Yes, that's a good symmetry argument. I would not connect infinity to ground because in 1 or 2 dimensions, the potential drop from the injection node to infinity is infinite. Instead ground the injection node, and all is well. -- Gene ________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 11:30:05 AM Subject: Re: [math-fun] The people you know... Connect the boundary of a large hypercubic resistor grid to "ground". Ground acts as a current source at zero potential. Now apply a voltage V to the node at the origin such that 1 amp is extracted there. By symmetry, the current flowing in each of the resistors connected to the origin node is 1/(2D) amp, where D is the dimension. The potential across these resistors is 1/(2D) volt since their resistance is 1 ohm. Now consider what happens when instead, voltage -V is applied to a neighboring node, say [1,0]. We have the same symmetrical distribution of currents, only the center is shifted and the current directions are reversed. On the other hand, the voltage across the resistor R connecting [0,0] to [1,0] is the same (in magnitude and sign) as it was before. Now superimpose these two current/voltage distributions. The voltage across R will be 2/(2D) and exactly one amp will flow out of [0,0] and into [1,0]. The equivalent resistance is therefore 1/D. Veit On Jan 28, 2011, at 1:51 PM, Eugene Salamin wrote:
I don't see the symmetry argument. Consider the 1-dimensional case; then R[1,0]
= 1, not 1/2. Additionally, I would be wary of injecting current into one node
without also extracting an equal current. Current must flow out to infinity.
In a 2 or more dimensional lattice, how can you be sure you have a unique solution without imposing boundary conditions at infinity?
-- Gene
________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 10:29:04 AM Subject: Re: [math-fun] The people you know...
R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2
R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580
R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789
You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2.
Veit
On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
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