# Find shape of minimum resistance between two points given a fixed volume of conductor

I am given a homogenous volume $$F$$ of isotropic conductor with resistivity $$\rho$$. I need to allow current to flow from Point A to B which are a distance $$L$$ away from each other. I can shape the conductor anyway I like.

Let's assume the conductor lies along the $$x$$-axis. We can assume the conductor has some longish shape and $$L \gg \sqrt{A(x)} , \forall x$$ where $$A(x)$$ is the cross-sectional area at point $$x$$ along the line joining A and B.

What is the shape of the conductor with minimal resistance i.e. what function $$A(x)$$ will give minimize resistance? Update: Assume contact areas will be as big as needed - meaning contact area at $$x=0$$ will be $$A(x=0)$$ and at $$x=L$$ will be $$A(x=L)$$.

I have shown my own answer below, but I'm not happy with it. Primarily because it starts off with assuming the optimum shape (cylinder?). Further, it doesn't help with a different case where, say, there are multiple sources and sinks for current. I'm looking for a more general method.

This makes mathematically no sense, point contacts to a solid medium have infinite resistance because of the $$1/r$$ current density, spreading out from such a point. It has a singularity. Formulated like this you cannot do rigorous mathematics.

Maybe you can change the points to circles, or disks on each side. Then at least a finite result for $$R$$ is possible and there must be a rigorously defined minimum! Still, if the disks are small in size they will dominate the total $$R$$ which would result in a shape more or less like this:

to minimize the spreading resistance. This means that the simplest solution, just trying an ellipsoid and putting it into the Euler-Lagrange equation to see if it is an extremum, will probably not work...

• You're right, I didn't specify contact resistances, but can it not be rescued by having contact areas be $A_{c1}=A(x=0)$ and $A_{c2}=A(x=L)$ ?? Commented Mar 14 at 15:07
• You then prohibit the shape that I've drawn (where The object's area $A(x=0)$ clearly extends beyound the area of the contact disk $A$. But that shape will not be needed if your contact disks automatically "adapt" to the area around them, so that's fine. Commented Mar 18 at 7:02
• Yeah, it will not be needed was my hope... I'm now looking for a better way to prove the optimality of cylinder, or a counterexample. Commented Mar 20 at 9:53
• The cylinder is optimal if the contact disks fit exactly on its ends, making them a bit smaller or larger would force us to deform the cylinder. And after that, using $R=\int\limits_0^L\frac{\rho\,dx}{A(x)}$ is only an approximation, because it is not valid for all possible 3D shapes. Commented Mar 20 at 21:33

I got one (unsatisfactory) way of solving it, which is basically the intuition that because $$R=\frac{\rho l}{A}$$, any reduction in cross-sectional area anywhere (say at $$x=k$$) cannot be compensated by merely adding that area elsewhere (say at $$x=m > k$$) — instead I have to scale the area at $$m$$ by the ratio I decreased it at $$k$$. Further intuition leads me to the obvious idea that a cylinder has to hence yield minimal resistance.

Nevertheless, with this intuition, I set out to prove below that the cylinder is indeed optimal.

A cylinder implies $$A(x)$$ is a constant, value equal to $$\frac{F}{L}$$. The resistance would then be $$R=\frac{\rho L}{F/L} = \frac{\rho L^2}{F}$$. I model the cylinder in the form of thin slices with area $$A(x)$$ and length $$dx$$.

I now reduce the cross-sectional area by $$\Delta A$$ between $$k. I now have spare conducting material of volume $$\Delta A \Delta x$$, which I will deposit between $$m, $$m>k$$. So between $$k$$ and $$k+\Delta x$$ the cross-sectional area is $$A-\Delta A$$, and between $$m$$ and $$m+\Delta x$$ the cross-sectional area is $$A+\Delta A$$.

Resistance is given by $$R=\int\limits_0^L\frac{\rho\,dx}{A(x)}$$ $$= \int\limits_{0}^{k}\frac{\rho\,dx}{F/L} + \int\limits_{k}^{k+\Delta x}\frac{\rho\,dx}{F/L - \Delta A} + \int\limits_{k+\Delta x}^{m}\frac{\rho\,dx}{F/L} + \int\limits_{m}^{m+\Delta x}\frac{\rho\,dx}{F/L + \Delta A} + \int\limits_{m+\Delta x}^L\frac{\rho\,dx}{F/L}$$ $$= \frac{\rho L}{F} \left( \int\limits_{0}^{k}dx + \int\limits_{k+\Delta x}^{m} dx + \int\limits_{m+\Delta x}^L dx \right) + \rho L \left( \int\limits_{k}^{k+\Delta x}\frac{dx}{F - L\Delta A} + \int\limits_{m}^{m+\Delta x}\frac{dx}{F + L\Delta A} \right)$$ $$=\frac{\rho L}{F} [ (k - 0) + (m - (k + \Delta x)) + (L - (m - \Delta x))] + \rho L \left[ \frac{k+ \Delta x - k}{F - L\Delta A} + \frac{m + \Delta x - m}{F + L\Delta A} \right]$$ $$= \frac{\rho L}{F} (L-2\Delta x) + \rho L \left[\frac{\Delta x}{F - L\Delta A} + \frac{\Delta x}{F + L\Delta A}\right]$$ $$= \frac{\rho L^2}{F} + \rho L \left[\frac{\Delta x}{F - L\Delta A} - \frac{2 \Delta x}{F} + \frac{\Delta x}{F + L\Delta A}\right]$$ Since we know $$\frac{\rho L^2}{F}$$ is the original resistance of the cylinder, we need to show that the remainder term $$R_{rem}$$ is positive. $$R_{rem} = \rho L \Delta x \left[\frac{1}{F - L\Delta A} - \frac{1}{F} - \frac{1}{F} + \frac{1}{F + L\Delta A}\right]$$ $$=\rho L \Delta x \left[ \frac{F-F+L\Delta A}{F(F-L\Delta A)} - \frac{F + L\Delta A - F}{F(F+L\Delta A)} \right]$$ $$=\frac{\rho L^2 \Delta x \Delta A}{F} \left[ \frac{1}{F-L\Delta A} - \frac{1}{F+L\Delta A} \right]$$ Since $$F>L\Delta A$$, that implies $$F-L\Delta A > 0$$. Also, $$\frac{1}{F-L\Delta A} > \frac{1}{F+L\Delta A}$$

Therefore $$R_{rem}>0$$, which applied for any $$0 and any $$\Delta x > 0$$ and $$\Delta A > 0$$.

Therefore the cylinder is the shape with the lowest resistance $$R = \frac{\rho L^2}{F}$$. Any change in the shape only increases the resistance.