Physics Friday 2

Infinite Resistor Grid

This problem was inspired by this XKCD comc:

Upon reading it, my first thought was that we could approach the problem with a discrete-space Fourier transform (as used in digital image processing theory), which would move from the infinite discrete space of the grid points of the array to the frequency domain (not actually “frequency”, but technically wavenumber as we are transforming space and not time coordinates, though I will still use 'frequency'), which would be continuous and finite:
If we have grid coordinates m,n and corresponding frequency coordinates u,v; then for a property of the grid nodes X_{m,n}, we have:
\tilde{X}(u,v)=\sum_{m=-\infty}^\infty\,\sum_{n=-\infty}^\infty\,X_{m,n}\cdot e^{-2\pi i(mu+nv)}
and the inverse transform:
X_{m,n}=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\tilde{X}(u,v)\cdot e^{2\pi i(mu+nv)}\,du\,dv

Consider a current I_{m,n} injected (or removed) at point (m,n). By Kirchoff’s first law (conservation of electric charge), I_{m,n} must equal the sum of the currents flowing out through the four resistors connecting to that node. Using Ohm’s Law V=IR and letting V_{m,n} be the potential at node (m,n) we find those currents are V_{m,n}-V_{m+1,n}, V_{m,n}-V_{m-1,n}, V_{m,n}-V_{m,n+1}, V_{m,n}-V_{m,n-1} (as R=1 Ω). Summing these,
I_{m,n}=4V_{m,n}-V_{m+1,n}-V_{m-1,n}-V_{m,n+1}-V_{m,n-1}

Using the inverse transform definition,
\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\tilde{I}(u,v)\cdot e^{2\pi i(mu+nv)}\,du\,dv
=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\tilde{V}(u,v)\cdot e^{2\pi i(mu+nv)}[4-e^{2\pi iu}-e^{-2\pi iu}-e^{2\pi in}-e^{-2\pi in}]\,du\,dv

For this to be true for any point (or frequencies), the integrands must be equal, and so:
\tilde{I}(u,v)=\tilde{V}(u,v)[4-e^{2\pi iu}-e^{-2\pi iu}-e^{2\pi in}-e^{-2\pi in}]
\tilde{I}(u,v)=2\tilde{V}(u,v)[2-\cos(2\pi u)-\cos(2\pi v)]

Putting in a current source of 1 A at the origin and a sink of -1 A at the point (M,N) gives us an I_{m,n}=\delta_{m,n}-\delta_{m-M,n-N}. Note that the equivalent resistance between (0,0) and (M,N) is equal to V_{0,0}-V_{M,N} for this current load. (V=I\cdot{R}_{eq}).

Transforming I_{m,n}=\delta_{m,n}-\delta_{m-M,n-N} to frequency,
\tilde{I}(u,v)=\sum_{m=-\infty}^\infty\,\sum_{n=-\infty}^\infty\,I_{m,n}\cdot e^{-2\pi i(mu+nv)}
=e^{-2\pi i(0\cdot u+0\cdot v)}-e^{-2\pi i(Mu+Nv)}
=1-e^{-2\pi i(Mu+Nv)}

Thus \tilde{V}(u,v)=\frac{1}{2}\frac{1-e^{-2\pi i(Mu+Nv)}}{2-\cos(2\pi u)-\cos(2\pi v)}

Using the inverse transform,
R_{eq}=V_{0,0}-V_{M,N}
=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\tilde{V}(u,v)\left[1-e^{2\pi i(Mu+Nv)}\right]\,du\,dv
=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\frac{1}{2}\frac{\left(1-e^{2\pi i(Mu+Nv)}\right)\left(1-e^{-2\pi i(Mu+Nv)}\right)}{2-\cos(2\pi u)-\cos(2\pi v)}\,du\,dv

Now, (1-e^{ix})(1-e^{-ix})=2-2\cos{x}, thus
R_{eq}=\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\frac{\left(1-\cos[2\pi(Mu+Nv)]\right)}{2-\cos(2\pi u)-\cos(2\pi v)}\,du\,dv

This gives our resistance as a (real) definite integral.

For the example in the comic, M=2, N=1, and with a lot of mathematical work, we can compute the integral to find R_{eq}=\frac{4}{\pi}-\frac{1}{2}, which is approximately 0.77324 Ω

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One Response to “Physics Friday 2”

  1. Roberto Miguez Says:

    Absolutely awesome. I’m really enjoying these articles. Great stuff.

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