# Debye screening in $\mathbb{R^d}$

Consider the Poisson-Boltzmann equation $$\nabla^2 V(r) = -\frac{1}{\epsilon_0}en\left(1 - e^{e V(r)/k_BT}\right)$$ which models the electrostatic potential in a spherically symmetric ideal gas of classical electrons embedded in a uniform neutralizing background with number density $$n$$. This equation is often used to introduce the concept of Debye screening by linearizing the electron density $$\nabla^2 V\approx -\lambda^{-2} V$$ where $$\lambda=\sqrt{\epsilon_0k_BT/e^2n}$$ is the screening length. Most textbooks (in plasma physics, at least) only consider the solution in $$\mathbb R^3$$, with the boundary condition $$V(\infty)=0$$, which is $$V_3(r) = c_3 e^{-r/\lambda}/r$$, with $$c_3$$ a constant. I am interested in the solution in $$\mathbb R^d$$, which has the form $$V_d(r) = c_d \frac{K_{\frac{d}{2}-1}(r/\lambda)}{r^{\frac{d}{2}-1}}$$ in which $$K_\nu(x)$$ is the modified Bessel function of the second kind. Playing around in Mathematica, I made a few curious observations that I'd like to solicit commentary on.

1. As $$d\to\infty$$, $$V_d(r)$$ takes the form of an infinite step. That is, in very high dimension the Debye-screened particles behave as if they were hard (hyper)spheres with diameter $$\lambda$$. I wonder if there is a geometric understanding of this by way of Gauss's Law? I have a hunch that there is a qualitative argument that can be made by comparing the surface area of a Gaussian hypersphere to its volume, but I can't articulate it.

2. For $$d<2$$ (including fractional $$d$$), $$V_d(0)$$ is finite. I take this to mean that in low dimension, like charges can overlap, since $$n(r)=n\exp[eV_d(r)/k_BT]$$. However, this seems in conflict with the expected physics of a collection of charges in, e.g., $$d=1$$. By Gauss's Law, the electric field of each electron is $$E=-e/2\epsilon_0$$, which provides a constant and repulsive force. What allows the electrons to pile up near each other despite this?

3. Regardless of the dimension, the asymptotic behavior as $$r\to\infty$$ is $$V_d(r)\sim c_d \sqrt{\frac{\pi}{2}}\frac{e^{-r/\lambda}}{r^{\frac{d-1}{2}}}$$ meaning that exponential screening is universal in some sense.

4. The above asymptotic solution coincides exactly with the full solution for $$d=1$$ and $$d=3$$. Is this a coincidence or is there something meaningful about this fact?

• The "infinite-step" behavior shows up for unscreened charges as well. As you've noted this has to do with how the surface area of a sphere grows with $d$. I'm confused about your point 2. Are you saying that $V_d(0)$ is constant in $d$ for $d<2$? I don't observe this playing around in Mathematica. – d_b Dec 27 '18 at 20:06
• @d_b "$V_d(0)$ is constant" was poor word choice. I meant to say "finite" and have edited the post. – Endulum Dec 27 '18 at 20:15
• Is the asymptotic solution exact for other odd values of $n$ as well? If so, it may be related to the fact that Huygen's principle only holds in odd dimensions; see here. – Michael Seifert Dec 27 '18 at 21:25
• @MichaelSeifert Empirically, it seems to only be 1 and 3 dimensions. Interesting link, though! – Endulum Dec 28 '18 at 16:01