I'm sure that many here are familiar with this famous problem that popped up on xkcd one day:

On this infinite grid of ideal one-ohm resistors, what's the equivalent resistance between the two marked nodes?

Nerd Sniping

I am intensely curious, however: what if instead of an infinite grid of ideal one-ohm resistors, we had an infinite lattice? (interestingly, I could not find a single image on google illustrating this) This brings up two interesting questions:

  • What is the equivalent resistance between two nodes that lie on the same plane of the lattice?
  • What is the equivalent resistance between two nodes that do not lie on the same plane of the lattice?

I've searched the site a fair bit, and this question does not seem to have been asked. I don't know anywhere near enough about electrical engineering and circuit physics to even begin tackling this problem, but I would love to see something akin to a solution.

  • $\begingroup$ There's a way to write down the resistance between two points on the 2D grid as an integral. My guess is that when you go to 3D the form of that integral is preserved. If this question is still unanswered tomorrow night I'll try to reconstruct those integrals for you. $\endgroup$ – DanielSank Mar 2 '15 at 5:14
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/10308/2451 and links therein. $\endgroup$ – Qmechanic Mar 2 '15 at 8:08

A good reference was given in an answer to a related question: Cserti 2000 (arXiv preprint, whose numbers I'll be referring to) solved a number of generalizations of the 2D lattice problem.

For a $d$-dimensional lattice, the resistance between the origin and the point $(l_1, \ldots, l_d)$ is given by eq. 18 in that paper: $$ R(l_1, \ldots, l_d) = R_0 \int_{-\pi}^\pi \frac{\mathrm{d}x_1}{2\pi} \cdots \int_{-\pi}^\pi \frac{\mathrm{d}x_d}{2\pi} \left(1 - \mathrm{e}^{\mathrm{i}(l_1x_1+\cdots+l_dx_d)}\right) \left(\sum_{i=1}^d (1 - \cos x_i)\right)^{-1}, $$ where $R_0$ is the resistance of a single resistor.

The author goes on to talk about the specific case of $d = 3$, where in eq. 35 it is shown that $$ R(l_1, l_2, l_3) = R_0 \int_{-\pi}^\pi \frac{\mathrm{d}x_1}{2\pi} \int_{-\pi}^\pi \frac{\mathrm{d}x_2}{2\pi} \int_{-\pi}^\pi \frac{\mathrm{d}x_3}{2\pi} \frac{1-\cos(l_1x_1+l_2x_2+l_3x_3)}{3-\cos x_1-\cos x_2-\cos x_3}. $$ Moreover, as one might expect with enough dimensions, the increase in the number of paths between points can grow fast enough to balance the increase in path length. Indeed, for $d = 3$ the resistance between two points asymptotes to a constant as the points diverge. The paper gives the constant in several forms, one of which is eq. 39: $$ R_\infty = \frac{\sqrt{3}-1}{96\pi^3} \Gamma^2\left(\frac{1}{24}\right) \Gamma^2\left(\frac{11}{24}\right) R_0 \approx 0.505 R_0. $$ Note that this means even if we stick to a single plane in a 3D lattice, the existence of the third dimension drastically alters the qualitative behavior as we move the two nodes apart, since in 2D $R_\infty$ diverges.

  • $\begingroup$ Just to make sure that I am reading this correctly: is the resistance above valid for any two points throughout the lattice? Or only with the distance between them that I posit in the question? $\endgroup$ – user991710 Mar 2 '15 at 15:59
  • $\begingroup$ Yes, the formulas hold for any points (one taken to be the origin). The knight's move case is $(l_1,l_2,l_3) = (0,1,2)$ in any order. $\endgroup$ – user10851 Mar 2 '15 at 17:11
  • $\begingroup$ Quite interesting. Thank you very much for your answer. :) $\endgroup$ – user991710 Mar 2 '15 at 19:33

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