# Electrical Resistance and Conformal Maps

I would like to elaborate on the observation in my April 2023 article, titled “Conformal Deformation of Conductors.” Imagine a current-conducting sheet: negligibly thin, homogeneous, and isotropic. Let us cut a square out of the sheet and measure the resistance, as in Figure 1. The following fact is both fundamental and almost trivial:

\[\boxed{ \begin{split} & \ \ \ \ \hbox{Squares of all sizes } \\ & \hbox{have the same resistance.} \end{split}} \tag1\]

Indeed, dilation of the square changes the distance that the current must travel, and by the same factor as the width; these two effects cancel each other out — but *only *in \(\mathbb{R} ^2\). In \(\mathbb{R} ^3\), for example, dilating a cube by a factor \(\lambda\) divides the resistance (between the opposite faces) by \(\lambda\), and in \(\mathbb{R}\) the resistance multiplies by \(\lambda\).

From now on, let the resistance of the square \(=1\) ohm. Geometrically, resistance is a *measure of elongation*: a rectangle whose resistance \(=1\) must then be a square (see Figure 2).

A classical theorem in complex analysis states that two annuli (see Figure 3) are conformally equivalent—i.e., they can be mapped onto one another by a conformal \(1-1\) map—if and only if they have the same ratio of radii. A more general theorem states that two doubly connected “annuli” (like those in Figure 4) are conformally equivalent if they have the same modulus: a certain number that is associated with the region. I would like to point out that that the modulus is simply the electrical resistance.

To rephrase these theorems: *Two annular regions \(A\) and \(A ^\prime\) (as in Figure 4) are conformally equivalent \((A \sim A ^\prime)\) if and only if they have the same electrical resistance between their inner and outer boundaries:*

\[R (A)= R (A ^\prime). \tag2\]

#### Idea of the Proof

In order to construct a conformal map \(A \leftrightarrow A ^\prime\), let us push the current by applying voltages \(V=0\) to the inner boundary and \(V=1\) to the outer boundary.^{1} For a large integer \(n\), consider the equipotential lines \(h_i\), \(i=0, \ldots n\) that are spaced by the potential difference \(1/n\) (see Figure 5); \(h_0\) is the inner boundary and \(h_n\) is the outer boundary. Fix an arbitrary line \(v_0\) of steepest descent of the electrostatic potential—the line of current—and let \(v_1\) be the line of steepest descent that is chosen so that the current through the channel \(v_0 v_1\) is \(1/n\). Continue adding current lines \(v_j\), as in Figure 5, and stop at \(j=m\) when the current through the channel \(v_mv_0\) becomes \(< 1/n\). This last channel plays no role in the limit of \(n \rightarrow \infty\).

We divided the annulus into \(n \times m\) infinitesimal curvilinear rectangles \(Q_{ij}\), which we enumerate by the rectangle’s layer \(i\), \(1\leq i\leq n\) and the channel \(j\), \(1\leq j\leq m\) (see Figure 5).

I claim that each curvilinear rectangle \(Q_{ij}\) *is a square in the limit of *\(n \rightarrow \infty\). Indeed, the resistance is

\[R (Q_{ij}) = \frac{{\rm voltage\ drop}}{\rm current} = \frac{1/n}{1/n} =1,\]

and a rectangle for which resistance \(=1\) is a square (as indicated in Figure 2).

What is the resistance of \(A\)? Each channel has resistance \(n\) (being a stack of \(n\) squares), and with \(m\) channels in parallel,

\[R( A)= \frac{n}{m},\]

ignoring a small error due to the resistance of the last channel \(v_mv_0\). The resistance therefore has an almost combinatorial meaning.

To construct the map \(A\leftrightarrow A ^\prime\), we divide \(A ^\prime\) into \(n \times m ^\prime\) squares \(Q ^\prime_{ij}\). If \(R(A)= R(A ^\prime)\), then \(m = m ^\prime\); this allows a \(1-1\) assignment of \(Q ^\prime_{ij}\) to \(Q_{ij}\). The result is a discrete conformal map since it takes squares to squares.

#### Showing the Converse

\(A\sim A ^\prime\) implies \(R(A)=R(A ^\prime)\). We divide \(A\) into “squares” \(Q_{ij}\) as before, with \(1\leq i\leq n\) and \(1\leq j \leq m\). The conformal equivalence induces a division of \(A ^\prime\) into “squares” (by conformality) with the same \(m ^\prime = m\) (since the map is \(1-1\)). Therefore, \(n/m = n/m ^\prime\) and \(R(A) = R(A ^\prime)\). In short, \((1)\) demonstrates that* the resistance is a conformal invariant*, as was already mentioned in my April 2023 article.

^{1} By doing so, we consider the solution of the Dirichlet problem in the annulus with prescribed boundary values \(0\) and \(1\).

*The figures in this article were provided by the author.*

### About the Author

#### Mark Levi

##### Professor, Pennsylvania State University

Mark Levi ([email protected]) is a professor of mathematics at the Pennsylvania State University.

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