Imagine that we have a potential in the form of: $U(r) \propto \frac{1}{r^n}$ in a 2D system, with the high concentration of particles interacting with the above potential. How do you find the screening length of this system? I am asking this question because I need to consider a cut-off radius for a simulation and considering such a cut-off radius could enhance my simulation speed a lot. I want to make sure that I am choosing the right value for that.



Two hints: Fourier transform and linearized Poisson-Boltzmann equation.

Linearizing a Poisson-Boltzmann type equation (typically expressed in terms of different types of charges in solution, here only using one such term; $e$: elementary charge, $c_0$: concentration constant, $\epsilon$: relative permittivity, $\epsilon_0$: permittivity of vacuum, $k_{\rm B}$: Boltzmann constant, and $T$: temperature): $$ \nabla^2 \phi = \frac{c_0\,e}{\epsilon\,\epsilon_0}\exp\left(-\frac{e\phi}{k_{\rm B}T}\right) \approx \frac{c_0\,e}{\epsilon\,\epsilon_0} \left( 1 - \frac{e\phi}{k_{\rm B}T} \right), $$ the screening term can be identified as the additional term applied to the potential $\phi$ besides $\nabla^2$: $$ \left( \nabla^2 + \frac{c_0\,e^2}{\epsilon\,\epsilon_0\,k_{\rm B}T} \right) \phi = \frac{c_0\,e}{\epsilon\,\epsilon_0}. $$

Fourier transformation of $U_{\rm screened}(r) \propto r^{-n} \exp(-k\,r)$ will allow for determining $k$ by comparison to the above equation. In 3D with the usual Coulomb interaction $\propto\!r^{-1}$, e.g., one would have for the Fourier transform: $\propto\!q^{-2} + k^{-2}$ (where $q$ is the wave vector and $k$ is the inverse screening length). $q^{-2}$ can be identified with $\nabla^2$ in Fourier space, and $$ k^{-1}_{{\rm 3D},n=1} \propto \sqrt{\frac{c_0\,e^2}{\epsilon\,\epsilon_0\,k_{\rm B}T}} . $$

For 2D, Fourier transformation of $U_{\rm screened}(r)=r^{-n}\exp(-kr)$ yields the following: $$ \int_0^{2\pi} {\rm d}\psi \int_0^\infty {\rm d}r \exp[iqr\cos(\psi)] \, r \, U_{\rm screened}(r) = 2\pi \int_0^\infty J_0(qr)\,r\,U_{\rm screened}(r) \, {\rm d}r = \frac{2\pi}{\sqrt{q^2+k^2}} , $$ for $n\ge 1$. $J_0$ is the Bessel function of first kind and order zero. Due to the square root, unfortunately no direct comparison to the linearized Poisson-Boltzmann equation can be made. For $q\rightarrow 0$ (corresponding to long wavelengths in real space), $\frac{1}{\sqrt{q^2+k^2}}\rightarrow k^{-1}$ and thus $$ k^{-1}_{\rm 2D,longrange} \approx \frac{c_0\,e^2}{2\pi\,\epsilon\,\epsilon_0\,k_{\rm B}T} , $$ interestingly not dependending on the value of $n\ge 1$.


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