In this arxiv paper (pg2) they introduce the concept of mean field scaling, which keeps the overall charge constant. To do this they introduce a factor of $1/N$ in the force term: $$\dot V_i =\frac{1}{N} \sum_{j\ne i} K(X_i-X_j)$$ I am confused with this is not a factor of $1/N^2$ since in the case of the electrostatic potential the force is proportional to to $q^2$ and to keep the overall charge constant, charge must scale as $1/N$ and therefore force as $1/N^2$. Thus my question is why do we have $1/N$ rather then $1/N^2$. (This paper is not the only place I have seen this)


I have also seen contradictory statements such as that in (Cercignani, 1988; pg 58) which says that we want the total force to remain constant, which therefore agrees with the above $1/N$ factor, but does not agree with the charge density remaining constant.

  • $\begingroup$ The mean field scaling corresponds to a scaling in which there are many particles making many collisions (interactions) of low intensity. In order to obtain a meaningful limit the kinetic and potential energy should be of the same order (if else, either you get a free effective theory, or you do not see the free motion anymore). Since the kinetic energy is a "one-particle function" (it is the sum of the $N$ one-particle contributions), while the interaction is a "two-particle function" (it is the sum of $N(N-1)$ contributions), the factor $1/N$ makes them of the same order when $N\to \infty$. $\endgroup$ – yuggib Dec 5 '17 at 14:47
  • $\begingroup$ @yuggib "meaningful limit the kinetic and potential energy" I disagree. The Boltzmann equation has such a limit between its kinetic and potential energies. $\endgroup$ – Quantum spaghettification Dec 5 '17 at 16:43
  • $\begingroup$ I apologize but I do not understand your comment. I am just saying that if the two objects do not scale properly, then you would either get a free effective theory, or a theory were the kinetic part is subleading with respect to the interaction, and thus you do not see it anymore in the effective theory. Such cases may also be interesting, but they do not yield what is called a mean field behavior. $\endgroup$ – yuggib Dec 5 '17 at 17:17

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