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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?

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  • $\begingroup$ 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. $\endgroup$ – d_b Dec 27 '18 at 20:06
  • $\begingroup$ @d_b "$V_d(0)$ is constant" was poor word choice. I meant to say "finite" and have edited the post. $\endgroup$ – Endulum Dec 27 '18 at 20:15
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    $\begingroup$ 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. $\endgroup$ – Michael Seifert Dec 27 '18 at 21:25
  • $\begingroup$ @MichaelSeifert Empirically, it seems to only be 1 and 3 dimensions. Interesting link, though! $\endgroup$ – Endulum Dec 28 '18 at 16:01

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