In discussions of the Vlasov equation it is often said (e.g. Cercignani, 1988; pg59) that we require rarefied gas with weak, long range interactions. I understand why we need weak interactions (since we take a scaling of $1/N$ for the force) but I can't see why we need the interaction to be long range.
Does anyone know why this is so? (a source would be helpful).
Answer
I understand why we need weak interactions (since we take a scaling of 1/N for the force) but I can't see why we need the interaction to be long range.
It is one of the assumptions required to create a limit where the Boltzmann collision operator goes to zero. More importantly, the physical reason they state as much is because the fluids handled by the Vlasov equation are typically ionized gases, namely plasmas. In a plasma, the particles are charged and respond to the collective fields of all the charged particles out to infinity (in principle/theory, but generally Debye shielding and quasi-neutrality keep the interactions more local).
Does anyone know why this is so?
The only real difference between the Boltzmann and Vlasov equations are the lack of a collision operator in the latter, namely, that the latter is considered time-reversible. Using things like ensemble averages (e.g., mean field theory) can change the Vlasov equation from a time-reversible to a time-irreversible equation, but that is an additional nuance.
In discussions of the Vlasov equation it is often said (e.g. Cercignani, 1988; pg59) that we require rarefied gas with weak, long range interactions.
If the particle-particle interactions (e.g., Coulomb collisions) are strong, then the Vlasov equation is no longer the appropriate approximation to use and requires a collision operator.
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