SIAM News Blog
SIAM News

# Electrical Resistance and Conformal Maps

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 $$1$$.