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The Brachistochrone

Source lecture(s): PHY622 Ch. 1

Intuition

Brachistos + chronos: shortest time. Given two points, what shape of frictionless wire gets a bead from one to the other fastest? The straight line is the shortest path and it is not the answer — it is worth pausing on why that is not a paradox. Minimising length and minimising time are different variational problems, and the curve that wins is the one that trades a steeper initial drop (buying speed early) against a longer path.

Setting up the functional

With the bead released from rest at the origin and \(y\) measured downward, energy conservation gives \(v = \sqrt{2gy}\). The time is

\[T[y] = \int \frac{ds}{v} = \int_0^X\sqrt{\frac{1 + y'^2}{2gy}}\,dx\]

This is a functional of the whole curve, and we want the \(y(x)\) that makes it stationary — exactly what the Euler–Lagrange equation is for.

The Beltrami shortcut

The integrand \(f(y,y') = \sqrt{(1+y'^2)/2gy}\) has no explicit \(x\). Whenever that happens the Euler–Lagrange equation admits a first integral (the Beltrami identity):

\[f - y'\frac{\partial f}{\partial y'} = \text{constant}\]

This is the variational analogue of energy conservation — a cyclic coordinate giving a conserved quantity — and it reduces a second-order ODE to first order. Carrying it out:

\[y\left(1 + y'^2\right) = 2a\]

The solution is a cycloid

Substituting \(y = a(1-\cos\theta)\) solves it, and integrating for \(x\) gives

\[\boxed{\;x = a(\theta - \sin\theta),\qquad y = a(1 - \cos\theta)\;}\]

the curve traced by a point on the rim of a rolling wheel. The constant \(a\) and the final parameter \(\theta_1\) are fixed by the endpoint, and the descent time comes out in closed form:

\[T = \sqrt{\frac{a}{g}}\;\theta_1\]

Note that the optimal curve overshoots below the endpoint when \(X\) is large enough (\(\theta_1 > \pi\)): the bead dives past the target depth, then climbs back up. Losing height to gain speed is worth it. No amount of intuition produces that; the variational calculation does.

The tautochrone bonus

The same cycloid has a property Huygens found earlier: a bead released anywhere on it reaches the bottom in the same time,

\[T = \pi\sqrt{a/g}\]

independent of starting height. Fastest-descent curve and isochronous curve are the same object, which is a genuine coincidence of the sort that makes people believe in mathematics. Huygens built pendulum clocks with cycloidal cheeks to exploit it; friction in the constraint defeated him.

The history

Johann Bernoulli posed it as a public challenge in 1696. Newton, then running the Mint, reportedly solved it in one evening and published anonymously; Bernoulli recognised the author at once — tanquam ex ungue leonem, "as the lion is known by its claw." Solutions also arrived from Leibniz, l'Hôpital and Jakob Bernoulli.

The generalisation of these ad-hoc solutions, by Euler and Lagrange, is the calculus of variations — and hence Hamilton's principle, Lagrangian mechanics, and every field theory since.

Common mistakes

  • Expecting the straight line. Shortest ≠ fastest. Different functional, different extremal.
  • Forgetting the bead starts from rest. If it enters with speed the functional changes and the cycloid is no longer optimal.
  • Missing the overshoot. For \(\theta_1 > \pi\) the optimal path dips below the endpoint.
  • Attacking the full Euler–Lagrange equation. With no explicit \(x\), use Beltrami — it is the difference between one line and a page.

Knowledge graph position

Prerequisites: calculus of variations, Euler–Lagrange. Leads to: Hamilton's principle, optimal control, Lagrangian mechanics.

Quiz

Q1 (conceptual). Why is the straight line not the fastest path?

Answer

It minimises length, not time. Time depends on speed as well as distance, and speed is bought by dropping early. A steeper initial descent gains speed sooner, and that gain more than repays the extra path length — up to the optimum, which the cycloid attains.

Q2 (conceptual). Why does the Beltrami identity apply here?

Answer

The integrand contains \(y\) and \(y'\) but no explicit \(x\). That translational symmetry in \(x\) yields a first integral $f - y'\partial f/\partial y' = $ const, dropping the Euler–Lagrange equation from second order to first — the variational version of a cyclic coordinate giving a conserved momentum.

Q3 (MCQ). The cycloid is also the tautochrone, meaning:

  • (a) it is the shortest path between the points
  • (b) the descent time to the lowest point is independent of the starting height
  • (c) it has constant curvature
  • (d) it is the path of a projectile
Answer

(b). Huygens' result, and the reason he attempted cycloidal pendulum clocks. That one curve is both the brachistochrone and the tautochrone is a genuine and celebrated coincidence.