By Mark Levi
Figure 1. Resistance—i.e., the necessary voltage to push through a unit of current—is measured between opposite sides (coated with a perfect conductor).
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).
Figure 2. If \(R=1\), the rectangle is a square.
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:
\[R (A)= R (A ^\prime). \tag2\]
Figure 3. Two annuli are conformally equivalent if and only if their radii have the same ratios.
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 \(n \rightarrow \infty\).
Figure 4. Two doubly connected regions are conformally equivalent precisely when they have the same resistance between their inner and outer boundaries.
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,
\[R( A)= \frac{n}{m},\]
Figure 5. Subdivision of \(A\) into infinitesimal squares \(Q_{ij}\). Concentric “horizontal” lines \(h_i\) are equipotentials. Steepest descent “vertical” current lines \(v_j\) are added in a counterclockwise direction until the last line \(v_m\). The “square” \(Q_{ij}\) is bounded by \(h_{i-1}, h_{i}\) and \(v_{j-1}, \ v_j\).
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.
Mark Levi ([email protected]) is a professor of mathematics at the Pennsylvania State University.