I have seen in Long/short-range interaction that in scattering theory $𝑟^{−𝑛}$ is a short range potential for $n>1$ and a long range potential for $n\leq1$.

Now, why do we say that van-der-Waals and dipole-dipole are long range if they have the form $𝑟^{−𝑛}$ with $n=3,6$ respectively?

  • $\begingroup$ But $1/r^n$ is long range. It has scale symmetry. $\endgroup$ Jan 23 '20 at 20:21
  • 1
    $\begingroup$ "Short" and "long" are conventional and inexact terms (and different conventions are used in different areas). In this case, the forces are long-range because they (approximately) obey a power law: short-range would be exponentially decreasing forces. $\endgroup$
    – NickD
    Jan 23 '20 at 20:23
  • $\begingroup$ Yes, not very clear. Thanks $\endgroup$
    – Kio
    Jan 23 '20 at 22:46
  • What constitutes a long range as opposed to short range interaction depends primarily on the specific problem under investigation.

  • One finds many definitions in the literature; usually criteria can be expressed through the 2-body interaction potential $V(r)$.

  • $V(r) \propto r^{-\alpha}$, with $\alpha \leq$ dimensions is considered a standard definition for long-range interactions. Dipolar interactions in a Bose-Einstein condensates are long-range by this definition.

  • Also, if $V(r)$ falls off slower than exponentially, correlations are then qualitatively different and are often considered as long range in some contexts. Van der Waals interactions fall into this category.


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