links for 2007-12-13

I suspect zero is correct as well. Since parallel resisters should halve the resistance, an infinite parallel circuit will rapidly converge on zero creating, essentially, a wire.

Re: precise clocks

NIST F-1 is a frequency standard, not a clock, and NIST doesn't provide the time for GPS. That comes from the US Naval Observatory.

Not an official spokesperson.

I'm no physicist, but this answer from Hit3K at xkcd forum seems to have been accepted:

10. On an infinite, two-dimensional, rectangular lattice of 1-ohm resistors, what is the resistance between two nodes that are a knight's move away?

R[m_, n_] := 1/(2Ï) Integrate[1/t (1 - ((t - I)/(t + I))^(m + n) ((t - 1)/(t + 1))^Abs[m - n]), {t, 0, â}]

R[1, 2]

(8 - Ï)/(2 Ï)

This problem is discussed in J. Cserti's 1999 arXiv preprint. It is also discussed in The Mathematica GuideBook for Symbolics, the forthcoming fourth volume in Michael Trott's GuideBook series, the first two of which were published just last week by Springer-Verlag. The contents for all four GuideBooks, including the two not yet published, are available on the DVD distributed with the first two GuideBooks.

Do you think my googling abilities will be enough to land me a job there?

Numbers are given. They are ideal _1_ ohm resistors.

""In this paper we demonstrate that a sufficiently advanced civilization could, in principal, manipulate the radius of the extra dimension to locally adjust the value of the cosmological constant.""

Are we supposed to take people who don't know the difference between "principle" and "principal" seriously?

Since parallel resisters should halve the resistance, an infinite parallel circuit will rapidly converge on zero creating, essentially, a wire.
That's what I thought at first too, but it turned out the resistors can't always be summed as if they were in parallel. Consider that the current has only four paths out of the starting node, and each of those links has a resistance of 1 ohm. Even if the entire rest of the network had an equivalent resistance of zero, the total resistance couldn't be less than 0.25 ohms. Essentially, much of a given path from start to finish overlaps a whole bunch of other paths, so they aren't in parallel.

For example, if you cut out two rows and the connections between them, I think the resistance between opposite nodes is 1/sqrt(3) ohms instead of zero, even though there are infinitely many paths for current to take. (After seeing the solution Eyal Ben David posted, suddenly I don't feel so badly about not getting the correct answer.)

This problem would actually be easier for an infinite dimensional lattice of resistors. Then, since there is an infinite valence at each node, there really are an infinite number of non-overlapping paths from start to finish, and the resistance would be zero. Conversely, in an infinite one-dimensional line of resistors, the resistance between adjacent resistors would be 1 ohm, I think.

Ahh, I see. Good point. Been a decade since I took electronics, sigh.

Despite the several learned commentators, the resistance is certainly not zero, as an example calculation for a different pair of points will show.

Along this line, my favorite nerd catcher was the American Mathematical Monthly problem that asked (more or less) "what is the minimum number of one ohm resistors that can be assembled to form a circuit with a total resistance of pi ohm to within one part in one million?"