## Saturday, January 1, 2011

### How I wasted New Year's Eve

Apparently I am blessed with the kind of brain susceptible to nerd sniping (by the way, the answer to that problem is $4/\pi-1/2$; see [Cserti], an elegant approach using discrete Green functions). On New Year's Eve, an ill-considered click brought this problem before my eyes:
Both ends of a thin flexible rod are joined to the same pivot. What is the angle between the rod's ends?
I had not used much of the mathematical technique learned in the university for ten years. I don't like writing. I have other, arguably more important things to do. Oh well.

This problem seems made for a demonstration of variational calculus. Accordingly, I need an expression for the potential energy of a bent rod to vary. The rod being thin, I can apply simple bending theory, which says that the flexural energy of a piece of rod is proportional to the square of its curvature. The shape the rod makes does not depend on the rod's material or length as long as the assumptions of simple bending theory hold. Also the shape will obviously be symmetrical about the line bisecting my angle of interest (I will make this line my $y$ axis). A final observation is that the rod's curvature at the pivot is zero, because the pivot rotates freely and nonzero curvature generates momentum. At this point physics ends and mathematics, as applied by physicists, begins.

Let $\theta$ be the angle between the tangent to the curve and the $x$ axis, and $s$ the natural parameter (i.e. arc length). Using the Frenet's formula $\dot{\bf{t}}=\kappa\bf{n}$ ($\bf{t}$ the tangent unit vector, $\bf{n}$ the normal unit vector, $\kappa$ the curvature), I obtain that $\kappa=\dot{\theta}$, so the main term to vary is $\dot\theta^2$. In addition, I have to constrain the curve's ends to meet; the curve being symmetrical about the $y$ axis, I worry only about the separation of the curve's ends along the $x$ axis: $\int\cos\theta ds=0$. Accordingly I shall vary $$\tag{*}{(*)}\int_{-1}^{1}\dot\phi^2+½\mu^2\cos2\phi ds$$ where I have introduced $\phi=\theta/2$, arbitrarily made the curve's length equal $2$ and where $\mu$ is the Lagrange multiplier. Equating the variation of (*) to zero yields the ODE for $\phi$: $$\dot\phi^2=\mu^2k^{-2}(1-k^2\sin^2\phi)$$ with $k$ a constant of integration. The curve's symmetry and my choice of axes give me the boundary condition $\phi(0)=0$, and I denote $\phi(-1)=\alpha$. The solution to this ODE is the elliptic integral of the first kind: $\mu s=k\,\mathrm{F}(\phi,k)$. Since the curvature at $s=\pm1$ is zero, $k=1/\sin\alpha$. Now I bring into play the end-meeting constraint: $$0=\int_{-1}^{1}\cos2\phi ds=(\mu/k)\int_{-\mu/k}^{\mu/k}2\mathrm{cn}^2u-1 du,$$ whence G&R 5.134.2 produces $$\mathrm{E}(\alpha,k)=(1-½k^2)\,\mathrm{F}(\alpha,k).$$ At this point I have enough equations to solve for $\alpha$ and need not bother about $\mu$ — not surprising considering that $\mu$ is really a dimensional quantity which arises because I fixed the length of the curve. The solution is implicitly determined by $$2\sin^2\alpha\,\mathrm{E}(\alpha,1/\sin\alpha)=(2\sin^2\alpha-1)\,\mathrm{F}(\alpha,1/\sin\alpha).$$ Mathematica fails to find $\alpha$ numerically from this equation, however, so I solved the two equations together to find $\alpha\approx65°21'$. $\alpha$ being one half of the tangent angle to one end of the rod, the angle between the rod's ends is $4\alpha-\pi$, approximately $81°25'$. ■