Electrical resistance across the diagonal of a square metal plate

Question

According to Thévenin's theorem, any graph built from wires with specified resistances and possessing a distinguished source and sink will be "equivalent" to a single resistor connecting the source to the sink.

1. What is the best mathematical generalization of Thévenin's theorem? Given a positive smooth function on a Riemannian manifold (which gives the resistance at every point), is there a nice operator which measures the resistance between a pair of points?

2. Given a square metal plate of uniform resistance, how can I compute the measured resistance from one corner to its opposite?

Answer 1

Answer to Question 2: this resistance is infinite. In a vicinity of one corner, any (quarter of) corona delimited by the circles of radii $r$ and $r+\delta r$ has a resistance proportional to $\delta r$ and to the inverse $\frac2{\pi r}$ of its width. Thus the overall resistance involves the integral $$\int_0\frac{dr}r,$$ which diverges.

The answer to Question 1 is essentially the same. The problem is that a point has an infinite capacity: There does not exist a solution of $\Delta V=0$ in a punctered domain $\Omega\setminus(P,Q)$ such that $V(P)=+1$, $V(Q)=-1$ and $V\equiv0$ along $\partial\Omega$. Actually, points are removable singularities in planar domains: suppose that $\Delta f=0$ in $\Omega\setminus(P)$, and that $f$ has a finite "electric energy", which is Dirchlet integral here, $$\int_{\Omega\setminus P}|\nabla f|^2dx<+\infty.$$ Then $f$ extends as a harmonic function in $\Omega$. In particular, the potential $V$ considered above should be harmonic in $\Omega$; then because of the boundary condition and the maximum principle, one should have $V\equiv0$, which contradicts the values at $P$ and $Q$.

Answer 2

I don't know if you can have something as fancy as Thévenin's theorem in a general Riemannian manifold, but as a physicist I'd say you'd be better off looking at generalizations of the conductivity, $\sigma=1/\rho$, where $\rho$ is the resistivity and can be measured as $\rho=R A/L$ for a (homogeneous) wire of cross-section $A$, length $L$, and resistance $R$. Then you can formulate a local version of Ohm's law as $$\mathbf{j}=\sigma\mathbf{E},$$ where $\mathbf{j}$ is the current density, related to the current $I$ flowing through a surface $\Sigma$ as $I=\int_\Sigma \mathbf{j}\cdot\textrm{d}\mathbf{A}$, and $\mathbf{E}$ is the electric field, related to the potential difference $V(\mathbf{x})-V(\mathbf{y})$ between arbitrary points $\mathbf{x},\mathbf{y} $ as the line integral $\Delta V=\int_{\mathbf{x}}^{\mathbf{y}} \mathbf{E}\cdot\textrm{d}\mathbf{r}$ (independent of the integration path).

As for the second question, you'd have to solve Laplace's equation $\nabla^2 V=0$ on a rectangular-box domain $[0,L]\times[0,L]\times[0,h]$, with appropriate boundary conditions - I'm not that sure there, but I'd set $V=0$ on $(0,0,z)$ and $V=V_0$ on $(L,L,z)$ to begin with; if that does not determine a unique solution then Neumann conditions on the rest of the boundary should do it. You then find the electric field $\mathbf{E}=-\nabla V$ to get the current density and integrate across an appropriate spanning surface, say the other diagonal plane.

I'll try and flesh this out and fill this in when I have time.

EDIT J. Cserti's paper Application of the lattice Green's function for calculating the resistance of an infinite networks of resistors, Am. J. Phys. 68 no. 10, pp 896 (2000), doi:10.1119/1.1285881, arXiv:cond-mat/9909120, solves the discretized problem of an infinite network of resistors on a square grid (including as a special case the Nerd Sniping xkcd problem). In the continuum limit that yields a solution to this problem.