Conformal equivalence of resistance

Question

Link to the question in the physics portal.

I'm currently working on a system that uses a logarithmic and a Schwarz-Christoffel transformation to calculate the resistance of a specific area. With resistance I mean the resistance that would apply to that area if it were between two equipotential irregular plates (which are defined by their vertices). In my case it is the magnetic reluctance but it is analogous to electrical resistance.

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I have multiple areas of the canonical domain that have to be translated to a resistance value in the physical domain(with the intermediate step of a logarithm). Is the resistance (defined analogous to electrical resitance) that can be calculated in the homogeneous field of the canonical domain directly related to the resitance in the phyisical domain? The geometry of a resistor in the canonical domain is a simple rectangle formed from four prevertices. In the physical domain it is far more complicated (see figure).

The measure of resistance can be formulated with the help of an "equivalent length". Could it be that this $\Lambda$ is not affected by the conformal tranformations? \begin{align} &R=\rho\dfrac{1}{\Lambda}\\ &\text{cylinder gives: } &\Lambda_\text{cyl}=\frac{2\pi \ell}{\log\frac{r_2}{r_1}}\\ &\text{cuboid gives: } &\Lambda_\text{cub}=\dfrac{A}{\ell} \end{align}

I have done some calculations by discretising the physical domain and comparing it to the resistance generated in the canonical domain. The results were comparable, maybe because of accident, maybe because of a strong mathematical implication, I can't say.

Question in short: Can the prevertices that define a rectangle in the canonical domain define the resistance value of the mapped rectangle in the phyisical domain. (With the intermidate step of a logarithm)

I use the SC-Toolbox for Matlab created by Tobin A. Driscoll.

EDIT: Addition

Lets say that we have a rectangle which has been rotated and translated (Möbius transformations if you want) so that its sides are perpendiculary oriented to the real and imaginary axis, one of its vertices is (0,0) and that it is in the 1st quadrant. We have to calculate its resistance.

Lets say the opposite vertex of (0,0) is $z=x+iy=|z|e^{i\varphi}$, the imaginary part of $z$ represents the length of the path, while the real part represents the projected area(the area is only visible in 3D, we are looking at it in 2D). We write $R'$ because of the 2D representation.

The resistance of this rectangle will therefore be: \begin{align} R'=\rho \frac{\mathfrak{Im}(z)}{\mathfrak{Re}(z)} \end{align} Which can be written as: \begin{align} R'&=\rho \frac{|z|\sin\varphi}{|z|\cos\varphi}=\rho \frac{\sin\varphi}{\cos\varphi}\\ R'&=\rho\tan\varphi \end{align}

The thing that the resistance is only dependent on the angle $\varphi$ and the definition of conformal transformations as angle preserving, provoked me to ask this question. I found a lot about space being conformally invariant, but in this case it is a function that is discussed.

I'm not sure how to proceed. I can prove that the transformations(maps) are conformal, but I don't know what to do about a function in the canonical space.

Answer 1

Given what you wrote and what this wikipedia page says (references can be found there as well), here is, I think, the mathematical interpretation (and answer to) of your question. However, in order to understand it you need to know more math than you do (you need some basic complex analysis, general topology and differential geometry). In short, yes, resistance is conformally invariant.

We consider open connected bounded subsets $\Omega$ of the complex plane ${\mathbb C}$. Fix in the boundary of $\Omega$ in two disjoint subsets $A$ and $B$. I will assume that the sets $A$ and $B$ are closed and connected. The triple $Q=(\Omega, A, B)$ is a condenser. (The physical meaning here is that the domain $\Omega$ is filled with a material that has constant conductivity, which is set to be 1. The unit electric charge is applied along the continuum $B$.)

Example. $\Omega$ is the open rectangle $$ R_{a,b}=\{(x,y): 0<x<a, 0<y<b\}. $$ Set $$ A=\{(0, y): 0\le y\le b\}, B= \{(a, y): 0\le y\le b\} $$ (the vertical sides of the rectangle).

Define the capacity of a condenser
$$ c(Q)=\inf \{E(u)| \ u:\bar{\Omega}\to {\mathbb R}, u|_A=0, u|_B=1\} $$
where we assume that functions $u$ are continuously differentiable in $\Omega$ and are continuous on its closure $\bar\Omega$. The quantity $E(u)$ is the energy: $$ E(u)=\iint_{\Omega} |du|^2 dxdy, $$ where $du$ is the differential of $u$. It is known that infimizing sequences $u_n$ for $c(Q)$ converge (uniformly on compacts in $\Omega$) to harmonic functions $h$ on $\Omega$ so that $$ E(h)=c(Q)= \lim_{n\to \infty} E(u_n). $$ However, the harmonic functions $h$ (in general) do not extend continuously to the closure $\bar\Omega$. Lastly, define the resistence of the condenser as $$ r(Q)= \frac{1}{c(Q)}. $$ This quantity is also know as the extremal length of the family of paths $\gamma: (0,1)\to \Omega$, such that $\gamma(t)$ accumulates to $A$ as $t\to 0$ and accumulates to $B$ as $t\to 1$.

Example. Suppose that $\Omega=R_{a,b}$, a rectangle with the sets $A$ $B$ chosen to be its vertical sides as above. Then $$ r(\Omega, A, B)= \frac{a}{b}. $$ The extreme function in this case is $h(x,y)=x/a$.

It is now well-known that for a general condenser $Q$, the capacity $c(Q)$ is a conformal invariant of $Q$, in the following sense. Suppose that $f: \bar\Omega\to \bar\Omega'$ is a homeomorphism which restricts to a conformal map $f: \Omega\to \Omega'$; define the new capacitor $$ Q'=(\Omega', f(A), f(B)). $$ Then $c(Q)=c(Q')$. The reason for this is the following. One can define energy of functions with respect not only to the Euclidean metric $g_0=dx^2 + dy^2$ but to an arbitrary Riemannian metric $g$ on $\Omega$: $$ E(u,g)= \iint_{\Omega} |d u|_g^2 dA_g, $$ where $dA_g$ is the area form of $g$ and $|du|_g$ is the norm of the differential of $u$ with respect to the Riemannian metric $g$. Then for a function $v: \Omega'\to {\mathbb R}$ and $u=v\circ f$, we clearly have $$ E(u, g)= E(v), $$ where $g=f^*(dx^2 + dy^2)$. Since $f$ is conformal in $\Omega$, the metric $g$ is conformal, i.e. $$ g= \lambda(z) g_0, $$ where $\lambda$ is a positive continuous function on $\Omega$. Lastly, one observes that $$ E(u, g)= E(u) $$ for every function $u$ on $\Omega$, since the $$ dA_g= \lambda^2 dxdy $$ while $$ |du|^2_g= \lambda^{-2} |du|^2. $$ The two scalar functions $\lambda^2, \lambda^{-2}$ cancel each other and integrals are the same since you are integrating equal forms.

Since capacity is conformally invariant, so is reciprocal $r(Q)$.