Harald Kirsch

genug Unfug.

2013-09-18

Brachistochrone (II.1, p.122)*

Contents

Exercise

The brachistochrone problem asks for the path to be taken by a bead on a wire from a higher position A to a lower position B such that the time is minimized. To keep things simple, the gravitational force is kept constant on the whole path and friction is ignored.

Zee has a solution in his book, but it is too terse for me to remember all the intermediate steps. Therefore I write them down here.

Idea

The idea is to use the Euler-Lagrange equations to find the path that minimizes the time. The notation used will be the one I used already in the solution to the shortest path problem.

Solution

We describe the path taken by the bead with the curve \begin{equation} K(r) = \begin{pmatrix}x(r) \\ y(r)\end{pmatrix} \in \mathbb{R}^2 \end{equation} with $r\in\mathbb{R}$ a free parameter. By $y$ we describe the vertical coordinate starting at point $A$ and pointing downwards. At a point $K(r)$, the bead has converted the potential energy of $m\, g \, y(r)$ into kinetic energy. Denoting the speed at $K(r)$ with $v(r)$, we get (in nonrelativistic terms) \begin{equation} m\, g \, y(r) = \frac{1}{2} m\, |v(r)|^2\,. \end{equation} Solving for the absolute value of the speed, we get \begin{equation} |v(r)| = \sqrt{2g\,y(r)}\,. \end{equation} To derive the length of the curve, it is sufficient to integrate the speed vector over time. To do so, we assume we know how our parameter $r$ depends on time $t$ to get \begin{equation} \int_0^{t_0} \sqrt{ \left(\frac{dx(r(t))}{dt}\right)^2 + \left(\frac{dy(r(t))}{dt}\right)^2 } dt \,. \end{equation} If we divide the integrand pointwise by the absolute value of the speed at this point, the resulting dimension is time and indeed we then integrate differential time to arrive at the total time. \begin{equation} \int_0^{t_0} \sqrt{\frac{1}{2g\,y(r(t))} \left( \left(\frac{dx(r(t))}{dt}\right)^2 + \left(\frac{dy(r(t))}{dt}\right)^2 \right) } dt \,. \end{equation} Now we replace $dt = \frac{dr}{r'(t)}$ and after canceling an $r'(t)$ in numerator and denominator we arrive at \begin{equation} \int_0^{r_0} \sqrt{\frac{1}{2g\,y(r)} \left( \left(\frac{dx(r)}{dr}\right)^2 + \left(\frac{dy(r)}{dr}\right)^2 \right) } dr \,. \end{equation} We further choose the parameter $r$ to actually be the $y$ coordinate such that we have the identities $y(r) = r$ and $\frac{dy(r)}{dr}=1$. This simplifies the previous equation further to \begin{equation} \int_0^{y_0} \sqrt{\frac{1}{2g\,y} \left( \left(\frac{dx}{dy}\right)^2 + 1 \right) } dy \,, \end{equation} where we additionall switched variables from $r$ to $y$.

Choosing the $y$ coordinate to parameterize the curve is what Zee proposes in his solution to the exercise as the simpler way. We could have tried $x$ as well, but indeed this seems to be much more complicated to solve.