boltzmann collision term and its role in demonstrating irreversibility in the context of the Boltzmann equation. It emphasizes deriving the H theorem to prove that the collision term is not symmetric under time reversal. To establish this, one should first identify an equilibrium distribution where the collision term equals zero. The conversation suggests showing that deviations from this equilibrium lead to a collision term with an opposite sign, facilitating a return to equilibrium. The analysis also notes the importance of constraining the vector q to ensure only the angle of relative velocity changes during collisions.
lalbatros
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Hello,

Here is the Boltzmann collision term as it is expressed on http://en.wikipedia.org/wiki/Boltzmann_equation" :

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Deriving the H theorem from this equation is the way to usually prove of the irreversibility of this term.
Of course, this collision term should not be symmetric by time reversal.
I guess it is easy to prove that without recourse to the H theorem.
How could I do (see) that?

Thanks,

Michel
 
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First find an equilibruim distribution feq such that the collision term is zero. Next show that the sign of the collision term is the opposite of any deviation of f from feq so that f restores back to feq.
To start off, show that the collision term is zero for feq~exp[-(p-p0)2]. Where p0 is a constant average momentum of the gas (in dimensionless units scaled by sqrt(2 m Eavg) ). p0could be set to zero for simplicity); m is mass of the particles.
Keep in mind that the vector q must be constrained so that only the angle of the relative velocity changes during a collision.
 
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