= Solution
A curve element has <arc length> $ds=\sqrt{1+z'^2}\,dx$ and mass $dm=\mu ds$. Adding its gravitational energies gives
$$
\boxed{V[z]=\mu\int_{-a/\sqrt2}^{a/\sqrt2}\Phi(z)\sqrt{1+z'^2}\,dx.}
$$
For the fixed-length constraint, use the augmented integrand $(\mu\Phi(z)-\lambda)\sqrt{1+z'^2}$. It has no explicit $x$ dependence, so the <Beltrami identity> gives
$$
\boxed{\frac{\mu\Phi(z)-\lambda}{\sqrt{1+z'^2}}=C.}
$$
The desired lower circular arc is
$$
z(x)=h-\sqrt{a^2-x^2},\qquad h=\frac a{\sqrt2}.
$$
It joins the required endpoints, subtends angle $\pi/2$, and has length $\pi a/2$. Its slope gives $\sqrt{1+z'^2}=a/(h-z)$. Therefore a <gravitational potential producing a circular hanging chain> is
$$
\boxed{\Phi(z)=\Phi_0+\frac{\kappa}{a/\sqrt2-z},\qquad\kappa>0,\quad z<a/\sqrt2.}
$$
Choosing $\lambda=\mu\Phi_0$ makes the first <integral> constant, $C=\mu\kappa/a$. Its <gravitational acceleration> is $-\Phi'(z)=-\kappa/(h-z)^2$, which points downward. An arbitrary additive potential constant changes no force.
This construction really gives a minimum. With $w=h-z>0$, the variable part of $V$ is $\mu\kappa\int ds/w$; the $\Phi_0$ term is fixed by the length. The <calibration for semicircular hyperbolic geodesics> uses $S(x,w)=\operatorname{arsinh}(x/w)$, whose <gradient> has <norm> $1/w$. The <Cauchy-Schwarz inequality> gives $\int ds/w\geq S(B)-S(A)$, with equality on $x^2+w^2=a^2$ traversed left to right. Thus the circular arc minimizes the energy among competitors of the same length and endpoints in the stated potential domain, rather than merely satisfying a necessary stationarity condition.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-q9-circular-chain.png]
{title=A circular hanging chain and the inverse-height gravitational potential that supports it}
Back to article page
